At every rung of the tower, size is deleted: sign, comparison, and the ordering of the integers are not channel quantities at any rung with two or more channels — the size wall, charted with its price on the Walls page. This page charts what survives the deletion. The demo rung is k = 7, Z/510,510, where every element is a 7-tuple of remainders mod 2, 3, 5, 7, 11, 13, 17 — small enough that every claim below is proved or checked exhaustively over a stated range. The chart: the skeleton that replaces order, the graded logic it supports, the gates that measure it, how far the measurements reach, what reading costs — and the one door that buys exact identification back.
With size deleted, the ordering of the integers is gone — and the loss is total, not partial. Test any relation by asking whether it equals the AND of its per-channel readings: less-than fails maximally — every channel's reading saturates, so the channels jointly retain nothing of the order (exhaustive at Z/210) — and arranging the ring on a circle instead of a line buys nothing, since the circular relation carries the line's order inside it.
The test has a communication reading. Split the channels into one versus the rest and lay the relation out as a table — rows for everything the chosen channel sees of the two elements, columns for everything the other channels see. A relation reads channel-by-channel exactly when every such table is one solid rectangle, some set of rows times some set of columns (proved) — in communication terms, the shape of a question that needs no conversation. Less-than's tables never are: not at any split of Z/30 (measured whole), and at the two-channel ring Z/6, where the whole table is 4 by 9, covering its yes-entries takes four rectangles — the most a table that size can need (computed exactly). Deciding order across a channel split is a conversation between the channels, and the ring runs no wire between them.
What passes the test is divisibility, and it reads exactly one bit per channel: zero here, or not. The zero pattern — which channels an element has died in — is the surviving skeleton: where you live, not how big you are.
That skeleton supports a logic. Write AND as ab, OR as a + b − ab, NOT as 1 − a: ring polynomials, so all three run channel-by-channel. On elements whose every residue is 0 or 1 — the zero patterns themselves — they are exact Boolean logic; every other element is a graded truth value, a residue that is neither 0 nor 1 being a channel only partly true.
Two measurements read the grade, and both are ring operations. Raising to the power 240 — the ring's universal exponent: for any element, powers past the first repeat with period dividing 240 — collapses each residue to 0 or 1 and answers “true to some degree?” channel by channel (mod 7, the graded residue 3 collapses to 3240 ≡ 1), while the mirrored map 1 − (1 − a)240 answers “fully true?” (the same 3 reads 0). Each is the other's reflection through NOT, exactly, and the pair behaves as the possibility and necessity of the modal logic S5 — its laws, taken in equation form (the graded elements carry no order for the textbook form to live in), hold exactly here — quantifying the degree of truth inside a channel.
The other quantifier — across the channels: compress “is this element zero?” into a single 0-or-1 answer inside the ring — does not exist: no polynomial computes it over any composite modulus (proved), while a single prime field has it as Fermat's 1 − ap−1. Degree quantifiers come free; one bit quantified across the channels costs the deleted window. It is the size wall's question again, asked of logic.
The pair turns measurement into certificates. Take any formula built from AND, OR and NOT over graded inputs, pushing the NOTs down onto the inputs first (De Morgan holds identically here, so the value does not change). Its honest verdict, channel by channel, is what the two measurements say of the formula evaluated in the ring.
Now measure each input instead — two bits per input — and evaluate the formula twice as plain Boolean arithmetic: once generously, reading every plain input by its some-degree bit and every negated one as “not fully true”, and once strictly (fully-true, “not true to any degree”). The verdict's two bits land between the two evaluations in every channel (proved): a generous 0 certifies the formula false there, a strict 1 certifies it fully true, and where the two agree the verdict is settled outright — from two bits per input, without revisiting the elements. The slack between them lives only in channels where some input was already graded; channels where every input is classical are always certified.
And one bit is not enough: the some-degree bit alone can certify only falsity, the fully-true bit alone only truth — read the some-degree bit as the whole answer and it errs at a nonzero rate, enumerated exactly. The two-sided decision needs the pair.
