Ceilings

Evaluations whose best possible score is derived rather than estimated: what the optimum is, which members of the family the evidence is worthless on, and what attaining it costs — every value in closed form, on families designed so that there is one.

An eval here is a triple: a task family with a prior over its instances, an evidence channel the solver is shown, and a score. Its ceiling is the Bayes-optimal expected score — what the best conceivable solver gets, with no solver required to exhibit it — and its floor is the prior's own best score with the evidence severed, computed by direct summation rather than set at a rhetorical one half. Each member of the family is a cell, and a cell is dead when ceiling = floor, which says the evidence buys nothing there, and interior when the ceiling strictly beats the floor without reaching certainty. A ceiling is tight when some computable procedure of stated cost meets it — an achiever; the Bayes posterior meeting the Bayes optimum is a definition and claims nothing, so the content is the closed form together with the achiever's price. Two scores are asked throughout: 0-1 accuracy, which pays for a correct answer, and log-loss, which reads the probability assigned to the truth and whose optimum is a least conditional entropy rather than a greatest score — “ceiling” below means the optimum under whichever score is being asked.

The eval that started this was not designed to carry its own ceiling. A grown world's history has an exactly computable posterior given its finished state, so induction on such worlds scores a compression-driven inducer against a literal Bayes value instead of against other solvers (the ledger of the knowable). Whether that was an accident of that one construction is the question the families below answer, and they are built from unrelated material.

That material is the archimedean deletion: rings of residues, which carry arithmetic and no notion of size. Fix N squarefree — a product of at least two distinct primes — and draw from Z/N uniformly: a single element x, or, in two of the four families below, an ordered pair or triple of distinct elements. A channel is one residue mod p; the solver is shown a proper subset of the channels, whose product is the modulus M, leaving the unknown cofactor c = N/M. What the evidence pins is a fiber: for a single element, the c ring elements sharing the shown residues, an arithmetic progression of step M; for a pair or a triple, one such progression per point. The task asked is always archimedean — a question about size or order, which the ring does not carry (the hiding lemma) — and it is always a single bit, so each eval attaches an exact number to a limit otherwise stated qualitatively. A cell here is one pair (ring, shown subset), the family is every cell, and a dial is the exact arithmetic condition naming which cells are dead.

Every family below is designed, and each ceiling is derived from that design: exact in rational arithmetic, proved by hand, and confirmed cell by cell across the stated range. That is what makes deadness a certificate here and its failure boundaries algebraic. It is also the scope. These are toy rings of at most a few tens of thousands of elements, and no number on this page is a ceiling for a benchmark built by anyone else.

The anatomy

The sign bit and the parity dial rule

Ask for the sign bit [2xN] — which half of the ring x fell in. The Bayes ceiling is exactly (c+1)/2c when the unknown cofactor is odd and exactly 1/2 when it is even, with no third behaviour. The proof is three lines: the fiber is an arithmetic progression, so its below-count is ⌈c/2 − r/M⌉ at shown residue r, which is c/2 for every r at even c and takes exactly two values at odd c. Because N is squarefree, c is even precisely when 2 divides N and channel 2 is among the unknowns, so this family's dial is a single arithmetic bit: a cell is dead exactly when its ring has a channel 2 and the solver is not shown it. At odd N, which has no channel 2 to withhold, there are no dead cells at all — every proper subset lifts strictly off the floor. The log-loss ceiling is the same dichotomy in entropy dress: the binary entropy H2((c+1)/2c) nats at odd c and log 2 at even. And the achiever costs a constant per query: reconstruct the shown residues into the single residue r = x mod M, which the Chinese remainder theorem does in one modular pass, then answer “below” iff 2r < M — one reconstruction and one comparison. This is the exact Bayes value of a qualitative fact the hiding lemma already carried, that the fiber's sign bias is at most half an element and exactly zero without channel 2.

Scope. The closed form is proved for every squarefree N and every proper divisor M; the confirmations are exhaustive over the 602 cells of the 57 rings built from the first six primes, largest N = 30030, exact in rationals with entropies to 1e-12 nats. 211 cells sit at the floor and all 211 are the 2-unknown cells; the achiever was enumerated on the 362 cells with N ≤ 2310. Uniform prior; 0-1 and log-loss scores. Toy scale.

verifier: explore_eval_ceiling.py

The threshold law rule

The sign bit is the predicate [xt] at the midpoint, so the dial can be asked at every threshold. Write s for t mod M; a fiber's below-count is ⌊t/M⌋ + [r < s], so exactly two fiber types exist and the ceiling is the no-evidence floor max(t, Nt)/N everywhere except where c is odd and t lies strictly inside the middle fiber window (mM, (m+1)M), m = (c−1)/2. Inside that window the ceiling is (c+1)/2c, independent of where in the window t sits. So the even half of the parity dial survives verbatim at every threshold — even c is dead at all of them — while the odd half deforms from a parity condition into a window condition. What does depend on t is the lift over the floor, the tent min(s, Ms)/N, maximal at the midpoint: the sign bit is the extremal member of its own family, sitting at the top of the tent. The achiever is unchanged and stays exactly tight.

