SHIFTS — a digit shift is a change of numeration
The object: the trailing Ostrowski numeration at an irrational α = [0; a₁, a₂, …] (OSTROWSKI), with denominators q_k, numerators p_k and remainders θ_k = q_k α − p_k, and the maps linear in its digits. The shift by stride r ≥ 1 sends n = Σ d_k q_k, the greedy string, to σ_r(n) = Σ d_k q_(k+r). The drop sends n to Σ_(k≥1) d_k q_(k−1), and ×m sends n to mn. A reader at lookahead c commits the output's depth-t tile having seen the input's depth-(t + c) tile, as in READING, and t₀ is the first depth with q_t ≥ 2. [a₁, …, a_P] is the purely periodic α whose quotients repeat that block, golden, silver and bronze are [1], [2] and [3], Euler's α is e − 2 = [0; 1, 2, 1, 1, 4, 1, 1, 6, …], e Euler's number, and the one-class numeration (P, a), a ≥ 2, is [0; (1^(P−1), a)^∞], the quotient a at every P-th position and 1 elsewhere. This page shows that a shift is a change of numeration: it takes α's digits as the digits of α_r = [0; a_(r+1), a_(r+2), …] and multiplies by −θ_(r−1), so it is a map of the circle R/Z with jumps at the cuts, the points −jα, j ≥ 1, the first q_t of which bound the depth-t tiles. Every map of the integers whose image point is a sum of the digits against weights that alternate in sign from some position on, with Σ cap_k |w_k| finite, cap_k the largest digit allowed at position k (OSTROWSKI: The circle numeration), the shifts, ×m and (where its sum converges) the drop among them, has its walls where The landing criterion puts them, which for the affine maps is the set OSTROWSKI: The containment criterion gives. Where some non-cut's image point is a cut, it reads at no lookahead from the depth where the shallowest such cut becomes a tile boundary, which makes it DISCONTINUOUS, and at a finite lookahead at every shallower depth. Whether a stride reads is then a fact about α's caps set against α_r's remainders, whether some non-cut's image point is a cut; the image carries the factor −θ_(r−1), and whether it is a unit never enters.
Where the unit went. At a purely periodic α the shift by the period and multiplication by the unit −θ_(P−1), whose norm is ±1, are one map, so a quadratic numeration cannot tell which of the two makes the period shift read. The change of numeration separates them. The shift by the period reads at lookahead 0 because α_P = α, and any stride whose rotated caps never fall below the original ones (The copy reader) copies its string the same way: every stride of [0; 2, 4, 8, …], and the strides ≡ 0 mod 3 of Euler's α, where no unit exists. The strides ≡ 1 mod 3 of Euler's α are DISCONTINUOUS for the same reason as the strides with r mod P odd of a one-class periodic numeration: a comb of digits set against the rotated numeration's caps codes a non-cut whose image point is a cut. The strides r ≥ 2, r ≡ 2 mod 3, of Euler's α read at lookahead 0 though their rotated caps fall: the one digit over its new cap spills a pair of digits, one up and one down, and the carry that follows never chains. Where the spill fills a cap over a nonzero, as at the class quotient 2, borrows from the digit below fix it under the borrow code's condition (S′), and the lookahead is then at most max(0, 3 − r). So the period shift reads by the copy reader's hypothesis, not because it is a unit's action, and that hypothesis holds where no unit exists.
MAP WEIGHTS w_k READS
n + ω, ω ≥ 0 θ_k (and ωα) LIPSCHITZ (OSTROWSKI)
mn, m ≥ 2 m θ_k DISCONTINUOUS: no lookahead
from t₀
σ_r, caps not falling θ_(k+r) lookahead 0 at every depth
σ_r, r ≡ 2 mod 3, θ_(k+r) lookahead 0 at every depth
class caps under (S)
σ_r, r ≡ 2 mod 3, θ_(k+r) lookahead ≤ max(0, 3 − r),
class caps under (S′) never MIDDLE
σ_r, one-class, θ_(k+r) DISCONTINUOUS at r mod P odd,
[0; (1^(P−1), a)^∞] LIPSCHITZ at r mod P even ≠ 0
the drop θ_(k−1) LIPSCHITZ or DISCONTINUOUS as
σ_(P−1) is, at purely periodic
α; DISCONTINUOUS where
the quotients eventually ascend
at every other step, strictly
infinitely often, and
Σ a_(k+1)/q_k converges, or
where a_(D+1)|θ_(D−1)| > 1
infinitely often
the drop, class caps θ_(k−1) LIPSCHITZ, lookahead ≤ 5 from
under (S′) at ℓ = −1 the depth with q_(t−1) ≥ b₁
the drop, class caps θ_(k−1) LIPSCHITZ, (S′) or not,
2 … 9 lookahead ≤ 21 at every depth
(rule: every class sequence)
the drop, class caps θ_(k−1) LIPSCHITZ, lookahead
2 … M ≤ 3M + 11 at every depth, and
≥ (3M + 8)/2 at depth 2 at one
sequence, M even, M ≥ 4
(theorem)
any of these, purely LIPSCHITZ or DISCONTINUOUS
periodic α
(S) and (S′) are the conditions of the block code and the borrow code
below. The script's counts on this page are those of shifts.py --full;
its default run checks smaller ranges.
The change of numeration
Tier: theorem.
Verifier: proof; shifts.py::section_change.
For every r ≥ 1 and k ≥ −1, θ_(k+r) = −θ_(r−1) θ⁽ʳ⁾_k, where θ⁽ʳ⁾, q⁽ʳ⁾ and p⁽ʳ⁾ are the remainders, denominators and numerators of α_r. In integers, q_(k+r) = q⁽ʳ⁾_k q_r + p⁽ʳ⁾_k q_(r−1), and p likewise. So σ_r(n)α ≡ −θ_(r−1) Σ d_k θ⁽ʳ⁾_k mod 1: the shift takes the string at α_r and multiplies by −θ_(r−1). The drop is the stride −1: its point is Σ_(k≥0) d_k θ_(k−1) mod 1, since θ₋₁ = −1 is an integer. The image point g_r(x) = Σ d_k(x) θ_(k+r) is a map of the circle, continuous at every non-cut and with two one-sided values at each cut.
Proof. Both sides obey u_(k+1) = a_(k+r+1) u_k + u_(k−1). At k = −1 they agree since θ⁽ʳ⁾₋₁ = −1, and at k = 0 since −θ_r/θ_(r−1) = α_r, the standard ratio of consecutive remainders. At r ≥ 0, a_(k+1) |θ_(k+r)| ≤ a_(k+1)/q_(k+r+1) ≤ 1/q_k, so the sum converges uniformly in the string, and a non-cut, whose coding is unique and continuous in the point, has a continuous image. The drop's sum converges wherever Σ a_(k+1)/q_k does: at every α with bounded quotients, and at Euler's α. A cut has two codings (OSTROWSKI: The circle numeration) and so two values, one from each side.
The script checks the integer identities with none off at eight α, r ≤ 8, 0 ≤ k ≤ 40, and the term bound at r = 0 … 8, 0 ≤ k ≤ 40 at seven of them, all but the one-class (4, 3) numeration.
