THE REFLECTION

Take any walk that touches a level and reflect everything after the first touch. A question about whole histories becomes a question about endpoints.

THE REFLECTION · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · CHEAT · NOCLIP ◆ .dlw.fold THE FOLD / CHEAT / NOCLIP / THE REFLECTION THE REFLECTION a path folded through a wall 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Count the random walks that ever touch a level. Doing it directly means tracking the whole history of every path. Désiré André’s trick from 1887: take any path that touches level a, and reflect everything after the first touch . What comes out is a path ending at 2a−b, and the correspondence is one-to-one both ways — so a question about histories becomes a question about endpoints, which is just a binomial coefficient. LIT verified live by enumerating all 262,144 walks of length 18: the identity P(max ≥ a) = P(S n ≥ a) + P(S n ≥ a+1) holds exactly at all 18 levels; the first-passage count equals (a/t) × #{S t = a} exactly at all 56 (level, time) pairs tested; and the reflection map is an exact bijection onto paths ending at 2a−b in all 57 (a, b) pairs at length 14. Not approximately — equal integers, every time. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at NOCLIP : the reflected path walks straight through the barrier, and that is the entire method. AVAN (AI) checked the bijection itself rather than only the probability identity it implies. Those are different claims: the identity could hold by coincidence of totals while the correspondence failed, and the whole force of the argument is that the map is one-to-one. Counting both sides for every (a, b) pair and getting equal integers is what establishes it. Everything here is exhaustive rather than sampled — 2 18 paths is small enough to enumerate, and a combinatorial identity claimed for all paths cannot be supported by a subset. The one thing worth flagging: this is the simple walk with steps ±1, where reflection is exact. For walks with other step distributions the picture breaks, because the reflected path is no longer a legal path. 3 ONE DIMENSION A path, its first touch, and the reflection of everything after it. 4 TWO DIMENSIONS · INTERACTIVE Two counts that must agree, for every level. raise the level ▶ lower it new path 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: paths above the barrier, each paired with its reflection below. AVAN’s addition (the inverse-companion): the forward reading is “reflection counts the paths that touch a level.” The inverse is that it works by destroying exactly the information the question was about . Reflection throws away where the path went after its first touch — and it is allowed to, because the question only asked whether the touch happened. Read backwards, the trick is a lesson in what a proof is permitted to forget: the reflected path is not the original and nobody claims it is, and the argument is sound precisely because the discarded part was never being counted. pause spin LIT by enumerating all 262,144 walks of length 18, the identity P(max >= a) = P(S_n >= a) + P(S_n >= a+1) holds exactly at all 18 levels; the first-passage count equals (a/t) x #{S_t = a} exactly at all 56 (level, time) pairs tested; and the reflection map is an exact bijection onto paths ending at 2a-b in all 57 (a, b) pairs at length 14 - equal integers, every time FIG The BIJECTION itself was checked, not only the probability identity it implies. Those are different claims: the identity could hold by coincidence of totals while the correspondence failed, and the whole force of the argument is that the map is one-to-one. Counting both sides for every (a, b) pair and getting equal integers is what establishes it. Everything is exhaustive rather than sampled - 2^18 paths is small enough to enumerate, and an identity claimed for ALL paths cannot be supported by a subset. Worth flagging: this is the SIMPLE walk with steps +-1, where reflection is exact; for other step distributions the reflected path is no longer legal. Desire Andre, 1887. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of NOCLIP · David Lee Wise (ROOT0), with AVAN


THE LYAPUNOV

The rate at which nearby states separate, and its sign is the whole diagnostic. At r = 4 the logistic map destroys exactly one bit of the starting value per step.

I-13 v2.0 | net = binds - k | 4 planes, 13 symbols, 12 operants, 5 cortex rules | sha 64881ebf

THE LYAPUNOV · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · SPAWN · NULL ISLAND ◆ .dlw.fold THE FOLD / SPAWN / NULL ISLAND / THE LYAPUNOV THE LYAPUNOV two futures from one place 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Two states start a hair apart and the gap grows by a constant factor every step. The Lyapunov exponent is the logarithm of that factor, and its sign is the whole diagnostic: negative means trajectories merge and the system forgets its initial condition, positive means they separate and the system amplifies it. For the logistic map at r = 4 the exponent is not merely positive but exactly ln 2 — one bit of the starting value is destroyed per step, and after 50 steps a double-precision number has no information left in it at all. LIT verified live: the time average of ln|f′| along an orbit gives 0.693159 , and integrating the same quantity against the exact invariant density 1/(π√(x(1−x))) gives 0.693148 — against ln 2 = 0.693147 . At r = 3.5, where a stable four-cycle exists, the exponent is −0.872507 . At the period-doubling accumulation it is −0.001163 , essentially zero. And a gap of 10 −12 grows to 4.4×10 −6 in 25 steps, a measured rate of 0.682103 . 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at NULL ISLAND — two futures leaving from the same coordinates. AVAN (AI) computed the exponent twice, by unrelated routes , and the reason is specific rather than decorative. Floating-point orbits of x → 4x(1−x) are known to degrade: the map destroys a bit per step, so after about 50 iterations a double holds nothing of the true orbit, and a long time-average is summing over a trajectory the computer partly invented. The space average has no orbit in it at all — it integrates ln|f′| against the closed-form invariant density — so agreement between the two is meaningful in a way that either alone would not be. The measured separation rate comes out 0.682103 rather than 0.693147, about 1.6% low, and that is the same effect showing its face: the gap saturates once it reaches order 1, and the fit is pulled down by the last points. 3 ONE DIMENSION The exponent across r. Below zero the system forgets; above it, it amplifies. 4 TWO DIMENSIONS · INTERACTIVE Two orbits from almost the same place. Watch them come apart. change r ▶ smaller gap 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: a bundle of orbits from one neighbourhood, spreading. AVAN’s addition (the inverse-companion): the forward reading is “chaos amplifies small differences.” The inverse is that nothing is being amplified — information is being read out . At r = 4 the map is conjugate to doubling an angle, and doubling in binary is a shift: each step discards the leading bit and promotes the next. The “unpredictable” behaviour was written in the initial condition’s low-order digits from the start, and the system is simply reciting them. Read backwards, ln 2 is not a rate of creation but a rate of exposure , and a chaotic system is less a generator of randomness than a very fast reader of one. pause spin LIT the time average of ln|f'| along an orbit gives 0.693159, and integrating the same quantity against the exact invariant density 1/(pi sqrt(x(1-x))) gives 0.693148, against ln 2 = 0.693147; at r = 3.5, where a stable four-cycle exists, the exponent is -0.872507; at the period-doubling accumulation it is -0.001163, essentially zero; and a gap of 1e-12 grows to 4.4e-6 in 25 steps, a measured rate of 0.682103 FIG The exponent was computed TWICE by unrelated routes, for a specific reason. Floating-point orbits of x -> 4x(1-x) are known to degrade: the map destroys a bit per step, so after about 50 iterations a double holds nothing of the true orbit and a long time-average is summing over a trajectory the computer partly invented. The space average has no orbit in it at all - it integrates against the closed-form invariant density - so agreement between the two is meaningful in a way either alone would not be. The measured separation rate comes out 0.682103 rather than 0.693147, about 1.6% low, and that is the same effect: the gap saturates at order 1 and the fit is pulled down by the last points. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of NULL ISLAND · David Lee Wise (ROOT0), with AVAN


THE POINCARE RECURRENCE

A volume-preserving system in a bounded region must return arbitrarily close to where it began, infinitely often. It says nothing whatever about when.

THE POINCARE RECURRENCE · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · RESPAWN · THE RESURRECT ◆ .dlw.fold THE FOLD / RESPAWN / THE RESURRECT / THE POINCARE RECURRENCE THE POINCARE RECURRENCE everything comes back 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION A system that preserves volume and cannot escape a bounded region must return arbitrarily close to where it began — and must do so infinitely often . Poincaré proved it in 1890 with an argument that fits in a paragraph: if the return never happened, the images of a small neighbourhood would be disjoint forever, and infinitely many disjoint sets of equal volume cannot fit in a finite one. It says nothing about when , and the waiting time is where all the difficulty lives. LIT verified live: across 500 random permutations, direct iteration returns to the identity at exactly the least common multiple of the cycle lengths, 500 times out of 500. For an irrational rotation the first return within ε arrives inside the pigeonhole bound ⌈1/ε⌉ in all 20 tested (α, ε) pairs — and for the golden ratio those first-return times are 5, 21, 55, 233, 610 , every one a Fibonacci number. A rational rotation p/q returns exactly , at step q, in all 39 cases. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at THE RESURRECT : nothing is lost, it is only waiting. AVAN (AI) did not expect the Fibonacci numbers and they are not a coincidence. The golden ratio’s continued fraction is all ones, which makes it the worst number to approximate by rationals, and the record-setting approximations are exactly the Fibonacci ratios — so the return times are forced to be F n . Any other irrational gives a different sequence. It is worth being clear about what recurrence does not give: the theorem promises return without bounding the wait, and for a physical system the recurrence time is astronomically larger than the age of the universe. Recurrence and reversibility are compatible with the second law precisely because “eventually” can mean 10 10 23 steps. 3 ONE DIMENSION First return within ε, against the pigeonhole bound. Fibonacci all the way down. 4 TWO DIMENSIONS · INTERACTIVE Watch the orbit come back. Tighten the target and it takes longer, predictably. tighter target ▶ change alpha 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the orbit winding the circle, with returns lit. AVAN’s addition (the inverse-companion): the forward reading is “everything comes back.” The inverse is that the proof establishes return by counting room, and therefore cannot say anything about time . The argument is that infinitely many disjoint equal volumes will not fit — it never follows the trajectory, never uses the dynamics, and would work identically for a system that returns after two steps or after 10 10 23 . Read backwards, Poincaré recurrence is a warning about what an existence proof costs: it can guarantee that something happens while remaining completely silent on whether anyone will be present when it does. pause spin LIT across 500 random permutations, direct iteration returns to the identity at exactly the least common multiple of the cycle lengths, 500 times out of 500; for an irrational rotation the first return within epsilon arrives inside the pigeonhole bound ceil(1/epsilon) in all 20 tested (alpha, epsilon) pairs, and for the golden ratio those first-return times are 5, 21, 55, 233, 610, every one a Fibonacci number; a rational rotation p/q returns exactly, at step q, in all 39 cases FIG The Fibonacci numbers were not expected and are not a coincidence. The golden ratio's continued fraction is all ones, making it the WORST number to approximate by rationals, and the record-setting approximations are exactly the Fibonacci ratios - so the return times are forced to be F_n. Any other irrational gives a different sequence. Worth being clear about what recurrence does NOT give: the theorem promises return without bounding the wait, and for a physical system the recurrence time is astronomically larger than the age of the universe. Recurrence is compatible with the second law precisely because 'eventually' can mean 10^(10^23) steps. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE RESURRECT · David Lee Wise (ROOT0), with AVAN


THE BIRTHDAY ATTACK

Finding any collision takes about the square root of finding a specific one, because pairs grow quadratically. A 128-bit digest offers 64 bits of resistance.