Scope. The closed form proved for every squarefree N, proper divisor M and threshold; confirmed exhaustively on the 40,914 (cell, threshold) slices of the 194 cells over 37 rings with N ≤ 500, every one of the 2,697 interior slices carrying the tent exactly, plus 2,508 boundary-sampled slices up to N = 2310. The achiever was enumerated on the 12,090 all-threshold slices with N ≤ 210 and 1,137 sampled above. Uniform prior. Toy scale.

verifier: explore_ceiling_anatomy.py

Which channels pay

The orientation ceiling and the inverted dial rule

Change the task shape. Take a uniform ordered triple of distinct points of Z/N, show all three residues mod M, and ask for the triple's cyclic orientation — a relation whose fibers are not intervals. Translating the base point to 0 leaves a difference pair (a, b) with orientation positive iff a < b as integers, and the fiber posterior is uniform on a product of ladders, so the positive count is a staircase threshold count on a c-by-c grid. A fiber whose two shown differences are distinct and nonzero leans by exactly (c+1)/2c toward the residue order; a fiber where they agree, or where one of them is zero, is exactly balanced, and at c = 2 the all-zero fiber holds no triples at all. Weighting the fiber classes gives the ceiling

12  +  c(M1)(M2)2(N1)(N2)\tfrac{1}{2} \;+\; \frac{c\,(M-1)(M-2)}{2\,(N-1)(N-2)}

which lifts off the floor exactly when M ≥ 3. The hiding here is total, and independently proved to be — the fraction of triples the shown channels determine is 0 on every cell, no fiber ever deciding (cyclic orientation is totally hidden) — and the skew inside that total hiding is nonetheless real, and priced. The dial inverts against the sign bit: the sole dead subset is {2}, where the sign bit lived in channel 2 alone. So does the channel ordering. Single channels obey the same formula at M = p, worth zero at p = 2 and strictly increasing in p, making the best single channel the largest prime read, where for the sign bit channel 2 was the only single channel worth anything at all. Which channels pay is a property of the task, not of the ring.

Scope. The ceiling proved for every squarefree N and proper divisor M; exhaustive on all 194 cells over the 37 rings with N ≤ 500, where 19 cells sit at the floor and all 19 are the subset-{2} cells. The reduction to difference pairs is checked against raw triple evidence at N = 30 over every proper divisor before the scan is read. The log-loss ceiling matches its own closed form to 1e-12 nats and the achiever — two residue subtractions and one comparison — is exactly tight. Uniform prior. Toy scale.

verifier: explore_ceiling_anatomy.py

The comparison ceiling and the constant dial rule

Ask the archimedean question of two hidden elements at once: [x < y] on a uniform ordered pair of distinct elements, with both residues mod M shown. A fiber with unequal shown residues is a product of two full c-ladders and leans by (c+1)/2c toward the residue order, the same unit again; a fiber with equal ones is exactly balanced, since as many ordered distinct pairs ascend as descend. Weighting them, the unknown cofactor cancels and the ceiling is

12  +  M12(N1)\tfrac{1}{2} \;+\; \frac{M-1}{2\,(N-1)}

against a floor of exactly 1/2. This family has no dead cells: every proper subset lifts, M = 2 included. That is a third dial shape — a constant one, where the sign bit's dial was parity and orientation's was inversion — and the value depends on the ring and the shown modulus alone, not on how much is hidden. The achiever reconstructs both residues, orders them, and answers a fixed way on a tie, at two reconstructions and one comparison; the best single channel is again the largest prime read. These rings do admit an exact comparator, but it buys exactness by importing an extra coordinate and reading every channel (the exact comparator and its cost law); this is the other end of the same question — what a proper subset alone is worth, with nothing imported.

Scope. The ceiling, the floor and the log-loss form proved; exhaustive on the 362 cells of the 49 rings with N ≤ 2310, with the achiever enumerated exactly on the 116 cells with N ≤ 210 and the entropy strictly below log 2 on every cell. Uniform prior on the pair. Toy scale.

verifier: explore_ceiling_dials.py

The skew unit and the price of doing less observation

Four dials over three task shapes, and one number underneath all of them. Every interior mechanism above reduces to a staircase threshold count on a c-ladder, and the per-fiber skew (c+1)/2c is what such a count yields: it is the sign bit's whole ceiling, the threshold family's interior value, the lean of an orientation fiber before dilution by the fraction of fibers that lean at all, and the lean of a comparison fiber before ties. The dials say which cells are alive and differ across the four; what a live cell is worth keeps reducing to the same unit.

The families also price their own cheaper solvers, by their own laws one level coarser. A solver that reads a single channel, or drops one, is the same eval at a smaller modulus — M = p for the single channel — so the cost of doing less is a difference of two instances of the family's own closed form rather than a separate measurement. At N = 2310 with channels {2, 3, 5} shown, the sign bit's joint ceiling is 39/77, the best single channel is worth 578/1155, and the gap is 1/165.