The landing criterion
Tier: criterion.
Verifier: proof; shifts.py::section_controls;
shifts.py::section_change; shifts.py::section_criterion.
Let f map the integers n ≥ 0 to themselves with f(n)α ≡ Σ d_k(n) w_k mod 1, where sign(w_k) = ε sign(θ_k) for one ε at every large k and Σ cap_k |w_k| converges, and let g(x) = Σ d_k(x) w_k. A non-cut x lands on a cut −sα when g(x) ≡ −sα. At a depth t with q_t ≥ 2, f reads at some lookahead iff no non-cut lands on a cut −jα with j ≤ q_t. So f reads at no lookahead exactly from the first depth t with q_t ≥ max(s, 2), where −sα is the shallowest cut a non-cut lands on: the depth where that cut's two codings part. The walls of f are g⁻¹(B) \ B, B the cuts, as for the affine maps, though g need not be continuous or injective.
Proof. Among the legal tails from a position k past which sign(w_i) = ε sign(θ_i), where no nonzero digit at k − 1 bars the cap at k, Σ d_i w_i is largest at the tail with the caps where ε θ_i > 0 and zeros elsewhere, and nowhere else. It is smallest at the other parity's. Where a nonzero digit at k − 1 bars the cap at k, the extreme that uses position k holds it one below its cap instead, the second form of OSTROWSKI: The circle numeration. Each extreme is legal, since a capped digit sits over a zero. These two are the cap-fillings, the tails of a cut's two codings. A non-cut's string is never eventually a cap-filling, so at every depth it has legal tails above and below its own. Truncated far out they are integers agreeing with x to that depth, with images on both sides of g(x). If g(x) is a cut of depth t, some −jα with j ≤ q_t, they lie in two output tiles, so no input tile about x commits one. Conversely, if f reads at no lookahead at depth t, the input tiles that fail shrink to a point x, every neighbourhood of it on one side holding integers in two output tiles. Were x a cut, the points on one side of it follow that side's coding for long, so their images all sit on one side of g's value there by the extremality above, in one tile. So x is a non-cut, g is continuous at x, and g(x) is a cut of depth t. The shifts and ×m have the weights θ_(k+r) and mθ_k, and the drop has θ_(k−1) wherever Σ a_(k+1)/q_k converges.
Over every legal tail of length 7 at four α and 72 alternating weight vectors, with the first digit free and with it held below its cap, the script finds the cap-filling alone at the maximum. Its exhaustive reader, run over n below two ranges, sets a depth t's least lookahead against each range's ceiling T − t + 1 there, T the range's depth, the largest T with q_T at most the range. Over 68 cells (five periodic α at the drop and strides 1 … 3, six one-class ones at the drop and strides 1 … 2P; 64 distinct, since [1, 2] is the one-class (2, 2) and its four cells are counted twice) it labels a depth TRACKING, climbing with the range as it does where no lookahead exists, when its least lookahead comes within 2 of the ceiling at both ranges or, at five or more depths short of the smaller range's depth, rises between the ranges, and by at least one less than their depths differ by; BOUNDED when it is the same at both ranges and at least 4 below the smaller range's ceiling; and otherwise leaves the depth undecided. The 691 depths it decides over the 68 cells all agree with the landing walk's prediction, TRACKING from its first depth at no lookahead and BOUNDED before it, and 85 are left undecided. Wherever the reader decides the depths around a predicted first depth at no lookahead, it begins to track exactly there: the one-class (4, 3) numeration at r = 1, 3, 5, 7 from t = 2, 3, 5, 7, the shallowest landed cuts being −α, −3α, −14α and −39α. At the one-class (4, 2) numeration it leaves every depth from the predicted one on undecided at the drop and at r = 1, 3, 5, 7, its least lookahead the same at both ranges and one below the smaller range's ceiling.
The copy reader
Tier: theorem.
Verifier: proof; shifts.py::section_copy.
If a_(k+1) ≤ a_(k+r+1) at every k ≥ 1 and a₁ ≤ a_(r+1) + 1, then σ_r(n)'s greedy string is n's shifted by r, at every n, so σ_r reads at lookahead 0 at every depth. The hypothesis holds at the period's multiples of every purely periodic α, at every stride of an α whose quotients never fall, and at the strides ≡ 0 mod 3 of Euler's α.
Proof. The shifted digit d_k sits at position k + r under the cap a_(k+r+1) ≥ a_(k+1) ≥ d_k (d₀ ≤ a₁ − 1 ≤ a_(r+1)). If d_k reaches that cap then a_(k+1) = a_(k+r+1) and d_k was capped before, so d_(k−1) = 0 and a capped digit sits over a zero; below position r every digit is 0. A legal string is the greedy one.
The script finds σ_r(n)'s string equal to n's shifted for every n < 20000 in 20 cases: golden and [0; 2, 4, 8, …] at r = 1 … 6, Euler's α at r = 3, 6, and three one-class numerations at r = P and 2P.
The landing walk
Tier: theorem.
Verifier: proof; shifts.py::section_controls;
shifts.py::section_criterion; shifts.py::section_uniform.
At a purely periodic α, whether a map of the landing criterion with weights m θ_(k+r) is DISCONTINUOUS, and from which depth it reads at no lookahead, is decided by a finite graph, a walk in the frame of FLUSH: The lift game with a single player, who chooses the string's digits. A string's sum Σ m d_j θ_(j+r) is λ = κ − sα, κ an integer, exactly when D_k = λ − Σ_(j<k) m d_j θ_(j+r), the value it still owes, stays at every k between the least and greatest legal tail sums from k, ends included. Written D_k = x θ_(k+r) + y θ_(k+r−1), (x, y) is a pair of integers and a step is X = y, Y = x − m d_k − a_(k+r+1) y. The map is DISCONTINUOUS iff the graph has a cycle through a state that is live (between the tail extremes, ends included), interior (strictly between them) and not (0, 0).
Proof. The frame repeats with the period. The purely periodic α is a quadratic irrational; α′ is its Galois conjugate, the other root of its quadratic, and a prime mark ′ on any quantity built from α and integers replaces α by α′, so θ′_j = q_j α′ − p_j. On the conjugate side, u = x ζ′_(k+r) + y, with ζ′_j = θ′_j/θ′_(j−1) > 1, steps as u ↦ u/ζ′ − md, a contraction, so on a cycle u is its fixed point and lies in [−U, 0], U = mAζ′_min/(ζ′_min − 1) with A the largest quotient, and u = 0 only at (0, 0). The real side bounds xζ + y, ζ = θ_(k+r)/θ_(k+r−1). Together they bound x and y, so every cycle lies in a finite box. A state equal to a tail extreme has one live successor, the cap-filling's, so a cycle is all extreme, a cut's coding, or all interior, a non-cut. The value's α-coefficient v falls along a walk by m d_k q_(k+r). A cycle placed ever deeper has |v| growing with its depth, which a falling v allows only on the negative side, so the landing is on a cut. Along any path u converges geometrically into [−U, 0], so an infinite interior path stays in the box widened by any slack from some depth on, and there it closes a cycle; a cycle placed after zeros codes a non-cut that lands, a landing string. The depth comes from the least s whose λ reaches an interior cycle.