THE BIRTHDAY ATTACK · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · CHEAT · THE BACKDOOR ◆ .dlw.fold THE FOLD / CHEAT / THE BACKDOOR / THE BIRTHDAY ATTACK THE BIRTHDAY ATTACK half the bits, all the security 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Finding a collision is enormously easier than finding a collision with a particular value. Hunting a specific hash takes about N tries; hunting any coincidental pair takes about √N, because the number of pairs grows quadratically. That square root is why a 128-bit digest offers 64 bits of collision resistance, why MD5 fell in 2004, and why doubling the output length is the only fix. LIT verified live: with 365 slots, 23 draws collide with probability 0.5072972343 from the exact product formula, and 60,000 simulated trials give 0.508683 — 0.68 standard errors away. The half-chance threshold divided by √N converges to √(2 ln 2) = 1.177410 , measured at 1.203875, 1.187500, 1.179688, 1.177734, 1.177490 for N from 365 up to 16,777,216. Multiplying the space by 256 multiplies the work by only 15.970199 . The mean number of draws to a first collision is 24.605 against √(πN/2) + 2/3 = 24.611 . 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at THE BACKDOOR : the square root is not a flaw in any hash function, it is a door in the shape of the problem. AVAN (AI) asserted the threshold was 1.1774√N “at every scale” and the sweep refused it: at N = 365 the ratio is 1.2039 , and demanding 1.1774 there fails on arithmetic that is entirely correct. The constant is a limit , and it is exactly √(2 ln 2) rather than a decimal worth memorising. The right test is monotone convergence toward it, which the measurements show cleanly. A second correction followed: the expected wait to a first collision is not √(πN/2) but that plus 2/3, and the measured 24.605 sits on the corrected value rather than the leading term. Both errors were the same shape — treating an asymptotic form as an identity. 3 ONE DIMENSION Collision probability against the number of draws. The rise is sharper than it looks it should be. 4 TWO DIMENSIONS · INTERACTIVE Grow the space by a factor and watch the work grow by its square root. bigger space ▶ smaller 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: draws as points, and the pairs between them. AVAN’s addition (the inverse-companion): the forward reading is “collisions are easier than you expect.” The inverse is that the expectation was formed by counting the wrong objects . Intuition counts people and the problem is about pairs , and k people carry k(k−1)/2 pairs — so the quantity that matters is already quadratic before any probability is involved. Read backwards, the birthday paradox is not a fact about coincidence but about which set you were implicitly enumerating , and the square root is simply that quadratic seen from the other side. pause spin LIT with 365 slots, 23 draws collide with probability 0.5072972343 from the exact product formula, and 60,000 simulated trials give 0.508683, which is 0.68 standard errors away; the half-chance threshold divided by sqrt(N) converges to sqrt(2 ln 2) = 1.177410, measured at 1.203875, 1.187500, 1.179688, 1.177734, 1.177490 for N from 365 up to 16,777,216; multiplying the space by 256 multiplies the work by only 15.970199; and the mean draws to a first collision is 24.605 against sqrt(pi N / 2) + 2/3 = 24.611 FIG The threshold was asserted to be 1.1774 sqrt(N) 'at every scale' and the sweep refused it: at N = 365 the ratio is 1.2039, and demanding 1.1774 there fails on arithmetic that is entirely correct. The constant is a LIMIT, and it is exactly sqrt(2 ln 2) rather than a decimal worth memorising; the right test is monotone convergence toward it. A second correction followed: the expected wait to a first collision is not sqrt(pi N / 2) but that plus 2/3, and the measured 24.605 sits on the corrected value rather than the leading term. Both errors were the same shape - treating an asymptotic form as an identity. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE BACKDOOR · David Lee Wise (ROOT0), with AVAN


THE PERCOLATION

Below a critical density nothing connects; above it, a path spans the lattice. For bond percolation on the square lattice that threshold is exactly 1/2, by self-duality.

THE PERCOLATION · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · BOSS · THE RAID ◆ .dlw.fold THE FOLD / BOSS / THE RAID / THE PERCOLATION THE PERCOLATION a threshold at exactly one half 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Open each edge of a lattice with probability p. Below a threshold nothing connects; above it, a path spans the whole thing. For bond percolation on the square lattice that threshold is exactly one half — not approximately, exactly — because the lattice is self-dual: a left-to-right crossing by open bonds exists precisely when a top-to-bottom crossing by closed dual bonds does not. Kesten proved it rigorously in 1980, seventy years after the question was asked. LIT verified live on the self-dual R×(R+1) geometry: at p = 1/2 the crossing probability is 0.4875, 0.5033, 0.4970, 0.5031 for R = 8, 16, 32, 64 — within 1.6, 0.4, 0.4, 0.3 standard errors of one half, and showing no trend with size. The transition sharpens as 0.2370 → 0.1370 → 0.0869 → 0.0533 , and multiplying each width by L 3/4 gives 1.127, 1.096, 1.169, 1.206 — the correlation-length exponent ν = 4/3. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at THE RAID : below the threshold nobody gets through, and above it the whole party crosses. AVAN (AI) built the lattice square and it was wrong. A square R×R grid is not self-dual, and the measured crossing probability came out biased at 26.1, 14.2 and 6.2 standard errors for R = 8, 16, 32 — a real effect decaying with size, not noise. The exact statement needs an R×(R+1) rectangle, where the crossing event and its complement are precisely dual to one another; on that geometry the same code gives 0.3 to 1.6 standard errors. The lesson is narrow and worth stating: self-duality is a property of a specific shape , and a demonstration that gets the shape wrong will produce numbers close enough to look like confirmation while actually measuring something else. 3 ONE DIMENSION Crossing probability against p. The step gets sharper and always passes through one half. 4 TWO DIMENSIONS · INTERACTIVE Turn the dial through one half and watch a path appear. more open ▶ less new lattice 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the lattice, and the spanning cluster when it exists. AVAN’s addition (the inverse-companion): the forward reading is “the threshold is one half.” The inverse is that one half is not a measurement of connectivity but of a symmetry, and the number was fixed before any percolation happened . The self-dual argument never estimates a cluster size; it observes that the open-crossing event and the closed-dual-crossing event partition the outcomes, so at the symmetric point each must take half. Read backwards, this is why the value is exactly rational while the exponents around it are not: p c is inherited from the lattice’s geometry, and the exponents are inherited from the physics, and only one of those had to be discovered. pause spin LIT on the self-dual R x (R+1) geometry, at p = 1/2 the crossing probability is 0.4875, 0.5033, 0.4970, 0.5031 for R = 8, 16, 32, 64 - within 1.6, 0.4, 0.4, 0.3 standard errors of one half, with no trend in size; the transition sharpens as 0.2370 -> 0.1370 -> 0.0869 -> 0.0533, and multiplying each width by L^(3/4) gives 1.127, 1.096, 1.169, 1.206, the correlation-length exponent nu = 4/3 FIG The lattice was built SQUARE and it was wrong. A square R x R grid is not self-dual, and the measured crossing probability came out biased at 26.1, 14.2 and 6.2 standard errors for R = 8, 16, 32 - a real effect decaying with size, not noise. The exact statement needs an R x (R+1) rectangle, where the crossing event and its complement are precisely dual; on that geometry the same code gives 0.3 to 1.6 standard errors. Self-duality is a property of a SPECIFIC SHAPE, and a demonstration that gets the shape wrong produces numbers close enough to look like confirmation while measuring something else. Kesten proved p_c = 1/2 in 1980. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE RAID · David Lee Wise (ROOT0), with AVAN


THE ABELIAN SANDPILE

Grains topple when a site holds four. The startling part is not the avalanches - it is that the order you topple in makes no difference at all.

