We need to compute f(3) where: - f(x) = -x for x < 0 - f(x) = f(x - f(x-1))/2 for x ≥ 0 Let me compute step by step. First, f(-1) = 1 (since f(x) = -x for x < 0, f(-1) = -(-1) = 1). f(0) = f(0 - f(-1))/2 = f(0 - 1)/2 = f(-1)/2 = 1/2. f(1) = f(1 - f(0))/2 = f(1 - 1/2)/2 = f(1/2)/2. Now I need f(1/2). f(1/2) = f(1/2 - f(-1/2))/2 = f(1/2 - 1/2)/2 = f(0)/2 = (1/2)/2 = 1/4. So f(1) = f(1/2)/2 = (1/4)/2 = 1/8. f(2) = f(2 - f(1))/2 = f(2 - 1/8)/2 = f(15/8)/2. Now I need f(15/8). f(15/8) = f(15/8 - f(7/8))/2. Need f(7/8): f(7/8) = f(7/8 - f(-1/8))/2 = f(7/8 - 1/8)/2 = f(6/8)/2 = f(3/4)/2. Need f(3/4): f(3/4) = f(3/4 - f(-1/4))/2 = f(3/4 - 1/4)/2 = f(1/2)/2 = (1/4)/2 = 1/8. So f(7/8) = f(3/4)/2 = (1/8)/2 = 1/16. Then f(15/8) = f(15/8 - 1/16)/2 = f(30/16 - 1/16)/2 = f(29/16)/2. Need f(29/16): f(29/16) = f(29/16 - f(13/16))/2. Need f(13/16): f(13/16) = f(13/16 - f(-3/16))/2 = f(13/16 - 3/16)/2 = f(10/16)/2 = f(5/8)/2. Need f(5/8): f(5/8) = f(5/8 - f(-3/8))/2 = f(5/8 - 3/8)/2 = f(2/8)/2 = f(1/4)/2. Need f(1/4): f(1/4) = f(1/4 - f(-3/4))/2 = f(1/4 - 3/4)/2 = f(-1/2)/2 = (1/2)/2 = 1/4. So f(5/8) = f(1/4)/2 = (1/4)/2 = 1/8. Then f(13/16) = f(5/8)/2 = (1/8)/2 = 1/16. Then f(29/16) = f(29/16 - 1/16)/2 = f(28/16)/2 = f(7/4)/2. Need f(7/4): f(7/4) = f(7/4 - f(3/4))/2. f(3/4) = 1/8 (computed above). f(7/4) = f(7/4 - 1/8)/2 = f(14/8 - 1/8)/2 = f(13/8)/2. Need f(13/8): f(13/8) = f(13/8 - f(5/8))/2. f(5/8) = 1/8 (computed above). f(13/8) = f(13/8 - 1/8)/2 = f(12/8)/2 = f(3/2)/2. Need f(3/2): f(3/2) = f(3/2 - f(1/2))/2. f(1/2) = 1/4 (computed above). f(3/2) = f(3/2 - 1/4)/2 = f(6/4 - 1/4)/2 = f(5/4)/2. Need f(5/4): f(5/4) = f(5/4 - f(1/4))/2. f(1/4) = 1/4 (computed above). f(5/4) = f(5/4 - 1/4)/2 = f(4/4)/2 = f(1)/2. f(1) = 1/8. So f(5/4) = (1/8)/2 = 1/16. f(3/2) = f(5/4)/2 = (1/16)/2 = 1/32. f(13/8) = f(3/2)/2 = (1/32)/2 = 1/64. f(7/4) = f(13/8)/2 = (1/64)/2 = 1/128. f(29/16) = f(7/4)/2 = (1/128)/2 = 1/256. f(15/8) = f(29/16)/2 = (1/256)/2 = 1/512. f(2) = f(15/8)/2 = (1/512)/2 = 1/1024. f(3) = f(3 - f(2))/2 = f(3 - 1/1024)/2 = f(3071/1024)/2. Hmm, this is getting complicated. Let me look for a pattern. So far: - f(0) = 1/2 - f(1/4) = 1/4 - f(1/2) = 1/4 - f(1) = 1/8 - f(3/4) = 1/8 - f(5/8) = 1/8 - f(1/4) = 1/4... Hmm wait, let me tabulate: - f(0) = 1/2 - f(1/4) = 1/4 - f(1/2) = 1/4 - f(5/8) = 1/8 - f(3/4) = 1/8 - f(13/16) = 1/16 - f(7/8) = 1/16 - f(1) = 1/8 - f(5/4) = 1/16 - f(3/2) = 1/32 - f(13/8) = 1/64 - f(7/4) = 1/128 - f(29/16) = 1/256 - f(15/8) = 1/512 - f(2) = 1/1024 Hmm, interesting. Let me guess a general formula. Note f(x) for x in [0, ...]: Let's think about it as follows. Actually, this is a known problem. Let me think about the structure. Claim: For 0 ≤ x, f(x) = f(x - f(x-1))/2. Let me try to find a pattern: maybe f(x) = 2^{-n} where n depends on x. Let's define g(x) = -log2(f(x)) so f(x) = 2^{-g(x)}. f(0) = 1/2: g(0) = 1. f(1/4) = 1/4: g = 2. f(1/2) = 1/4: g = 2. f(5/8) = 1/8: g = 3. f(3/4) = 1/8: g = 3. f(13/16) = 1/16: g = 4. f(7/8) = 1/16: g = 4. f(1) = 1/8: g = 3. Hmm, f(1) = 1/8 but f(15/16)? Let me compute more values to find pattern. Actually, let's think about this more carefully. Define h(x) = f(x-1) for x ≥ 0 (with f(x-1) known for x-1 < 0 as -(x-1) = 1-x). For x in [0, 1): x - 1 < 0, so f(x-1) = 1 - x. Then f(x) = f(x - (1-x))/2 = f(2x-1)/2. If 2x - 1 < 0, i.e., x < 1/2, then f(2x-1) = 1 - 2x, so f(x) = (1-2x)/2 for x ∈ [0, 1/2). Check: f(0) = 1/2. ✓. f(1/4) = (1 - 1/2)/2 = (1/2)/2 = 1/4 ✓. f(1/2) is the boundary: x = 1/2, 2x-1 = 0, f(0)/2 = (1/2)/2 = 1/4 ✓ matches f(1/2)=1/4. For x ∈ [1/2, 1): f(x) = f(2x-1)/2 where 2x-1 ∈ [0, 1). So on [1/2, 3/4): 2x-1 ∈ [0, 1/2), f(2x-1) = (1 - 2(2x-1))/2 = (3-4x)/2, so f(x) = (3-4x)/4. Check f(5/8): (3 - 4·5/8)/4 = (3 - 2.5)/4 = 0.5/4 = 1/8 ✓. f(3/4): (3-3)/4 = 0. Hmm, but we computed f(3/4) = 1/8. Let me recheck. Wait, at x = 3/4: 2x - 1 = 1/2, f(1/2) = 1/4, f(3/4) = (1/4)/2 = 1/8. But the formula (3-4x)/4 at x=3/4 gives 0. So the formula only holds on [1/2, 3/4), and at 3/4 we use continuity from the recursive definition... Actually f(x) = f(2x-1)/2 for x in [1/2, 1). At x = 3/4, f(3/4) = f(1/2)/2 = 1/8. The formula (3-4x)/4 gives limit as x → 3/4^-: 0. So there's a discontinuity! f(3/4) = 1/8 but limit from left is 0. Interesting, so f is discontinuous. Let me continue. For x ∈ [3/4, 1): f(x) = f(2x-1)/2, 2x-1 ∈ [1/2, 1). f(2x-1) = f(4x-3)/4 for 2x-1 ∈ [1/2, 3/4), i.e., x ∈ [3/4, 7/8). So f(x) = f(4x-3)/4. If 4x-3 < 1/2, i.e., x < 7/8, then f(4x-3) = (1-2(4x-3))/2 = (7-8x)/2, so f(x) = (7-8x)/8 for x ∈ [3/4, 7/8). Check f(13/16): (7 - 8·13/16)/8 = (7 - 6.5)/8 = 0.5/8 = 1/16 ✓. f(7/8): boundary, f(7/8) = f(3/4)/2 = (1/8)/2 = 1/16 ✓. And formula limit at 7/8: (7-7)/8 = 0. Again discontinuity. So the pattern: on interval [1 - 2^{-k}, 1 - 2^{-(k+1)})... hmm wait, let me define intervals. [0, 1/2): f(x) = (1-2x)/2, decreasing from 1/2 to 0. At x = 1/2: f = 1/4 (jump discontinuity: limit from left is 0, value is 1/4). [1/2, 3/4): f(x) = (3-4x)/4, decreasing from 3/4·... at x=1/2: (3-2)/4 = 1/4 ✓ continuous from the right at 1/2 (matches f(1/2) = 1/4). Decreasing to 0 at x → 3/4^-. At x = 3/4: f = f(1/2)/2 = 1/8 (jump: left limit 0, value 1/8). [3/4, 7/8): f(x) = (7-8x)/8, from (7-6)/8 = 1/8 at x=3/4 ✓ down to 0. At x = 7/8: f = f(3/4)/2 = 1/16. [7/8, 15/16): f(x) = (15-16x)/16: at x = 7/8: (15-14)/16 = 1/16 ✓ down to 0. So in general, on [1 - 2^{-k}, 1 - 2^{-(k+1)}): f(x) = (2^{k+1}-1 - 2^{k+1}x)/2^{k+1} = (1 - 2^{k+1}(x - (1 - 2^{-k})) ... let me just write: f(x) = (a - 2^{k+1}x)/2^{k+1} where a = 2^{k+1} - 1. At x = 1 - 2^{-k}: f = (2^{k+1} - 1 - 2^{k+1} + 2)/2^{k+1} = 1/2^{k+1}. Hmm wait let me recompute: 2^{k+1}x at x = 1-2^{-k} is 2^{k+1} - 2. So f = (2^{k+1} - 1 - (2^{k+1} - 2))/2^{k+1} = 1/2^{k+1}. At x → (1 - 2^{-(k+1)})^-: 2^{k+1}x → 2^{k+1} - 1, f → 0. And f(1 - 2^{-(k+1)}) = f(2(1-2^{-(k+1)}) - 1)/2 = f(1 - 2^{-k})/2 = (1/2^{k+1})/2 = 1/2^{k+2}. Consistent with the next interval's start value. So on [1 - 2^{-k}, 1 - 2^{-(k+1)}), f(x) = (1 - 2^{-k-1}·... hmm, let me simplify: f(x) = (2^{k+1}(1-x) - 1)/2^{k+1} = (1-x) - 2^{-(k+1)}. Oh nice! So f(x) = (1 - x) - 2^{-(k+1)} on that interval where k is such that x ∈ [1-2^{-k}, 1-2^{-(k+1)}). Check: x = 0 (k=0): f(0) = 1 - 0 - 1/2 = 1/2 ✓. x = 1/4 (k=1, since 1/4 ∈ [1/2, 3/4)): f = 1 - 1/4 - 1/4 = 1/2? That's wrong. Hmm wait, 1/4 is not in [1/2, 3/4). 1/4 ∈ [0, 1/2), k=0: f = 1 - 1/4 - 1/2 = 1/4 ✓. x = 1/2, k=1: f = 1 - 1/2 - 1/4 = 1/4 ✓. x = 5/8, k=1: f = 1 - 5/8 - 1/4 = 1/8 ✓. x = 3/4, k=2: f = 1 - 3/4 - 1/8 = 1/8 ✓. x = 13/16, k=2: f = 1 - 13/16 - 1/8 = 1/16 ✓. x = 7/8, k=3: f = 1 - 7/8 - 1/16 = 1/16 ✓. x = 1, k=∞... f(1) = 1/8. Hmm, 1 is not in any of these intervals. Let's compute f(1) directly: f(1) = f(1 - f(0))/2 = f(1/2)/2 = (1/4)/2 = 1/8 ✓. So f(x) = (1-x) - 2^{-(k+1)} for x ∈ [1-2^{-k}, 1-2^{-(k+1)}), k = 0, 1, 2, .... Now for x ∈ [1, 2): x - 1 ∈ [0, 1), f(x-1) = (1-(x-1)) - 2^{-(k'+1)} = (2-x) - 2^{-(k'+1)} where k' determined by x-1. Then f(x) = f(x - f(x-1))/2. For x ∈ [1, 2), f(x-1) ∈ ... hmm, as x-1 ranges over [0,1), f(x-1) takes values in (0, 1/2]. Actually f on [0,1) takes values: on [1-2^{-k}, 1-2^{-(k+1)}), from 1/2^{k+1} down to 0 (limit). So f(x-1) ∈ (0, 1/2]. So x - f(x-1) ∈ [x - 1/2, x). For x ∈ [1, 2), x - f(x-1) ∈ [1/2, 2). Hmm, this is getting complex. Let me just track f(3) computation, but maybe there's a pattern: f(1) = 1/8 = 2^{-3}, f(2) = 1/1024 = 2^{-10}. Let me compute f on [1, 2) more carefully, at least the values needed. Actually wait, I need f(3) = f(3 - f(2))/2 = f(3 - 1/1024)/2. So I need f at x = 3 - 1/1024, which is in [2, 3). Hmm, so I need f on [2, 3), which requires f on [1, 3/2) roughly... Let me think. For x ∈ [2, 3): f(x) = f(x - f(x-1))/2. x - 1 ∈ [1, 2). f(x-1) for x-1 ∈ [1, 2): what's the range? f(1) = 1/8, f(2) = 1/1024, and presumably f decreases... but with jumps. Let's figure out f on [1,2) first. Actually, let me reconsider. Let me define things on dyadic structure. It seems f maps into dyadic rationals when x is dyadic. Let me just carefully compute f(3). f(3) = f(3 - f(2))/2 = f(3 - 1/1024)/2. Let me denote x₀ = 3 - 1/1024 = 3071/1024. f(x₀) = f(x₀ - f(x₀ - 1))/2 = f(x₀ - f(2 - 1/1024))/2. Need f(2 - 1/1024) = f(2047/1024). f(2047/1024) = f(2047/1024 - f(1023/1024))/2. Need f(1023/1024) = f(1 - 1/1024). From the pattern on [0,1): 1 - 1/1024 = 1 - 2^{-10}, so k = 9: x ∈ [1 - 2^{-9}, 1 - 2^{-10}) = [511/512, 1023/1024). f(x) = (1-x) - 2^{-10}. So f(1023/1024) = 1/1024 - 1/1024 = 0. Hmm, that gives 0. Wait: (1 - x) - 2^{-(k+1)} with k=9: 2^{-10} = 1/1024. f(1023/1024) = (1 - 1023/1024) - 1/1024 = 1/1024 - 1/1024 = 0. Interesting. But wait, is that right? At the left endpoint of interval k=9, x = 511/512: f = (1/512) - 1/1024 = 1/1024. As x increases to 1023/1024, f → 0. At x = 1023/1024 exactly, f = 0 by formula. But also f(1 - 2^{-(k+1)}) = f(1-2^{-k})/2 = (1/2^{k+1})·(1/2)... hmm wait earlier I computed f(1 - 2^{-(k+1)}) = 1/2^{k+2}. Let me recheck with formula: at x = 1 - 2^{-(k+1)}, which is the left endpoint of interval k+1: f = (1-x) - 2^{-(k+2)} = 2^{-(k+1)} - 2^{-(k+2)} = 2^{-(k+2)} ✓. Good, consistent. So f(1023/1024) = f(1 - 2^{-10}) = 2^{-11} = 1/2048. Hmm wait, contradiction. Let me recheck. 1 - 2^{-10} = 1023/1024. Is 1023/1024 the left endpoint of interval k=10? Interval k=10 is [1 - 2^{-10}, 1 - 2^{-11}) = [1023/1024, 2047/2048). So x = 1023/1024 is the left endpoint of interval k=10, and the formula gives f(x) = (1-x) - 2^{-11} = 1/1024 - 1/2048 = 1/2048 ✓. I made an error: x ∈ [1-2^{-k}, 1-2^{-(k+1)}), k=9 interval is [1-2^{-9}, 1-2^{-10}) = [511/512, 1023/1024). Right, so 1023/1024 is NOT in interval k=9 (it's the excluded right endpoint); it's the left endpoint of interval k=10. Good: f(1023/1024) = 1/2048. OK so this dyadic structure: f(1 - 2^{-m}) = 2^{-(m+1)}. Now: f(2047/1024) = f(2047/1024 - f(1023/1024))/2 = f(2047/1024 - 1/2048)/2 = f(4094/2048 - 1/2048)/2 = f(4093/2048)/2. Need f(4093/2048) = f(2 - 3/2048). Hmm, this is in [1, 2)... wait no, 4093/2048 ≈ 1.9985, yes in [1, 2). This is getting messy. Let me think about f on [1, 2) generally. For x ∈ [1, 2): f(x) = f(x - f(x-1))/2. x - 1 ∈ [0, 1). f(x-1) = (1 - (x-1)) - 2^{-(k+1)} = (2 - x) - 2^{-(k+1)} where x - 1 ∈ [1 - 2^{-k}, 1 - 2^{-(k+1)}), i.e., x ∈ [2 - 2^{-k}, 2 - 2^{-(k+1)}). On such x-interval: f(x-1) = (2 - x) - 2^{-(k+1)}. Then x - f(x-1) = x - (2-x) + 2^{-(k+1)} = 2x - 2 + 2^{-(k+1)}. For x ∈ [2 - 2^{-k}, 2 - 2^{-(k+1)}): 2x - 2 ranges over [2 - 2^{1-k}... wait: at x = 2 - 2^{-k}: 2x-2 = 2 - 2^{1-k}. Hmm, let me compute: 2(2 - 2^{-k}) - 2 = 2 - 2^{1-k}. And at x → 2 - 2^{-(k+1)}: 2x - 2 → 2 - 2^{-k}. Hmm, that's weird: 2 - 2^{1-k} < 2 - 2^{-k}? For k ≥ 1: 2^{1-k} ≥ 2^{-k}, so 2 - 2^{1-k} ≤ 2 - 2^{-k}. Yes so range of 2x-2 is [2 - 2^{1-k}, 2 - 2^{-k}). Hmm OK so x - f(x-1) = 2x - 2 + 2^{-(k+1)} ∈ [2 - 2^{1-k} + 2^{-(k+1)}, 2 - 2^{-k} + 2^{-(k+1)}). Hmm, this is in [1, 2) for the values... let me sanity check with actual numbers. Take x = 1 (k: x - 1 = 0 = 1 - 2^0, so k = 0, interval [2-1, 2-1/2) = [1, 3/2)). f(0) = 1/2. f(1) = f(1 - 1/2)/2 = f(1/2)/2 = 1/8. With formula: f(x-1) = (2-x) - 2^{-1} = 1 - 1/2 = 1/2 at x=1 ✓. x - f(x-1) = 2x - 2 + 1/2 = 1/2 at x=1 ✓. f(1) = f(1/2)/2 = 1/8 ✓. Let me get f on [1, 3/2) explicitly: x - 1 ∈ [0, 1/2), f(x-1) = (1 - (x-1)) - 1/2 = 3/2 - x. So f(x) = f(x - (3/2 - x))/2 = f(2x - 3/2)/2. 2x - 3/2 ∈ [-3/2, -1/2) for x ∈ [1, 3/2). So 2x - 3/2 < 0, f(2x - 3/2) = 3/2 - 2x. So f(x) = (3/2 - 2x)/2 = 3/4 - x for x ∈ [1, 3/2). Check: f(1) = 3/4 - 1 = -1/4?? That contradicts f(1) = 1/8! Hmm, wait. Something's off. Let me recompute f(1). f(1) = f(1 - f(0))/2. f(0) = 1/2. So f(1) = f(1/2)/2. f(1/2) = 1/4. So f(1) = 1/8. Hmm. But with the interval formula: x = 1, x - 1 = 0. Which interval does 0 belong to? Interval k=0: [1-1, 1-1/2) = [0, 1/2). Yes, 0 ∈ [0, 1/2), k=0. f(0) = (1-0) - 1/2 = 1/2 ✓. So f(x-1) = (2-x) - 2^{-1} = 3/2 - x at x = 1: 1/2 ✓. f(1) = f(1 - 1/2)/2 = f(1/2)/2 = 1/8. But my formula said for x ∈ [1, 3/2), f(x) = f(2x - 3/2)/2 and 2x - 3/2 ∈ [-1/2, 1/2). Wait: x ∈ [1, 3/2) → 2x ∈ [2, 3) → 2x - 3/2 ∈ [1/2, 3/2). I made an arithmetic error. 2·1 - 3/2 = 1/2. Right! 2x - 3/2 ∈ [1/2, 3/2) for x ∈ [1, 3/2). I don't know why I wrote [-3/2, -1/2). Stupid error. OK so for x ∈ [1, 3/4... let me redo. For x ∈ [1, 3/2): f(x) = f(2x - 3/2)/2, with argument in [1/2, 3/2). For x ∈ [1, 5/4): 2x - 3/2 ∈ [1/2, 1): f(2x-3/2) = (1 - (2x - 3/2)) - 2^{-(k+1)} where 2x - 3/2 ∈ [1 - 2^{-k}, 1 - 2^{-(k+1)}). Hmm, I need to be careful. Let me just carefully build up. Alternatively, maybe there's a cleaner way to think: define F(x) = f(x). Let me look at the sequence of iterates for computing f at integers: f(0) = 1/2 f(1) = 1/8 f(2) = 1/1024 f(3) = ? 1/2 = 2^{-1}, 1/8 = 2^{-3}, 1/1024 = 2^{-10}. Exponents: 1, 3, 10. Differences: 2, 7. Hmm, 2 and 7... maybe next difference is related? 1, 3, 10: these are triangular numbers + ... 1 = T_1, 3 = T_2, 10 = T_4? T_1 = 1, T_2 = 3, T_3 = 6, T_4 = 10. Hmm, 1, 3, 10 skips 6. Alternatively 1, 3, 10, ... could be a(n) = a(n-1)·3 - ... 1·3 = 3, 3·3+1 = 10, 10·3 + 0 = 30? Hmm. Or central binomial-ish? C(2,1) = 2, no. Let me guess f(n) = 2^{-a(n)} with a(0) = 1, a(1) = 3, a(2) = 10. Hmm, maybe a(n) = (a(n-1)+... Let me compute f(3) exactly and see. Actually, let me be systematic. I'll compute f on [1, 2) piecewise, then [2, 3), then f(3). Actually hold on. Let me reconsider the structure. For x ∈ [0, 1), we found: f(x) = (1 - x) - 2^{-(k+1)} for x ∈ [1 - 2^{-k}, 1 - 2^{-(k+1)}), k ≥ 0. Equivalently: f(x) = (1-x) - 2^{-(k+1)} where k = number of leading 1's in binary expansion of x (after the point)... Since x ∈ [0,1), write x = 0.b₁b₂b₃... in binary. x ∈ [1-2^{-k}, 1-2^{-(k+1)}) means x = 0.111...1 0 ... with k ones then a zero. So k = number of initial consecutive 1s. For x ∈ [1, 2): x = 1.b₁b₂... Let me try to compute f on [1, 2). f(x) = f(x - f(x-1))/2. Case: x - 1 ∈ [0, 1/2) (x ∈ [1, 3/2)): f(x-1) = (2 - x) - 1/2 = 3/2 - x. f(x) = f(x - (3/2 - x))/2 = f(2x - 3/2)/2. For x ∈ [1, 3/2): 2x - 3/2 ∈ [1/2, 3/2). Subcase x ∈ [1, 5/4): 2x - 3/2 ∈ [1/2, 1). On [1/2, 3/4): f(y) = (1-y) - 1/4. On [3/4, 7/8): f(y) = (1-y) - 1/8. Etc. So f(2x - 3/2) = (1 - (2x - 3/2)) - 2^{-(k+1)} = (5/2 - 2x) - 2^{-(k+1)} where 2x - 3/2 ∈ [1 - 2^{-k}, 1 - 2^{-(k+1)}), i.e., x ∈ [5/4 - 2^{-k-1}, 5/4 - 2^{-(k+2)}). So for x ∈ [5/4 - 2^{-(k+1)}, 5/4 - 2^{-(k+2)}): f(x) = [(5/2 - 2x) - 2^{-(k+1)}]/2 = 5/4 - x - 2^{-(k+2)}. Check continuity: at x = 5/4 - 2^{-(k+1)} (left endpoint): f = 5/4 - (5/4 - 2^{-(k+1)}) - 2^{-(k+2)} = 2^{-(k+1)} - 2^{-(k+2)} = 2^{-(k+2)}. Hmm, and for the previous interval k-1, as x → (5/4 - 2^{-(k+1)})^-: f → 2^{-(k+1)} - 2^{-(k+2)}... wait no. For interval k-1: x ∈ [5/4 - 2^{-k}, 5/4 - 2^{-(k+1)}), f = 5/4 - x - 2^{-(k+1)}. As x → right end: f → 5/4 - (5/4 - 2^{-(k+1)}) - 2^{-(k+1)} = 0. And at left endpoint x = 5/4 - 2^{-k}: f = 2^{-k} - 2^{-(k+1)} = 2^{-(k+1)}. Consistent. So on [1, 5/4): f(x) = 5/4 - x - 2^{-(k+2)} where x ∈ [5/4 - 2^{-(k+1)}, 5/4 - 2^{-(k+2)}), k ≥ 0. At x = 5/4: need special handling? x = 5/4: 2x - 3/2 = 1. f(1) = 1/8. So f(5/4) = f(1)/2 = 1/16. And by formula with next interval... the pattern: f(5/4 - 2^{-m}) = 2^{-(m+1)}. Check: f(5/4 - 2^{-1}) = f(3/4) = 1/8 = 2^{-3} ✓ (m=1: 2^{-2}? hmm). Wait: f(3/4): x = 3/4 = 5/4 - 1/2, so m = 1, formula gives 2^{-(m+1)} = 2^{-2} = 1/4. But f(3/4) = 1/8. Hmm, let me recheck. f(x) = 5/4 - x - 2^{-(k+2)} on x ∈ [5/4 - 2^{-(k+1)}, 5/4 - 2^{-(k+2)}). At x = 5/4 - 2^{-(k+1)} (left endpoint of interval k): f = 2^{-(k+1)} - 2^{-(k+2)} = 2^{-(k+2)}. x = 3/4 = 5/4 - 1/2 = 5/4 - 2^{-1}: k+1 = 1, k = 0: f = 2^{-2} = 1/4. But earlier direct computation gave f(3/4) = 1/8. Contradiction! Let me recheck f(3/4). f(3/4) = f(3/4 - f(-1/4))/2 = f(3/4 - 1/4)/2 = f(1/2)/2 = (1/4)/2 = 1/8. Now with new formula: x = 3/4 ∈ [1, 3/2), so f(x) = f(2x - 3/2)/2 = f(0)/2 = (1/2)/2 = 1/4?? Wait, that's wrong. Let me recompute. x = 3/4: is x ∈ [1, 3/2)? 3/4 < 1! No! 3/4 is not in [1, 3/2). I confused myself. The interval [1, 5/4) etc. is for x, and 3/4 is in [0,1). Duh. So on [1, 5/4), f(x) = 5/4 - x - 2^{-(k+2)}. Let me verify with f(1): x = 1 = 5/4 - 1/4 = 5/4 - 2^{-2}, so k+1 = 2, k = 1: f(1) = 5/4 - 1 - 2^{-3} = 1/4 - 1/8 = 1/8 ✓. Check f(9/8): x = 9/8 = 5/4 - 1/8 = 5/4 - 2^{-3}, k = 2: f = 5/4 - 9/8 - 2^{-4} = 1/8 - 1/16 = 1/16. Let me verify directly: f(9/8) = f(9/8 - f(1/8))/2. f(1/8) = (1 - 1/8) - 1/2 = 3/8. So f(9/8) = f(9/8 - 3/8)/2 = f(6/8)/2 = f(3/4)/2 = (1/8)/2 = 1/16 ✓. So the structure on [1, 5/4) mirrors [0,1) but "centered" at 5/4 instead of 1: f(x) = (5/4 - x) - 2^{-(k+2)}, x ∈ [5/4 - 2^{-(k+1)}, 5/4 - 2^{-(k+2)}). Similarly there will be intervals [5/4, ...): x ∈ [5/4, 3/2): 2x - 3/2 ∈ [1, 2). Then f(x) = f(2x - 3/2)/2 where f on [1, 2) is being computed recursively... this suggests self-similar structure. Hypothesis: f(x) = (c - x)·something... Actually, notice on [0, 1): f(x) = (1 - x) - 2^{-(k+1)}. On [1, 5/4): f(x) = (5/4 - x) - 2^{-(k+2)}. Interesting: it seems f(x) = (A - x) - ε where the intervals accumulate at some point A from below, with f(A⁻) → 0, and f has jumps at dyadic points. Hmm wait, but actually the structure on [1, 2) is more complex because [1, 5/4) maps to [1/2, 1) ⊂ [0, 1), [5/4, ...) maps to [1, ...) etc. Let me continue the computation. For x ∈ [5/4, 3/2): 2x - 3/2 ∈ [1, 3/2). f(2x - 3/2) = f((2x - 3/2) - f(2x - 5/2))/2. Hmm, 2x - 5/2 ∈ [-1/2, 0) for x ∈ [5/4, 3/2)... wait: 2x - 5/2 at x = 5/4 is 0, at x → 3/2 is 1/2. Hmm: 2·(5/4) - 5/2 = 5/2 - 5/2 = 0. 2·(3/2) - 5/2 = 3 - 5/2 = 1/2. So 2x - 5/2 ∈ [0, 1/2). So for y = 2x - 3/2 ∈ [1, 3/2): f(y) = f(y - f(y - 1))/2, y - 1 ∈ [0, 1/2): f(y-1) = 1 - (y-1) - 1/2 = 3/2 - y... This recursion is getting complicated. Let me just say f on [1, 3/2): f(y) = f(2y - 3/2)/2 (as derived). And for y ∈ [5/4, 3/2): 2y - 3/2 ∈ [1, 3/2), so f(y) = f(2(2y-3/2) - 3/2)/4 = f(4y - 9/2)/4. 4y - 9/2 ∈ [1/2, 3/2) for y ∈ [5/4, 3/2). Hmm interesting. Let me instead try to compute f(3) by direct recursion with exact rational arithmetic, being smart about it. But the recursion tree might blow up. Let me think about what values I need. f(3) = f(3 - f(2))/2 = f(3071/1024)/2. Let me define a general approach: I'll compute f on [2, 3) via the relation f(x) = f(x - f(x-1))/2, needing f on [1, 2) and f on [1, 3)... it's self-referential but constructive since arguments decrease... wait, do they? x - f(x-1): since f(x-1) > 0 (presumably), argument < x. But is it ≥ 0? We need f(x-1) ≤ x for the argument to be in domain [we need argument to be... well actually the recursion f(x) = f(x - f(x-1))/2 is valid for x ≥ 0, and the argument x - f(x-1) could be negative, in which case f(argument) = -(argument) directly. Since f(t) for t ∈ [0,1) is at most 1/2, and generally f seems small, the argument stays ≥ ... whatever, let me just compute. Plan: compute f on the needed points using exact fractions. The needed points for f(3): f(3) = f(3 - f(2))/2. f(2) = 1/1024. So need f(3071/1024). f(3071/1024) = f(3071/1024 - f(2047/1024))/2. Need f(2047/1024). f(2047/1024) = f(2047/1024 - f(1023/1024))/2. Need f(1023/1024). f(1023/1024): 1023/1024 = 1 - 2^{-10}, which is left endpoint of interval k=10 in [0,1): f = 2^{-11} = 1/2048. f(2047/1024) = f(2047/1024 - 1/2048)/2 = f(4094/2048 - 1/2048)/2 = f(4093/2048)/2. f(4093/2048) = f(4093/2048 - f(2045/2048))/2. Need f(2045/2048). f(2045/2048) = f(2045/2048 - f(997/1024))/2. Need f(997/1024). 997/1024 = 1 - 27/1024. Hmm, 27/1024 is not a power of 2. 997/1024 ∈ which interval? 1 - 2^{-5} = 969/1024... wait 2^{-5} = 1/32 = 32/1024, so 1 - 1/32 = 992/1024. 1 - 2^{-6} = 1 - 16/1024 = 1008/1024. So 997/1024 ∈ [992/1024, 1008/1024) = [1 - 2^{-5}, 1 - 2^{-6}), k = 5. f(997/1024) = (1 - 997/1024) - 2^{-6} = 27/1024 - 16/1024 = 11/1024. Hmm OK this is doable but tedious. The tree: each f(x) for x ≥ 0 requires f(x-1) and then f(x - f(x-1)). The arguments decrease, but slowly, and we have chains. Let me think about whether there's a pattern that collapses this. Alternative: understand f on [1, 2) and [2, 3) structurally. Let me look at the structure more. On [0, 1): f(x) = (1-x) - 2^{-(k+1)}, intervals accumulating at 1, with k = index. Value at 1: f(1) = 1/8. On [1, 5/4): f(x) = (5/4 - x) - 2^{-(k+2)}, accumulating at 5/4, f(5/4) = 1/16. What about [5/4, 3/2)? Let's compute. For x ∈ [5/4, 3/2): f(x) = f(2x - 3/2)/2, 2x - 3/2 ∈ [1, 3/2). For y ∈ [1, 5/4): f(y) = (5/4 - y) - 2^{-(k+2)}, y ∈ [5/4 - 2^{-(k+1)}, 5/4 - 2^{-(k+2)}). So for x with 2x - 3/2 ∈ [1, 5/4 - 2^{-(k+1)}): f(x) = [(5/4 - (2x - 3/2)) - 2^{-(k+2)}]/2 = [9/4 - 2x - 2^{-(k+2)}]/2 = 9/8 - x - 2^{-(k+3)}. Condition: 2x - 3/2 ∈ [1 - ... hmm wait, y ∈ [1, 5/4 - 2^{-(k+1)}) — but hold on, for k=0 the interval is [5/4 - 1/2, 5/4 - 1/4) = [3/4, 1). That's not ⊂ [1, 5/4). I need to intersect with [1, 5/4). The intervals on [1, 5/4): k such that 5/4 - 2^{-(k+1)} ≥ 1, i.e., 2^{-(k+1)} ≤ 1/4, k+1 ≥ 2, k ≥ 1. So intervals: k ≥ 1: [5/4 - 2^{-(k+1)}, 5/4 - 2^{-(k+2)}) ⊂ [1, 5/4). And [1, 3/4)... no wait. For k=1: [5/4 - 1/4, 5/4 - 1/8) = [1, 9/8). k=2: [9/8, 17/16). Etc. So [1, 5/4) = [1, 9/8) ∪ [9/8, 17/16) ∪ [17/16, 33/32) ∪ ... Good. So on [1, 5/4): for x ∈ [5/4 - 2^{-(k+1)}, 5/4 - 2^{-(k+2)}), k ≥ 1: f(x) = (5/4 - x) - 2^{-(k+2)}. Now for x ∈ [5/4, 3/2): y = 2x - 3/2 ∈ [1, 3/2). Split: y ∈ [1, 5/4) vs y ∈ [5/4, 3/2). Subcase y ∈ [5/4 - 2^{-(k+1)}, 5/4 - 2^{-(k+2)}) (k ≥ 1), i.e., x ∈ [(5/4 - 2^{-(k+1)} + 3/2)/2, ...] = [17/8·(1/2)... let me compute: x = (y + 3/2)/2. y ∈ [5/4 - 2^{-(k+1)}, 5/4 - 2^{-(k+2)}) → x ∈ [11/8 - 2^{-(k+2)}, 11/8 - 2^{-(k+3)}). f(x) = f(y)/2 = [(5/4 - y) - 2^{-(k+2)}]/2 with y = 2x - 3/2: = [5/4 - 2x + 3/2 - 2^{-(k+2)}]/2 = [11/4 - 2x - 2^{-(k+2)}]/2 = 11/8 - x - 2^{-(k+3)}. So on x ∈ [11/8 - 2^{-(k+2)}, 11/8 - 2^{-(k+3)}), k ≥ 1: f(x) = 11/8 - x - 2^{-(k+3)}. Hmm interesting, so we get pieces accumulating at 11/8 from below. What about x ∈ [11/8, 3/2)? Then y = 2x - 3/2 ∈ [11/8 - ... wait, x ∈ [11/8, 3/2) → y ∈ [5/4, 3/2). Hmm wait: y = 2x - 3/2; at x = 11/8: y = 11/4 - 3/2 = 5/4. Yes. So y ∈ [5/4, 3/2), and we recurse further. I see a pattern: centers 1, 5/4, 11/8, ... each center c_{n+1} = ... 1, 5/4, 11/8: differences 1/4, 1/8. Next: 11/8 + 1/16 = 23/16, then 23/16 + 1/32 = 47/32, ... These are 1, 1+1/4, 1+1/4+1/8, 1+1/4+1/8+1/16, ... = 2 - 2^{-n}·... let me compute: 1 = 2 - 1; 5/4 = 2 - 3/4; 11/8 = 2 - 5/8; 23/16 = 2 - 9/16; 47/32 = 2 - 17/32. Hmm: 1, 3/4, 5/8, 9/16, 17/32: these are (2^{n+1}... let me see: 1 = 1, 3/4, 5/8, 9/16, 17/32: numerators 1,3,5,9,17 = 2^n + 1: 2^0+1 = 2? No. 1, 3, 5, 9, 17: differences 2, 2, 4, 8. Hmm, a_n = a_{n-1}... 1 = 2^1 - 1, 3 = 2^2 - 1, 5 = 2^2+1, 9 = 2^3 + 1, 17 = 2^4 + 1. Hmm, not clean. Let me recompute. Centers: c_0 = 1 (accumulation point of [0,1) pieces). Wait, actually let me re-examine. On [0, 1), pieces accumulate at 1 (from below). On [1, 5/4), pieces accumulate at 5/4. On [5/4, 11/8), pieces accumulate at 11/8. Hmm wait, I derived pieces on [5/4, 3/2)?? Let me recheck. I had: for x ∈ [5/4, 3/2): y = 2x - 3/2 ∈ [1, 3/2). For y ∈ [1, 5/4): f(x) = 11/8 - x - 2^{-(k+3)} on x ∈ [11/8 - 2^{-(k+2)}, 11/8 - 2^{-(k+3)}). Hmm wait, but x ∈ [5/4, 3/2) and x ∈ [11/8 - 2^{-(k+2)}, 11/8 - 2^{-(k+3)}). For k = 1: x ∈ [11/8 - 1/8, 11/8 - 1/16) = [5/4, 11/8 - 1/16). Hmm, 11/8 - 1/16 = 21/16 > 5/4 = 20/16. So interval k=1: [5/4, 21/16). k=2: [21/16, 43/32)... wait 11/8 - 2^{-4} = 11/8 - 1/16 = 21/16, and 11/8 - 2^{-5} = 11/8 - 1/32 = 43/32. So k=2: [21/16, 43/32). Hmm, 21/16 = 42/32, so [42/32, 43/32). Yes accumulating at 11/8 from below. OK so pieces of [5/4, 11/8) accumulate at 11/8. Then [11/8, 3/2) will have pieces accumulating at some further point. The centers: 5/4, 11/8, next? For x ∈ [11/8, 3/2): y = 2x - 3/2 ∈ [5/4, 3/2). Then f(y) = f(2y - 3/2)/4 = f(4x - 9/2)/4. 4x - 9/2 ∈ [1, 3/2) for x ∈ [11/8, 3/2). At x = 11/8: 4·11/8 - 9/2 = 11/2 - 9/2 = 1. At x = 3/2: 6 - 9/2 = 3/2. Yes. So f(x) = f(4x - 9/2)/4, 4x - 9/2 ∈ [1, 3/2). For the sub-range where 4x - 9/2 ∈ [1, 5/4 - 2^{-(k+1)}) (k ≥ 1): f(4x - 9/2) = (5/4 - (4x - 9/2)) - 2^{-(k+2)} = 23/4 - 4x - 2^{-(k+2)}. So f(x) = 23/16 - x - 2^{-(k+4)}. Condition: 4x - 9/2 ∈ [5/4 - 2^{-(k+1)}, 5/4 - 2^{-(k+2)}) → x ∈ [23/16 - 2^{-(k+3)}, 23/16 - 2^{-(k+4)}). So pieces on [23/16 - ..., ...) accumulating at 23/16 from below. 23/16 = 11/8 + 1/16 ✓ matches guess. So the "accumulation centers" on [1, 2): starting at 1: 1, 5/4, 11/8, 23/16, 47/32, ..., c_n = 2 - 2^{-n}·... let me compute: c_1 = 5/4, c_2 = 11/8, c_3 = 23/16, c_4 = 47/32. c_n = ? c_1 = 5/4 = (2^3 - 3)/4? 5/4: numerator 5 = 2^2+1, c_2 = 11/8: 11 = 2^3 + 3, c_3 = 23/16: 23 = 2^4 + 7, c_4 = 47/32: 47 = 2^5 + 15. Hmm: 5 = 4+1, 11 = 8+3, 23 = 16+7, 47 = 32+15: so c_n = (2^{n+1} + 2^n - 1)/2^n = (3·2^n - 1)/2^n = 3 - 2^{-n}. Check: c_1 = 3 - 1/2 = 5/2? No! 5/2 ≠ 5/4. Hmm. Let me recompute: c_1 = 5/4. 3 - 2^{-1} = 5/2. Nope. Let me just directly: c_0 = 1, c_1 = 5/4, c_2 = 11/8, c_3 = 23/16, c_4 = 47/32. c_n - c_{n-1}: 1/4, 1/8, 1/16, 1/32. So c_n = 1 + 1/4 + 1/8 + ... + 1/2^{n+1} = 1 + (1/2)(1 - 2^{-n}) = 3/2 - 2^{-(n+1)}·... let me just sum: 1 + Σ_{j=2}^{n+1} 2^{-j} = 1 + (2^{-1} - 2^{-(n+1)}) = 3/2 - 2^{-(n+1)}. Check: c_1 = 3/2 - 1/4 = 5/4 ✓. c_2 = 3/2 - 1/8 = 11/8 ✓. c_3 = 3/2 - 1/16 = 23/16 ✓. c_4 = 3/2 - 1/32 = 47/32 ✓. So accumulation points c_n = 3/2 - 2^{-(n+1)}, n = 0, 1, 2, ..., accumulating at 3/2. And f(c_n): f(1) = 1/8, f(5/4) = 1/16, f(11/8) = f(1)/4 = 1/32? Let me verify: f(11/8) = f(2·11/8 - 3/2)/2 = f(5/4)/2 = (1/16)/2 = 1/32. f(23/16) = f(4·23/16 - 9/2)/4 = f(1)/4 = 1/32? Wait: 4·23/16 = 23/4, minus 9/2 = 23/4 - 18/4 = 5/4. So f(23/16) = f(5/4)/4 = (1/16)/4 = 1/64. Hmm. Let me also get f(3/2): f(3/2) = f(3/2 - f(1/2))/2 = f(3/2 - 1/4)/2 = f(5/4)/2 = (1/16)/2 = 1/32. Hmm OK. So values at centers: f(1) = 1/8 = 2^{-3}, f(5/4) = 2^{-4}, f(11/8) = 2^{-5}, f(23/16) = 2^{-6}, f(47/32) = 2^{-7}... Check: f(23/16) = 1/64 = 2^{-6} ✓ (computed above). Let me double check f(11/8) = 1/32 = 2^{-5} ✓. And f(3/2) = 1/32 = 2^{-5}. Hmm, same as f(11/8). Interesting. Wait, I should double-check f(5/4) = 1/16: f(5/4) = f(5/4 - f(1/4))/2 = f(5/4 - 1/4)/2 = f(1)/2 = (1/8)/2 = 1/16 ✓. So on [1, 2), the picture: [1, 5/4) has pieces accumulating at 5/4 with f(5/4) = 2^{-4}; [5/4, 11/8) pieces accumulating at 11/8, f(11/8) = 2^{-5}; [11/8, 23/16) accumulating at 23/16, f(23/16) = 2^{-6}; ... accumulating at 3/2 overall, and f(3/2) = 2^{-5}. Hmm, and then what about [3/2, 2)? Let's compute f(7/4) = 1/128 (computed earlier). f(2) = 1/1024. Hmm, f(7/4) = 2^{-7}, f(2) = 2^{-10}. Let me now think about [3/2, 2): f(x) = f(x - f(x-1))/2, x - 1 ∈ [1/2, 1). f(x-1) = (2 - x) - 2^{-(k+1)} where x - 1 ∈ [1-2^{-k}, 1-2^{-(k+1)}), i.e., x ∈ [2 - 2^{-k}, 2 - 2^{-(k+1)}). So x - f(x-1) = 2x - 2 + 2^{-(k+1)}. For x ∈ [3/2, 2): 2x - 2 ∈ [1, 2), so argument ∈ [1 + 2^{-(k+1)}, 2 - 2^{-k} + 2^{-(k+1)}) hmm wait: x ranges over [2 - 2^{-k}, 2 - 2^{-(k+1)}) for fixed k ≥ 1 (since x ≥ 3/2 means k ≥ 1: [2-1/2, 2-1/4) = [3/2, 7/4) is k=1). For k=1: x ∈ [3/2, 7/4): argument = 2x - 2 + 1/4 ∈ [3/4·2 - ... at x = 3/2: 1 + 1/4 = 5/4. At x → 7/4: 2·7/4 - 2 + 1/4 = 7/2 - 2 + 1/4 = 5/4 + 1/4·... hmm 7/2 - 2 = 3/2, +1/4 = 7/4. So argument ∈ [5/4, 7/4). So f(x) = f(2x - 7/4)/2 for x ∈ [3/2, 7/4), argument ∈ [5/4, 7/4) ⊂ [1, 2). Hmm OK. It's like the same structure but shifted/scaled. This is getting really intricate. Let me just brute-force compute f(3) with exact arithmetic, but organize it well. Actually, wait. Let me look at the values at integers and see if the sequence 1, 3, 10 continues in a recognizable way, then verify by computing f(3). Known: f(0) = 1/2 = 2^{-1}, f(1) = 1/8 = 2^{-3}, f(2) = 1/1024 = 2^{-10}. Sequence of exponents: 1, 3, 10, ? Hmm, 1, 3, 10: a(n+1) = ? 3 = 1·3, 10 = 3·3 + 1. Maybe a(n+1) = 3a(n) + something. Or maybe a(n) relates to sums: 1 = 1, 3 = 1+2, 10 = 1+2+3+4? T_4 = 10. T_1 = 1, T_2 = 3, T_4 = 10. Indices 1, 2, 4: powers of 2! So a(n) = T_{2^{n-... hmm: a(0) = T_1 = 1, a(1) = T_2 = 3, a(2) = T_4 = 10, a(3) = T_8 = 36? So f(3) = 2^{-36} = 1/68719476736? That's a fun guess. T_{2^n} = 2^n(2^n+1)/2 = 2^{n-1}(2^n + 1). Check: n=0: 2^{-1}(2+1) = 3/2? No. Hmm, T_1 = 1: 1·2/2 = 1 ✓. T_2 = 3 ✓. T_4 = 10 ✓. T_8 = 36. So guess f(3) = 2^{-36}. Alternatively the pattern could be a(n) = T_{2^n}: a(0) = T_1 = 1 ✓, a(1) = T_2 = 3 ✓, a(2) = T_4 = 10 ✓, a(3) = T_8 = 36. But wait, I should double check whether indices are 1, 2, 4 or something else like 1, 2, 4, 7 (incrementing 1, 2, 3): then a(3) = T_7 = 28. Or 1, 2, 4, 8 (doubling): a(3) = 36. Need to verify by actual computation. Let me try to verify with a careful computation of f(3). Actually, let me first double-check f(2) = 1/1024 with the structural formulas, to make sure my earlier step-by-step was right. f(2) = f(2 - f(1))/2 = f(2 - 1/8)/2 = f(15/8)/2. 15/8 = 1.875 ∈ [23/16, 47/32)? 23/16 = 1.4375, 47/32 = 1.46875. No. 15/8 = 1.875. Where does 15/8 fall? It's in [3/2, 2). Which sub-piece? Let me use the recursion: f(15/8) = f(15/8 - f(7/8))/2 = f(15/8 - 1/16)/2 = f(29/16)/2. 29/16 = 1.8125. f(29/16) = f(29/16 - f(13/16))/2 = f(29/16 - 1/16)/2 = f(28/16)/2 = f(7/4)/2. f(7/4) = f(7/4 - f(3/4))/2 = f(7/4 - 1/8)/2 = f(13/8)/2. f(13/8) = f(13/8 - f(5/8))/2 = f(13/8 - 1/8)/2 = f(12/8)/2 = f(3/2)/2. f(3/2) = f(3/2 - f(1/2))/2 = f(3/2 - 1/4)/2 = f(5/4)/2 = (1/16)/2 = 1/32. So f(13/8) = 1/64, f(7/4) = 1/128, f(29/16) = 1/256, f(15/8) = 1/512, f(2) = 1/1024 ✓. Consistent with earlier. Now f(3) = f(3 - 1/1024)/2 = f(3071/1024)/2. I need f on [2, 3). Let me develop the structure on [2, 3) enough to evaluate at the specific points needed. f(x) = f(x - f(x-1))/2 for x ∈ [2, 3). x - 1 ∈ [1, 2). I need f on [1, 2) at various points, specifically at points of the form (needed x) - 1, and then f at (x - f(x-1)) which lies in... let's see. Let me first find f(3071/1024). 3071/1024 = 3 - 2^{-10}. Hmm, there might be a pattern: points like n - 2^{-m}. Let me compute f(2 - 2^{-m}) for various m, and f(3 - 2^{-m}). We have f(2 - 2^{-1}) = f(3/2) = 1/32 = 2^{-5}. f(2 - 2^{-2}) = f(7/4) = 2^{-7}. f(2 - 2^{-3}) = f(15/8) = 2^{-9}. Pattern: f(2 - 2^{-m}) = 2^{-(2m+3)}? Check: m=1: 2^{-5} ✓; m=2: 2^{-7} ✓; m=3: 2^{-9} ✓. Let me also compute f(2 - 2^{-4}) = f(31/16) to verify the pattern: f(31/16) = f(31/16 - f(15/16))/2. f(15/16) = (1 - 15/16) - 2^{-5} = 1/16 - 1/32 = 1/32. So f(31/16) = f(31/16 - 1/32)/2 = f(61/32)/2. f(61/32) = f(61/32 - f(29/32))/2. f(29/32): 29/32 = 1 - 3/32 ∈ [1 - 1/16, 1 - 1/32) = [31/32, ...) no wait: 1 - 3/32, is it ≥ 1 - 1/8 = 7/8 = 28/32? 29/32 ≥ 28/32 yes. Interval [1-2^{-3}, 1-2^{-4}) = [7/8, 15/16) = [28/32, 30/32). So 29/32 ∈ [28/32, 30/32), k=3: f(29/32) = (1 - 29/32) - 2^{-4} = 3/32 - 2/32 = 1/32. f(61/32) = f(61/32 - 1/32)/2 = f(60/32)/2 = f(15/8)/2 = 2^{-9}/2 = 2^{-10}. So f(31/16) = f(61/32)/2 = 2^{-11}. Pattern says m=4: 2^{-(2·4+3)} = 2^{-11} ✓. So conjecture: f(2 - 2^{-m}) = 2^{-(2m+3)} for m ≥ 1. Also f(2) = 2^{-10}, which fits... hmm, f(2) = 2^{-10}: if pattern continued to m → ∞, 2m+3 → ∞, but f(2) = 2^{-10} finite. So f(2) doesn't fit that pattern; f(2) is its own thing (jump discontinuity at 2, like at 1: f(1) = 1/8 while f(1 - 2^{-m}) = 2^{-(m+1)} → 0). Similarly at 1: f(1 - 2^{-m}) = 2^{-(m+1)}, and f(1) = 2^{-3}. Now I need f(3 - 2^{-10}). Let me find the structure on [2, 3). For x ∈ [2, 3): x - 1 ∈ [1, 2). Let me get a formula for f on [1, 2) at general points... that's hard due to the nested structure. Instead, let me set up: f(x) = f(x - f(x-1))/2. Let me compute the chain for x₀ = 3 - 2^{-10}: f(x₀) = f(x₀ - f(x₀ - 1))/2, x₀ - 1 = 2 - 2^{-10}, f(x₀ - 1) = f(2 - 2^{-10}) = 2^{-23} (by conjecture, m=10: 2m+3 = 23). So f(x₀) = f(3 - 2^{-10} - 2^{-23})/2 = f(3 - 2^{-10} - 2^{-23})/2. Hmm, then I need f at 3 - 2^{-10} - 2^{-23}, which requires f at 2 - 2^{-10} - 2^{-23} (subtract 1), etc. This could go deep. Let me think recursively. Define x₀ = 3 - 2^{-10}. Steps: x_{i+1} = x_i - f(x_i - 1), and f(x_i) = f(x_{i+1})/2, until x_i - 1 < 0 (then f(x_i) = -x_i... wait no, f(x_i) = f(x_i - f(x_i - 1))/2 always for x_i ≥ 0; the recursion bottoms out when the argument x_i - f(x_i - 1) < 0, then f(x) = -(x)). Hmm wait, actually careful: f(x) = f(x - f(x-1))/2. The argument x - f(x-1) might be ≥ 0, in which case we continue. The recursion for computing f(x₀): we need f(x₀ - 1) [recursion on smaller first argument... x₀ - 1 < x₀ ✓] and f(x₀ - f(x₀ - 1)) [argument < x₀ since f(x₀-1) > 0 ✓]. So it's well-founded (arguments strictly decrease), but the tree branches. However, along the "second" branch, arguments decrease by tiny amounts, so chains can be long. But there's structure. Let me reconsider. For x ∈ [2, 3): write u = x - 2 ∈ [0, 1). Hmm, is there self-similarity between [1, 2) and [2, 3)? The values of f on [1, 2) are much smaller than on [0, 1). Hmm. Alternatively, think of it as: f(x) for x ∈ [2, 3): x - 1 ∈ [1, 2), f(x-1) is tiny (≤ 1/32), so x - f(x-1) ∈ (x - 1/32, x], which is still in [2, 3) for x ≥ 2 + 1/32... So the recursion f(x) = f(x - f(x-1))/2 with argument still in [2,3) creates long chains: f(x) = f(x - Σ f(·))/2^k... Hmm, let me think about this differently. For x ∈ [2, 3), define the sequence: a₀ = x, a_{i+1} = a_i - f(a_i - 1). Then f(x) = f(a_{i})/2^i, and we stop when a_i - f(a_i - 1) < 0, i.e., ... hmm, but actually f(a_i) = f(a_{i+1})/2^i requires each step valid: f(a_i) = f(a_i - f(a_i - 1))/2. The bottom: when a_{i+1} < 0, f(a_{i+1}) = -a_{i+1} > 0. Wait but can a_{i+1} < 0 while a_i ∈ [2, 3)? a_{i+1} = a_i - f(a_i - 1). For a_i ∈ [2, 3), f(a_i - 1) ≤ f values on [1, 2) which are ≤ 1/8 (max at x=1... f on [1, 2): max is f(1) = 1/8? Actually on [1, 5/4), f decreases from 1/8; jump at 5/4 to 1/16; etc. So max on [1,2) is 1/8 at x = 1.) So for a_i ∈ [2, 3), a_{i+1} = a_i - f(a_i - 1) ≥ 2 - 1/8 > 0. So the chain never goes below 15/8... wait, but the chain decreases: a_i decreases? a_{i+1} = a_i - f(a_i - 1) < a_i since f > 0 (need f(a_i - 1) > 0: on [1, 2), f > 0 everywhere? f(1) = 1/8 > 0, and pieces are positive, approaching 0 at accumulation points from left, but at the points themselves positive. Is f ever 0 on [1,2)? The pieces (c - x) - 2^{-...} are > 0 on half-open intervals, = 0 only at right endpoints which are excluded. So f > 0 on [1, 2). OK.) So chain: a₀ = x ∈ [2, 3), a_{i+1} = a_i - f(a_i - 1) > a_i - 1/8, decreasing. Since decreasing and bounded below by... hmm, it could go below 2? If a_i ∈ [2, 15/8], then a_i - 1 ∈ [1, 7/8]... wait a_i - 1 ∈ [1, 7/8]?? No: a_i - 1 ∈ [1, 7/8] is empty. Let me redo: a_i ∈ [2, 15/8] → a_i - 1 ∈ [1, 7/8]. 7/8 < 1. Contradiction. a_i - 1 ∈ [1, 7/8]: impossible since 7/8 < 1. Ugh, I mean a_i - 1 ∈ [1, 7/8] is nonsense; if a_i ∈ [2, 15/8], a_i - 1 ∈ [1, 7/8]. Hmm 15/8 - 1 = 7/8. So a_i - 1 ∈ [1, 7/8]?? That's backwards: [1, 7/8] is empty because 7/8 < 1. But a_i - 1 ranges from 1 (at a_i = 2) to 7/8 (at a_i = 15/8). So a_i - 1 ∈ [7/8, 1]. Fine. So the chain a_i decreases from x, and f(a_i) = f(a_{i+1})/2. The chain continues while... we need f(a_i) = f(a_i - f(a_i - 1))/2 valid whenever a_i ≥ 0, which is always here. The chain "bottoms out" in the sense that eventually we express f(x) = f(a_N)/2^N where a_N is in a region where we can compute f directly (i.e., a_N < 0 → f = -a_N; or a_N ∈ [0, 1) with known piece; or a_N ∈ [1, 2) with known piece; etc.) Wait, but the chain a_{i+1} = a_i - f(a_i - 1) is only valid as "f(a_i) = f(a_{i+1})/2" — this is valid for all a_i ≥ 0. The chain decreases; where does it end? It ends when a_N < 0? But we showed a_{i+1} > a_i - 1/8 ≥ 2 - 1/8 > 0. So if x ∈ [2, 3), all a_i > 15/8... no wait: a_{i+1} > a_i - 1/8 only bounds the decrease per step. The chain could decrease from 3 down towards... it's decreasing, bounded below by 0, so converges. Hmm, but if it converges to L ≥ 2... then f(a_i - 1) → 0, so L is an accumulation point of f-values... The limit L satisfies L = L - f(L - 1) → f(L-1) = 0 → L - 1 ∈ {points where f = 0}. On [1, 2), f = 0 only in limit... f(y) → 0 as y → c_n from below, or... hmm, f never attains 0 on [1, 2)? At right endpoints of pieces (excluded) limit is 0. So f(L-1) = 0 impossible for L - 1 ∈ [1, 2). So L < 2 eventually? Hmm, wait, but if a_i enters [1, 2)... no, a_i ≥ 15/8 always if x ≥ 2? Hmm no: a_i decreases; from x ∈ [2, 3), steps of size ≤ 1/8 (actually ≤ f on [1,2) values which can be much smaller). The chain could pass below 2: e.g., x = 2, f(1) = 1/8, a₁ = 2 - 1/8 = 15/8 ∈ [1, 2). Then we switch to [1, 2) regime. OK here's the thing: computing f(3) exactly requires following this chain from x₀ = 3 - 2^{-10} down through [2, 3), possibly into [1, 2) and [0, 1), with exact rational arithmetic. The chain lengths could be long but with structure. Let me try to find the structure on [2, 3). Let me hypothesize the structure on [2, 3) mirrors [1, 2) in some scaled way. On [1, 2): we found accumulation centers c_n = 3/2 - 2^{-(n+1)}, n ≥ 0, with f(c_n) = 2^{-(n+3)}·... wait: f(1) = 2^{-3}, f(5/4) = 2^{-4}, f(11/8) = 2^{-5}, f(23/16) = 2^{-6}. So f(c_n) = 2^{-(n+3)}. And pieces: on [c_{n}, c_{n+1})... hmm wait, actually the pieces on [1, 5/4) = [c_0, c_1) accumulate at c_1 = 5/4: f(x) = (5/4 - x) - 2^{-(k+2)} for x ∈ [5/4 - 2^{-(k+1)}, 5/4 - 2^{-(k+2)}). Generalizing: on [c_{n-1}, c_n), pieces accumulate at c_n: f(x) = (c_n - x) - 2^{-(k + something)}. Let me verify for [5/4, 11/8) = [c_1, c_2): pieces f(x) = 11/8 - x - 2^{-(k+3)} for x ∈ [11/8 - 2^{-(k+2)}, 11/8 - 2^{-(k+3)}), k ≥ 1. So on [c_{n-1}, c_n): f(x) = (c_n - x) - 2^{-(k+n+2)} for x ∈ [c_n - 2^{-(k+n+1)}, c_n - 2^{-(k+n+2)}), k ≥ 1. Check n=1 ([5/4, 11/8)): f(x) = (11/8 - x) - 2^{-(k+3)}, x ∈ [11/8 - 2^{-(k+2)}, 11/8 - 2^{-(k+3)}) ✓ matches. Check n=0 ([1, 5/4)): f(x) = (5/4 - x) - 2^{-(k+2)}, x ∈ [5/4 - 2^{-(k+1)}, 5/4 - 2^{-(k+2)}) ✓ matches. And f(c_n) = f(3/2 - 2^{-(n+1)}) = 2^{-(n+3)} for n ≥ 0, and f(3/2) = 2^{-5}. Hmm wait, f(3/2) = 2^{-5} = 1/32, but f(c_2) = f(11/8) = 2^{-5} too. Interesting coincidence? c_2 = 11/8 = 3/2 - 1/8. Hmm. Let me double check f(3/2): f(3/2) = f(3/2 - f(1/2))/2 = f(3/2 - 1/4)/2 = f(5/4)/2 = (1/16)/2 = 1/32 ✓ = 2^{-5}. And f(11/8): f(11/8) = f(11/8 - f(3/8))/2. f(3/8) = (1 - 3/8) - 1/2 = 1/8. So f(11/8) = f(11/8 - 1/8)/2 = f(5/4)/2 = 1/32 ✓ = 2^{-5}. Yes both 2^{-5}. Interesting. So at 3/2 there's a "reset": f(3/2) = 2^{-5}, same as f(c_2). And then presumably on [3/2, 2) there's another cascade of centers. Recall f(7/4) = 2^{-7}, f(2) = 2^{-10}. Let me find centers on [3/2, 2). We have f(7/4) = 2^{-7}. 7/4 = 2 - 1/4. Guess centers on [3/2, 2): d_m = 2 - ... hmm. Let me compute a few values: f(13/8) = 2^{-6}, f(7/4) = 2^{-7}, f(15/8) = 2^{-9}, f(2) = 2^{-10}. 13/8 = 1.625, 7/4 = 1.75, 15/8 = 1.875, 2 = 2. Points: 3/2 = 1.5, 13/8 = 1.625, 7/4 = 1.75, 15/8 = 1.875, 2. These are 3/2 + 1/8·{0, 1, 2, 4}, then 2. Hmm: 1, 1+1/8, 1+2/8, 1+4/8... no: 3/2, 3/2+1/8, 3/2+2/8, 3/2+4/8. Increments: 1/8, 1/8, 1/8, then...? If pattern of doubling increments: next 3/2 + 8/8 = 2.5 > 2. Hmm, but we know f(2) = 2^{-10}. Alternatively centers: 13/8, 7/4, 15/8 with f-values 2^{-6}, 2^{-7}, 2^{-9}: hmm, and what about points between? Let me compute f(25/16) (25/16 = 1.5625, between 3/2 and 13/8): f(25/16) = f(25/16 - f(9/16))/2. f(9/16): 9/16 ∈ [1/2, 3/4): f = (1 - 9/16) - 1/4 = 7/16 - 4/16 = 3/16. So f(25/16) = f(25/16 - 3/16)/2 = f(22/16)/2 = f(11/8)/2 = (1/32)/2 = 1/64 = 2^{-6}. Hmm interesting, f(25/16) = 2^{-6} = f(13/8). Let me compute f(27/16) = 1.6875: f(27/16) = f(27/16 - f(11/16))/2. f(11/16) = (1 - 11/16) - 1/4 = 5/16 - 4/16 = 1/16. f(27/16) = f(27/16 - 1/16)/2 = f(26/16)/2 = f(13/8)/2 = 2^{-7}. Hmm so f(27/16) = 2^{-7}. And f(13/8) = 2^{-6}, f(7/4) = 2^{-7}. Let me compute f(7/4) again to be sure: f(7/4) = f(7/4 - f(3/4))/2 = f(7/4 - 1/8)/2 = f(13/8)/2 = 2^{-7} ✓. Hmm OK so I need a more systematic approach for [3/2, 2). Let me think about the general structure via the "chain" method. For x ∈ [3/2, 2): x - 1 ∈ [1/2, 1). On [1/2, 1): f(y) = (1-y) - 2^{-(k+1)}, y ∈ [1 - 2^{-k}, 1 - 2^{-(k+1)}), k ≥ 1. So f(x-1) = (2 - x) - 2^{-(k+1)} for x ∈ [2 - 2^{-k}, 2 - 2^{-(k+1)}), k ≥ 1. Chain step: x → x - f(x-1) = 2x - 2 + 2^{-(k+1)}. Let me define φ(x) = x - f(x - 1) (the argument). For x ∈ [2 - 2^{-k}, 2 - 2^{-(k+1)}), k ≥ 1: φ(x) = 2x - 2 + 2^{-(k+1)}. Note φ maps [2 - 2^{-k}, 2 - 2^{-(k+1)}) → [2 - 2^{1-k} + 2^{-(k+1)}, 2 - 2^{-k} + 2^{-(k+1)}) = [2 - (2^{2} - 1)·2^{-(k+1)}, 2 - (2 - 1)·2^{-(k+1)}·... let me compute: 2^{1-k} = 2^{2-(k+1)} = 4·2^{-(k+1)}. So φ(x) range: [2 - 4·2^{-(k+1)} + 2^{-(k+1)}, 2 - 2·2^{-(k+1)} + 2^{-(k+1)}) = [2 - 3·2^{-(k+1)}, 2 - 2^{-(k+1)}). So φ([2 - 2^{-k}, 2 - 2^{-(k+1)})) = [2 - 3·2^{-(k+1)}, 2 - 2^{-(k+1)}). Interesting! So if x ∈ [2 - 2^{-k}, 2 - 2^{-(k+1)}), then φ(x) ∈ [2 - 3·2^{-(k+1)}, 2 - 2^{-(k+1)}). Hmm, so the "k-index" of φ(x): φ(x) ∈ [2 - 3·2^{-(k+1)}, 2 - 2^{-(k+1)}). Note 3·2^{-(k+1)} = 1.5·2^{-k} = between 2^{-k} and 2^{-(k-1)}... Let me see where φ(x) falls relative to the grid {2 - 2^{-j}}. For k = 1: x ∈ [3/2, 7/4): φ(x) = 2x - 2 + 1/4 ∈ [2 - 3/4·... compute: at x = 3/2: φ = 3 - 2 + 1/4 = 5/4. At x → 7/4: φ → 7/2 - 2 + 1/4 = 7/4. So φ([3/2, 7/4)) = [5/4, 7/4). Note 5/4 = 2 - 3/4 ✓ (3·2^{-2} = 3/4), 7/4 = 2 - 2^{-2}·... 2 - 1/4 ✓. So for x ∈ [3/2, 7/4), φ(x) ∈ [5/4, 7/4) ⊂ [1, 2). And f(x) = f(φ(x))/2. Now φ(x) ∈ [5/4, 7/4): split into [5/4, 3/2) ∪ [3/2, 7/4). For x ∈ [3/2, 11/8): φ(x) ∈ [5/4, 11/8·... compute: φ(3/2) = 5/4, φ(11/8) = 2·11/8 - 2 + 1/4 = 11/4 - 2 + 1/4 = 5/2 - 2 + ... 11/4 + 1/4 = 12/4 = 3, minus 2 = 1. So φ(11/8) = 1?? Wait: φ(x) = 2x - 2 + 1/4. At x = 11/8: 22/8 - 16/8 + 2/8 = 8/8 = 1. Hmm, so φ(11/8) = 1, not 11/8. Let me recompute φ([3/2, 7/4)) endpoints: φ(3/2) = 2·(3/2) - 2 + 1/4 = 3 - 2 + 1/4 = 5/4 ✓. φ(7/4) = 7/2 - 2 + 1/4 = 3/2 + 1/4 = 7/4 ✓. So φ: [3/2, 7/4) → [5/4, 7/4), slope 2. φ(11/8) = 1? But 11/8 ∈ [3/2, 7/4), φ(11/8) should be in [5/4, 7/4). 2·11/8 = 11/4 = 2.75. 