The walk finds ×1 never DISCONTINUOUS and ×2, ×3 reading at no lookahead from t₀, landing on −α, at golden, silver and [1, 2], as OSTROWSKI: The affine maps requires. The drop reads at golden (its least lookahead 2, 1, 2, 1, … from depth 2), silver, bronze and the three odd-P one-class cells, and reads at no lookahead from t = 2 at the three even-P ones, [1, 2] = (2, 2) among them.
The uniform lookahead
Tier: theorem.
Verifier: proof; shifts.py::section_uniform.
At a purely periodic α, a map of the landing walk that is not DISCONTINUOUS reads at one lookahead at every depth: it is LIPSCHITZ, never MIDDLE (READING: The Lipschitz criterion). If the nonzero interior states of the box the proof builds carry a longest path of L steps, the lookahead at depth t is at most max(0, L + 1 + max(k_s − t, 1 − r)), with k_s = max(0, −r) the walk's first step. Off periodicity, wherever the sums converge, the same walk, started from every cut of depth t, decides that one depth exactly whenever every walk dies; at the class numerations with bounded class quotients The drop bounds the drop's walk at every depth at once.
Proof. The depth-t cuts are −jα, 1 ≤ j ≤ q_t. Lookahead c fails at depth t exactly when some input prefix of length t + c holds a depth-t cut strictly inside the hull of its tails' images: integers approach both ends of the hull, so they sit on both sides of the cut, in two output tiles, and two images in two tiles have a cut between them. An extreme state's one live successor is extreme, so a failing walk is interior at every step, and the lookahead at depth t is max(0, K + 1 − t), K the last step at which a walk from some κ − jα, j ≤ q_t, is interior. No such walk meets (0, 0), which owes 0: the value it owes has α-coefficient at most −j < 0. On the conjugate side u_k is λ′/θ′_(k+r−1) plus the past digits' term, which lies in [−U, 0] as in the landing walk, whose proof defines the primed (conjugate) quantities. Here λ′ = λ + j(α − α′) and |θ′_i| = q_i(α − α′) − θ_i, and once q_(k+r−1) ≥ q_t ≥ max(j, 2) the first term lies in [−1 − G, G⁺], with G = (H₀ + 1)/(2(α − α′) − 1) and G⁺ = max(0, H₀ − (α − α′))/(α − α′ − 1), H₀ the largest |Σ_(i≥k_s) m d_i θ_(i+r)| over the legal tails from the first step, which bounds |λ|. λ′ can be negative, a start below −j(α − α′) (at the drop of [0; (1, 1, 8, 2)^∞] a live start has λ′ ≈ −0.88); then the term is at most (H₀ − j(α − α′))/(j(α − α′) − 1) ≤ G⁺, since α′ < −1, α being the reciprocal of a reduced quadratic, makes α − α′ > 1. With the real side's bound, every interior state of such a walk lies in one finite box per residue of the step mod P from step max(k_s, t − r + 1) on. Its nonzero interior states carry no cycle, since one would make the map DISCONTINUOUS, so a failing walk spends at most L steps there.
The script walks every start with j ≤ 5000 at once. As positive controls it reproduces the exhaustive reader's lookahead columns for the drop, which at golden reads 0, 2, 1, 2, 1, … from depth 1 and at the one-class (3, 2) reads 4, 3, 2 repeating from depth 2, and at the one-class (3, 2) and (4, 2) at r = 1 some walk survives exactly from t = 2, the first depth at no lookahead the landing criterion predicts. Every interior state met past the box's step lies in the box. Over 221 cells that are not DISCONTINUOUS, 3194 depths, the box graph is acyclic and no depth passes the bound; the largest lookahead in any column is 10, the drop at the one-class (9, 2). Of the columns the script prints, which run to the depths with q_t ≤ 5000, whether every column repeats with the period is an open question. The exhaustive reader never reads above the walk and equals it at all 3133 depths where its range suffices.
The combs
Tier: theorem.
Verifier: proof; shifts.py::section_combs; shifts.py::section_e2;
shifts.py::section_uniform.
At the one-class numeration [0; (1^(P−1), a)^∞], a ≥ 2, every stride with r mod P odd is DISCONTINUOUS. At even P the comb of 1s at the odd positions lands on the cut −q_r α, and at odd P the comb with 1 at the positions 2, 4, …, P − 3 of each period and 2 at its class slot P − 1 lands on −q_(r+1) α, so the stride reads at no lookahead from the first depth with q_t ≥ max(q_r, 2), or max(q_(r+1), 2), at the latest. At every α whose quotients are 1 off one residue class mod 3 and at least 2 on it, every stride r ≡ 1 mod 3 is DISCONTINUOUS, at no lookahead from the first depth with q_t ≥ max(q_(c₀+r−1), 2) at the latest: the comb of 2s at the class positions c₀, c₀ + 3, …, c₀ the first where a 2 is legal, lands on −q_(c₀+r−1) α. Euler's α is one, with c₀ = 1.
Proof. At even P the quotient a_(2i+1) is 1, so θ_(2i) = θ_(2i+1) − θ_(2i−1), and the comb's image Σ θ_(k+r) over odd k, all at even indices, telescopes to −θ_r ≡ −q_r α. The comb is legal, since each 1 sits over a 0, and it holds 1 at the class slots, whose cap is a, so it is no cap-filling: a non-cut. At odd P write ξ = MP + r, M ≥ 0 a count of periods. The 1s give Σ θ_(ξ+2i) = θ_(ξ+P−2) − θ_(ξ+1), since each a_(ξ+2i+1) = 1 (with ρ = r mod P odd, ρ + 2i + 1 is even and lies in [ρ + 3, ρ + P − 2], below 2P and never P), and 2θ_(ξ+P−1) = θ_(ξ+P+1) − θ_(ξ+P−2), since a_(ξ+P) = a_(ξ+P+1) = 1. A period gives θ_(ξ+P+1) − θ_(ξ+1), and the periods telescope to −θ_(r+1). Two zeros open each period, so this comb is no cap-filling either. At the mod-3 numerations 2θ_j = θ_(j+2) − θ_(j−1) wherever a_(j+1) = a_(j+2) = 1, which r ≡ 1 mod 3 gives at every j = c + r with c a class position. Consecutive teeth cancel and the sum is −θ_(c₀+r−1). The comb has two zeros between teeth, so it is no cap-filling either. The landing criterion finishes both.
At Euler's α the exhaustive reader tracks from t = 2, 4, 7 at r = 1, 4, 7, the depths where −α, −7α and −71α become tile boundaries, as the combs require; its two ranges share the depth 13 there, and the reading climbs between them at every tracking depth (9 to 12 at r = 1, t = 2). The walk from every cut of depth t finds, at r = 1, 4, some walk surviving from depths 2 and 4 respectively, as the combs prove, and at r = 4 every walk dying at depths 2 and 3, so those depths read at a finite lookahead; and it finds that r = 2, 5, 8 read at lookahead 0 at every depth t ≤ 10, which the block code below proves at every depth; r = 3, 6 read 0 at every depth the reader reaches, and at every depth by the copy reader.