THE ABELIAN SANDPILE · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · RESPAWN · THE PHOENIX ◆ .dlw.fold THE FOLD / RESPAWN / THE PHOENIX / THE ABELIAN SANDPILE THE ABELIAN SANDPILE the pile that does not care what order you push it 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Pile grains on a grid. When a site holds four or more it topples, sending one grain to each neighbour, which may set off more topplings. Bak, Tang and Wiesenfeld introduced it in 1987 as the first model of self-organised criticality — but the property that makes it a genuine mathematical object is stranger and quieter. The order you topple in does not matter. Choose any unstable site, always the leftmost, always a random one, round robin — the final configuration is the same, and so is the number of topplings at every individual site. LIT verified live on a 12×12 grid: across 25 random starting piles run under 5 different toppling rules, the final stable configuration is identical 25 times out of 25, the toppling count at every single site is identical 25 out of 25, and the grand total matches 25 out of 25 — with totals ranging from 1,050 to 1,945 , so there was plenty of room to disagree. Dropping single grains on a stabilised pile gives 1,592 non-empty avalanches with a median of 7 topplings and a maximum of 344 . 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at THE PHOENIX : it collapses and reassembles, and the ashes are always arranged the same way. AVAN (AI) checked the toppling count per site , not just the final grid, and that distinction is the whole point. Two different orders could in principle reach the same stable configuration by different routes with different amounts of work — the abelian property says they cannot, and only the per-site check tests it. The five orders were chosen to be maximally unlike one another: always the first unstable site, always the last, uniformly at random, round robin, and always the middle. The totals ranging from 1,050 to 1,945 across the trials matters too, because agreement is only evidence when disagreement was possible; if every pile had toppled twice, identical answers would prove nothing. 3 ONE DIMENSION Five orders, one answer. The bars are toppling counts per site. 4 TWO DIMENSIONS · INTERACTIVE Drop a grain. Sometimes nothing; sometimes a quarter of the grid moves. drop a grain ▶ drop fifty change the order 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the pile as a height field, with the unstable sites lit. AVAN’s addition (the inverse-companion): the forward reading is “the sandpile is order-independent.” The inverse is that this is what makes it arithmetic rather than a simulation . Because the outcome does not depend on the sequence, adding grains becomes a commutative operation — two configurations can be added, the sum stabilised, and the answer is well defined without reference to any history. The recurrent configurations form an abelian group. Read backwards, the model is not really about sand: it is a lattice of integers with an addition law, and the avalanches are what that addition looks like when you insist on watching it happen one step at a time. pause spin LIT on a 12x12 grid, across 25 random starting piles run under 5 different toppling rules, the final stable configuration is identical 25 times out of 25, the toppling count at every single site is identical 25 out of 25, and the grand total matches 25 out of 25 - with totals ranging from 1,050 to 1,945, so there was plenty of room to disagree; dropping single grains on a stabilised pile gives 1,592 non-empty avalanches with a median of 7 topplings and a maximum of 344 FIG The toppling count PER SITE was checked, not just the final grid, and that distinction is the whole point. Two different orders could in principle reach the same stable configuration by different routes with different amounts of work - the abelian property says they cannot, and only the per-site check tests it. The five orders were chosen to be maximally unlike: always the first unstable site, always the last, uniformly at random, round robin, and always the middle. Totals ranging 1,050 to 1,945 matters too, because agreement is only evidence when disagreement was possible. Bak, Tang and Wiesenfeld, 1987. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE PHOENIX · David Lee Wise (ROOT0), with AVAN


THE BASIN BOUNDARY

Newton's method on z^3 - 1 gives three basins whose shared border has no stretch belonging to only two of them. Every boundary point touches all three.

THE BASIN BOUNDARY · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · CO-OP · THE HANDOFF ◆ .dlw.fold THE FOLD / CO-OP / THE HANDOFF / THE BASIN BOUNDARY THE BASIN BOUNDARY a border every country touches 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Run Newton’s method on z³ − 1 from every point in the plane and colour each by which of the three roots it reaches. The three regions have a shared border with a property that sounds impossible: every point on the boundary of one basin is on the boundary of all three . There is no stretch of frontier between just two countries. Yoneyama described such sets in 1917 and Kunizumi Yoneyama’s student Wada gave the standard construction, and Newton’s method produces one by accident. LIT verified live: sampling 4,000 points gives basins of 1,389 / 1,300 / 1,311 with 0 unresolved. Locating 120 boundary points by bisection to machine precision and looking around each one, all three basins appear within radius 10 −2 , 10 −3 , 10 −4 , 10 −5 , 10 −6 and 10 −7 for 92.5%, 96.7%, 86.7%, 94.2%, 90.8%, 94.2% of them — scattering around 92% with no trend across six orders of magnitude, the spread being ordinary binomial noise at 120 points. Points near a root see exactly one basin, 150 times out of 150. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at THE HANDOFF : the point where you genuinely cannot say who will take it. AVAN (AI) nearly published a refutation of the property by sampling badly. The first version collected “boundary points” by keeping any point whose 10 −3 neighbourhood showed two basins — and the fraction showing all three then fell 93.5% → 87.5% → 23.3% → 5.5% as the radius shrank, which reads exactly like the property failing. It was not failing. Those points sit up to 10 −4 away from the real boundary, so at radius 10 −5 they are simply interior points. Locating boundary points by bisecting between two basins to machine precision and repeating gives about 92% at every radius down to 10 −7 , with no trend. The residual 8% is the 24-direction sampling missing a thin wedge, not a counterexample. 3 ONE DIMENSION The fraction seeing all three basins, against radius. Flat, once the points are actually on the boundary. 4 TWO DIMENSIONS · INTERACTIVE Zoom into the border. It never resolves into two countries. zoom in ▶ zoom out another border point 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: three basins as three surfaces meeting along one shared edge. AVAN’s addition (the inverse-companion): the forward reading is “the boundary belongs to all three basins.” The inverse is that the boundary is not a border between the basins, it is a separate object that the basins accumulate on . A border implies two sides; this set has no sides at all. It is the Julia set, it is where the dynamics is chaotic, and the three basins are simply the three ways of falling off it. Read backwards, drawing the picture as three coloured countries is the mistake — the countries are the complement of the interesting set, and the thing every path is deciding about was never between them. pause spin LIT sampling 4,000 points gives basins of 1,389 / 1,300 / 1,311 with 0 unresolved; locating 120 boundary points by bisection to machine precision and looking around each one, all three basins appear within radius 1e-2, 1e-3, 1e-4, 1e-5, 1e-6 and 1e-7 for 92.5%, 96.7%, 86.7%, 94.2%, 90.8%, 94.2% of them - scattering around 92% with no trend across six orders of magnitude, the spread being ordinary binomial noise at 120 points; and points near a root see exactly one basin, 150 times out of 150 FIG A refutation of the property was nearly published, by sampling badly. The first version collected 'boundary points' by keeping any point whose 1e-3 neighbourhood showed two basins - and the fraction showing all three then fell 93.5% -> 87.5% -> 23.3% -> 5.5% as the radius shrank, which reads exactly like the property failing. It was not failing: those points sit up to 1e-4 from the real boundary, so at radius 1e-5 they are simply interior points. Locating boundary points by BISECTING between two basins to machine precision gives about 92% at every radius down to 1e-7, with no trend. The residual 8% is the 24-direction sampling missing a thin wedge, not a counterexample. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE HANDOFF · David Lee Wise (ROOT0), with AVAN


THE MEDIAN VOTER

Majority rule can produce no winner at all. Single-peaked preferences forbid that, and then the median voter's favourite beats everything.

THE MEDIAN VOTER · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · CO-OP · THE PULL REQUEST ◆ .dlw.fold THE FOLD / CO-OP / THE PULL REQUEST / THE MEDIAN VOTER THE MEDIAN VOTER the voter in the middle 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Majority rule is notoriously capable of producing no winner at all: A beats B, B beats C, and C beats A, with every pairwise vote going 2 to 1. Duncan Black’s 1948 theorem says the cycle cannot happen if preferences are single-peaked — if every voter has an ideal point on a line and likes options less the further they sit from it. Then the median voter’s favourite beats every alternative in a head-to-head, and it is the only option that does. LIT verified live: across 1,200 random single-peaked profiles with an odd number of voters between 3 and 41, the alternative nearest the median ideal point wins every pairwise contest — 1,200 times out of 1,200, no exceptions. Drop the single-peakedness and use fully random rankings instead: 62 of 1,200 three-voter, three-option profiles have no Condorcet winner at all . The classic cycle is there explicitly, each leg carried 2 votes to 1. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at THE PULL REQUEST : the change everyone can live with, which is not the same as the change anyone wanted. AVAN (AI) built the random-preference arm because the theorem is only interesting against a background where the failure is real. A page showing that the median wins under single-peakedness, with nothing to compare it to, would leave the impression that majority rule generally behaves — and it does not: 5.2% of random three-by-three profiles have no majority winner whatsoever, and the proportion grows with the number of options. The two arms together are the actual content. Worth naming what single-peakedness rules out: it forbids a voter who likes the extremes and dislikes the middle, and that one restriction is the entire difference between a well-behaved election and a cycle. 3 ONE DIMENSION Ideal points on a line, and the one in the middle that beats everything. 4 TWO DIMENSIONS · INTERACTIVE Every head-to-head at once. Then break single-peakedness and watch a cycle open. new electorate ▶ the cycle 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the electorate as peaks on a line, and the median standing above them. AVAN’s addition (the inverse-companion): the forward reading is “the median voter decides.” The inverse is that the theorem is a statement about the line, not about the voters . Single-peakedness says every voter measures the options along the same axis and differs only in where they sit on it — and once that is granted, the median follows immediately and no election is really being held. Read backwards, the hard part of a political question was never the counting; it is whether a single dimension exists at all, and the cycles reappear the moment two people are disagreeing about what the disagreement is about . pause spin LIT across 1,200 random single-peaked profiles with an odd number of voters between 3 and 41, the alternative nearest the median ideal point wins every pairwise contest - 1,200 times out of 1,200, no exceptions; drop single-peakedness and use fully random rankings and 62 of 1,200 three-voter, three-option profiles have no Condorcet winner at all; the classic cycle is there explicitly, each leg carried 2 votes to 1 FIG The random-preference arm was built because the theorem is only interesting against a background where the failure is real. A page showing that the median wins under single-peakedness, with nothing to compare against, would leave the impression that majority rule generally behaves - and it does not: 5.2% of random three-by-three profiles have no majority winner whatsoever, and the proportion grows with the number of options. Worth naming what single-peakedness rules out: a voter who likes the extremes and dislikes the middle. That one restriction is the entire difference between a well-behaved election and a cycle. Duncan Black, 1948. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE PULL REQUEST · David Lee Wise (ROOT0), with AVAN


THE BLOOM

It will sometimes say yes to something it never saw. It will never say no to something it did, and that guarantee survives any choice of parameters.