2.75 - 2 + 0.25 = 1.0. That's 1, not in [5/4, 7/4). Contradiction! Let me recompute. φ(x) = 2x - 2 + 2^{-(k+1)}. For k=1: 2^{-(k+1)} = 2^{-2} = 1/4. φ(x) = 2x - 2 + 1/4 = 2x - 7/4. At x = 3/2: 3 - 7/4 = 5/4 ✓. At x = 11/8: 11/4 - 7/4 = 1. Hmm! So φ(11/8) = 1. But φ should map [3/2, 7/4) to [5/4, 7/4)... φ(3/2) = 5/4, φ(7/4) = 2·7/4 - 7/4 = 7/4. φ is linear increasing (slope 2), so φ([3/2, 7/4)) = [5/4, 7/4). But φ(11/8) = 1 < 5/4?? 11/8 = 1.375. φ(1.375) = 2.75 - 1.75 = 1.0. But 11/8 ∈ [3/2, 7/4)?? 3/2 = 1.5, 11/8 = 1.375 < 1.5! No! 11/8 is NOT in [3/2, 7/4). 11/8 = 1.375 < 1.5. I keep making these dumb errors. OK so 11/8 ∈ [1, 3/2), handled by the earlier [1, 3/2) analysis. Fine. So φ: [3/2, 7/4) → [5/4, 7/4), i.e., f(x) = f(φ(x))/2 with φ(x) = 2x - 7/4. Check: f(7/4) = f(2·7/4 - 7/4)/2 = f(7/4)/2?? That's self-referential! Hmm wait: x = 7/4: x - 1 = 3/4, f(3/4) = 1/8, φ(7/4) = 7/4 - 1/8 = 13/8. And formula φ(x) = 2x - 7/4 at x = 7/4 gives 7/4. Discrepancy! Ah I see the issue: at x = 7/4, x - 1 = 3/4, which is the left endpoint of interval k=2 in [0,1) (i.e., [3/4, 7/8)). f(3/4) = 1/8 = (1 - 3/4) - 2^{-3} = 1/4 - 1/8 = 1/8 ✓. So f(x-1) = (2 - x) - 2^{-(k+1)} with k = 2 at x = 7/4: (2 - 7/4) - 2^{-3} = 1/4 - 1/8 = 1/8 ✓. So φ(7/4) = 7/4 - 1/8 = 13/8. But my formula said for x ∈ [2 - 2^{-k}, 2 - 2^{-(k+1)}) with k=1: x ∈ [3/2, 7/4) — 7/4 is NOT in this half-open interval (right endpoint excluded). At x = 7/4, k = 2: interval [2 - 1/4, 2 - 1/8) = [7/4, 15/8). φ(x) = 2x - 2 + 2^{-3} = 2x - 15/8. φ(7/4) = 7/2 - 15/8 = 28/8 - 15/8 = 13/8 ✓. So the map φ on [3/2, 2): for x ∈ [2 - 2^{-k}, 2 - 2^{-(k+1)}), k ≥ 1: φ(x) = 2x - 2 + 2^{-(k+1)}, mapping onto [2 - 3·2^{-(k+1)}, 2 - 2^{-(k+1)}). Hmm wait, for k=1: φ([3/2, 7/4)) = [2 - 3/4, 2 - 1/4) = [5/4, 7/4) ✓. For k=2: φ([7/4, 15/8)) = [2 - 3/8, 2 - 1/8) = [13/8, 15/8). Check: φ(x) = 2x - 15/8: φ(7/4) = 13/8 ✓, φ(15/8) = 15/4 - 15/8 = 30/8 - 15/8 = 15/8 ✓ (approached). For k=3: φ([15/8, 31/16)) = [2 - 3/16, 2 - 1/16) = [29/16, 31/16). φ(x) = 2x - 2 + 1/16 = 2x - 31/16: φ(15/8) = 15/4 - 31/16 = 60/16 - 31/16 = 29/16 ✓. So f(x) = f(φ(x))/2 where φ(x) = 2x - (2 - 2^{-(k+1)})·... hmm let me just note φ(x) = 2x - 2 + 2^{-(k+1)} for x ∈ I_k := [2 - 2^{-k}, 2 - 2^{-(k+1)}). And φ(I_k) = [2 - 3·2^{-(k+1)}, 2 - 2^{-(k+1)}). Note that φ(I_k) ⊂ [1, 2) for k ≥ 1 (since 2 - 3·2^{-(k+1)} ≥ 2 - 3/4 = 5/4 ≥ 1 for k ≥ 1). ✓. So for x ∈ [3/2, 2), f(x) = f(φ(x))/2, φ(x) ∈ [5/4, 2) ⊂ [1, 2). Now I need f on [1, 2) at arbitrary points of the form produced. On [1, 2), similarly: for x ∈ [1, 3/2): x - 1 ∈ [0, 1/2), f(x-1) = (1 - (x-1)) - 1/2 = 3/2 - x. φ(x) = x - (3/2 - x) = 2x - 3/2. φ([1, 3/2)) = [1/2, 3/2). For x ∈ [3/2, 7/4): x - 1 ∈ [1/2, 3/4), f(x-1) = (2 - x) - 1/4, φ(x) = 2x - 9/4, φ([3/2, 7/4)) = [3/4, 5/4). For x ∈ [7/4, 15/8): x - 1 ∈ [3/4, 7/8), f(x - 1) = (2 - x) - 1/8, φ(x) = 2x - 17/8. φ([7/4, 15/8)) = [7/8, 9/8). In general for x ∈ [1 + (1 - 2^{-k}), 1 + (1 - 2^{-(k+1)})) = [2 - 2^{-k}, 2 - 2^{-(k+1)}), k ≥ 1: φ(x) = 2x - 2 + 2^{-(k+1)}, φ(x) ∈ [2 - 3·2^{-(k+1)}, 2 - 2^{-(k+1)}). Hmm interesting — same formula as before but that was for [3/2, 2)... wait no. Hold on: [1, 3/2) is x - 1 ∈ [0, 1/2) which is "k=0" interval [0, 1/2) = [1 - 2^0, 1 - 2^{-1}). φ(x) = 2x - 3/2 for x ∈ [1, 3/2). φ([1, 3/2)) = [1/2, 3/2). For x ∈ [3/2, 7/4): x - 1 ∈ [1/2, 3/4) = [1 - 2^{-1}, 1 - 2^{-2}), k=1: φ(x) = 2x - 2 + 2^{-2} = 2x - 7/4. φ([3/2, 7/4)) = [2 - 3/4, 2 - 1/4) = [5/4, 7/4). Hmm wait, that contradicts what I wrote a line ago (φ([3/2, 7/4)) = [3/4, 5/4)). Let me recompute directly: x ∈ [3/2, 7/4) → x - 1 ∈ [1/2, 3/4) → f(x-1) = (1 - (x-1)) - 2^{-2} = (2 - x) - 1/4. At x = 3/2: f(1/2) = 1/4 = (2 - 3/2) - 1/4 = 1/2 - 1/4 = 1/4 ✓. φ(x) = x - (2 - x - 1/4) = 2x - 7/4. At x = 3/2: 3 - 7/4 = 5/4. At x → 7/4: 7/2 - 7/4 = 7/4. So φ([3/2, 7/4)) = [5/4, 7/4). Earlier I wrongly wrote [3/4, 5/4). OK so φ([3/2, 7/4)) = [5/4, 7/4) ⊂ [1, 2). Good, so this matches the general formula: for x ∈ [2 - 2^{-k}, 2 - 2^{-(k+1)}), k ≥ 1: φ(x) = 2x - 2 + 2^{-(k+1)}, φ(x) ∈ [2 - 3·2^{-(k+1)}, 2 - 2^{-(k+1)}). And for x ∈ [1, 3/2) ("k = 0"): φ(x) = 2x - 3/2 = 2x - 2 + 1/2 ✓ consistent with 2^{-(0+1)} = 1/2. φ([1, 3/2)) = [2 - 3/2, 2 - 1/2) = [1/2, 3/2) ✓ consistent. So unified: for x ∈ [2 - 2^{-k}, 2 - 2^{-(k+1)}), k ≥ 0 (where k=0 gives [1, 3/2)): φ(x) = 2x - 2 + 2^{-(k+1)}, f(x) = f(φ(x))/2. φ(x) ∈ [2 - 3·2^{-(k+1)}, 2 - 2^{-(k+1)}). Now the key: iterate φ. For x ∈ [1, 2), the orbit under φ: where does it go? φ(x) ∈ [1/2, 2) always. If φ(x) ∈ [1/2, 1), we can compute f directly (known pieces). If φ(x) ∈ [1, 2), continue. Note that φ(x) - 1 = 2x - 3 + 2^{-(k+1)}... Let me parametrize x = 2 - s where s ∈ (0, 1]. Then x ∈ [2 - 2^{-k}, 2 - 2^{-(k+1)}) ↔ s ∈ (2^{-(k+1)}, 2^{-k}]. φ(x) = 2(2 - s) - 2 + 2^{-(k+1)} = 2 - 2s + 2^{-(k+1)}. So s' = 2s - 2^{-(k+1)} where s ∈ (2^{-(k+1)}, 2^{-k}]. Interesting: s' = 2s - 2^{-(k+1)}. If s = 2^{-k}·t... let me write s = 2^{-k}·u, u ∈ (1/2, 1]. Then s' = 2^{1-k}u - 2^{-(k+1)} = 2^{-(k+1)}(4u - 1)·... hmm: 2^{1-k} = 2^{-(k+1)}·4. So s' = 2^{-(k+1)}(4u - 1). With u ∈ (1/2, 1]: 4u - 1 ∈ (1, 3]. So s' ∈ (2^{-(k+1)}, 3·2^{-(k+1)}]. So s' ∈ (2^{-(k+1)}, 3·2^{-(k+1)}]. Which k' does s' correspond to? s' ∈ (2^{-(k'+1)}, 2^{-k'}]. If s' ∈ (2^{-(k+1)}, 2^{-k}], then k' = k. If s' ∈ (2^{-k}, 3·2^{-(k+1)}], then k' = k - 1 (since 2^{-(k-1)} = 2·2^{-k} = 4·2^{-(k+1)} ≥ 3·2^{-(k+1)}). So k' ∈ {k-1, k}. In terms of u: s' = 2^{-(k+1)}(4u-1). If 4u - 1 ∈ (1, 2], i.e., u ∈ (1/2, 3/4], then s' ∈ (2^{-(k+1)}, 2^{-k}·... 2·2^{-(k+1)} = 2^{-k}], so k' = k, and u' = s'/2^{-k} = (4u-1)/2. If 4u - 1 ∈ (2, 3], u ∈ (3/4, 1], then s' ∈ (2^{-k}, 3·2^{-(k+1)}], k' = k - 1, u' = s'/2^{-(k-1)} = (4u - 1)/4. So the dynamics on u ∈ (1/2, 1] with level k: - If u ∈ (1/2, 3/4]: k' = k, u' = (4u - 1)/2 = 2u - 1/2. - If u ∈ (3/4, 1]: k' = k - 1, u' = (4u - 1)/4 = u - 1/4. And f(x) = f(φ(x))/2 each step. Also note: when does φ(x) drop below 1? φ(x) = 2 - s' ≥ 1 ↔ s' ≤ 1. s' ≤ 1 ↔ ... we need s' ≤ 1, i.e., 2^{-(k+1)}(4u - 1) ≤ 1. Hmm, this depends on k. Actually φ(x) ∈ [1, 2) always in this analysis? φ(x) = 2 - s' where s' ∈ (2^{-(k+1)}, 3·2^{-(k+1)}]. Since k ≥ 0: s' could be up to 3·2^{-(k+1)}. For k = 0: s' ∈ (1/2, 3/2]. If s' > 1, φ(x) < 1. So for k = 0 (x ∈ [1, 3/2)): s = 2 - x ∈ [1/2, 1), u = s/2^0 = s ∈ [1/2, 1). Hmm wait, k=0: s ∈ (2^{-1}, 2^0] = (1/2, 1]. u = s. If u ∈ (3/4, 1]: k' = -1, meaning s' ∈ (2^0, 3·2^{-1}] = (1, 3/2], so φ(x) = 2 - s' ∈ [1/2, 1) — drops into [1/2, 1) ⊂ [0, 1), computable. If u ∈ (1/2, 3/4]: k' = 0, u' = 2u - 1/2 ∈ (1/2, 1]. Stay in [1, 3/2). OK so now the dynamics: starting from x ∈ [1, 2), we iterate: u → u' with rules above, and k → k' ∈ {k, k-1}, and f halves each step, until s' > 1 (φ < 1), i.e., k' = -1, or... wait for k ≥ 1, s' ≤ 3·2^{-(k+1)} ≤ 3/4 < 1, so φ(x) > 5/4 ≥ 1: never drops below 1 while k ≥ 1. Only from k = 0 can we exit to [0, 1). So: starting at level k, the u-dynamics: while u ∈ (3/4, 1]: u → u - 1/4, k → k - 1. While u ∈ (1/2, 3/4]: u → 2u - 1/2, k stays. Note u → u - 1/4 decreases u by 1/4 each step (and k decreases). u → 2u - 1/2 maps (1/2, 3/4] → (1/2, 1]. Hmm interesting. Let me now track f(2) computation in this framework to validate: f(2): hmm, x = 2 is special (x - 1 = 1, f(1) = 1/8). Let me instead validate with x = 15/8: x = 15/8 = 2 - 1/8: s = 1/8 = 2^{-3}, k = 3, u = s/2^{-3} = 1. u = 1 ∈ (3/4, 1]: u → 3/4, k → 2. s' = 2^{-2}·(3/4) = 3/16. φ = 2 - 3/16 = 29/16 ✓ (matches earlier: φ(15/8) = 29/16 ✓). Next: x = 29/16: s = 3/16, k = 2 (s ∈ (2^{-3}, 2^{-2}] = (1/8, 1/4] ✓), u = (3/16)/(1/4) = 3/4. u = 3/4 ∈ (1/2, 3/4]: u → 2·(3/4) - 1/2 = 1, k stays 2. s' = 2^{-2}·1 = 1/4. φ = 2 - 1/4 = 7/4 ✓ (matches φ(29/16) = 7/4 ✓). Next: x = 7/4: s = 1/4, k = 2, u = 1. u ∈ (3/4, 1]: u → 3/4, k → 1. s' = 2^{-2}·3/4 = 3/16?? Wait, k' = 1, so s' should be 2^{-k'}·u' = 2^{-1}·(3/4) = 3/8. Hmm, let me recompute. Rule: u ∈ (3/4, 1]: s' = 2^{-(k+1)}(4u - 1). At u = 1: s' = 2^{-(k+1)}·3 = 3·2^{-(k+1)}. k' = k - 1, u' = s'/2^{-k'} = 3·2^{-(k+1)}/2^{-(k-1)} = 3·2^{-2} = 3/4. OK so u' = 3/4, k' = k-1. For x = 7/4: k = 2 → k' = 1, u' = 3/4, s' = 2^{-1}·3/4 = 3/8. φ = 2 - 3/8 = 13/8 ✓ (matches φ(7/4) = 13/8 ✓). I made an error above saying s' = 2^{-2}·(3/4); correct: u' = 3/4, s' = 2^{-k'}·u' = 2^{-1}·(3/4) = 3/8. Good. Next: x = 13/8: s = 3/8, k = 1 (s ∈ (1/4, 1/2] ✓), u = (3/8)/(1/2) = 3/4. u = 3/4 ∈ (1/2, 3/4]: u → 1, k stays 1. s' = 2^{-1}·1 = 1/2. φ = 2 - 1/2 = 3/2 ✓ (φ(13/8) = 3/2 ✓). Next: x = 3/2: s = 1/2, k = 1, u = 1. Hmm wait, 3/2 = 2 - 1/2, s = 1/2 = 2^{-1}, k = 1, u = 1. But actually x = 3/2: x - 1 = 1/2, f(1/2) = 1/4, φ(3/2) = 3/2 - 1/4 = 5/4. Via formula: u = 1 ∈ (3/4, 1]: u' = 3/4, k' = 0, s' = 2^0·(3/4) = 3/4. φ = 2 - 3/4 = 5/4 ✓. Next: x = 5/4: s = 3/4, k = 0 (s ∈ (1/2, 1]), u = 3/4. u ∈ (1/2, 3/4]: u → 1, k stays 0. s' = 1·1 = 1. φ = 2 - 1 = 1. Hmm, φ(5/4) = 1? Check directly: x = 5/4: x - 1 = 1/4, f(1/4) = 1/4, φ = 5/4 - 1/4 = 1 ✓. Next: x = 1: x - 1 = 0, f(0) = 1/2, φ(1) = 1 - 1/2 = 1/2. Then f(1/2) is on [0,1): direct. So chain from 15/8: 15/8 → 29/16 → 7/4 → 13/8 → 3/2 → 5/4 → 1 → 1/2. And f(15/8) = f(1/2)/2^7 = (1/4)/128 = 1/512 ✓. Now let's do f(3) = f(3 - 1/1024)/2 = f(x₀)/2, x₀ = 3 - 2^{-10}. Now I need the dynamics for x ∈ [2, 3). Similar analysis: for x ∈ [2, 3), x - 1 ∈ [1, 2). Need f on [1, 2) at point x - 1. For y ∈ [1, 2) (y = x - 1): f(y) = f(φ(y))/2, φ(y) = 2y - 2 + 2^{-(k+1)} for y ∈ [2 - 2^{-k}, 2 - 2^{-(k+1)}), k ≥ 0. So f(x - 1) = f(φ(x-1))/2. Hmm, but for the chain from x₀, I need f(x - 1) as a value (to compute the step), not just halving. Let me think again. Chain for f(x₀), x₀ ∈ [2, 3): a₀ = x₀, a_{i+1} = a_i - f(a_i - 1), f(x₀) = f(a_N)/2^N where a_N < 0 (bottom out)... but wait, we showed a_{i+1} > a_i - 1/8 ≥ 2 - 1/8 > 0 hmm, but that was assuming a_i ∈ [2, 3) and f(a_i - 1) ≤ 1/8. Actually as the chain decreases below 2 into [1, 2), then f(a_i - 1) has a_i - 1 ∈ [0, 1) with f up to 1/2, so steps can be bigger. Anyway, the chain: a₀ = 3 - 2^{-10}. Step 1: a₀ - 1 = 2 - 2^{-10}. Need f(2 - 2^{-10}). From the [1,2) analysis: f(2 - 2^{-m}) = 2^{-(2m+3)}? We verified m = 1..4. Let me re-derive via the u-dynamics to be sure, for general m. x = 2 - 2^{-m}: s = 2^{-m}, k = m, u = 1. Dynamics: u = 1 → (3/4, 1] rule: u' = 3/4, k' = m - 1. Then u = 3/4: (1/2, 3/4] rule: u' = 2·(3/4) - 1/2 = 1, k' = m - 1. Then u = 1 → u' = 3/4, k' = m - 2. Then u = 3/4 → u' = 1, k' = m - 2. ... alternating: each pair of steps reduces k by 1: (u=1,k) → (u=3/4,k-1) → (u=1,k-1) → (u=3/4,k-2) → (u=1,k-2) → ... Wait let me redo: start (u=1, k=m). Step: u=1 ∈ (3/4,1] → (u=3/4, k=m-1). Step: u=3/4 ∈ (1/2,3/4] → (u=1, k=m-1). Step: u=1 → (u=3/4, k=m-2). Step: → (u=1, k=m-2). So after 2 steps, k decreases by 1. Starting k=m, we reach k=0 after 2m steps. At k=0: hmm, when k=0 and u ∈ (3/4, 1]: exit to [0,1) region: s' = 2^0·... let me recompute: k=0, u=1: s' = 3·2^{-1} = 3/2 > 1, so φ = 2 - 3/2 = 1/2 ∈ [0, 1). Exit! Let me recount: starting (1, m). Steps: (1,m) → (3/4, m-1) → (1, m-1) → (3/4, m-2) → (1, m-2) → ... → (3/4, 0) → (1, 0) → exit. From (1, m) to (1, 0): 2m steps. Then from (1, 0): one more step exits to φ = 1/2. Total steps: 2m + 1. Then f(1/2) = 1/4. So f(2 - 2^{-m}) = f(1/2)/2^{2m+1} = 2^{-2}/2^{2m+1} = 2^{-(2m+3)} ✓✓. Great, conjecture confirmed: f(2 - 2^{-m}) = 2^{-(2m+3)}, m ≥ 1. (Also check m=1: f(3/2) = 2^{-5} ✓.) Now, f(2) itself: x = 2: x - 1 = 1, f(1) = 2^{-3}. φ(2) = 2 - 1/8 = 15/8. f(2) = f(15/8)/2 = 2^{-9}/2 = 2^{-10} ✓. Now the main event: f(x₀), x₀ = 3 - 2^{-10} = 3 - s₀, s₀ = 2^{-10}. I need the dynamics on [2, 3). For x ∈ [2, 3): a_{i+1} = x - f(x - 1). Let me set up similar machinery. Write x = 3 - s, s ∈ (0, 1]. x - 1 = 2 - s. Need f(2 - s) for s ∈ (0, 1]. Hmm, f on [1, 2) in terms of s = 2 - y (y ∈ [1, 2)): we have the (u, k) parametrization: s ∈ (2^{-(k+1)}, 2^{-k}], u = s·2^k ∈ (1/2, 1], and dynamics as above. The value: f(2 - s) = f(exit)/2^{steps}. Let me figure out f(2 - s) for general s. From the dynamics: starting (u₀, k₀) with s = 2^{-k₀}u₀: - If u ∈ (3/4, 1]: go to (3/4, k-1) [1 step] - If u ∈ (1/2, 3/4]: go to (2u - 1/2, k) [1 step] Exit happens from k=0 (s ∈ (1/2, 1]) when we'd compute s' > 1. Specifically from state (u, 0): if u ∈ (3/4, 1]: s' = 2^{-1}(4u - 1) ∈ (1, 3/2], φ = 2 - s' ∈ [1/2, 1): exit with φ ∈ [1/2, 1). If u ∈ (1/2, 3/4]: s' = 2^{-1}(4u-1) ∈ (1/2·... 4u - 1 ∈ (1, 2], s' ∈ (1/2, 1], φ = 2 - s' ∈ [1, 3/2). Hmm wait, that's not exit; φ ∈ [1, 3/2) ⊂ [1, 2). Hmm, so from k=0, u ∈ (1/2, 3/4]: φ(x) = 2 - s' with s' ∈ (1/2, 1], so φ ∈ [1, 3/2). But that's y ∈ [1, 2) again — no exit! But wait, k=0 and u ∈ (1/2, 3/4] means s ∈ (1/2, 3/4], x = 2 - s ∈ [5/4, 3/2). And φ maps [5/4, 3/2) → [1, 3/2). Right: earlier we had for x ∈ [1, 3/2): φ(x) = 2x - 3/2 ∈ [1/2, 3/2). The part x ∈ [5/4, 3/2) maps to φ ∈ [1, 3/2). OK. So exit to [0, 1) only happens from x ∈ [1, 5/4) (s ∈ (3/4, 1]) via u ∈ (3/4, 1] at k=0: φ = 2 - (4u-1)/2 with u ∈ (3/4, 1] → 4u - 1 ∈ (2, 3] → φ ∈ [1/2, 1). Precisely: x = 2 - u (u ∈ (3/4, 1]), φ(x) = 2x - 3/2 = 2(2 - u) - 3/2 = 5/2 - 2u ∈ [1/2, 1). So: f on [1, 2): f(x) = f(φ^T(x))/2^T where φ^T(x) ∈ [1/2, 1). Let me now count steps T(s) as function of s = 2 - x, and figure out the exit point. Dynamics summary (state (u, k), s = 2^{-k}u): 1. u ∈ (1/2, 3/4]: (u, k) → (2u - 1/2, k). Note 2u - 1/2 ∈ (1/2, 1]. 2. u ∈ (3/4, 1]: (u, k) → (3/4, k-1) [when k ≥ 1]. At k = 0: 1'. u ∈ (1/2, 3/4]: → (2u - 1/2, 0). (stays in [1, 2)) 2'. u ∈ (3/4, 1]: exit: φ = 5/2 - 2u ∈ [1/2, 1). Hmm wait, rule 2 at k=0: (u, 0) → (3/4, -1) means s' = 2^{1}·(3/4)·... k' = -1: s' ∈ (2^0, 2^1]?? Let me recompute: rule 2: s' = 2^{-(k+1)}(4u - 1). At k=0, u ∈ (3/4, 1]: s' = (4u-1)/2 ∈ (1, 3/2]. φ = 2 - s' ∈ [1/2, 1). Exit ✓. Let me understand rule 1 dynamics on u: u → 2u - 1/2 for u ∈ (1/2, 3/4]. Fixed point: u = 2u - 1/2 → u = 1/2. So iterating rule 1 from u ∈ (1/2, 3/4]: u increases towards... 2u - 1/2 > u ↔ u > 1/2. So u increases, eventually exceeding 3/4, then rule 2 applies. Hmm, but this could take many steps if u close to 1/2. In terms of dyadic: u = 1/2 + ε: u' = 1/2 + 2ε. So ε doubles each step. Starting ε₀, after t steps: ε = 2^t ε₀. Exit rule 1 when 1/2 + 2^t ε₀ > 3/4, i.e., 2^t ε₀ > 1/4. Hmm OK. But note: our points of interest are dyadic rationals, so this terminates. Now, the outer chain for x₀ = 3 - 2^{-10}: a₀ = 3 - 2^{-10}. a₀ - 1 = 2 - 2^{-10}. f(a₀ - 1) = f(2 - 2^{-10}) = 2^{-23} (m = 10: 2m+3 = 23). a₁ = a₀ - 2^{-23} = 3 - 2^{-10} - 2^{-23}. a₁ - 1 = 2 - 2^{-10} - 2^{-23}. Need f(2 - 2^{-10} - 2^{-23}). Hmm. So I need f on [1, 2) at points 2 - s with s = 2^{-10} + 2^{-23}, etc. This is getting into heavy detail. Let me set up the (u, k) machinery to compute f(2 - s) for the specific s values. General: f(2 - s), s ∈ (0, 1]. Find k with s ∈ (2^{-(k+1)}, 2^{-k}], u = s/2^{-k} = s·2^k ∈ (1/2, 1]. Then iterate rules until exit, counting steps T, and track exit point z = φ^T ∈ [1/2, 1) (or z = 1/2? can z = 1/2 exactly? u = 1, k = 0: z = 5/2 - 2 = 1/2. Yes.) Then f(2 - s) = f(z)/2^T, and f(z) for z ∈ [1/2, 1): f(z) = (1 - z) - 2^{-(k_z + 1)} where z ∈ [1 - 2^{-k_z}, 1 - 2^{-(k_z+1)}), k_z ≥ 1. Hmm wait, but z ∈ [1/2, 1): need k_z ≥ 1 since z ≥ 1/2 = 1 - 2^{-1}: z ∈ [1/2, 1) → z ∈ [1 - 2^{-k_z}, 1 - 2^{-(k_z+1)}) for unique k_z ≥ 1. f(z) = (1-z) - 2^{-(k_z+1)}. OK let me now carefully trace. Let me define the outer chain: a₀ = 3 - 2^{-10}, a_{i+1} = a_i - f(a_i - 1), and f(a_i) = f(a_{i+1})/2 as long as a_i ≥ 2 hmm, actually f(a) = f(a - f(a-1))/2 for any a ≥ 0; the chain relation f(a_i) = f(a_{i+1})/2 holds regardless, as long as a_i ≥ 0. We bottom out when a_i < 0 (f(a_i) = -a_i) — but as discussed, a_i stays ≥ 2 - 1/8 hmm, actually while a_i ∈ [2, 3), steps are ≤ max f on [1, 2) = 1/8; when a_i ∈ [1, 2) steps ≤ max f on [0,1) = 1/2; when a_i ∈ [0, 1), steps ≤ 1/2. The chain from a₀ = 3 - 2^{-10} ≈ 2.999 will decrease; each step subtracts f(a_i - 1) which is positive but tiny. Total decrease needed to get below 2: about 1 - 2^{-10}. Steps are tiny (like 2^{-23}), so many steps! Unless the step sizes grow. Hmm wait. Let me reconsider. As a_i decreases from ~3 towards 2, a_i - 1 decreases from ~2 towards 1, and f(a_i - 1) varies over f on [1, 2): tiny values (≤ 2^{-10}?). Hmm, f on [1, 2) has max f(1) = 2^{-3} = 1/8, but on [1.9, 2), f is like 2^{-9}·small... Let me think: f(2 - s) = f(z)/2^T where T ≈ ? For s small (x close to 2), T is large: T(s) ≈ 2k where k ≈ log2(1/s)... e.g., s = 2^{-m}: T = 2m + 1. And f(z) ≤ 1/4. So f(2 - s) ≈ 2^{-T-2} with T ≈ 2 log2(1/s), so f(2 - s) ≈ s²·2^{-4}·... for s = 2^{-m}: f = 2^{-2m-3} = s²·2^{2m}·... s² = 2^{-2m}, so f = s²/8. Interesting: f(2 - s) ≈ s²/8 for dyadic s. So the outer chain from 3 - 2^{-10}: steps of size f(a_i - 1) ≈ s_i²/8 where s_i = 3 - a_i ∈ (0, 1): wait no. Let me define s_i = 3 - a_i. Then a_i - 1 = 2 - s_i. f(a_i - 1) = f(2 - s_i) ≈ s_i²/8. And s_{i+1} = s_i + f(2 - s_i) ≈ s_i + s_i²/8. Starting s₀ = 2^{-10} ≈ 0.001: s grows slowly: ds/di = s²/8. To reach s = 1 (a = 2): di = 8 ds/s² → i ≈ 8(1/s₀ - 1) ≈ 8·1024 ≈ 8192 steps. Hmm, that's a lot of steps, but maybe there's structure making it tractable exactly. The total halving: f(3) = f(a_N)/2^N where N = total steps to bottom out. If N ≈ 8192 + (steps in [1,2)) + (steps in [0,1))... this suggests f(3) = 2^{-N-ish}, and N is roughly 8·1024 = 8192... Hmm, but my earlier guess was f(3) = 2^{-36} = 2^{-T_8}. T_8 = 36. That's way off from 8192. Let me re-examine. Hmm wait, maybe I misestimated. Let me reconsider: f(2 - s) ≈ s²/8? For s = 2^{-1} = 1/2: f(3/2) = 2^{-5} = 1/32. s²/8 = (1/4)/8 = 1/32 ✓. For s = 2^{-2}: f(7/4) = 2^{-7} = 1/128. s²/8 = (1/16)/8 = 1/128 ✓. s = 2^{-3}: f(15/8) = 2^{-9} = 1/512 = (1/64)/8 ✓. Great, so f(2 - s) = s²/8 exactly for s = 2^{-m}. Interesting! So conjecture: f(2 - s) = s²/8 for all s ∈ (0, 1]?? Let me test with a non-dyadic-ish point: s = 3/4: x = 5/4, f(5/4) = 1/16. s²/8 = (9/16)/8 = 9/128 ≠ 1/16 = 8/128. Close but not equal. Hmm. So not exact in general. s = 3/8: x = 13/8, f = 2^{-6} = 1/64 = 16/1024. s²/8 = (9/64)/8 = 9/512 = 18/1024. Not equal (16 vs 18). Hmm. So f(2 - s) = s²/8 holds for s = 2^{-m} but not generally. The outer chain will involve general s. Hmm. OK so let me reconsider. Maybe I should think about the outer chain more cleverly. Outer chain: a₀ = 3 - 2^{-10}, a_{i+1} = a_i - f(a_i - 1). Let s_i = 3 - a_i ∈ [2^{-10}, ...). f(a_i - 1) = f(2 - s_i). Let me get exact formula for f(2 - s) for s ∈ (0, 1]. Using the (u,k) dynamics: s ∈ (2^{-(k+1)}, 2^{-k}], u = 2^k s. Dynamics: - Rule A (u ∈ (1/2, 3/4]): u → 2u - 1/2 (k fixed). In terms of s: s → 2s - 2^{-(k+1)}. [since s' = 2^{-(k+1)}(4u - 1) = 2s·... let me verify: s' = 2^{-(k+1)}(4u - 1) = 2^{-(k+1)}(4·2^k s - 1) = 2^{-(k+1)}(2^{k+2} s - 1) = 2s - 2^{-(k+1)}.] ✓ - Rule B (u ∈ (3/4, 1], k ≥ 1): s → s' with u' = 3/4, k' = k - 1: s' = 2^{-(k-1)}·(3/4) = 3·2^{-(k+1)}·... 2^{-(k-1)} = 2^{-(k+1)}·4, so s' = 3·2^{-(k+1)}. In terms of s: s ∈ (3·2^{-(k+1)}, 2^{-k}] → s' = 3·2^{-(k+1)}. - Exit (k=0, u ∈ (3/4, 1], i.e., s ∈ (3/4, 1]): φ = 2 - s' where s' = (4u-1)/2 = (4s - 1)/2, so φ = 2 - (4s-1)/2 = (5 - 4s)/2. Check s=1: φ = 1/2 ✓; s → 3/4+: φ → (5-3)/2 = 1. ✓. And each step halves f. Also need: exit value f(φ) where φ ∈ [1/2, 1): f(φ) = (1 - φ) - 2^{-(k_φ+1)}. Alternatively, maybe I should find closed form for f(2 - s). Let me think about what the dynamics do to s. Let me re-express: x = 2 - s, s ∈ (0, 1]. k: 2^{-(k+1)} < s ≤ 2^{-k}. Rule A: s ∈ (2^{-(k+1)}, 3·2^{-(k+1)}] → s' = 2s - 2^{-(k+1)}. Rule B: s ∈ (3·2^{-(k+1)}, 2^{-k}] → s' = 3·2^{-(k+1)}, k' = k-1. Hmm, let me define t = 2^{k+1} s ∈ (1, 2]. Then: Rule A: t ∈ (1, 3/2] → t' = 2t - 2 (k fixed). Rule B: t ∈ (3/2, 2] → t' = 3, k' = k - 1. Hmm, t' = 2^{k'+1} s' = 2^k · 3·2^{-(k+1)} = 3/2·... let me compute: s' = 3·2^{-(k+1)}, k' = k-1: t' = 2^{k'}+1... t' = 2^{(k-1)+1} s' = 2^k·3·2^{-(k+1)} = 3/2. So t' = 3/2 (and then Rule A applies to t=3/2: t → 1). Interesting. So in t-coordinates: t ∈ (1, 2]: - t ∈ (1, 3/2]: t → 2t - 2 ∈ (0, 1]... wait, if t = 1.2: t' = 0.4?? That's out of range (1, 2]. Hmm, I think I need to be careful: after Rule A, t' = 2t - 2 which for t ∈ (1, 3/2] gives t' ∈ (0, 1]. But t' must be in (1, 2] for the parametrization... contradiction means k changes? Let me recheck. Rule A: s → s' = 2s - 2^{-(k+1)}, k' = k. t' = 2^{k+1} s' = 2^{k+1}(2s - 2^{-(k+1)}) = 2·2^{k+1}s - 1 = 2t - 1. Ah, t' = 2t - 1, not 2t - 2. Let me recompute: t = 2^{k+1} s. s' = 2s - 2^{-(k+1)}. t' = 2^{k+1}·s' = 2^{k+1}·2s - 1 = 2t - 1. Yes t' = 2t - 1 ∈ (1, 2] for t ∈ (1, 3/2]. ✓. Rule B: s ∈ (3·2^{-(k+1)}, 2^{-k}]: t ∈ (3/2, 2]. s' = 3·2^{-(k+1)}, k' = k - 1: t' = 2^{k'}... t' = 2^{(k-1)+1}·3·2^{-(k+1)} = 2^k·3·2^{-(k+1)} = 3/2. So t-dynamics: t ∈ (1, 2]: - t ∈ (1, 3/2]: t → 2t - 1 (k same) - t ∈ (3/2, 2]: t → 3/2 (k decreases by 1) with s = t·2^{-(k+1)}. So overall: f(2 - s) where s = t·2^{-(k+1)}, t ∈ (1, 2]: Each step: f halves. Steps: while t ∈ (3/2, 2]: t → 3/2, k → k-1. Then t = 3/2 → 1 (Rule A: 2·3/2 - 1 = 2? Wait: 2t - 1 at t = 3/2: 3 - 1 = 2. Hmm, 2·(3/2) - 1 = 2. But earlier I said t = 3/2 → t' = 1?? Let me recheck with the earlier example. Earlier: x = 7/4: s = 1/4, k = 2, u = 1, t = 2^{k+1} s = 8·(1/4) = 2. Step: t = 2 ∈ (3/2, 2]: t → 3/2, k → 1. So s' = 3/2·2^{-2} = 3/8 ✓ (matches φ(7/4) = 13/8, s' = 3/8 ✓). Next: x = 13/8: s = 3/8, k = 1 (s ∈ (1/4, 1/2] ✓), t = 2^2·(3/8) = 3/2. Step: t = 3/2 ∈ (1, 3/2]: t → 2t - 1 = 2, k same = 1. s' = 2·2^{-2} = 1/2 ✓ (φ(13/8) = 3/2, s' = 1/2 ✓). Next: x = 3/2: s = 1/2, k = 1, t = 4·(1/2) = 2. Step: t = 2 → 3/2, k → 0. s' = (3/2)·2^{-1} = 3/4 ✓ (φ(3/2) = 5/4, s' = 3/4 ✓). Next: x = 5/4: s = 3/4, k = 0, t = 2·(3/4) = 3/2. Step: t = 3/2 ∈ (1, 3/2]: t → 2, k = 0. s' = 2·2^{-1} = 1. φ = 1 ✓. Next: x = 1: s = 1, k = 0, t = 2. Step: k = 0, t = 2 ∈ (3/2, 2]: exit. φ = 2 - s' = 2 - (4s - 1)/2 = ... s=1: φ = 2 - 3/2 = 1/2 ✓. OK so the t-dynamics: t ∈ (1, 2]: - t ∈ (1, 3/2]: t → 2t - 1 ∈ (1, 2] - t ∈ (3/2, 2]: t → 3/2 (with k decrement) So from any t ∈ (1, 2]: if t ∈ (3/2, 2], one step to t = 3/2. From t = 3/2: → 2 (one step). From t = 2: → 3/2 (k decrement) or exit (if k = 0). From t ∈ (1, 3/2): t → 2t - 1, i.e., in terms of binary: t = 1 + ε, ε ∈ (0, 1/2]: t' = 1 + 2ε. So ε doubles until > 1/2, i.e., t ∈ (3/2, 2], then → 3/2 → 2 → ... So the cycle: t = 2 ↔ 3/2: 2 → 3/2 → 2 → 3/2 → ..., with k decreasing every 2 steps (t=2 step decrements k). Precisely: t = 3/2 → t = 2 (k same); t = 2 → t = 3/2 (k-1). So each "period" of 2 steps: k decreases by 1. Starting: s = t₀·2^{-(k₀+1)}. Steps to reach k = 0: depends. Let me count total steps T and exit φ. Case t₀ ∈ (1, 3/2): t₀ = 1 + ε₀, ε₀ ∈ (0, 1/2). Steps to get ε > 1/2: ε_i = 2^i ε₀, need 2^i ε₀ > 1/2... but wait when ε_i ∈ (1/2·... t = 1 + ε; Rule A applies when t ∈ (1, 3/2], i.e., ε ∈ (0, 1/2]. When ε > 1/2 (t ∈ (3/2, 2]), Rule B: t → 3/2. Hmm, but note t = 1 + ε with ε ∈ (1/2, 1]: one step to t = 3/2. Then 3/2 → 2 → 3/2 → 2..., k decreasing by 1 every second step. Hmm, so let me define: total steps from (t₀, k₀) to exit: Let me think of it as: the dynamics quickly reach the 2-cycle {3/2, 2}, then each 2 steps decrement k, until k = 0, then exit at the t=2 step. Actually wait: when k = 0 and t = 2: exit. When k = 0 and t = 3/2: t → 2 (Rule A, k stays 0), then exit. When k = 0, t ∈ (1, 3/2]: t → 2t - 1... but hold on, with k = 0, s = t/2 ∈ (1/2, 1]. x = 2 - s ∈ [1, 3/2). Rule A at k=0: s' = 2s - 1/2, φ = 2 - s' = 2 - 2s + 1/2 = 5/2 - 2s ∈ [1/2, 3/2). Hmm: s ∈ (1/2, 3/4] → φ ∈ [1, 3/2). s ∈ (3/4, 1) → Rule B exit: φ = (5 - 4s)/2 ∈ (1/2, 1). Hmm OK wait, I realize Rule A can apply at k = 0 without exiting: t ∈ (1, 3/2], k=0: t → 2t - 1 ∈ (1, 2], k = 0. Then if t ∈ (3/2, 2], Rule B at k = 0: exit. So exit happens at first Rule B application when k = 0. Let me now just carefully count for our specific chain. Actually, let me now set up the outer chain properly. Outer chain: s_i = 3 - a_i, s₀ = 2^{-10}. Step: s_{i+1} = s_i + f(2 - s_i). I need f(2 - s) for s = 2^{-10}, then s = 2^{-10} + f(2 - 2^{-10}) = 2^{-10} + 2^{-23}, etc. So I need f(2 - s) for a sequence of s values. Let me find the pattern of f(2 - s) for s ∈ (0, 1] in terms of the (t, k) parametrization, get a formula, then iterate the outer chain. General formula attempt: given s ∈ (0, 1], k₀ = floor stuff: k₀ such that s ∈ (2^{-(k₀+1)}, 2^{-k₀}], t₀ = 2^{k₀+1} s ∈ (1, 2]. Dynamics in steps (each halving f): Phase 1: while t ∈ (1, 3/2]: t → 2t - 1. (Rule A) Phase 2: t ∈ (3/2, 2]: t → 3/2, k → k - 1. (Rule B) Then t = 3/2 → 2 (Rule A), then t = 2 → 3/2 (Rule B, k-1), alternating, k decreasing every 2 steps, until k = 0, then the Rule B step at k = 0 exits. Wait, but careful: Phase 2 from t ∈ (3/2, 2] goes to t = 3/2 with k decremented. Then t = 3/2 (Rule A) → t = 2, k same. Then t = 2 (Rule B) → t = 3/2, k-1. So yes: after reaching t ∈ (3/2, 2] (call it "high"), steps alternate: high → 3/2 (k-1) → 2 (k) → 3/2 (k-1) → 2 (k) → ... Each pair of steps decrements k by 1. Exit when at k = 0 we apply Rule B: exit φ = 2 - (4s_current... hmm, exit φ depends on current s at k=0: φ = 5/2 - 2s where s ∈ (3/4, 1] (t ∈ (3/2, 2], k = 0: s = t/2 ∈ (3/4, 1]). Alternatively φ = (5 - 4s)/2? Let me recompute: Rule B: s' = 3·2^{-(k+1)}; at k=0: s' = 3/2; φ = 2 - s' = 1/2. Wait, that's different from what I had! Hmm, contradiction. Let me recompute Rule B at k=0. Rule B: s ∈ (3·2^{-(k+1)}, 2^{-k}], k ≥ 1 normally. The derivation: u ∈ (3/4, 1]: s' = 2^{-(k+1)}(4u - 1). With u = 2^k s: 4u - 1 = 2^{k+2}s - 1. s' = 2^{-(k+1)}(2^{k+2}s - 1) = 2s - 2^{-(k+1)}. Hmm! So Rule B gives s' = 2s - 2^{-(k+1)} too?! Wait, I think I conflated. Let me redo this carefully. φ(x) = x - f(x - 1). For x = 2 - s (y = x - 1 = 1 - ... hmm wait. x ∈ [1, 2): x - 1 ∈ [0, 1). Let me redo: x ∈ [1, 2), x - 1 ∈ [0, 1) = y. f(y) = (1 - y) - 2^{-(k+1)} for y ∈ [1 - 2^{-k}, 1 - 2^{-(k+1)}). y = x - 1 = 1 - s. So y ∈ [1 - 2^{-k}, 1 - 2^{-(k+1)}) ↔ s ∈ (2^{-(k+1)}, 2^{-k}]. f(y) = (1 - y) - 2^{-(k+1)} = s - 2^{-(k+1)}. φ(x) = x - f(x-1) = (2 - s) - (s - 2^{-(k+1)}) = 2 - 2s + 2^{-(k+1)}. So s' = 2 - φ(x) = 2s - 2^{-(k+1)}. OH WAIT. This is the formula for φ for x ∈ [1, 2)! I previously derived φ(x) = 2x - 2 + 2^{-(k+1)} — check: 2x - 2 + 2^{-(k+1)} = 2(2-s) - 2 + 2^{-(k+1)} = 2 - 2s + 2^{-(k+1)} ✓ consistent. Good. So s' = 2s - 2^{-(k+1)} for ALL s ∈ (2^{-(k+1)}, 2^{-k}] — both Rules A and B?! But then what distinguishes Rule A and Rule B? The k' of the new point: s' determines new k' by which interval s' falls into. Hmm, so I overcomplicated: there's only ONE rule: s' = 2s - 2^{-(k+1)}, k = current level; then re-compute k' from s'. Let me recheck with examples: x = 7/4: s = 1/4, k = 2 (s ∈ (1/8, 1/4]). s' = 2·(1/4) - 1/8 = 1/2 - 1/8 = 3/8 ✓ (φ = 2 - 3/8 = 13/8 ✓). New k': s' = 3/8 ∈ (1/4, 1/2] → k' = 1 ✓. x = 13/8: s = 3/8, k = 1. s' = 2·(3/8) - 1/4 = 3/4 - 1/4 = 1/2 ✓ (φ = 3/2 ✓). k': s' = 1/2 ∈ (1/4, 1/2] → k' = 1. x = 3/2: s = 1/2, k = 1. s' = 1 - 1/4 = 3/4 ✓ (φ = 5/4 ✓). k': 3/4 ∈ (1/2, 1] → k' = 0. x = 5/4: s = 3/4, k = 0. s' = 3/2 - 1/2 = 1 ✓ (φ = 1 ✓). k': s' = 1 ∈ (1/2, 1] → k' = 0. x = 1: s = 1, k = 0. s' = 2 - 1/2 = 3/2. φ = 1/2 ✓ (φ(1) = 1 - f(0) = 1 - 1/2 = 1/2 ✓). s' = 3/2 > 1: exit to [0, 1) ✓. x = 15/8: s = 1/8, k = 3. s' = 1/4 - 1/16 = 3/16 ✓ (φ = 29/16 ✓). k' = 2 (3/16 ∈ (1/8, 1/4]). x = 29/16: s = 3/16, k = 2. s' = 3/8 - 1/8 = 1/4 ✓ (φ = 7/4 ✓). k' = 2 (1/4 ∈ (1/8, 1/4]). Great, so single rule: s' = 2s - 2^{-(k+1)}, k' = level(s'). So the dynamics on s ∈ (0, 1]: s' = 2s - 2^{-(k+1)} where k = level(s) (s ∈ (2^{-(k+1)}, 2^{-k}]). In terms of t = 2^{k+1}s ∈ (1, 2]: s' = 2s - 2^{-(k+1)}. Hmm, t' = 2^{k'+1}s'. If k' = k: t' = 2^{k+1}(2s - 2^{-(k+1)}) = 2t - 1. When is k' = k? s' ∈ (2^{-(k+1)}, 2^{-k}] ↔ 2s - 2^{-(k+1)} ∈ (2^{-(k+1)}, 2^{-k}] ↔ s ∈ (2^{-k}, 2^{-k} + 2^{-(k+2)}·...] hmm: 2s ∈ (2^{-k+1}·... let me solve: s' > 2^{-(k+1)} ↔ s > 2^{-k}. But s ≤ 2^{-k} (level k). So s' > 2^{-(k+1)} ↔ s > 2^{-k}: impossible except s = 2^{-k} gives s' = 2^{-k}·... s = 2^{-k}: s' = 2^{-k+1} - 2^{-(k+1)} = 2^{-(k+1)}(4 - 1)·... 2^{-k+1} = 2^{-(k+1)}·4. So s' = 3·2^{-(k+1)} ∈ (2^{-(k+1)}, 2^{-k}] ✓ (3·2^{-(k+1)} ≤ 2^{-k} = 4·2^{-(k+1)} ✓). So s = 2^{-k} (t = 2): k' = k, t' = 2t - 1 = 3. Hmm, t' = 3 > 2?! That breaks the parametrization. Hmm: t' = 2^{k'+1}s' = 2^{k+1}·3·2^{-(k+1)} = 3. But t should be in (1, 2]... Contradiction. Let me recheck: s = 2^{-k} (i.e., s = 2^{-k}, the right endpoint of level k, e.g., x = 7/4: s = 1/4, k = 2). s' = 3·2^{-(k+1)} = 3/8. Level of 3/8: 3/8 ∈ (1/4, 1/2]: k' = 1 = k - 1. So k' = k - 1, not k! Let me redo: s' = 3·2^{-(k+1)} = 3·2^{-(k+1)}. Is 3·2^{-(k+1)} ≤ 2^{-k}? 3·2^{-(k+1)} vs 2^{-k} = 2·2^{-(k+1)}: 3 > 2, so 3·2^{-(k+1)} > 2^{-k}. So s' ∈ (2^{-k}, 2^{-(k-1)}]·... s' > 2^{-k} means level k' = k - 1 ✓ (since s' ≤ 2^{-(k-1)} = 4·2^{-(k+1)} ✓). OK so I made an arithmetic error: 2^{-k} = 2·2^{-(k+1)}, not 4·. Let me redo: s ∈ (2^{-(k+1)}, 2^{-k}] = (2^{-(k+1)}, 2·2^{-(k+1)}]. s' = 2s - 2^{-(k+1)} ∈ (2·2^{-(k+1)} - 2^{-(k+1)}, 4·2^{-(k+1)} - 2^{-(k+1)}] = (2^{-(k+1)}, 3·2^{-(k+1)}]. So s' ∈ (2^{-(k+1)}, 3·2^{-(k+1)}]. Levels: - If s' ∈ (2^{-(k+1)}, 2·2^{-(k+1)}] = (2^{-(k+1)}, 2^{-k}]: k' = k. Corresponds to s ∈ (2^{-k}, ... wait: s' ≤ 2^{-k} ↔ 2s - 2^{-(k+1)} ≤ 2·2^{-(k+1)} ↔ 2s ≤ 3·2^{-(k+1)} ↔ s ≤ (3/2)·2^{-(k+1)} = 3·2^{-(k+2)}. - If s' ∈ (2^{-k}, 3·2^{-(k+1)}]: k' = k - 1. ↔ s ∈ (3·2^{-(k+2)}, 2·2^{-(k+1)} = 2^{-k}]. In t-coordinates (t = 2^{k+1}s ∈ (1, 2]): - t ∈ (1, 3/2]: s' ∈ level k, t' = 2^{k+1}s' = 2^{k+1}(2s - 2^{-(k+1)}) = 2t - 1 ∈ (1, 2]. - t ∈ (3/2, 2]: k' = k - 1, t' = 2^{k'}... t' = 2^{(k-1)+1}s' = 2^k s'. s' = 2s - 2^{-(k+1)} = 2^{-(k+1)}(2t - 1)·... s = 2^{-(k+1)}t, so s' = 2·2^{-(k+1)}t - 2^{-(k+1)} = 2^{-(k+1)}(2t - 1). t' = 2^k·2^{-(k+1)}(2t-1) = (2t-1)/2 ∈ (1, 3/2]. OK so this matches my earlier Rules A/B (roughly): - t ∈ (1, 3/2]: t → 2t - 1 (k same). [2t-1 ∈ (1, 2]] - t ∈ (3/2, 2]: t → (2t-1)/2 = t - 1/2 ∈ (1, 3/2] (k-1). Hmm interesting, so earlier I had Rule B as t → 3/2 which is wrong except... let me recheck with example: x = 7/4: s = 1/4, k = 2, t = 2^{3}·(1/4) = 2. Rule: t = 2 ∈ (3/2, 2]: t' = t - 1/2 = 3/2, k' = 1. s' = 2^{-(k'+1)}t' = 2^{-2}·(3/2) = 3/8 ✓. And x = 15/8: s = 1/8, k = 3, t = 16·(1/8) = 2. t' = 3/2, k' = 2. s' = 2^{-3}·(3/2) = 3/16 ✓. x = 29/16: s = 3/16, k = 2, t = 8·(3/16) = 3/2. Rule A: t' = 2·(3/2) - 1 = 2, k' = 2. s' = 2^{-3}·2 = 1/4 ✓. x = 13/8: s = 3/8, k = 1, t = 4·(3/8) = 3/2. t' = 2, k' = 1. s' = 2^{-2}·2 = 1/2 ✓. x = 3/2: s = 1/2, k = 1, t = 4·(1/2) = 2. Rule B: t' = 3/2, k' = 0. s' = 2^{-1}·(3/2) = 3/4 ✓. x = 5/4: s = 3/4, k = 0, t = 2·(3/4) = 3/2. Rule A: t' = 2, k' = 0. s' = 2^{-1}·2 = 1 ✓. x = 1: s = 1, k = 0, t = 2. Rule B at k = 0: s' = 2^{-(0+1)}·(2t-1) = 2·... s' = (2t-1)/2 = (4-1)/2 = 3/2 > 1: exit, φ = 2 - 3/2 = 1/2 ✓. So dynamics: t ∈ (1, 2], k ≥ 0: - t ∈ (1, 3/2]: (t, k) → (2t - 1, k) - t ∈ (3/2, 2]: (t, k) → (t - 1/2, k - 1); if k = 0: exit with s' = (2t-1)/2 ∈ (1, 3/2], φ = 2 - s' ∈ [1/2, 1). Now, f(x) = f(φ(x))/2 per step. Let me now compute f(2 - s) for the needed points, i.e., total steps to exit and exit point. Given (t₀, k₀), dynamics: Note: t = 3/2 → 2 (Rule A); t = 2 → 3/2 (Rule B, k-1); t ∈ (1, 3/2) → 2t - 1 ∈ (1, 2). Let me think of t in binary: t = 1.b₁b₂b₃... Rule A: t → 2t - 1: shifts left: 1.b₂b₃... (drops b₁ = 0). Hmm: t ∈ (1, 3/2] means b₁ = 0. 2t - 1 = b₂b₃... in binary as 1.b₂b₃...? Let me verify: t = 1.01₂ = 1.25: 2t - 1 = 1.5 = 1.1₂. Yes: t = 1.0b₂b₃..., 2t - 1 = 1.b₂b₃... So Rule A removes a leading 0 after the point. Rule B: t ∈ (3/2, 2]: t = 1.1b₂b₃...: t - 1/2 = 1.0b₂b₃...: replaces leading 1 with 0. And k decreases. Exit (k=0, t ∈ (3/2, 2]): s' = (2t-1)/2 = t - 1/2·... φ = 2 - s'. Hmm wait s' = t - 1/2 computed above: (2t-1)/2 = t - 1/2. So φ = 2 - (t - 1/2) = 5/2 - t. With t ∈ (3/2, 2]: φ ∈ [1/2, 1). ✓ (matches earlier: φ = 5/2 - 2s at k=0? Let me check: earlier for k=0, Rule B: I computed φ = (5-4s)/2 = 5/2 - 2s. With t = 2^{k+1}s = 2s (k=0): 5/2 - t = 5/2 - 2s ✓.) So in binary, the dynamics on t = 1.b₁b₂b₃... (t ∈ (1, 2]): - If b₁ = 0 (t ∈ (1, 3/2]): t → 1.b₂b₃... (shift left; k unchanged) - If b₁ = 1 (t ∈ (3/2, 2]): t → 1.0b₂b₃... (b₁: 1 → 0; k decreases by 1) - Exit when k = 0 and b₁ = 1: φ = 5/2 - t. Interesting. So the process: each step either shifts out a 0, or flips a leading 1 to 0 (and decrements k). Equivalently: think of t - 1 = 0.b₁b₂b₃... ∈ (0, 1]. Steps: shift left dropping b₁ (if b₁ = 0, k same; if b₁ = 1, k-1). So each step shifts out one bit: 0-bits don't decrement k, 1-bits decrement k. Exit when k reaches 0 AND we shift out a 1... hmm wait exit condition: k = 0 and b₁ = 1 (before shift). Then φ = 5/2 - t where t = 1.1b₂b₃... Hmm, let me recount: exit happens at a Rule B step with k = 0. At that moment t = 1.1b₂b₃..., φ = 5/2 - t = 3/2 - 0.b₂b₃...·... t = 1.1b₂b₃... = 3/2 + 0.0b₂b₃... = 3/2 + (t - 3/2). φ = 5/2 - t = 1 - (t - 3/2) = 1 - 0.0b₂b₃... So φ = 1 - (fractional part beyond the leading 1). Hmm, nice. Let me now also handle: what if t hits exactly values like 3/2 or 2? t = 3/2 = 1.1: b₁ = 1, so Rule B applies (t ∈ (3/2, 2] includes 3/2? t ∈ (3/2, 2] — 3/2 is excluded! Rule A: t ∈ (1, 3/2] includes 3/2. So t = 3/2: Rule A: 2t - 1 = 2. OK: t = 1.1000... shifts out the 1? Rule A is for b₁ = 0. t = 3/2 = 1.01111...? Hmm, binary representation ambiguity: 3/2 = 1.1₂ = 1.0111...₂. Since Rule A applies at t = 3/2, effectively t = 3/2 behaves as 1.01111...: b₁ = 0, shift: t → 1.111... = 2. ✓ matches 2t - 1 = 2. Good. Similarly t = 2 = 1.111...? or 10.0? t = 2: Rule B: t → 3/2. In the b-form: t = 2 = 1.1111...₂: b₁ = 1: flip to 0: 1.0111... = 3/2 ✓. So canonical: t ∈ (1, 2] use representation without terminating (use recurring 1s): t = 1.b₁b₂... with infinitely many bits (no terminating zeros). Rule A: t ∈ (1, 3/2]: b₁ = 0 → shift. Rule B: t ∈ (3/2, 2]: b₁ = 1 → flip to 0 (equivalent to shift after flipping? 1.1b₂b₃... → 1.0b₂b₃...; hmm not a shift). Hmm, actually wait: Rule B: t → t - 1/2: 1.1b₂b₃... → 1.0b₂b₃.... Then next step: b₁ = 0 → shift: 1.b₂b₃.... So Rule B followed by Rule A = shift out the 1. So combined: each bit shifted out: 0-bits cost 1 step (k same), 1-bits cost 2 steps (k-1)?? Hmm no wait: Rule B itself is a step (halving f once). Let me recount. Sequence of steps: at each step, look at b₁ of current t: - b₁ = 0: one step: shift left. (t → 1.b₂b₃...) - b₁ = 1: one step: t → 1.0b₂b₃... (k decreases by 1). Then the new b₁ = 0, so next step shifts: t → 1.b₂b₃... So each bit of the original t - 1 costs: 1 step if bit = 0; 2 steps if bit = 1. And k decreases by 1 for each 1-bit. Exit: when k = 0 and current b₁ = 1: exit step, φ = 5/2 - t, and this exit step also counts as a halving step (f(x) = f(φ)/2^{steps including exit step}). Hold on, I need to double check whether exit happens exactly when a 1-bit is encountered at k = 0, or whether the process could encounter k = 0 with b₁ = 0 and keep going. Yes: k = 0, b₁ = 0: Rule A applies, shift, k stays 0. So the process continues at k = 0 shifting out 0-bits until a 1-bit appears, then exit. So total: starting with (t₀, k₀), let the binary expansion of t₀ - 1 = 0.b₁b₂b₃... (choosing the representation with infinitely many 1s... hmm, need care: for t₀ - 1 dyadic like 1/2 = 0.1000... = 0.0111..., which representation? The dynamics: t = 3/2 (t-1 = 1/2): Rule A applies (t = 3/2 ∈ (1, 3/2]), so t = 3/2 behaves like 0.0111...: b₁ = 0. So yes, use the non-terminating representation (recurring 1s).) Process: shift out bits one by one. 0-bits: 1 step each. 1-bits: 2 steps each (flip+shift), decrement k. When k reaches 0: continue shifting 0-bits (1 step each) until hit a 1-bit: exit (that 1-bit costs 1 step — the exit step; no shift after). Hmm wait, let me re-examine: at k ≥ 1, b₁ = 1: step: t → 1.0b₂b₃..., k → k-1. This is ONE halving step. Then b₁ = 0 now: step: shift: t → 1.b₂b₃...: SECOND halving step. So a 1-bit at k ≥ 1 costs 2 steps. At k = 0, b₁ = 1: exit: ONE halving step, φ = 5/2 - t. But hmm, wait: what if k reaches 0 in the middle? k decrements when a 1-bit is flipped. If after flipping, k = 0, and the remaining bits after shift... continue shifting 0s (each 1 step) until a 1 appears → exit (1 step). Also special: what if k₀ = 0 initially and t₀ has b₁ = 0: shift 0s until 1-bit: exit. Let me now recompute f(2 - 2^{-m}) with this framework: s = 2^{-m}, k₀ = m, t₀ = 2^{m+1}·2^{-m} = 2. t₀ = 2 = 1.111... (all 1s). Bits: all 1. Process: flip (k: m → m-1), shift: that's 2 steps per 1-bit, k decreases per 1-bit. After m 1-bits (2m steps), k = 0, and remaining bits all 1: next 1-bit at k=0: exit (1 step). Total 2m + 1 steps ✓ (matches earlier). Exit point: at exit, t = 2 (all remaining bits 1): φ = 5/2 - 2 = 1/2 ✓. f(2 - 2^{-m}) = f(1/2)/2^{2m+1} = (1/4)/2^{2m+1} = 2^{-2m-3} ✓✓. Now, the outer chain: a₀ = 3 - 2^{-10}. f(a₀) = f(a₀ - f(a₀ - 1))/2 = f(a₁)/2, a₁ = a₀ - f(2 - 2^{-10}) = 3 - 2^{-10} - 2^{-23}. So s₁ = 2^{-10} + 2^{-23} (where s_i = 3 - a_i). f(a₁) = f(a₁ - f(a₁ - 1))/2: a₁ - 1 = 2 - s₁, need f(2 - s₁), s₁ = 2^{-10} + 2^{-23}. Compute f(2 - s₁): s₁ = 2^{-10}(1 + 2^{-13}). Level: s₁ ∈ (2^{-11}, 2^{-10}] → k₀ = 10. t₀ = 2^{11}·s₁ = 2(1 + 2^{-13}) = 2 + 2^{-12}. Hmm, t₀ = 2 + 2^{-12} > 2! Out of range (1, 2]?! Problem! Let me recheck: t = 2^{k+1}s should be ∈ (1, 2] for s ∈ (2^{-(k+1)}, 2^{-k}]. s₁ = 2^{-10} + 2^{-23}. 2^{-10} = 1024·2^{-20}... let me compare with 2^{-10}: s₁ = 2^{-10} + tiny > 2^{-10}. So s₁ > 2^{-10} = 2^{-k} for k = 10. So level: s₁ ∈ (2^{-10}, 2^{-9}]? Is s₁ ≤ 2^{-9}? 2^{-9} = 2·2^{-10} = 2^{-10} + 2^{-10} >> s₁ ✓. So k₀ = 9! Because s₁ slightly exceeds 2^{-10}. I see — so the level is k₀ = 9, t₀ = 2^{10}·s₁ = 2^{10}(2^{-10} + 2^{-23}) = 1 + 2^{-13}. t₀ = 1 + 2^{-13} = 1.0000000000000111...₂ (binary: 1 followed by 12 zeros then 1s? 2^{-13} = 0.0000000000001₂ (13 places). So t₀ - 1 = 0.0000000000001111...₂? Hmm: 2^{-13} in binary: 0. followed by 12 zeros, then 1, then recurring 0s: 0.0000000000001000... Or non-terminating: 0.00000000000001111... Which representation do we use? The rule: at each step, if t ∈ (1, 3/2] i.e. b₁ = 0: shift. The ambiguity only matters at dyadic points. t₀ - 1 = 2^{-13}: representations 0.0000000000001 (terminating) or 0.0000000000000111... (non-terminating). First bit b₁ = 0 either way. Shifting out 0s: after 12 shifts, t = 1.1 (= 3/2) or t = 1.0111... = 1.1 hmm: 0.0000000000001: after 12 shifts: 1.0 hmm wait shifting t: t → 2t - 1 each shift. t₀ = 1 + 2^{-13}. After 1 shift: 1 + 2^{-12}. ... After 12 shifts: 1 + 2^{-1} = 3/2. Then t = 3/2: Rule A (since 3/2 ∈ (1, 3/2]): shift: t = 2. Then t = 2 = 1.111...: Rule B: t → 3/2, k → k-1. Etc. So with the terminating representation: t₀ - 1 = 0.[12 zeros]1[zeros...]: the bits after position 13 are all 0. Hmm, but then we'd shift forever through 0s without decrementing k — that can't be right since f should bottom out. Let me re-examine. Ah, I see the subtlety: if all remaining bits are 0, then t → 1 asymptotically... but t stays > 1 and keeps shifting: t = 1 + 2^{-j}, j increasing. These are valid points (x = 2 - s with s = 2^{-(k+1)}t...). Hmm wait, but each shift is a halving step, and there are infinitely many? That would make f = 0. Hmm! Wait, but that contradicts f > 0. Let me recheck with a concrete example. Take s = 3·2^{-(k+1)} + δ... hmm, let me take a concrete point: x = 2 - s with s = 0.6 (non-dyadic, k=0 since s ∈ (1/2, 1]): t₀ = 2s = 1.2, t₀ - 1 = 0.2 = 0.0011001100...₂ recurring. Fine, non-dyadic works: shift out bits: 0,0,1,1,0,0,1,1,...: 0-bits 1 step, 1-bits 2 steps + k decrement. k₀ = 0! So at the first 1-bit (position 3), exit. Let me verify concretely: x = 1.4 = 2 - 0.6. x - 1 = 0.4 ∈ [0, 1/2): f(0.4) = (1 - 0.4) - 1/2 = 0.1. φ(x) = 1.4 - 0.1 = 1.3. f(1.4) = f(1.3)/2. x = 1.3: x - 1 = 0.3, f(0.3) = 0.7 - 0.5 = 0.2. φ = 1.1. f(1.3) = f(1.1)/2. x = 1.1: x - 1 = 0.1, f(0.1) = 0.9 - 0.5 = 0.4. φ = 0.7. f(1.1) = f(0.7)/2. 0.7 ∈ [1/2, 3/4): f(0.7) = (1-0.7) - 1/4 = 0.05. So f(1.4) = 0.05/8 = 0.00625. Framework: t₀ = 1.2, k₀ = 0. Steps: t = 1.2: b₁ = 0 (1.2 < 1.5): shift: t = 1.4 (2·1.2 - 1). t = 1.4: b₁ = 0: shift: t = 1.8. t = 1.8: b₁ = 1 (1.8 > 1.5): k = 0 → exit: φ = 5/2 - 1.8 = 0.7 ✓!! Total steps: 3, f = f(0.7)/2³ = 0.05/8 ✓✓. So exit at the first 1-bit after k reaches 0. For k₀ = 0: exit at first 1-bit in t₀ - 1. Now back to the dyadic concern: t₀ - 1 = 2^{-13} (bits: twelve 0s then 1 then all 0s — in terminating representation). With k₀ = 9: we shift out 0s: 12 steps (t: 1 + 2^{-13} → 1 + 2^{-12} → ... → 1 + 2^{-1} = 3/2). At t = 3/2 = 1.1₂: Rule A applies (3/2 ≤ 3/2): shift: t = 2 = 1.111...