Scope. The recurrence of the skew unit is an observation across the four instances, each of whose ceilings is separately proved and confirmed at the ranges the blocks above state. The self-similarity of the single-channel law is a rule, proved and exhaustive on the same cells. Toy scale.

verifiers: explore_eval_ceiling.py, explore_ceiling_anatomy.py, explore_ceiling_dials.py

Whether the evidence is worthless

Deadness is score-relative rule

At the midpoint threshold — the sign bit — the two scores agree about which cells are dead, both naming exactly the even-c ones, and that agreement is a coincidence of that one threshold. Across the rest of the family they diverge sharply. Call the set of thresholds at which a cell is dead its floor set, one per score. The log-loss floor set is the thin set of thresholds divisible by M — where a threshold on a fiber boundary makes every fiber the same type, so the posterior does not move at all — while the 0-1 floor set is nearly everything, every threshold outside the odd-c middle window. Off that window the posterior does move; it simply never crosses one half, so a decision score cannot spend what a log score is paid for. At any threshold outside the window and not divisible by M, then, the two scores disagree outright about whether the evidence is worth anything: “the evidence is worthless” is not one statement but one per score. What survives the choice of score is the structural core, the thresholds divisible by M, where posterior equals prior. Robustness across a family of priors is what makes a statement of this kind portable — the same test the strongest amnesia grade applies to a forgotten datum, flatness at every weight rather than at a named one (the four grades).

Scope. Proved, and exhaustive on the threshold family's 40,914 slices: 4,599 in the log floor set against 38,217 in the 0-1 floor set, entropies to 1e-12 nats. The structural core's independence of the prior is the window law below. Toy scale.

verifiers: explore_ceiling_anatomy.py, explore_ceiling_dials.py

The window law and the resonance rule

The prior was the one knob held fixed above. Replace the uniform prior by the geometric tilt, P(x) proportional to θx for a positive rational θ, and re-derive the threshold family from scratch under it; within a fiber the weights become a geometric ladder in q = θM, and every value — ceiling, floor and conditional entropy — stays closed-form. A fiber's below-mass exceeds one half exactly when its below-count exceeds

Q=logq1+qc2Q^{*} = \log_{q} \frac{1 + q^{c}}{2}

and since a cell offers exactly two fiber types, one apart, its interior thresholds are exactly one fiber window of M−1 thresholds located at ⌊Q*⌋ — unless Q* is itself an integer, in which case the window is empty. That exception is where the uniform prior lives: Q*c/2, which lands on an integer exactly when c is even, and that is where the window collapses to nothing. No scanned tilt off uniform put Q* on an integer, and whether any rational tilt can is open — so across the grid there is one live window at every tilt, and the law names which. The parity dial is therefore a resonance of θ = 1 rather than an invariant of the anatomy — every tilt off it frees the even cells. What does not move with the prior is the structural core: a threshold divisible by M leaves every fiber the same type whatever the tilt, so those cells stay dead at every θ and remain the whole log-loss floor set throughout, while the 0-1 floor set goes on adding everything outside the current window. A single fixed threshold is the brittle object: the sign bit dies at every tilt scanned, its even-c half structurally, since the midpoint threshold sits on a fiber boundary there and is dead at every prior in the tilt family, and its odd-c half at an algebraic death boundary, 2qm = 1 + qc with m = (c−1)/2 the middle fiber's index, whose root at c = 3 is the reciprocal golden ratio. The threshold family self-repairs where its extremal member does not: the window slides toward the bottom fiber as q → 0 and toward the top as q → ∞, and at no tilt on the grid is it empty.

Scope. The window law, the resonance and the sign bit's fragility proved; exhaustive on the 116 cells with N ≤ 210 across a nine-point θ grid at every threshold — 1,044 sweeps — plus 234 spot points up to N = 2310. The θ = 1 grid point reproduces the uniform law exactly, and the reflection identity, that tilting by 1/θ and reflecting the threshold leaves the ceiling unchanged, holds throughout. All 39 even-c cells are interior at each of the eight tilts and none at uniform; every sign-bit cell is dead at each of the eight. The threshold family is the only one re-derived under tilt — the orientation and comparison evals above stand at the uniform prior only. Three things are open and none is claimed above: whether a prior outside this one-parameter family also yields a single window, whether any tilt off uniform can put Q* on an integer at all, and what the other two task shapes do under a tilted prior. Toy scale.

verifier: explore_ceiling_dials.py

What these families make exact is a statement about the phrase “remaining headroom”. Headroom is a gap between a ceiling and a score, and the ceiling moves with the score asked and with the prior assumed — enough, in these families, for two scores to disagree about whether the evidence is worth anything, and for a family's dead cells to be an artifact of one prior. Where the family is designed, both dependencies are written down: the dial, the window, and the boundary each of them fails at.