The block code
Tier: criterion.
Verifier: proof; shifts.py::section_block; shifts.py::section_sharp.
Let a_j = 1 off the class indices J_i = 3i + ψ, ψ ∈ {−1, 0, 1} the phase, with class quotients b_i = a_(J_i) ≥ 2 (Euler's α is ψ = −1, b_i = 2i), and let r = 3ℓ + 2. Write B_i for the input cap of the digit the shift puts in block i's third slot: b_(i−ℓ), except b₀ − 1 at ψ = 1, where the first class digit is digit 0, and B_i = 0 where that slot gets no class digit. The block code below is σ_r(n)'s greedy string at every n iff at every block i with B_(i−1), B_(i−2) ≥ 2
(S) ⌊B_i/2⌋ + max(1, ⌊(B_(i−1) − 1)/2⌋) < b_i,
and the blocks below those pass a finite enumeration of their legal inputs. (S) at every block, those below included, is sufficient on its own. Then output digit j reads the input at most at j + 2 − r ≤ j, so σ_r reads at lookahead 0 at every depth. Euler's α satisfies (S) at every block at every ℓ ≥ 1, and at ℓ = 0 at every block but the first, which the enumeration passes; every constant class quotient b ≥ 3, even b included, satisfies it at every block and every ℓ; b = 2 alone fails.
Proof. Cut the output into blocks (J_i − 1, J_i, J_i + 1), caps (b_i, 1, 1). At every phase the shift sends the class digit at J_h − 1 to the third slot of block h + ℓ and J_h, J_h + 1 to the first two of block h + ℓ + 1, so the shifted string holds x_i, y_i ∈ {0, 1} and e_i ≤ B_i there, and the phase moves only the bottom. Since a_(J_i+1) = a_(J_i+2) = 1, 2q_(J_i+1) = q_(J_i+2) + q_(J_i−1): spill ⌊e_i/2⌋ pairs, one digit up a block and one down, leaving β_i = e_i mod 2, and let X_i = x_i + ⌊e_i/2⌋ + ⌊e_(i−1)/2⌋. Then carry once, c_(i+1) = y_i β_i by q_(J_i) + q_(J_i+1) = q_(J_i+2), set Z_i = X_i + c_i, and output (Z_i, 0, 0) if y_i = β_i = 1, (Z_i − 1, 0, 1) if y_i = 1, β_i = 0 and Z_i ≥ 1, (0, 1, 0) if y_i = 1 and Z_i = 0, and (Z_i, 0, β_i) if y_i = 0. Each case keeps the value. The carry reads y_i and β_i and never Z_i, so no carry chains, and block i's output reads the shifted string on J_i − 3 … J_i + 1 alone. The output is legal, so greedy, iff its first digit o_i is at most b_i, strictly when block i − 1 ends in a nonzero. Legality of the input ties the slots: x_i = 1 forces e_(i−1) = 0, and c_i = 1 forces x_i = 0 and e_(i−1) odd below its cap. Over the legal inputs, o_i is at most ⌊B_i/2⌋ + max(1, ⌊B_(i−1)/2⌋) in general, ⌊B_i/2⌋ + ⌊B_(i−1)/2⌋ through a carry, and ⌊B_i/2⌋ + max(1, ⌊(B_(i−1) − 1)/2⌋) over a nonzero, which needs e_(i−1) odd, or even below its cap under y = 1, or 0 under x_i = 1. The third bound is at least the others less one, and each is attained by a legal input once the two blocks below, i − 1 and i − 2, carry caps ≥ 2, so (S) is exactly legality there. The enumeration runs the lowest blocks, where fewer inputs exist.
The script finds the block code equal to the greedy string at every n < 200000 and on 500 random 60-digit strings at Euler's α, r = 2, 5, 8, 11, at [0; (1, 3, 1)^∞] and [0; (1, 5, 1)^∞] at r = 2, 5, and at two random class sequences drawn under the stricter condition (G), (S) with max(1, ⌊B_(i−1)/2⌋) as its second term, with none off, and the walk from every cut reads 0 at every depth to t = 10 or more at each of those cells but Euler's α, which the uniform lookahead walks at r = 2, 5, 8. Its controls flag the code with the carry removed at Euler's α and the full code at [0; (1, 2, 1)^∞], r = 2, where (S) fails. At the other two phases, the constant 3 at r = 2 and 5 and two random sequences, one at r = 2 and one at r = 5, each drawn so the enumeration finds no violation in its blocks, the phase's code is the greedy string at every n < 200000 and on 500 strings, 0 off, and the walk reads 0 at every depth to t = 9 … 13. A finite search over the block states fed by the first 60 input digits finds a reachable violation, with a legal witness the comparison flags, at all 24 random sequences over 2 … 5, and none at [0; (1, 4, 1)^∞] and [0; (1, 6, 1)^∞], where (G) fails, both read at 0. Fed only the digits up to a block whose two blocks below carry class digits, it agrees with (S) at that block at 3000 random cases at phase ψ = −1.
The borrow code
Tier: theorem.
Verifier: proof; shifts.py::section_borrow; shifts.py::section_sharp.
In the setting of the block code, if at every block
(S′) ⌊B_i/2⌋ + max(1, ⌊B_(i−1)/2⌋) ≤ b_i,
then σ_r reads at lookahead at most max(0, 3 − r) at every depth, the lowest blocks decided by the enumeration: 0 at r ≥ 5, and at r = 2 exactly 1 at the depths J_i − 1 of the blocks i with B_(i−1), B_(i−2) ≥ 2 where (S) fails, and at the depths J_i − 1 of the lower blocks where the enumeration finds a violation, 0 at every other. So it is never MIDDLE. Every class sequence with caps in {2, 3} satisfies (S′) at every ℓ, the constant 2 and the Thue–Morse word 2 + tm(i) among them; b_i = 2 with B_i or B_(i−1) at 4 or more fails it.
Proof. Under (S′) the first two maxima of the block code's proof are within the cap, so its output breaks legality only as o_i = b_i over a last digit 1 of block i − 1. Fix every such block at once, running up the blocks with a pending unit. Add the unit to block i's first digit, turning (0, 1, 0) into (0, 0, 1) by q_(J_i−1) + q_(J_i) = q_(J_i+1). If now o_i = b_i over a 1, rewrite b_i q_(J_i−1) + q_(J_i−2) = q_(J_i): the 1 below becomes 0 and the block's first two digits become 0, 1, its third unchanged. A block left (0, 1, 1) waits: if the block above violates over its 1, that rewrite consumes it; if not, q_(J_i) + q_(J_i+1) = q_(J_i+2) clears it and sends the unit up. The unit arrives only where the block below ended 1 and this block did not violate, so its first digit was at most b_i − 1 and now sits over a 0, legal. A (0, 1, 0) turned (0, 0, 1) can make the block above violate, with the one kind the run fixes, and only where (S) fails. For the block below was the case y = 1, β = 0, Z = 0, so its e is 0 and it sends no carry, and the violating first digit is at most Z_i = x_i + ⌊e_i/2⌋ ≤ 1 + ⌊B_i/2⌋, while (S′) makes b_i at least that, so 1 + ⌊B_i/2⌋ = b_i, where (S) fails. So the output is legal, the greedy string. Every change runs upward but one, block i − 1's last digit, decided by block i's first digit, which reads the shifted string to J_i + 1, three places above it. So output digit j reads the input at most at j + 3 − r, and at j + 2 − r unless j = J_i − 2 at a block with a legal violating input. There lowering e_i, one input place beyond the depth J_i − 1, removes the violation and moves the digit, so that depth needs exactly one more place at r = 2.