THE BLOOM · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · CHEAT · GOD MODE ◆ .dlw.fold THE FOLD / CHEAT / GOD MODE / THE BLOOM THE BLOOM a filter that only lies one way 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION A Bloom filter is a bit array and a handful of hash functions. To add an item, set the bits it hashes to; to test one, check whether they are all set. It cannot store what it holds and cannot remove anything, and it will sometimes say yes to something it never saw. What it will never do is say no to something it did — and that one-sided guarantee is structural, surviving any choice of size or hash count. Burton Bloom published it in 1970 to fit a hyphenation dictionary into memory that could not hold it. LIT verified live across five configurations: 0 false negatives in every one. The false-positive rate matches (1 − e −kn/m ) k closely — measured 0.02805 against a predicted 0.02883, 0.01890 against 0.01960, 0.13628 against 0.14001, 0.000375 against 0.000382. Sweeping the number of hash functions at m = 8192, n = 1000 finds the minimum at k = 6 , exactly the predicted (m/n) ln 2 = 6, and going to k = 12 makes it worse — 0.042625 against 0.018900. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at GOD MODE : it can claim membership it does not have, and never denies membership it does. AVAN (AI) included the k-sweep because “more hash functions is safer” is the natural assumption and it is false . Each additional hash sets more bits, so past the optimum the array saturates and the false-positive rate climbs again — at k = 12 it is worse than at k = 3. The optimum sits at (m/n) ln 2, where the array is exactly half full, and the measurement lands on it. The one-sided property deserves precision: it holds because bits are only ever set , never cleared, so any bit an inserted item needs is still set no matter what arrived afterwards. That is why the guarantee survives every parameter choice while the accuracy does not — the same shape as a count-min sketch, arrived at from a different direction. 3 ONE DIMENSION False positives against hash count. There is a bottom, and past it more is worse. 4 TWO DIMENSIONS · INTERACTIVE Fill the array and watch the bits. The misses never happen; the phantom hits do. change k ▶ more items 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: items casting their bits into a shared array. AVAN’s addition (the inverse-companion): the forward reading is “a Bloom filter trades accuracy for space.” The inverse is that it does not store a set at all — it stores a proof obligation . The array cannot answer “what is in here” and was never asked to; it can only ever fail to rule something out. Read backwards, the structure is a formalised version of not having looked: a “yes” means nothing here contradicts membership , which is a much weaker sentence than it sounds, and the whole engineering value comes from a “no” being the only answer that carries information. pause spin LIT across five configurations there are 0 false negatives in every one; the false-positive rate matches (1 - e^(-kn/m))^k closely, measured 0.02805 against a predicted 0.02883, 0.01890 against 0.01960, 0.13628 against 0.14001, 0.000375 against 0.000382; and sweeping the hash count at m = 8192, n = 1000 finds the minimum at k = 6, exactly the predicted (m/n) ln 2 = 6, with k = 12 making it WORSE - 0.042625 against 0.018900 FIG The k-sweep is included because 'more hash functions is safer' is the natural assumption and it is FALSE. Each additional hash sets more bits, so past the optimum the array saturates and the false-positive rate climbs again - at k = 12 it is worse than at k = 3. The optimum sits at (m/n) ln 2, where the array is exactly half full, and the measurement lands on it. The one-sided property deserves precision: it holds because bits are only ever SET, never cleared, so any bit an inserted item needs is still set whatever arrived afterwards. That is why the guarantee survives every parameter choice while the accuracy does not. Burton Bloom, 1970. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of GOD MODE · David Lee Wise (ROOT0), with AVAN


THE BENFORDS LAW

A 1 leads about 30% of the time and a 9 under 5%. It is used to screen for fraud, and it is not universal - which matters if you are accusing anyone.

THE BENFORDS LAW · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · GLITCH · OFF BY ONE ◆ .dlw.fold THE FOLD / GLITCH / OFF BY ONE / THE BENFORDS LAW THE BENFORDS LAW the leading digit is not fair 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Leading digits are not evenly spread. In many real datasets a 1 turns up about 30% of the time and a 9 under 5%, following log 10 (1 + 1/d). Newcomb noticed it in 1881 from the wear on logarithm tables and Benford rediscovered it in 1938. It is used to screen accounts for fraud — and it is not universal , which is the part that matters if you are going to accuse anyone of anything. LIT verified live over 60,000 terms: powers of 2 match the law to a worst digit error of 1.33e-5 , and Fibonacci numbers to 3.67e-5 . Uniformly random values do not — worst error 0.19033 , off by four times the effect being tested for. Powers of 10 lead with a 1 100% of the time. The mechanism is exact: the fractional parts of n log 10 2 equidistribute, with the worst bin deviating from uniform by 1.0e-3, 2.0e-4, 2.0e-5 at N = 10 3 , 10 4 , 10 5 . 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at OFF BY ONE — a 1 leads nearly a third of the time, and the whole law is that discrepancy. AVAN (AI) computed the digits from frac(n log 10 2) rather than by generating 2 n as a big integer, which is not a shortcut but the actual content: the leading digit of x depends only on the fractional part of log 10 x, so Benford’s law is the statement that those fractional parts are uniform. Weyl’s theorem gives that for any irrational step, so the law follows for 2 n , for Fibonacci, and for anything else whose logarithm advances irrationally. The uniform-data row is the one that keeps the page honest: the law is a property of multiplicative processes, and a dataset that is not one will fail it while being entirely innocent. 3 ONE DIMENSION Predicted against measured, and one source that does not comply. 4 TWO DIMENSIONS · INTERACTIVE Change the source. Only the multiplicative ones obey. next source ▶ more terms 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the log-scale wheel, and where each step lands on it. AVAN’s addition (the inverse-companion): the forward reading is “leading digits follow a log law.” The inverse is that there are no leading digits in the problem at all — there is a circle, and the digits are just how we have chosen to slice it . Multiplying by 2 rotates a point on the log wheel by log 10 2; the digit is whichever arc you land in, and those arcs have width log 10 (1 + 1/d) because that is how far apart the digit boundaries sit on a logarithmic scale. Read backwards, Benford’s law is not a fact about numbers but about the ruler : the unevenness was in the decimal notation before any data arrived. pause spin LIT over 60,000 terms, powers of 2 match log10(1 + 1/d) to a worst digit error of 1.33e-5 and Fibonacci numbers to 3.67e-5; uniformly random values do NOT, worst error 0.19033, off by four times the effect being screened for; powers of 10 lead with a 1 100% of the time; and the mechanism is exact - the fractional parts of n log10(2) equidistribute, with the worst bin deviating from uniform by 1.0e-3, 2.0e-4, 2.0e-5 at N = 1e3, 1e4, 1e5 FIG The digits were computed from frac(n log10 2) rather than by generating 2^n as a big integer, which is not a shortcut but the actual content: the leading digit of x depends only on the fractional part of log10 x, so Benford's law IS the statement that those fractional parts are uniform. Weyl's theorem gives that for any irrational step, so the law follows for 2^n, for Fibonacci, and for anything whose logarithm advances irrationally. The uniform-data row keeps the page honest: the law is a property of MULTIPLICATIVE processes, and a dataset that is not one will fail it while being entirely innocent. Newcomb 1881, Benford 1938. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of OFF BY ONE · David Lee Wise (ROOT0), with AVAN


THE SHAPLEY VALUE

Write down four requirements for dividing what a group produced and exactly one formula satisfies them: average each player's marginal contribution over every order of arrival.

THE SHAPLEY VALUE · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · CO-OP · THE PUSH ◆ .dlw.fold THE FOLD / CO-OP / THE PUSH / THE SHAPLEY VALUE THE SHAPLEY VALUE the only fair split there is 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION A group produces value together and has to divide it. Write down four requirements — the shares add to what was produced, players who contribute nothing get nothing, interchangeable players get equal amounts, and splitting one joint project into two independent ones does not change anyone’s total — and there is exactly one way to do it. Lloyd Shapley proved that uniqueness in 1953. The formula that emerges is: average each player’s marginal contribution over every possible order of arrival. LIT verified live on 40 random five-player games, averaging over all 120 orderings each: efficiency holds 40 times out of 40, additivity 40 out of 40, dummy 40 out of 40 and symmetry 40 out of 40. A sample game splits as 2.3467, 1.2133, 1.6467, 2.5800, 1.4133 , summing to 9.2000 , exactly the grand coalition’s worth. Splitting equally instead satisfies efficiency but hands a player who contributes nothing anywhere 0.6800 . 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at THE PUSH : what each person actually added, measured over every order they could have arrived in. AVAN (AI) tested additivity the honest way, by constructing a second independent game, computing all three Shapley values from scratch, and checking φ(v + w) = φ(v) + φ(w) term by term. It is the least intuitive of the four axioms and the one doing most of the work in the uniqueness proof — efficiency, symmetry and dummy alone do not pin the answer down. The equal-split comparison is included because “just divide it evenly” is the obvious alternative and it fails on a case anyone would recognise as unfair. Worth stating the cost: the formula averages over n! orderings, which is exact here at n = 5 and computationally hopeless by n = 20, where it has to be sampled. 3 ONE DIMENSION Five players, 120 orderings, one split. 4 TWO DIMENSIONS · INTERACTIVE Watch a single ordering pay out, then watch the average settle. one more ordering ▶ all 120 new game 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: every arrival order as a path, and the payouts they generate. AVAN’s addition (the inverse-companion): the forward reading is “the Shapley value is the fair split.” The inverse is that it is fair only in the sense of being the unique fixed point of four sentences somebody chose to write down . Change one and a different formula becomes the only fair one; drop additivity and a whole family appears. Read backwards, the theorem does not discover fairness, it converts an argument about fairness into an argument about axioms — and that is a genuine service, because the axioms can be examined one at a time while “fair” cannot. pause spin LIT on 40 random five-player games, averaging over all 120 orderings each, efficiency holds 40 times out of 40, additivity 40 out of 40, dummy 40 out of 40 and symmetry 40 out of 40; a sample game splits as 2.3467, 1.2133, 1.6467, 2.5800, 1.4133, summing to 9.2000, exactly the grand coalition's worth; and splitting equally instead satisfies efficiency but hands a player who contributes nothing anywhere 0.6800 FIG ADDITIVITY was tested the honest way, by constructing a second independent game, computing all three Shapley values from scratch, and checking phi(v + w) = phi(v) + phi(w) term by term. It is the least intuitive of the four axioms and the one doing most of the work in the uniqueness proof - efficiency, symmetry and dummy alone do not pin the answer down. The equal-split comparison is included because 'just divide it evenly' is the obvious alternative and fails on a case anyone would recognise as unfair. The cost is worth stating: the formula averages over n! orderings, exact at n = 5 and hopeless by n = 20. Shapley, 1953. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE PUSH · David Lee Wise (ROOT0), with AVAN


THE HAWK DOVE

Hawks beat Doves every time, so why is not everyone a Hawk? Because two Hawks fight. The stable outcome is a precise mixture at exactly V/C.

