Conflicting Supervision Moves Commitment, Not Capability: A 12.29σ arrangement effect that is exactly zero under a convention-agnostic score

Wenhui Chen

cs.AISep 23, 2026

Conflicting Supervision Moves Commitment, Not Capability: A 12.29σ arrangement effect that is exactly zero under a convention-agnostic score

Authors: Wenhui Chen

Organizations: University of Macau

Abstract

"Train a model on the same problems written under two incompatible conventions, both correct, and ask what the ordering of that data writes into the parameters. The learning-rate schedule is not a background condition for that question. It is the averaging operator, and it decides the answer. We prove a bound in which the arrangement and the schedule enter the ordering effect as separate multiplied factors: the arrangement only as a block period, the schedule only as how much weight the endpoint can place on any one moment of the run. A decaying schedule cannot put a large step size and an uncontracted remainder at the same moment; a constant one does exactly that at the last step. That decay moderates ordering effects has been reported in pretraining; the mechanism, the separation, and a controlled measurement of both halves are ours. Ten orderings of one corpus, one budget, everything but the path held fixed, run twice under families differing in lr_scheduler_type and nothing else: at a constant rate the interior spans 0.2221 in allocation, 11.63 contrast floors, monotone in how blocked the arrangement is. Under the single cosine every published arm uses, the same ten arms occupy two distinguishable states where their own resolution would allow about ten. "Order matters" and "order does not matter" are the two ends of one knob. What the path writes is which convention the model commits to, and no exact-match benchmark can see it. Across twelve arms acc_A+acc_B is constant to within 9.7% while the allocation share runs 0.04 to 0.87, so the 12.29-sigma arrangement switch this paper measures is exactly zero under a convention-agnostic metric. That conservation is quoted from the decayed family throughout, the constant-rate one being a noisier place to read it. Marking the convention in the prompt collapses the switch and reaches 87.5% of the union ceiling."

Figures & tables

Panel of Figure 1 — Figure 1: The paper as a map: one instrument, one wall, and the three terms that get past it. Read left to right. Every box carries a measured verdict rather than a description , which is the property that decides whether a map of this kind is worth printing: change the experiments and this picture changes, because two of its boxes would say the opposite of what they say and the wall’s three exits would be a different three. The instrument (Definition 2 ) is what makes the rest expressible, since the quantity every term here moves is invisible to the score a benchmark reports. The three exits are not alternatives to one another: the schedule decides whether an ordering survives to the endpoint at all, the stopping phase is the one term that survives either schedule, and the key is a second channel rather than more bandwidth on this one. Magnitudes are in contrast floors and are the body’s, not recomputed here.

Panel of Figure 2 — Figure 2: The space of data paths, and its image under each of the two schedules. (a) A data path is a monotone lattice path: at each optimiser step the trainer consumes a row from $A$ or from $B$ . There are $\binom{1728}{864}$ of them: the size of the space sampled, not a quantity anything here transmits. shuf hugs the diagonal, a ladder rung is a coarser staircase, blocked is the corner. (b) and (c) Its image, in the read-out’s own coordinates (Definition 2 ), every arm of the synthetic conflict, on the three seeds the two families share. The two panels differ in one field of the trainer configuration and in nothing else : same corpus, same paths, same budget, same seeds, same step. In both, the arms lie in a narrow capability band and spread along it, so capability is the band’s width and allocation is its length. Under the single cosine every published arm uses, the nine ladder arms fall in one cell of the $2\hat{\sigma}$ resolution of Definition 3 and blocked in another: two states, with eight empty cells between them where the range would support about ten, the eight interior arms spanning $0.042$ of share, $2.2$ contrast floors, below anything this design can resolve. Under a constant rate those same eight span $0.2221$ , $11.63$ floors, and $L{=}54$ leaves them for the corner. The difference between the panels is Proposition 1 : the endpoint is the schedule-weighted average of what the path visited, so a schedule that decays to zero has averaged the ordering away by the time anyone scores it. Axes are square and identical across the two panels, so both bands are true $45^{\circ}$ strips and the pictures can be compared by eye. Source: constlr_ladder_table .

Panel of Figure 3 — Figure 3: The division ladder under the single cosine, with the measured allocation at every rung (synthetic conflict, Qwen2.5-7B, three seeds; § 4.3 ). From one optimiser step to a quarter of the epoch, arrangement does not move the allocation: the whole interior of the ladder sits at the averaging limit. That flatness is the schedule’s and not the path’s : the identical ladder at a constant rate spans $11.63$ contrast floors in the same interior. Under this schedule the only escapes are at the top, stopping the alternation mid-swing (§ 5 ), and off the ladder entirely, via a query-time key (§ 5.1 ). Source: constlr_ladder_table .

Panel of Figure 4 — Figure 4: The instrument: one corpus that disagrees with itself. The two training halves are disjoint in problems and no single row disagrees with itself: the disagreement is a property of the corpus. The evaluation is the opposite, one held-out set scored twice against mutually exclusive conventions, so a sample can be credited to at most one. That is what makes the sum a probability of solving and the ratio a policy. Drawn for nat ; cf2 is the same construction at $864$ rows per half.