At 36 cells, the Thue–Morse caps, the constant 2 and two random (S′) sequences at each phase, r = 2, 5, 8, the script finds the borrow code equal to the greedy string at every n < 200000 and on 500 strings, 0 off, and the walk's least lookahead within max(0, 3 − r) at every depth it reaches, to t = 9 … 14; at r = 2 it is 1 exactly at the search's violating depths (the constant 2 at t = 7, 10, 13 at the phase of Euler's α), 0 elsewhere. Its controls flag the code with the rewrites removed at [0; (1, 2, 1)^∞], r = 2, and the full code where (S′) fails, at the class quotients alternating 2 and 9, r = 5, whose least lookahead reads 3, 2, 1 at t = 8, 9, 10.
Open: off periodicity, does σ_r with r ≡ 2 mod 3 read at a class sequence failing (S′), where a borrow can reach a 0 and chain down? Does σ_r with r ≡ 0 mod 3 read at a class sequence that falls, where the copy reader is silent: the primes with adjacent pairs swapped, 3, 2, 7, 5, 13, 11, …, or the Thue–Morse word? At a periodic sequence the uniform lookahead makes each LIPSCHITZ or DISCONTINUOUS.
The even residues
Tier: theorem.
Verifier: proof; shifts.py::section_even.
At the one-class numeration [0; (1^(P−1), a)^∞], a ≥ 2, the landing walk of a stride with ρ = r mod P even and nonzero has no nonzero interior cycle: at odd P it has no nonzero cycle at all, and at even P only the cap-filling of the even positions, which is extreme. Index by j = k + r, so the digit's cap is a at j ≡ ρ − 1 and the quotient under its weight θ_j is a at j ≡ −1, mod P. A cycle is then a periodic legal string z over every j in Z, and its state (x, y) at an index j ≡ 0 mod P, re-indexed to j = 0, is fixed by two readings of the one string, the future on the real side, F = Σ_(j≥0) z_j θ_j, and the past on the conjugate side, C = Σ_(j<0) z_j |θ′_j|:
x = (F − C)/(α − α′), F = xα − y, C = xα′ − y.
Proof. The period is a unit: θ_(j+P) = ηθ_j with η = −θ_(P−1), an algebraic identity (the change of numeration at α_P = α), so it holds for the conjugates at every j in Z, and every θ′_j is negative. One period summed geometrically makes C the conjugate of F. Three bounds finish.
x < 1. F is at most the cap-filling of the even positions, and C > 0 at a nonzero state, while α − α′ > 1. At even P every even position has cap 1 and quotient 1, and the cap-filling telescopes to 1. At odd P a class slot changes parity each period, and it falls short of 1 by (a − 1)|η|(1 − |θ_(ρ−1)|)/(1 − η²).
x > −1. In C − F omit the even j ≥ 0, whose weights are negative, and charge a digit at an even j < 0 to its odd right neighbour, which legality leaves a unit below its cap and whose weight is larger. With the quotients in place of the caps, the odd sums telescope to α on the future and |α′| on the past, so C − F ≤ α − α′ + T. T compares the odd positions of the digits' class residue c = ρ − 1 with those of the quotients' residue −1, each a geometric sum in |η|. With R_c = |θ′_c| / |θ′_(P−1)|, at even P every class slot is odd and
T = −(a − 1)(1 + |η| − |θ_c| − R_c)/(1 − |η|) < 0,
since |θ_c| ≤ 1 − α < 1/2 and R_c ≤ 1/2. At odd P an even class slot leaves a remainder, and T(1 − η²) ≤ (a − 1)(|θ_c| + (1 + |η|)R_c − 1 − η²), negative at P ≥ 5: c ≤ P − 4 gives R_c ≤ 1/3 with |θ_c| < 1/2 and |η| < 1/2, and c = P − 2 gives R_c ≤ 2/3, |θ_c| < 1/q_(P−1) ≤ 1/5 and |η| < 1/q_P ≤ 1/13. At P = 3 that bound is positive, and the remainder is taken exactly: at each past class slot j ≡ c it is a|θ′_j| − |θ′_(j+1)| where positive, the quotient under θ_j is 1 (c ≠ P − 1), so |θ′_(j+1)| = |θ′_j| + |θ′_(j−1)|, and in units of |θ′_(P−1)|, as R_c is measured, the remainder is max(0, (a − 1)R_c − R_(c−1)), not the bound's (a − 1)R_c. With it, at P = 3 exactly, T = (a² − a√(a² + 2a + 2) − 2)/(4(a + 1)) < 0.
The census. At x = 0 and a nonzero state, F = C = −y is a positive integer at most 1. At odd P there is none, and at even P F reaches 1 only on the cap-filling of the even positions. Which multiple of P the cycle's period is never enters.
As a positive control the script finds nonzero strings with integer x and y at the odd residue ρ = 1 of (3, 2) and (4, 2). It checks x(α − α′) against F − C summed directly, and x against an exact integer form of the same coordinate, on 227 random strings. Over all 3,563,549 legal cyclic strings of length a multiple of P, up to 18, a cycle counted at each multiple of its period, at eight cells it finds every x in (−1, 1), both bounds held, and the strings with integer x and y exactly the zero string and, at even P, the cap-filling of the even positions. At 1900 cells, P ≤ 40, it finds T at most −1/4, the P = 3 limit, and every hand bound it encodes true, the rest of the bounds read by hand. At P = 10 … 16 and a = 2 and 7 the walk finds no interior cycle, and at odd P no nonzero state on an infinite walk in the box.
The parity law
Tier: theorem; rule (that the combs' cuts are the shallowest landed,
verified P = 2 … 9, a = 2 … 5, r ≤ 8).
Verifier: proof; shifts.py::section_parity; shifts.py::section_combs;
shifts.py::section_even.
At the one-class numeration [0; (1^(P−1), a)^∞], a ≥ 2, the shift by r reads at lookahead 0 when r ≡ 0 mod P, is DISCONTINUOUS when r mod P is odd, and is LIPSCHITZ when r mod P is even and nonzero.
The first case is the copy reader, the second the combs and the third the even residues with the uniform lookahead. In the third the rotated caps fall, since the class slot moves onto a cap-1 position, and yet nothing lands on a cut: the copy reader's hypothesis is sufficient and not necessary. The landing walk decides all 352 cells of P = 2 … 9, a = 2 … 5, r = 1 … 2P with none off. That the combs' cuts are the shallowest landed is a rule: the walk finds −q_r α at even P and −q_(r+1) α at odd P at all 112 DISCONTINUOUS cells of that range with r ≤ 8.