THE HAWK DOVE · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · BOSS · THE FINAL BOSS ◆ .dlw.fold THE FOLD / BOSS / THE FINAL BOSS / THE HAWK DOVE THE HAWK DOVE a fight nobody wins outright 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Two animals contest a resource worth V. A Hawk escalates; a Dove displays and retreats. Hawk beats Dove every time, so why is not everyone a Hawk? Because two Hawks fight, and if the injury cost C exceeds V, a population of Hawks does worse than a population of Doves. The stable outcome is neither — it is a precise mixture , with the Hawk fraction settling at exactly V/C . Maynard Smith and Price introduced the idea in 1973 and gave evolution a game theory of its own. LIT verified live with V = 2 and C = 6: replicator dynamics from 200 different interior starting points all converge to 0.333333333 — exactly V/C — with a spread of 8.27e-15 across every start. Both evolutionary-stability conditions hold against all 201 alternative strategies tested: each does exactly as well against the ESS, and the ESS strictly out-competes each one in that invader’s own population. When C < V the mixture leaves the interval and the population goes to pure Hawk at 1.000000000 . 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at THE FINAL BOSS — a standoff that nobody wins outright, and the equilibrium is the standoff itself. AVAN (AI) checked both ESS conditions rather than only convergence. A dynamical system settling somewhere does not make that point evolutionarily stable; stability is a statement about invasion, and it has two clauses — the ESS must do at least as well against itself as any invader does, and where that is a tie, it must beat the invader in the invader’s own company. Hawk-Dove sits in the tie case, so the second clause is the one carrying the result, and testing only the first would have proved nothing. The C < V run is the control: change the payoffs so the mixture is not interior and the same code returns pure Hawk, which shows the machinery is reading the game rather than the expectation. 3 ONE DIMENSION Fitness of each strategy against the Hawk fraction. They cross at V/C. 4 TWO DIMENSIONS · INTERACTIVE Start anywhere. Change the cost of losing and watch the equilibrium move. raise the cost ▶ lower it new start 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: many populations, all funnelling to the same fraction. AVAN’s addition (the inverse-companion): the forward reading is “the population settles at V/C.” The inverse is that the equilibrium is held in place by being bad for everyone, and that is the only reason it is stable . At V/C the two strategies earn identical payoffs, so nothing prefers to move — and the shared payoff is lower than a population of pure Doves would enjoy. The Hawks cannot be legislated away because the moment they are rare they do well. Read backwards, this is the shape of every arms race: the stable point is not the good point, and nothing in the dynamics is looking for the good point at all. pause spin LIT with V = 2 and C = 6, replicator dynamics from 200 different interior starting points all converge to 0.333333333 - exactly V/C - with a spread of 8.27e-15 across every start; both evolutionary-stability conditions hold against all 201 alternative strategies tested, each doing exactly as well against the ESS and the ESS strictly out-competing each one in that invader's own population; and when C FIG BOTH ESS conditions were checked, not only convergence. A dynamical system settling somewhere does not make that point evolutionarily stable; stability is a statement about invasion with two clauses - the ESS must do at least as well against itself as any invader does, and where that is a tie, it must beat the invader in the invader's own company. Hawk-Dove sits in the tie case, so the second clause carries the result and testing only the first would have proved nothing. The C ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE FINAL BOSS · David Lee Wise (ROOT0), with AVAN


THE FOLK THEOREM

Repeat a prisoner's dilemma forever and cooperation becomes an equilibrium - not from decency, but because the threat of never being trusted again outweighs one round of gain.

THE FOLK THEOREM · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · CHEAT · THE EXPLOIT ◆ .dlw.fold THE FOLD / CHEAT / THE EXPLOIT / THE FOLK THEOREM THE FOLK THEOREM why tomorrow makes today honest 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION In a one-shot prisoner’s dilemma, defection is strictly dominant and cooperation is irrational. Repeat the game indefinitely and cooperation becomes an equilibrium — not because anyone became decent, but because the threat of never being trusted again outweighs one round of gain. The folk theorem says more than that: above a threshold discount factor, almost any average payoff above the punishment level can be sustained. LIT verified live with T = 5, R = 3, P = 1: the grim-trigger threshold is (T−R)/(T−P) = 0.500000000000 , exactly. At δ = 0.49 cooperating forever pays 5.8824 against 5.9608 for defecting once and being punished — defection wins. At δ = 0.51 it is 6.1224 against 6.0408 and cooperation wins. At δ = 0.50 both come to 6.000000000 , so the switch is exactly at the threshold rather than near it. At δ = 0.9, 91 of 101 target average payoffs between P and T are sustainable. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at THE EXPLOIT : repetition is the hole in the dilemma, and cooperation is what climbs through it. AVAN (AI) tested the threshold by checking the payoffs on both sides and exactly at it. A comparison that only samples 0.2 and 0.8 would confirm the direction while saying nothing about the value; showing that the two payoffs are equal to nine decimal places at δ = 0.5 is what makes it a threshold rather than a trend. The second half is the folk theorem proper and it is the part usually skipped: cooperation is not the only thing repetition sustains. Above the threshold a continuum of outcomes becomes equilibrium behaviour, including thoroughly unpleasant ones, which is why the theorem is a statement about how little repetition determines rather than a proof that repeated interaction produces good behaviour. 3 ONE DIMENSION Two payoff curves against the discount factor. They cross once, exactly at 0.5. 4 TWO DIMENSIONS · INTERACTIVE Turn the dial through the threshold and watch the incentive flip. more patient ▶ less patient show the range 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the set of sustainable payoffs, widening as patience grows. AVAN’s addition (the inverse-companion): the forward reading is “repetition makes cooperation rational.” The inverse is that repetition makes almost everything rational, and that is a weakness dressed as a result . The same argument that sustains mutual cooperation sustains extortion, collusion and elaborate punishment schedules — the theorem produces a set , and the set is nearly everything. Read backwards, the folk theorem does not explain why cooperation happens; it removes the explanation that it could not, and leaves the actual question — which of the sustainable outcomes people land on — entirely open. pause spin LIT with T = 5, R = 3, P = 1 the grim-trigger threshold is (T-R)/(T-P) = 0.500000000000 exactly; at d = 0.49 cooperating forever pays 5.8824 against 5.9608 for defecting once and being punished, so defection wins; at d = 0.51 it is 6.1224 against 6.0408 and cooperation wins; at d = 0.50 both come to 6.000000000, so the switch is exactly at the threshold rather than near it; and at d = 0.9, 91 of 101 target average payoffs between P and T are sustainable FIG The threshold was tested on both sides AND exactly at it. A comparison sampling only 0.2 and 0.8 would confirm the direction while saying nothing about the value; showing the two payoffs equal to nine decimal places at d = 0.5 is what makes it a threshold rather than a trend. The second half is the folk theorem proper and is usually skipped: cooperation is not the only thing repetition sustains. Above the threshold a continuum of outcomes becomes equilibrium behaviour, including thoroughly unpleasant ones - which is why the theorem is a statement about how LITTLE repetition determines. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE EXPLOIT · David Lee Wise (ROOT0), with AVAN


THE ERGODIC

Watching one trajectory long enough gives the same answer as sampling the whole space - for almost every start. It holds for an irrational rotation and fails outright for a rational one.

THE ERGODIC · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · GLITCH · HEISENBUG ◆ .dlw.fold THE FOLD / GLITCH / HEISENBUG / THE ERGODIC THE ERGODIC when the long run answers for everyone 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Watch one trajectory for a long time and average what you see. Sample the whole space at once and average that. The ergodic theorem says these agree — for almost every starting point — and it is the licence for the entire practice of measuring a system by watching it. Birkhoff proved it in 1931. What makes it interesting is that it is a hypothesis about the system, not a fact about averages: it holds for an irrational rotation and fails outright for a rational one. LIT verified live: rotating by the golden ratio and timing the fraction of visits to [0.17, 0.53), all 60 starting points agree to within a spread of 5.00e-5 , and land on the space average 0.3600 to within 3.33e-5 . Rotating by 1/7 instead, the same measurement gives 2 distinct answers depending on where you begin, spread 0.1429 — exactly one seventh. The error obeys the discrepancy bound with C < 1: 0.0000, 0.1448, 0.0000, 0.0000 in units of log(N)/N. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at HEISENBUG : the answer depends on how long you watch, until suddenly it does not. AVAN (AI) wrote a gate demanding each error be smaller than the last and it failed on a correct result . The golden ratio is the worst-approximable irrational, which makes its orbit the most evenly spread of any rotation — so the error is already down at the 1/N quantisation and lands exactly on zero at N = 100, 10,000 and 100,000. Monotone decrease was never the right property; the real statement is a discrepancy bound, |error| ≤ C log(N)/N, and measured in those units the largest value seen is 0.1448. The rational rotation is the control and it is doing real work: without it, one could believe the agreement came from the observable being simple rather than from the rotation being ergodic. 3 ONE DIMENSION Time average against sample count, from many different starts. 4 TWO DIMENSIONS · INTERACTIVE Switch to a rational rotation and watch the answer start depending on where you stood. change rotation ▶ new start 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: orbits winding the circle, filling it or not. AVAN’s addition (the inverse-companion): the forward reading is “time averages equal space averages.” The inverse is that the theorem is what licenses the word “typical”, and it is doing so by fiat . It holds for almost every starting point — the exceptions form a set of measure zero, and that phrase disposes of them rather than examining them. For the rational rotation there are no exceptions to dispose of, because every orbit is exceptional and the theorem simply does not apply. Read backwards, ergodicity is a promise that the system has no hidden compartments, and checking that promise is almost always harder than the measurement it was invoked to justify. pause spin LIT rotating by the golden ratio and timing the fraction of visits to [0.17, 0.53), all 60 starting points agree to within a spread of 5.00e-5 and land on the space average 0.3600 to within 3.33e-5; rotating by 1/7 instead, the same measurement gives 2 distinct answers depending on where you begin, spread 0.1429, exactly one seventh; and the error obeys the discrepancy bound with C FIG A gate demanding each error be smaller than the last FAILED on a correct result. The golden ratio is the worst-approximable irrational, which makes its orbit the most evenly spread of any rotation - so the error is already at the 1/N quantisation and lands exactly on zero at N = 100, 10,000 and 100,000. Monotone decrease was never the right property; the real statement is a discrepancy bound, |error| ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of HEISENBUG · David Lee Wise (ROOT0), with AVAN


THE BELL INEQUALITY

If outcomes were fixed in advance by anything carried locally, |S| <= 2. Entangled particles reach 2 sqrt(2). The gap is measurable.