The two measurements are the ends of a family. For each divisor m of 240, raise to the power m and ask “fully true?” of the result: the answer, channel by channel, is one bit — does the residue's multiplicative order divide m? — a zero residue, which has no order, reading no (proved). At m = 1 this is the fully-true bit; at m = 240, where every order divides, the some-degree bit; between them sit the rest — one gate per divisor, twenty in all here, all distinct (computed), ordered as the divisors are and combining as gcds do: the AND of two gates is exactly the gate at their gcd (proved).
Repetition is this family in disguise: “fully true?” of a AND a AND … AND a, m copies evaluated in the ring, is the order-divides-m gate, so what repeating an input adds to its two bits is its order and nothing else.
And the family decides what the pair cannot. The formula a AND NOT a, evaluated in the ring, is fully true exactly at residues satisfying x − x² = 1 — primitive sixth roots of unity wherever a channel has them (mod 7 and mod 13 here), plus one degenerate point in the mod-3 channel — while its strict evaluation from the two bits per input is identically zero: the pair can never certify it. The order gates decide it in every channel with zero errors (proved): order divides 6 and neither 2 nor 3 — except in the mod-3 channel, where the equation's one truth point is the channel's only graded residue, read straight off the family's two ends: some degree, not fully.
With two inputs the gates read words. Take any product of powers of a and b — ab, a²b, ab−1 — and ask the order gates of it. Within each channel, the full profile of answers measures the two positions exactly up to one shared rescaling: a second pair of residues answers every gate of every word identically precisely when it is (au, bu) for a single invertible exponent u applied to both (proved — a fact about cyclic groups, and not special to two inputs); across channels the rescalings are independent, one per window, as everything here is. What a second input genuinely adds is relative position: where a generates a channel's nonzero residues, which power of a the residue of b is — pinned exactly.
The reach then has a clean boundary: a yes/no question about the pair is decidable by words precisely when its answer is constant under that shared rescaling (proved). Every multiplicative relation aibj = 1 passes. Additive questions are blind — with a crystallographic exception. Ask “does a + b = c?”: at c = 0, where b = −a is multiplication in disguise, the question is decided outright; on every other line the words see only points of multiplicative order 1, 2, 3 or 6 — c = ±2 keeps (±1, ±1), c = 1 keeps the primitive sixth roots, c = −1 the cube roots, and every remaining line is invisible in full (proved, with census across the primes to 100). Orders 1, 2, 3, 4, 6 — counting c = 0's quarter-turn points — is the same short list a plane crystal's rotations obey, by the same arithmetic.
One word deserves singling out: the gates of ab−1 grade equality — channel by channel, for invertible residues, whether a and b sit at the same position around the multiplicative cycle at each of the twenty resolutions, exact equality at the bottom (proved) — a graded “how equal” in a ring with no “how close”.
Every wall on this page ends at one door. The reads so far draw on a fixed alphabet — powers of the inputs and of their NOTs. But this ring divides totally: every nonzero residue inverts channel by channel, the inverse of 0 is 0, division never refuses. Let that inverse nest freely with NOT and the walls fall in one line: NOT(a−1) · (NOT a)−1 · a works out to the constant −1 at every residue except 0 and 1, and to 0 at those two (proved) — a constant, conjured from a variable.
From −1, NOT and multiplication reach every other constant, and with constants in hand “is x equal to c?” is one order gate of xc−1: every residue pinned outright, 0 and 1 being the original pair's own bits. Identification, not measurement. Each wall above — the pair's, the order gates', the words' — is a statement about what the alphabet withholds: total division buys back exact identification.
Short of that door, the words' reach has an exact map, and the crystal list of the lines is one row of it. Ask the same question of any curve — any polynomial condition on the pair, not just a line: which of its points do the words see? By the boundary above, exactly those whose whole rescaling orbit stays on the curve, and one criterion sorts them (proved). Take a point whose two coordinates are invertible — the residues that have orders. Each term of the curve, evaluated there, lands somewhere on one cycle of length d — d the least common multiple of those two orders — and collecting coefficients slot by slot turns the curve into a word of length d. The point is seen exactly when that word either is zero outright — the curve contains the pair's whole power cycle — or is a codeword of a classical error-correcting code: the cyclic code that the d-th cyclotomic polynomial generates.