₂ (now the representation becomes all-1s? Hmm!). Wait: 3/2 → 2 via Rule A (2t - 1 = 2). Then t = 2: Rule B: t → 3/2, k → k - 1. Then 3/2 → 2 (Rule A), 2 → 3/2 (Rule B, k-1), ... So once we hit t = 3/2 or 2, we cycle 3/2 → 2 → 3/2 → 2..., decrementing k every 2 steps, until k = 0, then at t = 2 (b₁ = 1): exit: φ = 5/2 - 2 = 1/2. So dyadic t values: the process does hit the cycle and decrements k reliably. The "all 0s after position 13" worry was wrong because at t = 3/2 Rule A gives t = 2 (i.e., the recurring-1s representation kicks in: 1.1000... → shift → 1.000... hmm, that would give t = 1, stuck). Hmm wait, now I'm confused: t = 3/2: is it 1.1000... or 1.0111...? If 1.1000...: b₁ = 1 → Rule B. If 1.0111...: b₁ = 0 → Rule A. Which is right? Rule A applies for t ∈ (1, 3/2] — INCLUSIVE of 3/2. Rule B for t ∈ (3/2, 2] — exclusive. So t = 3/2 → Rule A → t = 2. So the dynamics treats 3/2 as 1.0111.... Then t = 2 = 1.111... (must be, since 2 is in (3/2, 2], b₁ = 1): Rule B → 3/2, k-1. OK. So for dyadic t₀ (t₀ - 1 = dyadic), the orbit: t₀ = 1 + 2^{-j} (j ≥ 1): shifts (j - 1 steps) → t = 3/2 → 2 → 3/2 → 2 → ... So: reaching t = 3/2, then cycles 3/2 ↔ 2, k decreasing every cycle (2 steps), exit at t = 2 when k = 0. Let me now carefully recount steps for t₀ = 1 + 2^{-13}, k₀ = 9: - Shifts from t = 1 + 2^{-13} to t = 3/2: t = 1 + 2^{-13} → 1 + 2^{-12} → ... → 1 + 2^{-1} = 3/2. That's 12 steps. - Then cycle: 3/2 → 2 (1 step) → 3/2 (1 step, k-1) → 2 (1 step) → 3/2 (1 step, k-1) → ... Each 2 steps: k decreases 1. From k = 9 to k = 0: 9 decrements: 18 steps. But careful with the exact sequence and where exit happens. Sequence: t = 3/2 (k = 9): step → t = 2 (k = 9). Step → t = 3/2 (k = 8). Step → t = 2 (k = 8). Step → t = 3/2 (k = 7). ... So at t = 2 with k = 0: exit. Count: after 12 shifts: t = 3/2, k = 9. Steps to reach t = 2, k = 0: pattern: (3/2, 9) → (2, 9) → (3/2, 8) → (2, 8) → (3/2, 7) → ... → (2, 1) → (3/2, 0) → (2, 0): exit at (2, 0). From (3/2, 9) to (2, 0): 2·9 = 18 steps. Hmm: (3/2,9)→(2,9): 1; →(3/2,8): 2; →(2,8): 3; ... Let me count: states with k=9: (3/2,9) [start], (2,9). k=8: (3/2,8), (2,8). ... k=0: (3/2,0), (2,0). Steps from (3/2,9) to (2,0): from (3/2, 9) to (3/2, 8): 2 steps. To (3/2, 0): 18 steps. To (2, 0): 19 steps. Exit at (2,0): the exit step itself — hmm, is exit a step? The transition at (2, 0): Rule B with k=0 → exit, and this transition is a halving step: f(x) = f(φ)/2. So total steps: 12 + 19 = 31? Let me recount. Hmm wait, I need to be careful: when we're at (t, k) and apply a rule, that's one step (one halving). Exit: at state (2, 0), applying Rule B exits: 1 step, φ = 1/2. Steps: 12 (shifts) + [from (3/2, 9) to (2, 0) exclusive of exit]: (3/2,9) → (2,9) → (3/2,8) → (2,8) → ... → (3/2,0) → (2,0): how many transitions? From (3/2,9) to (3/2,0): 2 per k-decrement × 9 = 18. Then (3/2,0) → (2,0): 1 more: 19. Then exit step at (2,0): 1. Total: 12 + 19 + 1 = 32 steps. Then f = f(1/2)/2^{32} = 2^{-2}/2^{32} = 2^{-34}. So f(2 - s₁) = 2^{-34} where s₁ = 2^{-10} + 2^{-23}. Hmm interesting. Let me double check by another method: f(2 - s₁) where s₁ = 2^{-10} + 2^{-23}. Note f(2 - 2^{-10}) = 2^{-23}. And f(2 - s₁) = 2^{-34}. Hmm. Sanity check the formula f(2 - s) ≈ s²/8: s₁ ≈ 2^{-10}: s₁²/8 ≈ 2^{-23}. But we got 2^{-34}. That's way off!! Something's wrong. Hmm. Let me recheck. Let me recompute the steps for f(2 - 2^{-10}) with the same careful counting: s = 2^{-10}, k₀ = 10, t₀ = 2^{11}·2^{-10} = 2. t₀ = 2 = 1.111...: b₁ = 1: Rule B (k ≥ 1): t → 3/2, k → 9: 1 step. Then (3/2, 9) → (2, 9) → (3/2, 8) → ... → (2, 0) → exit. From (3/2, 9): 19 steps to (2, 0) (as counted: 18 to (3/2,0), +1 to (2,0)), then exit: 1 step. Total: 1 + 19 + 1 = 21 = 2m + 1 with m = 10 ✓. Good: f(2 - 2^{-10}) = f(1/2)/2^{21} = 2^{-23} ✓. Now for s₁: 12 + 19 + 1 = 32 steps: f = 2^{-2-32} = 2^{-34}. Hmm, but the quadratic approximation suggested ≈ 2^{-23}·ish. Let me sanity-check with a small case where I can compute directly. Let me compute f(2 - s) for s = 2^{-1} + 2^{-4} = 9/16? Hmm, or let me pick something where direct computation is feasible: s = 3/4·... let me use the framework to compute f(2 - s) for s = 1/2 + 1/8 = 5/8, and verify by direct recursion. Framework: s = 5/8: level: s ∈ (1/2, 1] → k₀ = 0. t₀ = 2s = 5/4 = 1.25. t₀ - 1 = 1/4 = 0.01₂. Bits: b₁ = 0, b₂ = 1, then all 0s (terminating) — but recurring-1s: 0.001111... Process at k₀ = 0: shift out b₁ = 0 (1 step): t = 1.5. t = 3/2: Rule A: t = 2 (1 step). t = 2: Rule B, k = 0: exit (1 step): φ = 1/2. Total 3 steps: f = f(1/2)/2³ = 2^{-5} = 1/32. Direct: x = 2 - 5/8 = 11/8. f(11/8) = 1/32 ✓ (computed earlier!). Another: s = 1/2 + 1/16 = 9/16: x = 2 - 9/16 = 23/16. Framework: k₀ = 0, t₀ = 2s = 9/8 = 1.125, t₀ - 1 = 1/8 = 0.001₂: b₁ = 0, b₂ = 0, b₃ = 1. Steps: shift (t = 1.25), shift (t = 1.5), then t = 3/2 → 2, then exit at (2, 0): total 4 steps: f = f(1/2)/2⁴ = 2^{-6} = 1/64. Direct: f(23/16) = 1/64 ✓ (computed earlier!). Another: s = 1/2 + 1/32 = 17/32: x = 47/32. Framework: t₀ = 17/16, t₀ - 1 = 1/16 = 0.0001₂: 3 shifts → t = 3/2, then → 2, exit: 5 steps: f = 2^{-7} = 1/128. Direct: f(47/32) = ? f(47/32) = f(47/32 - f(15/32))/2. f(15/32): 15/32 ∈ [7/16, 3/8]?? Let me see: 15/32 = 0.46875 ∈ [1/4, 1/2): f = (1 - 15/32) - 1/2 = 17/32 - 16/32 = 1/32. So f(47/32) = f(47/32 - 1/32)/2 = f(46/32)/2 = f(23/16)/2 = (1/64)/2 = 1/128 ✓✓. So the framework works. Note: these examples have k₀ = 0 (s ∈ (1/2, 1]) — the "1 + 2^{-j}" t-pattern: j - 1 shifts, then 3/2 → 2, exit: j + 1 steps total. f = 2^{-(j+3)}: for s = 1/2 + 2^{-j}·... hmm interesting: f(2 - (1/2 + 2^{-j})) = 2^{-j-3}? Check j=3 (s = 5/8): 2^{-6}? But we got 1/32 = 2^{-5}. Hmm: s = 1/2 + 1/8: 1/8 = 2^{-3}: f = 2^{-5}. j = 3: 2^{-j-2} = 2^{-5} ✓? j=4 (s = 9/16): 2^{-6} ✓. j=5: 2^{-7} ✓. So f(2 - 1/2 - 2^{-j}) = 2^{-j-2}. Hmm OK. Now let me also double check the s₁ = 2^{-10} + 2^{-23} computation differently, since k₀ = 9 there (s₁ ∈ (2^{-10}, 2^{-9}]). Wait, actually, hold on: s₁ = 2^{-10} + 2^{-23}: s₁ ∈ (2^{-10}, 2^{-9}]: 2^{-10} < s₁ ≤ 2^{-9} ✓ (s₁ barely above 2^{-10}). So k₀ = 9 ✓. t₀ = 2^{k₀+1} s₁ = 2^{10}(2^{-10} + 2^{-23}) = 1 + 2^{-13} ✓. Dynamics: t = 1 + 2^{-13}, k = 9. Shift 0-bits: t - 1 = 2^{-13} = 0.[12 zeros]1...: shifts: each shift doubles t - 1: after 12 shifts: t - 1 = 2^{-1}: t = 3/2, k = 9 (12 steps, k unchanged since all shifted bits were 0). Hmm wait: shifting out b₁ when b₁ = 0: t → 2t - 1: t - 1 doubles. t₀ - 1 = 2^{-13}: after 1 shift: 2^{-12}... after 12 shifts: 2^{-1}: t = 3/2. ✓. But careful: when t - 1 = 2^{-1}, t = 3/2: Rule A: t → 2t - 1 = 2. So the "1-bit" at position 13 gets shifted out via Rule A: t = 1.0...01 (bit at position 13) — after 12 shifts of the leading zeros, t = 1.1 = 3/2, and then Rule A shifts out this 1-bit: t = 2. But Rule A does NOT decrement k (only Rule B decrements k)! Hmm!! Wait, this is important: at t = 3/2, Rule A applies (not Rule B), so k does NOT decrement. Then t = 2: Rule B: t → 3/2, k - 1. So the cycle 3/2 → 2 → 3/2 → 2... with k decrementing at each 2 → 3/2 transition. So from (3/2, 9): → (2, 9) [1 step] → (3/2, 8) [1 step] → (2, 8) → (3/2, 7) → ... → (2, 1) → (3/2, 0) → (2, 0) → exit. From (3/2, 9), reaching (3/2, 0): 9 k-decrements × 2 steps = 18 steps. Then (3/2, 0) → (2, 0): 1 step. Then exit at (2, 0): 1 step. Total from (3/2, 9): 20 steps. Plus 12 initial shifts: 32 steps total. f(2 - s₁) = f(1/2)/2^{32} = 2^{-34} ✓ (matches my earlier count). Hmm, wait, but let me double-check the claim "Rule A doesn't decrement k". Rule A: t ∈ (1, 3/2]: t' = 2t - 1, k' = k. Yes — because s' = 2s - 2^{-(k+1)} and s' ∈ (2^{-(k+1)}, 2^{-k}] when t ∈ (1, 3/2] (shown earlier: s' ≤ 2^{-k} ↔ t ≤ 3/2 ✓). So k stays. ✓. And Rule B: t ∈ (3/2, 2]: s' ∈ (2^{-k}, 3·2^{-(k+1)}] ⊂ (2^{-k}, 2^{-(k-1)}]: k' = k - 1 ✓. OK so now: f(2 - s₁) = 2^{-34}, s₁ = 2^{-10} + 2^{-23}. Hmm wait, but now I'm worried about the approximation intuition (f ≈ s²/8), but fine — the exact computation is what it is. Actually hmm, let me reconsider: f(2 - 2^{-10}) = 2^{-23} but f(2 - 2^{-10} - ...) hmm, s₁ = 2^{-10} + 2^{-23} > 2^{-10}: f(2 - s₁) = 2^{-34} < f(2 - 2^{-10}) = 2^{-23}. So f dropped a lot when s increased slightly. Because the level k dropped from 10 to 9, hmm, no wait: k₀ went from 10 (for s = 2^{-10}) to 9 (for s₁)? s = 2^{-10}: level k₀ = 10 (s ∈ (2^{-11}, 2^{-10}]). s₁ = 2^{-10} + ε: level 9. And t₀: for s = 2^{-10}: t₀ = 2^{11}·2^{-10} = 2. For s₁: t₀ = 2^{10}s₁ = 1 + 2^{-13}. Interesting: so f(2 - s) is very sensitive: at s = 2^{-10}: 2^{-23}; just above: ≈ 2^{-34}·... wait hmm, that seems discontinuous, but f is discontinuous everywhere on this set, fine. Hmm wait, actually let me double-check f(2 - s₁) via direct-ish computation. Let me verify with a smaller analogous case. Analogue: s = 2^{-1} + 2^{-3}? Let me pick the pattern: s₁ = 2^{-m} + f(2 - 2^{-m})·... the outer chain step: s_{i+1} = s_i + f(2 - s_i). For m = 1: s = 1/2: f(3/2) = 2^{-5} = 1/32. s' = 1/2 + 1/32 = 17/32. f(2 - 17/32) = f(47/32) = 1/128 = 2^{-7} (computed above ✓ via direct). Framework for s = 17/32: level: 17/32 ∈ (1/2, 1] → k₀ = 0. t₀ = 2·17/32 = 17/16 = 1 + 1/16: t₀ - 1 = 2^{-4}: 3 shifts → t = 3/2 (3 steps), → t = 2 (Rule A, 1 step: total 4), exit (1 step: total 5): f = f(1/2)/2⁵ = 2^{-7} ✓✓ matches direct computation. Now, the analogous "k₀ ≥ 1" case: let me construct a small test: s = 1/4 + δ where δ makes level 2→... hmm, s = 1/4: level 2 (s ∈ (1/8, 1/4]), t₀ = 8·(1/4) = 2. f(2 - 1/4) = f(7/4) = 2^{-7}. Outer step: s' = 1/4 + 2^{-7} = 1/4 + 1/128. Level: s' ∈ (1/8, 1/4]? 1/4 + 1/128 > 1/4. So level 1: s' ∈ (1/4, 1/2] ✓. t₀ = 4·(1/4 + 1/128) = 1 + 1/32. Framework: t = 1 + 2^{-5}, k₀ = 1. Shifts: 4 shifts → t = 3/2 (k = 1). Then (3/2, 1) → (2, 1) → (3/2, 0) → (2, 0) → exit. Steps: 4 + 2 + 1 + 1 = 8. f = f(1/2)/2⁸ = 2^{-10}. Direct verification: x = 2 - (1/4 + 1/128) = 2 - 33/128 = 223/128. f(223/128) = f(223/128 - f(95/128))/2. f(95/128): 95/128 = 0.7421875 ∈ [3/4, 7/8): f = (1 - 95/128) - 1/8 = 33/128 - 16/128 = 17/128. So f(223/128) = f(223/128 - 17/128)/2 = f(206/128)/2 = f(103/64)/2. f(103/64) = f(103/64 - f(39/64))/2. f(39/64): 39/64 = 0.609375 ∈ [1/2, 3/4): f = (1 - 39/64) - 1/4 = 25/64 - 16/64 = 9/64. f(103/64) = f(103/64 - 9/64)/2 = f(94/64)/2 = f(47/32)/2 = (1/128)/2 = 1/256. So f(223/128) = (1/256)/2 = 1/512 = 2^{-9}. Hmm! Framework said 2^{-10}, direct says 2^{-9}. Discrepancy! Let me recheck. Direct recomputation: x = 223/128 = 1.7421875. x - 1 = 95/128 = 0.7421875. Level of 95/128 in [0,1): [3/4, 7/8) = [96/128, 112/128). 95/128 < 96/128! So 95/128 ∈ [1/2, 3/4) = [64/128, 96/128). f(95/128) = (1 - 95/128) - 1/4 = 33/128 - 32/128 = 1/128. Redo: f(223/128) = f(223/128 - 1/128)/2 = f(222/128)/2 = f(111/64)/2. f(111/64) = f(111/64 - f(47/64))/2. f(47/64): 47/64 ∈ [1/2, 3/4): f = (1 - 47/64) - 1/4 = 17/64 - 16/64 = 1/64. f(111/64) = f(111/64 - 1/64)/2 = f(110/64)/2 = f(55/32)/2. f(55/32) = f(55/32 - f(23/32))/2. f(23/32): 23/32 ∈ [1/2, 3/4): f = (1 - 23/32) - 1/4 = 9/32 - 8/32 = 1/32. f(55/32) = f(55/32 - 1/32)/2 = f(54/32)/2 = f(27/16)/2. f(27/16) = f(27/16 - f(11/16))/2. f(11/16) = (1 - 11/16) - 1/4 = 5/16 - 4/16 = 1/16. f(27/16) = f(27/16 - 1/16)/2 = f(26/16)/2 = f(13/8)/2 = (1/64)/2 = 1/128. So f(55/32) = 1/256, f(111/64) = 1/512, f(223/128) = 1/1024 = 2^{-10} ✓✓. I made an arithmetic error before (95/128 vs 96/128 boundary). Now direct gives 2^{-10}, matching framework. So framework validated again. Let me also recheck the earlier "f(2 - s₁) = 2^{-34}" — hmm wait, in this test case: s = 1/4 + 2^{-7}: k₀ = 1, t₀ = 1 + 2^{-5}: framework steps: 4 shifts (t: 1+2^{-5} → 1+2^{-4} → 1+2^{-3} → 1+2^{-2} → 1+2^{-1} = 3/2: that's 4 shifts ✓), then (3/2,1) → (2,1): 1 step [5], (2,1) → (3/2,0): 1 step [6], (3/2,0) → (2,0): 1 step [7], exit: 1 step [8]. Total 8 steps: f = 2^{-2}/2^{8} = 2^{-10} ✓✓ matches. So the general pattern for the outer chain: s_{i+1} = s_i + f(2 - s_i). Let me now compute the outer chain from s₀ = 2^{-10} step by step using the framework. Each f(2 - s_i) requires the framework analysis of s_i. Let me define the framework computation precisely: Given s ∈ (0, 1], find k (s ∈ (2^{-(k+1)}, 2^{-k}]), t = 2^{k+1}s ∈ (1, 2]. Write t - 1 in binary (non-terminating for dyadics): t - 1 = 0.b₁b₂b₃... Process: current state (t, k). - If t = 3/2 exactly: Rule A: t → 2. (1 step, k same) - Else if t ∈ (1, 3/2): t → 2t - 1. (1 step, k same) - If t ∈ (3/2, 2]: if k ≥ 1: t → t - 1/2, k → k - 1 (1 step); if k = 0: exit, φ = 5/2 - t (1 step). - Note t = 2: → 3/2 (k-1) or exit. f(2 - s) = f(φ)/2^{T}, T = total steps, φ = exit point ∈ [1/2, 1), and f(φ) = (1 - φ) - 2^{-(K+1)} where φ ∈ [1 - 2^{-K}, 1 - 2^{-(K+1)}). Hmm, actually for the outer chain, the exit point φ was 1/2 in the cases above (exit at t = 2: φ = 1/2). Exit at t ∈ (3/2, 2): φ = 5/2 - t ∈ (1/2, 1). Depends on bits. This is getting complicated, but let me push through the outer chain for f(3). Actually, wait. Let me reconsider — maybe there's a pattern to the outer chain that collapses. Outer chain: s₀ = 2^{-10}. s_{i+1} = s_i + f(2 - s_i). f(2 - s₀) = 2^{-23}. s₁ = 2^{-10} + 2^{-23}. f(2 - s₁) = 2^{-34} (computed: k₀ = 9, t₀ = 1 + 2^{-13}, steps 32, exit φ = 1/2, f = 2^{-2-32} = 2^{-34}). s₂ = s₁ + 2^{-34} = 2^{-10} + 2^{-23} + 2^{-34}. f(2 - s₂): level: s₂ ∈ (2^{-10}, 2^{-9}] → k₀ = 9. t₀ = 2^{10}s₂ = 1 + 2^{-13} + 2^{-24}. Hmm. t₀ - 1 = 2^{-13} + 2^{-24} = 0.0000000000001000000000001₂ (bit at positions 13 and 24) — non-terminating: positions 13 and 24 have 1s, rest 0s... but non-terminating representation: since it's dyadic (finite), the canonical for our dynamics: hmm. The dynamics at exactly-representable points: we need to be careful. t - 1 = 2^{-13} + 2^{-24}: shifting: after 12 shifts: t - 1 = 2^{-1} + 2^{-12} → t = 3/2 + 2^{-12} ∈ (3/2, 2): Rule B! (Not Rule A, since t > 3/2.) Rule B: t → t - 1/2 = 1 + 2^{-12}, k → 8. So the bit at position 13 (a 1-bit) got "flipped" via Rule B, decrementing k. Then t = 1 + 2^{-12}: shift 11 more times: t = 1 + 2^{-1}·... t - 1 = 2^{-12}: after 11 shifts: 2^{-1}: t = 3/2: Rule A: t → 2. Then (2, 8): Rule B: → (3/2, 7): ... cycle until (2, 0): exit. Let me count: initial: (1 + 2^{-13} + 2^{-24}, k=9). Step 1-12: shifts (12 steps): t = 3/2 + 2^{-12}, k = 9. [shifted out twelve 0-bits] Hmm wait: t₀ - 1 = 2^{-13} + 2^{-24}. Shifting: t - 1 → 2(t-1) - 0? Rule A: t' = 2t - 1: t' - 1 = 2(t - 1). So t - 1 doubles each shift: after 12 shifts: t - 1 = 2^{-1} + 2^{-12}: t = 3/2 + 2^{-12} > 3/2: next rule is B. Step 13: Rule B: t → t - 1/2: t - 1 = 2^{-12}·... t = 1 + 2^{-12}, k → 8. (1 step) Steps 14-24: 11 shifts: t - 1 = 2^{-1}: t = 3/2, k = 8. (11 steps) Step 25: Rule A: t = 3/2 → 2. (1 step, k = 8) Step 26: Rule B: (2, 8) → (3/2, 7). (1 step) Steps: cycle: (3/2,7) → (2,7) → (3/2,6) → (2,6) → ... → (3/2,0) → (2,0) → exit. From (3/2, 7): to (3/2, 0): 14 steps; to (2, 0): 15 steps; exit: +1: 16 steps from (3/2,7). Total: 12 + 1 + 11 + 1 + 1 + 16 = 42 steps. Exit at t = 2: φ = 1/2. f(2 - s₂) = 2^{-2}/2^{42} = 2^{-44}. Hmm wait let me recount: steps 1-12: 12. Step 13 (Rule B): 13. Steps 14-24: 11 shifts: 24. Step 25 (Rule A at 3/2): 25. Step 26 (Rule B at (2,8)): 26. Then from (3/2, 7) [after step 26]: need to get to (2,0) and exit: (3/2,7) → (2,7): 27; → (3/2,6): 28; → (2,6): 29; → (3/2,5): 30; → (2,5): 31; → (3/2,4): 32; → (2,4): 33; → (3/2,3): 34; → (2,3): 35; → (3/2,2): 36; → (2,2): 37; → (3/2,1): 38; → (2,1): 39; → (3/2,0): 40; → (2,0): 41; exit: 42. Total 42 steps ✓. f(2 - s₂) = f(1/2)/2^{42} = 2^{-44}. s₃ = s₂ + 2^{-44} = 2^{-10} + 2^{-23} + 2^{-34} + 2^{-44}. Hmm, I see a pattern emerging: s_i = 2^{-10} + Σ (increments), increments: 2^{-23}, 2^{-34}, 2^{-44}, ... and f-values: 2^{-23}, 2^{-34}, 2^{-44}, ... Let me guess: s₄ = s₃ + 2^{-54}? And the exponents: 23, 34, 44, 54: differences 11, 10, 10. Hmm. Let me compute f(2 - s₃): s₃ = 2^{-10} + 2^{-23} + 2^{-34} + 2^{-44}. Level: s₃ ∈ (2^{-10}, 2^{-9}]: k₀ = 9. t₀ = 2^{10} s₃ = 1 + 2^{-13} + 2^{-24} + 2^{-34} + 2^{-54}·... wait: 2^{10}·2^{-34} = 2^{-24}, 2^{10}·2^{-44} = 2^{-34}. So t₀ - 1 = 2^{-13} + 2^{-24} + 2^{-34}·... hold on: 2^{10}s₃ = 2^{10}(2^{-10} + 2^{-23} + 2^{-34} + 2^{-44}) = 1 + 2^{-13} + 2^{-24} + 2^{-34}. So t₀ - 1 = 2^{-13} + 2^{-24} + 2^{-34}. Dynamics: shifts double t-1: after 12 shifts: t - 1 = 2^{-1} + 2^{-12} + 2^{-22}: t = 3/2 + 2^{-12} + 2^{-22} > 3/2: Rule B: t → 1 + 2^{-12} + 2^{-22}, k → 8. (13 steps so far) Shifts: t-1 doubles: after 11 shifts: t - 1 = 2^{-1} + 2^{-11}: t = 3/2 + 2^{-11} > 3/2: Rule B: t → 1 + 2^{-11}, k → 7. (13 + 11 + 1 = 25 steps) Shifts: 10 shifts: t - 1 = 2^{-1}: t = 3/2: Rule A: t → 2. (25 + 10 + 1 = 36 steps) Then (2, 7): Rule B → (3/2, 6): 37. Then from (3/2,6): cycle to (2,0): (3/2,6)→(2,6):38; →(3/2,5):39; →(2,5):40; →(3/2,4):41; →(2,4):42; →(3/2,3):43; →(2,3):44; →(3/2,2):45; →(2,2):46; →(3/2,1):47; →(2,1):48; →(3/2,0):49; →(2,0):50; exit: 51. Total 51 steps: f(2 - s₃) = 2^{-2}/2^{51} = 2^{-53}. Hmm, so increments: 2^{-23}, 2^{-34}, 2^{-44}, 2^{-53}. Exponents: 23, 34, 44, 53: differences 11, 10, 9. Interesting! Next difference 8?: 61? Let me verify pattern by computing f(2 - s₄) where s₄ = 2^{-10} + 2^{-23} + 2^{-34} + 2^{-44} + 2^{-53}. t₀ - 1 = 2^{10}(s₄ - 2^{-10}) = 2^{-13} + 2^{-24} + 2^{-34} + 2^{-44}·... wait: 2^{10}·2^{-44} = 2^{-34}, 2^{10}·2^{-53} = 2^{-43}. Hmm: s₄ - 2^{-10} = 2^{-23} + 2^{-34} + 2^{-44} + 2^{-53}. t₀ - 1 = 2^{-13} + 2^{-24} + 2^{-34} + 2^{-43}. Dynamics: 12 shifts: t - 1 = 2^{-1} + 2^{-12} + 2^{-22} + 2^{-31}: Rule B: → 1 + 2^{-12} + 2^{-22} + 2^{-31}, k = 8. (13) 11 shifts: t - 1 = 2^{-1} + 2^{-11} + 2^{-20}: Rule B: → 1 + 2^{-11} + 2^{-20}, k = 7. (25) 10 shifts: t - 1 = 2^{-1} + 2^{-9}·... hmm: 2^{-11+10}·... let me recompute: t - 1 = 2^{-11} + 2^{-20}: after 10 shifts: 2^{-1} + 2^{-10}: t = 3/2 + 2^{-10}: Rule B: → 1 + 2^{-10}, k = 6. (25 + 10 + 1 = 36) 10 shifts: t - 1 = 2^{-10} → after 9 shifts: 2^{-1}: t = 3/2: Rule A: → 2, k = 6. (36 + 9 + 1 = 46) (2, 6) → (3/2, 5): 47. From (3/2, 5): to (2, 0): (3/2,5)→(2,5):48; →(3/2,4):49; →(2,4):50; →(3/2,3):51; →(2,3):52; →(3/2,2):53; →(2,2):54; →(3/2,1):55; →(2,1):56; →(3/2,0):57; →(2,0):58; exit: 59. Total 59 steps: f(2 - s₄) = 2^{-61}. Increments: 23, 34, 44, 53, 61: differences 11, 10, 9, 8. So next: 61 + 7 = 68, then 68+6 = 74, 74+5=79, 79+4=83, 83+3=86, 86+2=88, 88+1=89, then +0? Hmm interesting. Let me guess the pattern continues: increments exponents e_j: 23, 34, 44, 53, 61, 68, 74, 79, 83, 86, 88, 89, 89, 89, ...? Hmm, or does it stop? Hmm wait, but this outer chain needs to continue until s_i reaches... the chain a_i = 3 - s_i decreases; we need a_i to drop below 2, i.e., s_i > 1, to move to the next phase. But s_i = 2^{-10} + tiny increments: the increments are tiny (2^{-23} etc.). The chain: s_{i+1} = s_i + f(2 - s_i). For s_i ≈ 2^{-10}, f ≈ 2^{-e_i} with e_i ≈ 23-89: increments are minuscule. How does s ever grow to 1?? Wait, I think I need to reconsider. Hmm. The increments 2^{-23}, 2^{-34}, ... are the f-values, but as s_i grows, f(2 - s_i) changes. The pattern I'm seeing (increments decreasing) suggests the chain stalls?? But f(3) = f(a_N)/2^N requires the chain to bottom out (reach negative argument or [0,1) computable region). Hmm wait, no! Let me re-examine: the outer chain relation: f(a_i) = f(a_{i+1})/2 where a_{i+1} = a_i - f(a_i - 1). This is valid for a_i ≥ 0. The chain bottoms out when a_{i+1} < 0 (f = -a_{i+1}) — but a_{i+1} = a_i - f(a_i - 1) < 0 requires f(a_i - 1) > a_i. For a_i ∈ [2, 3): f(a_i - 1) ≤ 1/8 < 2: impossible. For a_i ∈ [1, 2): f(a_i - 1) ≤ 1/2: a_{i+1} < 0 requires a_i < 1/2 and f(a_i - 1) > a_i: a_i ∈ [0, 1/2): a_i - 1 ∈ [-1, -1/2): f = -(a_i - 1) = 1 - a_i > a_i ↔ a_i < 1/2 ✓. So for a_i ∈ [0, 1/2): a_{i+1} = a_i - (1 - a_i) = 2a_i - 1 < 0 ✓. So bottoming out happens only when a_i < 1/2. Wait, hmm: a_i ∈ [0, 1/2): a_i - 1 < 0: f(a_i - 1) = 1 - a_i. a_{i+1} = 2a_i - 1 < 0: f(a_{i+1}) = 1 - 2a_i. And f(a_i) = f(a_{i+1})/2 = (1-2a_i)/2 = 1/2 - a_i ✓ (consistent with f on [0, 1/2): (1 - a_i) - 1/2 ✓). So the chain must reach [0, 1/2) to bottom out. Starting from a₀ = 3 - 2^{-10}: the chain decreases by f(a_i - 1) each step. Total decrease needed: from ~3 to <1/2: about 2.5. But the increments f(a_i - 1) are tiny (2^{-23} initially)!! With ~8000+ steps... hmm, but actually as a_i decreases, f(a_i - 1) changes. Let me reconsider. Oh wait. I think I misidentified the chain. Let me recompute: f(x) = f(x - f(x - 1))/2 for x ≥ 0. For x = 3: f(3) = f(3 - f(2))/2 = f(3 - 2^{-10})/2. ✓. Then f(3 - 2^{-10}) = f((3 - 2^{-10}) - f(2 - 2^{-10}))/2 = f(3 - 2^{-10} - 2^{-23})/2. ✓ — this is what I had: s₁ = 2^{-10} + 2^{-23}. And f(3 - s₁) = f(3 - s₁ - f(2 - s₁))/2: s₂ = s₁ + f(2 - s₁) = s₁ + 2^{-34}. So the chain: s_{i+1} = s_i + f(2 - s_i), f(3 - s_i) = f(3 - s_{i+1})/2, valid while 3 - s_i ≥ 2? No — valid while 3 - s_i ≥ 0, but f(2 - s_i) requires 2 - s_i ≥ ... the recursion f(x) = f(x - f(x-1))/2 needs f(x - 1) which for x - 1 < 0 is -(x-1) = 1 - x. So for x ∈ [2, 3): f(x - 1) = f(2 - s) with 2 - s ∈ [1, 2): fine. For x ∈ [1, 2): x - 1 = 1 - s ∈ [0, 1): f(x-1) = f on [0,1). For x ∈ [0, 1): x - 1 < 0: f = 1 - x. So the chain from x₀ = 3 - 2^{-10}: as long as x ∈ [2, 3), steps subtract f(2 - s) where s = 3 - x ∈ (0, 1]. The f(2 - s) values: for s tiny (≈2^{-10}), f ≈ 2^{-23} (tiny). So the chain moves very slowly at first. As s grows, f(2 - s) grows. Once s gets larger (say s ∈ (1/2, 1], x ∈ [1.5, 2)): f(2 - s) = f values on [1, 3/2): up to 1/8. Then x drops below 2: s > 1: x ∈ [1, 2): steps subtract f(1 - s') where s' = 2 - x ∈ [0, 1): up to 1/2. Then x ∈ [0, 1): steps subtract 1 - x: quickly bottom out. So the chain does bottom out, but the number of steps while x ∈ [2, 3) is huge: s needs to grow from 2^{-10} to 1 via ds/steps = f(2 - s) ≈ s²/8 (for the generic/dyadic points... hmm, but our points are special). Hmm wait, but actually hold on. Let me reconsider: is it really true that f(2 - s) ≈ s²/8 for the chain points? The chain points s_i are specific dyadic-ish values. From the computations: s₀ = 2^{-10}: f = 2^{-23} = (2^{-10})²/8 ✓. s₁ = 2^{-10} + 2^{-23}: f = 2^{-34}. (s₁)²/8 ≈ 2^{-23}·... s₁² ≈ 2^{-20}: /8 = 2^{-23}. But actual f = 2^{-34}!! 11 orders smaller! So for these special points, f is much smaller than s²/8. Hmm!! So the chain increments are much smaller than generic. That changes the step count dramatically. But wait — why is f(2 - s₁) so small? Because of the level structure: s₁ = 2^{-10} + 2^{-23} has level 9 (just above 2^{-10}), and t₀ = 1 + 2^{-13} — the binary expansion of t₀ - 1 has a 1-bit way out at position 13, requiring many shifts... The number of steps T ≈ 2k₀ + (stuff) + bit-positions. For f = 2^{-2-T}: T = 32 for s₁. For generic s (with "random" binary expansion), T ≈ (number of bits shifted) ≈ position of the (k₀+1)-th 1-bit in t₀ - 1 or so... Hmm. OK here's the thing: this problem (I recall) is a known competition/puzzle problem, and the answer f(3) might be something like 2^{-36} or similar "nice" number... but my analysis suggests the chain has thousands of steps, giving f(3) = 2^{-(huge)}. Hmm, wait, but actually let me reconsider whether the chain really needs s to grow to 1. Hold on, wait. Let me recheck the relation f(x) = f(x - f(x-1))/2 for x ∈ [2, 3). x - f(x - 1): f(x-1) = f(2 - s) which is TINY (like 2^{-34}). So x - f(x-1) ≈ x: the argument barely decreases! So f(x) = f(x - tiny)/2: and f(x - tiny) ≈ f(x)·... this is a functional equation that's nearly f(x) = f(x)/2 — contradiction unless... hmm, no: it's f(x) = f(x - tiny)/2 exactly, and f(x - tiny) is evaluated at a slightly smaller point where f is slightly... hmm, but f is wildly discontinuous, so f(x - tiny) can be very different from f(x). Actually the point: f(x) for x ∈ [2, 3): f(x) = f(x - f(x-1))/2. If f(x - 1) is tiny, then x - f(x-1) is very close to x, and we iterate. The values f along the chain halve each step. For the chain to be consistent, f must decrease by factor 2 per tiny step — i.e., f has this crazy structure where f(x - f(x-1)) = 2f(x). OK. So indeed the chain from 3 - 2^{-10} will take many steps to descend to 2, and the number of steps is determined by the structure. The increments: f(2 - s_i) = 2^{-e_i}: I found e_i: 23, 34, 44, 53, 61, ... Let me understand this sequence better, because the total number of outer steps N (which gives f(3) = f(final)/2^{N + inner steps}) depends on summing many increments. Hmm, wait. Actually, let me reconsider. Let me recompute the increments more carefully and find the pattern. s₀ = 2^{-10}. f(2 - s₀) = 2^{-23}. [T = 21 = 2·10 + 1, f = 2^{-2-21}] s₁ = 2^{-10} + 2^{-23}. f(2 - s₁) = 2^{-34}. [T = 32] s₂ = s₁ + 2^{-34} = 2^{-10} + 2^{-23} + 2^{-34}. f(2 - s₂) = 2^{-44}. [T = 42] s₃ = s₂ + 2^{-44}. f(2 - s₃) = 2^{-53}. [T = 51] s₄ = s₃ + 2^{-53}. f(2 - s₄) = 2^{-61}. [T = 59] Hmm wait, let me double-check s₃ and s₄ computations. s₃ = 2^{-10} + 2^{-23} + 2^{-34} + 2^{-44}. t₀ - 1 = 2^{10}(s₃ - 2^{-10}) = 2^{-13} + 2^{-24} + 2^{-34}. Bits of t₀ - 1: 1 at positions 13, 24, 34. Dynamics trace: k₀ = 9, t - 1 = 2^{-13} + 2^{-24} + 2^{-34}. - 12 shifts (12 steps): t - 1 = 2^{-1} + 2^{-12} + 2^{-22}. t = 1.5 + 2^{-12} + 2^{-22} ∈ (3/2, 2): Rule B. - Rule B (1 step): t - 1 = 2^{-12} + 2^{-22}, k = 8. [13 steps] - 11 shifts (11 steps): t - 1 = 2^{-1} + 2^{-11}·... 2(2^{-12} + 2^{-22}) = 2^{-11} + 2^{-21}; after 11 shifts: 2^{11}(2^{-12} + 2^{-22}) = 2^{-1} + 2^{-11}. t = 1.5 + 2^{-11}: Rule B. [24 steps] - Rule B (1 step): t - 1 = 2^{-11}, k = 7. [25 steps] - 10 shifts: t - 1 = 2^{-1}: t = 3/2. [35 steps] - Rule A (t = 3/2): t = 2. [36 steps] - Rule B: (2, 7) → t = 3/2, k = 6. [37] - Now cycle: (3/2, 6) → (2, 6) [38] → (3/2, 5) [39] → (2,5) [40] → (3/2,4) [41] → (2,4) [42] → (3/2,3) [43] → (2,3) [44] → (3/2,2) [45] → (2,2) [46] → (3/2,1) [47] → (2,1) [48] → (3/2,0) [49] → (2,0) [50] → exit [51]. Total T = 51, exit at t = 2: φ = 1/2: f(2 - s₃) = 2^{-2}·2^{-51} = 2^{-53} ✓. s₄ = 2^{-10} + 2^{-23} + 2^{-34} + 2^{-44} + 2^{-53}. t₀ - 1 = 2^{10}(s₄ - 2^{-10}) = 2^{-13} + 2^{-24} + 2^{-34} + 2^{-43}. Bits at positions: 13, 24, 34, 43. - 12 shifts [12]: t - 1 = 2^{-1} + 2^{-12} + 2^{-22} + 2^{-31}: Rule B [13]: t - 1 = 2^{-12} + 2^{-22} + 2^{-31}, k = 8. - 11 shifts [24]: t - 1 = 2^{-1} + 2^{-11} + 2^{-20}: Rule B [25]: t - 1 = 2^{-11} + 2^{-20}, k = 7. - 10 shifts [35]: t - 1 = 2^{-1} + 2^{-10}·... 2^{10}(2^{-11} + 2^{-20}) = 2^{-1} + 2^{-10}: t = 1.5 + 2^{-10}: Rule B [36]: t - 1 = 2^{-10}, k = 6. - 9 shifts [45]: t - 1 = 2^{-1}: t = 3/2: Rule A [46]: t = 2, k = 6. - Rule B [47]: (3/2, 5). - Cycle: (3/2,5)→(2,5)[48]→(3/2,4)[49]→(2,4)[50]→(3/2,3)[51]→(2,3)[52]→(3/2,2)[53]→(2,2)[54]→(3/2,1)[55]→(2,1)[56]→(3/2,0)[57]→(2,0)[58]→exit[59]. Total T = 59: f = 2^{-61} ✓. So increments e_i: 23, 34, 44, 53, 61, and the s-values accumulate. Let me get the general pattern. It seems like the 1-bits of (s_i - 2^{-10})·2^{10} = t₀ - 1 march leftward: positions 13, 24, 34, 43, ... The next bit positions: the increments to t₀ - 1: each new increment 2^{-e}·2^{10} = 2^{-(e-10)}: e = 23 → position 13; e = 34 → position 24; e = 44 → position 34; e = 53 → position 43. So bit positions p_j: 13, 24, 34, 43, ...: differences 11, 10, 9: next: 43 + 8 = 51, then 51 + 7 = 58, 58 + 6 = 64, 64 + 5 = 69, 69 + 4 = 73, 73 + 3 = 76, 76 + 2 = 78, 78 + 1 = 79, 79 + 0 = 79?? Hmm, when differences hit 0... or maybe the pattern is different. Wait, but hold on: the increments f(2 - s_i) = 2^{-e_i} are added to s_i, and t₀ - 1 = 2^{10}(s_i - 2^{-10})·... but CAREFUL: this assumed level k₀ = 9, i.e., s_i ∈ (2^{-10}, 2^{-9}]. As increments accumulate, eventually s_i exceeds 2^{-9}, changing the level! The increments are tiny (2^{-23} etc.), so s_i stays ≈ 2^{-10} for a long time... but wait, the bits accumulate: t₀ - 1 = Σ 2^{-p_j}: as bits march left (p decreasing), eventually the leading bit gets close to position 1: t₀ - 1 ≈ 2^{-1} + ...: then t₀ ≈ 3/2, and further increments push t₀ over 3/2, 2, etc. Then t₀ > 2?? But t₀ must be ≤ 2 for level 9. If t₀ > 2, i.e., s_i > 2^{-9}·... s_i = 2^{-10}t₀: t₀ > 2 ↔ s_i > 2^{-9}: level changes to 8. Hmm OK. So the process: s_i accumulates bits; the "bit positions" in t₀ - 1 (with the current level scaling) march left; when they cross thresholds, the level changes... This is reminiscent of... hmm. Actually WAIT. Let me step back and think about what f(3) should be, given the answer might be nice. Let me reconsider. Hmm, let me reconsider the problem. Actually, let me look at this from a different angle: the known sequence f(0) = 2^{-1}, f(1) = 2^{-3}, f(2) = 2^{-10}. If the pattern of exponents is 1, 3, 10, then next might be... 1, 3, 10: these are the "central binomial" related? C(0,0)... Hmm: 1, 3, 10, 35?? (C(2n, n)·: C(0,0)=1, C(2,1)=2, C(4,2)=6: no.) 1, 3, 10: a(n) = (3^n + ...)/...: 3^0 = 1, 3^1 = 3, 3^2 = 9 ≠ 10. Hmm. 1, 3, 10, 33? a(n+1) = 3a(n) + 1: 1, 4, 13: no. a(n+1) = 3a(n) + n: 1, 3+1=... hmm: a(1) = 3a(0) + 0 = 3; a(2) = 3·3 + 1 = 10; a(3) = 3·10 + 2 = 32? Or a(n+1) = 3a(n) + (n+1): 1→3 (3·1+0), 3→10 (3·3+1), 10→32 (3·10+2)? So f(3) = 2^{-32}? Or the triangular-with-doubling-index guess: T_1, T_2, T_4, T_8: 1, 3, 10, 36: f(3) = 2^{-36}. Or: 1, 3, 10, ...: differences 2, 7: second differences 5: next difference 12: a(3) = 22? Or differences 2, 7: 7 = 2·2+3, next = 2·7+3 = 17: a(3) = 27? Meh. I recall this problem! I think this is from a Putnam or a Monthly problem... "f(x) = -x for x < 0, f(x) = f(x - f(x-1))/2 for x ≥ 0" hmm. Hmm, I have some memory of a similar problem where f(1) = 1/8... Actually let me just carefully continue the computation and get the exact answer. The structure is becoming clear; let me formalize the outer chain. Let me re-derive the dynamics more cleanly, because I want to make sure about the exact rules, then simulate the outer chain symbolically. Framework recap: For x ∈ [1, 2), x = 2 - s, s ∈ (0, 1]: - level k ≥ 0: s ∈ (2^{-(k+1)}, 2^{-k}]. (Note: s = 1 ↔ x = 1: k = 0, t = 2.) - t = 2^{k+1} s ∈ (1, 2]. - Step: s' = 2s - 2^{-(k+1)}, i.e., φ = 2 - s'. New level k' from s'. - f(x) = f(φ)/2. For x ∈ [2, 3): x = 3 - s, s ∈ (0, 1]: - f(x) = f(x - f(x - 1))/2, x - 1 = 2 - s: f(x - 1) = f(2 - s) via [1,2) framework. - So the outer step: s' = s + f(2 - s), f(3 - s) = f(3 - s')/2. And f(2 - s) = f(z)/2^{T(s)} where z = exit ∈ [1/2, 1), T(s) = number of inner steps. For the outer chain, the inner exit z: in all cases computed, exit happened at t = 2 giving z = 1/2, f(z) = f(1/2) = 1/4 = 2^{-2}. Exit at t ∈ (3/2, 2) would give z = 5/2 - t ∈ (1/2, 1), f(z) = (1 - z) - 2^{-(K+1)}. Hmm. When does exit happen at t < 2? Exit is Rule B at k = 0: t ∈ (3/2, 2]. If the current t (at k = 0) is in (3/2, 2), exit there: z = 5/2 - t. In our chain so far, at k = 0 we always had t = 2 exactly (because the last inner bit was at "position 1" i.e. t - 1 = 2^{-1}·... hmm, because after shifting, t - 1 = 2^{-1} exactly (dyadic termination), then Rule A: t = 2, then Rule B exit at t = 2.) When would t at k=0 be strictly between 3/2 and 2? If t - 1 ∈ (1/2, 1) at k = 0, i.e., the bits don't align to give exactly 2^{-1}. This happens when t₀ - 1 has 1-bits at various positions such that after processing, t - 1 ∈ (2^{-1}, 1) at k = 0. Hmm, e.g., t₀ - 1 = 2^{-1} + 2^{-3} with k₀ = 0: t = 1.625: Rule B immediately (k=0): exit: z = 5/2 - 1.625 = 0.875. So yes possible. In our outer chain, the s_i are sums of distinct powers of 2 (dyadic with sparse bits), so t₀ - 1 similarly. Whether exit is at t = 2 or t < 2 depends on the bit pattern. So far (bits at positions 13, 24, 34, 43, ...) exits were at t = 2. Let me continue the chain and watch for changes. Actually, let me carefully re-derive the relationship between the outer chain's s_i and the inner dynamics, to find the general step. Outer chain: s_{i+1} = s_i + f(2 - s_i). Let me write s_i = 2^{-10} + σ_i where σ_i = accumulated increments (tiny). Level of s_i: since σ_i << 2^{-10} for a while: level 9 (s_i ∈ (2^{-10}, 2^{-9}]) — as long as σ_i < 2^{-10}, i.e., t₀ = 2^{10}s_i = 1 + 2^{10}σ_i ∈ (1, 2]. Let υ_i = 2^{10}σ_i = t₀ - 1 (for level 9). Then: υ₀ = 0 (s₀ = 2^{-10}: level 10 though! Careful: s₀ = 2^{-10} exactly: level 10, t = 2.) Hmm, s₀ = 2^{-10} is special (level 10, t₀ = 2). Then s₁ = 2^{-10} + 2^{-23}: level 9, υ₁ = 2^{-13}. s₂ = s₁ + 2^{-34}: υ₂ = 2^{-13} + 2^{-24}. s₃: υ₃ = 2^{-13} + 2^{-24} + 2^{-34}. s₄: υ₄ = 2^{-13} + 2^{-24} + 2^{-34} + 2^{-43}. s₅ = s₄ + 2^{-61}: υ₅ = 2^{-13} + 2^{-24} + 2^{-34} + 2^{-43} + 2^{-51}. Predicted bit positions: 13, 24, 34, 43, 51, then +7 = 58, +6 = 64, +5 = 69, +4 = 73, +3 = 76, +2 = 78, +1 = 79, +0 = 79?? Hmm, that last one is weird. Let me actually compute f(2 - s₅) properly rather than guess. s₅: level 9 (υ₅ = Σ_{p ∈ {13,24,34,43,51}} 2^{-p} < 2^{-12}·... obviously < 1). t = 1 + υ₅. Inner dynamics: t - 1 = 2^{-13} + 2^{-24} + 2^{-34} + 2^{-43} + 2^{-51}, k₀ = 9. - 12 shifts [12]: t - 1 = 2^{-1} + 2^{-12} + 2^{-22} + 2^{-31} + 2^{-39}: t ∈ (3/2, 2): Rule B [13]: t - 1 = 2^{-12} + 2^{-22} + 2^{-31} + 2^{-39}, k = 8. - 11 shifts [24]: t - 1 = 2^{-1} + 2^{-11} + 2^{-20} + 2^{-28}: Rule B [25]: t - 1 = 2^{-11} + 2^{-20} + 2^{-28}, k = 7. - 10 shifts [35]: t - 1 = 2^{-1} + 2^{-10} + 2^{-18}: Rule B [36]: t - 1 = 2^{-10} + 2^{-18}, k = 6. - 9 shifts [45]: t - 1 = 2^{-1} + 2^{-9}: Rule B [46]: t - 1 = 2^{-9}, k = 5. - 8 shifts [54]: t - 1 = 2^{-1}: t = 3/2: Rule A [55]: t = 2, k = 5. - Rule B [56]: t = 3/2, k = 4. - Cycle: (3/2,4)→(2,4)[57]→(3/2,3)[58]→(2,3)[59]→(3/2,2)[60]→(2,2)[61]→(3/2,1)[62]→(2,1)[63]→(3/2,0)[64]→(2,0)[65]→exit[66]. T = 66: f(2 - s₅) = 2^{-68}. Hmm, so increment exponent e₅ = 68, not 61+7=68 ✓!! Wait: I predicted next bit position 51 → e = 51 + 10 = 61?? Hmm no: e_i relates to the NEW bit added: s₅ = s₄ + f(2 - s₄) = s₄ + 2^{-61}: bit position in υ: 61 - 10 = 51 ✓ (υ₅ has bit at 51 ✓). And f(2 - s₅) = 2^{-68}: so e₆ = 68: next bit at position 68 - 10 = 58 ✓ matches the predicted pattern (43 + 8 = 51, 51 + 7 = 58)! So the pattern: f(2 - s_i) = 2^{-e_i}, with e_i: 23, 34, 44, 53, 61, 68, ... and bit positions p_i = e_i - 10: 13, 24, 34, 43, 51, 58, ...: differences 11, 10, 9, 8, 7, ...: so p_j = 13 + Σ_{d=11 down}: p: 13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 78, 79, 79, ... hmm when the difference reaches 0: p stays? Let's think: differences 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0?? If difference becomes 0, the bit position stops advancing — that would mean the increment exponent stops growing: e stuck. Hmm, that seems odd; more likely the pattern changes qualitatively (e.g., exit no longer at t = 2, or level changes, or the bits start interacting/carrying). Wait, hmm, actually let me reconsider. Why are the bit positions marching left by decreasing amounts? Let me re-derive the increment formula. Claim: if s (level 9) has υ = t - 1 = Σ 2^{-p_j} (bits at positions p₁ < p₂ < ...), then the inner dynamics give T and exit. In the cases above: T = 21·... let me tabulate: - υ = 2^{-13} (bits: {13}): T = 32, exit t = 2. - υ = 2^{-13} + 2^{-24} ({13, 24}): T = 42, exit t = 2. - υ = {13, 24, 34}: T = 51, exit t = 2. - υ = {13, 24, 34, 43}: T = 59, exit t = 2. - υ = {13, 24, 34, 43, 51}: T = 66, exit t = 2. T: 32, 42, 51, 59, 66: differences 10, 9, 8, 7: next T = 72 (diff 6), 77 (5), 81 (4), 84 (3), 86 (2), 87 (1), then 87 (0)?? And f = 2^{-T-2}: e = T + 2: 34, 44, 53, 61, 68, 74, 79, 83, 86, 88, 89, 89?? Hmm the "89 repeated" is suspicious. Let me compute the case υ = {13, 24, 34, 43, 51, 58} (i.e., after adding increment 2^{-68}): s₆ = s₅ + 2^{-68}: υ₆ = 2^{-13} + 2^{-24} + 2^{-34} + 2^{-43} + 2^{-51} + 2^{-58}. Inner: t - 1 = υ₆, k₀ = 9: - 12 shifts [12]: t-1 = 2^{-1} + 2^{-12} + 2^{-22} + 2^{-31} + 2^{-39} + 2^{-46}: Rule B [13]: t-1 = 2^{-12} + 2^{-22} + 2^{-31} + 2^{-39} + 2^{-46}, k=8. - 11 shifts [24]: t-1 = 2^{-1} + 2^{-11} + 2^{-20} + 2^{-28} + 2^{-35}: Rule B [25]: t-1 = 2^{-11} + 2^{-20} + 2^{-28} + 2^{-35}, k=7. - 10 shifts [35]: t-1 = 2^{-1} + 2^{-10} + 2^{-18} + 2^{-25}: Rule B [36]: t-1 = 2^{-10} + 2^{-18} + 2^{-25}, k=6. - 9 shifts [45]: t-1 = 2^{-1} + 2^{-9} + 2^{-16}: Rule B [46]: t-1 = 2^{-9} + 2^{-16}, k=5. - 8 shifts [54]: t-1 = 2^{-1} + 2^{-8}: Rule B [55]: t-1 = 2^{-8}, k=4. - 7 shifts [62]: t-1 = 2^{-1}: t = 3/2: Rule A [63]: t = 2, k=4. - Rule B [64]: (3/2, 3). - Cycle: (3/2,3)→(2,3)[65]→(3/2,2)[66]→(2,2)[67]→(3/2,1)[68]→(2,1)[69]→(3/2,0)[70]→(2,0)[71]→exit[72]. T = 72: f(2 - s₆) = 2^{-74}. So e₇ = 74, bit position 64 = 58 + 6 ✓ pattern continues. Predicted next positions: 69 (+5), 73 (+4), 76 (+3), 78 (+2), 79 (+1), then +0?? Position 79 + 0 = 79: the next bit would be AT position 79 again?? That would double the bit at 79 → carry! Interesting. Let me carefully compute when the differences reach small values. Let me formalize. Let me define the state as the set of bit positions P = {p₁ < p₂ < ... < p_m} of υ = t - 1 (level 9, k₀ = 9 — valid while υ < 1, which holds since s < 2^{-9} ↔ υ < 1). Inner dynamics from (t - 1 = Σ_{p∈P} 2^{-p}, k = 9): Phase structure: repeatedly: shift j times until the leading bit crosses position 1 (i.e., t - 1 ≥ 1/2), then Rule B removes the 1/2 (k decrements), continue with remaining bits shifted. Let me think of it as: process bits from smallest position (largest exponent) — wait, bits at positions p: 2^{-p}: smaller p = bigger. The dynamics: shifts multiply by 2: bit at position p moves to position p - 1. Rule B: when there's a bit at position... hmm, Rule B triggers when t - 1 > 1/2, i.e., when the leading bit is at position 1 (t - 1 ∈ (1/2, 1)): Rule B: t → t - 1/2: removes the bit at position 1 (subtracts 2^{-1}), and k decrements. But careful: Rule B at t - 1 ∈ (1/2, 1): t - 1 = 2^{-1} + (rest): subtracting 1/2 leaves (rest). And Rule A at exactly t - 1 = 1/2 (t = 3/2): t → 2t - 1: t - 1 = 1: t = 2: i.e., the bit at position 1 gets "shifted" to position 0 (t - 1 = 1) WITHOUT k decrement. Then at t = 2 (t - 1 = 1 = 2^0·... i.e., "bit at position 0"): Rule B: t - 1 → 1/2 (bit at position 1), k decrement. Hmm, this t=2 ↔ t=3/2 cycle: t = 3/2: Rule A: t = 2: t - 1: 1/2 → 1 (shifted left); t = 2: Rule B: t - 1: 1 → 1/2 (bit flipped from position 0 to position 1), k--. Hmm OK. So in terms of bit positions: shifting moves bits left (p → p - 1). A bit reaching position 1: Rule B moves it to... no wait: Rule B subtracts 1/2: the bit at position 1 is REMOVED (it doesn't move). k decrements. But a bit reaching position 1 exactly when... hmm, Rule A vs Rule B: Rule A applies when t - 1 ≤ 1/2 (bit at position ≥ 2, or exactly at position 1 with nothing else? t - 1 = 1/2 exactly: Rule A: shift: bit moves to position 0: t = 2). Let me re-trace: bits at positions p₁ < p₂ < ...: the leading (leftmost) bit is p₁. Case: t - 1 < 1/2 (p₁ ≥ 2): Rule A shifts: all bits p → p - 1. (1 step) Case: t - 1 = 1/2 (p₁ = 1, only bit): Rule A: bit 1 → 0: t = 2: t - 1 = 1 (bit at position 0). (1 step) Case: t - 1 ∈ (1/2, 1): p₁ = 1 and other bits: Rule B: remove bit at position 1: remaining bits stay. k--. (1 step) Case: t - 1 = 1 (t = 2: "bit at position 0"): Rule B: t - 1 → 1/2: bit 0 → 1. k--. (1 step) Exit: Case t - 1 ∈ (1/2, 1] with k = 0: exit (1 step), z = 5/2 - t. So the algorithm on the multiset of bit positions, with k: - If min position ≥ 2: shift all (p → p-1). [1 step] - If positions = {1} exactly (t-1 = 1/2): bit 1 → 0. [1 step] [state: {0}] - If positions = {0} (t-1 = 1): bit 0 → 1. [1 step, k--] [state: {1}] - If min position = 1 and |positions| > 1 (t - 1 ∈ (1/2, 1)): remove bit 1. [1 step, k--] - If min position = 0 and |positions| > 1?? t - 1 ≥ 1 + something > 1: impossible (t ≤ 2 → t - 1 ≤ 1). - Exit: when k = 0 and we're about to apply a k-- rule (bit at position 1 with company, or bit at position 0): exit instead. Hmm wait, also: what if min position = 0 and |positions| = 1: t - 1 = 1: handled: bit 0 → 1, k-- (or exit if k=0). And min position = 1, |positions| = 1: t = 3/2: Rule A: bit 1 → 0: state {0}, k same. So the pair {1} → {0} → {1} → {0}...: alternates, with k-- only on {0} → {1} transition. Exit when {0} with k = 0 → exit. Hmm wait: exit condition: k = 0 and Rule B applies. {0} → {1} is Rule B: so exit when state = {0} (or bit at 1 with company) and k = 0. Hold on: state {1} (t = 3/2): Rule A: NOT exit even if k = 0. State {1, ...}: Rule B: exit if k = 0. State {0}: Rule B: exit if k = 0. In our chain: after processing all the "real" bits, the state became {0} or {1} etc. with some k, then cycled {1} → {0} → {1} → ... until k = 0 with state {0}: exit, z = 5/2 - 2 = 1/2. ✓ consistent with exits at t = 2 (state {0}). Wait, exit at state {0}: t = 2: z = 5/2 - 2 = 1/2 ✓. Exit at state {1, p, ...}: t = 3/2 + ...: z = 5/2 - t ∈ (1/2, 1). Exit at state... can exit happen at state {0, ...}? t - 1 > 1: impossible. OK so now the full inner algorithm: given bits P (positions ≥ 2 initially, since υ < 1/2 in our cases? hmm, υ = t - 1 could be ≥ 1/2 if s large... in our chain υ is tiny) and k₀ = 9: While true: - If all bits at positions ≥ 2: shift all bits left by 1 (p → p - 1). [step] (If P empty... hmm, P empty means t = 1: but t ∈ (1, 2]: t = 1 impossible. P empty can't happen since υ > 0.) - When min position = 1: - If P = {1}: shift: P = {0}. [step] - Else (P = {1, ...}): if k > 0: remove bit 1: P = P \ {1}. [step, k--]. If k = 0: exit. [step] - When min position = 0: - If P = {0}: if k > 0: P = {1}. [step, k--]. If k = 0: exit. [step] - Else (P = {0, ...}): impossible. Wait, there's an issue: after "shift all bits" when min position was 2: bits at position 2 → 1. If there were multiple bits, e.g., P = {2}: shift → {1}. P = {2, 5}: shift → {1, 4}. Then min = 1 with company: Rule B: remove 1: {4}, k--. Then continue shifting. Hmm wait, but that's not exactly right: after removing bit 1 from {1, 4}: {4}: shift 3 times: {3}, {2}, {1}: then {1} alone: shift → {0}: then {0}: k-- → {1}: cycle... Let me re-trace s₂'s computation with this algorithm: P = {13, 24}, k = 9. - Shifts: 12 steps: P = {1, 12}. (bits 13→1, 24→12) - Min = 1, company: k = 9 > 0: remove 1: P = {12}, k = 8. [13 steps] - Shifts: 11 steps: P = {1}. [24 steps] - P = {1}: shift: P = {0}. [25 steps] - P = {0}: k = 8 > 0: P = {1}, k = 7. [26 steps] - P = {1}: shift: {0}. [27] - {0}: k=7: {1}, k=6. [28] - ... alternating: each 2 steps k decreases by 1. From k=7 at step 26 hmm wait let me recount. At step 25: P = {0}, k = 8. Step 26: P={1}, k=7. Step 27: {0}. Step 28: {1}, k=6. ... So {0} with k = j: appears at steps: k=8 at 25, k=7 at 27, k=6 at 29, ..., k=0 at 25 + 2·8 = 41. Exit at state {0}, k=0: step 42. ✓ T = 42 ✓ matches! Great, the algorithm works. Exit at {0}, k = 0: t = 2: z = 1/2, f(z) = 2^{-2}. So: f(2 - s) = 2^{-2 - T} where T = number of steps, when exit at t = 2. Now the outer chain: s_{i+1} = s_i + 2^{-2 - T(s_i)}. And the bits P_i of υ_i = 2^{10}(s_i - 2^{-10}) (while level 9). We have the pattern: T values: 21 (for s₀, special), 32, 42, 51, 59, 66, 72, ...: differences 10, 9, 8, 7, 6, ...: T_j = ? Let me get the formula: T as function of P = {p₁ < p₂ < ... < p_m} and k₀ = 9: From the traces: P = {13}: shifts 12 → {1}: shift → {0} [13 steps]: then {0},{1} cycle from k=9: exit. {0} first reached at step 13 with k = 9. Then each 2 steps k--: exit at {0}, k=0: steps 13 + 2·9 = 31: T = 32 ✓. P = {13, 24}: 12 shifts [12]: {1, 12}: remove 1 [13]: {12}, k=8: 11 shifts [24]: {1}: shift [25]: {0}: k=8: cycle: exit at {0},k=0: 25 + 2·8 = 41: T = 42 ✓. P = {13, 24, 34}: [12 shifts]: {1,12,23}: remove [13]: {12,23}, k=8: [11 shifts]: {1,12}: remove [24]: {11}, k=7: [10 shifts]: {1}: shift [34+1=35]: {0}: 35 + 2·7 = 49: T = 51?? Hmm wait, earlier trace for P={13,24,34} gave T = 51. Let me recompute via formula: {0} reached at step: 12 + 1 + 11 + 1 + 10 + 1 = 36 with k = 7. Then exit at {0}, k = 0: 36 + 2·7 = 50: T = 51 ✓. OK. General pattern: bits {13, 24, 34, 43, ...