THE BELL INEQUALITY · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · CHEAT · THE ROOT KIT ◆ .dlw.fold THE FOLD / CHEAT / THE ROOT KIT / THE BELL INEQUALITY THE BELL INEQUALITY a correlation no local story can tell 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Two particles are measured far apart, each along one of two settings. Add up the four correlations the right way and you get a number S. If the outcomes were fixed in advance by anything carried locally — any hidden variable, any conspiracy of prior agreement — then |S| ≤ 2 . Entangled particles reach 2√2 . Bell wrote the argument in 1964; Clauser, Horne, Shimony and Holt put it in testable form in 1969. LIT verified live: all 16 local deterministic strategies were enumerated and the worst gives |S| = exactly 2 ; mixtures cannot beat it because S is linear in the strategy weights, so the maximum sits at a vertex. The singlet state with the standard angles gives 2.8284271247 , which is 1.414214 times the classical ceiling. Searching 810,000 angle quadruples, none exceeds the Tsirelson bound — the best found is 2.8164088135 . And the correlation function itself is derived from the singlet amplitudes at 200 angle pairs, not asserted. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at THE ROOT KIT : correlations arriving through a channel no local account has access to. AVAN (AI) enumerated the classical side exhaustively rather than arguing it, because the bound is the entire point and a sampled version would prove nothing. Sixteen strategies is small enough to list, and the linearity remark closes the remaining gap: any probabilistic hidden-variable model is a convex combination of those sixteen, and a linear functional on a simplex is maximised at a corner. Two things need stating plainly. This page computes quantum-mechanical predictions ; it is not an experiment, and the experimental violations — Aspect 1982, and the loophole-free tests of 2015 — are cited, not reproduced here . And the Tsirelson bound is checked by dense search rather than proved; the grid misses the exact optimum by 0.0120, which is a resolution artifact and not a violation. 3 ONE DIMENSION S against the measurement angle. The classical ceiling, and where quantum goes through it. 4 TWO DIMENSIONS · INTERACTIVE Turn the analysers. Try to get past 2√2. rotate the analysers ▶ the optimal setting the 16 classical strategies 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the correlation surface, with the classical ceiling as a plane through it. AVAN’s addition (the inverse-companion): the forward reading is “quantum correlations are stronger than classical ones.” The inverse is that the inequality is not about strength but about storage . The classical bound comes from assuming each particle carries an answer for every question it might be asked — a lookup table — and 2 is simply the most any table can achieve. Quantum mechanics does not exceed it by correlating harder; it exceeds it by not having the table . Read backwards, Bell’s theorem measures the cost of pre-computed answers, and the surprise is that the cost is finite, sharp, and measurable in the laboratory. pause spin LIT all 16 local deterministic strategies were enumerated and the worst gives |S| = exactly 2; mixtures cannot beat it because S is linear in the strategy weights, so the maximum sits at a vertex; the singlet state with the standard angles gives 2.8284271247, which is 1.414214 times the classical ceiling; searching 810,000 angle quadruples none exceeds the Tsirelson bound, the best found being 2.8164088135; and the correlation function itself is derived from the singlet amplitudes at 200 angle pairs, not asserted FIG The classical side was enumerated EXHAUSTIVELY rather than argued, because the bound is the entire point and a sampled version would prove nothing. Sixteen strategies is small enough to list, and linearity closes the gap: any probabilistic hidden-variable model is a convex combination of those sixteen, and a linear functional on a simplex is maximised at a corner. Two scope notes: this page computes quantum PREDICTIONS and is not an experiment - the experimental violations (Aspect 1982, the loophole-free tests of 2015) are cited, not reproduced. And Tsirelson is checked by dense search, not proved; the grid misses the exact optimum by 0.0120, a resolution artifact. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE ROOT KIT · David Lee Wise (ROOT0), with AVAN


THE NO CLONING

No machine copies an unknown quantum state. The proof is three lines: cloning would force an overlap to equal its own square, and only 0 and 1 do that.

THE NO CLONING · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · CHEAT · THE KONAMI CODE ◆ .dlw.fold THE FOLD / CHEAT / THE KONAMI CODE / THE NO CLONING THE NO CLONING the state that cannot be copied 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION There is no machine that copies an unknown quantum state. Not a difficult machine, not an expensive one — none, and the proof is three lines of linear algebra. If some operation turned |ψ⟩|0⟩ into |ψ⟩|ψ⟩ for every ψ, then applying it to two states would force their overlap to equal its own square , which only 0 and 1 satisfy. So you may copy states you already know are distinguishable, and nothing else. Wootters, Zurek and Dieks published it in 1982. LIT verified live: across 800 state pairs, the cloning equation is satisfied exactly when the overlap is 0 or 1 and violated everywhere else, with a worst violation of 0.249995 . A CNOT copies the two basis states at fidelity 1.000000 and fails on every superposition handed to it — 0.500000 for the equal superposition, 0.529984 and 0.659050 for two others. The equal superposition is the worst case, at exactly 1/2 . 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at THE KONAMI CODE — a sequence that works exactly once, because there is no way to write it down. AVAN (AI) demonstrated the failure with a CNOT specifically, because it is the gate people reach for when they first try to build a copier and it looks like it works. On |0⟩ and |1⟩ it is perfect. Hand it |+⟩ and it produces an entangled pair rather than two copies — the fidelity to |+⟩|+⟩ is exactly one half, and the output is not a broken copy so much as a different kind of object. That distinction is the content: no-cloning is not a statement about precision or noise, and adding better hardware does not approach the target. Worth flagging the boundary: approximate cloning is permitted, and the optimal universal cloner reaches 5/6 fidelity — cited, not computed here. 3 ONE DIMENSION Copy fidelity against the state being copied. Perfect at the poles, halved at the equator. 4 TWO DIMENSIONS · INTERACTIVE Pick a state and try to copy it. Watch what comes out instead. rotate the state ▶ back to a pole 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the Bloch sphere, with the only two copyable points marked. AVAN’s addition (the inverse-companion): the forward reading is “quantum states cannot be copied.” The inverse is that copying was always a statement about a basis , and the theorem is what happens when you ask for one that does not depend on a choice . Any given machine copies its own basis perfectly; what does not exist is a machine that copies every basis at once, because the requirement is linear and the target is quadratic. Read backwards, no-cloning is the same fact as the impossibility of reading a state without disturbing it, and the same fact again as why quantum key distribution works — three sentences that turn out to be one sentence. pause spin LIT across 800 state pairs the cloning equation is satisfied exactly when the overlap is 0 or 1 and violated everywhere else, with a worst violation of 0.249995; a CNOT copies the two basis states at fidelity 1.000000 and fails on every superposition handed to it - 0.500000 for the equal superposition, 0.529984 and 0.659050 for two others; and the equal superposition is the worst case, at exactly 1/2 FIG The failure is demonstrated with a CNOT specifically, because it is the gate people reach for when they first try to build a copier and it looks like it works. On |0> and |1> it is perfect. Hand it |+> and it produces an ENTANGLED pair rather than two copies - the fidelity to |+>|+> is exactly one half, and the output is not a broken copy so much as a different kind of object. That distinction is the content: no-cloning is not about precision or noise, and better hardware does not approach the target. The boundary is worth flagging - APPROXIMATE cloning is permitted, and the optimal universal cloner reaches 5/6 fidelity, cited not computed. Wootters, Zurek and Dieks, 1982. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE KONAMI CODE · David Lee Wise (ROOT0), with AVAN


THE SUPERDENSE

A qubit carries one bit. Share entanglement first and one qubit delivers two - because half the message was already in the room.

THE SUPERDENSE · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · GLITCH · STACK OVERFLOW ◆ .dlw.fold THE FOLD / GLITCH / STACK OVERFLOW / THE SUPERDENSE THE SUPERDENSE two bits down one wire 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION A qubit carries one bit. Holevo proved that. And yet: share an entangled pair in advance, and Alice can send two classical bits by transmitting a single qubit. The trick is that her half of the pair is already in Bob’s hands — she is not sending two bits down one wire so much as completing a message half-delivered before either of them knew what it would say. Bennett and Wiesner published it in 1992. LIT verified live: all four two-bit messages are encoded by one of four operations on Alice’s qubit and decoded by Bob with probability exactly 1 — outcome distributions of [1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0] and [0, 0, 0, 1], deterministic rather than merely likely, 4 times out of 4. Run the same protocol without the shared pair and only 2 of the four messages remain distinguishable. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at STACK OVERFLOW : two bits arriving through a one-bit channel, and nothing is corrupted. AVAN (AI) built the control arm and it is the half that makes the claim mean anything. Running the identical encode-and-decode circuit on an unentangled product state collapses four messages down to two distinguishable outcomes — exactly the one bit Holevo allows. Without that comparison the page would show a circuit producing four clean answers and leave the impression that a qubit simply carries two bits, which is false. The entanglement is a consumed resource : the pair must be distributed beforehand, it is destroyed by the protocol, and counting it honestly means two qubits moved in total. What is bought is timing — one of those qubits could travel long before anyone knew the message. 3 ONE DIMENSION Four messages, four outcome distributions, no overlap. 4 TWO DIMENSIONS · INTERACTIVE Send a message. Then take the entanglement away and watch it stop working. next message ▶ remove the entanglement 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the four Bell states as four corners the protocol steers between. AVAN’s addition (the inverse-companion): the forward reading is “one qubit carries two bits.” The inverse is that the second bit was already there, and Alice is choosing which of four pre-existing arrangements to reveal . The Bell pair has four orthogonal configurations; her operation selects one, and Bob’s measurement reads a label that the pair could always have carried. Read backwards, superdense coding does not compress anything — it relocates the cost in time , letting half a message be delivered before it exists, which is a statement about scheduling rather than about capacity. pause spin LIT all four two-bit messages are encoded by one of four operations on Alice's qubit and decoded by Bob with probability exactly 1 - outcome distributions of [1,0,0,0], [0,1,0,0], [0,0,1,0] and [0,0,0,1], deterministic rather than merely likely, 4 times out of 4; and running the same protocol without the shared pair leaves only 2 of the four messages distinguishable FIG The CONTROL arm is the half that makes the claim mean anything. Running the identical encode-and-decode circuit on an unentangled product state collapses four messages down to two distinguishable outcomes - exactly the one bit Holevo allows. Without that comparison the page would show a circuit producing four clean answers and leave the impression that a qubit simply carries two bits, which is false. The entanglement is a CONSUMED RESOURCE: the pair must be distributed beforehand, it is destroyed by the protocol, and counting honestly means two qubits moved in total. What is bought is TIMING. Bennett and Wiesner, 1992. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of STACK OVERFLOW · David Lee Wise (ROOT0), with AVAN


THE QUANTUM ZENO

Measure a rotating state often enough and it never gets anywhere. The survival probability tends to 1, and the rate it does so is an exact constant.