The drop
Tier: theorem; rule (that the drop reads at lookahead at most S + 8,
periodic or not, every phase, at every class numeration with class
quotients at most M, verified M = 2 … 9 over every sequence at once,
and with class quotients in one band [m, m + s], verified s = 2, 3, 4,
m = 2 … 24, 40, 41, 80, 81).
Verifier: proof; shifts.py::section_drop; shifts.py::section_wide;
shifts.py::section_drop_class; shifts.py::section_drop_graph;
shifts.py::section_drop_band; shifts.py::section_drop_estimates;
shifts.py::section_drop_potential.
At a purely periodic α of period P, whether a stride r ≥ −1 is DISCONTINUOUS depends on r mod P alone, so the drop, the stride −1, is DISCONTINUOUS exactly when σ_(P−1) is. At the one-class numeration it is DISCONTINUOUS at even P and LIPSCHITZ at odd P, and at a numeration with one repeated quotient (P = 1) it reads. Off periodicity two strings decide it for the numerations below. The ascent comb is the string with d_j = a_j at j = j₀ + 2, j₀ + 4, …, j₀ ≥ 0, and zeros elsewhere. When the quotients ascend at every other step from j₀ + 2 on, a_j ≤ a_(j+1) at each of those j and strictly infinitely often, it is legal and no cap-filling, and it lands on the cut −q_(j₀)α. Where Σ a_(k+1)/q_k also converges, so that the landing criterion covers the drop, the drop then reads at no lookahead from the first depth with q_t ≥ max(q_(j₀), 2) at the latest. Every [0; 1, γ₁, 1, γ₂, …] with bounded γ_i ≥ 1 and γ_i ≥ 2 recurring is such a numeration, periodic or not, at j₀ = 1, with the cut −α and the depth 2. The string at Euler's α: 1 at position 1 and the cap 2μ at position 3μ − 2 for every μ ≥ 2 lands on −α, so the drop reads at no lookahead from t = 2 there, though from no position on do the quotients ascend at every other step. These strings are landings, so they can show the drop DISCONTINUOUS and never that it reads. Wherever Σ a_(k+1)/q_k converges the landing criterion still characterizes it: the drop is DISCONTINUOUS exactly when some non-cut lands on a cut. Past the criterion large quotients decide it: if a_(D+1)|θ_(D−1)| > 1 at infinitely many D, which a_(D+1) ≥ q_D + q_(D−1) ensures, the sum diverges and the drop reads at no lookahead from t₀. At a class numeration, a_j = 1 off J_i = 3i + ψ at any phase ψ, the drop is LIPSCHITZ wherever every consecutive pair of class quotients satisfies
(S′) at ℓ = −1 ⌊b_(i+1)/2⌋ + max(1, ⌊b_i/2⌋) ≤ b_i,
that is b_(i+1) ≤ 2⌈b_i/2⌉ + 1,
periodic or not: it reads at lookahead at most 5 at every depth with q_(t−1) ≥ b₁ (at every depth with q_t ≥ 2 at ψ = 1), at most 4 where the block code's (S) holds at ℓ = −1, and at a finite lookahead at the depths below. Every constant class quotient satisfies (S′), and so does every sequence that rises by at most 1, the Thue–Morse words over {2, 3} and {4, 5} among them; Euler's α, rising by 2 after each even quotient, fails it by one at every pair, and its drop is DISCONTINUOUS. (S′) is sufficient and not necessary: boundedness suffices. Wherever the class quotients are at most M the drop is LIPSCHITZ, periodic or not, at every phase, at lookahead at most 3M + 11 at every depth, and no bound grows slower than linearly in M: at M = 2R + 2, R ≥ 1, the climbing word, the class quotients 2, 2, 2, 4, 6, …, 2R + 2 and then 2 at ψ = 0, reads at lookahead at least 3R + 7 = (3M + 8)/2 at depth 2. So the largest, over the class numerations with class quotients at most M, of the drop's least lookahead is Θ(M). The climbing word follows Euler's α's class quotients 2, 4, 6, … to 2R + 2; Euler's α climbs without end, and its drop is DISCONTINUOUS. A sharper bound, S + 8 at every depth, S the longest path in nodes of the graph below, holds wherever the class quotients are at most M, verified over every such sequence at once at M = 2 … 9, S = 1, 1, 7, 7, 10, 10, 13, 13, and wherever they all lie in one band [m, m + s], verified at s = 2, 3, 4, m = 2 … 24, 40, 41, 80, 81, most at sizes the first family never reached. S bounds the walk, not the drop's own lookahead. A failure at (b, b′) needs b′ ≥ 2⌈b/2⌉ + 2, so a run of failing pairs climbs 2, 4, 6, …, no bounded sequence fails at every pair, and the longest run R is ⌊(M − 2)/2⌋ over [2, M] and, in a band, 1, 0 at even, odd m (s = 2), 1 (s = 3), 2, 1 (s = 4). S reads 1 + 3R in every band run from m = 5 on, and 7, 1, 3 (s = 2), 7, 7, 3 (s = 3), 10, 7, 7 (s = 4) at m = 2, 3, 4; over [2, M] it reads 1 at R = 0 and 4 + 3R after, the same with a window reaching 9 quotients back or 12 ahead over M ≤ 6, and 3, 3, 9, 9, 12 with one reaching 2 back: a pattern. It is open how the lookahead grows in M between (3M + 8)/2 and 3M + 11, whether raising every class quotient by 2 can raise it at any phase at a depth with q_(t−1) ≥ b₁ (q_t ≥ 2 at ψ = 1), which other numerations with a convergent sum carry a landing, and, where none does, whether the drop is LIPSCHITZ or MIDDLE; where the sum diverges with a_(D+1)|θ_(D−1)| ≤ 1 from some D on, nothing here decides it.
Proof. The landing walk sees r only through the quotient a_(k+r+1), the caps a_(k+1) and the frame (θ_(k+r), θ_(k+r−1)). Moving r by P leaves the first two alone and multiplies the frame, the value owed and both tail extremes by the period's unit η = −θ_(P−1) alike (the change of numeration at α_P = α), swapping the extremes when η < 0, so no state's label and no step changes. The one-class cases are then the combs at P − 1 odd and the even residues at P − 1 even; at P = 1 its graph is σ_P's, never DISCONTINUOUS since the copy reader reads σ_P at every depth, though the two maps' lookaheads differ. The comb: a_j θ_(j−1) = θ_j − θ_(j−2), so the image telescopes to −θ_(j₀). A digit at its cap has a zero below it, the other parity, so the string is legal, and a strict ascent leaves a digit under its cap, so it is no cap-filling. At Euler's α the truncation below 3μ − 2 misses −α by E_μ = μ θ_(3μ−4) − (μ − 1) θ_(3μ−3): at μ = 2 it is 2θ₂ − θ₃ = 2α − 1, and a step adds 2μ θ_(3μ−3), which a_(3μ−2) = a_(3μ) = 1 and a_(3μ−1) = 2μ rewrite as E_(μ+1). Both terms are at most μ/q_(3μ−3), so E_μ → 0. The teeth sit three apart, alternating parity, so the string is no cap-filling. The landing criterion finishes all three. Past it, |θ_(D−1)| lies between 1/(q_D + q_(D−1)) and 1/q_D, so a_(D+1) ≥ q_D + q_(D−1) gives a_(D+1)|θ_(D−1)| > 1, which gives a_(D+1)/q_D > 1. The inputs b·q_D and (b + 1)·q_D, b < a_(D+1), are single digits over zeros and share every input tile of depth at most D. Their drops' points b·θ_(D−1) and (b + 1)·θ_(D−1) sweep past a whole turn as b runs, so some pair sits on the two sides of −α, on an arc of length |θ_(D−1)|. At a depth t ≤ D − 1 with q_t ≥ 2 the cuts lie at least |θ_(t−1)| apart, so −α is the one cut on that arc and the two drops lie in two output tiles. A D past t + c from the infinite set then defeats lookahead c at depth t.