}: each bit p_j (j ≥ 2) causes: shifts + 1 removal. The leading bit 13: 12 shifts + then... Let me define: after the leading bit reaches position 1 (12 shifts from 13), each subsequent bit p_j: (p_{j-1} - 1) shifts... hmm, let me think again: Bits p₁ = 13 < p₂ < p₃ < ...: - 12 shifts: p₁ → 1, others → p_j - 12. - Remove bit 1 (k--): now bits {p_j - 12}. - (p₂ - 12 - 1) shifts: p₂ - 12 → 1, others → p_j - 12 - (p₂ - 13). - Remove bit 1 (k--): bits shifted... - ... continue until last bit p_m → 1: remove? or if alone... Hmm wait: in the traces, the LAST bit reaching position 1: when it's alone at {1}: Rule A (shift to {0}), NOT removal. Then cycle {0}/{1} with k--, exit at {0}, k=0. Hmm, but wait: in the P = {13, 24} trace: after removing bit from {1, 12}: {12}, k = 8. Shifts: 11: {1}. Then {1} alone: shift → {0} [Rule A]. So the last bit goes 1 → 0 via Rule A (no k--), then {0} → {1} via Rule B (k--), alternating, exit at {0} with k = 0. So: k decrements: once per bit except the last one (bits 1..m-1 each decrement once via "remove at position 1"), plus the cycle decrements k from (k₀ - (m-1)) down to 0: that's (k₀ - (m-1)) more decrements hmm: wait: after processing m-1 bits (removals), k = k₀ - (m-1). Then last bit: reaches {1}, Rule A → {0}: then cycle: {0}→{1} (k--)→{0}→... exit at {0}, k=0. Total k--: (m - 1) + (k₀ - (m-1)) = k₀. ✓ (k goes from k₀ to 0.) Total steps: shifts + (m-1) removals + cycle steps. Shifts: 12 (for p₁ = 13) + Σ_{j=2}^{m} (p_j - p_{j-1} - 1)·... hmm: after removing bit at 1, the next bit is at position p₂ - 12 (it shifted 12 times): to bring it to position 1: (p₂ - 12 - 1) shifts. Then it's at 1: but wait, when the next bit arrives at position 1, are there other bits? Yes (p₃, ...). So it gets REMOVED (Rule B, k--). Hmm, unless it's the last bit. So shifts total: 12 + Σ_{j=2}^{m} (p_j - p_{j-1} - 1) = 12 + (p_m - p₁) - (m - 1) = 12 + p_m - 13 - m + 1 = p_m - m. Hmm: 12 + Σ_{j=2}^m (p_j - p_{j-1} - 1) = 12 + (p_m - p₁) - (m-1) = 12 + p_m - 13 - m + 1 = p_m - m. Check P = {13, 24}: p_m = 24, m = 2: shifts = 22: 12 + 11 = 23?? Hmm: 12 + (24 - 13 - 1) = 12 + 10 = 22. And formula p_m - m = 24 - 2 = 22 ✓. Removals: m - 1 steps. Wait: bits p₁..p_{m-1} get removed when reaching position 1 (each 1 step). Bit p_m reaches position 1 alone: Rule A: 1 step → {0}. Cycle: from {0}, k = k₀ - (m-1): steps: 2 per k-- until k=0 at {0}: steps: 2(k₀ - m + 1) - 1 to reach {0}, k=0, then exit 1 step. Hmm: {0} (k = j) → [1 step] {1} (k = j-1) → [1 step] {0} (k = j-1). So from {0}, k = K: reaching {0}, k = 0 takes 2K steps; exit + 1: total 2K + 1. Total T = [p_m - m] + [m - 1] + 1 + 2(k₀ - m + 1) + 1 = p_m - m + m - 1 + 1 + 2k₀ - 2m + 2 + 1 = p_m + 2k₀ - 2m + 3. Hmm wait let me recount: shifts: p_m - m. Removals: (m - 1). Rule A for last bit: +1. Cycle: 2(k₀ - (m-1)) + 1 (including exit step). T = (p_m - m) + (m - 1) + 1 + 2(k₀ - m + 1) + 1 = p_m - m + m - 1 + 1 + 2k₀ - 2m + 2 + 1 = p_m + 2k₀ - 2m + 3. Check P = {13}, m=1, k₀=9: T = 13 + 18 - 2 + 3 = 32 ✓. P = {13, 24}, m=2: T = 24 + 18 - 4 + 3 = 41?? But actual T = 42. ✗ Hmm. Let me retrace P = {13, 24}, k₀ = 9: - 12 shifts [12]: {1, 12}. - Rule B: remove 1 [13]: {12}, k = 8. - 11 shifts [24]: {1}. - Rule A [25]: {0}, k = 8. - Cycle: {0}(k=8) →{1}(k=7) [26] → {0} [27] → {1}(k=6) [28] → {0} [29] → {1}(k=5) [30] → {0} [31] → {1}(k=4) [32] → {0} [33] → {1}(k=3) [34] → {0} [35] → {1}(k=2) [36] → {0} [37] → {1}(k=1) [38] → {0} [39] → {1}(k=0) [40] → {0} [41] → exit [42]. Hmm wait: {0} with k=8: step 25. {0}→{1}: k--, so {1} with k=7 at step 26. {1}→{0}: k=7 at step 27. {0}(k=7)→{1}(k=6): 28. {0}(k=6): 29. So {0} with k=j at step 25 + 2(8 - j): k=0: step 25 + 16 = 41. Exit: step 42. T = 42 ✓. Formula check: shifts: 12 + 11 = 23. Removals: 1. Rule A: 1. Cycle from {0}, k=8: 2·8 + 1 = 17 steps (including exit). Total: 23 + 1 + 1 + 17 = 42 ✓. My formula above was wrong; recount: shifts = 12 + Σ_{j=2}^m (p_j - p_{j-1} - 1). For {13, 24}: 12 + (24 - 13 - 1) = 12 + 10 = 22?? But trace shows 12 + 11 = 23 shifts! Discrepancy: after removing bit at 1 (state {12}), we need 11 shifts to bring 12 → 1. (p₂ - p₁ - 1) = 24 - 13 - 1 = 10. But actual: 11. Hmm: bit p₂ = 24 was at position 24 initially; it shifted 12 times (to 12), then 11 more times (to 1): total 23 = 24 - 1 ✓. So shifts for bit j to reach position 1: p_j - 1. Total shifts = Σ... but bits shift together: total shift steps = (p_m - 1) if we only count until last bit reaches position 1? Bits all shift simultaneously; the process: shift until SOME bit hits position 1; the shifts happen one at a time; total shifts until the last bit reaches 1: p_m - 1. But wait — when a bit reaches position 1 with company, it's removed; other bits continue shifting. Total shift steps: each shift decrements all positions by 1: bit p_j reaches 1 after p_j - 1 shifts: at that moment it's removed (if company) or Rule A'd. The last bit p_m reaches 1 after p_m - 1 shifts. So total shifts = p_m - 1. Redo: T = (p_m - 1) [shifts] + (m - 1) [removals] + 1 [Rule A: {1} → {0}] + [2(k₀ - (m - 1)) + 1] [cycle incl. exit]. = p_m - 1 + m - 1 + 1 + 2k₀ - 2m + 2 + 1 = p_m + 2k₀ - m + 2. Check {13}, m=1, k₀=9: 13 + 18 - 1 + 2 = 32 ✓. {13,24}, m=2: 24 + 18 - 2 + 2 = 42 ✓. {13,24,34}, m=3: 34 + 18 - 3 + 2 = 51 ✓. {13,24,34,43}, m=4: 43 + 18 - 4 + 2 = 59 ✓. {13,24,34,43,51}, m=5: 51 + 18 - 5 + 2 = 66 ✓. {13,24,34,43,51,58}, m=6: 58 + 18 - 6 + 2 = 72 ✓. So T = p_m + 2k₀ - m + 2, where p_m is the LARGEST bit position (rightmost bit), m = number of bits, k₀ = 9 (level). [Note: this assumed exit at {0} (t = 2), which requires... the last bit reaching position 1 ALONE (i.e., p_m - 1 shifts bring p_m to 1, and all other bits already removed — true since p_m is largest: other bits reached 1 earlier and were removed — wait, unless two bits coincide! If bits coincide (carry), things differ. Also if the last bit reaches position 1 exactly when... hmm, bits are distinct here.) Also note: this assumed k₀ ≥ m - 1 (enough k to remove m-1 bits and still have the cycle reach 0 — actually cycle handles k from k₀ - (m-1) down to 0: fine as long as k stays ≥ 0 during removals: need k₀ ≥ m - 1... if k hits 0 during removals, exit happens early at a "remove" step: different formula. In our case k₀ = 9 and m grows... eventually m > 10!! Then the formula changes! Hmm, wait: when m - 1 = k₀ = 9: after 9 removals k = 0: the 10th bit arriving at position 1 with company → Rule B at k = 0 → EXIT at t ∈ (3/2, 2): different exit point! Let me keep this in mind.) So the increments: Δ_i = f(2 - s_i) = 2^{-2 - T_i} = 2^{-(p_m + 2k₀ - m + 4)} = 2^{-(p_m - m + 22)} (k₀ = 9). And the new bit position (in υ = t - 1 coordinates): the increment Δ added to s: new bit in υ = 2^{10}Δ at position: (p_m - m + 22) - 10 = p_m - m + 12. So: if current bits are {p₁, ..., p_m} (p_m largest), the new bit is at position q = p_m - m + 12, and the new largest bit is p_{m+1} = max(p_m, q). Since q = p_m - m + 12: if m < 12: q > p_m: new bit is the new largest: p_{m+1} = q. If m = 12: q = p_m: COINCIDES with p_m: carry! If m > 12: q < p_m: new bit is NOT the largest (it's inserted inside)! Interesting!! So the pattern changes at m = 12. Let me carefully track. Bits: p₁ = 13; p_{j+1} = p_j - j + 12 for j < 12 (m = j at each step: careful with indices). Let me re-index: state after i-th increment: bits B_i = {p_1, ..., p_{m_i}} with m_i = i (starting: B_1 = {13} after first increment? hmm: s₁ has bit {13}: m = 1. Then increment computed from B_1: new bit q = p - m + 12 = 13 - 1 + 12 = 24 ✓. B_2 = {13, 24}. New bit: q = 24 - 2 + 12 = 34 ✓. B_3 = {13, 24, 34}. q = 34 - 3 + 12 = 43 ✓. B_4: q = 43 - 4 + 12 = 51 ✓. B_5: q = 51 - 5 + 12 = 58 ✓. B_6: q = 58 - 6 + 12 = 64 ✓. B_7: q = 64 - 7 + 12 = 69. B_8: q = 69 - 8 + 12 = 73. B_9: q = 73 - 9 + 12 = 76. B_10: q = 76 - 10 + 12 = 78. B_11: q = 78 - 11 + 12 = 79. B_12: q = 79 - 12 + 12 = 79!! At m = 12: bits B_12 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 78, 79}: the new bit q = 79 coincides with existing bit at 79: CARRY: 2^{-79} + 2^{-79} = 2^{-78}: but there's already a bit at 78: carry again: 78: 2× → 77: nothing at 77: so B_13 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 77}. Hmm wait: bits at 79 and 78: adding 2^{-79}: 79 becomes 2·2^{-79} = 2^{-78}: merges with existing 2^{-78}: total 2·2^{-78} = 2^{-77}: new bit at 77. So B_13 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 77}: m = 11, largest p = 77. But hold on: I need to double check the increment formula when m = 12: the increment Δ = 2^{-(p_m - m + 22)} = 2^{-(79 - 12 + 22)} = 2^{-89}. And the new bit in υ at position 89 - 10 = 79 ✓. So yes: at m = 12, the increment 2^{-89} adds bit at 79, causing carries: 79 + 79 → 78 (merge with existing) → 77. But WAIT: I need to double-check the T formula for m = 12 — the concern above: removals = m - 1 = 11 > k₀ = 9! So during removals, k would hit 0 before all bits removed: EXIT happens at a removal step (Rule B at k=0 with t ∈ (3/2, 2)): different exit point z ∈ (1/2, 1), different f(z)! Hmm! So actually the formula T = p_m + 2k₀ - m + 2 is only valid for m ≤ k₀ + 1 = 10 (removals m - 1 ≤ 9 = k₀). For m = 11: removals = 10 > 9: the 10th removal would occur at k = 0 → exit before. Let me recompute! Ugh, I conflated. Let me redo carefully. k₀ = 9 (level 9). Bits m of them. Process: bit j (j = 1..m) arrives at position 1: bits 1..m-1 arrive WITH company (assuming distinct positions, no coincident arrivals) → Rule B (k--). Bit m arrives alone → Rule A. For this to work, need k ≥ 1 at each of the first m-1 arrivals: k after j-1 removals: k₀ - (j - 1) ≥ 1 ↔ j ≤ k₀. So bits j = 1..k₀ can be removed (j ≤ 9); bit j = 10 arriving with company at k = 0: EXIT (Rule B at k=0): z = 5/2 - t where t = 3/2 + (remaining bits beyond position 1). So for m ≥ 11 (i.e., at least 10 bits arrive with company — the 10th arrival triggers exit at k=0): the process exits early, at the 10th bit's arrival! Let me recompute T for m ≥ 11: exit happens when bit 10 (the 10th smallest... wait, bits arrive in order of position: smallest position (leftmost) first. Bits sorted: p₁ < p₂ < ... < p_m: arrivals: p₁ first, ..., p_m last. Arrivals with company: p₁, ..., p_{m-1}. p_m arrives alone (Rule A) — IF the process gets that far. But if m - 1 ≥ 10: the 10th arrival-with-company (bit p₁₀) happens at k = 0: EXIT. So for m ≥ 11: exit at bit p₁₀'s arrival: t - 1 = 2^{-1} + (bits p₁₁, ..., p_m shifted appropriately): z = 5/2 - t = 1 - (t - 3/2) = 1 - (bits after removal... let me compute: at exit, t = 3/2 + 2^{-(p₁₁ - p₁₀)}·... hmm: when p₁₀ arrives at position 1: bits p₁₁..p_m are at positions p_{j} - (p₁₀ - 1). So t - 1 = 2^{-1} + Σ_{j=11}^m 2^{-(p_j - p₁₀ + 1)}. z = 5/2 - t = 1 - Σ_{j≥11} 2^{-(p_j - p₁₀ + 1)}. And T = [shifts: p₁₀ - 1] + [removals: 9] + [exit step: 1] = p₁₀ - 1 + 9 + 1 = p₁₀ + 9. Hmm wait: also the k-- from removals: 9 removals consume k from 9 to 0 ✓. So for m ≥ 11: T = p₁₀ + 9, and exit z = 1 - Σ_{j=11}^m 2^{-(p_j - p₁₀ + 1)}. Then f(2 - s) = f(z)/2^T. f(z) for z ∈ [1/2, 1): z = 1 - w where w = Σ_{j≥11} 2^{-(p_j - p₁₀ + 1)} ∈ (0, 1/2]: f(z) = (1 - z) - 2^{-(K+1)} = w - 2^{-(K+1)} where K = level of z: z ∈ [1 - 2^{-K}, 1 - 2^{-(K+1)}) ↔ w ∈ (2^{-(K+1)}, 2^{-K}]. OK!! So now I need to carefully recompute the outer chain from m = 10 or so. Let me first double check: for m = 10: removals = 9 = k₀: OK: 9 removals (k: 9 → 0), then bit p₁₀ = p_m arrives alone: Rule A: {0}, k = 0: then Rule B at {0}, k = 0: EXIT. T = (p_m - 1) + 9 + 1 [Rule A] + 1 [exit] = p_m + 10. Via the old formula: T = p_m + 2k₀ - m + 2 = p_m + 18 - 10 + 2 = p_m + 10 ✓ consistent! Good: so old formula valid for m ≤ 10. For m = 11: old formula would give T = p_m + 18 - 11 + 2 = p_m + 9. New analysis: exit at p₁₀'s arrival: T = p₁₀ + 9. Since p₁₀ < p₁₁ = p_m: T = p₁₀ + 9 < p_m + 9. So different! Let me now carefully recompute the chain around m = 10, 11, 12. Bits progression: B_m: {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 78, 79} for m = 1..12 as computed: B_10 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 76}, B_11 = B_10 ∪ {78}, B_12 = B_11 ∪ {79}. Increments: for m ≤ 10: Δ = 2^{-(p_m + 10)} (since T = p_m + 10, Δ = 2^{-T-2} = 2^{-(p_m + 12)}... wait: Δ = 2^{-2-T} = 2^{-(p_m + 12)}. Hmm: earlier: T = p_m + 2k₀ - m + 2 = p_m + 20 - m·... k₀ = 9: T = p_m + 18 - m + 2 = p_m + 20 - m. For m = 10: T = p_m + 10 ✓. Δ = 2^{-(T + 2)} = 2^{-(p_m + 22 - m)}. New bit position: q = (p_m + 22 - m) - 10 = p_m + 12 - m ✓ (matches earlier: q = p_m - m + 12 ✓). So increments for m = 1..10 (adding bit to go from B_m to B_{m+1}): m=1: p_m=13: Δ = 2^{-(13+21)} = 2^{-34}: new bit q = 13 + 11 = 24 ✓. m=2: p=24: Δ = 2^{-(24+20)} = 2^{-44}: q = 24 + 10 = 34 ✓. m=3: p=34: Δ = 2^{-(34+19)} = 2^{-53}: q = 34+9 = 43 ✓. m=4: p=43: Δ = 2^{-(43+18)} = 2^{-61}: q = 51 ✓. m=5: p=51: Δ = 2^{-(51+17)} = 2^{-68}: q = 58 ✓. m=6: p=58: Δ = 2^{-75}: q = 64 ✓. m=7: p=64: Δ = 2^{-(64+15)} = 2^{-81}: q = 69 ✓. m=8: p=69: Δ = 2^{-(69+14)} = 2^{-83}: q = 73 ✓. m=9: p=73: Δ = 2^{-(73+13)} = 2^{-85}: q = 76 ✓. Wait, hold on: m=9: B_9 = {13,24,34,43,51,58,64,69,73}: increment computed from m=9 state: Δ = 2^{-(p_m + 22 - m)} = 2^{-(73 + 13)} = 2^{-86}: q = 73 + 12 - 9 = 76 ✓. Hmm: 73 + 22 - 9 = 86: Δ = 2^{-86}, q = 76 ✓. (I wrote 85 above by mistake; recompute: p_m + 22 - m = 73 + 22 - 9 = 86. And T = p_m + 20 - m = 84: Δ = 2^{-86} ✓.) Let me redo the table cleanly. For state B_m (m bits), the increment Δ_m = 2^{-(p_m + 22 - m)} (valid m ≤ 10), new bit at position q_m = p_m + 12 - m. m=1: B_1 = {13}: Δ₁ = 2^{-(13+21)} = 2^{-34}: q = 24. [T = 32: Δ = 2^{-34} ✓] m=2: B_2 = {13,24}: Δ₂ = 2^{-(24+20)} = 2^{-44}: q = 34. [T = 42 ✓] m=3: B_3: Δ₃ = 2^{-(34+19)} = 2^{-53}: q = 43. [T = 51 ✓] m=4: Δ₄ = 2^{-(43+18)} = 2^{-61}: q = 51. [T = 59 ✓] m=5: Δ₅ = 2^{-(51+17)} = 2^{-68}: q = 58. [T = 66 ✓] m=6: Δ₆ = 2^{-(58+16)} = 2^{-74}: q = 64. [T = 72 ✓] m=7: Δ₇ = 2^{-(64+15)} = 2^{-79}: q = 69. m=8: Δ₈ = 2^{-(69+14)} = 2^{-83}: q = 73. m=9: Δ₉ = 2^{-(73+13)} = 2^{-86}: q = 76. m=10: B_10 = {13,24,34,43,51,58,64,69,73,76}: Δ₁₀ = 2^{-(76+12)} = 2^{-88}: q = 76 + 12 - 10 = 78. Wait: q = p_m + 12 - m = 76 + 2 = 78. But m = 10 state: T = p_m + 20 - m = 76 + 10 = 86: Δ = 2^{-88} ✓: new bit at 88 - 10 = 78 ✓. B_11 = {13,24,34,43,51,58,64,69,73,76,78}: m = 11. Now m = 11: the formula changes! For m = 11 ≥ 11: exit at 10th arrival: T = p₁₀ + 9 where p₁₀ = 10th smallest bit = 76. T = 76 + 9 = 85. Exit z = 1 - Σ_{j=11}^{11} 2^{-(p_j - p₁₀ + 1)} = 1 - 2^{-(78 - 76 + 1)} = 1 - 2^{-3} = 7/8. f(z) = f(7/8): 7/8 = 1 - 2^{-3}: level: [1 - 2^{-3}, 1 - 2^{-4}) = [7/8, 15/16): K = 3: f(7/8) = (1 - 7/8) - 2^{-4} = 1/8 - 1/16 = 1/16 = 2^{-4}. So Δ₁₁ = f(2 - s₁₁) = f(z)/2^T = 2^{-4}/2^{85} = 2^{-89}. New bit position: q = 89 - 10 = 79. B_12 = {13,...,78, 79}: m = 12 ✓ (matches earlier expectation: bit at 79). m = 12: B_12 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 78, 79}. Exit at 10th arrival: p₁₀ = 76: T = 85. Exit z = 1 - [2^{-(78-76+1)} + 2^{-(79-76+1)}] = 1 - [2^{-3} + 2^{-4}] = 1 - 3/16 = 13/16. f(13/16): 13/16 = 0.8125: level: [1 - 2^{-2}, 1 - 2^{-3}) = [3/4, 7/8) = [12/16, 14/16): K = 2: f = (1 - 13/16) - 2^{-3} = 3/16 - 2/16 = 1/16 = 2^{-4}. Δ₁₂ = 2^{-4}/2^{85} = 2^{-89}. New bit: q = 79. But bit at 79 already exists! Carry: 2^{-79}·2 → position 78: exists → carry → 77: B_13 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 77}: m = 11. (Largest bit 77.) Hmm interesting. m went 12 → 11. Let me continue. m = 11: B_13 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 77}: p₁₀ = 76, p₁₁ = 77. T = p₁₀ + 9 = 85. Exit z = 1 - 2^{-(77-76+1)} = 1 - 2^{-2} = 3/4. f(3/4) = 1/8 = 2^{-3} (computed early on: f(3/4) = 1/8 ✓). Δ₁₃ = 2^{-3}/2^{85} = 2^{-88}. New bit: q = 88 - 10 = 78. B_14 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 77, 78}: m = 12. m = 12: B_14 = {..., 76, 77, 78}: p₁₀ = 76: T = 85. Exit z = 1 - [2^{-(77-76+1)} + 2^{-(78-76+1)}] = 1 - [2^{-2} + 2^{-3}] = 1 - 3/8 = 5/8. f(5/8) = 1/8 = 2^{-3} ✓ (computed early: f(5/8) = 1/8 ✓). Δ₁₄ = 2^{-3}/2^{85} = 2^{-88}. New bit q = 78: exists! Carry: 78 → 77: exists → 76: exists → 75: B_15 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 75}: m = 10. Hmm wait: bits 76, 77, 78 all exist: adding 78: 78×2 → 77: 77×2 → 76: 76×2 → 75: nothing at 75: B_15 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 75}. m = 10, largest 75. Hmm, wait, but hold on: I should double-check the "exit at 10th arrival" analysis — specifically whether the exit z computation is right, and also whether the bits p₁₁, p₁₂ etc. positions shift correctly. Let me re-derive: state: bits at positions p₁ < ... < p_m, k₀ = 9. Process: shifts bring p₁ to position 1: at that time, bits p₂..p_m are at p_j - p₁ + 1. Remove bit at 1 (k--). Continue: shift p₂ - p₁ - 1 more times: p₂ at 1: bits p₃.. at p_j - p₂ + 1. Remove (k--). ... After 9 removals (bits p₁..p₉): k = 0. Shifts bring p₁₀ to position 1: bits p₁₁..p_m at p_j - p₁₀ + 1. State t - 1 = 1/2 + Σ_{j≥11} 2^{-(p_j - p₁₀ + 1)}: Rule B at k=0: EXIT: z = 5/2 - t = 1 - Σ_{j≥11} 2^{-(p_j - p₁₀ + 1)} ✓. T = shifts (p₁₀ - 1) + removals (9) + exit (1) = p₁₀ + 9 ✓. OK so continuing the chain. Hmm, wait, but now with B_15 having m = 10 ≤ 10, the formula reverts: Δ = 2^{-(p_m + 22 - m)} with p_m = 75, m = 10: Δ = 2^{-(75 + 12)} = 2^{-87}: q = 75 + 12 - 10 = 77. Hmm wait: but hold on: m = 10 formula: T = p_m + 20 - m = 75 + 10 = 85: Δ = 2^{-87}: new bit at 87 - 10 = 77 ✓. B_16 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 75, 77}: m = 11. Hmm wait, B_15 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 75}: adding bit at 77: B_16 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 75, 77}: m = 11: p₁₀ = 75, p₁₁ = 77. m=11: T = p₁₀ + 9 = 84. Exit z = 1 - 2^{-(77 - 75 + 1)} = 1 - 2^{-3} = 7/8. f(7/8) = 2^{-4}. Δ₁₆ = 2^{-4-84} = 2^{-88}: q = 78: B_17 = {13,24,34,43,51,58,64,69,73,75,77,78}: m = 12. m=12: p₁₀ = 75: T = 84. z = 1 - [2^{-(77-75+1)} + 2^{-(78-75+1)}] = 1 - [2^{-3} + 2^{-4}] = 13/16: f = 2^{-4} (as before). Δ = 2^{-88}: q = 78: CARRY: 78 → 77 → 76: B_18 = {13,24,34,43,51,58,64,69,73,75,76}: m = 11. Hmm, wait: 78 exists → 77 exists → 76: free: B_18 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 75, 76}: m = 11. m=11: p₁₀ = 75, p₁₁ = 76: T = 75 + 9 = 84: z = 1 - 2^{-(76-75+1)} = 1 - 1/4 = 3/4: f(3/4) = 2^{-3}: Δ = 2^{-87}: q = 77: B_19 = {13,24,34,43,51,58,64,69,73,75,76,77}: m = 12. m=12: p₁₀ = 75: T = 84: z = 1 - [2^{-2} + 2^{-3}] = 5/8: f = 2^{-3}: Δ = 2^{-87}: q = 77: carry: 77→76→75→74: B_20 = {13,24,34,43,51,58,64,69,73,74, ...wait. Hold on: B_19 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 75, 76, 77}. Adding bit at 77: 77 exists → carry to 76: exists → carry to 75: exists → carry to 74: free: B_20 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 74, 75, 76}?? Hmm wait: no: when bit at 77 doubles, it merges: positions 77 and 76 and 75 had bits: adding 2^{-77}: now 77 has 2 bits → carries to 76: 76 had 1 → now 2 → carries to 75: 75 had 1 → 2 → carries to 74: 74 had 0 → 1. Net: bits at 77, 76, 75 removed; bit at 74 added. B_20 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 74}: m = 10. Hmm OK wait, wait. Let me recount B_20: B_19 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 75, 76, 77} (12 bits). Add 2^{-77}: 77: 2 bits → carry → 76: 2 bits → carry → 75: 2 bits → carry → 74: 1 bit. Result: {13, 24, 34, 43, 51, 58, 64, 69, 73, 74}: the bits at 75, 76, 77 are gone (consumed), bit at 74 added: m = 10 ✓. m=10: B_20 = {13,24,34,43,51,58,64,69,73,74}: p_m = 74: Δ = 2^{-(74 + 22 - 10)} = 2^{-86}: q = 74 + 12 - 10 = 76: B_21 = {13,24,34,43,51,58,64,69,73,74,76}: m = 11. m=11: p₁₀ = 74, p₁₁ = 76: T = 74 + 9 = 83: z = 1 - 2^{-(76-74+1)} = 1 - 2^{-3} = 7/8: f = 2^{-4}: Δ = 2^{-87}: q = 77: B_22 = {13,24,34,43,51,58,64,69,73,74,76,77}: m = 12. m=12: p₁₀ = 74: T = 83: z = 1 - [2^{-3} + 2^{-4}] = 13/16: f = 2^{-4}: Δ = 2^{-87}: q = 77: carry: 77 → 76 → 75: B_23 = {13,24,34,43,51,58,64,69,73,74,75,76}?? Hmm wait: B_22 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 74, 76, 77}: add 77: 77 has 1 → 2: carry → 76: 1 → 2: carry → 75: 0 → 1. Result: {13, 24, 34, 43, 51, 58, 64, 69, 73, 74, 75}: bits at 76, 77 gone, 75 added: m = 11. B_23 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 74, 75}: m = 11. m=11: p₁₀ = 74, p₁₁ = 75: T = 83: z = 1 - 2^{-(75-74+1)} = 1 - 1/4 = 3/4: f = 2^{-3}: Δ = 2^{-86}: q = 76: B_24 = {13,24,34,43,51,58,64,69,73,74,75,76}: m = 12. m=12: p₁₀ = 74: T = 83: z = 1 - [2^{-2} + 2^{-3}] = 5/8: f = 2^{-3}: Δ = 2^{-86}: q = 76: carry: 76 → 75 → 74 → 73?? B_24 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 74, 75, 76}: add 76: 76→77·...: 76: 1→2: carry → 75: 1→2: carry → 74: 1→2: carry → 73: 1→2: carry → 72: 0→1. Result: {13, 24, 34, 43, 51, 58, 64, 69, 72}: m = 9. Hmm!! Wait: 73 had a bit (yes, 73 ∈ B_24): so carry chain: 76, 75, 74, 73 all had bits: all consumed, new bit at 72. B_25 = {13, 24, 34, 43, 51, 58, 64, 69, 72}: m = 9. Hmm, this is getting complicated — the bit set is evolving in a nontrivial way, kind of like an odometer/carry dynamics. Let me organize: I'll track the state as (sorted bit positions), and the rule: Given bits P (positions of 1-bits of υ = t - 1, where s = 2^{-10}(1 + υ), level 9 — valid while υ < 1... hmm, actually valid while s ∈ (2^{-10}, 2^{-9}] i.e. υ ∈ (0, 1]): - Let m = |P|, and let p₁ < ... < p_m. - If m ≤ 10: T = p_m + 20 - m: Δ = 2^{-(T+2)}: new bit at position q = T + 2 - 10 = p_m + 12 - m. [exit at t = 2, z = 1/2] - If m ≥ 11: T = p₁₀ + 9: z = 1 - Σ_{j≥11} 2^{-(p_j - p₁₀ + 1)}: f(z) = w - 2^{-(K+1)} with w = 1 - z, K = level of z (w ∈ (2^{-(K+1)}, 2^{-K}]): Δ = f(z)/2^T: new bit at q = log2(1/Δ). Hmm wait, but actually, I realize the exit z analysis needs care when m ≥ 11: the bits p₁₁, ..., p_m at exit: positions p_j - p₁₀ + 1 ≥ 2 (since p_j ≥ p₁₀ + 1: position ≥ 2). w = Σ 2^{-(p_j - p₁₀ + 1)} ≤ Σ_{i≥2} 2^{-i}·... if bits are consecutive: p_j = p₁₀ + (j - 10): w = Σ_{i=2}^{m-10+1} 2^{-i} < 1/2. So w ∈ (0, 1/2): z = 1 - w ∈ (1/2, 1): fine. This is getting intricate but tractable. However, I realize this could go on for MANY steps (the bits need to march from position ~13 down to... hmm, wait: actually what's the termination condition for the outer chain? The outer chain continues while x = 3 - s ∈ [2, 3), i.e., s ≤ 1. As bits march left and accumulate, υ grows: eventually υ → 1: s → 2^{-9}: then level drops to 8, etc. This is like an odometer filling up. Hmm, the total number of outer steps could be enormous. Hmm, wait. Let me reconsider. Hmm, actually, let me reconsider the problem statement: f(x) = f(x - f(x-1))/2. Wait, maybe I should double-check whether the problem is f(x) = f(x - f(x-1))/2 or something else. The problem: "f(x) = -x for x<0; f(x) = f(x - f(x-1))/2 for x≥0. Compute f(3)." Yes. Hmm, and we have f(0) = 1/2, f(1) = 1/8, f(2) = 1/1024. Let me see... this is actually a known problem! I believe this is from the 2010(or so) "Internet puzzle" or Putnam... Actually I recall this exact problem: it's Problem 4 from the 2014 MIT Integration Bee? No... Hmm. I recall the sequence 1/2, 1/8, 1/1024... hmm, 2^{-1}, 2^{-3}, 2^{-10}: exponents 1, 3, 10: these are the values a(n) where a(n) = C(2n, n) - ... hmm: 1, 3, 10: hmm, wait: 1 = 2^0, 3 = 2^1 + 1, 10 = 2^3 + 2: or 1, 3, 10, 33, 104? a(n+1) = 3a(n) + 3^{...}: 1→3: ×3; 3→10: ×3+1; 10→33? Hmm. Actually maybe: a(n) = (3^n - 1)/2 + ...: (3-1)/2 = 1 ✓ (n=1); (9-1)/2 = 4 ✗. Or a(n) = T_n + ...: T_1 = 1, T_2 = 3, T_4 = 10, T_8 = 36, T_16 = 136: pattern T_{2^{n-1}}? For n = 1: T_1 = 1 ✓; n = 2: T_2 = 3 ✓; n = 3: T_4 = 10 ✓; n = 4: T_8 = 36. So f(3) = 2^{-36}? Hmm, T_{2^{n-1}} = 2^{n-2}(2^{n-1}+1): n=1: 2^{-1}·3 = 1.5?? Let me just compute: T_1 = 1, T_2 = 3, T_4 = 10, T_8 = 36, T_16 = 136. If f(n) = 2^{-T_{2^{n-1}}}: f(0) = 2^{-T_1} = 2^{-1} ✓; f(1) = 2^{-T_2} = 2^{-3} ✓; f(2) = 2^{-T_4} = 2^{-10} ✓; f(3) = 2^{-T_8} = 2^{-36}. Hmm, that's a cute pattern: T_{2^k} = 2^{k-1}(2^k + 1) = 2^{2k-1} + 2^{k-1} = (4^k + 2^k)/2. Check: k=0: (1+1)/2 = 1 ✓; k=1: (4+2)/2 = 3 ✓; k=2: (16+4)/2 = 10 ✓; k=3: (64+8)/2 = 36. So f(3) = 2^{-36} = 1/68719476736. Alternatively, maybe the exponent sequence is 1, 3, 10, 29 (a(n+1) = 3a(n) - n·...): 3 = 3·1, 10 = 3·3+1, 29 = 3·10 - 1? Meh. Or 1, 3, 10, 31: a(n+1) = 3a(n) + n·: 1→3 (+0·3?); hmm: 3·1 + 0 = 3; 3·3 + 1 = 10; 3·10 + 2 = 32? or 3·10 + 1 = 31. Given the structure I'm uncovering (the huge step counts), let me sanity check by computing the total number of halvings in f(2) via my framework and see if it matches 10 + (inner steps): f(2) = f(15/8)/2 [1 outer step]. f(15/8) = f(29/16)/2 [2]. f(29/16) = f(7/4)/2 [3]. f(7/4) = f(13/8)/2 [4]. f(13/8) = f(3/2)/2 [5]. f(3/2) = f(5/4)/2 [6]. f(5/4) = f(1)/2 [7]. f(1) = f(1/2)/2 [8]. f(1/2) = f(0)/2 [9]. f(0) = f(-1/2)/2 [10]. f(-1/2) = 1/2. So f(2) = (1/2)/2^{10} = 2^{-11}?? WAIT. That contradicts f(2) = 2^{-10}!! Let me recount. f(0) = f(0 - f(-1))/2 = f(-1)/2 = 1/2. ✓ (f(-1) = 1.) Hmm: f(0) = 1/2. f(1/2) = f(1/2 - f(-1/2))/2 = f(1/2 - 1/2)/2 = f(0)/2 = 1/4 ✓. f(1) = f(1 - f(0))/2 = f(1 - 1/2)/2 = f(1/2)/2 = 1/8 ✓. f(5/4) = f(5/4 - f(1/4))/2 = f(5/4 - 1/4)/2 = f(1)/2 = 1/16 ✓. f(3/2) = f(3/2 - f(1/2))/2 = f(3/2 - 1/4)/2 = f(5/4)/2 = 1/32 ✓. f(13/8) = f(13/8 - f(5/8))/2 = f(13/8 - 1/8)/2 = f(3/2)/2 = 1/64 ✓. f(7/4) = f(7/4 - f(3/4))/2 = f(7/4 - 1/8)/2 = f(13/8)/2 = 1/128 ✓. f(29/16) = f(29/16 - f(13/16))/2 = f(29/16 - 1/16)/2 = f(7/4)/2 = 1/256 ✓. f(15/8) = f(15/8 - f(7/8))/2 = f(15/8 - 1/16)/2 = f(29/16)/2 = 1/512 ✓. f(2) = f(2 - f(1))/2 = f(2 - 1/8)/2 = f(15/8)/2 = 1/1024 ✓. So the chain for f(2): 2 → 15/8 → 29/16 → 7/4 → 13/8 → 3/2 → 5/4 → 1 → 1/2 → 0 → -1/2: that's 10 halvings from f(2) down: f(2) = f(-1/2)/2^{10} = (1/2)/2^{10} = 2^{-11}?? But direct computation gives 2^{-10}! Let me recount the halvings: f(2) = f(15/8)/2 [1]. f(15/8) = f(29/16)/2 [2]. f(29/16) = f(7/4)/2 [3]. f(7/4) = f(13/8)/2 [4]. f(13/8) = f(3/2)/2 [5]. f(3/2) = f(5/4)/2 [6]. f(5/4) = f(1)/2 [7]. f(1) = f(1/2)/2 [8]. f(1/2) = f(0)/2 [9]. f(0) = f(-1/2)/2 [10]. f(-1/2) = 1/2 (no halving: direct). So f(2) = 2^{-10}·(1/2) = 2^{-11}??? But the step-by-step values: f(15/8) = 1/512 = 2^{-9}, f(2) = 2^{-10}. And f(29/16) = 2^{-8}, f(7/4) = 2^{-7}, f(13/8) = 2^{-6}, f(3/2) = 2^{-5}, f(5/4) = 2^{-4}, f(1) = 2^{-3}, f(1/2) = 2^{-2}, f(0) = 2^{-1}, f(-1/2) = 2^0 = 1?? WAIT: f(-1/2) = -(-1/2) = 1/2. Hmm: f(x) = -x for x < 0: f(-1/2) = 1/2 = 2^{-1}. And f(0) = f(-1)/2 = 1/2: f(-1) = 1: f(0) = 1/2 ✓. So the chain: f(-1/2) = 2^{-1}. f(0) = f(-1/2)/2 = 2^{-2}?? But f(0) = 1/2 = 2^{-1}!!! CONTRADICTION! Ah, I see my error: f(0) = f(0 - f(-1))/2 = f(-1)/2: f(-1) = 1: f(0) = 1/2. The chain from 0 goes to -1 (not -1/2): 0 → 0 - f(-1) = -1. So f(0) = f(-1)/2 = 1/2 ✓. Then f(1/2) = f(0)/2: chain 1/2 → 1/2 - f(-1/2) = 0: f(1/2) = f(0)/2 = 1/4 ✓. f(1) → 1 - f(0) = 1/2: f(1) = f(1/2)/2 ✓. f(5/4) → 5/4 - f(1/4) = 1: f(5/4) = f(1)/2 ✓. f(3/2) → 3/2 - f(1/2) = 5/4 ✓. f(13/8) → 13/8 - f(5/8) = 3/2 ✓. f(7/4) → 7/4 - f(3/4) = 13/8 ✓. f(29/16) → 29/16 - f(13/16) = 7/4 ✓. f(15/8) → 15/8 - f(7/8) = 29/16 ✓. f(2) → 2 - f(1) = 15/8 ✓. So chain: 2 → 15/8 → 29/16 → 7/4 → 13/8 → 3/2 → 5/4 → 1 → 1/2 → 0 → -1: 10 steps: f(2) = f(-1)/2^{10} = 1/2^{10} = 2^{-10} ✓✓. Phew: f(-1) = 1 = 2^0. OK so the "bottom" value is f(-1) = 1, and f(2) = 2^{-10} ✓. Good, consistent: exponents: f(0) = 2^{-1} [1 step: f(0) = f(-1)/2], f(1) = 2^{-3} [3 steps: 1 → 1/2 → 0 → -1], f(2) = 2^{-10} [10 steps]. So exponent = number of steps in the chain from n down to negative. a(0) = 1, a(1) = 3, a(2) = 10: number of steps. And f(3) = 2^{-a(3)} where a(3) = number of steps in the chain from 3. So a(3) = total number of halvings = (outer steps in [2,3)) + (steps from 2 down... wait, no: the chain from 3: 3 → 3 - f(2) = 3 - 2^{-10} → ... all the way down to negative. The chain passes through [2, 3), [1, 2), [0, 1), then negative. Total steps a(3). So my framework: outer chain in [2,3): steps counted; then when the chain enters [1,2) (i.e., a_i < 2), continue with the [1,2) chain (which itself is the "inner" dynamics); then [0,1); then negative. Note: the chain from 2 took 10 steps: 2 → 15/8 (1 step in [2, 3)? No: 2 → 15/8: 15/8 < 2: so from 2, ONE step lands in [1, 2). Then 9 more steps to -1.) So a(2) = 1 + 9 where the 9 steps: 15/8 → 29/16 → 7/4 → 13/8 → 3/2 → 5/4 → 1 → 1/2 → 0 → -1: that's 10 transitions: hmm: 15/8 → 29/16 [1] → 7/4 [2] → 13/8 [3] → 3/2 [4] → 5/4 [5] → 1 [6] → 1/2 [7] → 0 [8] → -1 [9]. So from 15/8: 9 steps. Total from 2: 10 ✓. So a(3) = (steps in [2,3) from 3 - hmm, from x₀ = 3 - 2^{-10}... wait, actually the chain from 3: first step: 3 → 3 - f(2) = 3 - 2^{-10}. Then steps within [2, 3): from 3 - 2^{-10} down to some point < 2. Then steps within [1, 2), then [0, 1), then negative. So a(3) = 1 + N_{[2,3)} + N_{[1,2)} + N_{[0,1)} + 1·... hmm: let me define: the chain x₀ = 3, x₁ = 3 - 2^{-10}, ..., until x_j < 0. a(3) = j. The relation f(3) = f(x_j)/2^j = (-x_j)/2^j — hmm, but -x_j is not necessarily a nice number... BUT: wait: in the f(2) case, the chain ended at -1 exactly, giving f(2) = 1/2^{10}. For f(3): the chain ends at some negative point -w: f(3) = w/2^{a(3)}. For f(3) to be a power of 2, w should be a power of 2 (or the product is). Hmm. In the f(2) chain: ended at -1: w = 1 = 2^0. For f(1): chain: 1 → 1/2 → 0 → -1: ends at -1: f(1) = 1/8 ✓. f(0): 0 → -1: f(0) = 1/2 ✓. Interesting: chains from integers end at -1?! For f(3): the chain will end at some negative point: is it -1? Hmm, maybe not necessarily; but if the structure is self-similar, maybe. Actually, from the framework: the inner dynamics exit to z ∈ [1/2, 1), then the [0,1) chain: from z: z → z - f(z - 1) = z - (1 - z)·... for z ∈ [1/2, 1): z - 1 ∈ [-1/2, 0): f(z-1) = 1 - z. So z → 2z - 1 ∈ [0, 1). Hmm. Chain in [0, 1): x → x - f(x - 1) = x - (1 - x) = 2x - 1 (for x ∈ [0, 1/2): x - 1 < -1/2: wait: x ∈ [0, 1): x - 1 ∈ [-1, 0): f(x-1) = 1 - x. So x → 2x - 1. For x ∈ [0, 1/2): 2x - 1 < 0: bottom out: f(x) = (1 - 2x)/2 ✓ matches f(x) = (1-2x)/2 on [0, 1/2). For x ∈ [1/2, 1): 2x - 1 ∈ [0, 1): continue. So in [0, 1): the chain: x → 2x - 1 → 2(2x-1) - 1 = 4x - 3 → ... : x_j = 2^j x - (2^j - 1): bottom out when 2^j x - (2^j - 1) < 0 ↔ x < 1 - 2^{-j}. So for dyadic x = 1 - 2^{-r}·... hmm: x ∈ [1/2, 1): x = 1 - 2^{-r}·m... For x = 1 - 2^{-r} (r ≥ 1): chain: x → 1 - 2^{-r+1} → ... → 1 - 2^{-1} = 1/2 → 2·(1/2) - 1 = 0 → -1. Steps from x = 1 - 2^{-r} to -1: r + 1 steps (r steps to reach 1/2: 1 - 2^{-r} → 1 - 2^{-r+1} → ... → 1 - 2^{-1}: that's r - 1 steps? Hmm: from 1 - 2^{-r}: after 1 step: 1 - 2^{-(r-1)}: after r - 1 steps: 1 - 2^{-1} = 1/2: after r steps: 0: after r + 1 steps: -1. So r + 1 steps to -1.) In the f(2) chain: entered [0,1) at 1/2 (from 1 → 1/2): then 1/2 → 0 → -1: 2 steps. Total from 1: 1 → 1/2 [1] → 0 [2] → -1 [3]: 3 steps ✓ a(1) = 3. OK so now the key question: the total step count a(3) = 1 [3 → 3 - 2^{-10}] + N₃ [steps from 3 - 2^{-10} within [2, 3) — i.e., until chain drops below 2] + N₂ [steps within [1, 2)] + N₁ [steps within [0, 1)] + 1 [final step to negative]. Hmm wait, the final step: from x ∈ [0, 1/2): x → 2x - 1 < 0: that step is counted in N₁? Let me define N₁ = steps from entering [0,1) until reaching negative. And the value at the end: f(-w) = w. But CAREFUL: f(3) = f(x_end)/2^{total} = (-x_end)/2^{total}. For f(3) to be 2^{-a}: need -x_end = 2^{total - a}·... i.e., the end point matters! In the f(2) chain: end at -1: f(2) = 1/2^{10}: total = 10 = a(2) with f = 2^{-10} ✓ (end value 1 = 2^0). For f(3): end point: the chain in [0,1): enters at some z ∈ [1/2, 1) (from the [1,2) exit), then z → 2z - 1 → ... until negative. If z = 1 - 2^{-r}·(odd)·... hmm: z = 5/2 - t at exit: z = 1 - w where w = bits: w dyadic: z = 1 - 2^{-c}·odd. The chain in [0,1): x → 2x - 1: x_end = 2^j z - (2^j - 1) ∈ [-1, 0): x_end = 2^j(z - 1) + ... hmm: x_end = 2^j z - 2^j + 1 = 2^j(z - 1) + 1 = 1 - 2^j w·... with w = 1 - z: x_end = 1 - 2^j w. Bottom out: smallest j with 2^j w > 1. If w = 2^{-c} (single bit): j = c + 1: x_end = 1 - 2 = -1 ✓. So if the exit z has w = 2^{-c} (single bit!), the chain ends at -1 and f(3) = 1/2^{total} = 2^{-a(3)}. In our chain so far, exits were at z = 1/2 (w = 1/2 = 2^{-1}: single bit ✓) for the m ≤ 10 cases, and z = 1 - w with w = multiple bits for m ≥ 11 cases: e.g., z = 7/8 (w = 1/8: single bit ✓), z = 13/16 (w = 3/16 = 2^{-3} + 2^{-4}: TWO bits!), z = 3/4 (w = 1/4 ✓ single), z = 5/8 (w = 3/8: two bits: 2^{-2} + 2^{-3}). Hmm so for multi-bit w, the end point is not -1: x_end = 1 - 2^j w: e.g., w = 3/16: j = 5 (2^5·3/16 = 6 > 1; 2^4·3/16 = 3 < 1... wait: bottom out when 2^j w > 1: j: 2^j·3/16 > 1 ↔ 2^j > 16/3: j = 3: 8·3/16 = 1.5 > 1 ✓: x_end = 1 - 1.5 = -0.5. So f contribution: 0.5 = 2^{-1}: f(3) = 2^{-1}/2^{total}: still a power of 2! Since w is a sum of distinct powers of 2 (dyadic), 2^j w = integer·... hmm: 2^j w with w = Σ 2^{-p}: 2^j w = Σ 2^{j - p}: integer if j ≥ max p: and the bottom-out j: smallest j with 2^j w > 1: hmm, 2^j w might not be integer at bottom-out j. E.g., w = 2^{-3} + 2^{-5}: 2^j w > 1: j = 4: 2 + 1/2 = 2.5: x_end = 1 - 2.5 = -1.5: f(x_end) = 1.5: NOT a power of 2!! Then f(3) = 1.5/2^{total} = 3/2^{total+1}: not a power of 2. Hmm! But the problem asks to compute f(3) — answer might be a non-power-of-2! But f(0), f(1), f(2) are powers of 2... Hmm, but maybe f(3) is too. Let me reconsider: maybe w ends up being a single bit. Hmm, actually wait. Let me reconsider the whole approach. The number of steps is potentially huge (thousands+), and the bit-carry dynamics is complex. Let me look for a smarter way — perhaps the exact self-similar structure. Let me reconsider. Actually, let me look at this from the perspective of known results. I'm fairly confident this is a known problem: "f(x) = -x if x < 0; f(x) = f(x - f(x-1))/2 if x ≥ 0; find f(3)". Hmm... I have a vague memory that the answer is 2^{-36}... or maybe f(3) = 2^{-something big}. Hmm. Let me think about the exponent sequence differently: a(0) = 1, a(1) = 3, a(2) = 10. Hmm: 1, 3, 10: a(n+1) = a(n) + something: 2, 7: hmm: 2 = 2^1, 7 = 2^3 - 1: next: 2^7 - 1?? a(3) = 10 + 127 = 137?? Hmm: pattern: a(n+1) - a(n) = 2^{a(n)-1} - 1: check: a(1) - a(0) = 2 = 2^{1-1} - 1 = 0?? No. Differences: a(1) - a(0) = 2; a(2) - a(1) = 7. Hmm 2 = 2^1, 7 = 2^3 - 1 = 2^{a(1)} - 1. And 2 = 2^{a(0)} = 2^1 = 2 ✓!! So maybe a(n+1) - a(n) = 2^{a(n)} - 1: check n=0: 2^1 - 1 = 1 ≠ 2. ✗. Try a(n+1) - a(n) = 2^{a(n)-1}: n=0: 2^0 = 1 ≠ 2. ✗. Hmm: 2, 7: maybe 2 = 3 - 1, 7 = 3^2 - 2·... meh. 1, 3, 10: these are partial sums of 1, 2, 7: and 1, 2, 7: hmm: 1, 2, 7: b(n+1) = ... 2 = 2·1, 7 = 3·2 + 1, next = 4·7 + 2 = 30?? then a(3) = 40? Or 1, 2, 7, 20: b: differences of b: 1, 5: hmm. 1, 2, 7: b(n) = (3^{n-1} + 1)/2·...: n=1: (1+1)/2 = 1 ✓; n=2: (3+1)/2 = 2 ✓; n=3: (9+1)/2 = 5 ✗. Hmm: 1, 2, 7, ?: OEIS A003462-ish? 1, 2, 7, 20, 61: b(n) = 3b(n-1) + ... 2 = 3·1 - 1; 7 = 3·2 + 1; 20 = 3·7 - 1; 61 = 3·20 + 1: alternating: hmm: 1, 2, 7, 20, 61, 182: (3^n + 1)/2: 1, 2, 5, 14, 41 ✗. (3^n - 1)/2: 1, 4, 13 ✗. 1, 2, 7: hmm, maybe: 1, 2, 7, 23, 76·... b(n+1) = 3b(n) + f(n)? Alternatively 1, 3, 10, 33, 109·...: a(n+1) = 3a(n) + n·: 3·1 + 0 = 3 ✓; 3·3 + 1 = 10 ✓; 3·10 + 2 = 32; hmm or 3a(n) + a(n-1)·... Alternatively 1, 3, 10, 35: a(n+1) = 3a(n) + T(n)? Hmm, alternatively — maybe I should look at the structure: a(n) = number of steps from n to negative. The chain from n: n → n - f(n-1) → ... The steps within [k, k+1) for the chain from n+1 relate to the chain from n somehow (self-similarity). If f on [k, k+1) is a scaled copy of f on [0, 1) or [1, 2)... Let me look at the chains: - From 1: 1 → 1/2 → 0 → -1: 3 steps. Steps in [0, 1): from 1/2: 1/2 → 0 → -1: 2 steps. Plus 1 step in [1, 2). - From 2: 2 → 15/8 → 29/16 → 7/4 → 13/8 → 3/2 → 5/4 → 1 → 1/2 → 0 → -1: 10 steps. Steps in [1, 2): from 15/8 to 1: 15/8 → 29/16 → 7/4 → 13/8 → 3/2 → 5/4 → 1: 7 steps? Let me count: 15/8 → 29/16 [1] → 7/4 [2] → 13/8 [3] → 3/2 [4] → 5/4 [5] → 1 [6]: 6 steps. Then [0,1): 1/2 → 0 → -1: hmm wait: 1 → 1/2 is a step: 1 ∈ [1, 2) → 1/2 ∈ [0, 1): so steps: in [1,2): from 15/8 to... the step 1 → 1/2 starts in [1,2). Let me just count total: 10 ✓. Structure: from 2: 1 step in [2, 3)·... no: 2 → 15/8: 2 is the start; the step goes from 2 to 15/8: I'll count it as the [2,3) → [1,2) transition. Then 6 steps within [1, 2) (15/8 → ... → 1 → hmm, the step 1 → 1/2: from 1: lands in [0,1)). So steps "within [1,2)" including the exit step: 15/8 → 29/16 → 7/4 → 13/8 → 3/2 → 5/4 → 1 → 1/2: 7 steps. Then [0, 1): 1/2 → 0 → -1: 2 steps. Total: 1 + 7 + 2 = 10 ✓. From 1: 1 → 1/2 (1 step, [1,2) → [0,1)), then 1/2 → 0 → -1: 2 steps: total 3 ✓. Interesting: so a(2) = 10 = 1 + 7 + 2 and a(1) = 1 + 2. The "[1,2) portion" from 2: 7 steps; from 1: 0 steps within [1,2) (just the exit step... hmm, 1 → 1/2 is the exit step: so "within [1,2)" = 0 + exit). Hmm. Alternatively: think of g(x) = number of steps from x to negative. g satisfies: g(x) = 1 + g(x - f(x-1)) for x ≥ 0, g(x) = 0 for x < 0. Then a(n) = g(n). And f(x) = 2^{-g(x)}·(-x_end)·... hmm, f(x) = f(x_end)/2^{g(x)} where x_end < 0: f(x_end) = -x_end. If x_end = -2^{-c}: f(x) = 2^{-g(x)-c}. Hmm OK. Let me now think about whether there's an exact self-similarity that gives g nicely. Observation: the map on [1, 2): x → φ(x) = x - f(x - 1) with f(x-1) = (2 - x) - 2^{-(k+1)}: φ(x) = 2x - 2 + 2^{-(k+1)}: affine with slope 2 on each level-piece. The number of steps from x ∈ [1, 2) to exit [1, 2)·... to reach [0, 1): g₁(x) = T(x) as computed (inner steps), and exit point z(x). Similarly the map on [2, 3): x → x - f(x - 1): f(x-1) = f(2 - s): complicated. Hmm, let me think about the outer chain differently: the outer chain: s_{i+1} = s_i + f(2 - s_i). We found f(2 - s) for the chain points = 2^{-e} with the bit-dynamics. The chain continues until s exceeds 1 (x = 3 - s < 2). Given increments ~2^{-e} with e ~ 80-90, and s needs to grow from ~2^{-10} to 1... that would take ~2^{80} steps?!?! That can't be right — f(3) would be unimaginably tiny. Hmm wait, but no: as s grows, the increments f(2 - s) grow too. The increments are ~2^{-e} where e depends on the bit structure of s. Hmm. Wait, wait. Let me reconsider. Hmm, actually I wonder if I've made an error: let me re-examine the claim f(2 - s) ≈ s²/8 vs the chain values 2^{-34} etc. Let me directly compute f(2 - s₁) for s₁ = 2^{-10} + 2^{-23} by a DIFFERENT method to double-check: s₁ = 2^{-10} + 2^{-23}: x = 2 - s₁ = 2 - 2^{-10} - 2^{-23}. x - 1 = 1 - 2^{-10} - 2^{-23}. Level in [0, 1): 1 - 2^{-10} - 2^{-23} ∈ [1 - 2^{-10}, 1 - 2^{-9})? 1 - 2^{-10} - 2^{-23} < 1 - 2^{-10}: so it's in [1 - 2^{-9}, 1 - 2^{-10}) = [511/512·... hmm: [1 - 2^{-9}, 1 - 2^{-10}): K = 9. f(x - 1) = (1 - (x-1)) - 2^{-10} = (2^{-10} + 2^{-23}) - 2^{-10} = 2^{-23}. φ(x) = x - f(x-1) = 2 - 2^{-10} - 2^{-23} - 2^{-23} = 2 - 2^{-10} - 2^{-24}. Next: x' = 2 - 2^{-10} - 2^{-24}: x' - 1 = 1 - 2^{-10} - 2^{-24}: level 9 again: f(x' - 1) = 2^{-10} + 2^{-24} - 2^{-10} = 2^{-24}. φ(x') = 2 - 2^{-10} - 2^{-24} - 2^{-24} = 2 - 2^{-10} - 2^{-23}. Hmm!! x'' = 2 - 2^{-10} - 2^{-23} = x: we're in a CYCLE: x → 2 - 2^{-10} - 2^{-24} → 2 - 2^{-10} - 2^{-23} → ... Wait, that contradicts the framework! Let me recheck. x = 2 - s₁, s₁ = 2^{-10} + 2^{-23}. f(x - 1) = f(1 - s₁) = f(2^{-...}): x - 1 = 1 - 2^{-10} - 2^{-23}. Hmm: 1 - 2^{-10} - 2^{-23}: is this in [1 - 2^{-9}, 1 - 2^{-10})? 1 - 2^{-9} = 1 - 2^{-10} - 2^{-10}: since 2^{-10} > 2^{-23}: 1 - 2^{-10} - 2^{-23} > 1 - 2^{-9} ✓ (because 2^{-10} + 2^{-23} < 2^{-10} + 2^{-10} = 2^{-9}). And 1 - 2^{-10} - 2^{-23} < 1 - 2^{-10} ✓. So level K = 9: f(x-1) = (1 - (x - 1)) - 2^{-10} = s₁ - 2^{-10} = 2^{-23} ✓. x → x - 2^{-23} = 2 - 2^{-10} - 2^{-24}. Call x' : s' = 2^{-10} + 2^{-24}. f(x' - 1) = s' - 2^{-10} = 2^{-24}: x' → x' - 2^{-24} = 2 - 2^{-10} - 2^{-25}: s'' = 2^{-10} + 2^{-25}. Hmm wait: 2^{-24} - 2^{-24}·... x' = 2 - 2^{-10} - 2^{-24}: x' - f(x'-1) = 2 - 2^{-10} - 2^{-24} - 2^{-24} = 2 - 2^{-10} - 2^{-23}. Hold on: 2^{-24} + 2^{-24} = 2^{-23}. So x'' = 2 - 2^{-10} - 2^{-23} = x. So the chain cycles: x → x' → x → x' → ... INFINITELY! f(x) = f(x')/2, f(x') = f(x)/2: implies f(x) = f(x)/4: f(x) = 0?! Hmm!!! That's a contradiction — unless f(x) = 0. But f > 0 assumed... Hmm, wait: but this suggests f(x) = 0 for x = 2 - 2^{-10} - 2^{-23}?? But wait, is that consistent? Let me recheck the framework computation for s₁: I claimed f(2 - s₁) = 2^{-34} via the bit-dynamics. But this direct computation suggests f(2 - s₁) = 0 (infinite loop)! One of them is wrong. Let me recheck. Framework: s₁ = 2^{-10} + 2^{-23}: level 9 (s₁ ∈ (2^{-10}, 2^{-9}]): t₀ = 2^{10}·s₁ = 1 + 2^{-13}. Hmm wait: 2^{10}·(2^{-10} + 2^{-23}) = 1 + 2^{-13} ✓. Rule: t ∈ (1, 3/2]: shift: t → 2t - 1. Here's the thing: the shift rule: s' = 2s - 2^{-(k+1)}: check with s₁: s' = 2s₁ - 2^{-10} = 2^{-9} + 2^{-22} - 2^{-10} = 2^{-10} + 2^{-10} + 2^{-22} - 2^{-10}·... hmm: 2^{-9} = 2·2^{-10}: s' = 2·2^{-10} + 2^{-22} - 2^{-10} = 2^{-10} + 2^{-22}. Direct computation: x = 2 - s₁: φ(x) = 2 - s' where s' = s₁ + f(x - 1)·... hmm wait: φ(x) = x - f(x - 1) = 2 - s₁ - 2^{-23} = 2 - (s₁ + 2^{-23}): s' = s₁ + 2^{-23} = 2^{-10} + 2^{-23} + 2^{-23} = 2^{-10} + 2^{-22} ✓✓ consistent! s' = 2s₁ - 2^{-10} ✓. OK so both agree so far: s' = 2^{-10} + 2^{-22}. Level of s': s' ∈ (2^{-10}, 2^{-9}] ✓ level 9. t' = 2^{10}s' = 1 + 2^{-12}. Continue: s'' = 2s' - 2^{-10} = 2^{-9} + 2^{-21} - 2^{-10} = 2^{-10} + 2^{-21}. Hmm: but the direct computation said: x' = 2 - s': f(x' - 1) = f(1 - s'): 1 - s' = 1 - 2^{-10} - 2^{-22}: level 9: f = s' - 2^{-10} = 2^{-22}: φ(x') = 2 - s' - 2^{-22} = 2 - (2^{-10} + 2^{-22} + 2^{-22}) = 2 - 2^{-10} - 2^{-21}: s'' = 2^{-10} + 2^{-21} ✓ consistent. Continue: s''' = 2s'' - 2^{-10} = 2^{-10} + 2^{-20}. ... s^{(j)}: the exponent: -22, -21, -20, ...: each shift: s → 2s - 2^{-10}: the "bit" at 2^{-10}·(1 + 2^{-j}): hmm: s^{(j)} = 2^{-10}(1 + 2^{-(13-j)})·... check: s₁ = 2^{-10}(1 + 2^{-13}): s' = 2^{-10}(1 + 2^{-12}) ✓: s'' = 2^{-10}(1 + 2^{-11}): ...: after 12 shifts: s = 2^{-10}(1 + 2^{-1}) = 3·2^{-11} = 1.5·2^{-10} > 2^{-10}·... t after 12 shifts: t = 1 + 2^{-1} = 3/2: ✓ matches framework (t reaches 3/2 after 12 shifts). At t = 3/2: Rule A: t → 2: s → 2^{-10}·2 = 2^{-9}·... wait: s' = 2s - 2^{-10}: s = 3·2^{-11}: s' = 3·2^{-10} - 2^{-10} = 2^{-10}·2 = 2^{-9}. Hmm: s' = 2^{-9}: but 2^{-9} is the BOUNDARY: s' = 2^{-9} exactly: level: s ∈ (2^{-(k+1)}, 2^{-k}]: s' = 2^{-9} ∈ (2^{-9}, 2^{-8}]?? NO: 2^{-9} ∈ (2^{-9}, 2^{-8}] is false: 2^{-9} is not > 2^{-9}. s' = 2^{-9} ∈ (2^{-10}, 2^{-9}]: level 9! (right endpoint). Hmm: so level stays 9: t' = 2^{10}·2^{-9} = 2. So: s = 2^{-9} exactly: level 9, t = 2: Rule B: t → t - 1/2 = 3/2: s' = 2s - 2^{-10} = 2^{-8} - 2^{-10} = 3·2^{-10}: level: 3·2^{-10} ∈ (2^{-10}, 2^{-9}] ✓: level 9: t' = 2^{10}·3·2^{-10} = 3 ✓... WAIT: t' = 3 > 2?! Contradiction: t must be ≤ 2. Hmm: t' = 2^{k'+1}s' = 2^{10}·3·2^{-10} = 3?? That's wrong: level 9: t = 2^{10}s: s = 3·2^{-10}: t = 3?? But t should be ∈ (1, 2]!! 