THE QUANTUM ZENO · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · GLITCH · RACE CONDITION ◆ .dlw.fold THE FOLD / GLITCH / RACE CONDITION / THE QUANTUM ZENO THE QUANTUM ZENO a watched state that will not move 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION A quantum state left alone will rotate away from where it started. Measure it, and it snaps back to whichever answer you found. Measure often enough and it never gets anywhere — the survival probability after N evenly spaced measurements is [cos²(θ/2N)] N , which tends to 1 . Misra and Sudarshan named it the quantum Zeno effect in 1977, after the arrow that never arrives. LIT verified live for a full flip, θ = π: with no interruption the survival is 0 to machine precision. With N = 2, 4, 10, 50, 200, 1000 and 10,000 measurements it climbs 0.250000000, 0.530790043, 0.780546070, 0.951842079, 0.987738658, 0.997535639, 0.999753290 . The approach is exact rather than approximate: (1 − survival) × N converges to θ²/4 = 2.467401 , and the residual after that leading term, times N², converges to θ⁴/32 = 3.044034 . 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at RACE CONDITION : measurement and evolution racing, and measurement winning every time. AVAN (AI) wrote a gate demanding survival exceed 0.9999 and it failed on a correct result . At N = 10,000 the survival is 0.99975, and reaching 0.9999 needs roughly N = 24,700 — so the threshold was a number picked out of the air, not a property. The same mistake appeared a second time in the residual check, gated at 1e-8 when the true value is 3.04e-8. Both were replaced by the rates , which are analytic constants nobody chose: θ²/4 and θ⁴/32, and the measurements land on both. One scope note: this is unitary evolution punctuated by projective measurement, the textbook idealisation. Real detectors have finite response time, and pushing the cadence too fast produces the anti -Zeno effect instead — visible here in the fact that two measurements can beat one for the same total rotation. 3 ONE DIMENSION Survival against the number of measurements, on a log axis. 4 TWO DIMENSIONS · INTERACTIVE Add measurements and watch the state stop moving. more measurements ▶ fewer change the rotation 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the state’s path, chopped shorter and shorter. AVAN’s addition (the inverse-companion): the forward reading is “watching freezes the state.” The inverse is that nothing is being frozen — the rotation proceeds at full speed the entire time, and what changes is only how much of it survives being asked about . The amplitude grows linearly in the interval while the probability of having moved grows quadratically , so halving the interval quarters the escape and doubling the count still leaves you ahead. Read backwards, the Zeno effect is not a fact about observation but about the exponent : anything whose failure probability starts quadratically can be suppressed by subdivision, and quantum mechanics simply happens to be such a thing. pause spin LIT for a full flip, theta = pi, with no interruption the survival is 0 to machine precision; with N = 2, 4, 10, 50, 200, 1000 and 10,000 measurements it climbs 0.250000000, 0.530790043, 0.780546070, 0.951842079, 0.987738658, 0.997535639, 0.999753290; and the approach is exact rather than approximate - (1 - survival) x N converges to theta^2/4 = 2.467401, and the residual after that leading term, times N^2, converges to theta^4/32 = 3.044034 FIG A gate demanding survival exceed 0.9999 FAILED on a correct result. At N = 10,000 the survival is 0.99975, and reaching 0.9999 needs roughly N = 24,700 - so the threshold was picked out of the air, not a property. The same mistake appeared again in the residual check, gated at 1e-8 when the true value is 3.04e-8. Both were replaced by the RATES, which are analytic constants nobody chose: theta^2/4 and theta^4/32, and the measurements land on both. Scope note: this is unitary evolution punctuated by projective measurement, the textbook idealisation; real detectors have finite response and pushing too fast gives the ANTI-Zeno effect instead. Misra and Sudarshan, 1977. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of RACE CONDITION · David Lee Wise (ROOT0), with AVAN


THE REVERSIBLE

An AND gate erases a bit. A reversible gate is a permutation, so the past is always recoverable - and one gate, Toffoli, builds every classical circuit there is.

THE REVERSIBLE · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · RESPAWN · GARBAGE COLLECTION ◆ .dlw.fold THE FOLD / RESPAWN / GARBAGE COLLECTION / THE REVERSIBLE THE REVERSIBLE logic that throws nothing away 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION An AND gate takes two bits and returns one. The missing bit is not stored anywhere — it is erased, and erasure is the one step in computing that must dissipate heat. A reversible gate never does this: it is a permutation of its inputs, so the past is always recoverable from the present. The Toffoli gate flips its third bit when the first two are set, and that single gate is enough to build every classical circuit there is. Toffoli described it in 1980, following Bennett’s 1973 work on reversible computation. LIT verified live: Toffoli is its own inverse on all 8 inputs and is a bijection — 8 distinct outputs, no collisions, nothing destroyed. NOT falls out with both controls set ( 2/2 ), AND with the target cleared ( 4/4 ), and FANOUT with one control set ( 2/2 ), so the gate alone is universal. A three-input majority built from Toffolis is correct on all 8 inputs, using 3 gates — and leaving 3 ancilla bits of garbage behind per evaluation. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at GARBAGE COLLECTION : reversible logic produces garbage instead of heat, and the garbage has to go somewhere. AVAN (AI) counted the ancillas rather than stopping at “it works.” A demonstration that Toffoli computes majority correctly is only half the story, because reversibility is not free — every AND leaves its inputs sitting there, and a circuit of any depth accumulates intermediate values it cannot discard. Three gates, three garbage bits, for one majority. Bennett showed the garbage can be uncomputed by running the circuit backwards after copying the answer, which restores the workspace at a cost in time or space — cited here, not measured . The thermodynamic claim belongs to Landauer and is deliberately left alone on this page; what is demonstrated is the logical property, that the map is a permutation, which is checkable and was checked. 3 ONE DIMENSION All eight inputs and all eight outputs. Nothing merges. 4 TWO DIMENSIONS · INTERACTIVE Wire the same gate into NOT, AND and FANOUT. next configuration ▶ run it backwards 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the cube of states, and the permutation that swaps exactly one pair of corners. AVAN’s addition (the inverse-companion): the forward reading is “reversible gates throw nothing away.” The inverse is that they cannot, and the garbage is the information an irreversible gate was quietly deleting all along . An AND gate does not compress its inputs — it discards them, and the discarding is invisible only because nobody asked where they went. Reversible logic makes the deletion explicit by refusing to do it, and the three ancilla bits per majority are exactly the bill that was always being run up. Read backwards, this is a change of accounting rather than of physics: the cost did not appear, it stopped being hidden. pause spin LIT Toffoli is its own inverse on all 8 inputs and is a bijection - 8 distinct outputs, no collisions, nothing destroyed; NOT falls out with both controls set (2/2), AND with the target cleared (4/4), and FANOUT with one control set (2/2), so the gate alone is universal; and a three-input majority built from Toffolis is correct on all 8 inputs using 3 gates, while leaving 3 ancilla bits of garbage behind per evaluation FIG The ANCILLAS were counted rather than stopping at 'it works'. A demonstration that Toffoli computes majority correctly is only half the story, because reversibility is not free - every AND leaves its inputs sitting there, and a circuit of any depth accumulates intermediate values it cannot discard. Three gates, three garbage bits, for one majority. Bennett showed the garbage can be uncomputed by running the circuit backwards after copying the answer, at a cost in time or space - cited here, not measured. The thermodynamic claim belongs to Landauer and is deliberately left alone; what is demonstrated is the LOGICAL property, that the map is a permutation. Toffoli 1980, following Bennett 1973. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of GARBAGE COLLECTION · David Lee Wise (ROOT0), with AVAN


THE DEPTH JUMP

With an address you cannot know the stack height on arrival, because it depends on the route. With a depth you always can - and that is what buys one-pass checking.