At a class numeration the drop is the stride r = 3ℓ + 2 at ℓ = −1, which puts input digit J_h − 1 in block h − 1's third slot and J_h, J_h + 1 in block h's first two, so B_i = b_(i+1) in the block code's notation. The block code's and the borrow code's rewrites are identities of the output numeration, and each legality tie joins adjacent places, so both proofs run unchanged but for the reach: output digit j reads the input to j + 3 under (S) and to j + 4 under (S′). The bottom is the drop's own. q₀ = 1, and q₁ = 1 at ψ ≤ 0, so drop(n) = Δ(n′) + ω, where n′ is the input with digit 1 (and digit 2 at ψ = 0) set to 0, ω = d₁ (+ d₂) is at most b₁ at ψ ≤ 0 (at ψ = 0, d₂ = b₁ forces d₁ = 0) and at most 1 at ψ = 1, where digit 1's cap is a₂ = 1, and Δ is the code on n′'s shifted string. The lowest block's first digit is then at most ⌊b₁/2⌋ against its cap b₀ − 1 at ψ = 1, and at most 1 + ⌊b₂/2⌋ against b₁ over a zero at ψ ≤ 0, both within the cap by (S′) at the lowest pair, so no enumeration is needed. n ↦ n + ω reads at lookahead 1 at every depth with q_t ≥ 2 and q_(t−1) ≥ ω (OSTROWSKI: The affine maps), and composed with the code that gives the bound.
Past (S′) the walk is finite-state. Its state at step k is the landing walk's pair (x, y) at r = −1, the value still owed being D_k = xθ_(k−1) + yθ_(k−2). The future form y − xφ_k, φ_k = −θ_(k−1)/θ_(k−2), lies in the hull of the legal tails' sums in the frame's units. The past form y + xq_(k−1)/q_(k−2) is D_k's α-coefficient over q_(k−2), which is −N_k/q_(k−2), N_k = j + Σ_(1≤i<k) d_i q_(i−1) the cut's own j plus the past digits' drop, so it lies in [−C_k − j/q_(k−2), −j/q_(k−2)], C_k the legal pasts' largest such sum over q_(k−2); at k ≥ t + 8 a depth-t cut has j/q_(k−2) ≤ 1/8, since q_(m+2) ≥ 2q_m. With the class quotients at most M, both ranges are bounded, and closed interval enclosures of φ_k, the tail extremes, q_(k−1)/q_(k−2) and C_k built from a window a_(k−6) … a_(k+9) of the quotients (the future past the window by an interval fixed point, the past below it by weights from a ratio in [0, 1] and a remainder below q_(k−7)) admit finitely many states. The windows of the family (2 … M at the class places, 1 elsewhere, or one band [m, m + s], the tail's fixed point read off the band) with their admitted states are the nodes of one graph and the walk's steps, the window shifted, its edges, so every interior state of a walk from a depth-t cut at a step k ≥ t + 8 is a node. If sweeping away the nodes with no successor S times leaves none, no such walk is interior past step t + 7 + S, and the identity there, the lookahead at depth t being max(0, K + 1 − t), of The uniform lookahead, which holds wherever the sums converge, puts the lookahead at depth t at most S + 8. The enclosures are conservative, so a surviving node would prove nothing; none survives at M = 2 … 9 nor at the bands run.
The estimates. Write ν = max(|x|, |y|) for a state's height and call an edge that lowers it a fall. At a class numeration a window's past holds at most three class quotients, b at the class place k or one or two steps before it and b′, b″ the ones before that, so the ratio q_(k−1)/q_(k−2) and C_k are rational functions of them, C_k a maximum over finitely many legal strings. Over one denominator in b = 2 + u, b′ = 2 + v, b″ = 2 + w every coefficient of each margin has one sign, so at every class quotient at once: at the class place C_k ≤ b + 4/5 and the ratio is at least (2b′ + 1)/(b′ + 1); one step past it C_k + 1/8 < the ratio, and two steps past it C_k + 1/8 < the ratio + 1, both by 1/100 more. A state with x > 0 has |y| > x, the past form being at most 0 and the ratio above 1, and (−X, y ≥ 1) rises on every edge, so only (−X, −Y), X > Y ≥ 1, falls. The last two estimates admit no state with x ≤ −1, y ≤ 0 off the class place but (−1, 0) two steps past it, so falls leave the class place only, and at Y = 1: Y ≥ 2 would need 2X > bY − 1 ≥ 2b − 1, which the past form with the first two estimates forbids. At ν = 1 the states (0, −1), (−1, −1), (−1, 0), (−1, 1) carry paths of at most 6 nodes and no cycle. The script certifies the estimates, its control catching the class place's C_k ≤ b + 4/5 planted at b + 1/2, which fails at b = b′ = b″ = 2.
The potential. Call a state at the class place with ν ≥ 2 and an edge into it a vertex, and the three steps to the next class place a leg: digits d₀, d₁ ≤ 1 and d₂ ≤ c, c the next class quotient, land on x′ = −x + (b + 1)y + d₀ − d₁, y′ = y − 2x′ − d₁ − d₂. At a vertex (−X, y), y ≤ 0, d₀ is 0, its edge coming from (x₀, −X), x₀ ≥ 0, by the positive digit x₀ + X − y; its past form gives X < b at y ≤ −1 and X ≤ 3(b − 3/40)/5, so X ≤ b − 2 with X ≥ 2, at y = −1. Put Φ = X − b at (−X, −1), X − bY + 1 at (−X, −Y) with Y ≥ 2, −bY/2 at (x ≥ 0, −Y), −X at (−X, 0) and 2 − 2X − (2b + 1)y at (−X, y ≥ 1). The map, the cap on d₂ (c ≥ y + 2X′ − y′ where x′ = −X′ < 0) and these bounds make Φ fall by at least 1 along every leg, at each of the fourteen pairs of types a leg joins, and put Φ ≤ −2 at every vertex. A fall from a vertex lands on (−1, b − X), at ν = b − X ≥ 2, so no path re-enters ν = 1, where the six-node bound holds, and a cycle would close a walk of legs along which Φ falls: there is none, and nothing survives at any M.