3·2^{-10} vs the level-9 interval (2^{-10}, 2^{-9}]: 3·2^{-10} = 1.5·2^{-9} > 2^{-9}!!! So 3·2^{-10} is NOT in level 9: it's in (2^{-9}, 2^{-8}]: LEVEL 8. I made an error: 3·2^{-10} ∈ (2^{-9}, 2^{-8}]: 2^{-9} = 2·2^{-10} < 3·2^{-10} ✓ and 3·2^{-10} ≤ 2^{-8} = 4·2^{-10} ✓. So level 8: t' = 2^9·3·2^{-10} = 3/2 ✓ (not 3). Phew. OK so I had the level boundary wrong: s = 2^{-9} is level 9 (since interval is (2^{-10}, 2^{-9}], i.e., s ∈ (2^{-(k+1)}, 2^{-k}] with k = 9: (2^{-10}, 2^{-9}]: YES s = 2^{-9} ∈ level 9 ✓). And s = 3·2^{-10} ∈ (2^{-9}, 2^{-8}]: level 8 ✓. Hmm OK so let me redo the framework trace for s₁ CAREFULLY: s₁ = 2^{-10} + 2^{-23} = 2^{-10}(1 + 2^{-13}): level 9 ✓ (s₁ ∈ (2^{-10}, 2^{-9}]: 2^{-10} < 2^{-10}(1 + 2^{-13}) ≤ 2^{-9} ✓). t₀ = 2^{10}s₁ = 1 + 2^{-13} ✓. Step (Rule A, t ∈ (1, 3/2)): t → 2t - 1, k stays 9: t = 1 + 2^{-12}, s = 2^{-10}(1 + 2^{-12}). ... 12 steps: t = 1 + 2^{-1} = 3/2: s = 2^{-10}·(3/2) = 3·2^{-11}. Step 13: t = 3/2: Rule A (t ∈ (1, 3/2] inclusive): t → 2: s = 2^{-10}·2 = 2^{-9}. (k = 9) Step 14: t = 2: Rule B (t ∈ (3/2, 2]): t → 3/2, k → 8: s = 2^{-9}·... hmm: with k = 8: t = 2^{9}s: s = t·2^{-9} = (3/2)·2^{-9} = 3·2^{-10} ✓. Step 15: t = 3/2: Rule A: t → 2: s = 2·2^{-9} = 2^{-8}. (k = 8) Step 16: t = 2: Rule B: t → 3/2, k → 7: s = 3·2^{-9}·... s = (3/2)·2^{-8} = 3·2^{-9}. Step 17: t = 3/2 → 2: s = 2^{-7}·... s = 2·2^{-8} = 2^{-7}. (k = 7) Step 18: Rule B: k → 6: s = 3·2^{-8}. ... Hmm wait, this is different from what I had before! Before, I had the cycle {1} ↔ {0} in bit-positions which corresponded to t = 3/2 ↔ t = 2, and I said k decrements every 2 steps ✓ (steps 14, 16, 18: k: 9→8→7→6 ✓). And s doubles every 2 steps (s: 2^{-9} → 2^{-8} → 2^{-7}·...). Hmm wait: s after step 13: 2^{-9}: step 14: 3·2^{-10}: step 15: 2^{-8}: step 16: 3·2^{-9}·... wait: step 15: s = 2^{-8}, step 16: s = 3·2^{-9}, step 17: s = 2^{-7}, step 18: s = 3·2^{-8}·... Hmm, hmm: but wait: this contradicts the earlier analysis: s is GROWING rapidly now (doubling every 2 steps): s: ... 3·2^{-11} [step 12] → 2^{-9} [13] → 3·2^{-10} [14] → 2^{-8} [15] → 3·2^{-9} [16] → 2^{-7} [17] → ... So s grows: after step 13 + 2j: s = 2^{-9+j}: s reaches ~1 when j = 9: s = 2^0 = 1: step 13 + 18 = 31: s = 1: x = 1: hmm: x = 2 - s = 1. And f(1) = 1/8: so f(x) at that point = 1/8: and f(2 - s₁) = f(1)/2^{31} = 2^{-3-31} = 2^{-34} ✓✓!! Wait: but earlier I computed via the cycle {0},{1} that exit happens at k = 0 with state {0} (t = 2): exit z = 5/2 - t = 1/2. Now I'm getting: the chain reaches x = 1 (s = 1) at step 31 and continues INTO [0, 1)·... hmm, no wait: exit means the argument φ(x) < 1: the chain from x = 2 - s: φ(x) = 2 - s': when does s' > 1 (φ < 1)? s' > 1: from the growth: s = 1 at step 31: step 32: s' = 3·2^{-1} = 1.5 > 1: φ = 2 - 1.5 = 0.5 ✓: exit z = 1/2 at step 32 ✓✓ (T = 32 ✓ matches!). Hmm OK wait, but hold on: at step 31: s = 1: x = 2 - s = 1: but the level/t framework: s = 1: level: s ∈ (2^{-1}, 2^0]: level 0: t = 2s = 2. And k = 9 - 8 = 1?? Let me recount k: k decremented at steps 14, 16, 18, 20, 22, 24, 26, 28, 30: that's 9 decrements: k = 0 at step 30. Hmm: at step 30: k = 0. Step 31: t = 3/2 → 2 (Rule A, k stays 0): s = 1: hmm: with k = 0: t = 2^{k+1}s = 2s: s = 1 ↔ t = 2 ✓. Step 32: t = 2, k = 0: Rule B → EXIT ✓: z = 5/2 - t·... z = 2 - s' with s' = 2s - 2^{-(k+1)} = 2 - 1/2 = 3/2: z = 1/2 ✓. GREAT: so T = 32 ✓, exit z = 1/2, f(2 - s₁) = f(1/2)/2^{32} = 2^{-34} ✓. The framework is consistent with the direct computation. But WAIT: then the direct computation above (x → x' → x cycle) — that was for the OUTER chain step from a₁ = 3 - s₁, which requires f(a₁ - 1) = f(2 - s₁) = 2^{-34}: a₂ = 3 - s₁ - 2^{-34}: s₂ = s₁ + 2^{-34} ✓ consistent with before. But in my "different method check" I wrote: "x = 2 - s₁: x - 1 = 1 - s₁: level 9: f(x-1) = s₁ - 2^{-10} = 2^{-23}: φ(x) = 2 - s₁ - 2^{-23}": s' = s₁ + 2^{-23} = 2^{-10} + 2^{-22}: and then I said "x' = 2 - 2^{-10} - 2^{-24}" — WRONG: 2^{-22} ≠ 2^{-24}·... s' = 2^{-10} + 2^{-22}: I mistakenly wrote 2^{-24}. Let me redo: s₁ + 2^{-23} = 2^{-10} + 2^{-23} + 2^{-23} = 2^{-10} + 2^{-22} ✓. Then next: f(x' - 1) = s' - 2^{-10} = 2^{-22}: φ = 2 - s' - 2^{-22} = 2 - 2^{-10} - 2^{-21}: s'' = 2^{-10} + 2^{-21}: NOT a cycle: the exponent decreases: -22 → -21 → -20 → ... ✓ consistent with the framework (bit marching left). Phew: no contradiction: my "cycle" was based on an arithmetic slip (2^{-23} + 2^{-23} = 2^{-22}, not 2^{-24}). OK so the framework stands. f(2 - s₁) = 2^{-34} ✓. Now, back to the outer chain and the total step count a(3). This is going to be a long computation, but it's structured. Let me set it up very carefully. The outer chain: s_i (i = 0, 1, 2, ...): s₀ = 2^{-10}: s_{i+1} = s_i + f(2 - s_i): and each outer step contributes 1 halving: f(3 - s_i) = f(3 - s_{i+1})/2·... WAIT, no: hmm: the outer chain: f(3 - s) = f((3 - s) - f(2 - s))/2 = f(3 - s - f(2 - s))/2 = f(3 - (s + f(2 - s)))/2 = f(3 - s')/2 ✓. So each outer step: 1 halving, s → s + f(2 - s). ✓. The outer chain runs while x = 3 - s ≥ 2, i.e., s ≤ 1. When s > 1: x < 2: switch to [1, 2) chain (the "inner" dynamics of [1,2)): and then [0, 1), then negative. So a(3) = N_outer + a(2 - s_final + 1)·... hmm: let me define: the chain enters [1, 2) at x* = 3 - s* (first s > 1): then continues per the [1,2) dynamics: x → φ(x) = 2x - 2 + 2^{-(k+1)} (level k of s = 2 - x): T* steps to exit to z* ∈ [1/2, 1), then [0,1) chain: z* → 2z* - 1 → ... to negative. Hmm, BUT: note that the [1,2) dynamics is exactly what the "inner framework" computes: f(2 - s) = f(z)/2^T: the chain from x = 2 - s: T steps to z. So a(2 - s) = T + a(z) where a(z) = steps from z ∈ [1/2, 1) to negative: z → 2z - 1 → ...: if z = 1 - w: steps: until negative: if w = 2^{-c}: z = 1 - 2^{-c}: chain: z → 1 - 2^{-c+1} → ... → 1/2 → 0 → -1: c + 1 steps: end at -1. So the plan: 1. Compute the outer chain: s₀ = 2^{-10} → s₁ → s₂ → ... until s > 1. Count N_outer steps. [This is the hard part: could be MANY steps.] 2. At s* (first > 1): x* = 3 - s* ∈ [1, 2): compute the [1,2) chain: T* steps to z* ∈ [1/2, 1). 3. From z*: c* + 1 steps to -1 (if w* = 2^{-c*} single bit; else different end point). 4. a(3) = 1 [step 3 → 3 - 2^{-10}] + N_outer + T* + (steps from z* to negative). Hmm wait: also the first step: f(3) = f(3 - f(2))/2: 3 → 3 - 2^{-10}: that's 1 outer step already (s₀ = 2^{-10} is after 1 step). Then s₀ → s₁ → ...: each +1 step. Let me index: s values after each step: s^{(0)} = 2^{-10} (after step 1), s^{(i+1)} = s^{(i)} + f(2 - s^{(i)}) (after step i + 2)... Let me just say the outer chain: steps 1, 2, 3, ...: step 1: s: 0 → 2^{-10} (i.e., x: 3 → 3 - 2^{-10}). Steps 2..: s → s + f(2 - s). Continue while s ≤ 1. When s^{(N)} > 1: that step N landed at x = 3 - s^{(N)} < 2: total outer steps: N. Then continue in [1, 2) etc. So total a(3) = N_outer + T* + N_{[0,1)} where N_outer counts steps until x < 2 (including the step that lands in [1, 2)). Hmm, let me recount with the f(2) example: chain from 2: step 1: 2 → 15/8 (lands in [1,2)): N_outer = 1: then [1,2) chain: 15/8 → ... → 1/2: T* = 7: then [0,1): 1/2 → 0 → -1: 2 steps: total: 1 + 7 + 2 = 10 ✓. Good. So now the main task: compute the outer chain from s = 2^{-10} until s > 1, counting steps, using the bit-dynamics framework. And also the [1,2) exit and [0,1) chain. Hmm, wait: but actually, hold on: the outer chain steps: s → s + f(2 - s): the increments f(2 - s) are ~2^{-34}, 2^{-44}, ...: TINY. And s needs to grow from 2^{-10} to > 1?! With increments ~2^{-34}? That's ~2^{24}·... no wait: the increments grow as s grows: f(2 - s) ≈ s²/8 generically: so ds/dstep = s²/8: steps ≈ 8/s₀ = 8·2^{10} = 8192. Hmm: so ~8000 outer steps. But the exact count depends on the fine structure. Hmm, and the T* and [0,1) parts are small. So a(3) ≈ 8200?? Then f(3) = 2^{-a(3)}-ish. That's not a "nice" number like 2^{-36}. Hmm. Hmm wait, but that contradicts my earlier guess that f(3) = 2^{-36}·... Let me re-examine: is it really true that the outer chain takes ~8000 steps? Let me sanity check with f(2): the outer chain from 2: step: 2 → 2 - f(1) = 15/8: the "s" for [1,2): s = 2 - 15/8 = 1/8: and the [1,2) chain from s = 1/8: increments: f(2 - s)·... hmm, the [1,2) chain: x → φ(x): s → 2s - 2^{-(k+1)}: the increments in s: s' - s = s - 2^{-(k+1)}: for s = 1/8 (level 3): increment = 1/8 - 1/16 = 1/16: so s: 1/8 → 3/16 → 1/4 → ... wait: s' = 2s - 2^{-(k+1)}: 1/8 → 2/8 - 1/16 = 3/16 → 6/16 - 1/16 = 5/16 → 10/16 - 1/16 = 9/16 → hmm: 9/16: level 0: s' = 2·9/16 - 1/2 = 9/8 - 1/2 = 5/8?? But s must stay ≤ 1: 5/8 ✓: → 2·5/8 - 1/2 = 5/8?? Hmm: 10/8 - 4/8 = 6/8 = 3/4: → 2·3/4 - 1/2 = 1: s = 1: x = 1: hmm: s = 1: level: s ∈ (1/2, 1]: level 0: t = 2: Rule B: exit: s' = 2 - 1/2 = 3/2: z = 1/2 ✓. So the [1,2) chain from x = 15/8: s: 1/8 → 3/16 → 5/16 → 9/16 → 3/4 → 1 → exit: 6 steps + ... hmm: earlier I counted 7 steps from 15/8 to 1/2: 15/8 → 29/16 → 7/4 → 13/8 → 3/2 → 5/4 → 1 → 1/2: 7 steps ✓: s values: 1/8, 3/16, 1/4, 3/8, 1/2, 3/4, 1, (3/2): 7 transitions ✓. OK so in the f(2) case: the [1,2) portion took 7 steps: s grew from 1/8 to 1 via doubling-ish: ~log2 growth: 7 ≈ log2(8) ✓. By analogy: the outer chain from 3: s grows from 2^{-10} to 1: but the increments are NOT s - 2^{-(k+1)} ≈ s: they're f(2 - s) ≈ s²/8: quadratic: so ~8/s steps: ~8192: yeah. Hmm, so a(3) ≈ 8192 + small. That gives f(3) ≈ 2^{-8200}: not nice. Hmm!! But wait: f(2) = 2^{-10} = 2^{-T_4}: and T_4 = 10: hmm: and if a(3) = T_8 = 36: no: 36 ≠ 8200. So the T_{2^n} guess must be wrong. Hmm. Hmm, wait: let me reconsider. Actually hold on: let me recheck the increment size claim f(2 - s) ≈ s²/8. We verified: f(2 - 2^{-m}) = 2^{-2m-3}: s = 2^{-m}: s²/8 = 2^{-2m-3} ✓ EXACT for s = 2^{-m}. And for s = 2^{-10} + tiny: f = 2^{-34}·... wait: f(2 - s₁) = 2^{-34} where s₁ = 2^{-10} + 2^{-23}: s₁²/8 ≈ 2^{-23}: but f = 2^{-34}: MUCH smaller. Hmm!! So for these chain points, f(2 - s) is much smaller than s²/8. Interesting. Hmm, so which is it: does the outer chain increment by ~s²/8 (implying ~8000 steps) or by much tinier amounts (implying way MORE steps)?? The increments we computed: 2^{-23}, 2^{-34}, 2^{-44}, 2^{-53}, 2^{-61}, 2^{-68}, 2^{-74}, 2^{-79}, 2^{-83}, 2^{-86}, 2^{-88}, 2^{-89}, 2^{-89}, 2^{-88}, ...: the increments DECREASE initially! The exponents: 23, 34, 44, 53, 61, 68, 74, 79, 83, 86, 88, 89, 89, 88, 87, ...: hmm: so increments shrink to ~2^{-89} and then what? The bit dynamics: bits march left and merge... Hmm, let me reconsider. The bits (in υ = 2^{10}(s - 2^{-10})): started at position 13 and marching left (13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 78, 79): they march left and PILE UP near position ~77-79, with carries propagating. The increments are 2^{-e} where e = (rightmost bit position stuff): as bits pile up at the left end (small positions), the "rightmost bit" p_m... hmm wait, I need to re-examine: the increment exponent relates to p_m (the RIGHTMOST = largest position = smallest bit) or p₁₀ (10th bit)... Hold on, from the formulas: for m ≤ 10 bits: T = p_m + 20 - m: p_m = LARGEST position (rightmost bit): increment Δ = 2^{-(p_m + 22 - m)}: new bit at position p_m + 12 - m: which is LARGER position (further right): q = p_m + 12 - m > p_m iff m < 12. So new bits are added further right, at ever-closer positions (the gap shrinks: 11, 10, 9, ...). When the gap would be ≤ 0 (m ≥ 12): bits pile up and carry. Then for m ≥ 11: T = p₁₀ + 9: p₁₀ = 10th smallest position: the increment Δ = f(z)/2^T: with f(z) = 2^{-4} or 2^{-3}·...: Δ ≈ 2^{-(p₁₀ + 13)} or so: new bit at position ≈ p₁₀ + 3: hmm. This is a complex combinatorial dynamic. The bits are marching left (positions decreasing over time: 13 was the initial, new bits appear at 24, 34, ...: INCREASING positions: bits added further RIGHT (smaller values). Hmm wait: positions: 13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 78, 79: increasing: bits added at ever-larger positions = ever-smaller values 2^{-p}. But then carries: when the position gap closes (at 79), adding bits at 79 causes carry: bits at 79, 78, 77, 76... merge leftward: REDUCING positions: net effect? Hmm, hmm. Let me reconsider: actually wait: I think I need to recheck the m ≥ 11 analysis: is the "exit at 10th arrival" right? Let me recheck with a concrete trace: B = {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 78} (m = 11), k₀ = 9. Trace: bits shift left; bit 13 arrives at position 1 after 12 shifts: at that point bits at {1, 12, 23, 32, 40, 47, 53, 58, 62, 65, 67}. Bit at 1 with company: Rule B: remove: k = 8: bits {12, 23, 32, 40, 47, 53, 58, 62, 65, 67}. 11 shifts: bit 12 → 1: bits {1, 12, 21, 29, 36, 42, 47, 51, 54, 56}: remove: k = 7: {12, 21, 29, 36, 42, 47, 51, 54, 56}. 11 shifts: {1, 10, 18, 25, 31, 36, 40, 43, 45}: remove: k = 6: {10, 18, 25, 31, 36, 40, 43, 45}. 10 shifts: {1, 9, 16, 22, 27, 31, 34, 36}: remove: k = 5: {9, 16, 22, 27, 31, 34, 36}. 9 shifts: {1, 8, 14, 19, 23, 26, 28}: remove: k = 4: {8, 14, 19, 23, 26, 28}. 8 shifts: {1, 7, 12, 16, 19, 21}: remove: k = 3: {7, 12, 16, 19, 21}. 7 shifts: {1, 6, 10, 13, 15}: remove: k = 2: {6, 10, 13, 15}. 6 shifts: {1, 5, 8, 10}: remove: k = 1: {5, 8, 10}. 5 shifts: {1, 4, 6}: remove: k = 0: {4, 6}. 4 shifts: {1, 3}: Rule B at k = 0: EXIT: t - 1 = 2^{-1} + 2^{-3}: t = 1.625: z = 5/2 - t = 0.875 = 7/8 ✓ (matches my formula: z = 1 - 2^{-(78-76+1)} = 1 - 2^{-3} = 7/8 ✓). Steps: 12 + 1 + 11 + 1 + 11 + 1 + 10 + 1 + 10 + 1 + 9 + 1 + 9 + 1 + 8 + 1 + 8 + 1 + 7 + 1 + 7 + 1 + 6 + 1 + 6 + 1 + 5 + 1 + 5 + 1 + 4 + 1 = let me add: shifts: 12+11+11+10+10+9+9+8+8+7+7+6+6+5+5+4 = let me sum: 12+11=23, +11=34, +10=44, +10=54, +9=63, +9=72, +8=80, +8=88, +7=95, +7=102, +6=108, +6=114, +5=119, +5=124, +4=128. Removals: 9 (bits 13,24,34,43,51,58,64,69,73: 9 removals: k: 9→8→...→0 ✓). Exit: 1. Total T = 128 + 9 + 1 = 138??? Hmm!! But my formula said T = p₁₀ + 9 = 76 + 9 = 85. CONTRADICTION! Let me recheck. Formula: T = p₁₀ + 9: p₁₀ = 10th smallest bit = 76: T = 85. But trace gives 138. Let me see where the formula went wrong: I claimed "T = shifts (p₁₀ - 1) + removals (9) + exit (1)": shifts = p₁₀ - 1 = 75. But the trace has 128 shifts! Because: after each removal, the NEXT bit is at position (p_{j+1} - p_j·...) hmm: the bits don't shift one at a time to position 1 sequentially with p₁₀ total shifts: NO: bit p₁ = 13 takes 12 shifts to reach 1: then removal: then bit p₂ = 24: it has already shifted 12 times (now at 12): needs 11 more shifts to reach 1: total shifts so far: 23. Then bit p₃ = 34: at 34 - 23 = 11: needs 10 more: total 33. Hmm: the shifts to bring bit p_j to position 1: p_j - 1 total (it shifts every step until it arrives). So bit p₁₀ = 76 arrives after 75 shifts TOTAL. But the trace shows: after 9 removals, the shifts total: 12 + 11 + 11 + 10 + 10 + 9 + 9 + 8 + 8 = let me sum: 12+11=23, +11=34, +10=44, +10=54, +9=63, +9=72, +8=80, +8=88: 88 shifts for 9 arrivals?? Bit 9 (p₉ = 73) arrives after 72 shifts: ✓ (73 - 1 = 72 ✓). Bit 10 (p₁₀ = 76) should arrive after 75 shifts: but the trace shows after 9th removal (at shift-count 72), we shift 6 more?? Hmm wait, I think I mis-traced. Let me redo the trace carefully. Bits: {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 78}. Shift count 0: positions {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 78}. After 12 shifts (total shifts 12): positions {1, 12, 22, 31, 39, 46, 52, 57, 61, 64, 66}. Hmm: 24 - 12 = 12 ✓; 34 - 12 = 22 ✓; 43 - 12 = 31 ✓; 51 - 12 = 39; 58 - 12 = 46; 64 - 12 = 52; 69 - 12 = 57; 73 - 12 = 61; 76 - 12 = 64; 78 - 12 = 66. ✓ (I made errors above.) Remove bit at 1 (k=8): {12, 22, 31, 39, 46, 52, 57, 61, 64, 66}. Shift 11 more (total 23): bit 12 → 1: positions {1, 11, 20, 28, 35, 41, 46, 50, 53, 55}. Remove (k=7): {11, 20, 28, 35, 41, 46, 50, 53, 55}. Shift 10 (total 33): bit 11 → 1: {1, 10, 18, 25, 31, 36, 40, 43, 45}. Remove (k=6): {10, 18, 25, 31, 36, 40, 43, 45}. Shift 9 (total 42): bit 10 → 1: {1, 9, 16, 22, 27, 31, 34, 36}. Remove (k=5): {9, 16, 22, 27, 31, 34, 36}. Shift 8 (total 50): {1, 8, 14, 19, 23, 26, 28}. Remove (k=4): {8, 14, 19, 23, 26, 28}. Shift 7 (total 57): {1, 7, 12, 16, 19, 21}. Remove (k=3): {7, 12, 16, 19, 21}. Shift 6 (total 63): {1, 6, 10, 13, 15}. Remove (k=2): {6, 10, 13, 15}. Shift 5 (total 68): {1, 5, 8, 10}. Remove (k=1): {5, 8, 10}. Shift 4 (total 72): {1, 4, 6}. Remove (k=0): {4, 6}. Shift 3 (total 75): {1, 3}. Rule B, k=0: EXIT: t - 1 = 1/2 + 1/8: t = 5/8 + 1 = 1.625: z = 5/2 - 1.625 = 0.875 = 7/8 ✓. Total shifts: 75 = p₁₀ - 1 = 76 - 1 ✓✓. Removals: 9. Exit: 1. T = 75 + 9 + 1 = 85 ✓✓ formula confirmed (my earlier trace had arithmetic errors). OK good: T = p₁₀ + 9 = 85 for that state, exit z = 7/8, f(z) = f(7/8) = 1/16 = 2^{-4}: Δ = 2^{-89} ✓ (as before). All right. So now let me carefully set up the full outer chain computation. The state: the bit set P of υ = 2^{10}(s - 2^{-10}) — hmm wait, but this parametrization assumes s ∈ (2^{-10}, 2^{-9}] (level 9). As the chain progresses, s grows: when s exceeds 2^{-9}, the level changes to 8: then υ should be re-parametrized: υ = 2^9(s - 2^{-9})?? Hmm: the framework is in terms of (s, k, t): t = 2^{k+1}s: for level k: t - 1 = 2^{k+1}s - 1. Hmm, but for the chain, s changes level as it grows: s: 2^{-10} → ... → 2^{-9} → ... → 2^{-8} → ... → 1. Hmm, so I've been computing with level 9 (s ∈ (2^{-10}, 2^{-9}]). The bits: υ = t - 1 = 2^{10}s - 1: s = 2^{-10}(1 + υ): υ ∈ (0, 1]. When υ reaches 1: s = 2^{-9}: level 8 boundary... Actually s = 2^{-9} is still level 9 (right endpoint). s > 2^{-9}: level 8: then t = 2^9 s = 2(1 + υ)/2·... hmm: s = 2^{-10}(1 + υ): for s ∈ (2^{-9}, 2^{-8}]: t = 2^9·s = (1 + υ)/2: t - 1 = (υ - 1)/2. So when υ crosses 1, the new υ' = (υ - 1)/2: i.e., bits shift right by 1 (positions +1) and... υ' = (υ-1)/2: υ = 1 + 2υ'·...: so υ ≥ 1 means: hmm: υ = 1 + ε: υ' = ε/2: the "1" bit (position 0) is dropped and remaining bits shift right by 1. But wait: can υ exceed 1 with our increments? υ = 2^{10}(s - 2^{-10}): s ≤ 2^{-9} ↔ υ ≤ 1. The chain: s grows past 2^{-9} when υ > 1. In the m ≥ 11 regime: the dynamics: T = p₁₀ + 9 and increments add bits at position ≈ p₁₀ + 3-ish: with carries: the bits near the left (small positions) accumulate and carry leftward... hmm: carries move bits to SMALLER positions (larger values): eventually a carry could push a bit to position ≤ 0: υ ≥ 1: level change! Hmm, wait: carry to position 0 means υ ≥ 1: s > 2^{-9}: level 8. Hmm, but actually: also the k₀ changes: level 8: k₀ = 8: and the bit dynamics re-runs with k₀ = 8: bits (in new υ' = (υ - 1)/2·...) hmm. Hmm, hold on. Wait: but also: when level changes 9 → 8, do the "removals" change? k₀ = 8: only 8 removals before k = 0: exit at 9th arrival (p₉): T = p₉ + 8: and z = 1 - Σ_{j ≥ 10} 2^{-(p_j - p₉ + 1)}: f(z) = w - 2^{-(K+1)}. OK. This is quite involved, but let's carefully do it. Actually, WAIT. Let me reconsider the problem: maybe there's a much simpler pattern that I'm missing. Let me compute f on a grid and look for the pattern of the outer chain more cleverly. Hmm. Alternatively: think of g(x) = -log2 f(x) (for the chain points where f is a power of 2 times the end value). We have the functional equation: f(x) = f(x - f(x-1))/2: so g(x) = g(x - f(x-1)) + 1 (when f(x - f(x-1)) is a power of 2). Hmm, let me instead consider the sequence of integer values: a(0) = 1, a(1) = 3, a(2) = 10. Let me look at the structure of a(2) = 10 = 1 + 7 + 2: hmm: 1 = steps in [2,3): 7 = steps in [1,2): 2 = steps in [0,1). Hmm: a(1) = 3 = 1 + 0 + 2: 1 step [1,2)→[0,1)·... hmm: 1 = steps in [1, 2) (the exit step 1 → 1/2): 2 = steps in [0,1). a(0) = 1 = 1 step [0,1) (0 → -1). Let me define: h₁(s) = number of steps for the [1,2) chain starting at x = 2 - s to exit to [0,1) (i.e., T(s)), and the exit point z(s). Then: a(1) = 1 + h₁(1) + h₀(z(1)): x = 1: s = 1: h₁(1): s = 1: level 0: t = 2: Rule B: exit: T = 1: z = 1/2: hmm: wait: x = 1: the chain: 1 → 1 - f(0) = 1/2: 1 step: lands in [0,1). So h₁(1) = 1 (the framework: t = 2, k₀ = 0: Rule B: exit: T = 1 ✓: z = 1/2 ✓). Then h₀(1/2): 1/2 → 0 → -1: 2 steps. a(1) = 1 + 2 = 3 ✓. a(2) = 1 + h₁(1/8) + h₀(z): h₁(1/8) = 7 (computed: T = 7 for s = 1/8: check: s = 1/8: level 3: t = 16·(1/8) = 2: hmm: T for s = 2^{-m}: 2m + 1: m = 3: T = 7 ✓). z = 1/2: h₀ = 2: a(2) = 1 + 7 + 2 = 10 ✓. a(3) = N_outer + h₁(s*) + h₀(z*): where N_outer = outer steps (from 3, until landing in [1, 2)): s* = 3 - x* where x* = first chain point < 2: s* ∈ (1, 1 + f(2 - s**)]·... hmm: s* = s_last + f(2 - s_last) where s_last ≤ 1 < s*: s* ∈ (1, ~1.13]. So I need: N_outer and s*, then h₁(s*), z*, h₀(z*). Alternatively — hmm, wait: actually, maybe I can think of the outer chain as analogous to the [1,2) chain: is there a self-similarity: the outer chain s → s + f(2 - s) vs the [1,2) chain s → 2s - 2^{-(k+1)}? Hmm, the [1,2) chain: x → x - f(x - 1) where x - 1 ∈ [0, 1): f(x-1) = (1 - (x-1)) - 2^{-(K+1)} = s - 2^{-(K+1)}: so s' = s + (s - 2^{-(K+1)}) = 2s - 2^{-(K+1)}. The outer chain: x → x - f(x - 1) where x - 1 ∈ [1, 2): f(x - 1) = f(2 - s). So the outer chain increment: f(2 - s) = the [1,2)-f-value. If f(2 - s) were ≈ c·s²: the outer chain: s' = s + c s²: the continuous approximation: ds/dn = c s²: n ≈ 1/(c s₀): with c = 1/8: s₀ = 2^{-10}: n ≈ 8·1024 = 8192. But the ACTUAL increments at the chain points are much smaller than s²/8 (as seen: 2^{-34} vs s²/8 ≈ 2^{-23} for s ≈ 2^{-10}): so the chain takes MORE than 8192 steps?? Hmm, but as the bits evolve, the increments might grow. Ugh. Hmm, wait: actually, let me reconsider. Let me recompute f(2 - s) for a few GENERIC small s to understand the typical size. Take s = 3·2^{-4} = 3/16 (x = 13/8): f = 2^{-6}: s²/8 = (9/256)/8 = 9/2048 ≈ 2^{-7.8}: f = 2^{-6} vs 2^{-7.8}: same ballpark. s = 2^{-10}: f = 2^{-23} = s²/8 ✓ exactly. s = 2^{-10} + 2^{-23}: f = 2^{-34} << s²/8 ≈ 2^{-23}. So the chain points are special (their bit structure makes f much smaller). Interesting: the chain from 3 visits points where f(2 - s) is anomalously small! Hmm, this is like a "Cantor-like" structure. OK let me just carefully continue the bit-dynamics computation. Let me re-define everything cleanly. SETUP: The outer chain: s^{(0)} = 2^{-10}. s^{(i+1)} = s^{(i)} + f(2 - s^{(i)}). The chain continues while s ≤ 1 (x ≥ 2); N_outer = number of steps until s > 1 (the step count includes the step that crosses). Hmm wait, actually: also need care: the step that lands in [1,2): x* = 3 - s* ∈ [1, 2): then h₁(s*') where s*' = 2 - x* = s* - 1 ∈ (0, 1]: yes: s*' = s* - 1. The increments: f(2 - s) for the chain points: computed via the (k, t) framework. But NOTE: as s grows through levels, the framework's k₀ = level(s) changes: 9, 8, 7, ..., 0. And the "bit set" P is the bit set of υ = t - 1 = 2^{k₀+1}s - 1. Hmm, wait, no: hold on: the framework: s ∈ (2^{-(k+1)}, 2^{-k}]: t = 2^{k+1}s ∈ (1, 2]: υ := t - 1 = 2^{k+1}s - 1 ∈ (0, 1]. The dynamics: shift: s' = 2s - 2^{-(k+1)}: t' = 2t - 1 (k same) — valid when t ≤ 3/2 i.e. υ ≤ 1/2; Rule B: t' = t - 1/2, k' = k - 1 — valid when t ∈ (3/2, 2] i.e. υ ∈ (1/2, 1]; then re-level: new t' = 2^{k'+1}s' etc. In terms of υ (with the CURRENT level's scaling): shift: υ → 2υ (if 2υ ≤ 1/2·... hmm: t = 1 + υ: shift: t' = 2t - 1 = 1 + 2υ: υ' = 2υ: valid if t' ∈ (1, 2] with same k: υ' = 2υ ≤ 1: i.e., the new point still level k: t' ∈ (1, 2] ✓: but the RULE was: shift applies when t ∈ (1, 3/2] i.e. υ ≤ 1/2: so υ ≤ 1/2: υ' = 2υ ≤ 1 ✓: and new level: t' ∈ (1, 2]: level k ✓ (t' ≤ 2 ↔ υ' ≤ 1 ✓). OK: so shift: υ → 2υ (for υ ≤ 1/2). Rule B (υ > 1/2): t' = t - 1/2: s' = t'·2^{-k}·... hmm: t' = 1 + (υ - 1/2): with new level k - 1: t'' = 2^{k}·s' where s' = 2^{-(k+1)}t': hmm: s' = 2^{-(k+1)}t' = 2^{-(k+1)}(1/2 + υ): new level k' = k - 1: new t = 2^{k'+1}s' = 2^{k}(2^{-(k+1)}(1/2 + υ)) = (1/2 + υ)/2 = 1/4 + υ/2: new υ = t - 1 = υ/2 - 3/4?? That's negative for υ < 3/2·... υ ≤ 1: υ/2 - 3/4 < 0: CONTRADICTION! Hmm, I think I'm confusing myself. Let me recompute: Rule B: t ∈ (3/2, 2]: t' = t - 1/2 ∈ (1, 3/2]. k' = k - 1. s' = 2^{-(k'+1)}t' = 2^{-k}t'. Check with s: s = 2^{-(k+1)}t: s' should equal 2s - 2^{-(k+1)}: 2s - 2^{-(k+1)} = 2^{-(k+1)}(2t - 2^{-(k+1)}·2^{k+1}) = 2^{-(k+1)}(2t - 2): hmm: that's 2^{-(k+1)}·2(t-1) = 2^{-k}(t - 1). And 2^{-k}t' = 2^{-k}(t - 1/2). CONTRADICTION: 2^{-k}(t-1) ≠ 2^{-k}(t - 1/2)!! Let me recheck: s' = 2s - 2^{-(k+1)}: with s = 2^{-(k+1)}t: s' = 2^{-k}t - 2^{-(k+1)} = 2^{-(k+1)}(2t - 1). Hmm: 2^{-k}t = 2^{-(k+1)}·2t: so s' = 2^{-(k+1)}(2t - 1) ✓ (t' = 2t - 1 in SAME-level scaling). Now the NEW level k' of s': if t ∈ (1, 3/2]: s' = 2^{-(k+1)}(2t-1) with 2t - 1 ∈ (1, 2]: so s' ∈ (2^{-(k+1)}, 2^{-k}]: level k ✓: new t = 2t - 1 ✓. If t ∈ (3/2, 2]: 2t - 1 ∈ (2, 3]: s' ∈ (2^{-k}, 3·2^{-(k+1)}]: level: s' ∈ (2^{-k}, 2^{-(k-1)}] (since 3·2^{-(k+1)} ≤ 2^{-k+1} = 4·2^{-(k+1)} ✓): level k - 1: new t = 2^{k'}+1·... new t = 2^{(k-1)+1}s' = 2^k·2^{-(k+1)}(2t - 1) = (2t-1)/2 = t - 1/2 ✓✓. OK good: so Rule B: new t = t - 1/2 (in level k-1 scaling) ✓. I had it right: t' = t - 1/2, k' = k - 1. So in υ terms: shift (t ∈ (1, 3/2], i.e., υ ∈ (0, 1/2]): υ → 2υ (level same). Rule B (υ ∈ (1/2, 1]): new level k - 1, new t = t - 1/2 = 1/2 + υ: new υ = t - 1 = υ - 1/2 ∈ (0, 1/2]. ✓ (I previously wrote garbage; correct: new υ = υ - 1/2.) So the dynamics in υ: - υ ≤ 1/2: υ → 2υ (k same). [1 step] - υ > 1/2: υ → υ - 1/2 (k → k-1). [1 step] - Exit: k = 0 and υ > 1/2 (Rule B with k = 0): exit: z = 5/2 - t = 1/2 - υ·... t = 1 + υ: z = 5/2 - 1 - υ = 3/2 - υ. Check: υ = 1: z = 1/2 ✓; υ = 1/2+: z → 1 ✓ (z ∈ [1/2, 1)). OH NICE: this is much cleaner: the υ-dynamics: υ doubles while ≤ 1/2; when > 1/2, subtract 1/2 and decrement k; exit when k = 0 and υ > 1/2: z = 3/2 - υ. In bit terms: υ = 0.b₁b₂b₃...: doubling = left shift; "υ > 1/2" = b₁ = 1: subtract 1/2 = remove the leading 1: so: shift out bits: each 0-bit: 1 step; each 1-bit: 1 step (removal) + k--; exit at first 1-bit encountered when k = 0: z = 3/2 - υ_current. And when a 1-bit is removed, the remaining bits are unchanged (just the leading 1 dropped): so the process: read υ's binary digits from the left: 0s cost 1 step each; 1s cost 1 step each and decrement k; when k = 0 and we read a 1: exit with z = 3/2 - (current υ) where current υ = 0.(remaining bits starting with 1)·... hmm wait: at exit, υ = 0.1b...: z = 3/2 - υ. Hold on, but this is EXACTLY the "bit-dynamics" I had before (shifts and removals), but now cleaner: the "state" is just the bit string of υ and k. Let me re-verify with s = 2^{-10}: level 10: hmm: s = 2^{-10} ∈ (2^{-11}, 2^{-10}]: level 10: t = 2^{11}·2^{-10} = 2: υ = 1. Bits of υ = 1: "1" (i.e., 0.111...? or 1.0?): υ = 1 = 0.1111... in our representation? Hmm: υ ∈ (0, 1]: υ = 1: bits: all 1s (0.111...). Process: first bit 1: k = 10 > 0: remove: k = 9: υ = 0.111...: bit 1: remove: k = 8... wait: but that's not right: υ = 1: remove leading 1: υ = 0.111... = 1 again?! Infinite loop? Hmm: υ = 1: 0.111...₂ = 1: remove leading 1 (subtract 1/2): 0.0111... = 1/2: then k = 9: υ = 1/2: ≤ 1/2: double: 1: hmm: 0.0111... doubled = 0.111... = 1: then > 1/2: remove: 1/2: k = 8: ... So the cycle: υ = 1 → (remove, k--) → 1/2 → (double) → 1 → (remove, k--) → 1/2 → ...: 2 steps per k--: from k = 10: exit when k = 0 and υ = 1 > 1/2: z = 3/2 - 1 = 1/2 ✓. Steps: 2k₀ + 1 = 21 ✓ (T = 21 for s = 2^{-10} ✓✓). So the υ-dynamics: bits of υ (with the convention: dyadic υ use the terminating representation? or non-terminating?): υ = 1: as "0.111..." (non-terminating) — the process reads: 1 (remove, k--): then υ = 0.0111... hmm wait: subtracting 1/2 from 0.111... gives 0.0111...: reading bits: the leading bit is now 0: double: 0.111...: leading 1: remove: so per k--: 2 steps ✓. Equivalently: think of υ's bits as a queue: the process: while true: if υ ≤ 1/2: υ *= 2 (step). else: if k > 0: υ -= 1/2 (step, k--). else: exit (step): z = 3/2 - υ. In bit terms: υ = 0.b₁b₂b₃...: - b₁ = 0: shift left (step): now reading b₂. - b₁ = 1, k > 0: remove leading 1 (step, k--): now υ = 0.b₂b₃... (shift left AND the new leading bit is b₂). - b₁ = 1, k = 0: exit (step): z = 3/2 - υ. So: reading bits left to right: 0s: 1 step each. 1s (before k exhausted): 1 step each, k--. When k = 0: keep reading 0s (1 step each) until a 1: exit (1 step): z determined by remaining bits: z = 3/2 - 0.1r (where r = remaining bits after that 1). Hmm wait: at exit: υ = 0.1r: z = 3/2 - υ = 3/2 - (1/2 + 0.0r) = 1 - 0.0r. So z = 1 - 0.0r where 0.1r was the remaining string at exit. I.e., if υ = 0.1r then z = 1 - 0.0r = 0.1r̄·... hmm: 1 - 0.0r: in bits: 0.0r complemented? Let me just note: z = 1 - (υ - 1/2) = 3/2 - υ ✓. Number of 1s read before exit: exactly k₀ (k decrements from k₀ to 0), then the (k₀+1)-th 1 triggers exit. Hmm: unless the string has ≤ k₀ ones: then we'd read all bits... if υ has exactly j ≤ k₀ ones: after reading all bits: υ = 0 (all bits consumed)? Then the process: υ = 0: double: 0: forever?! Hmm: υ = 0 means t = 1: not in range. Can this happen? υ = 0.000...: only if υ = 0: impossible (υ > 0). But dyadic υ: e.g., υ = 0.101 (terminating): bits: 1, 0, 1, then 0s forever: the process reads: 1 (k--), 0, 1 (k--), 0, 0, 0, ...: if k exhausted (k = 0) before the second 1: exit at second 1 ✓. If k₀ ≥ 2: after reading "101": υ = 0.000... = 0: hmm: then double: still 0: infinite loop with f = 0?! Wait: but dyadic υ with terminating representation: e.g., υ = 1/2 = 0.1000...: process: b₁ = 1: remove (k--): υ = 0: then doubling 0 forever: f = 0??? But earlier: s = 2^{-10}·... hmm: υ = 1/2 case: s = 2^{-(k+1)}·(3/2): e.g., k = 0: s = 3/4: x = 5/4: f(5/4) = 1/16 ≠ 0. Let me trace: s = 3/4: level 0: t = 2s = 3/2: υ = 1/2. Process: υ = 1/2 ≤ 1/2: double: υ = 1: step [1]. υ = 1 > 1/2: k = 0: exit: z = 3/2 - 1 = 1/2: step [2]. T = 2: f(5/4) = f(1/2)/4 = 1/16 ✓✓. AH I see: at υ = 1/2: b₁ = 0 in the NON-terminating representation (0.0111...): so the process reads 0 (double), then 1s. So the convention: dyadic υ uses NON-terminating representation (recurring 1s). Then: υ = 1/2 = 0.0111...: bits: 0, 1, 1, 1, ...: process: 0: double (step): υ = 0.111... = 1: 1: k--·...: for k = 0: exit at the first 1: z = 3/2 - 1 = 1/2: T = 2 ✓. So with non-terminating representation, every dyadic υ has infinitely many 1s: the process always exits ✓. And υ = 0.101 (dyadic) = 0.1001111...·... hmm: 0.101 = 0.100111...? Let me verify: 0.1001111... = 1/2 + 0/4 + 0/8 + 1/16 + 1/32 + ... = 1/2 + (1/16)/(1 - 1/2) = 1/2 + 1/8 = 5/8 ≠ 1/2 + 1/8 = 0.101 ✓ (0.101 = 1/2 + 1/8 = 5/8 ✓). So the non-terminating form of 0.101 is 0.100111... OK so now the outer chain, cleanly: State: s (or equivalently, level k₀ and the bit string of υ = 2^{k₀+1}s - 1). Step: compute f(2 - s) = f(z)/2^T where [T, z] from the υ-process: T = (# steps), z = exit point. Then s → s + f(2 - s). But expressing s + increment in the (k₀, υ) coordinates is awkward since the increment is a power of 2 times f(z)·... Since f(z) is itself a power of 2 (f(z) = 2^{-something}) when z is a chain-relevant dyadic·... hmm, f(z) for z ∈ [1/2, 1): f(z) = (1-z) - 2^{-(K+1)}: for z = 3/2 - υ_exit: 1 - z = υ_exit - 1/2 = 0.0r: f(z) = 0.0r - 2^{-(K+1)}: this is a difference of powers/dyadics: e.g., z = 7/8: f = 1/16 ✓ power of 2. z = 13/16: f = 3/16 - 1/8 = 1/16 ✓. z = 3/4: f = 1/4 - 1/8 = 1/8 ✓. z = 5/8: f = 3/8 - 1/4 = 1/8 ✓. Hmm interesting: in these cases f(z) = 2^{-4} or 2^{-3}: powers of 2. Because 0.0r - 2^{-(K+1)}: r starts at some position: 0.0r = 2^{-c}·(1 + ...): hmm: if r = "111...": 0.0r = 2^{-c}: f(z) = 2^{-c} - 2^{-(K+1)}: and K is such that z ∈ level K: 1 - z ∈ (2^{-(K+1)}, 2^{-K}]: 1 - z = 2^{-c}: so K + 1 = c: f(z) = 2^{-c} - 2^{-c} = 0??? Hmm: no: z = 1 - 2^{-c}: 1 - z = 2^{-c} ∈ (2^{-(K+1)}, 2^{-K}]: 2^{-c} ≤ 2^{-K} ↔ K ≤ c: and 2^{-c} > 2^{-(K+1)} ↔ c < K + 1 ↔ c ≤ K: so K = c: f(z) = 2^{-c} - 2^{-(c+1)} = 2^{-(c+1)} ✓ power of 2 (this matches: z = 7/8: c = 3: f = 2^{-4} ✓; z = 3/4: c = 2: f = 2^{-3} ✓; z = 1/2: c = 1: f = 2^{-2} ✓). If r is not all 1s: 1 - z = 0.0r with r having 0s: then 1 - z = 2^{-c}(1 + 0.something): f(z) = (1-z) - 2^{-(K+1)}: K = c (as above, since 1 - z ∈ (2^{-(c+1)}·...): hmm: 1 - z = 2^{-c}(1 + w) with w ∈ [0, 1): level K: 1 - z ∈ (2^{-(K+1)}, 2^{-K}]: 2^{-c}(1 + w) ≤ 2^{-K}: if w < 1: 2^{-c}(1+w) < 2^{-(c-1)}: so K ∈ {c - 1, c}·... hmm: (1 + w) ∈ [1, 2): 2^{-c}(1+w) ∈ [2^{-c}, 2^{-(c-1)}): so 1 - z ∈ [2^{-c}, 2^{-c+1}): K: 1 - z ∈ (2^{-(K+1)}, 2^{-K}]: if 1 - z ∈ [2^{-c}, 2^{-c+1}): K = c - 1 (since (2^{-c}, 2^{-c+1}] = level c - 1) — with boundary care: 1 - z = 2^{-c} exactly (w = 0): level: 2^{-c} ∈ (2^{-c}, 2^{-c+1}]? No: 2^{-c} is not > 2^{-c}: so level c - 1 requires 1 - z > 2^{-c}: for 1 - z = 2^{-c}: level c: hmm: (2^{-(c+1)}, 2^{-c}]: yes 2^{-c} is the right endpoint of level c: K = c ✓ (matches above). For 1 - z ∈ (2^{-c}, 2^{-c+1}): K = c - 1: f(z) = (1-z) - 2^{-c} = 2^{-c}(1 + w) - 2^{-c} = 2^{-c}w: NOT a power of 2 in general! Hmm: so if the exit has r with 0s in it (w > 0), then f(z) is not a power of 2: e.g., z such that 1 - z = 2^{-3} + 2^{-5} (r = "101"): f(z) = 2^{-3} + 2^{-5} - 2^{-3} = 2^{-5}?? hmm: K: 1 - z = 0.0101: ∈ (2^{-4}, 2^{-3}]: level K = 3: f = (1-z) - 2^{-4} = 2^{-3} + 2^{-5} - 2^{-4} = 2^{-5}·(4 + 1 - 2) = 3·2^{-5}: not a power of 2. Hmm wait: level: 1 - z = 0.0101₂ = 5/32: 5/32 ∈ (2^{-5}·...: level K: 5/32 ∈ (2^{-(K+1)}, 2^{-K}]: 5/32 ∈ (1/16, 1/8]?? 5/32 = 0.15625: (0.0625, 0.125]: no. (0.125, 0.25]: yes: K = 2: f = 5/32 - 2^{-3} = 5/32 - 4/32 = 1/32 ✓ power of 2! Hmm interesting. Let me recompute: f(z) = (1 - z) - 2^{-(K+1)}: z = 1 - 5/32: 1 - z = 5/32: K = 2 (5/32 ∈ (2^{-3}, 2^{-2}] = (4/32, 8/32] ✓): f = 5/32 - 2^{-3} = 5/32 - 4/32 = 1/32 = 2^{-5} ✓. Another: 1 - z = 0.01001 = 1/8 + 1/32 = 5/32·... wait: 0.01001 = 2^{-3} + 2^{-6} = 9/64: level: 9/64 ∈ (2^{-4}·... (1/8, 1/4] = (8/64, 16/64] ✓ K = 2: f = 9/64 - 1/8 = 9/64 - 8/64 = 1/64 = 2^{-6} ✓ power of 2! Interesting: f(z) = (1-z) - 2^{-(K+1)}: the leading term of (1 - z) is 2^{-(position of first 1)}: and 2^{-(K+1)} = the largest power of 2 strictly less than 1 - z·... hmm: K is the level: 2^{-(K+1)} < 1 - z ≤ 2^{-K}: so f(z) = (1-z) - 2^{-(K+1)}: subtracting the largest power of 2 that's < (1-z)·... hmm: 2^{-(K+1)} is the largest dyadic power strictly less than 1 - z? If 1 - z = 2^{-c}·(odd): hmm: e.g., 1 - z = 5/32: largest power of 2 less than 5/32: 4/32 = 2^{-3} ✓: f = 5/32 - 4/32 = 1/32 ✓. So f(z) = (1 - z) - 2^{-⌊log2(1-z)⌋·...}·: f(z) = (1-z) minus its leading power of 2: = the "tail" of 1 - z after removing the leading 1-bit! So f(z) = 0.[bits of 1-z after the first 1]: i.e., if 1 - z = 0.1r (r = remaining bits): f(z) = 0.0r·... no wait: 1 - z = 2^{-c} + tail: f(z) = tail: in bits: 1 - z = 0.0...01r (first 1 at position c): f(z) = 0.0...00r: the bits after position c. Interesting: so f(z) = the bits of (1 - z) after the leading 1, i.e., f(z) = frac part scaled: f(z) = 2^{c}·(1-z) - 1)·2^{-c}·... = ((2^{c}(1-z)) - 1)·2^{-c}: where 2^c(1-z) ∈ [1, 2): f(z) = (2^c(1-z) - 1)·2^{-c}. And exit: z = 3/2 - υ_exit: 1 - z = υ_exit - 1/2: υ_exit = 0.1r: 1 - z = 0.0r: first 1 of (1 - z) at position 2 iff r starts with 0·... hmm: 0.0r: r = b₂b₃...: 1 - z = 2^{-2}b₂ + 2^{-3}b₃ + ...: the first 1 in (1-z) is at position 2 if b₂ = 1: else later. So f(z) = bits of r after its first 1. And then f(2 - s) = f(z)/2^T: and the increment to s: Δ = f(2 - s) = f(z)·2^{-T}. OK so now everything is in terms of bit strings. Let me now define the outer chain in bit terms. s = 2^{-(k₀+1)}(1 + υ): the level k₀ and bit string υ. Hmm: but the increments: s + Δ: Δ = f(z)/2^T: messy in (k₀, υ) coordinates. Alternatively: track s directly in binary! s ∈ (0, 1]: binary: s = 0.b₁b₂b₃... The level k₀ = position of first 1 (b_{k₀} = 1·... s ∈ (2^{-(k+1)}, 2^{-k}] ↔ first 1 at position k+1: hmm: s = 0.0...01...: first 1 at position j: s ∈ [2^{-j}, 2^{-j+1}): level: s ∈ (2^{-(k+1)}, 2^{-k}] ↔ 2^{-(k+1)} < s ≤ 2^{-k}: first 1 at position k+1 (then s > 2^{-(k+1)} requires more 1s after, or s = 2^{-k} = 0.0...01000... has first 1 at position k+1 ✓ with s = 2^{-k} ≤ 2^{-k} ✓ level k ✓). So: k₀ = (position of first 1) - 1. And υ = 2^{k₀+1}s - 1: shift s left by k₀ + 1 positions (bring first 1 to just before the point·... hmm: 2^{k₀+1}s ∈ (1, 2]: υ = 2^{k₀+1}s - 1 = 0.b'... where b' = bits after the first 1.) So: s = 0.b₁b₂...: first 1 at position p = k₀ + 1: υ = 0.b_{p+1}b_{p+2}... (bits after the first 1). The υ-process reads υ's bits: which are s's bits from position p+1 on. The process: reads bits of s after the first 1: 0s cost 1 step; 1s cost 1 step and k--; when k = 0 and a 1 is read: exit: z = 3/2 - υ_current·... and the exit uses the remaining bits. Number of 1s available: the process reads until it has consumed k₀ ones (k: k₀ → 0) plus the exit 1: total ones read: k₀ + 1·... no: the ones read with k-- : k₀ of them: then exit at the next 1: so the process reads: all bits up to and including the (k₀+1)-th 1 after position p. Wait: k-- happens per 1 read: starting k = k₀: reads 1s: after k₀ ones: k = 0: then continues reading 0s (1 step each) until the next 1: exit. So: T = (# of 0s read) + (# of 1s read) = (number of bits read) + (extra for... hmm: each bit read = 1 step: 0s: 1 step: 1s: 1 step: exit 1: 1 step: so T = number of bits read + 1?? Let me recount: reading a bit: if 0: step (double). If 1 (k > 0): step (remove: k--). If 1 (k = 0): step (exit). So every bit read costs 1 step: T = (number of bits read from υ) where the last bit read is the exit-1. Hmm: but wait: the "remove" step: does it read the next bit? The remove: υ → υ - 1/2: the leading 1 is removed: the new leading bit is the next bit: but that next bit hasn't been "read" yet — the process then examines it: if 0: double (step)·... Hmm, so in fact: T = number of bits examined, where examining the final 1 (at k=0) is the exit step. Let me re-verify with s = 2^{-10}: p = 11: υ = bits after position 11: s = 0.(10 zeros)1(then 0s): υ = 0.000... (all 0s?!) Hmm: s = 2^{-10} = 0.(10 zeros)1: first 1 at position 11: υ = bits after: all 0: υ = 0: but υ should be 1 (t = 2)!! CONTRADICTION. Ah: the issue: s = 2^{-10}: first 1 at position 11·... wait: 2^{-10} = 0.0000000001: the 1 is at position 10+1 = 11? 2^{-1} = 0.1 (position 1): 2^{-2} = 0.01 (position 2): 2^{-10} = position 10: 0.(9 zeros)1. Hmm: 2^{-10}: 1 at position 10. So p = 10: k₀ = p - 1 = 9?? But s = 2^{-10} should be level 10 (s ∈ (2^{-11}, 2^{-10}])! Hmm: level k: s ∈ (2^{-(k+1)}, 2^{-k}]: s = 2^{-10}: 2^{-10} ≤ 2^{-k} ↔ k ≤ 10: 2^{-10} > 2^{-(k+1)} ↔ k + 1 > 10 ↔ k ≥ 10: so k = 10 ✓. And first 1 at position 10: hmm: so k₀ = 10 = position of first 1: NOT position - 1. I had it backwards: s = 2^{-j}: first 1 at position j: level: s ∈ (2^{-(k+1)}, 2^{-k}]: 2^{-j} ∈ level j (right endpoint: 2^{-j} ≤ 2^{-k} ↔ k ≤ j: and 2^{-j} > 2^{-(k+1)} ↔ k ≥ j): k = j ✓. So level k₀ = position of first 1 = p. And for s with first 1 at position p and MORE 1s later (s > 2^{-p}): s ∈ (2^{-p}, 2^{-p+1}]·... s = 0.0...01...: with first 1 at p: s ≥ 2^{-p}: s ∈ [2^{-p}, 2^{-p+1}): level: s ∈ (2^{-(k+1)}, 2^{-k}]: s ∈ [2^{-p}, 2^{-p+1}): if s > 2^{-p}: level p - 1: if s = 2^{-p}: level p. Hmm: so level = p - 1 for s > 2^{-p}, and p for s = 2^{-p} exactly. Equivalent: level k₀ = p - 1 + [s = 2^{-p}]·... ugh, boundary: fine: the framework: t = 2^{k₀+1}s ∈ (1, 2]: for s = 2^{-p}: t = 2: υ = 1: bits "111...". For s > 2^{-p} (first 1 at p): t = 2^p·s·... hmm: k₀ = p - 1: t = 2^{p}s·... 2^{k₀+1} = 2^p: t = 2^p s ∈ (1, 2): υ = 2^p s - 1 = 0.(bits of s after position p). OK so: for s > 2^{-p} (strictly): υ = bits of s after position p. For s = 2^{-p} exactly: υ = 1 (equivalently bits "111..." — consistent with "bits after position p" being 000... → but we use non-terminating: s = 2^{-p} = 0.(p-1 zeros)1(then recurring 0s)·... hmm: the non-terminating representation of s = 2^{-p}: 0.(p-1 zeros)0111...: first 1 at position p+1?? That changes everything. Ugh: representation matters. Let me avoid the ambiguity: use the terminating representation for s (bits after the last 1 are 0), and handle υ = 0 (all bits after first 1 are 0) as the special case υ = 1·... hmm: no: let me think again: s = 2^{-p}: k₀ = p: t = 2^{p+1}·2^{-p} = 2: υ = 1. Whereas "bits of s after position p" = 000... = 0. The discrepancy: because for s = 2^{-p}, the level is p (not p - 1): t = 2^{p+1}s: the first 1 lands at position·... 2^{p+1}s = 2: υ = 1. In terms of bits: s = 0.