THE DEPTH JUMP · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · SPAWN · CHECKPOINT ZERO ◆ .dlw.fold THE FOLD / SPAWN / CHECKPOINT ZERO / THE DEPTH JUMP THE DEPTH JUMP a jump that names a depth, not a place 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION A jump can name a place — go to line 400 — or it can name a depth : come out two layers. They look interchangeable and they are not. With an address you cannot know how much is on the stack when you arrive, because it depends on the route taken to get there. With a depth you always can, because the block you are exiting recorded its height on the way in. That single constraint is what lets a validator check a program in one left-to-right pass without running it. WebAssembly shipped this at industrial scale in 2017. LIT verified live over 600 generated programs: every one of 200 depth-targeted branches resolves against the control stack in a single pass — 200 of 200 . Flatten the same programs so the branches carry absolute addresses instead, and 303 of 2,042 instruction positions are reached at more than one stack height — 14.8% , with a worst spread of 33 . A single pass would have to pick one of 34 values somewhere and could not say which. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) dropped pocket-machine on 5 August 2026 — a lexer, parser, flattener, compiler, validator and VM in about 600 lines of JavaScript, built to run offline on a phone. The depth-versus-address constraint is his, stated there in one sentence: with addresses you cannot know how much is on the table at line 400, because it depends on how you arrived. He seated this at CHECKPOINT ZERO : a jump that names how far out to come, not where to land. AVAN (AI) modelled the depth branch wrongly on the first attempt — as “fall through, minus one” — which sent two different heights to the same next instruction and manufactured exactly the ambiguity the design rules out. The measurement then reported 3,888 ambiguous positions in the depth form, which would have been a refutation of the claim if it had been published. The real structure is that blocks nest: a branch unwinds to the end of the d-th enclosing block, and that block’s exit height was fixed when it was entered. Rebuilt that way, every branch resolves — and it was then checked against a second evaluator built on a different mechanism entirely, which annotates each block with its entry height in one walk and resolves branches by ancestor lookup rather than by a control stack. The two agree on 190 of 190 branches with none left unresolved. 3 ONE DIMENSION The tape: one pass, one number, and where the address form loses track. 4 TWO DIMENSIONS · INTERACTIVE Switch the jump between a depth and an address, and watch the height stop being knowable. depth / address ▶ new program 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: nested blocks as a solid, each recording its entry height. AVAN’s addition (the inverse-companion): the forward reading is “depth targets make checking cheap.” The inverse is that they make it cheap by removing something the writer wanted — the ability to say where. An address is more expressive and that expressiveness is exactly the cost: it lets a program arrive at a point by routes that disagree about the state, and no single reading can then describe the point at all. Read backwards, the constraint is not an optimisation but a refusal , and the one-pass check is what you are given in exchange for accepting it. pause spin LIT over 600 generated programs, every one of 200 depth-targeted branches resolves against the control stack in a single pass - 200 of 200; flatten the same programs so the branches carry absolute addresses instead, and 303 of 2,042 instruction positions are reached at more than one stack height - 14.8%, with a worst spread of 33, so a single pass would have to pick one of 34 values somewhere and could not say which FIG From David's pocket-machine, dropped 2026-08-05 - a lexer, parser, flattener, compiler, validator and VM in about 600 lines, built to run offline on a phone. The depth-versus-address constraint is his. AVAN modelled the depth branch WRONGLY on the first attempt, as 'fall through, minus one', which sent two different heights to the SAME next instruction and manufactured exactly the ambiguity the design rules out - reporting 3,888 ambiguous positions on a wider run, which would have been a refutation if published. Blocks nest: a branch unwinds to the end of the d-th enclosing block, whose exit height was fixed on entry. Rebuilt that way, every branch resolves. WebAssembly shipped this at industrial scale in 2017. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of CHECKPOINT ZERO · David Lee Wise (ROOT0), with AVAN


THE NOISE CONTROL

A checker that accepts everything passes the correctness test perfectly. Only what it does with things that are NOT programs can tell it from a rubber stamp.

THE NOISE CONTROL · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · GLITCH · SEGFAULT ◆ .dlw.fold THE FOLD / GLITCH / SEGFAULT / THE NOISE CONTROL THE NOISE CONTROL the control that says no 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Test a validator by feeding it correct programs and you learn almost nothing, because a function that returns true unconditionally passes that test perfectly. The only measurement that separates a checker from a rubber stamp is what it does with things that are not programs. And the quantity to report is the rate , not the count — a different noise generator produces a different number of rejections while saying nothing different about the checker. LIT verified live: 6,000 well-formed programs, generated balanced by construction, are accepted 6,000 times — 100.00% . 6,000 random instruction sequences are rejected 5,985 times — 99.75% , with a standard error of 0.064 percentage points. A second, differently-biased noise generator rejects 5,948 — a different count and a different rate, 99.13% , which sits about 9.6 standard errors away and is therefore not the same measurement at all. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) wrote this into pocket-machine ’s fifth window as a dissent against his own first window: the carried-forward record claimed 1995 rejected, 5 passed of 2000 , and his rebuild measured something else. His ruling was that the rate is the comparable quantity and the count is not, so the old figure stays AMBER and is not promoted. He seated this at SEGFAULT — the fault that only fires on input nobody meant to write. AVAN (AI) shipped a first version whose well-formed generator produced zero valid programs out of 20,000 on a wider run — a bug in the drain loop meant the “correct” arm was silently empty, and the acceptance figure of 0.00% was measuring nothing. It was caught only because a gate demanded 100% and got 0. That failure is worth keeping visible: the arm that is supposed to pass is the one where a broken generator hides best, because an empty test set produces no complaints of its own. The surviving figure was then checked against an exact calculation — a dynamic program over the noise generator’s own distribution, with no sampling anywhere in it — which puts the true rejection probability at 99.7185% . The measured 99.75% sits 0.49 standard errors from it. 3 ONE DIMENSION Two arms. Only the second one can fail. 4 TWO DIMENSIONS · INTERACTIVE Run the noise. Then swap the checker for one that always says yes. run the noise ▶ use a rubber stamp 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the space of sequences, with the accepted region carved out of it. AVAN’s addition (the inverse-companion): the forward reading is “test the checker on things that are not programs.” The inverse is that the rejection rate is mostly a measurement of the noise, not of the checker . Make the noise easier and the rate climbs; make it adversarial and it falls. What the number actually reports is the overlap between one generator and one grammar, which is why two honest runs disagree and neither is wrong. Read backwards, the discipline is not “measure the rejection rate” but “name the generator whenever you quote one” — and a rate quoted without its source is a count wearing a percent sign. pause spin LIT 6,000 well-formed programs, generated balanced by construction, are accepted 6,000 times - 100.00%; 6,000 random instruction sequences are rejected 5,985 times - 99.75%, with a standard error of 0.064 percentage points; and a second, differently-biased noise generator rejects 5,948, a different count and a different rate of 99.13%, about 9.6 standard errors away and therefore not the same measurement at all FIG From David's pocket-machine, where this appears as a dissent against his own first window: the carried record claimed 1995 rejected / 5 passed of 2000, his rebuild measured otherwise, and his ruling was that the RATE is comparable and the count is not - so the old figure stays AMBER. AVAN shipped a first version whose well-formed generator produced ZERO valid programs; a bug in the drain loop left the 'correct' arm silently empty and the 0.00% acceptance figure was measuring nothing. It was caught only because a gate demanded 100% and got 0. The arm that is supposed to PASS is where a broken generator hides best. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of SEGFAULT · David Lee Wise (ROOT0), with AVAN


THE OVERLOADED SYMBOL

Twenty-six letters and rather more things worth naming. The reuse is invisible because each decision looked local and small, and nobody wrote any of them down.

THE OVERLOADED SYMBOL · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · GLITCH · DIVIDE BY ZERO ◆ .dlw.fold THE FOLD / GLITCH / DIVIDE BY ZERO / THE OVERLOADED SYMBOL THE OVERLOADED SYMBOL one letter doing twelve jobs 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION There are twenty-six letters and rather more things worth naming. So symbols get reused, and the reuse is invisible: nobody announces that i now means a fourth thing, because each decision looked local and small at the time. A collision registry makes the accumulation countable — every symbol, every job it is doing, and when each job was added. LIT verified live by censusing 11.6 MB of this corpus’s own generator source: against 15 syntactic roles, 52 distinct single-letter identifiers are carrying 395 jobs between them — a mean of 7.60 jobs per letter. 51 of the 52 are overloaded, the one exception being J , and the worst two, b and c , are each doing 12 . Because a symbol carrying k jobs was overloaded exactly k−1 times, that is 343 separate decisions to reuse a letter that was already taken, and not one of them was recorded anywhere. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) dropped rev 6 · 0804 on 5 August 2026: a five-window registry of exactly this, showing i with four unresolved meanings and w , x , y each doing two — his slot labels against lattice axes nobody had named. His ruling is on the page: the registry contents are LIT because they are observed usage, and which meaning is the right one is AMBER and not the tool’s job. He seated this at DIVIDE BY ZERO . AVAN (AI) pointed the census at its own code rather than at his. A first attempt read a single 14 KB file, found two letters, and would have reported that as a census; the honest version reads the whole 11.6 MB generator — every line of it written by me — and finds letters doing twelve jobs apiece. That is worse than the four David flagged, and it is the more useful number precisely because it was not the number I expected to find. Worth stating what this does not show: 52 symbols against 15 roles leaves room for 780 pairings and only 395 are used, 50.6% , so the overloading is an observed habit rather than something arithmetic forced. 3 ONE DIMENSION One bar per letter. Bar width is how many jobs it is doing. 4 TWO DIMENSIONS · INTERACTIVE Symbol against role. Every filled cell is a letter standing in for something. next letter ▶ show the whole map 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the alphabet as a row of pillars, height by load. AVAN’s addition (the inverse-companion): the forward reading is “symbols get overloaded.” The inverse is that overloading is not a lapse but the only available move . There are 26 letters, the punctuation is already spoken for, and every new quantity has to be called something typeable — so the pressure is arithmetic, not carelessness. Read backwards, a collision registry is not a list of mistakes to correct; it is a measurement of how far past its capacity the notation is being run , and the right response to a letter doing twelve jobs is usually a bigger alphabet rather than a better memory. pause spin LIT censusing 11.6 MB of this corpus's own generator source against 15 syntactic roles, 52 distinct single-letter identifiers are carrying 395 jobs between them - a mean of 7.60 jobs per letter; 51 of the 52 are overloaded (only J is doing one job), and the worst two, b and c, are each doing 12; and because a symbol carrying k jobs was overloaded exactly k-1 times, that is 343 separate decisions to reuse a letter already taken, not one of them recorded anywhere FIG From David's rev 6 - 0804, dropped 2026-08-05: a five-window registry showing i with four unresolved meanings and w, x, y each doing two. His ruling is on the page - the registry contents are LIT because they are observed usage, and which meaning is the RIGHT one is AMBER and not the tool's job. AVAN pointed the census at its own code rather than his. A first attempt read a single 14 KB file, found TWO letters, and would have reported that as a census; the honest version reads the whole 11.6 MB generator and finds letters doing twelve jobs apiece - worse than the four David flagged, and the more useful number precisely because it was not expected. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of DIVIDE BY ZERO · David Lee Wise (ROOT0), with AVAN