The box. A walk's states satisfy both forms with its own sequence's exact tails, φ_m = 1/(a_m + φ_(m+1)), V_m = φ_m W_(m+1) and W_m = φ_m(a_(m+1) + V_(m+1)), the two tail extremes being V_k and W_k (less φ_k where a nonzero digit at k − 1 bars the cap at k). Where a_m = 1, φ_m φ_(m+1) = 1 − φ_m, so φ_(k+1)φ_(k+2) < 1/2 and φ_(k+2)φ_(k+3) ≤ 1/3 past a class place k, and bφ_k < 1 there. Unrolling one class period, the largest V one step past a class place, v₁, is at most M/2 + (4 + v₁)/20, so v₁ ≤ (10M + 4)/19; at the class place bW_k < 1 + v₁ and bV_k < (4 + v₁)/5. With the past form these bound Φ below, by 2 − M at (−X, −1), −v₁ at (−X, −Y), −(1 + v₁)/2 at (x ≥ 0, −Y), −(3/5)(M + 37/40) at (−X, 0) and −(101v₁ + 44)/200 at (−X, y ≥ 1), so Φ lies in [−L, −2], L = max(M − 2, (3/5)(M + 37/40)) < M. Φ falling by at least 1 per leg, a walk holds at most M − 2 vertices; with its first state, the six at ν = 1 and at most two either side of its vertices, three steps apart, at most 3M + 3 interior states, which gives 3M + 11. The bound is on the exact tails. The interval fixed point the graph's windows use for V one step past a class place drops the tie φ_(k+1)φ_(k+2) = 1 − φ_(k+1) and exceeds v₁ (5.1381 against 4.9474 at M = 9), so the graph as built is read against the bound and never covered by it. The script reads every leg at four families against the map, the caps, the bounds on X, Φ's range and the eight of the fourteen drops met there, the other six by the proof alone, its S at most 25 against 3M + 3, and a sample of exact tails, the 8⁵ class words of five places at M = 9 each followed by the constant 2, under all three bounds (V one step past a class place at most 4.3496); its controls catch Φ negated at (−X, 0), x′ one high in the map, every drop's bound one high, a digit cap removed, Φ's range narrowed to −2 and a planted step back to ν = 1.
The chain. At the climbing word M = 2R + 2, R ≥ 1, and q₂ = 2. At a class place k whose quotient is 2X, with c = 2X + 2 and φ = φ_(k+3) < 1/c at the next one, the states (−X, −1) at k, (−1, X) at k + 1 and (X, −(X + 1)) at k + 2 are interior: through φ_m φ_(m+1) = 1 − φ_m each margin reduces to c − 2X − 1 + V_(k+3) > (X + 1)φ, which holds as (X + 1)φ < 1/2. From (−X, −1) the digits 0 (forced by the digit below), 0 and the cap c reach (−(X + 1), −1) at k + 3. From the cut −2α the walk holds (−2, −2) at step 1, and the digits 1, 0, 1, 0, 1, 0, 1, 0, 1, 0 reach (1, −2) at step 11; those margins reduce to W₅ > 1, true as a₆ = 2, and to the chain's inequality at X = 1, c = 4. So the walk is interior through (R, −(R + 1)) at step 3R + 8, and the lookahead at depth 2 is at least 3R + 7 = (3M + 8)/2, a factor 2 + 3/(3R + 7) under the upper bound's 6R + 17. The walk's last interior step from a depth-2 cut is exactly 3R + 8 at R = 1 … 9, all 207 chain states interior, so the lookahead there is exactly 3R + 7; its control reads the walk's column at a fixed word over the quotients 2 and 24, whose largest lookahead, 19, it must print.
The script finds the drop's graph and σ_(P−1)'s one, their sets of states on infinite walks, interior states and cycles equal, at 37 purely periodic α (39 runs, the one-class (2, 2) and (3, 2) repeating [1, 2] and [1, 1, 2]; at golden, silver and bronze, P = 1, the two walks use one frame and agree by construction), 1373 states on infinite walks over the runs, and the one-class drop DISCONTINUOUS at exactly the 16 even-P cells of 32, P ≤ 9, a ≤ 5, landing on −α. The comb's truncations telescope exactly, over 60 teeth, at six sequences: three with the quotient 1 at every odd position and a word over {1, 2} at the even ones (the Fibonacci word, Thue–Morse and a random word), one random with the even positions ascending, one ascending from j₀ = 9 onto −32403α, and the one-class (4, 3). At each, raising the comb's last strict tooth gives a legal string on the cut's other side; at golden, where no ascent is strict, the comb is a cap-filling and the drop reads. E_μ holds exactly at μ ≤ 60. The exhaustive reader tracks from t = 2 at the three sequences over {1, 2} and decides no depth against it. At the numeration with a₁ = 2, a_(D+1) = q_D + q_(D−1) at D = 4, 8, …, 24 and every other quotient 1, the pair at each D sits on the two sides of −α, its inputs parting at position D and its drops at 0 = t₀ − 1. At 18 class numerations, the Thue–Morse words over {2, 3} and {4, 5}, the constants 2 and 3 and two random (S′) sequences at each phase, the code of Δ(n′) is the greedy string and Δ(n′) + ω the drop at every n < 200000 and on 500 random 60-digit strings, 0 off. The drop's least lookahead, read by the walk to t = 9 … 14, is at most 5 at every depth with q_(t−1) ≥ b₁ (with q_t ≥ 2 at ψ = 1), and 5, the bound, at t = 12 of one random (S′) sequence at ψ = −1; at the six cells under (S) it is at most 3, one below its bound. Its controls find the walk at Euler's α surviving to the last step it runs from t = 2, where the landing string above defeats every lookahead, and the one-class (3, 3) at a finite lookahead at every depth, and flag the code with the rewrites removed at the constant 2, ψ = −1, and with digit 1 left unfolded at ψ = 1 at block 0 of the constant 2. The graph has 2, 56, 1062, 5824, 27125, 83412, 235445 and 545280 nodes at M = 2 … 9 and none survives; over the 81 bands none survives either, their nodes about linear in m (31400 at [5, 9], 326625 at [81, 85]), and s ≥ 5, its windows growing as (s + 1)⁶, is not run. Its controls: the one-class (2, 2), whose drop is DISCONTINUOUS, keeps surviving nodes; every true φ_k, tail extreme, ratio and C_k at 7 ≤ k ≤ 200 lies in its enclosure at four class numerations over {2, 4} and 2 … 9 (4 × 1164 quantities), and at random sequences over five bands from [5, 9] to [81, 84] at every phase (15 × 1164), where the band shifted 4 down puts 155 to 259 of each 1164 outside (the true C_k is the same programme run on the true quotients, so its indices have no independent check); and at five class numerations over {2, 4} every interior state of the exact walk from j ≤ 10⁵ at k ≥ 7 is among its window's states, 68396 in all, the graph's steps from it the walk's; 13 of them lie at k ≥ t + 8, the steps the bound uses, all nodes. At the purely periodic class numerations of class period length at most 2 and class quotients 2 … 9, at every phase (216 cells, 192 distinct numerations, 108 cells failing (S′)), the exact walk agrees: no box graph has a cycle, and no nonzero state of its box lies on an infinite walk, as at the one-class numerations of odd P; its controls find a cycle at the one-class (2, 2) drop, none at the constant 3, and, at a class period, one at the constant 2's σ₁.