(p-1 zeros)1 000...: multiplying by 2^{p+1}: 10.000...: t = 2: υ = 1 = 0.111...: AH: so the recurring-1s representation appears naturally: t - 1 = 1: represented as 0.111...: because t ∈ (1, 2]: υ = t - 1 ∈ (0, 1]: υ = 1: bits 111... So: the υ-process on υ = 1 reads: 1 (k--), then υ = 1/2 = 0.1000... hmm: but 1/2 ≤ 1/2: double: 1: read 1 (k--)·...: so it cycles: consistent with 2 steps per k-- ✓ (the trace earlier). Fine: the υ = 1 case is handled by the "0.111..." representation: reads: 1, 1, 1, ...: each 1: k--, 1 step: 1 step per k--, NOT 2!! WAIT: contradiction with earlier: earlier I traced: υ = 1: remove (1 step, k--): υ = 1/2: then υ = 1/2 ≤ 1/2: DOUBLE (1 step): υ = 1: so 2 steps per k--. But with the 0.111... representation: reading "1": remove: k--: 1 step: then the next bit is 1 again (0.111...): remove: k--: 1 step. Hmm: which is right? The issue: after removing the leading 1 from 0.111...: the value is 0.0111... = 1/2: is the new leading bit 0 or 1? 0.0111...: leading bit 0: so double: 0.111... = 1: that's 2 steps. OR: the non-terminating representation of 1/2 is 0.0111...: leading bit 0: double: yes 2 steps. Hmm: but alternatively 1/2 = 0.1000...: leading bit 1: remove: k--: 1 step: then 0: double·... The rule (from the actual dynamics): υ ≤ 1/2: double. υ = 1/2: LE = so double: 1 step: υ = 1. So the dynamics at υ = 1/2: DOUBLES (1 step). So the representation: υ = 1/2 behaves as 0.0111... (leading 0): DOUBLE. ✓: this matches "Rule A applies at t = 3/2 (inclusive)": t = 3/2 ↔ υ = 1/2: Rule A (double). ✓. So: canonical representation: dyadic values use the representation with recurring 1s (0.0111... for 1/2, 0.111... for 1, etc.): at υ = 1/2 exactly: the bit is 0 (then 111...). So: reading: the process on bits: read b: if b = 0: double (1 step). If b = 1: if k > 0: remove (1 step, k--). If k = 0: exit (1 step): z = 3/2 - υ_remaining. And the subtlety: after doubling, we re-read: the bits shift: after doubling 0.0111...: 0.111...: the new first bit is 1 ✓. So the process reads the canonical (recurring-1) bit string from left to right, one bit per step, with the exit at the first 1 read when k = 0. Let me re-verify s = 2^{-10}: level 10: υ = 1 = 0.111...: k = 10: read 1: k = 9 [1 step]: read 1: k = 8 [2]: ... read 1: k = 0 [10 steps]: read 1: k = 0: exit [11 steps]: T = 11??? But earlier: T = 21 for s = 2^{-10}!! CONTRADICTION. Hmm. Let me recheck via the actual dynamics: s = 2^{-10}: level 10: t = 2: υ = 1: t = 2 ∈ (3/2, 2]: RULE B: t → t - 1/2 = 3/2: k → 9: 1 step. t = 3/2 ∈ (1, 3/2]: RULE A: t → 2t - 1 = 2: k = 9: 1 step. t = 2: RULE B: t → 3/2: k → 8: 1 step. ... So: t = 2: Rule B: t = 3/2: Rule A: t = 2: Rule B: ...: 2 steps per k--: from k = 10: exit when k = 0 and t = 2 (Rule B): steps: 2·10 + 1 = 21 ✓. So the correct bit-reading: υ = 1: represented as 0.111...: read first bit 1: REMOVE (k--): υ = 0.0111... = 1/2: NOW: the next bit: 0 (canonical for 1/2): DOUBLE: υ = 0.111... = 1: read 1: REMOVE (k--): υ = 1/2: DOUBLE: ...: so: per k--: 2 steps (1-remove + 1-double). The reading: bits: 1, 0, 1, 0, 1, 0, ...: hmm: so the canonical representation of υ = 1 for the process: "111..." but after each removal the remainder is re-canonicalized: 1/2 → "0111..." → after double: "111...". So effectively the process on υ = 1 reads the pattern (1,0) repeatedly. Equivalently: the string "1111..." is read as: 1 (remove), then the value 1/2 is read as 0 (double), then 1 (remove), 0 (double), ...: i.e., the recurring-1s representation interacts with removals to give the (1,0) pattern. Hmm: this is because "remove leading 1" from 0.111... gives 0.0111...: and then reading THAT: first bit 0: double: 0.111...: So the reading is: 1, 0, 1, 0, 1, 0, ...: I see: so the process is: maintain the CURRENT value υ and k: step: if υ ≤ 1/2: υ = 2υ (1 step). elif k > 0: υ = υ - 1/2, k-- (1 step). else: exit: z = 3/2 - υ (1 step). The bit-reading is just a way to see it: but the canonical representation must be recomputed after each step. The cleanest: simulate the value υ exactly (it's always dyadic here) as a rational. Let me re-express: υ ∈ (0, 1], dyadic. Step: - if υ ≤ 1/2: υ ← 2υ. - elif k > 0: υ ← υ - 1/2; k ← k - 1. - else (υ > 1/2, k = 0): exit, z = 3/2 - υ. Each step: 1 halving. Now, in terms of the DYADIC value υ = m/2^j (odd m): "υ ≤ 1/2" ↔ m ≤ 2^{j-1}: doubling: υ = m/2^{j-1}. Removal: υ - 1/2 = (m - 2^{j-1})/2^j: hmm: m > 2^{j-1}: new m' = m - 2^{j-1} (odd? m odd, 2^{j-1} even (j ≥ 2 since υ > 1/2 ≥ 2^{-1}: υ = m/2^j > 1/2 → j ≥ 2·... if j = 1: υ = m/2: m odd: υ ∈ (1/2, 1]: m = 1: υ = 1/2 not > 1/2: so m ≥ 3 impossible since υ ≤ 1: so j ≥ 2 for υ > 1/2 ✓): so m' = m - 2^{j-1}: odd - even = odd ✓: new υ = m'/2^j: same denominator. So the process on (m, j, k) [υ = m/2^j, m odd, 0 < m ≤ 2^j]: - if m ≤ 2^{j-1}: m stays, j ← j - 1. [double] - elif k > 0: m ← m - 2^{j-1} [k--] (note: new m odd, and new m ≤ 2^{j-1}: since m ≤ 2^j: m - 2^{j-1} ≤ 2^{j-1} ✓: so after removal, next step doubles.) - else exit. So the pattern: removals are always followed by doubles: each 1-bit costs 2 steps (remove + double) except the exit 1 (1 step) and... wait: unless after removal m = 2^{j-1} exactly?? m' = m - 2^{j-1} = 2^{j-1}·... m' odd: 2^{j-1} odd only if j = 1: excluded: so m' < 2^{j-1} strictly: hmm: m' ≤ 2^{j-1} - 1·... wait: m ≤ 2^j - 1 (m odd): hmm: m can be 2^j? m ≤ 2^j and odd: m ≤ 2^j - 1: so m' = m - 2^{j-1} ≤ 2^{j-1} - 1 < 2^{j-1} ✓. So after removal: m' < 2^{j-1}: next step doubles. But: hmm: EXCEPT: m' could equal 2^{j-1}·... no: shown m' ≤ 2^{j-1} - 1. Hmm: but wait: what if m = 2^j (υ = 1)? m must be odd·... υ = 1: m = 2^j: NOT odd. Ugh: υ = 1 = 2^j/2^j: representation: m = 2^{j-1}/2^{j-1}: still even·... υ = 1: represent as m = 1, j = 0: υ = 1/1: j = 0: then "m ≤ 2^{j-1} = 2^{-1}": 1 ≤ 1/2? No: so removal: m ← m - 2^{-1}?? Broken. Let me handle υ = 1 specially: υ = 1 > 1/2: k > 0: υ ← 1/2, k--. [Then υ = 1/2 ≤ 1/2: double: υ = 1·...] So: general step on dyadic υ ∈ (0, 1): - if υ ≤ 1/2: υ ← 2υ. - elif υ = 1: υ ← 1/2, k--. - elif k > 0: υ ← υ - 1/2, k--. - else: exit. Hmm: but note υ = 1: υ - 1/2 = 1/2 ✓ same thing. OK so no special case needed: υ = 1: removal gives 1/2 ✓. But in (m, j) coords: υ = 1: m = 1, j = 0: υ > 1/2: removal: υ ← 1/2: (m, j) = (1, 1) ✓: m - 2^{j-1} = 1 - 1/2: broken in integer coords but fine in rationals. Let me just use rationals or handle j = 0 case: for j = 0: υ = 1 (m = 1): removal → 1/2. OK. So now: the total steps: T = 2·(#1s read) + (#0s read) + 1·... hmm: let me re-derive: reading bits of υ (canonical recurring-1s for dyadics): 0-bits: 1 step (double). 1-bits: 1 step (remove, k--) — and then the FOLLOWING step is a double (since after removal υ < 1/2... wait no: after removal, υ' = υ - 1/2 ∈ (0, 1/2]: if υ' = 1/2: next step: double (υ = 1/2 ≤ 1/2 ✓): if υ' < 1/2: next step: double ✓. So after each removal: a double follows: BUT: the double is "reading the next 0-bit"?? Hmm: υ' = 0.0r (r = remaining bits): the double: 0.r: so the double corresponds to skipping the leading 0. So: reading "1" costs 1 step (remove) + the next double (1 step) which skips a 0. So: for the bit string b₁b₂... (canonical): cost: bᵢ = 0: 1 step. bᵢ = 1: 1 step (remove) + 1 step (the forced double after) — but the forced double consumes·... hmm: does the forced double "consume" the next bit? υ' = 0.0r: double → 0.r: the new leading bit is r's first bit: so the double consumed the 0 (position 2 of 0.0r). So: a "1" in the string costs 2 steps: the remove + the double of the following bit. But if the following bit is 1 (i.e., string "11..."): remove: υ = 0.0(1...): double: υ = 0.1...: then remove: so "11" = 3 steps for two 1s? Hmm: string "11": step 1: remove first 1: υ = 0.01... wait I need to be careful. Let me think again with the actual value dynamics, it's cleaner. The process: (υ, k): υ ≤ 1/2: double. υ > 1/2, k > 0: υ -= 1/2, k--. υ > 1/2, k = 0: exit, z = 3/2 - υ. Claim: T (total steps) and exit z depend on υ's bits as follows: let υ = 0.b₁b₂b₃... (canonical recurring-1s). The process reads bits until it has seen (k₀ + 1) ones: the (k₀+1)-th one triggers exit. Steps: each ZERO bit before that point: +1. Each ONE bit before the final one: +2 (remove + subsequent double). The final one (the (k₀+1)-th): +1 (exit). Hmm wait: but after a remove, the forced double — is that always exactly 1 step? After remove: υ' ∈ (0, 1/2]: double: υ'' = 2υ' ∈ (0, 1]: if υ'' = 1: hmm: υ'' = 1 means υ' = 1/2: then continue. If υ'' ≤ 1/2: double again? NO: the process re-examines: υ'' ≤ 1/2: double: yes! The process is a loop: each iteration is ONE step (either double or remove or exit). So after a remove, the next step: υ' ≤ 1/2 → double. After that double: υ'' = 2υ': if υ'' ≤ 1/2: double again (another step). So consecutive 0s cost 1 step each ✓. So: T = Σ over bits read: 0s: 1 each; 1s (except exit): 2 each; exit 1: 1. Hmm: wait: why 2 for a 1? The remove (1 step) + the double that follows (1 step): the double corresponds to the 0 that follows the 1 in the string 0.1r → 0.0r·... hmm: after remove: υ = 0.0r: the double: 0.r: so the remove + double together consume "10"·... no: consume "1" and then the "0"?? Let me test: υ = 0.101 (canonical: 0.100111...): bits: 1, 0, 0, 1, 1, 1, ... Trace: υ = 0.100111... > 1/2: remove: υ = 0.000111... = 0.000111·... hmm: 0.100111... - 0.1 = 0.000111...: k-- [1 step]. υ = 0.000111... ≤ 1/2: double: 0.00111... [2 steps]. ≤ 1/2: double: 0.0111... [3]. ≤ 1/2: double: 0.111... [4]. > 1/2: remove: 0.0111... = 1/2: k-- [5]. ≤ 1/2: double: 0.111... = 1 [6]. > 1/2: remove: 1/2: k-- [7]. double: 1 [8]. ... So bits read: 1 [steps 1-2: remove+double], 0 [step 3·... hmm: steps 2,3,4 were doubles: the doubles consumed the 0s: bit-string: 1,0,0,1,1,1,...: reading: "1": remove (1) + the next double (2) — but the double at step 2 consumed... υ went 0.000111 → 0.00111: that consumed one 0. Then step 3: another 0. Step 4: another 0: then υ = 0.111...: hmm: wait: original υ = 0.100111...: bits: 1,0,0,1,1,1,...: after remove (bit 1 gone): 0.000111...·... hold on: 0.100111...: remove leading 1: 0.000111...: bits: 0,0,0,1,1,1: hmm: original bits after the first 1: 0,0,1,1,1,...: so after removal: 0.00 111...: I wrote 0.000111 by mistake: it's 0.00111...: (bits: 0,0,1,1,1,...). OK: trace: remove [1]: 0.00111...: double [2]: 0.0111...: double [3]: 0.111...: hmm: 0.00111 doubled = 0.0111 (consumed one 0): doubled again = 0.111 (consumed the second 0): so steps 2, 3 consumed the two 0s: then υ = 0.111...: remove [4]: 0.0111... = 1/2·... wait: 0.111... = 1: remove: 1 - 1/2 = 1/2 = 0.0111...: hmm: 1/2 canonical: 0.0111...: double [5]: 0.111... = 1: remove [6]: 1/2: double [7]: 1: ... So: bits: "1": remove [1] + the following "0"s are consumed by doubles [2, 3]: then next "1": remove [4]: then: 1/2 → double [5]: remove [6]: double [7]: remove [8]·...: So the reading: each 0: 1 step (double). Each 1 (non-final): 1 step (remove). After a remove, if the remainder is < 1/2·... the doubles consume 0s at 1 step each ✓. The special case: remove when remainder = 1/2 exactly: then double: 1: remove: double: 1: ...: 2 steps per 1. This is the "0.111..." tail (recurring 1s): each 1 in a recurring-1 tail costs 2 steps (remove + double). Hmm: so the cost of a 1-bit: if the bits after it start with 0: the remove (1 step) then doubles consume the 0s (1 step per 0). If the bits after are all 1s (recurring tail): each 1 costs 2 steps. Actually, cleanest: T = number of steps of the process = (number of doubles) + (number of removes) + 1 (exit). And: doubles = number of 0-bits consumed + (number of removes that left exactly 1/2·...) hmm. Let me think of it via the VALUE: each step either doubles υ (removing a factor 2 from denominator·... in (m,j): double: j -= 1: remove: m -= 2^{j-1})·... Alternative clean formulation: the process on (m, j): υ = m/2^j (m odd, or υ = 1 special): - while υ ≤ 1/2: double: in (m,j): j -= 1. [each = 1 step] - υ > 1/2: remove: m -= 2^{j-1}, k--. [1 step] (Now m < 2^{j-1}, i.e., υ < 1/2·... υ = m/2^j < 1/2 ✓: so next: doubles.) Hmm: so the process: alternating: doubles until υ > 1/2, then a remove. The removes: each removes the leading 1. The doubles: each shifts left. Total steps = total shifts + total removes + 1. Now: υ = 0.b₁b₂...: the leading 1 is at position c₁ (b_{c₁} = 1, b_i = 0 for i < c₁): doubles: c₁ - 1 shifts bring υ to 0.1...: then remove: 1 step. Then the remainder: 0.0b_{c₁+1}...: hmm: υ - 2^{-c₁}·... wait: υ = 2^{-c₁} + rest: remove: υ ← rest: hmm: rest = υ - 1/2?? NO: the remove happens when υ > 1/2: υ = 0.1... at that point: remove subtracts 1/2: υ ← (0.1r - 0.1) = 0.0r: so: the leading 1 must be AT position 1 when removed: the doubles bring the leading 1 to position 1: then remove: the bits after position 1 (r) remain, shifted: υ = 0.0r: then doubles bring r's leading 1 to position 1: etc. So: total: the ones in υ at positions c₁ < c₂ < ... < c_{k₀+1} (the first k₀+1 one-positions): - Leading 1 at c₁: (c₁ - 1) doubles + 1 remove: c₁ steps. - Next 1 at c₂ (position in ORIGINAL string): after the remove, the string is 0.0(bits from c₁+1 on): the leading 1 of the remainder is at original position c₂: current position: c₂ - c₁: doubles needed: (c₂ - c₁) - 1: then remove: 1: total: c₂ - c₁ steps. - ... - j-th one (at c_j): c_j - c_{j-1} steps. - The (k₀+1)-th one: at c_{k₀+1}: brought to position 1: (c_{k₀+1} - c_{k₀}) - 1 doubles + 1 exit step: c_{k₀+1} - c_{k₀} steps. Total: T = c_{k₀+1} (the position of the (k₀+1)-th one!!). WOW: clean: T = position of the (k₀+1)-th 1-bit in υ. Check: s = 2^{-10}: υ = 1 = 0.111...: ones at positions 1, 2, 3, ...: k₀ = 10: (k₀+1)-th one at position 11: T = 21?? Hmm: position of the 11th one = 11: T = 21? CONTRADICTION: T should = position = 11. Hmm!! Let me recheck: s = 2^{-10}: earlier established T = 21 = 2·10 + 1. Via ones-positions: υ = 0.111...: ones at 1,2,3,...: 11th one at position 11. But T = 21. So the formula T = c_{k₀+1} is WRONG for recurring-1s. Because: in the recurring-1 tail, each remove leaves exactly 1/2: costing remove + double = 2 steps per one, not 1. So the formula holds when the ones are "isolated" (each followed by a 0 eventually): T = c_{k₀+1} when c_{k₀+1} is a genuine position with the tail after it not all-1s·... hmm: more precisely: T = c_{k₀+1} + (number of ones among c₁..c_{k₀} that are "in a recurring-1 tail"·...). Ugh. Let me re-derive: steps: for the j-th one (j = 1..k₀+1): the steps to go from "just after (j-1)-th remove" to "just after j-th remove/exit": = (gap) + 1 where gap = c_j - c_{j-1} - 1 (doubles), EXCEPT when the (j-1)-th remove left exactly υ = 1/2·... hmm: when does a remove leave 1/2? υ before remove = 1/2 + 1/2 = 1: i.e., υ = 1: recurring-1 tail: 0.111... So: if the remainder after the (j-1)-th one is all-1s recurring (i.e., c_{j-1} onwards the string is 111...·... i.e., υ = (2^{-c_{j-1}+1}·... υ = 0.b...0111... with ones from c_{j-1} on: then after removing the one at c_{j-1}: remainder = 0.(c_{j-1}-1 zeros)111...: hmm wait I need to be careful: if the string from position c_{j-1} to infinity is all 1s: υ = (stuff) + 2^{-(c_{j-1}-1)}: then after remove: υ' = υ - 2^{-1}·... no: υ' = υ - 1/2 (the leading 1 was at position 1 after doubling·...). Hmm, let me set up cleanly with actual values: υ = Σ_{j} 2^{-c_j} (ones at positions c_j, plus possibly the recurring tail). Dyadic υ: the recurring-1s representation means: υ = A + 2^{-(d-1)} where A = Σ 2^{-c_j} over isolated ones·... ugh. Let me just say: υ dyadic: υ = m/2^j: the canonical bit string: the ones at positions·... υ = 0.b₁b₂...b_{j}111... (b_j = 0 unless m = 2^j·... hmm: if υ = m/2^j with m odd: canonical: b_i for i ≤ j: then 111... from position j+1: wait: m/2^j: bits: the last 1 among b_1..b_j is at position j (m odd: b_j = 1): then recurring 1s?? NO: m/2^j = 0.b₁...b_j 000...: canonical non-terminating: 0.b₁...b_{j-1} (b_j - 1) 111...: i.e., decrement the last 1 to 0 and append 111...: e.g., υ = 5/8 = 0.101 = 0.100111...: ✓ (b = 1,0,0 then 111...). OK here's the thing: the process dynamics are on VALUES: I'll just compute T and z directly from the value recursion, but for the outer chain I need to do this MANY times (thousands of steps?!). That's not feasible by hand without finding the pattern. Let me look at the increments found so far and the pattern of bit positions: p_j (positions of ones in υ): 13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 78, 79, then carries...: and T values: 21, 32, 42, 51, 59, 66, 72, 78·... hmm wait: earlier T values: for m ≤ 10 bits: T = p_m + 20 - m: with p_m = LARGEST position: m bits: hmm: but now I derived: T = position of the (k₀+1)-th one: k₀ = 9: T = c₁₀ (the 10th one-position)!! Check: bits {13, 24, 34, 43, 51, 58, 64, 69, 73, 76}: 10 ones: 10th one at 76: T = 76?? But earlier: T = p_m + 20 - m = 76 + 10 = 86. CONTRADICTION! Hmm!! Let me recheck: earlier trace for B_10 = {13,24,34,43,51,58,64,69,73,76} (m = 10, k₀ = 9): I claimed T = p_m + 20 - m = 86. Let me re-derive via the ones-positions: ones at 13, 24, 34, 43, 51, 58, 64, 69, 73, 76: PLUS the recurring tail? υ = Σ 2^{-p}: dyadic: canonical: the last one (76) becomes: 2^{-76} → 2^{-77} + 2^{-78} + ...: υ = 2^{-13} + ... + 2^{-75} + 2^{-77} + 2^{-78} + ... (recurring 1s from 77). So ones at: 13, 24, 34, 43, 51, 58, 64, 69, 73, 75, 77, 78, 79, 80, ... (all positions ≥ 77). k₀ = 9: the (k₀+1) = 10th one: positions sorted: 13 [1], 24 [2], 34 [3], 43 [4], 51 [5], 58 [6], 64 [7], 69 [8], 73 [9], 75 [10]·... wait: with the canonical representation, the one at 76 became ones at 77, 78, ...: so positions: 13, 24, 34, 43, 51, 58, 64, 69, 73, 77, 78, ...: hmm: the 10th one is at 77. T = 77 + (recurring-tail adjustments)? Hmm: so which is right: T = 86 or T = 77? Let me DIRECTLY trace the value dynamics for υ = 2^{-13} + 2^{-24} + 2^{-34} + 2^{-43} + 2^{-51} + 2^{-58} + 2^{-64} + 2^{-69} + 2^{-73} + 2^{-76}, k₀ = 9: υ₀ = 2^{-13}·(1 + tiny). Steps: double 12 times [12 steps]: υ = 2^{-1} + 2^{-13} + 2^{-23} + ... (each term doubled 12 times: 2^{-1} + 2^{-12}·... wait: 2^{-24}·2^{12} = 2^{-12}: υ = 2^{-1} + 2^{-12} + 2^{-22} + 2^{-31} + 2^{-39} + 2^{-46} + 2^{-52} + 2^{-57} + 2^{-61} + 2^{-64}. υ > 1/2: remove: υ = 2^{-12} + 2^{-22} + ... [13 steps], k = 8. Double 11 times [24]: υ = 2^{-1} + 2^{-11} + 2^{-20} + 2^{-28} + 2^{-35} + 2^{-41} + 2^{-46} + 2^{-50} + 2^{-53}: remove [25]: k = 7. Double 10 [35]: υ = 2^{-1} + 2^{-10} + 2^{-18} + 2^{-25} + 2^{-31} + 2^{-36} + 2^{-40} + 2^{-43}: remove [36]: k = 6. Double 9 [45]: υ = 2^{-1} + 2^{-9} + 2^{-16} + 2^{-22} + 2^{-27} + 2^{-31} + 2^{-34}: remove [46]: k = 5. Double 8 [54]: υ = 2^{-1} + 2^{-8} + 2^{-14} + 2^{-19} + 2^{-23} + 2^{-26}: remove [55]: k = 4. Double 7 [62]: υ = 2^{-1} + 2^{-7} + 2^{-12} + 2^{-16} + 2^{-19}: remove [63]: k = 3. Double 6 [69]: υ = 2^{-1} + 2^{-6} + 2^{-10} + 2^{-13}: remove [70]: k = 2. Double 5 [75]: υ = 2^{-1} + 2^{-5} + 2^{-8}: remove [76]: k = 1. Double 4 [80]: υ = 2^{-1} + 2^{-4}: remove [81]: k = 0. Double 3 [84]: υ = 2^{-1} + 2^{-1}·... hmm: 2^{-4} doubled 3 times = 2^{-1}: υ = 2^{-1} + 2^{-1} = 1: remove?? k = 0: hmm: υ = 1 > 1/2: k = 0: EXIT [85]: z = 3/2 - 1 = 1/2. T = 85?!? Hmm: but earlier bit-position analysis said 86, and this trace says 85. And the ones-position formula: 10th one at 77: T = 77?? All different! Let me recount this trace carefully. υ₀ = 2^{-13} + 2^{-24} + 2^{-34} + 2^{-43} + 2^{-51} + 2^{-58} + 2^{-64} + 2^{-69} + 2^{-73} + 2^{-76}. Step 1-12: doubles [12 steps]: υ = 2^{-1} + 2^{-12} + 2^{-22} + 2^{-31} + 2^{-39} + 2^{-46} + 2^{-52} + 2^{-57} + 2^{-61} + 2^{-64}. (13→1, 24→12, 34→22, 43→31, 51→39, 58→46, 64→52, 69→57, 73→61, 76→64 ✓.) Step 13: υ > 1/2 (2^{-1} + ...): remove: υ = 2^{-12} + 2^{-22} + 2^{-31} + 2^{-39} + 2^{-46} + 2^{-52} + 2^{-57} + 2^{-61} + 2^{-64}. k = 8. Steps 14-24: 11 doubles: υ = 2^{-1} + 2^{-11} + 2^{-20} + 2^{-28} + 2^{-35} + 2^{-41} + 2^{-46} + 2^{-50} + 2^{-53}. [24 steps] (12→1: 11 doubles ✓; 22→11; 31→20; 39→28; 46→35; 52→41; 57→46; 61→50; 64→53 ✓.) Step 25: remove: υ = 2^{-11} + 2^{-20} + 2^{-28} + 2^{-35} + 2^{-41} + 2^{-46} + 2^{-50} + 2^{-53}. k = 7. Steps 26-35: 10 doubles: υ = 2^{-1} + 2^{-10} + 2^{-18} + 2^{-25} + 2^{-31} + 2^{-36} + 2^{-40} + 2^{-43}. [35] Step 36: remove: k = 6. υ = 2^{-10} + 2^{-18} + 2^{-25} + 2^{-31} + 2^{-36} + 2^{-40} + 2^{-43}. Steps 37-45: 9 doubles: υ = 2^{-1} + 2^{-9} + 2^{-16} + 2^{-22} + 2^{-27} + 2^{-31} + 2^{-34}. [45] Step 46: remove: k = 5. υ = 2^{-9} + 2^{-16} + 2^{-22} + 2^{-27} + 2^{-31} + 2^{-34}. Steps 47-54: 8 doubles: υ = 2^{-1} + 2^{-8} + 2^{-14} + 2^{-19} + 2^{-23} + 2^{-26}. [54] Step 55: remove: k = 4. υ = 2^{-8} + 2^{-14} + 2^{-19} + 2^{-23} + 2^{-26}. Steps 56-62: 7 doubles: υ = 2^{-1} + 2^{-7} + 2^{-12} + 2^{-16} + 2^{-19}. [62] Step 63: remove: k = 3. υ = 2^{-7} + 2^{-12} + 2^{-16} + 2^{-19}. Steps 64-69: 6 doubles: υ = 2^{-1} + 2^{-6} + 2^{-10} + 2^{-13}. [69] Step 70: remove: k = 2. υ = 2^{-6} + 2^{-10} + 2^{-13}. Steps 71-75: 5 doubles: υ = 2^{-1} + 2^{-5} + 2^{-8}. [75] Step 76: remove: k = 1. υ = 2^{-5} + 2^{-8}. Steps 77-80: 4 doubles: υ = 2^{-1} + 2^{-4}. [80] Step 81: remove: k = 0. υ = 2^{-4}. Steps 82-84: 3 doubles: υ = 2^{-1}. [84] Step 85: υ = 1/2 ≤ 1/2: DOUBLE: υ = 1. [85] Step 86: υ = 1 > 1/2: k = 0: EXIT. [86] T = 86 ✓✓. (I made an error above at step 84-85: υ = 2^{-1}: that's ≤ 1/2: so DOUBLE: υ = 1: then exit.) Great: T = 86 ✓ matches the earlier formula (p_m + 20 - m = 76 + 10 = 86 ✓). So the ones-positions formula needs the recurring-tail correction: the 10th one (in canonical form) is at position 77: but the steps: the ones at 13, 24, ..., 73 (9 ones) each cost (gap + 1): and the last "one" (originally at 76, canonically spread to 77, 78, ...): the process: after the 9th remove (of the one at 73): remainder: 2^{-3}·... hmm: υ after step 76: 2^{-4}: doubles 3 times [to 2^{-1}]: double [to 1]: exit: so the last stretch: from 2^{-4} (which is the canonical tail of the one at 76): 2^{-4} = 0.0001 = 0.0000111...: the ones at 5, 6, 7, ... (canonical): the 10th one at position 5·... hmm: canonical ones: 13, 24, 34, 43, 51, 58, 64, 69, 73, 75, 77, 78, ... wait: canonical of υ: the last genuine one is at 76: canonical: 2^{-76} = 2^{-77} + 2^{-78} + ...: so ones: 13, 24, 34, 43, 51, 58, 64, 69, 73, 77, 78, 79, ...: hmm: but ALSO: is 75 a one? υ has 2^{-73} + 2^{-76}: canonical: ...01101000... hmm: the bits: positions 73: 1, 76: 1: canonical keeps 73, and 76 → recurring from 77: so ones: 13, 24, 34, 43, 51, 58, 64, 69, 73, 77, 78, 79, 80, ...: the 10th one is at 77. Steps accounting: 9 removes for ones at 13..73 [steps: (13-1) + 1 + (24-13-1) + 1 + ... = 73 - hmm: 12 doubles + 1 remove, 10 doubles + 1 remove, 10 + 1, 9 + 1, 8 + 1, 7 + 1, 6 + 1, 5 + 1·... let me see: gaps: c₁ = 13: 12 doubles + 1: then c₂ = 24: after remove, the remainder's leading one: 2^{-11} (from 2^{-24} shifted 13): 10 doubles + 1 remove: hmm: 24 - 13 = 11: doubles: 10: then remove: 11 steps: total per one: (gap) steps where gap = c_j - c_{j-1} (c₀ = 0·... c₁ - 0 = 13: 12 doubles + 1 remove = 13 ✓). So 9 ones: steps = c₉ = 73. Then the 10th one (canonical at 77): remainder after 9th remove: 2^{-3}·...: hmm: υ after removing 9 ones: the remaining value: original υ minus the 9 ones = 2^{-76}·... hmm: but shifted: after step 73: υ = 2^{-3}·(the tail): the one at 76 has been shifted 73 times: 2^{-76+73} = 2^{-3}: so remainder = 2^{-3} + (canonical recurring): 2^{-3} = 0.001 = 0.000111...: ones at 4, 5, 6, ...: hmm: canonical: 2^{-3}: bits: 0.001000...: canonical: 0.000111...: ones at 4, 5, 6, ...: so the 10th one is at position 4 (in the current shifted frame): steps: 3 doubles [to 0.1] + 1 remove·... but wait: υ = 2^{-3}: ≤ 1/2: double: 2^{-2}: double: 2^{-1}: hmm: 2^{-1} ≤ 1/2: double: 1: then > 1/2: k = 0: exit. So: 3 doubles + 1 exit = 4 steps: total T = 73 + 4·... hmm: 73 + 4 = 77?? But actual T = 86!! Argh: discrepancy: let me recount: after the 9th remove (one at 73): what is υ? Trace: step 76 was the remove with k: 1 → 0: υ = 2^{-5} + 2^{-8}. Hmm: so after 8 removes: υ = 2^{-5} + 2^{-8}: the 8th remove was at step 76·... let me recount which ones were removed: removes at steps 13, 25, 36, 46, 55, 63, 70, 76, 81: 9 removes: k: 9 → 0 ✓. The 9th remove at step 81: before it: υ = 2^{-1} + 2^{-4}: remove: υ = 2^{-4}. So after 9 removes: υ = 2^{-4}: this is the tail: the ones at original positions: the 9 ones removed correspond to original ones at 13, 24, 34, 43, 51, 58, 64, 69, 73 ✓ (9 ones): remaining: 2^{-76} shifted by 81 - hmm: 2^{-76} doubled 77 times = 2^{-4}·... hmm: 76 - 77 = -1?? Let me recompute: total doubles before step 81: steps 1-12 (12), 14-24 (11), 26-35 (10), 37-45 (9), 47-54 (8), 56-62 (7), 64-69 (6), 71-75 (5), 77-80 (4): total doubles: 12+11+10+9+8+7+6+5+4 = 72. So the one at 76 is now at 76 - 72 = 4 ✓: υ = 2^{-4} ✓. Then: doubles: 2^{-4} → 2^{-3} → 2^{-2} → 2^{-1} → 1: 4 doubles [steps 82-85]: exit [86]. Total: 81 + 4 + 1 = 86 ✓. So: the accounting: 9 ones: each costs (gap): c₁ = 13: 13 steps (12 doubles + remove): then c₂ - c₁ = 11 steps (10 doubles + remove): c₃ - c₂ = 10: ...: ones at 13, 24, 34, 43, 51, 58, 64, 69, 73: gaps: 13, 11, 10, 9, 8, 7, 6, 5, 4: sum = 73: + 9 removes?? Hmm: I'm double counting: "13 steps" for the first one INCLUDES the remove: 12 doubles + 1 remove = 13 = c₁ ✓. Second one: 10 doubles + 1 remove = 11 = c₂ - c₁ ✓. So 9 ones: Σ = c₉ = 73 steps. Then the tail 2^{-4} (from the one at 76, after 72 doubles): 4 doubles + exit = 5 steps: T = 73 + 5 = 78??? But actual T = 86!! Hmm: discrepancy of 8: let me recount the trace: removes at steps: 13, 25, 36, 46, 55, 63, 70, 76, 81: after 9th remove (step 81): steps 82-85: 4 doubles: step 86: exit. Sum: 86 ✓. But formula: c₉ = 73: hmm: step of 9th remove: 81: not 73!! Let me recount: the 9th remove at step 81: but c₉ = 73: contradiction: the removes: 1st at 13 = c₁ ✓: 2nd at 25: c₂ - c₁ + c₁ = c₂?? 13 + 11 = 24: but actual 2nd remove at step 25!! Hmm: 24 vs 25: let me recount the 2nd remove: trace: steps 14-24: 11 doubles: υ = 2^{-1} + 2^{-11} + ...: remove at step 25. Hmm: so 2nd remove at 25: but formula says 24. Let me recount the doubles needed: after 1st remove: υ = 2^{-12} + 2^{-22} + ...: to bring leading one (at 12) to position 1: 11 doubles: steps 14-24: then remove at 25 ✓. So the 2nd one costs: 11 doubles + 1 remove = 12 steps: 13 + 12 = 25 ✓. So the cost per one j: (c_j - c_{j-1} - 1) doubles + 1 remove = c_j - c_{j-1} steps?? For j = 2: c₂ - c₁ = 11: but cost = 12. Hmm: c₂ - c₁ = 24 - 13 = 11: cost 12: so cost = c_j - c_{j-1} + 1?? For j = 1: cost 13 = c₁ - 0 + ... c₁ = 13: cost 13 = c₁ + 0: hmm: c₁ - c₀ with c₀ = 0: 13: +1 = 14 ≠ 13. I see the issue: after a remove, the remainder is 0.0r: the leading 1 of the remainder is at position (c_{j+1} - c_j) in the current frame: to bring it to position 1: (c_{j+1} - c_j - 1) doubles: + 1 remove: cost: c_{j+1} - c_j. Hmm: for j = 2: the remainder after removing one at 13: the one at 24 is now at 24 - 13 = 11: doubles to bring to 1: 10: + remove: 11 = c₂ - c₁ ✓. But trace shows 11 doubles + remove = 12. Contradiction: trace: after 1st remove (step 13): υ = 2^{-12} + 2^{-22} + ...: the leading one at position 12: to position 1: 11 doubles: steps 14, 15, ..., 24: that's 11 steps: then remove at 25. Hmm: position 12 → position 1: 11 doubles ✓: so cost = 11 doubles + 1 remove = 12: but c₂ - c₁ = 11. The discrepancy: c₂ = 24: after 13 steps (12 doubles + remove), the one at 24 has been doubled 12 times: at position 12: needs 11 more doubles: total doubling for one at c_j by the time of its remove: c_j - 1: ✓ (24 - 1 = 23 = 12 + 11 ✓): removes: 2: total steps: 23 + 2 = 25 ✓. So: steps up to j-th remove: (c_j - 1) + j·... hmm: c_j - 1 doubles + j removes = c_j + j - 1. Check j = 2: 24 + 1 = 25 ✓. j = 9: 73 + 8 = 81 ✓✓ (9th remove at step 81 ✓). So: j-th remove at step c_j + j - 1. The (k₀+1)-th one: if it's a remove (k₀+1 ≤ k₀?? no: removes happen for j = 1..k₀: the (k₀+1)-th one is the EXIT: exit at step: (c_{k₀+1} - 1) + k₀ + 1 = c_{k₀+1} + k₀. Hmm: but for the exit: the process brings the (k₀+1)-th one to position 1 (c_{k₀+1} - 1 - (already-done doubles)·...). Total steps T = c_{k₀+1} + k₀ where c_{k₀+1} = position of the (k₀+1)-th one. Check: 10th one (k₀ = 9): canonical positions: ones at 13, 24, 34, 43, 51, 58, 64, 69, 73, 77, 78, ...: 10th one at 77: T = 77 + 9 = 86 ✓✓✓!! And check s = 2^{-10}: υ = 1: canonical ones at 1, 2, 3, ...: k₀ = 10: 11th one at position 11: T = 11 + 10 = 21 ✓✓!! So: T = c_{k₀+1} + k₀, where c_j = position of the j-th 1 in the canonical (recurring-1s) binary expansion of υ. And the exit: z = 3/2 - υ_exit where υ_exit = the value when exit: after bringing the (k₀+1)-th one to position 1: υ = 1/2 + (tail): z = 1 - tail·... tail = bits after the (k₀+1)-th one: z = 1 - (2^{-(stuff)}). So f(2 - s) = f(z)/2^T: with f(z) = (1 - z) - 2^{-(K+1)} = tail - 2^{-(K+1)}: where tail ∈ (0, 1/2]: hmm: if tail = 2^{-d} exactly (all 1s after): 1 - z = 2^{-d}: K = d: f(z) = 2^{-d} - 2^{-(d+1)} = 2^{-(d+1)}. If the (k₀+1)-th one is the "last" one and the canonical tail is all 1s: i.e., the (k₀+1)-th one is at position c and bits c+1, c+2, ... all 1s: then υ_exit = 2^{-1} + 2^{-2} + ... = 1: z = 1/2: f(z) = 1/4 = 2^{-2}: f(2-s) = 2^{-2-T}. So now: the outer chain: s^{(i+1)} = s^{(i)} + f(2 - s^{(i)}): with f(2 - s) = 2^{-(T + 2 + ν)} where ν accounts for the tail structure: specifically: let the (k₀+1)-th one be at position c (canonical): and the bits after: the exit z: 1 - z = tail where tail = Σ_{j > c} 2^{-(j - c + 1)} (the bits after the (k₀+1)-th one, re-based): f(z) = tail - 2^{-(K+1)}: K = level of z: determined by tail. Since our s values are sums of powers of 2 (with the increments being powers of 2 — wait, are the increments always powers of 2? f(z) = tail - 2^{-(K+1)}: tail is a dyadic sum: if tail has multiple ones: f(z) = tail - leading power: = the rest: could have multiple ones: NOT a power of 2 in general!! Hmm: e.g., exit z with 1 - z = 2^{-3} + 2^{-5}: f(z) = (2^{-3} + 2^{-5}) - 2^{-4} = 2^{-5}·(4 + 1 - 2) = 3·2^{-5}: not a power of 2. So the increments could be non-powers-of-2: the s values would then be non-dyadic-sums·... they're still dyadic rationals but with multiple bits per increment. OK: so I really need to just track s as an exact dyadic rational through this process. The number of outer steps is potentially large. Hmm. Hmm, let me reconsider: how large is N_outer really? The increments: f(2 - s) ≈ ?: the increment relates to T ≈ position of the (k₀+1)-th one of υ: for s ≈ 2^{-10}(1 + υ): υ's ones: the chain adds bits to υ: the increments in υ-coordinates: δυ = 2^{k₀+1}·Δ = 2^{10}·f(2-s): with f(2-s) = 2^{-(T+2+ν)}: δυ = 2^{8 - T - ν}: T = c_{k₀+1} + k₀: so δυ ≈ 2^{-(c + k₀ - 8)}: the new bit lands at position ≈ c + k₀ - 8 in υ: while the (k₀+1)-th one is at position c: so new bits land ≈ k₀ - 8 positions to the RIGHT of the (k₀+1)-th one. With k₀ = 9: 1 position right?? Hmm: that contradicts the earlier finding (bits marching right by ~10 positions then slowing). Hmm: earlier: increments added bits at positions p_m + 12 - m (for m ≤ 10): with the 10th one at 76: hmm. Hmm wait, I think the resolution: the (k₀+1)-th one position c vs the RIGHTMOST bit p_m: these differ. The increments: new bit at position ≈ T + 2 + ν - 10: T = c + k₀ = c + 9: new bit at ≈ c + 1 + ν·... with ν ≥ 0: so new bits land just right of the 10th one. And the 10th one position c: as bits accumulate around c, carries push bits left, c decreases... The dynamics: bits pile up just right of the 10th one, carry, the 10th one moves left, T decreases?? Hmm: T = c + 9: as c decreases: T decreases: increments GROW (Δ = 2^{-(T + stuff)}): positive feedback: the chain accelerates. This makes sense with the continuous approximation: once s is not anomalously small-valued·... the increments ≈ s²/8: hmm. OK let me just carefully simulate. Strategy: track s exactly as a dyadic rational, but using the structure: I'll track υ (bits) and level k₀, and simulate the υ-process symbolically for each increment. But the number of outer steps: let me estimate: increments Δ ≈ s²/8: from s = 2^{-10}: to s = 1: N ≈ ∫8 ds/s² = 8/s₀ = 8192. Hmm: but the early increments are much SMALLER than s²/8 (2^{-34} vs 2^{-23}): so more steps: but as the dynamics "canonize" (bits spread out), increments approach s²/8: hmm. 8192+ steps: each requiring a bit-process analysis: infeasible by hand UNLESS the structure simplifies. Let me look for the simplification. KEY INSIGHT ATTEMPT: Let me look at the υ-process differently: the chain: s → s + f(2-s): and f(2 - s) = f(z)/2^T: Let me consider the MAP in terms of the "g-value" (step count function): define G(x) = number of steps from x to bottom (negative). Then f(x) = 2^{-G(x)}·(-x_end) where x_end = the negative endpoint. Hmm. Alternatively: let me think about what the outer chain looks like in terms of s: the increments f(2 - s): for the [1,2) framework: the value f(2 - s) as a function of s: Let me tabulate f(2 - s) for s = 2^{-m}: 2^{-2m-3}: and the claim: f(2 - s) = s²/8 for s = 2^{-m}: What about s = 2^{-m}·(1 + 2^{-j}) for large j: computed: s = 2^{-10}(1 + 2^{-13}): f = 2^{-34}: vs s²/8 = 2^{-20}(1 + 2^{-13})²/8 ≈ 2^{-23}: so f is MUCH smaller: f ≈ ?: 2^{-34}: hmm: 2^{-34} = 2^{-20}·2^{-14}: ≈ s²·2^{-14}/8·... hmm: interesting: 2^{-34}: let me see: T = 32: f = 2^{-2-32} = 2^{-34}: T = c₁₀ + 9: υ = 2^{-13}: ones at 13, then canonical recurring from 14: ones: 13, 14, 15, 16, ...: 10th one at position 23: T = 23 + 9 = 32 ✓✓. So the ones-positions formula gives T = 32 ✓. So: T = c_{k₀+1} + k₀: and f(2-s) = f(z)·2^{-T}: f(z) = 2^{-2} when the tail after the (k₀+1)-th one is all 1s (z = 1/2): i.e., when the (k₀+1)-th one is the last genuine one and it's·... hmm: z = 1/2 ↔ υ_exit = 1 ↔ after bringing the (k₀+1)-th one to position 1, the rest is all 1s: i.e., the bits after the (k₀+1)-th one (in canonical form) are all 1s. For υ = 2^{-13} (canonical: 2^{-13} = 0.(12 zeros)0111...: ones at 13, 14, 15, ...): 10th one at 23: bits after: all 1s ✓: z = 1/2: f = 2^{-34} ✓. For υ = 2^{-13} + 2^{-24} (canonical: ones at 13, 14, ..., 23, 24?? hmm: 2^{-13} + 2^{-24}: bits: 1 at 13, 1 at 24: canonical: 1 at 13, 1 at 24: the recurring: 2^{-24} canonical → ones at 25, 26, ...: so ones: 13, 24, 25, 26, ...: 10th one: 13 [1], 24 [2], 25 [3], ..., 30 [10]: c₁₀ = 30: T = 39?? But earlier: T = 42 for this case!! Hmm: discrepancy: earlier B_2 = {13, 24}: T = 42. Let me recheck via trace: υ = 2^{-13} + 2^{-24}: - 12 doubles [12]: υ = 2^{-1} + 2^{-12}: remove [13]: υ = 2^{-12}: k = 8. - 11 doubles [24]: υ = 2^{-1}: ≤ 1/2?? 2^{-1} = 1/2: double [25]: υ = 1: remove?? k = 8 > 0: remove [26]: υ = 1/2: k = 7. - double [27]: υ = 1: remove [28]: k = 6. - double [29]: remove [30]: k = 5. - double [31]: remove [32]: k = 4. - double [33]: remove [34]: k = 3. - double [35]: remove [36]: k = 2. - double [37]: remove [38]: k = 1. - double [39]: remove [40]: k = 0. - double [41]: υ = 1 > 1/2, k = 0: exit [42]. T = 42 ✓. Via formula: canonical ones: 2^{-13} + 2^{-24}: hmm: I said canonical of 2^{-24} is ones at 25, 26, ...: but that's WRONG: canonical (recurring 1s) means: 2^{-24} = 2^{-25} + 2^{-26} + ...: so bits: 1 at 13, and ones at 25, 26, 27, ...: sorted: 13, 25, 26, 27, ...: 10th one: 13 [1], 25 [2], 26 [3], 27 [4], 28 [5], 29 [6], 30 [7], 31 [8], 32 [9], 33 [10]: c₁₀ = 33: T = 33 + 9 = 42 ✓✓!! (I previously mis-listed the canonical ones.) So the formula T = c_{k₀+1} + k₀ is confirmed, with canonical recurring-1s representation. So now the outer chain in canonical-bit terms: s: level k₀ (first genuine one at position k₀+1 in s's canonical form·... hmm: wait: s's own canonical representation vs υ's: υ = 2^{k₀+1}s - 1: for s with first one at position p = k₀+1 (genuine, i.e., s > 2^{-p}): υ = bits of s after position p (in TERMINATING form) — and then υ's canonical form: the terminating bits of s after p, with the last 1 converted to 0111... Hmm: if s is a "nice" dyadic like 2^{-10} + 2^{-23}·...: s = 0.(10 zeros)1(12 zeros)1: hmm: 2^{-11} + 2^{-24}: wait: s = 2^{-10} + 2^{-23}: first one at position 10·... hold on: 2^{-10}: the 1 is at position 10. Hmm: but earlier I established level k₀ for s = 2^{-10} + 2^{-23}: s ∈ (2^{-10}, 2^{-9}]: level 9: first one at position 10: k₀ = 9 = p - 1 ✓ (s > 2^{-10} strictly: yes: s = 2^{-10} + 2^{-23} > 2^{-10}). And υ = 2^{10}s - 1 = 2^{-13} ✓ (bits of s after position 10: the one at 23: shifted: 23 - 10 = 13 ✓). OK so here's the clean setup: s = 2^{-p}(1 + υ) where p = k₀ + 1 = position of s's leading one, υ = 0.(bits of s after position p, in the terminating representation). [If s = 2^{-p} exactly: level k₀ = p: special: υ = 1: handled as ones at all positions.] The increment: Δ = f(2 - s) = f(z)/2^T: T = c_{k₀+1} + k₀ where c_j = positions of ones in υ's CANONICAL form: and z: exit point: f(z) = (1-z) - 2^{-(K+1)}. And then s ← s + Δ: υ ← υ + 2^p Δ (if the level doesn't change: i.e., if 2^pΔ keeps s ≤ 2^{-p+1}·... s + Δ ≤ 2^{-(p-1)}: Δ is tiny so yes for a long time). 2^p Δ = 2^{p} f(z)/2^T: with T = c + k₀ = c + p - 1 (c = c_{k₀+1}): 2^p Δ = f(z)·2^{p - c - p + 1} = f(z)·2^{1 - c}. So: υ ← υ + f(z)·2^{1-c} where c = position of the (k₀+1)-th one of υ (canonical), and f(z) = tail-based value ∈ (0, 1/2]: f(z) = 2^{-2} if the tail after c is all 1s; else f(z) = (bits after the (k₀+1)-th one) - (leading power): hmm. Let me simplify f(z): at exit: υ_exit = 1/2 + tail_bits: where tail_bits = the canonical bits of υ after position c, re-based: tail_bits = Σ_{j > c} b_j 2^{-(j - c + 1)}: hmm: υ after bringing the (k₀+1)-th one to position 1: υ = 0.1(b_{c+1}b_{c+2}...): so tail_bits = 0.0(b_{c+1}b_{c+2}...)·... i.e., 1 - z = υ_exit - 1/2 = 0.0b_{c+1}b_{c+2}... f(z) = (1-z) - 2^{-(K+1)} = [0.0b_{c+1}b_{c+2}...] - 2^{-(K+1)}. If b_{c+1}b_{c+2}... = all 1s: 1 - z = 0.0111... = 1/2: f(z) = 1/2 - 1/4 = 1/4 ✓ (z = 1/2: K = 1: f = 2^{-2} ✓). If the bits after c are: first 1 at position d (i.e., b_{c+1}...b_{d-1} = 0, b_d = 1): 1 - z = 2^{-(d-c+1)}·(1 + more): f(z) = (1-z) - 2^{-(K+1)}: K: 1 - z ∈ (2^{-(K+1)}, 2^{-K}]: 1 - z ∈ [2^{-(d-c+1)}, 2^{-(d-c)}): so K + 1 = d - c + 1 - 1 = d - c: hmm: 1 - z ≥ 2^{-(d-c+1)}: and < 2^{-(d-c)}: level K = d - c - 1: f(z) = (1-z) - 2^{-(d-c)}: So f(z) = [0.0 b_{c+1} ... b_d ...] - 2^{-(d-c)}: the bits: 1 - z = Σ_{j ≥ d} b_j 2^{-(j-c+1)}: leading term 2^{-(d-c+1)}·... hmm: wait: 1 - z ∈ [2^{-(d-c+1)}, 2^{-(d-c)}): hmm: if b_d = 1 is the first one after c: 1 - z ≥ 2^{-(d-c+1)}: with equality iff all later bits are 0 — but canonical has recurring 1s: 1 - z > 2^{-(d-c+1)} strictly?? No: canonical: bits after d: if d is the LAST one (terminating position): canonical makes bits d+1, d+2, ... all 1: so 1 - z = 2^{-(d-c+1)} + 2^{-(d-c+2)} + ... = 2^{-(d-c)}·... hmm: Σ_{i ≥ d-c+1} 2^{-i} = 2^{-(d-c)}: so 1 - z = 2^{-(d-c)}: level: 2^{-(d-c)} ∈ (2^{-(K+1)}, 2^{-K}]: K = d - c: f(z) = 2^{-(d-c)} - 2^{-(d-c+1)} = 2^{-(d-c+1)}. Hmm: so f(z) = 2^{-(d - c + 1)} where d = position of the first one after the (k₀+1)-th one (in canonical form), assuming d exists (bits after c not all 1s). And if all 1s after c: f(z) = 2^{-2} (d "at infinity"). Let me double check with an example: υ = 2^{-13} + 2^{-24}: canonical ones: 13, 25, 26, 27, ...: k₀ = 9: 10th one: c = 33 (ones: 13 [1], 25 [2], 26 [3], ..., 33 [10] ✓). Bits after 33: all 1s (canonical recurring): f(z) = 2^{-2}: T = 33 + 9 = 42 ✓: Δ = 2^{-44} ✓✓ matches!! Another: B_11 case: υ = 2^{-13} + ... + 2^{-78} (ones at 13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 78): canonical: ones at those positions, plus recurring after 78: i.e., ones: 13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 78, 79, 80, ...: 10th one: 76 [count: 13(1), 24(2), 34(3), 43(4), 51(5), 58(6), 64(7), 69(8), 73(9), 76(10)]: c = 76: bits after 76: 78 (0 at 77, 1 at 78): first one after c=76: d = 78: f(z) = 2^{-(78-76+1)} = 2^{-3}: T = 76 + 9 = 85: Δ = 2^{-3-85} = 2^{-88}?? Hmm: earlier I computed for B_11: Δ₁₁ = 2^{-89} (with exit z = 7/8, f(z) = 2^{-4}). DISCREPANCY: 2^{-88} vs 2^{-89}. Let me recheck: exit: υ_exit: bring 10th one (at 76) to position 1: υ_exit = 0.1 + bits after 76 re-based: bits: 77: 0, 78: 1, 79: 1 (canonical recurring from 78: bits 78, 79, 80, ... all 1): so after c = 76: bits: 0, 1, 1, 1, ...: υ_exit = 0.10111... = 1/2 + 0.0111...·... hmm: 0.10111...: = 1/2 + 1/8 + 1/16 + ... = 1/2 + 1/4 = 3/4: z = 3/2 - 3/4 = 3/4?? Hmm: but earlier I computed z = 7/8 for this case!! Let me recheck: earlier trace: exit with t - 1 = 1/2 + 1/8: t = 1.625: z = 5/2 - t = 0.875 = 7/8. Now I get z = 3/4. Contradiction: let me recheck the trace: B_11 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 78}: the trace: after 9 removes: υ = 2^{-4} + 2^{-6} (the ones at 76 and 78, shifted: 76 - 72 = 4, 78 - 72 = 6: with 72 = total doubles). Then: doubles: 2^{-4} + 2^{-6} → 2^{-3} + 2^{-5} → 2^{-2} + 2^{-4} → 2^{-1} + 2^{-3}: (3 doubles: steps 82, 83, 84): υ = 2^{-1} + 2^{-3} > 1/2: k = 0: exit: z = 3/2 - (1/2 + 1/8) = 3/2 - 5/8 = 7/8 ✓. So exit: υ_exit = 1/2 + 1/8: z = 7/8: 1 - z = 1/8: f(z) = 1/8 - 2^{-4} = 1/16 = 2^{-4} ✓ (z = 7/8: K = 3: f = (1/8) - 2^{-4} = 1/16 ✓). Via my formula: c = 76 (10th one): bits after 76: 77: 0, 78: 1, then recurring: so "first one after c": d = 78: 1 - z = bits after 76 re-based: 0.01 111...·... hmm: 1 - z = 0.0(b_{77}b_{78}b_{79}...) = 0.0(0111...) = 0.001111... hmm: 0.001111... = 1/8?? 0.001111... = 2^{-3} + 2^{-4} + ... = 1/4·... Σ_{i≥3} 2^{-i} = 1/4: hmm: 2^{-3} + 2^{-4} + 2^{-5} + ... = 1/4: so 1 - z = 1/4?? But direct: 1 - z = 1/8. CONTRADICTION. Let me recompute: υ_exit = 1/2 + 1/8: z = 3/2 - υ_exit = 3/2 - 5/8 = 12/8 - 5/8 = 7/8: 1 - z = 1/8 ✓. Via bits: υ_exit = 0.10111...: hmm: 0.10111... = 1/2 + 0.001111·... wait: 0.10111...: bits after the point: 1, 0, 1, 1, 1, ...: value = 2^{-1} + 2^{-3} + 2^{-4} + 2^{-5} + ... = 1/2 + 1/8·... Σ_{i≥3} 2^{-i} = 2^{-2} = 1/4: so total = 1/2 + 1/4 = 3/4?? But υ_exit = 5/8!! Hmm: 0.10111...: positions: b₁ = 1 (2^{-1}), b₂ = 0, b₃ = 1, b₄ = 1, b₅ = 1, ...: value = 1/2 + 1/8 + 1/16 + 1/32 + ... = 1/2 + 1/8 + (1/16)(1/(1-1/2)) = 1/2 + 1/8 + 1/8 = 3/4. But the trace: υ = 2^{-1} + 2^{-3}: = 1/2 + 1/8 = 5/8: and 5/8 = 0.101 (terminating): canonical: 0.100111...: AH: so at the exit moment, υ_exit = 5/8: canonical form 0.100111...: the bits after the leading 1: 0, 0, 1, 1, 1, ...: so 1 - z = υ_exit - 1/2 = 1/8: canonical bits of (1 - z): 0.000111...: I see: my formula confusion: 1 - z = υ_exit - 1/2: υ_exit = 0.1 b'₂ b'₃ ... where b' = the canonical bits of υ_exit after the leading 1: 1 - z = 0.0b'₂b'₃...: for υ_exit = 5/8 = 0.100111...: b' = 00111...: 1 - z = 0.000111... = 1/8·... hmm: 0.000111... = 2^{-4} + 2^{-5} + ... = 1/8 ✓. OK. So: 1 - z = 0.0(b'₂b'₃...): and the bits b' = bits of υ after position c (brought to front): υ's bits after the (k₀+1)-th one: b_{c+1} b_{c+2} ...: υ_exit = 0.1 b_{c+1} b_{c+2} ... (as VALUES: these are the bits of υ after position c, with position c's bit at the 2^{-1} place). Hmm: but wait: υ_exit = 0.1 b_{c+1} b_{c+2}...: for our case: bits after 76: b₇₇ = 0, b₇₈ = 1, b₇₉ = 1, ... (canonical recurring after 78): υ_exit = 0.1 0 1 1 1 ... = 1/2 + 0.0111...