Coding theory then writes the menu. That code's minimum weight is exactly the least prime factor of d (proved, in every characteristic), so a curve written in M terms can hold seen points of order d — short of containing their cycles whole — only if the least prime factor of d is at most M. Lines are the three-term row, their unit coefficients and constant term cutting “least prime factor at most 3” down to the crystal list above; each prime number of terms seats a species that fewer terms cannot; and the first arrival past the lines is order five, needing five terms, where a conic's budget of six first fits it: the pentagon conic y² + xy + x + y + 1 = 0 holds exactly one seen orbit of order-5 pairs at every prime p ≡ 1 (mod 5), and none elsewhere (proved; censused at p = 11, 31, 41, 61) — a degree-2 curve seeing torsion no line can.
How much degree an order demands is itself a law:
| orbit order | first degree to see it |
|---|---|
| even | 1 |
| odd prime q | ~ √q |
| odd composite | (least prime factor) − 1 |
Every even order is reached at degree 1 — a line already sees it. An odd prime order q needs degree on the scale of its square root — never below √(2q) − 2, never above 2√q + 1 (proved; the exact value censused through q = 149, the pentagon's conic the smallest instance). And every other odd order pays its least prime factor nearly whole — degree one less than that prime, no shortcut (proved for every order with at most two prime factors, counted with multiplicity, and beyond that wherever the order's two smallest primes sit far enough apart; the remaining band — three or more factors, the two smallest close together — is not fully closed). The hardest case fell last: two close prime factors, where a cheaper curve would amount to a degenerate cover of the order's residue grid beating the channels' own — and no such cover exists, at any pair of primes (proved).
The gap between the two odd rows is a composite tax: at the same least prime factor, order 5 falls to a conic while 25 and 35 hold out to degree 4; order 13 is seen at degree 5 while 143 waits for 10. The degree at which a curve first sees a torsion orbit reads whether the orbit's order is prime — the alphabet that cannot see size administers a primality test.
And reading has a price list. Keep the alphabet of the words — products of powers, inverses included, NOT not yet — and count the gates a decision needs. Every “does the order divide m?” is one gate by construction, the graded equality reads included; pinning an order exactly costs more. “Is the order of x exactly d?” takes one gate per distinct prime of d, plus one (proved, both directions: each prime of d forces a gate of its own, no gate serves two primes, and one more must accept d itself) — except at the top, where “does x generate the channel?” drops the extra gate.
That is the single-residue case of the full ledger: deciding any stable orbit — cut out its cycle, then read its order — costs exactly the equations the cut needs plus one gate per distinct prime of the order. Both halves are forced: each prime extorts a gate of its own (proved), and a program skimping on the cut would need its remaining gates to form a covering system with one class per prime modulus, which CRT never permits (proved); the one conceivable escape — covers built from repeated or composite moduli — finds no witness in exhaustive census everywhere swept, the largest pitting all 337 million structured programs of up to four gates against a three-equation orbit, none cheaper (and where a single equation cuts the cycle, the law is proved outright).
Two currencies now price one locus: the pentagon orbit, five terms to cut out, is decided by exactly three gates — two cutting the cycle, one reading the prime (proved; at p = 11 and 31 every one- and two-gate program fails, exhaustively) — deciding a locus and writing its equation are different costs, neither convertible into the other.
The questions words cannot decide are the table's infinite rows: a gate's bit never separates two pairs linked by the shared rescaling, so a question that splits an orbit — “does a + b = 3?” in the mod-7 channel — has no finite gate count at all. The walls of this page sit in the same table, each priced in a different currency — the alphabet a read requires rather than the gates it spends.
The door is on the table too: with NOT joining the alphabet, total division decides every question at exactly one gate, and the whole price moves into the formula handed to the gate — where it cannot be hidden. Only so many formulas exist under any given operation count (counted exactly), so as the channel prime grows, the costliest residue's identification formula grows without bound: size, deleted as data, returns as cost, and which residue you pin carries a tax with no ceiling. The six-operation formula above that conjures −1 is optimal — at p = 5 and p = 7 no shorter formula reaches any constant beyond the pair's own 0 and 1 (computed exactly) — and a single ring gate runs all seven channels in parallel (by construction), so every per-channel price here is already the ring's.