·... 0.10111... = 3/4: but actual υ_exit = 5/8!!! CONTRADICTION AGAIN. Hmm: what's wrong: the bits of υ after position 76: υ = ones at 13, 24, ..., 76, and 78, with canonical recurring ones from 79: so bits: b₇₆ = 1, b₇₇ = 0, b₇₈ = 1, b₇₉ = 1, b₈₀ = 1, ...: υ_exit should be: bring the one at 76 to position 1: i.e., υ_exit = 2^{75}·(the part of υ from position 76 on) = 2^{75}(2^{-76} + 2^{-78} + Σ_{j≥79} 2^{-j}) = 2^{-1} + 2^{-3} + 2^{-4}·... hmm: Σ_{j≥79} 2^{75-j} = Σ_{i≥4} 2^{-i} = 1/8: so υ_exit = 1/2 + 1/8 + 1/8 = 3/4?! But the trace says υ_exit = 5/8: hmm: the trace: after 9 removes: υ = 2^{-4} + 2^{-6}: then 3 doubles: 2^{-1} + 2^{-3}: exit. The 3 doubles: steps 82, 83, 84: hmm: but wait: the canonical recurring ones: in the trace, υ was handled as a VALUE: 2^{-4} + 2^{-6} = 1/16 + 1/64 = 5/64: doubles: 5/32, 5/16, 5/8: υ = 5/8 = 0.101: exit: z = 3/2 - 5/8 = 7/8 ✓. The VALUE 5/64 = 2^{-4} + 2^{-6}: canonical: 0.000101 (terminating) = 0.000100111...: bits: ones at 4, 6 (terminating) → canonical: 4, then 7, 8, 9, ... (since 2^{-6} = Σ_{j≥7} 2^{-j}): so canonical bits: b₄ = 1, b₅ = 0, b₆ = 0, b₇ = 1, b₈ = 1, ... AH I SEE MY ERROR: υ = 2^{-13} + ... + 2^{-76} + 2^{-78}: the canonical form: the LAST genuine one is at 78: canonical: ones at 13, ..., 76 (genuine), and 78 → recurring from 79: WAIT: canonical means: the terminating representation has its last 1 at 78: the canonical (recurring 1s) replaces the last 1: 2^{-78} = Σ_{j≥79} 2^{-j}: so canonical ones: 13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 79, 80, 81, ... — NOT 78!! I previously wrote "78, 79, 80..." — wrong: the one AT 78 is replaced by ones at 79+. Hmm: no wait: canonical representation of the VALUE: the value has bits: 1 at 13, 24, ..., 76, 78 (terminating): to get recurring-1s form: subtract 2^{-78}: value = ...+ 2^{-78}: = ... + 2^{-79} + 2^{-80} + ...: so bits: 78: 0, 79: 1, 80: 1, ...: canonical ones: 13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 79, 80, ... So the 10th one: 76 ✓ (positions 13, 24, 34, 43, 51, 58, 64, 69, 73, 76: 10 ones ✓): c = 76: bits after: b₇₇ = 0, b₇₈ = 0, b₇₉ = 1, b₈₀ = 1, ...: υ_exit = 0.1 0 0 1 1 1 ... = 1/2 + 2^{-4} + 2^{-5} + ... = 1/2 + 1/8 = 5/8 ✓✓!! Great: matches the trace: z = 3/2 - 5/8 = 7/8 ✓. And 1 - z = 1/8: canonical bits of 1/8: 0.001 = 0.000111...: f(z) = (1-z) - 2^{-(K+1)}: 1 - z = 1/8: level: 1/8 ∈ (2^{-4}, 2^{-3}]: K = 3: f = 1/8 - 2^{-4} = 1/16 ✓. General formula: after the (k₀+1)-th one at position c: let d = position of the NEXT one (the (k₀+2)-th): then: 1 - z = 0.0...0 1 ...: hmm: 1 - z = υ_exit - 1/2 = bits after c, re-based: = Σ_{j > c} b_j 2^{-(j - c + 1)}·... the first one after c is at d: 1 - z = 2^{-(d-c+1)}·(1 + tail): hmm: with canonical tail (recurring 1s after d if d is last): 1 - z ∈ [2^{-(d-c)}, 2^{-(d-c+1)})·... ugh: 1 - z ≥ 2^{-(d-c+1)}: with equality iff bits after d are all 0: but canonical: recurring 1s: so if d is the last canonical one: bits after d all 1: 1 - z = 2^{-(d-c+1)} + 2^{-(d-c+2)} + ... = 2^{-(d-c)}: exactly. So: 1 - z = 2^{-(d-c)}·(1 - ξ) hmm: let me just say: 1 - z ∈ [2^{-(d-c+1)}, 2^{-(d-c)}]: with the value determined by the bits after d. f(z) = (1 - z) - 2^{-(K+1)}: K: 1 - z ∈ (2^{-(K+1)}, 2^{-K}]: - If 1 - z = 2^{-(d-c)} exactly (d last, recurring): 1 - z = 2^{-(d-c)}: K = d - c: f(z) = 2^{-(d-c)} - 2^{-(d-c+1)} = 2^{-(d-c+1)}. - If 1 - z ∈ (2^{-(d-c+1)}, 2^{-(d-c)}) (more ones after d): K = d - c: f(z) = (1-z) - 2^{-(d-c+1)}: = the bits of (1-z) after its leading one·... = 0.[bits of 1-z after position d-c+1]: hmm: f(z) = (1 - z) - 2^{-(d-c+1)}: and 1 - z = 2^{-(d-c+1)}·(1 + w) with w = the re-based tail after d: f(z) = 2^{-(d-c+1)}·w. OK. So now the outer chain algorithm (all in terms of υ's canonical bits): Given: level k₀ (fixed at 9 for now), υ's canonical bit-one-positions: c₁ < c₂ < c₃ < ... (infinitely many, since canonical recurring). Compute: c = c_{k₀+1} (the (k₀+1)-th one: k₀+1 = 10th). T = c + k₀. Compute f(z): based on c' = c_{k₀+2} (the next one): d = c': - If the ones after c are exactly "all positions from some point" — hmm: the canonical form has recurring 1s at the end: so after the LAST genuine one, all positions are ones. Let me define: the genuine ones: g₁ < g₂ < ... < g_m (from the terminating representation of υ), then recurring ones at positions > g_m·... hmm: NO: canonical: ones: g₁, ..., g_{m-1}, then g_m + 1, g_m + 2, ... (the last genuine one g_m is replaced). So canonical ones: g₁, ..., g_{m-1}, g_m + 1, g_m + 2, g_m + 3, ... So: c_{k₀+1}: if k₀ + 1 ≤ m - 1: c = g_{k₀+1}: genuine. Else: c = g_m + (k₀ + 1 - m + 1)·... hmm: the canonical ones after g_{m-1}: positions g_m + 1, g_m + 2, ...: the (k₀+1)-th one: if k₀ + 1 > m - 1: i.e., m ≤ k₀: c = g_m + (k₀ + 2 - m): [count: ones: m - 1 genuine + ones at g_m + 1 (that's the m-th), g_m + 2 ((m+1)-th), ..., g_m + (k₀+2-m) ((k₀+1)-th) ✓]. And then: the "next one" d: if c is genuine (c = g_{k₀+1} ≤ g_{m-1}): d = next canonical one: g_{k₀+2} (genuine) or g_m + 1·... if c is in the recurring tail: d = c + 1: and then 1 - z = 2^{-1}·... hmm: if c and d = c + 1 and all ones after: 1 - z = 0.0111... = 1/2: z = 1/2: f(z) = 1/4. Hmm wait: if c is in the recurring tail (all ones from c onward): υ_exit = 0.111... = 1: z = 1/2: f(z) = 1/4 ✓. If c genuine with next genuine one at d (no recurring interference): 1 - z = (bits between c and d are 0) 2^{-(d-c)} + recurring·... hmm: bits after c: zeros until d, then 1 at d, then: if d is the last genuine one (d = g_m): recurring ones after: bits after d: all 1s: 1 - z = 2^{-(d-c+1)} + 2^{-(d-c+2)} + ... = 2^{-(d-c)}: f(z) = 2^{-(d-c)} - 2^{-(d-c+1)} = 2^{-(d-c+1)}·... wait: K: 1 - z = 2^{-(d-c)}: level: 2^{-(d-c)} ∈ (2^{-(K+1)}, 2^{-K}]: K = d - c: f(z) = 2^{-(d-c)} - 2^{-(d-c+1)} = 2^{-(d-c+1)} ✓. So: f(z) = 2^{-(d - c + 1)} where d = the position of the next one after c (canonical), and the bits between c and d are 0 (which they are, by definition of "next one"), and I need the bits after d: hmm: the formula f(z) = 2^{-(d-c+1)} assumed all ones after d: if there are more genuine ones after d: 1 - z = 2^{-(d-c+1)}(1 + stuff): f(z) = 2^{-(d-c+1)}·stuff·... Ugh: in general: 1 - z = 0.[zeros](1 at d)(bits after d): f(z) = (1-z) - 2^{-(d-c)}: hmm: K = d - c - 1?? Let me recompute: 1 - z: first one at position (d - c + 1) [re-based: the bit at υ-position d is at 1-z-position d - c + 1]: 1 - z ∈ [2^{-(d-c+1)}, 2^{-(d-c)}): level K: (2^{-(K+1)}, 2^{-K}] ∋ 1 - z: 1 - z < 2^{-(d-c)}: so K + 1 ≥ d - c: 1 - z ≥ 2^{-(d-c+1)}: K + 1 ≤ d - c + 1: K ∈ {d - c - 1, d - c}: if 1 - z = 2^{-(d-c+1)} exactly: K = d - c: f = 2^{-(d-c+1)} - 2^{-(d-c)}·... hmm: 2^{-(K+1)} = 2^{-(d-c+1)}: f = 0?! Hmm: contradiction: let me recompute: 1 - z = 2^{-(d-c+1)}: e.g., c = 76, d = 79 (as in the B_11 example: wait: in that example: c = 76 (genuine), next one: canonical ones after 76: 79 (since the last genuine is 78: canonical replaces 78 with 79, 80, ...): d = 79: 1 - z = ?: bits after 76: 77: 0, 78: 0, 79: 1, 80: 1, ...: 1 - z = 2^{-3} + 2^{-4} + ... = 1/4?? But actual 1 - z = 1/8!! ARGH. Hmm: actual: υ_exit = 5/8: 1 - z = υ_exit - 1/2 = 1/8 ✓: bits: υ_exit = 0.100111...: bits after the leading 1: 0, 0, 1, 1, 1: so 1 - z = 0.00111... = 2^{-3} + 2^{-4} + ... = 1/4?? But 1 - z = 1/8!! 0.001111... = 1/8 + 1/16 + 1/32 + ... = 1/4. Hmm: but 1 - z = 1/8: 1/8 = 0.001 (terminating) = 0.000111... (canonical). So the bits of 1 - z: 0.000111...: NOT 0.001111...: so: 1 - z = υ_exit - 1/2: υ_exit = 5/8 = 0.100111...: minus 1/2 = 0.000111... = 1/8 ✓. So: υ_exit - 1/2: the bits: υ_exit = 0.1 b'₂ b'₃ b'₄...: υ_exit - 1/2 = 0.0 b'₂ b'₃ b'₄...: with b' = 0, 0, 1, 1, 1, ...: so 1 - z = 0.000111...: But hold on: the bits b'₂ b'₃...: these are the bits of υ_exit after its leading 1: υ_exit's bits: I claimed: υ_exit = 0.1 b_{c+1} b_{c+2}...: with b = υ's canonical bits: b₇₇ = 0, b₇₈ = 0, b₇₉ = 1, b₈₀ = 1: so υ_exit = 0.1 0 0 1 1 1... = 1/2 + 1/16 + 1/32 + ... = 1/2 + 1/8 = 5/8 ✓✓!! (I previously miscomputed: 0.100111... = 1/2 + 2^{-4} + 2^{-5} + ... = 1/2 + 1/8 = 5/8 ✓: I earlier wrote "1/2 + 2^{-3}+2^{-4}+..." which was wrong: the bits are 1,0,0,1,1,1: so 2^{-1} + 2^{-4} + 2^{-5} + ... ✓.) So: 1 - z = 0.0 b_{c+1} b_{c+2} ... = 0.000111... = 1/8 ✓✓. So the formula: 1 - z = Σ_{j > c} b_j 2^{-(j - c + 1)}: first one at d: 1 - z = 2^{-(d - c + 1)}·(1 + Σ_{j > d} b_j 2^{-(j-d)}): and f(z) = (1-z) - 2^{-(K+1)}: with 1 - z ∈ [2^{-(d-c+1)}, 2^{-(d-c)}): K = d - c - 1 if 1 - z > 2^{-(d-c+1)}·... hmm: (2^{-(K+1)}, 2^{-K}]: 1 - z ∈ [2^{-(d-c+1)}, 2^{-(d-c)}): if 1 - z ∈ [2^{-(d-c+1)}, 2^{-(d-c)}): then K = d - c - 1 (since 1 - z < 2^{-(d-c)} = 2^{-((d-c-1)+1)}·... hmm: 1 - z < 2^{-(d-c)}: and 1 - z ≥ 2^{-(d-c+1)}: K: 2^{-(K+1)} < 1 - z ≤ 2^{-K}: so 2^{-K} ≥ 1 - z ≥ 2^{-(d-c+1)}: K ≤ d - c + 1: and 2^{-(K+1)} < 2^{-(d-c)}: K + 1 > d - c: K ≥ d - c: hmm: so K = d - c?? Let me test: c = 76, d = 79: d - c = 3: K should = 3: z = 7/8: 1 - z = 1/8: 1/8 ∈ (2^{-4}, 2^{-3}]: K = 3 ✓✓. So K = d - c: f(z) = (1 - z) - 2^{-(d-c+1)}: with 1 - z = 2^{-(d-c+1)} + (bits after d): f(z) = bits after d, re-based: f(z) = Σ_{j > d} b_j 2^{-(j - c + 1)} = 2^{-(d-c+1)}·Σ_{j>d} b_j 2^{-(j-d)}. So: f(z)·2^{1-c} (the increment to υ): δυ = 2^{p}Δ = f(z)·2^{1-c}·... wait: earlier: δυ = 2^p Δ = f(z)·2^{p - T} = f(z)·2^{p - c - k₀}: with k₀ = p - 1: δυ = f(z)·2^{1 - c} ✓. So: δυ = 2^{1-c}·f(z) = 2^{1-c}·Σ_{j>d} b_j 2^{-(j-c+1)} = Σ_{j > d} b_j 2^{-j}·2^{1-c+c-1}·... let me compute: 2^{1-c}·2^{-(j-c+1)} = 2^{-j}: so δυ = Σ_{j > d} b_j 2^{-j}!!! WOW: BEAUTIFUL: δυ = the bits of υ after position d (the (k₀+2)-th one), i.e., υ's tail after d. I.e.: υ ← υ + tail(υ, d) where d = position of the (k₀+2)-th one of υ (canonical), tail(υ,d) = Σ_{j > d} b_j 2^{-j}. Hmm wait: but careful: this assumed the level doesn't change and k₀ = p - 1 etc. Also: what if the (k₀+1)-th one is in the recurring tail (all ones from c on): then d = c + 1: tail = Σ_{j > c+1} 2^{-j} = 2^{-(c+1)}: δυ = 2^{-(c+1)}: hmm: check with s = 2^{-10} (υ = 1: special: level 10): hmm: υ = 1: let me handle: υ = 1: canonical: ones at 1, 2, 3, ...: k₀ = 10: c = 11th one = position 11: d = 12: tail = Σ_{j>12} 2^{-j} = 2^{-12}: δυ = 2^{-12}: then υ ← 1 + 2^{-12}?? But υ must stay ≤ 1!! CONTRADICTION. Hmm: right: for s = 2^{-10}: level 10: the increment: Δ = 2^{-23}: δυ = 2^{11}·Δ = 2^{-12}: but υ = 1: adding 2^{-12}: υ > 1: LEVEL CHANGE: s = 2^{-11}(1 + υ): υ = 1 + 2^{-12}: s = 2^{-11}(2 + 2^{-12}) = 2^{-10} + 2^{-23} ✓: new level: s ∈ (2^{-10}, 2^{-9}]: level 9: new υ = 2^{10}s - 1 = 2^{-13}. OK: so when υ overflows 1: re-level: υ_new = (υ_old - 1)/2·... : new level k₀ - 1: υ_new = 2^{k₀}·s - 1 = 2^{k₀}·2^{-(k₀+1)}(1 + υ_old) - 1 = (1 + υ_old)/2 - 1 = (υ_old - 1)/2. ✓: υ_new = (υ_old - 1)/2: bits: drop the leading 1 (position·... υ_old = 1 + ε: υ_new = ε/2: bits shift right by 1). So the algorithm in υ-space (with level k₀ tracked, and re-leveling when υ ≥ 1·... hmm: when υ = 1 exactly: hmm: s = 2^{-(k₀+1)}(1+1) = 2^{-k₀}: level: s = 2^{-k₀}: level k₀ (right endpoint of level k₀·... wait: level k: s ∈ (2^{-(k+1)}, 2^{-k}]: s = 2^{-k₀}: level k₀ ✓ (same level!): so υ = 1 is still level k₀. Only υ > 1 forces level k₀ - 1: but υ > 1 shouldn't happen if we re-level at the right time: hmm: υ ∈ (0, 1]: υ = 1 ↔ s = 2^{-k₀}: level k₀: fine: υ > 1 impossible (s ≤ 2^{-k₀} ↔ υ ≤ 1). But the increments could push υ above 1: then re-level: υ ← (υ-1)/2, k₀ ← k₀ - 1. So the FULL algorithm: State: level k₀, canonical bit-string of υ ∈ (0, 1] (dyadic, with recurring-1s canonical form). Increment: 1. Find c = position of the (k₀+1)-th one; d = position of the (k₀+2)-th one (canonical). 2. δυ = Σ_{j > d} b_j 2^{-j} (tail after d). 3. υ ← υ + δυ (this may cause carries; recompute canonical form). If υ > 1: re-level: k₀ ← k₀ - 1, υ ← (υ - 1)/2. [Also: the increment Δ to s: Δ = δυ·2^{-(k₀+1)}·... with the CURRENT (pre-increment) k₀: Δ = 2^{-(k₀+1)}δυ.] Hmm wait: Δ = f(2 - s) = f(z)/2^T: and δυ = 2^{k₀+1}Δ: so Δ = 2^{-(k₀+1)}δυ ✓ (with k₀ = current level, pre-increment). And the step count: each outer step = 1 halving. Also T = c + k₀ (not needed for the count: each outer step is 1 halving regardless). Hold on: wait: but actually: hmm: the outer chain: f(3 - s) = f(3 - s - Δ)/2: each outer step halves ONCE. So N_outer = number of increments until s > 1. But ALSO: hmm: wait: no!! Wait: I need to double check: the outer chain: f(x) = f(x - f(x-1))/2: ONE halving per outer step ✓. And f(2 - s) computation (the increment) is just the VALUE f(2-s) — used to advance s — but the halvings for f(3 - s) come only from the outer steps, NOT from the inner computation. ✓. So: f(3) = f(x_end)/2^{N_total}: N_total = N_outer + (steps from x* = 3 - s* (first < 2) down to negative) = N_outer + G(x*). And G(x*) for x* ∈ [1, 2): = T(s*) + G(z*): where T(s*) = c* + k₀* (the inner steps: = number of halvings to reach z* ∈ [1/2, 1)): G(z*) = steps from z* to negative: z* ∈ [1/2, 1): z* → 2z* - 1 → ... : G(z*) = (number of doublings to exceed 1/2) + 1: for z* = 1 - w: G(z*) = min{j: 2^j w > 1} + 1·... hmm: z* → 2z* - 1: iterating: z*_j = 2^j z* - (2^j - 1) = 1 - 2^j w·... wait: 2^j z* - (2^j - 1): with z* = 1 - w: 2^j - 2^j w - 2^j + 1 = 1 - 2^j w ✓: bottom out when 1 - 2^j w < 0: smallest such j: j* = floor(log2(1/w)) + 1 = position of w's leading one: j* = (position of first 1 in w) [if w = 0.0...01...: first one at position q: 2^q w ≥ 1: j* = q]. So G(z*) = q where w = 0.(q-1 zeros)1...: G(z*) = q + 1?? Let me check: z* = 1/2: w = 1/2: q = 1: chain: 1/2 → 0 → -1: G = 2. Hmm: 1/2 → 2·(1/2) - 1 = 0 [1 step]: 0 → 2·0 - 1 = -1 [2 steps]: G(1/2) = 2 ✓: q = 1: G = q + 1 ✓. z* = 3/4: w = 1/4: q = 2: chain: 3/4 → 1/2 → 0 → -1: G = 3 ✓ = q + 1. ✓. And the final value: x_end = 1 - 2^{j*}w: if w = 2^{-q} exactly: x_end = 1 - 2 = -1: f(3) = 1/2^{N_total}. If w has more bits: x_end = 1 - 2^q w = -(2^q w - 1) = -(the tail of w after its leading one): so f(3) = (2^q w - 1)/2^{N_total}: where (2^q w - 1) = tail of w: hmm: this is exactly f(z*)-related: f(z*) = w - 2^{-(q+1)}·... hmm: earlier: f(z) = (1-z) - 2^{-(K+1)} = w - 2^{-(q+1)}·... with K = q - 1·... ugh: anyway: the final multiplier is (2^q w - 1): the bits of w after its leading 1: re-based. So f(3) = [tail of w re-based]/2^{N_total}. Hmm interesting: so f(3) = (tail of w)/2^{N_total} where w = 1 - z*: and z* is the [1,2)-exit: 1 - z* = the bits of υ* after c* (the (k₀+1)-th one), re-based... OK: so now: the computation plan: (A) Simulate the outer chain: υ, k₀: increments δυ = tail(υ, d): until s = 2^{-(k₀+1)}(1 + υ) exceeds 1. Count N_outer. Also track: the FINAL state (s* just above 1: i.e., when s crosses 1: hmm: actually the outer chain runs while x = 3 - s ≥ 2: i.e., s ≤ 1: the chain: s values: s^{(0)} = 2^{-10} (after step 1)·... hmm: wait: s ≤ 1: the outer step from x = 3 - s (s ≤ 1): lands at 3 - s - Δ: if s + Δ > 1: lands in [1, 2). So the outer chain: s^{(i+1)} = s^{(i)} + Δ_i: while s^{(i)} ≤ 1: N_outer = number of i with s^{(i)} ≤ 1 (i.e., the number of outer steps taken from within [2, 3)): the landing point x* = 3 - s^{(N)}·... hmm: indices: let me say: the chain starts at x = 3 (s = 0): step 1: x₁ = 3 - 2^{-10} (s₁ = 2^{-10}): steps continue: x_{i+1} = x_i - f(x_i - 1): while x_i ≥ 2 (s_i ≤ 1): N_outer = the largest i such that x_i ≥ 2: then x_{N_outer + 1} ∈ [1, 2): and f(3) = f(x_{N_outer+1})/2^{N_outer + 1}·... hmm: no wait: f(3) = f(x₁)/2: f(x₁) = f(x₂)/2: ...: f(3) = f(x_{j})/2^{j}: chain until x_j < 0: hmm: no: until we can compute f(x_j) directly: x_j < 0: f = -x_j: total steps j = N_total: f(3) = (-x_{N_total})/2^{N_total}. So: N_total = (steps in [2,3): from x₀ = 3: x₀ ≥ 2: steps while current x ≥ 2·... including the step that lands in [1,2)) + G(x_land). Let me re-define cleanly: N_3 = number of steps from x = 3 until landing in [1, 2) (i.e., count steps: x → x - f(x-1): each step: if the STARTING x ≥ 2: it's an "outer" step). Then G(x_land): steps from x_land ∈ [1,2) to negative. N_total = N_3 + G(x_land). f(3) = (-x_final)/2^{N_total}. Now: the outer steps: x ∈ [2, 3): s = 3 - x ∈ (0, 1]: step: s ← s + f(2 - s): f(2-s) = 2^{-(k₀+1)}·δυ·... with δυ as derived. So: s^{(0)} = 0 (x = 3): step 1: s ← 2^{-10} (since f(2) = 2^{-10}·... hmm: wait: step 1: x₀ = 3: f(x₀ - 1) = f(2) = 2^{-10}: s¹ = 2^{-10} ✓: this is the special first step: f(2) = 2^{-10}: corresponds to: υ = 1, level 10: c = 11th one at position 11: d = 12: δυ = tail = 2^{-12}: Δ = 2^{-11}δυ = 2^{-23}: but the actual first step: Δ = f(2) = 2^{-10}: NOT 2^{-23}!!! DISCREPANCY: the first outer step: f(2) = 2^{-10}: but the υ-algorithm gives Δ = 2^{-23}?? What's wrong? AH: I see: the first step: f(3) = f(3 - f(2))/2: f(2) is the value at the INTEGER 2: 2 is a special point (the [1,2) framework: x = 2: s = 0: not in (0, 1]: the framework s ∈ (0, 1] excludes s = 0: x = 2 is the boundary: f(2) = f(2 - f(1))/2: f(1) = 2^{-3}: f(2) = f(2 - 2^{-3})/2 = f(15/8)/2: and f(15/8) = 2^{-9}: f(2) = 2^{-10} ✓. So the first step is special: from x = 3: s: 0 → 2^{-10}·... hmm: but actually: the increment from s = 0: Δ = f(2 - 0) = f(2) = 2^{-10}: s ← 2^{-10}. Then subsequent steps: s ∈ (0, 1]: use the υ-algorithm. s = 2^{-10}: level 10: υ = 1: canonical ones at 1, 2, 3, ...: c = 11th one = 11: d = 12: δυ = Σ_{j>12} 2^{-j} = 2^{-12}: Δ = 2^{-11}·2^{-12} = 2^{-23}: s ← 2^{-10} + 2^{-23} ✓✓ matches!! (s = 2^{-10} + 2^{-23} ✓.) Great. So the algorithm works, with the first step special (Δ = 2^{-10}: s: 0 → 2^{-10}). Hmm wait: but also: υ = 1 at level 10: the increment δυ = 2^{-12}: υ ← 1 + 2^{-12}: OVERFLOW: re-level: k₀ ← 9: υ ← (1 + 2^{-12} - 1)/2 = 2^{-13} ✓✓ (s = 2^{-10}(1 + 2^{-13}) ✓: level 9 ✓). So the algorithm: Initialize: step 1: s = 2^{-10}; state: k₀ = 10, υ = 1. [N = 1] Loop: while s ≤ 1: - canonical ones of υ: c = (k₀+1)-th one position, d = (k₀+2)-th. - δυ = tail(υ, d) = Σ_{j>d} b_j 2^{-j}. - Δ = 2^{-(k₀+1)}δυ. s ← s + Δ. [N += 1: but careful: count the step when we MOVE from s: the step count: each iteration of the loop = 1 outer step: so N increments by 1 each loop iteration: and the loop runs while the CURRENT s ≤ 1: hmm: the step FROM s (with s ≤ 1) lands at s + Δ: if s + Δ ≤ 1: still in [2, 3): continue: if s + Δ > 1: landed in [1, 2): stop.] - υ ← υ + δυ: if υ > 1: k₀ ← k₀ - 1: υ ← (υ - 1)/2. [re-level] Wait: hmm: but actually the re-leveling condition: υ > 1: but ALSO: the increment Δ added to s might push s over the level boundary 2^{-k₀}: s ≤ 2^{-k₀} ↔ υ ≤ 1 ✓ consistent. Hmm: but WAIT: there's an issue: the re-level: when υ → υ + δυ > 1: new level k₀ - 1: new υ = (υ + δυ - 1)/2: but ALSO: what if υ + δυ > 2?? Then s > 2^{-(k₀-1)}: double re-level: δυ is tiny (tail bits): can't jump that much... unless υ was already close to 1 hmm: υ ≤ 1: υ + δυ ≤ 1 + δυ: δυ < 2^{-d} ≤ small: new υ = (υ - 1 + δυ)/2 < δυ/2: small: fine: single re-level suffices (υ_new < 1 ✓ as long as δυ < 2: yes). But hmm: actually: also: what if υ + δυ = 1 exactly: s = 2^{-k₀}: still level k₀: υ stays 1: fine. Let me now also double check the δυ formula derivation once more with the B_11 → increment: υ (level 9) = ones at 13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 78 (genuine, terminating): canonical: 13, 24, 34, 43, 51, 58, 64, 69, 73, 76, then 79, 80, 81, ... (78 replaced by recurring from 79). (k₀+1) = 10th one: 76. (k₀+2) = 11th one: 79. δυ = tail after 79 = Σ_{j > 79} b_j 2^{-j} = Σ_{j ≥ 80} 2^{-j} = 2^{-79}. So δυ = 2^{-79}: Δ = 2^{-10}·2^{-79} = 2^{-89} ✓✓ MATCHES (Δ₁₁ = 2^{-89} ✓)!! And the previous: B_10 = ones 13..76 (10 genuine): canonical: 13, 24, ..., 73, then 77, 78, ... (76 replaced): 10th one: 77: 11th one: 78: δυ = tail after 78 = Σ_{j≥79} 2^{-j} = 2^{-78}: Δ = 2^{-88} ✓✓ (matches Δ₁₀ = 2^{-88} ✓). And B_9 = ones 13..73 (9 genuine): canonical: 13, ..., 69, then 74, 75, ... (73 replaced): 10th one: 74: 11th: 75: δυ = tail after 75: Σ_{j≥76} 2^{-j} = 2^{-75}: Δ = 2^{-85} ✓✓ (Δ₉ = 2^{-85}·... wait: earlier I had Δ₉ = 2^{-86}. Hmm: discrepancy! Let me recheck: B_9 = {13, 24, 34, 43, 51, 58, 64, 69, 73}: m = 9 genuine ones. Canonical: ones at 13, 24, 34, 43, 51, 58, 64, 69, and then 74, 75, 76, ... (the genuine 73 replaced by recurring from 74). 10th one: 74. 11th one: 75. δυ = Σ_{j > 75} b_j 2^{-j} = Σ_{j≥76} 2^{-j} = 2^{-75}. Δ = 2^{-10}δυ = 2^{-85}. Earlier (via T = p_m + 20 - m): T = 73 + 11 = 84: Δ = 2^{-2-84} = 2^{-86}. CONFLICT: 2^{-85} vs 2^{-86}. Let me recompute via the direct trace: υ = 2^{-13} + 2^{-24} + 2^{-34} + 2^{-43} + 2^{-51} + 2^{-58} + 2^{-64} + 2^{-69} + 2^{-73}, k₀ = 9: - 12 doubles [12]: υ = 2^{-1} + 2^{-12} + 2^{-22} + 2^{-31} + 2^{-39} + 2^{-46} + 2^{-52} + 2^{-57} + 2^{-61}: remove [13]: k=8: υ = 2^{-12} + ... + 2^{-61}. - 11 doubles [24]: υ = 2^{-1} + 2^{-11} + 2^{-20} + 2^{-28} + 2^{-35} + 2^{-41} + 2^{-46} + 2^{-50}: remove [25]: k=7. - 10 doubles [35]: υ = 2^{-1} + 2^{-10} + 2^{-18} + 2^{-25} + 2^{-31} + 2^{-36} + 2^{-40}: remove [36]: k=6. - 9 doubles [45]: υ = 2^{-1} + 2^{-9} + 2^{-16} + 2^{-22} + 2^{-27} + 2^{-31}: remove [46]: k=5. - 8 doubles [54]: υ = 2^{-1} + 2^{-8} + 2^{-14} + 2^{-19} + 2^{-23}: remove [55]: k=4. - 7 doubles [62]: υ = 2^{-1} + 2^{-7} + 2^{-12} + 2^{-16}: remove [63]: k=3. - 6 doubles [69]: υ = 2^{-1} + 2^{-6} + 2^{-10}: remove [70]: k=2. - 5 doubles [75]: υ = 2^{-1} + 2^{-5}: remove [76]: k=1: υ = 2^{-5}. - 4 doubles [80]: υ = 1/2: hmm: 2^{-5} → 2^{-4} → 2^{-3} → 2^{-2} → 2^{-1}: that's 4 doubles: [80]: υ = 2^{-1} ≤ 1/2: double [81]: υ = 1: k=1 > 0: remove [82]: υ = 1/2: k=0. - double [83]: υ = 1: k = 0: exit [84]: z = 1/2. T = 84: f(2-s) = 2^{-2}/2^{84} = 2^{-86}. So Δ = 2^{-86}: but the tail formula gave 2^{-85}. CONTRADICTION: let me recheck the tail formula derivation. The tail formula: δυ = Σ_{j > d} b_j 2^{-j}: derived: δυ = 2^{1-c} f(z): f(z) = Σ_{j > d} b_j 2^{-(j - c + 1)}: δυ = 2^{1-c}Σ_{j>d} b_j 2^{-(j-c+1)} = Σ_{j>d} b_j 2^{1-c-j+c-1} = Σ_{j>d} b_j 2^{-j} ✓: the algebra is right: so f(z) must be wrong for this case: f(z) = Σ_{j > d} b_j 2^{-(j-c+1)}: check with the trace: exit at z = 1/2: f(z) = 1/4 = 2^{-2}: Σ_{j > d} b_j 2^{-(j-c+1)}: c = 74 (10th canonical one), d = 75 (11th): bits after 75: all 1s (canonical recurring): Σ_{j ≥ 76} 2^{-(j - 74 + 1)} = Σ_{j≥76} 2^{-(j-73)} = Σ_{i≥3} 2^{-i} = 1/4 ✓✓: f(z) = 1/4 ✓: δυ = Σ_{j>d} b_j 2^{-j} = Σ_{j≥76} 2^{-j} = 2^{-75}: Δ = 2^{-(k₀+1)}δυ = 2^{-10}·2^{-75} = 2^{-85}. But the direct trace says Δ = 2^{-86}!! So there's an inconsistency: Δ = f(z)/2^T: T = 84: f(z) = 2^{-2}: Δ = 2^{-86}. Via δυ: Δ = 2^{-(k₀+1)}δυ: δυ = 2^{-75}: Δ = 2^{-85}. So the relation δυ = 2^{k₀+1}Δ: 2^{10}·2^{-86} = 2^{-76} ≠ 2^{-75}. So δυ ≠ 2^{k₀+1}Δ: ERROR somewhere. Let me recheck: Δ = f(2 - s) = f(z)/2^T: T = c + k₀: c = position of the (k₀+1)-th one: = 74: k₀ = 9: T = 83?? But the trace says T = 84!! Hmm: T = c + k₀ = 74 + 9 = 83 ≠ 84. CONTRADICTION: earlier I verified T = c_{k₀+1} + k₀ with the B_10 case (c = 77: T = 86 ✓). Let me recheck B_9: c = 10th canonical one: canonical ones: 13, 24, 34, 43, 51, 58, 64, 69, 74, 75, 76, ...: 10th one: 74: T should be 74 + 9 = 83: but trace gives 84. Hmm!! Let me recount the trace: removes at steps 13, 25, 36, 46, 55, 63, 70, 76, 82: that's 9 removes: then exit at 84. Let me recount the last part: after the 9th remove·... wait: removes: 13 (k 9→8), 25 (8→7), 36 (7→6), 46 (6→5), 55 (5→4), 63 (4→3), 70 (3→2), 76 (2→1), 82 (1→0): 9 removes ✓: then: υ = 1/2 (after remove at 82): double [83]: υ = 1: exit [84]: T = 84 ✓. Via formula: c₁ = 13: 1st remove at step 13 = c₁ + 0 ✓ (j-th remove at c_j + j - 1: j=1: 13 ✓: j=2: c₂ + 1 = 25 ✓ (c₂ = 24): j=3: 34+2 = 36 ✓: j=4: 43+3 = 46 ✓: j=5: 51+4 = 55 ✓: j=6: 58+5 = 63 ✓: j=7: 64+6 = 70 ✓: j=8: 69+7 = 76 ✓: j=9: 73+8 = 81?? But the 9th remove is at step 82!!! Hmm: c₉ = 73: 73 + 8 = 81: but trace says the 9th remove is at 82. Let me recount the trace at the 9th remove: after the 8th remove (step 76): υ = 2^{-5}·... trace: step 76: remove: υ = 2^{-5}: k = 1. Steps 77-80: 4 doubles: υ = 2^{-1}: [80]. Step 81: υ = 1/2 ≤ 1/2: double: υ = 1 [81]. Step 82: υ = 1 > 1/2: k = 1 > 0: remove: υ = 1/2: k = 0 [82]. Steps 83: double: υ = 1 [83]. Step 84: exit. So the 9th remove at 82: but the 9th ONE (canonical): the genuine ones: 13, 24, 34, 43, 51, 58, 64, 69, 73: the 9th genuine one: 73: canonical: the last genuine (73) is replaced by 74, 75, ...: so canonical ones: 13, 24, 34, 43, 51, 58, 64, 69, 74, 75, ...: the 9th CANONICAL one: 74!! (Not 73!) Because canonical replaces 73 with 74. Hmm: so the 9 removes correspond to canonical ones at 13, 24, 34, 43, 51, 58, 64, 69, 74: c₉ = 74: 9th remove at step c₉ + 8 = 82 ✓✓!! And the 10th canonical one: 75: the EXIT: T = c₁₀ + 9 = 75 + 9 = 84 ✓✓!!! So the formula T = c_{k₀+1} + k₀ holds with c_j = CANONICAL one-positions ✓. My error above: I used c₁₀ = 74 instead of 75. Canonical ones: 13, 24, 34, 43, 51, 58, 64, 69, 74, 75, 76, ...: 10th = 75 ✓: T = 84 ✓: and d = 11th one = 76: δυ = tail after 76 = Σ_{j≥77} 2^{-j} = 2^{-76}: Δ = 2^{-10}·2^{-76} = 2^{-86} ✓✓✓ MATCHES!! So: the CORRECT algorithm: canonical ones: c₁ < c₂ < ... (canonical = recurring-1s: the last genuine one g_m is replaced by g_m + 1, g_m + 2, ...): c = c_{k₀+1}, d = c_{k₀+2}: δυ = Σ_{j > d} b_j 2^{-j} (the canonical bits after d: which, if d is in the recurring tail: Σ_{j>d} 2^{-j} = 2^{-d}: i.e., δυ = 2^{-d}). In our chain so far: the (k₀+2)-th one d has always been in the recurring tail (since the number of genuine ones m ≤ k₀ + 1 = 10·... hmm: as long as m ≤ 10: the 10th and 11th ones are in the tail: δυ = 2^{-d}: d = c_{k₀+2}). So while m (genuine ones) ≤ 10: δυ = 2^{-c_{k₀+2}}: and c_{k₀+1}, c_{k₀+2}: if m ≤ 9: c_{k₀+1} = g_m + (k₀ + 2 - m)·... hmm: canonical ones: g₁, ..., g_{m-1}, then g_m + 1, g_m + 2, ...: so c_j = g_j for j ≤ m - 1: c_j = g_m + (j - m + 1) for j ≥ m. So: if m ≤ 10 = k₀ + 1: c_{k₀+1} = g_m + (k₀ + 2 - m): d = c_{k₀+2} = g_m + (k₀ + 3 - m): δυ = 2^{-d} = 2^{-(g_m + k₀ + 3 - m)}: Δ = 2^{-(k₀+1)}δυ = 2^{-(g_m + 2k₀ + 4 - m)}: with k₀ = 9: Δ = 2^{-(g_m + 22 - m)} ✓✓ MATCHES the earlier formula (Δ = 2^{-(p_m + 22 - m)} ✓). And the new genuine one added: at position d - ... : δυ = 2^{-d}: adds a genuine one at position d = g_m + k₀ + 3 - m: (in level-9 coords: position p_new = g_m + 12 - m ✓ matches earlier q = p_m + 12 - m ✓). So everything is consistent. The chain: while m ≤ 10: new bit at g_m + 12 - m (level-9 υ-coords): the bits: 13, then 24, 34, 43, 51, 58, 64, 69, 73, 76, 78, 79·... wait: with carries: when the new bit position coincides with an existing bit: carry. Hmm OK. So now I need to carefully simulate this whole thing: the outer chain until s > 1: with the state: genuine bits of υ (level-9 coords: but when re-leveling happens: level drops: coords change: υ ← (υ-1)/2: bits shift right by 1: positions +1: and k₀ ← 8: etc.) When does re-leveling happen? υ → 1: s → 2^{-9}·... hmm: υ = 1 ↔ s = 2^{-k₀-1}·2 = 2^{-k₀}·... wait: s = 2^{-(k₀+1)}(1 + υ): υ = 1: s = 2^{-k₀}: the boundary of level k₀ - 1·... hmm: level: s = 2^{-k₀}: level k₀ (right endpoint). υ > 1: s > 2^{-k₀}: level k₀ - 1. In the bit dynamics: bits march LEFT (positions decrease: 13, then new bits at 24, 34, 43, ...: positions INCREASE: hmm: new bits at larger positions = smaller values: but then carries push them left·...). Hmm: wait: actually: υ grows by δυ each step: υ = Σ 2^{-g_j}: the bits at positions 13, 24, 34, ...: new bits at g_m + 12 - m: as m grows: 24, 34, 43, 51, 58, 64, 69, 73, 76, 78, 79(?): then carries·...: the positions of NEW bits approach g_m: piling up: carries: when two bits at the same position: merge: position - 1: so bits start moving LEFT (to smaller positions = bigger values): υ grows towards 1: when the carry chain reaches position 1·... υ ≥ 1: re-level! Hmm: but wait: υ = Σ 2^{-g_j}: bits at positions ≥ 13: the carries: as bits pile up at position q and carry leftward: eventually bits reach small positions: υ → close to 1: hmm: but actually υ = Σ_{j} 2^{-g_j} with g₁ = 13: υ < 2^{-12}·... if all bits are at positions ≥ 13: υ < 2^{-12}: TINY: υ ≪ 1: re-leveling is FAR away?! Hmm: that contradicts·... wait: no: υ starts at 2^{-13}: the bits pile up: but the carry: new bit at position d: if position d is occupied: carry: d - 1: if occupied: d - 2: ...: the carry propagates LEFT: the leftmost bit is at 13: as bits pile: the positions fill in: 13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 78, 79: then next new bit at 79 + something: carry: 79 occupied: → 78: occupied: → 77: free: bits: 13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 77, (78, 79 gone): m = 11: hmm: wait: this changes m: 12 bits → 11 bits: the dynamics continues with m = 11: new bit at g_m + 12 - m·... wait: the formula δυ = 2^{-c_{k₀+2}} required m ≤ k₀+1 = 10: with m = 11 ≥ 11: the formula changes: c_{k₀+1} = 10th canonical one: genuine ones m = 11: canonical: g₁..g₁₀ genuine, g₁₁ replaced by recurring: c₁₀ = g₁₀ (genuine!): d = c₁₁ = g₁₁ + 1 (canonical): δυ = tail after d = Σ_{j > g₁₁ + 1} b_j 2^{-j} = 2^{-(g₁₁+1)} (recurring tail): Δ = 2^{-10}·2^{-(g₁₁+1)}·... Hmm OK: so for m ≥ 11: δυ = 2^{-(g₁₁ + 1)}?? Wait: careful: d = c_{k₀+2} = c₁₁ = the 11th canonical one: with m = 11 genuine: canonical: g₁, ..., g₁₀, then g₁₁ + 1, g₁₁ + 2, ...: c₁₁ = g₁₁ + 1: δυ = tail after d: the canonical bits after position g₁₁ + 1: all 1s (recurring): δυ = Σ_{j ≥ g₁₁+2} 2^{-j} = 2^{-(g₁₁+1)} ✓. And this adds a genuine bit at position g₁₁ + 1·... which is: just right of g₁₁: likely occupied?? g₁₁ + 1 vs existing bits: hmm. This is getting complicated but it's mechanical. However: the number of steps: the bits: I need to track until υ ≈ 1: υ = Σ 2^{-g_j}: bits pile up: carries: positions decrease·...: υ grows: 1 - υ·... hmm: υ grows towards 1? Let's see: the bits: 13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 78, 79 → carry → ..., 77: bits: 13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 77: m = 11: new bit at g₁₁ + 1 = 78: bits: ..., 77, 78: m = 12: hmm: with m = 12: c₁₀ = g₁₀ (genuine: the 10th genuine): d = c₁₁ = g₁₁ (genuine: 11th genuine): δυ = tail after g₁₁: = the canonical bits after g₁₁: hmm: canonical: g₁, ..., g₁₁ genuine, g₁₂ replaced: bits after g₁₁: position g₁₁ + 1, ..., g₁₂ - 1: 0s, then g₁₂: 0, then recurring: δυ = Σ_{j > g₁₁} b_j 2^{-j} = Σ_{j ≥ g₁₂+1}·... hmm: canonical bits: after g₁₁: zeros until g₁₂ - 1, then AT g₁₂: hmm: wait: the canonical replacement: the LAST genuine bit g_m (= g₁₂ = 78): replaced by recurring ones at 79, 80, ...: so bits after g₁₁ = 77: position 78: 0 (since 78 was replaced!), 79: 1, 80: 1, ...: δυ = Σ_{j≥79} 2^{-j} = 2^{-78}. Hmm: δυ = 2^{-78}: adds a bit at 78: which is FREE (78 was vacated by the canonical replacement·... wait: no: 78 is a GENUINE bit: g₁₂ = 78: the canonical representation doesn't remove the genuine bit from the VALUE: the VALUE has a bit at 78: the canonical form is just a different way of writing it!! Hmm: NO: I think I conflated: let me re-examine. υ's VALUE: υ = Σ_{j ∈ G} 2^{-j} where G = set of genuine bit positions (terminating representation). The canonical form: replace the largest position g_m: 2^{-g_m} = Σ_{j > g_m} 2^{-j}: so canonical ones: G \ {g_m} ∪ {g_m + 1, g_m + 2, ...}. The δυ formula: δυ = Σ_{j > d} b_j 2^{-j} where b = CANONICAL bits and d = c_{k₀+2} (the (k₀+2)-th canonical one). Case m = 12 (genuine ones g₁ < ... < g₁₂), k₀ = 9: canonical ones: g₁, ..., g₁₁, g₁₂ + 1, g₁₂ + 2, ...: c₁₀ = g₁₀: c₁₁ = g₁₁: d = c₁₁ = g₁₁: δυ = Σ_{j > g₁₁} b_j 2^{-j}: canonical bits after g₁₁: positions g₁₁+1, ..., g₁₂ - 1: 0: position g₁₂: 0 (replaced!): positions > g₁₂: 1: so δυ = Σ_{j > g₁₂} 2^{-j} = 2^{-g₁₂}. So δυ = 2^{-g₁₂}: adds a genuine bit AT g₁₂: which is OCCUPIED (g₁₂ is the largest genuine bit!): CARRY: 2^{-g₁₂} + 2^{-g₁₂} = 2^{-(g₁₂ - 1)}: if g₁₂ - 1 occupied: carry further... OK so now I understand the dynamics fully: The rule: δυ = 2^{-(largest genuine position)} when m = 12·... hmm: wait: in the m = 12 case: δυ = 2^{-g₁₂} = the SMALLEST bit: doubling the smallest bit: carry propagates left through consecutive occupied positions. Hmm interesting: so the dynamics: bits: {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 77, 78} (m = 12): δυ = 2^{-78}: carry: 78 → 77 occupied → 76 occupied → 75 free: bits become {13, 24, 34, 43, 51, 58, 64, 69, 73, 75}: m = 10. Hmm wait: 2^{-78} added: 78: 2 bits → carry → 77: 2 bits → carry → 76: 2 bits → carry → 75: 1 bit: and 78, 77, 76 vacated: bits: {13, 24, 34, 43, 51, 58, 64, 69, 73, 75}: m = 10 ✓ (matches my earlier B_15 computation ✓). OK so now let me carefully define the increment rule for all cases: State: genuine bits g₁ < ... < g_m, level k₀. Canonical ones: g₁, ..., g_{m-1}, g_m + 1, g_m + 2, ... c = c_{k₀+1} (the (k₀+1)-th canonical one): - if k₀ + 1 ≤ m - 1: c = g_{k₀+1} - else: c = g_m + (k₀ + 2 - m) d = c_{k₀+2}: - if k₀ + 2 ≤ m - 1: d = g_{k₀+2} - else: d = g_m + (k₀ + 3 - m) δυ = Σ_{j > d} b_j 2^{-j} (canonical bits after d): - if d ≤ m - 1 index-wise... hmm: let me think: if d is a genuine one (d = g_j for some j ≤ m - 1): the canonical bits after d: zeros at positions d+1..(next genuine - 1)·...: hmm: positions after d: the genuine bits g_{j+1}, ..., g_{m-1} remain as canonical ones: position g_m: 0: positions > g_m: 1s. So δυ = Σ_{i > j, i ≤ m-1} 2^{-g_i} + 2^{-g_m}·... wait: Σ_{positions > d} b_pos 2^{-pos} = Σ_{i=j+1}^{m-1} 2^{-g_i} + Σ_{pos > g_m} 2^{-pos} = Σ_{i=j+1}^{m-1} 2^{-g_i} + 2^{-g_m}. - if d is in the recurring tail (d = g_m + r, r ≥ 1): bits after d: all 1s: δυ = 2^{-d}. So: - If m ≤ k₀ + 1: both c and d in the recurring tail: c = g_m + (k₀+2-m), d = g_m + (k₀+3-m): δυ = 2^{-d} = 2^{-(g_m + k₀ + 3 - m)}. - If m = k₀ + 2: c = g_{k₀+1} = g_{m-1} (genuine), d = c_{k₀+2} = g_m + 1 (recurring tail): δυ = 2^{-(g_m + 1)}. - If m ≥ k₀ + 3: c = g_{k₀+1} (genuine), d = g_{k₀+2} (genuine): δυ = Σ_{i=k₀+3}^{m-1} 2^{-g_i} + 2^{-g_m}. With k₀ = 9: - m ≤ 10: δυ = 2^{-(g_m + 12 - m)}. - m = 11: δυ = 2^{-(g₁₁ + 1)}. - m ≥ 12: δυ = 2^{-g₁₂} + 2^{-g₁₃} + ... + 2^{-g_{m-1}} + 2^{-g_m}. Let me sanity check the m = 11 case with the earlier computation: B_11 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 78} (m = 11): δυ = 2^{-(78+1)} = 2^{-79}: Δ = 2^{-10}·2^{-79} = 2^{-89} ✓✓ matches Δ₁₁ = 2^{-89} ✓ (and the new bit: at 79: 79 free: bits {13,...,78, 79}: m = 12 ✓ matches B_12 ✓). And m = 12: B_12 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 78, 79}: δυ = Σ_{i=13}^{11}·... m = 12: δυ = 2^{-g₁₂} (the sum Σ_{i=k₀+3}^{m-1} is empty: k₀+3 = 12 = m: so δυ = 2^{-g_m} = 2^{-79}): Δ = 2^{-89}: new bit at 79: OCCUPIED: carry: 79 → 78: occupied → 77: free: bits: {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 77}: m = 11 ✓ matches B_13 ✓✓ (Δ₁₂ = 2^{-89} ✓ matches earlier). And m = 11 (B_13 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 77}): δυ = 2^{-(77+1)} = 2^{-78}: new bit at 78: free: bits {13, ..., 77, 78}: m = 12 ✓ matches B_14 ✓ (Δ₁₃ = 2^{-88} ✓). And m = 12 (B_14 = {13,24,34,43,51,58,64,69,73,76,77,78}): δυ = 2^{-g₁₂} = 2^{-78}: carry: 78 → 77 → 76 → 75: bits {13, 24, 34, 43, 51, 58, 64, 69, 73, 75}: m = 10 ✓ matches B_15 ✓ (Δ₁₄ = 2^{-88} ✓). And m = 10 (B_15 = {13,24,34,43,51,58,64,69,73,75}): δυ = 2^{-(75 + 12 - 10)} = 2^{-77}: new bit at 77: free: {13,...,75,77}: m = 11 ✓ matches B_16 ✓ (Δ = 2^{-87}·... hmm: Δ = 2^{-10}δυ = 2^{-87}: earlier I said Δ₁₅ = 2^{-87} ✓✓). The algorithm is confirmed. Now: the outer chain runs until s > 1: i.e., 2^{-(k₀+1)}(1 + υ) > 1: υ > 2^{k₀+1} - 1: for k₀ = 9: υ > 2^{10} - 1 = 1023?? But υ ≤ 1 always!! CONTRADICTION!! WAIT: what?! s = 2^{-(k₀+1)}(1 + υ) with υ ∈ (0, 1]: s ≤ 2^{-(k₀+1)}·2 = 2^{-k₀}: s ≤ 2^{-k₀} ≈ 2^{-9}·... so s NEVER exceeds 2^{-k₀}: the level k₀ = 9 caps s at 2^{-9}. For s > 1: need k₀ = 0: s ∈ (1/2, 1]: so the level must DESCEND from 9 to 0: re-leveling happens when υ > 1: υ → (υ-1)/2: the bits: the leading bit at position 1 (value ≥ 1/2)·... υ > 1 means the bits filled up to position 1·... So: the process: bits pile up: carries move bits left: eventually a bit reaches position 1: υ ≥ 1: re-level: bits shift right by 1 (positions +1): level drops: continue. So the bits GRADUALLY march left through positions, and each re-level shifts them right by 1: net: the bit-front advances leftward by 1 per re-level: 9 re-levels to go from level 9 to level 0·... and then s ∈ (1/2, 1]: more increments push s > 1: exit outer chain. So the total number of outer steps: the bits start at position 13 (level 9: υ ≈ 2^{-13}): and must fill up to υ = 1 (bit at position 1·... actually υ = 0.111... = 1: ALL positions 1, 2, 3, ... filled?? Hmm: υ > 1 impossible: υ = 1: all-ones: hmm: υ approaches 1 as bits fill in: re-level when υ + δυ > 1: i.e., υ > 1 - δυ: bits: nearly all positions filled: the carries: the bit dynamics is like an odometer: bits fill in from the right: carries: the count of steps: like counting in binary!! Let me think: the increments δυ: each increment: δυ = (roughly) the value of the tail: as bits fill up: this is EXACTLY like a binary counter: the state υ: increments: δυ = 2^{-(g_m + 12 - m)} (m ≤ 10) or similar: the increment SIZE adapts: it's like: each increment adds the smallest possible "quantum"·... Hmm: let me look at the increments in s-space: Δ = 2^{-(k₀+1)}δυ: and in s-space: s = 2^{-(k₀+1)}(1 + υ): the increments Δ: and the level drops when s crosses 2^{-k₀}·... Let me re-parametrize: let me define σ = 2^{k₀+1}·s = 1 + υ ∈ (1, 2]: the "normalized s". Increments: Δσ = 2^{k₀+1}Δ = δυ. Re-level: when σ > 2: k₀ ← k₀ - 1: σ ← σ/2. The dynamics of σ ∈ (1, 2]: increments δσ = δυ = (per the rules): and the genuine bits: σ = 1 + Σ 2^{-g_j}: hmm: σ's bits: the leading 1 at position 0 (value 1): plus bits at g_j. The increment rules in terms of σ: σ = 1 + υ: canonical bits of υ: hmm: let me redo: the ones of υ: canonical: g₁, ..., g_{m-1}, g_m + 1, ...: and σ = 1 + υ: bits: position 0: 1: positions g₁, ..., g_m: 1 (genuine): canonical σ: 1 + (canonical υ): σ's canonical ones: 0, g₁, ..., g_{m-1}, g_m + 1, g_m + 2, ... The rule: c = (k₀+1)-th one of υ = (k₀+2)-th one of σ: d = (k₀+3)-th one of σ: δσ = Σ_{j > d} (σ's canonical bits) 2^{-j}. With k₀ = 9 initially: c = 11th one of σ: d = 12th one of σ. Hmm: let me recount: υ's (k₀+1)-th one = σ's (k₀+2)-th one (since σ has the extra leading 1 at position 0). So: c = c_{k₀+2}(σ), d = c_{k₀+3}(σ). With k₀ = 9: c = 11th, d = 12th one of σ. And the re-level: σ > 2 → σ ← σ/2, k₀ ← k₀ - 1: then the indices shift: (k₀+2)-th one of the NEW σ: the new σ = old σ/2: bits shifted right by 1: the ones: position 1 (from the old position-0 one), then old ones + 1: hmm: and the count: the new c = (k₀'+2)-th one of new σ = (k₀+1)-th one of new σ = ... the old σ's leading 1 (position 0) becomes position 1 in new σ: so new σ's ones: 1, g₁+1, g₂+1, ...: the (k₀+1)-th one of new σ (k₀' = k₀ - 1: k₀'+2 = k₀+1): = the (k₀+1)-th one: 1 [1st], g₁+1 [2nd], ...: the (k₀+1)-th: g_{k₀}+1: hmm: interesting. This is like a shift-register/counter. Let me now think about the TOTAL number of steps. The dynamics: σ increments by δσ = tail value: and re-levels halve σ. The process ends when s > 1: s = 2^{-(k₀+1)}σ: s > 1 ↔ σ > 2^{k₀+1}: with the re-leveling keeping σ ≤ 2: s ≤ 2^{-(k₀+1)}·2 = 2^{-k₀}: so s > 1 requires k₀ = 0 and σ > 2·... wait: k₀ = 0: s = σ/2 ∈ (1/2, 1]: s > 1 impossible?!?! CONTRADICTION: s ≤ 2^{-k₀}: for k₀ = 0: s ≤ 1: s > 1 NEVER happens: the outer chain NEVER leaves [2, 3)?!?! That can't be right! Hmm: wait: something's off: the level k₀ = 0: s ∈ (1/2, 1]: the increments: Δ = f(2 - s): for s ∈ (1/2, 1]: 2 - s ∈ [1, 3/2): f(2-s) ∈ (0, 1/8]: hmm: f on [1, 3/2): max f(1) = 1/8: so Δ ≤ 1/8: and s ≤ 1: s + Δ ≤ 1.125: s CAN exceed 1!! But the framework said s ≤ 2^{-k₀} = 1: the increments push s beyond the level boundary: s > 1: then s is NOT in level 0: s ∈ (1, 2]: hmm: s > 1: x = 3 - s < 2: the outer chain ENDS (lands in [1, 2)). But in the υ-framework: s ∈ (0, 1] was assumed: the framework breaks when s > 1. So: the outer chain: s grows: level descends 9, 8, ..., 0: at level 0: s ∈ (1/2, 1]: increments: Δ = f(2 - s): for s ∈ (1/2, 1]: 2 - s ∈ [1, 3/2): the [1,2)-framework: 2 - s = x': s' = 2 - x' = s: hmm: f(2 - s) with the argument 2 - s ∈ [1, 3/2): i.e., the [1,2)-framework with parameter s' = s ∈ (0, 1/2]·... wait: the [1,2) framework: x' = 2 - s': s' = 2 - x': for x' ∈ [1, 3/2): s' = 2 - x' ∈ (1/2, 1]: hmm: s' ∈ (1/2, 1]: level k' = 0. So f(2 - s) for s ∈ (1/2, 1]: level 0: t = 2s ∈ (1, 2]: the υ = t - 1 = 2s - 1 ∈ (0, 1]: process: k₀ = 0: c = 1st one: exit at the 1st one·... T = c₁ + 0 = c₁: the exit: z = 3/2 - υ_exit: δυ (increment to υ = 2s - 1): hmm: the increment to s: Δ: δυ_s = 2Δ·... Hmm: wait: I need to redo the framework for the s-increment when s ∈ (1/2, 1] (level 0): the increment: Δ = f(2 - s) = f(z)/2^T: T = c₁ (position of the FIRST one of υ = 2s - 1·... hmm: υ = t - 1: t = 2s: υ = 2s - 1: and the process: k₀ = 0: the first one of υ triggers exit: T = c₁ + k₀ = c₁: hmm: wait: the formula T = c_{k₀+1} + k₀: k₀ = 0: T = c₁ ✓. And z = 3/2 - υ_exit: υ_exit = 0.1(bits after c₁): 1 - z = 0.0(bits after c₁): f(z) = (1 - z) - 2^{-(K+1)}: K = d - c₁ where d = the second one: hmm: as before: f(z) = Σ_{j > d} b_j 2^{-(j - c₁ + 1)}: and Δ = f(z)·2^{-T} = f(z)·2^{-c₁}. Then: s ← s + Δ: and υ_s := 2s - 1 ← υ_s + 2Δ: δ(υ_s) = 2Δ = f(z)·2^{1-c₁} = Σ_{j > d} b_j 2^{1 - c₁ - (j - c₁ + 1)} = Σ_{j>d} b_j 2^{-j} ✓ same formula: δυ = tail after d (the (k₀+2)-th = 2nd one). So the formula is uniform: υ_s (the appropriate normalized quantity) increments by the tail after the (k₀+2)-th one·... but CAREFUL: at level 0: the normalization: υ_s = 2s - 1 (NOT 2^{1}·s·... hmm: level k₀: s ∈ (2^{-(k₀+1)}, 2^{-k₀}]: t = 2^{k₀+1}s: υ = t - 1: level 0: s ∈ (1/2, 1]: t = 2s: υ = 2s - 1 ✓: consistent: υ = 2^{k₀+1}s - 1 ✓.) So the uniform rule: υ = 2^{k₀+1}s - 1 ∈ (0, 1]: canonical ones of υ: c = (k₀+1)-th, d = (k₀+2)-th: δυ = Σ_{j>d} b_j 2^{-j}: s ← s + 2^{-(k₀+1)}δυ: re-level if υ + δυ > 1: υ ← (υ + δυ - 1)/2, k₀ ← k₀ - 1. And the process ENDS when s > 1: which requires k₀ = 0 and s + Δ > 1: i.e., υ + δυ > 1 with k₀ = 0: the bits fill to the top at level 0. OK!! Now: let me figure out the step count. The dynamics in terms of the bits: this looks like a binary counter where the "quantum" is adaptive. Let me understand the bit dynamics at level k₀: Genuine bits of υ: g₁ < ... < g_m. The canonical ones: g₁, ..., g_{m-1}, g_m + 1, g_m + 2, ... The (k₀+1)-th and (k₀+2)-th canonical ones determine the increment: δυ = (canonical bits after d): and then the increment is ADDED: with carries. Case m ≤ k₀ + 1: δυ = 2^{-(g_m + k₀ + 3 - m)}: a single bit at position q = g_m + k₀ + 3 - m: added at q: if q is occupied (q ≤ g_m and q ∈ G·... q > g_m? q = g_m + (k₀+3-m): with m ≤ k₀+1: k₀+3-m ≥ 2: q ≥ g_m + 2: q > g_m: NOT occupied (g_m is the largest): so no carry: m ← m + 1: bits: g₁, ..., g_m, q. Wait: q ≥ g_m + 2: so the new bit is at least 2 to the right of the largest: fine: m increments. Hmm: but wait: m ≤ k₀ + 1: after adding: m + 1 ≤ k₀ + 2: next step: case m ≤ k₀+1 or m = k₀ + 2. So: starting from m = 1 (bit at 13: at level 9: hmm: level 9: k₀ = 9: the first state: after step 1: s = 2^{-10}: level 10!! Hmm: level 10: k₀ = 10: υ = 1: special: bits: υ = 1: hmm: the genuine representation of υ = 1: position 0?? υ = 1 = 2^0: the "bit at position 0": hmm: but our framework: υ ∈ (0, 1]: υ = 1: bits: canonical: ones at 1, 2, 3, ...: genuine: position 0: hmm: awkward: let me handle υ = 1 as: genuine bit at position 0·... Actually: the first step (from s = 2^{-10}, level 10, υ = 1): c = (k₀+1)-th = 11th canonical one = 11: d = 12: δυ = 2^{-12}: υ ← 1 + 2^{-12} > 1: re-level: k₀ ← 9: υ ← 2^{-13}: ✓ (matches: s = 2^{-10} + 2^{-23}·... hmm: wait: Δ = 2^{-(k₀+1)}δυ = 2^{-11}·2^{-12} = 2^{-23} ✓: s = 2^{-10} + 2^{-23} ✓✓). So after step 2: state: k₀ = 9: bits: {13}: m = 1. Then: steps with m = 1..10: each adds a bit at q = g_m + 12 - m (no carry): so: m=1: g = {13}: add at 13 + 11 = 24: {13, 24}: step 3: Δ = 2^{-10}·2^{-24} = 2^{-34} ✓. m=2: add at 24 + 10 = 34: {13, 24, 34}: Δ = 2^{-44} ✓. m=3: add at 43: Δ = 2^{-53} ✓. m=4: add at 51: Δ = 2^{-61} ✓. m=5: add at 58: Δ = 2^{-68} ✓. m=6: add at 64: Δ = 2^{-74} ✓. m=7: add at 69: Δ = 2^{-79} ✓. m=8: add at 73: Δ = 2^{-83} ✓. m=9: add at 76: Δ = 2^{-86} ✓. m=10: add at 78: Δ = 2^{-88} ✓. m=11 (bits {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 78}): case m = k₀+2 = 11: δυ = 2^{-(g₁₁ + 1)} = 2^{-79}: add at 79: free: bits {·, 78, 79}: m = 12: Δ = 2^{-89} ✓. m=12 (bits {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 78, 79}): case m ≥ k₀+3 = 12: δυ = Σ_{i=k₀+3}^{m-1} 2^{-g_i} + 2^{-g_m} = (empty sum: k₀+3 = 12 = m: so just) 2^{-g₁₂} = 2^{-79}: add at 79: OCCUPIED: carry: 79, 78 occupied: → 77: free: bits {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 77}: m = 11: Δ = 2^{-89} ✓. Now m = 11: bits {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 77}: δυ = 2^{-(77+1)} = 2^{-78}: add at 78: free: bits {·, 77, 78}: m = 12: Δ = 2^{-88} ✓. m = 12: bits {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 77, 78}: δυ = 2^{-g₁₂} = 2^{-78}: carry: 78 → 77 → 76 → 75: bits {13, 24, 34, 43, 51, 58, 64, 69, 73, 75}: m = 10: Δ = 2^{-88} ✓. m = 10: bits {·, 73, 75}: δυ = 2^{-(75 + 12 - 10)} = 2^{-77}: add at 77: free: {·, 75, 77}: m = 11: Δ = 2^{-87} ✓. m = 11: bits {13, 24, 34, 43, 51, 58, 64, 69, 73, 75, 77}: δυ = 2^{-78}: add at 78: free: m = 12: bits {·, 77, 78}: Δ = 2^{-87} ✓. m = 12: bits {13, 24, 34, 43, 51, 58, 64, 69, 73, 75, 77, 78}: δυ = 2^{-78}: carry: 78 → 77 → 76 → 75 → 74: bits {13, 24, 34, 43, 51, 58, 64, 69, 73, 74}: m = 10: Δ = 2^{-87} ✓. m = 10: bits {·, 73, 74}: δυ = 2^{-(74+2)} = 2^{-76}: add at 76: free: bits {·, 74, 76}: m = 11: Δ = 2^{-86} ✓. m = 11: bits {13, 24, 34, 43, 51, 58, 64, 69, 73, 74, 76}: δυ = 2^{-77}: add at 77: free: m = 12: bits {·, 76, 77}: Δ = 2^{-86} ✓. m = 12: bits {13, 24, 34, 43, 51, 58, 64, 69, 73, 74, 76, 77}: δυ = 2^{-g₁₂} = 2^{-77}: add at 77: occupied: carry: 77 → 76 → 75 → 74 → 73 occupied → 72: bits {13, 24, 34, 43, 51, 58, 64, 69, 72}: m = 9: Δ = 2^{-86} ✓. Hmm wait: let me recompute: bits {13, 24, 34, 43, 51, 58, 64, 69, 73, 74, 76, 77}: add 2^{-77} at 77: 77 occupied: carry to 76: occupied: carry to 75: free? bits: 73, 74, 76, 77: 75 is FREE: so carry: 77 → 76 → 75: bits: {13, 24, 34, 43, 51, 58, 64, 69, 73, 74, 75}: m = 11. Hmm: I made an error: 75 is free (bits at 74 and 76: not 75). Let me redo: bits: 13, 24, 34, 43, 51, 58, 64, 69, 73, 74, 76, 77. Add at 77: occupied → merge → 76: occupied → merge → 75: free → new bit at 75. Result: {13, 24, 34, 43, 51, 58, 64, 69, 73, 74, 75}: m = 11. Hmm: but wait: earlier when the bits were {·, 74, 76, 77}: I need to be careful: adding 2^{-77}: the bit at 77: 2 bits at 77 → value 2^{-76}: merges with bit at 76: 2 bits at 76 → 2^{-75}: 75 free: new bit at 75: bits: 13, 24, 34, 43, 51, 58, 64, 69, 73, 74, 75 ✓ m = 11. Hmm: but earlier I wrote "bits {13, 24, 34, 43, 51, 58, 64, 69, 72}: m = 9" — that was wrong: I incorrectly carried through 74 and 73. Let me redo: 74 is occupied: but the carry: 77 → 76 → 75 (free): STOPS at 75. Bits: {13, 24, 34, 43, 51, 58, 64, 69, 73, 74, 75}: m = 11. Hmm: wait: but in my earlier analysis (B_24 area) I had: "add 76: 76→75→74→73→72": with bits {13, 24, 34, 43, 51, 58, 64, 69, 73, 74, 75, 76}: adding at 76: occupied → 75: occupied → 74: occupied → 73: occupied → 72: free: result {13, 24, 34, 43, 51, 58, 64, 69, 72}: m = 9. That was for a different bit set (with 73, 74, 75, 76 all occupied). OK. So the dynamics is like a binary counter: bits fill in gaps and carry. The pattern: the bits: {13, 24, 34, 43, 51, 58, 64, 69, 73, ...}: the gaps between consecutive bits: 11, 10, 9, 8, 7, 6, 5, 4, ...: and then the fill/carry dynamics around the right end. Let me carefully simulate. I'll denote the bit set B (positions), level k₀ = 9, and track the increments. Let me re-verify the increment rule (level 9, k₀ = 9: (k₀+1) = 10th and (k₀+2) = 11th canonical ones): - If m ≤ 10: δυ = 2^{-(g_m + 12 - m)} (single bit, no carry since q = g_m + 12 - m ≥ g_m + 2). - If m = 11: δυ = 2^{-(g₁₁ + 1)} (single bit at g₁₁ + 1: could carry if g₁₁ + 1 occupied: g₁₁ + 1 > g₁₁: not occupied (g₁₁ is largest): no carry ✓). - If m ≥ 12: δυ = Σ_{i=12}^{m-1} 2^{-g_i} + 2^{-g_m} (multi-bit: the bits at positions g₁₂, ..., g_m: PLUS: hmm: Σ_{i = k₀+3}^{m-1} 2^{-g_i} + 2^{-g_m} with k₀+3 = 12: so δυ = Σ_{i=12}^{m} 2^{-g_i}: ALL the bits from the 12th smallest onward!!). Hmm wait: for m = 12: δυ = 2^{-g₁₂} + 2^{-g_m} = 2·2^{-g₁₂}?? Since m = 12: g_m = g₁₂: δυ = Σ_{i=12}^{12} 2^{-g_i} = 2^{-g₁₂}: just one term ✓ (I mis-stated: the sum Σ_{i=k₀+3}^{m-1} + 2^{-g_m}: for m = 12: Σ_{i=12}^{11} (empty) + 2^{-g₁₂} = 2^{-g₁₂} ✓). For m = 13: δυ = Σ_{i=12}^{12} 2^{-g_i} + 2^{-g₁₃} = 2^{-g₁₂} + 2^{-g₁₃}. Let me double check this case rule: m ≥ k₀ + 3 = 12: c = c₁₀ = g₁₀ (genuine: since m - 1 ≥ 11 ≥ 10 ✓): d = c₁₁ = g₁₁ (genuine: since m - 1 ≥ 11 ✓): δυ = canonical bits after position g₁₁: the genuine bits g₁₂, ..., g_{m-1} (canonical ones ✓): position g_m: canonical: replaced by recurring: bits at positions > g_m: all 1: hmm: and positions between g_{m-1} and g_m: 0: and AT g_m: 0 (replaced): so bits after g₁₁: {g₁₂, ..., g_{m-1}} ∪ {g_m + 1, g_m + 2, ...}: δυ = Σ_{i=12}^{m-1} 2^{-g_i} + Σ_{j > g_m} 2^{-j} = Σ_{i=12}^{m-1} 2^{-g_i} + 2^{-g_m} ✓. So for m ≥ 12: δυ = Σ_{i=12}^{m} 2^{-g_i}·... (combining): δυ = Σ_{i=12}^{m-1} 2^{-g_i} + 2^{-g_m} = Σ_{i=12}^{m} 2^{-g_i} ✓. So: the increment: for m ≥ 12: δυ = the sum of the bits from the 12th one onward (all the "small" bits below a threshold): adding this to υ: the bits g₁₂, ..., g_m get doubled: carry chain: from g_m upward (leftward through consecutive occupied positions): g_m: 2 bits → carry to g_m - 1: if occupied → continue: until a free position: the free position gets the bit: and positions g_m .. (free - 1) get vacated... wait: careful: adding Σ_{i=12}^m 2^{-g_i} to υ which has bits at g₁, ..., g_m: the positions g₁₂, ..., g_m each gain an extra bit: each becomes 2 bits → carries: hmm: but the carries interact: e.g., bits at positions 5, 6, 7 (consecutive): each doubled: 5: 2 bits → 4: hmm: this is just: υ_new = υ + δυ = υ + (bits g₁₂..g_m) = g₁..g₁₁ single + g₁₂..g_m doubled: doubling a bit at position p: 2·2^{-p} = 2^{-(p-1)}: so the doubled bits move left by 1: if the left neighbor is also doubled: they merge: e.g., bits at 5 and 6 doubled: → bits at 4 and 5: if 4 was free: fine: bits at 4, 5: hmm: but ALSO the original bits at 5, 6 are gone (they moved). Wait: υ_new = Σ_{i≤11} 2^{-g_i} + Σ_{i=12}^m 2·2^{-g_i} = Σ_{i≤11} 2^{-g_i} + Σ_{i=12}^m 2^{-(g_i - 1)}: so the bits g₁₂..g_m shift LEFT by one position (each doubled). Then: if some g_i - 1 collides with g_{i-1} (for i = 12: g₁₂ - 1 vs g₁₁: if g₁₂ = g₁₁ + 1: collision: carry further) or with another shifted bit: consecutive bits g_i, g_{i+1} with g_{i+1} = g_i + 1: shifted: g_i - 1, g_{i+1} - 1 = g_i: distinct ✓ no collision among shifted bits: collisions only with the stationary bits g₁..g₁₁ (if g₁₂ - 1 = g₁₁) or... hmm: and with each other: shifted bits: g_i - 1 all distinct ✓. So: for m ≥ 12: the update: bits g₁₂, ..., g_m each shift left by 1: then: if g₁₂ - 1 = g₁₁ (i.e., bits were consecutive at the boundary): carry: g₁₁ gets doubled: shifts to g₁₁ - 1: cascade: continues left through consecutive stationary bits: until a gap: the cascade bit lands in the gap: and the stationary bits it passed through are consumed·... hmm: wait: careful: if g₁₂ - 1 = g₁₁: the shifted bit lands ON g₁₁: g₁₁ now has 2 bits: → 2^{-(g₁₁-1)}: if g₁₁ - 1 = g₁₀: lands on g₁₀: doubled: → continue: until a free position p: the bit lands at p: and ALL the stationary bits from g₁₀·... hmm: the bits that got doubled and moved: they're consumed: so: e.g., bits {10, 11, 13, 14} (m = 4: but with k₀ = 9 this example is off: let me not worry): update shifts 13, 14 → 12, 13: 12 free? bits {10, 11}: 12 free ✓: new bits: {10, 11, 12, 13}: hmm: but wait: 13 was vacated and re-filled by the shifted 14 ✓: fine. If bits {11, 12, 14, 15} (k₀ = 9: m = 4: hmm: but the rule needs m ≥ 12: whatever: illustrative): update shifts 14, 15 → 13, 14: 13 free (bits 11, 12): new: {11, 12, 13, 14} ✓. If bits {12, 13, 14, 15}: shift 14, 15 → 13, 14: 13 occupied (stationary): cascade: 13 doubled → 12: occupied: doubled → 11: free: result: hmm: let me do it carefully: υ = 2^{-12} + 2^{-13} + 2^{-14} + 2^{-15}: δυ = 2^{-14} + 2^{-15} (the bits from the 12th one·... in this toy example: the "12th one" would be some position: never mind: suppose δυ = 2^{-14} + 2^{-15}): υ_new = 2^{-12} + 2^{-13} + 2^{-13} + 2^{-14} + 2^{-14} + 2^{-15} = 2^{-12} + 2^{-12} + 2^{-13} + 2^{-15}·... let me just: 2^{-13} + 2^{-13} = 2^{-12}: so 2^{-12} + 2^{-12} = 2^{-11}: υ_new = 2^{-11} + 2^{-14} + 2^{-14}·... hmm: 2^{-14} + 2^{-14} = 2^{-13}: υ_new = 2^{-11} + 2^{-13} + 2^{-15}·... hmm: I'm confusing myself: υ_new = 2^{-12} + 2^{-13} + 2^{-14} + 2^{-15} + 2^{-14} + 2^{-15} = 2^{-12} + 2^{-13} + 2^{-13} + 2^{-14} + 2^{-14} + 2^{-15} = 2^{-12} + 2^{-12} + 2^{-13} + 2^{-15} = 2^{-11} + 2^{-13} + 2^{-15}. Bits: {11, 13, 15}: m = 3. Interesting: the consecutive run {12, 13, 14, 15} became {11, 13, 15}: the run "compressed": leading bit moved left by 1, and the run shortened·... hmm: {12,13,14,15} + shift the last two left: {12,13} + {13,14} = merges: 13 doubled → 12: doubled → 11: so 12, 13, 14 consumed: 11 added: and 15: stayed (it shifted to 14: which merged with the shifted 14·... ugh: let me just: {12,13,14,15}: add {14,15}: → 12:1, 13:1+? hmm: 13: original 1: 14 shifted from 15·... I'll do it by value: υ = 2^{-12}(1 + 2^{-1} + 2^{-2} + 2^{-3}) = 2^{-12}·(15/8). δυ = 2^{-14} + 2^{-15} = 2^{-12}(2^{-2} + 2^{-3}) = 2^{-12}·(3/8). υ_new = 2^{-12}·(18/8) = 2^{-12}·(9/4) = 2^{-11}·(9/8) = 2^{-11}(1 + 1/8): bits: {11, 14}: hmm: 9/8 = 1 + 2^{-3}: so υ_new = 2^{-11} + 2^{-14}: bits {11, 14}·... hmm: my earlier computation gave {11, 13, 15}: WRONG: correct: {11, 14}: υ_new = 2^{-11} + 2^{-14} ✓ (2^{-11}·9/8 = 2^{-11} + 2^{-14} ✓). OK whatever: the point: the update rule: υ ← υ + δυ with δυ = Σ_{i=12}^m 2^{-g_i} (for m ≥ 12): equivalently: the bits g₁₂..g_m shift left by one WITH carries cascading. Fine: I'll just do exact rational/bit arithmetic in the simulation. Now: the simulation will be long. The bits: {13, 24, 34, 43, 51, 58, 64, 69, 73, ...}: the dynamics: fills in the gaps between the rightmost bits, carries propagate: the bits slowly march LEFT (the front advances): when the front reaches position 1·... υ → 1: re-level: all bits shift right by 1 (positions +1), level drops. Hmm: wait: does the front really march left? Let's see: the bits {13, 24, 34, ...}: the increments add bits near the right end (positions ~76-79): carries propagate left: e.g., the carry chains: 79 → 78 → 77: the bits at 78, 79 get consumed: a bit appears at 77: NET: the rightmost cluster moves LEFT. As this repeats: the cluster erodes leftward: the gaps between bits fill: eventually the bits become consecutive: {13, 24, 34, 43, 51, 58, 64, 69, 73, 74, 75, 76, ...}: consecutive runs form: carries propagate through runs: the run tips move left·... Actually: let me reconsider: the increments δυ for m ≥ 12: Σ_{i≥12} 2^{-g_i}: the rightmost cluster (bits ≥ g₁₂) DOUBLES: so the cluster value doubles: the cluster: bits at positions p, p+1, ..., q (a consecutive run, say): value = 2^{-p}(1 - 2^{-(q-p+1)})·... doubling: 2^{-p+1}(1 - 2^{-(q-p+1)}): = 2^{-(p-1)} - 2^{-q}·... hmm: 2·Σ_{j=p}^{q} 2^{-j} = 2^{-(p-1)} - 2^{-q}: so the doubled cluster = bits at p-1 (with the run p-1..q-1·...): 2^{-(p-1)} - 2^{-q} = Σ_{j=p-1}^{q-1} 2^{-j}: so the run SHIFTS left by 1: {p, ..., q} → {p-1, ..., q-1}: if p-1 is adjacent to the next stationary bit (g₁₁ = p - 1?): then merge: carry further. So: the dynamics: the cluster (bits from g₁₂ on) shifts left by 1 each step; when it bumps into g₁₁: they merge: g₁₁ doubles: shifts left: bumps g₁₀: etc.: so the cluster ABSORBS the stationary bits one by one, growing, and its left edge marches left. Each "absorption"·... hmm: when the cluster {p-1, ..., q-1} includes p-1 = g₁₁: then the cluster = {g₁₁, ..., q-1}: and the next step: the "12th bit" is now g₁₁·... hmm: the indices shift: the cluster grows: the threshold "12th bit" moves. Let me re-think: the rule: δυ = Σ_{i=12}^{m} 2^{-g_i}: the bits from the 12th one onward: these double. After doubling: they shift left by 1: possibly merging with g₁₁ (if consecutive): if merge: g₁₁ doubles too·... but wait: the merge means: the new bit at g₁₁ adds to the existing: 2 bits at g₁₁: → 1 bit at g₁₁ - 1: if g₁₁ - 1 = g₁₀: merge: → ... : so the cluster, when consecutive with the stationary bits, cascades through them: the cascade stops at the first gap: the bit lands there: and all the stationary bits from the cascade are consumed (absorbed into the cluster). So over time: the cluster absorbs g₁₁, then later g₁₀, etc.: the cluster grows: and its left edge: after absorbing g₁₁ (which was at position p): the cluster's left edge: p - 1·... hmm: each absorption moves the left edge left by 1 and the right edge·... the right edge also shifts left by 1 each step (the doubling): so the cluster maintains its width, drifting left, absorbing the stationary bits as it passes·... wait: no: the stationary bits are to the LEFT of the cluster: the cluster drifts left: when its left edge reaches a stationary bit: absorbs it: the cluster's width grows by 1·... hmm: absorbing: the stationary bit at position p joins the cluster: the cluster then spans [p-1, ...]: hmm. Let me just simulate numerically-ish. Ugh, this is a long haul: but let's do it. Actually: let me first estimate the total step count to make sure this is sane: the cluster drifts left from position ~76 down to position 1: that's ~75 steps of drift at level 9: plus the level changes: after reaching position 1: υ = 1-ish: re-level: bits shift right by 1: level 8: and the process repeats: the bits now at positions 14, 25, 35, ... +1·...: hmm: wait: after re-leveling: the bits are (old positions + 1): and the dynamics continues at level 8: the (k₀+1) = 9th and 10th ones now matter: hmm. Actually: hold on: let me reconsider: the re-level: υ ← (υ - 1)/2: if υ is very close to 1: (υ-1)/2 is very small·... hmm: υ → 1: υ = 1 - ε: (υ - 1)/2 = -ε/2?? NEGATIVE?! That's wrong: υ > 1 triggers re-level: υ = 1 + δ: (υ - 1)/2 = δ/2: δ = the overflow amount: tiny. So: when the bits fill up to position 1 (υ ≥ 1·... υ = 1 means bits at ALL positions 1, 2, 3, ...·... hmm: υ = 1: the bits: full: 0.111...: but our υ = Σ 2^{-g_j}: bits at positions g_j: for υ = 1: need bits at all positions ≥ 1: infinitely many: impossible with finitely many genuine bits: so υ < 1 always: but υ can EXCEED 1 via carries: e.g., bits at positions 1, 2, 3, ..., k: υ = 1 - 2^{-k}: an increment doubling the cluster: 2(1 - 2^{-k})·... if the cluster = all bits: υ_new = 2υ - (the leading bit handling)·... hmm: bits at 1..k: υ = 1 - 2^{-k}: increment δυ = Σ_{i=12}^m 2^{-g_i} = the cluster: hmm: if the cluster is bits at positions j..k (j ≤ 12·... the "12th one" is at some position: the bits 1..k: the 12th one is at position 12: cluster = bits 12..k: δυ = 2^{-12}(1 - 2^{-(k-11)}): υ_new = (1 - 2^{-k}) + 2^{-12} - 2^{-(k+1)}·... = 1 - 2^{-k} + 2^{-12} - 2^{-k-1}·... hmm: 1 - 2^{-k} - 2^{-(k+1)} + 2^{-12}: hmm: if k ≥ 12: 2^{-12} ≥ 2^{-k} + 2^{-(k+1)}·... for k = 12: 1 - 2^{-12} - 2^{-13} + 2^{-12} = 1 - 2^{-13} < 1: no overflow: hmm: for k > 12: 1 - 2^{-k} - 2^{-(k+1)} + 2^{-12} > 1 iff 2^{-12} > 2^{-k} + 2^{-(k+1)} = 3·2^{-(k+1)}: iff 2^{k-11} > 3: k ≥ 13: 2² = 4 > 3 ✓: so for k ≥ 13: overflow ✓. OK: so when the bits fill positions 1..k with k ≥ 13: the increment overflows: υ > 1: re-level: υ_new = (υ - 1)/2: the bits: hmm: υ_new = (1 - 2^{-k} - 2^{-(k+1)} + 2^{-12} - 1)/2 = (2^{-12} - 2^{-k} - 2^{-(k+1)})/2: = 2^{-13} - 2^{-(k+1)} - 2^{-(k+2)}: = bits: hmm: 2^{-13} - 2^{-(k+1)} - 2^{-(k+2)}: if k ≥ 13: = 2^{-13}(1 - 2^{k-12}·...)·...: positive: bits: {13, 14, ..., k+1} minus·... hmm: 2^{-13} - 2^{-(k+1)} - 2^{-(k+2)}: for k = 13: 2^{-13} - 2^{-14} - 2^{-15} = 2^{-15}·(4 - 2 - 1) = 2^{-15}: single bit at 15. Hmm interesting: so after overflow: the bits RESET to a small cluster: {15}: and level 8: and the dynamics restarts with k₀ = 8: the (k₀+1) = 9th one: etc. So the process: at each level: the bits start as a sparse set: {13, 24, 34, 43, 51, 58, 64, 69, 73} + evolving cluster: the cluster drifts left: absorbs: eventually the bits fill 1..k: overflow: reset: level drops: and the bits·... hmm: wait: but at the reset: the bits become just {15}?? What happened to the left bits {13, 24, ...}?? OH WAIT: I think I misunderstood: the bits fill up: the cluster absorbs the stationary bits: the stationary bits {13, 24, 34, 43, 51, 58, 64, 69, 73}: as the cluster drifts left from ~76: it absorbs 73: then 69: then 64: ... : each absorption: the cluster's left edge jumps·... hmm: absorbing a stationary bit at position p when the cluster is at [p+1, ...]: the cascade: the cluster shifts to [p, ...]: wait: hmm: the cluster shift: [p+1, q] → [p, q-1]: so the cluster's left edge moves onto p·...: but the stationary bit AT p: hmm: if the cluster is [p+1, q] and there's a stationary bit at p: then after shifting: the cluster occupies [p, q-1]: and the stationary bit at p: MERGES: 2 bits at p → cascade: [p-1, q-1]: and if a stationary bit at p-1: merge: [p-2, q-1]·...: so consecutive stationary bits get absorbed in one step: gaps require multiple steps. Hmm: wait: that's not right either: the cascade: 2 bits at p → 1 bit at p-1: the cluster [p, q-1] becomes [p-1, q-1]: so absorbing the stationary bit at p extends the cluster left by 1 (net: the cluster covers p-1..q-1 after absorbing the bit at p·... hmm: the cluster was [p+1, q]: after the update: [p, q-1]: merging with the bit at p: [p-1, q-1]: so the cluster now spans [p-1, q-1]: width: (q-1) - (p-1) + 1 = q - p + 1: same as before ([p+1, q]: width q - p): hmm: width grew by 1 ✓ (absorbed the bit at p). So: each step: the cluster shifts left by 1: each shift: if the position to the left is occupied by a stationary bit: absorb: the cluster shifts left by 1 more (net 2 for that step): wait: no: the cascade: after the shift to [p, q-1] with a stationary bit at p: the merge: [p-1, q-1]: so in ONE step: the cluster moved from [p+1, q] to [p-1, q-1]: left edge moved by 2: but the width: from q - p to q - p: hmm: [p-1, q-1]: width = q - p + 1: hmm: I keep messing up: [p+1, q]: number of bits: q - p: [p-1, q-1]: number of bits: q - p + 1: so width increased by 1 (absorbed) and the whole cluster moved left by 2 (its left edge: p+1 → p-1) while the right edge moved left by 1 (q → q-1). OK: so the cluster: each step: right edge -1: left edge: -1, or -2 (if absorbing): the cluster widens as it absorbs: the stationary bits at 73, 69, 64, 58, 51, 43, 34, 24, 13: gaps: 73→69: gap 3: 69→64: gap 4: ...: as the cluster passes each: absorbing. Hmm: wait: but actually: the cluster doesn't absorb ALL stationary bits: only when it reaches them. The cluster starts around [76, 79]·... and drifts left. Let me re-examine: actually: hold on: is the "cluster" really all bits from g₁₂ on? The threshold: the 12th one (k₀+3 = 12 for k₀ = 9): as the cluster absorbs bits and m changes: the threshold (12th one) is relative to the current bit set: hmm: the bits: {13, 24, 34, 43, 51, 58, 64, 69, 73} ∪ cluster: the 12th bit: if the cluster has c bits: the 12th bit is the (12 - 9 = 3rd) cluster bit·... hmm: 9 stationary + cluster bits: the 12th smallest: the 3rd smallest of the cluster (if the cluster bits are all > 73: yes: cluster bits at ≥ 74·...): hmm: so the "cluster" for the doubling = bits from the 12th one onward = the cluster minus its 2 smallest bits?! Hmm: no: wait: the doubling: δυ = Σ_{i=12}^m: the bits g₁₂, ..., g_m: if the cluster is [74, 79] (bits 74..79: 6 bits: m = 9 + 6 = 15: the 12th bit: 9 stationary + 3rd cluster bit = 76: so δυ = bits 76, 77, 78, 79 = 2^{-76}(1 - 2^{-4}): the doubling applies to the RIGHT PART of the cluster: the leftmost 2 cluster bits (74, 75) are NOT doubled! Hmm!! So it's not the whole cluster: the bits from the 12th one onward: the 12th one: as the dynamics proceeds: the 12th one is always the (12th) bit: with 9 stationary bits (positions 13..73) + cluster: the 12th one = the 3rd cluster bit. Hmm: wait: unless the cluster absorbs some stationary bits: then the counts change. OK: I really need to just simulate carefully. Let me do it. I'll track: level k₀ (starts 9), the sorted bit list, and count steps. Let me also double check the earlier steps (m ≤ 11 cases) once more and then continue methodically. Also: I realize the increments Δ = 2^{-(k₀+1)}δυ = 2^{-10}δυ (at level 9): and the step count: +1 per increment. SIMULATION (level 9: k₀ = 9: threshold indices: c = 10th one, d = 11th one, doubling = bits from the 12th one onward): Step 2 state: bits {13}: m = 1: [after step 2: s = 2^{-10} + 2^{-23}] Let me list states and increments (δυ and Δ = 2^{-10}δυ): S2: {13}: m=1: m ≤ 10: δυ = 2^{-(13 + 12 - 1)} = 2^{-24}: add at 24: S3: {13, 24}: Δ₃ = 2^{-34}·... hmm wait: indices: let me call the states S_i after i steps: S_1: s = 2^{-10} (level 10: υ = 1): step 2: S_2 = {13} (level 9): the increment from S_2 gives S_3: etc. Let me tabulate: the increment from state S (bits B, level 9): δυ per rules: new bits B'. S_2 = {13}: m=1: δυ = 2^{-(g₁ + 12 - 1)} = 2^{-(13+11)} = 2^{-24}: B' = {13, 24}: (this increment happened at step 3·... ugh: let me just number the steps: step i takes state S_i to S_{i+1}: S_1 = (level 10, υ=1): S_2 = (level 9, {13}): S_3 = (level 9, {13,24}): ... The increments: step 2: Δ = 2^{-23} (S_1 → S_2). Step 3: Δ = 2^{-34} (S_2 → S_3). Step 4: Δ = 2^{-44}: S_4 = {13, 24, 34}: step 5: Δ = 2^{-53}: S_5 = {·, 43}: step 6: Δ = 2^{-61}: S_6 = {·, 51}: step 7: Δ = 2^{-68}: S_7 = {·, 58}: step 8: Δ = 2^{-74}: S_8 = {·, 64}: step 9: Δ = 2^{-79}: S_9 = {·, 69}: step 10: Δ = 2^{-83}: S_10 = {·, 73}: step 11: Δ = 2^{-86}: S_11 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 76}: (m=10): step 12: Δ = 2^{-88}: S_12 = {·, 76, 78} (m=11): step 13: Δ = 2^{-89}: S_13 = {·, 78, 79} (m=12): step 14: m=12: δυ = 2^{-79}: add at 79: carry: 79, 78 occupied → 77: S_14 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 76, 77} (m=11): Δ = 2^{-89}: step 15: m=11: δυ = 2^{-78}: add at 78: free: S_15 = {·, 77, 78} (m=12): Δ = 2^{-88}: step 16: m=12: δυ = 2^{-78}: carry: 78 → 77 → 76 → 75: S_16 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 75} (m=10): Δ = 2^{-88}: step 17: m=10: δυ = 2^{-(75 + 12 - 10)} = 2^{-77}: add at 77: S_17 = {·, 75, 77} (m=11): Δ = 2^{-87}: step 18: m=11: δυ = 2^{-78}: add at 78: S_18 = {·, 77, 78} (m=12): Δ = 2^{-87}: step 19: m=12: δυ = 2^{-78}: carry: 78 → 77 → 76 → 75 → 74: S_19 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 74} (m=10): Δ = 2^{-87}: step 20: m=10: δυ = 2^{-(74+2)} = 2^{-76}: add at 76: S_20 = {·, 74, 76} (m=11): Δ = 2^{-86}: step 21: m=11: δυ = 2^{-77}: add at 77: S_21 = {·, 76, 77} (m=12): Δ = 2^{-86}: step 22: m=12: δυ = 2^{-77}: add at 77: occupied: carry: 77 → 76 → 75: S_22 = {13, 24, 34, 43, 51, 58, 64, 69, 73, 74, 75} (m=11): Δ = 2^{-86}: step 23: m=11: δυ = 2^{-76}: add at 76: S_23 = {·, 75, 76} (m=12): Δ = 2^{-86}: step 24: m=12: δυ = 2^{-76}: carry: 76 → 75 → 74 → 73 → 72: S_24 = {13, 24, 34, 43, 51, 58, 64, 69, 72} (m=9): Δ = 2^{-86}: [Check step 24: bits {13, 24, 34, 43, 51, 58, 64, 69, 73, 74, 75, 76}: add 2^{-76}: 76 occupied → 75 occupied → 74 occupied → 73 occupied → 72 free: bits {13, 24, 34, 43, 51, 58, 64, 69, 72}: m=9 ✓.] step 25: m=9: δυ = 2^{-(72 + 12 - 9)} = 2^{-75}: add at 75: S_25 = {·, 72, 75} (m=10): Δ = 2^{-85}: step 26: m=10: δυ = 2^{-(75 + 2)} = 2^{-77}: add at 77: S_26 = {·, 75, 77} (m=11): Δ = 2^{-87}: Hmm wait: m=10: δυ = 2^{-(g₁₀ + 12 - 10)} = 2^{-(g₁₀ + 2)}: g₁₀ = 75: δυ = 2^{-77} ✓. step 27: m=11: δυ = 2^{-(g₁₁+1)} = 2^{-78}: add at 78: S_27 = {·, 77, 78} (m=12): Δ = 2^{-88}: step 28: m=12: δυ = 2^{-78}: carry: 78→77→76→75: S_28 = {13, 24, 34, 43, 51, 58, 64, 69, 72, 76} (m=10): Δ = 2^{-88}: [Check: bits {13, 24, 34, 43, 51, 58, 64, 69, 72, 75, 77, 78}: add 2^{-78}: 78 occupied → 77 occupied → 76: free? bits: 72, 75: 76 free ✓: new bit at 76: and 78, 77 vacated: bits {13, 24, 34, 43, 51, 58, 64, 69, 72, 75, 76}: m = 11!! Hmm: wait: 75 is still there: the cascade: 78 (occupied) → 77 (occupied) → 76 (FREE): the bit lands at 76: bits: {13, 24, 34, 43, 51, 58, 64, 69, 72, 75, 76}: m = 11. I wrote m=10 above: ERROR: let me redo step 28: S_27 = {13, 24, 34, 43, 51, 58, 64, 69, 72, 75, 77, 78} (m=12): add 2^{-78}: cascade: 78 → 77 → 76 (free): S_28 = {13, 24, 34, 43, 51, 58, 64, 69, 72, 75, 76}: m = 11.] Let me restart the tabulation from step 24 with more care. I'll write the bits as a sorted list and simulate the carry carefully each time. Rules recap (level 9): - m ≤ 10: δυ = 2^{-(g_m + 12 - m)}: add bit at position q = g_m + 12 - m (always free since q ≥ g_m + 2): m += 1. - m = 11: δυ = 2^{-(g₁₁ + 1)}: add at q = g₁₁ + 1 (free since > g₁₁): m += 1. - m ≥ 12: δυ = Σ_{i=12}^{m} 2^{-g_i}: i.e., the bits g₁₂, ..., g_m all double (shift left by 1), with carries cascading left through occupied positions. For m ≥ 12, the update: all bits from g₁₂ onward shift left by 1; then resolve collisions: a shifted bit landing on an occupied position doubles it, shifting further left, etc. Let me carefully re-simulate from S_24: S_24 = {13, 24, 34, 43, 51, 58, 64, 69, 72}: m = 9. Step 25: m=9 ≤ 10: q = 72 + 12 - 9 = 75: S_25 = {13, 24, 34, 43, 51, 58, 64, 69, 72, 75}: m=10. Step 26: m=10: q = 75 + 2 = 77: S_26 = {13, 24, 34, 43, 51, 58, 64, 69, 72, 75, 77}: m=11. Step 27: m=11: q = 77 + 1 = 78: S_27 = {13, 24, 34, 43, 51, 58, 64, 69, 72, 75, 77, 78}: m=12. Step 28: m=12: bits g₁₂..g_m = {78} doubles: 78 → 77: occupied → 76: free: S_28 = {13, 24, 34, 43, 51, 58, 64, 69, 72, 75, 76}: m=11. [78 and 77 vacated, 76 added: wait: 77 occupied: the shifted bit from 78 lands on 77: 77 now has 2 bits: → 76: 76 free: bit at 76: so bits: 72, 75, 76: and 77, 78 gone ✓.] Step 29: m=11: q = 76 + 1 = 77: S_29 = {13, 24, 34, 43, 51, 58, 64, 69, 72, 75, 76, 77}: m=12. Step 30: m=12: {77} doubles → 76: occupied → 75: occupied → 74: free: S_30 = {13, 24, 34, 43, 51, 58, 64, 69, 72, 74}: m=10. [77, 76, 75 vacated: 74 added ✓] Step 31: m=10: q = 74 + 2 = 76: S_31 = {13, 24, 34, 43, 51, 58, 64, 69, 72, 74, 76}: m=11. Step 32: m=11: q = 77: S_32 = {·, 74, 76, 77}: m=12. Step 33: m=12: {77} → 76: occupied → 75: free: S_33 = {13, 24, 34, 43, 51, 58, 64, 69, 72, 74, 75}: m=11. [76, 77 vacated: 75 added: bits: 72, 74, 75 ✓] Step 34: m=11: q = 76: S_34 = {·, 74, 75, 76}: m=12. Step 35: m=12: {76} → 75: occ → 74: occ → 73: free: S_35 = {13, 24, 34, 43, 51, 58, 64, 69, 72, 73}: m=10. [76, 75, 74 vacated: 73 added ✓] Step 36: m=10: q = 73 + 2 = 75: S_36 = {·, 72, 73, 75}: m=11. Step 37: m=11: q = 76: S_37 = {·, 73, 75, 76}: m=12. Step 38: m=12: {76} → 75: occ → 74: free: S_38 = {13, 24, 34, 43, 51, 58, 64, 69, 72, 73, 74}: m=11. [75, 76 vacated: 74 added: bits: 72, 73, 74 ✓] Step 39: m=11: q = 75: S_39 = {·, 73, 74, 75}: m=12. Step 40: m=12: {75} → 74: occ → 73: occ → 72: occ → 71: free: S_40 = {13, 24, 34, 43, 51, 58, 64, 69, 71}: m=9. [75, 74, 73, 72 vacated: 71 added ✓] Step 41: m=9: q = 71 + 3 = 74: S_41 = {·, 71, 74}: m=10. Step 42: m=10: q = 74 + 2 = 76: S_42 = {·, 71, 74, 76}: m=11. Step 43: m=11: q = 77: S_43 = {·, 74, 76, 77}: m=12. Step 44: m=12: {77} → 76: occ → 75: free: S_44 = {13, 24, 34, 43, 51, 58, 64, 69, 71, 74, 75}: m=11. Step 45: m=11: q = 76: S_45 = {·, 74, 75, 76}: m=12. Step 46: m=12: {76} → 75: occ → 74: occ → 73: free: S_46 = {13, 24, 34, 43, 51, 58, 64, 69, 71, 73}: m=10. [74, 75, 76 vacated: 73 added ✓] Step 47: m=10: q = 73 + 2 = 75: S_47 = {·, 71, 73, 75}: m=11. Step 48: m=11: q = 76: S_48 = {·, 73, 75, 76}: m=12. Step 49: m=12: {76} → 75: occ → 74: free: S_49 = {13, 24, 34, 43, 51, 58, 64, 69, 71, 73, 74}: m=11. Step 50: m=11: q = 75: S_50 = {·, 73, 74, 75}: m=12. Step 51: m=12: {75} → 74: occ → 73: occ → 72: free: S_51 = {13, 24, 34, 43, 51, 58, 64, 69, 71, 72}: m=10. [73, 74, 75 vacated: 72 added ✓] Step 52: m=10: q = 72 + 2 = 74: S_52 = {·, 71, 72, 74}: m=11. Step 53: m=11: q = 75: S_53 = {·, 72, 74, 75}: m=12. Step 54: m=12: {75} → 74: occ → 73: free: S_54 = {13, 24, 34, 43, 51, 58, 64, 69, 71, 72, 73}: m=11. Step 55: m=11: q = 74: S_55 = {·, 72, 73, 74}: m=12. Step 56: m=12: {74} → 73: occ → 72: occ → 71: occ → 70: free: S_56 = {13, 24, 34, 43, 51, 58, 64, 69, 70}: m=9. [71, 72, 73, 74 vacated: 70 added ✓] I see a pattern: the "right block" {70, 71, 72, ...} region: the bits near 69-79 churn: and occasionally the leftmost of the right block decrements: 73 → 72 → 71 → 70·...: the stationary bits {13, 24, 34, 43, 51, 58, 64, 69} are being approached: the right block's left edge: started at 73 (step 10): now at 70 (step 56): decrements by 1 roughly every ~13-14 steps? From the trace: left edge: 73 at steps 10-24ish: 72 at step 24: 71 at step 40: 70 at step 56: so ~16 steps per decrement. The block must march left: 70 → 69 (absorb!) → then continue: 68, ...: down to·... hmm: but wait: as the block marches left: the gap to the next stationary bit (69) shrinks: absorb at 69: then 64: gaps: 70→69: 1 step-ish: 69→64: gap 5: etc. Hold on: but wait: as the block's left edge approaches the stationary bits: the block ABSORBS them: the block widens: and the dynamics: the block = consecutive bits: the update (m ≥ 12): the bits from the 12th one onward double: hmm: but if the block is wide (say bits 65..79: 15 bits: plus 8 stationary: m = 23: the 12th one: the 4th block bit (8 stationary + 4): position 68: so the doubling applies to bits 68..79: they shift to 67..78: colliding with the stationary bit at 67?? no: 67 is free (stationary bits: 64: gap): the shifted block: 67..78: and the stationary bits below 68: 65, 66, 67·... hmm wait: the block was 65..79: the bits 65, 66, 67 (block's left part) DON'T double: they stay: the doubled part: 68..79 → 67..78: collision at 67 (block had 67): cascade: 67 → 66 (occupied): → 65 (occupied): → 64 (STATIONARY bit!): → absorb: 64 → 63: free: so the whole block 65..79 becomes 63..78?! Hmm: let me redo: bits: stationary {13, 24, 34, 43, 51, 58, 64} + block {65, ..., 79}: m = 7 + 15 = 22: the 12th one: 7 stationary + 5th block bit = 69: doubling: bits 69..79 → 68..78: collisions: 68 was occupied (block): cascade: 68 → 67 (occupied): → 66 (occ): → 65 (occ): → 64 (stationary: occupied): → 63: free: so: bits 65..68 (un-doubled block bits) + the cascade: hmm: let me carefully compute the value: υ = Σ_{stationary} + Σ_{j=65}^{79} 2^{-j}: δυ = Σ_{j=69}^{79} 2^{-j}: υ_new = Σ_{stat} + Σ_{j=65}^{68} 2^{-j} + 2Σ_{j=69}^{79} 2^{-j} = Σ_{stat} + Σ_{j=65}^{68} 2^{-j} + Σ_{j=68}^{78} 2^{-j} = Σ_{stat} + 2^{-65} + 2^{-66} + 2^{-67} + 2·2^{-68} + Σ_{j=69}^{78}·... hmm: 2^{-68} appears twice (from Σ_{65}^{68} and from Σ_{68}^{78}): 2·2^{-68} = 2^{-67}: so: Σ_{65}^{68} + Σ_{68}^{78} = 2^{-65} + 2^{-66} + 2^{-67} + 2^{-67} + Σ_{69}^{78} = 2^{-65} + 2^{-66} + 2^{-66} + Σ_{69}^{78} = 2^{-65} + 2^{-65} + Σ_{69}^{78} = 2^{-64} + Σ_{j=69}^{78} 2^{-j}·... hmm wait: that's not right: 2^{-65} + 2^{-66} + 2^{-67} + [2^{-68} + 2^{-68}] + Σ_{69}^{78}·... let me recompute: Σ_{j=65}^{68} 2^{-j} + Σ_{j=68}^{78} 2^{-j}: = (2^{-65} + 2^{-66} + 2^{-67} + 2^{-68}) + (2^{-68} + 2^{-69} + ... + 2^{-78}): = 2^{-65} + 2^{-66} + 2^{-67} + 2^{-67} + 2^{-69} + ... + 2^{-78}: = 2^{-65} + 2^{-66} + 2^{-66} + 2^{-69} + ... + 2^{-78}: = 2^{-65} + 2^{-65} + Σ_{69}^{78}: = 2^{-64} + Σ_{j=69}^{78} 2^{-j}. So υ_new = Σ_{stat} + 2^{-64} + Σ_{j=69}^{78}: but 2^{-64} collides with the stationary bit at 64!! The stationary set includes 64: so 2^{-64} + 2^{-64} = 2^{-63}: υ_new = Σ_{stat \ {64}} + 2^{-63} + Σ_{j=69}^{78}: bits: stationary (minus 64) + {63} + {69..78}·... Hmm: wait: that gives bits 63, then 69..78: the block JUMPED: from 65..79 to 69..78 plus a bit at 63?! That seems weird: let me recompute: original bits: {13, 24, 34, 43, 51, 58, 64} ∪ {65, ..., 79}: υ_new = Σ_{stat} + Σ_{65}^{68} + 2Σ_{69}^{79}: 2Σ_{j=69}^{79} 2^{-j} = Σ_{j=68}^{78} 2^{-j}. υ_new = Σ_{stat} + Σ_{j=65}^{68} 2^{-j} + Σ_{j=68}^{78} 2^{-j} = Σ_{stat} + 2^{-65} + 2^{-66} + 2^{-67} + 2·2^{-68} + Σ_{69}^{78} 2^{-j} = Σ_{stat} + 2^{-65} + 2^{-66} + 2^{-67} + 2^{-67} + Σ_{69}^{78} = Σ_{stat} + 2^{-65} + 2^{-66} + 2^{-66} + Σ_{69}^{78} = Σ_{stat} + 2^{-65} + 2^{-65} + Σ_{69}^{78} = Σ_{stat} + 2^{-64} + Σ_{69}^{78} 2^{-j}. Stat includes 2^{-64}: total at 64: 2·2^{-64} = 2^{-63}: υ_new = [Σ_{stat} - 2^{-64}] + 2^{-63} + Σ_{j=69}^{78} 2^{-j}. So new bits: {13, 24, 34, 43, 51, 58, 63} ∪ {69, ..., 78}: m = 7 + 10 = 17. WOW: so the block 65..79 (15 bits) became: bit at 63 (absorbing the stationary 64: net: the stationary 64 and the block's left part 65-68 merged into a single bit at 63!) and the block's right part shifted to 69..78. Hmm: interesting: the block SHATTERED: because the doubling only applied to the right part (from the 12th one). Hmm OK: so the dynamics is more complex: the "block" doesn't move coherently: the 12th-one threshold: the doubling applies to bits from the 12th one onward: with 8-9 stationary bits + a block: the threshold sits INSIDE the block: the block's left part is stationary-ish·... Let me re-examine: with stationary bits {13, 24, 34, 43, 51, 58, 64, 69} (8 bits: after absorbing 73, the remaining stationary: 13, 24, 34, 43, 51, 58, 64, 69: 8 bits) + block {70, ...}: the 12th one: 8 stationary + 4th block bit: so the doubling applies from the 4th block bit onward: the block's left 3 bits stay: the rest shift left by 1: colliding with the left 3 bits: cascading: absorbing the stationary bits when reached·... Ugh: OK: the dynamics: each step: the block's right part shifts left by 1: the block compresses against its left part / stationary bits: carries cascade: the net effect (from the trace steps 24-56): the block's left edge decrements by 1 every ~16 steps: and the block's structure: bits like {70, 71·...}: hmm: from the trace: S_56 = {13, 24, 34, 43, 51, 58, 64, 69, 70}: the "block" is just {69, 70}?? wait: 69 is stationary and 70: the bits: {13, 24, 34, 43, 51, 58, 64, 69, 70}: m = 9. So the block has been fully absorbed/collapsed: the bits near 69-79 have churned and the leftmost advanced to 70: adjacent to the stationary 69. Continuing the simulation (I'll streamline: the pattern from steps 24-56: the "churning block" advances its left edge leftward: from 72 (step 24) to 70 (step 56): 16 steps for 2 positions: 8 steps per position: hmm: steps 24→40: 72→71: 16 steps: 40→56: 71→70: 16 steps: hmm: 16 steps per position.) Hmm wait: steps 24, 40, 56: differences 16, 16: and the left edge: 72 (at step 24), 71 (step 40), 70 (step 56): so ~16 steps per left-edge decrement. Next: 69-adjacent: S_56 = {13, 24, 34, 43, 51, 58, 64, 69, 70}: the left edge (70) is ADJACENT to the stationary 69!! So the next cascade will absorb 69: the dynamics changes. Let me continue the simulation carefully from S_56: S_56 = {13, 24, 34, 43, 51, 58, 64, 69, 70}: m = 9. Step 57: m=9: q = 70 + 12 - 9 = 73: S_57 = {·, 69, 70, 73}: m=10. Step 58: m=10: q = 73 + 2 = 75: S_58 = {·, 70, 73, 75}: m=11. Step 59: m=11: q = 76: S_59 = {·, 73, 75, 76}: m=12. Step 60: m=12: {76} → 75: occ → 74: free: S_60 = {13, 24, 34, 43, 51, 58, 64, 69, 70, 73, 74}: m=11. Step 61: m=11: q = 75: S_61 = {·, 73, 74, 75}: m=12. Step 62: m=12: {75} → 74: occ → 73: occ → 72: free: S_62 = {13, 24, 34, 43, 51, 58, 64, 69, 70, 72}: m=10. [73, 74, 75 → 72 ✓] Step 63: m=10: q = 72 + 2 = 74: S_63 = {·, 70, 72, 74}: m=11. Step 64: m=11: q = 75: S_64 = {·, 72, 74, 75}: m=12. Step 65: m=12: {75} → 74: occ → 73: free: S_65 = {13, 24, 34, 43, 51, 58, 64, 69, 70, 72, 73}: m=11. Step 66: m=11: q = 74: S_66 = {·, 72, 73, 74}: m=12. Step 67: m=12: {74} → 73: occ → 72: occ → 71: free: S_67 = {13, 24, 34, 43, 51, 58, 64, 69, 70, 71}: m=10. [72, 73, 74 → 71 ✓] Step 68: m=10: q = 71 + 2 = 73: S_68 = {·, 70, 71, 73}: m=11. Step 69: m=11: q = 74: S_69 = {·, 71, 73, 74}: m=12. Step 70: m=12: {74} → 73: occ → 72: free: S_70 = {13, 24, 34, 43, 51, 58, 64, 69, 70, 71, 72}: m=11. [73, 74 → 72 ✓] Step 71: m=11: q = 73: S_71 = {·, 71, 72, 73}: m=12. Step 72: m=12: {73} → 72: occ → 71: occ → 70: occ → 69: STATIONARY: occ → 68: free: S_72 = {13, 24, 34, 43, 51, 58, 64, 68}: m=8. [69, 70, 71, 72, 73 all consumed: new bit at 68: bits: {13, 24, 34, 43, 51, 58, 64, 68} ✓] So at step 72: the stationary 69 got absorbed: bits {13, 24, 34, 43, 51, 58, 64, 68}: m = 8. Pattern: the block churns near the leftmost stationary bit + a few: each "absorption" of a stationary bit takes: from step 56 (adjacent to 69) to step 72: 16 steps. Similar to before. And now the front is at 68: adjacent to·... the next stationary: 64: gap 4 (65, 66, 67 free): so the block must march 68 → 67 → 66 → 65 → then absorb 64: 4 decrements + absorption: at ~16 steps each: ~80 steps: hmm: but the decrement rate might change as m is smaller (m = 8: the thresholds: the 12th one: with 7 stationary + block: the 5th block bit: hmm). Let me continue (and let me re-derive the increment rules for m ≤ 10 with m = 8: q = g_m + 12 - m = 68 + 4 = 72·...): S_72 = {13, 24, 34, 43, 51, 58, 64, 68}: m = 8. Step 73: m=8: q = 68 + 12 - 8 = 72: S_73 = {·, 68, 72}: m=9. Step 74: m=9: q = 72 + 3 = 75: S_74 = {·, 68, 72, 75}: m=10. Step 75: m=10: q = 75 + 2 = 77: S_75 = {·, 72, 75, 77}: m=11. Step 76: m=11: q = 78: S_76 = {·, 75, 77, 78}: m=12. Step 77: m=12: {78} → 77: occ → 76: free: S_77 = {13, 24, 34, 43, 51, 58, 64, 68, 72, 75, 76}: m=11. Step 78: m=11: q = 77: S_78 = {·, 75, 76, 77}: m=12. Step 79: m=12: {77} → 76: occ → 75: occ → 74: free: S_79 = {13, 24, 34, 43, 51, 58, 64, 68, 72, 74}: m=10. [75, 76, 77 → 74 ✓] Step 80: m=10: q = 74 + 2 = 76: S_80 = {·, 72, 74, 76}: m=11. Step 81: m=11: q = 77: S_81 = {·, 74, 76, 77}: m=12. Step 82: m=12: {77} → 76: occ → 75: free: S_82 = {13, 24, 34, 43, 51, 58, 64, 68, 72, 74, 75}: m=11. Step 83: m=11: q = 76: S_83 = {·, 74, 75, 76}: m=12. Step 84: m=12: {76} → 75: occ → 74: occ → 73: free: S_84 = {13, 24, 34, 43, 51, 58, 64, 68, 72, 73}: m=10. [74, 75, 76 → 73 ✓] Step 85: m=10: q = 73 + 2 = 75: S_85 = {·, 72, 73, 75}: m=11. Step 86: m=11: q = 76: S_86 = {·, 73, 75, 76}: m=12. Step 87: m=12: {76} → 75: occ → 74: free: S_87 = {13, 24, 34, 43, 51, 58, 64, 68, 72, 73, 74}: m=11. Step 88: m=11: q = 75: S_88 = {·, 73, 74, 75}: m=12. Step 89: m=12: {75} → 74: occ → 73: occ → 72: occ → 71: free: S_89 = {13, 24, 34, 43, 51, 58, 64, 68, 71}: m=9. [72, 73, 74, 75 → 71 ✓] Step 90: m=9: q = 71 + 3 = 74: S_90 = {·, 68, 71, 74}: m=10. Step 91: m=10: q = 76: S_91 = {·, 71, 74, 76}: m=11. Step 92: m=11: q = 77: S_92 = {·, 74, 76, 77}: m=12. Step 93: m=12: {77} → 76: occ → 75: free: S_93 = {·, 71, 74, 75}: m=11. Step 94: m=11: q = 76: S_94 = {·, 74, 75, 76}: m=12. Step 95: m=12: {76} → 75: occ → 74: occ → 73: free: S_95 = {13, 24, 34, 43, 51, 58, 64, 68, 71, 73}: m=10. Step 96: m=10: q = 75: S_96 = {·, 71, 73, 75}: m=11. Step 97: m=

算f(3): f(x) = -x for x<0; f(x) = f(x - f(x-1))/2 for x≥0.