Panel of Figure 5 — Figure 5: The one thing that reads wrongly in prose. The read-out of Figure 4 as a change of coordinates: capability is constant along the dashed anti-diagonals, allocation along the dotted rays. A movement along a ray therefore changes capability at fixed allocation and a movement along a diagonal does the reverse, and an undecomposed accuracy cannot tell the two apart although their meanings are opposite.

	cf2 (flagship)	nat	status
what the two conventions are
construction	boxed answer vs. that $+1$	numeral vs. spelled out	—
is either wrong?	yes, by construction	neither	claim is about nat
rows per condition	$864$	$720$	—
learning rate	$3{\times}10^{-5}$	$1{\times}10^{-5}$	nat ran colder
how strong an instrument each is

Table 1: Two corpora, and which one carries which claim. The philosophical claim of this paper is a claim about nat , and nat is the weaker instrument ; we print the pair rather than leave it to be assembled from four sections. What travels between them is every qualitative statement the paper defends: capability conserved, allocation moved, interior flat, mirror antisymmetric at fixed

C

. What does not travel is size, and the direction is against us, so the headline figures in the abstract are cf2 figures and a reader rebuilding this on a both-correct conflict should expect two thirds the capability, and, once the step size is matched (§ 7 , branch NI-1 ), about seven eighths the install. The last block is the arms nat does not have; they are not claimed for it. § 7 gives the full accounting, including the one asymmetry the construction itself creates and the budget gate that measures how much of the size gap is the step size. Source: review_statistics , nat_replication .

build	control $D$	conflict $D$	conflict $\sigma$	verdict
v1 algebra pool, $1$ epoch	$+0.0461$ ( $2.41\sigma$ )	$+0.0020$	$0.10$	positive control fails
v2 pooled skills, $3$ epochs	$+0.0017$ ( $0.09\sigma$ )	$-0.1425$	$7.46$	control inert, switch fires
v3 the learning budget, lr $3{\times}10^{-5}$	$-0.0028$ ( $0.15\sigma$ )	$\mathbf{-0.2346}$	$\mathbf{12.29}$	control inert, switch fires

Table 2: The instrument, across its three builds.

D

is the arrangement contrast on the last-seen convention, signed :

D=\mathrm{acc}(\text{last seen})_{\textsc{blocked}}-\mathrm{acc}(\text{last seen})_{\textsc{shuf}}

with the sources in the published order, so the switch is negative. Elsewhere the same quantity is quoted as a span between two arms and is therefore positive;

-0.2346

here and

+0.2275

in § 4.3 are the same measurement at eight and at three seeds, not two results with different signs. The control is the same construction with both halves under one convention, so it must stay inert; the paragraph below reads v1, where it did not. v1 and v2 are three-seed builds; v3 and its control carry eight. Sources: the conflict_switch , conflict_switch2 and conflict_switch2b3lr3e5 verdicts.

arm	$\mathrm{acc}_{A}$	$\mathrm{acc}_{B}$	capability	share of $B$
$A$ only	$0.483$	$0.018$	$0.502$	$0.04$
$B$ only	$0.064$	$0.423$	$0.487$	$0.87$
shuf	$0.310$	$0.218$	$0.528$	$0.41$
purerand	$0.303$	$0.224$	$0.527$	$0.43$
$L=1\ldots 54$	$0.272$ – $0.318$	$0.216$ – $0.265$	$0.519$ – $0.537$	$0.41$ – $0.49$
$\text{A}\!\to\!\text{B}$ ( blocked )	$0.067$	$0.445$	$0.513$	$0.87$

Table 3: The conservation decomposition (synthetic conflict, Qwen2.5-7B, three seeds). Capability

\mathrm{acc}_{A}{+}\mathrm{acc}_{B}

is flat across every arrangement; only the allocation moves. Source: the cf2b3lr3e5 / ctl2b3lr3e5 result families (

3

epochs, lr

3{\times}10^{-5}

n{=}200

), named because a second conflict family at a different budget also sits in the result base, and the two must not be pooled.

Panel of Figure 6 — Figure 6: Conflict makes the policy stochastic, not the model weak. Per-problem commitment at $k{=}4$ samples on the synthetic conflict, three seeds, $200$ problems each. Bars within an arm run $0/4$ to $4/4$ . The committed arms are bimodal (the model either answers a problem the same way every time or cannot answer it) while the mixed arms move that mass into the middle without changing the unsolved fraction (grey, leftmost bar in each group). Same corpus, same volume, same budget; only the data path differs. Source cf2b3lr3e5 per-problem arrays.

transition	count	frac	transition	count	frac
$A\to B$ (re-labelled)	$179$	$0.298$	$U\to B$ (newly solved)	$26$	$0.043$
$U\to U$ (never solved)	$119$	$0.198$	$A\to U$ (lost)	$28$	$0.047$
$B\to B$ (stable)	$104$	$0.173$	$U\to A$ (newly solved)	$15$	$0.025$
$A\to A$ (stable)	$28$	$0.047$	$B\to U$ (lost)	$16$	$0.027$
$B\to A$ (re-labelled)	$7$	$0.012$	ties, either side	$78$	$0.130$

Table 4: Per-problem transitions from shuf to blocked (synthetic conflict, three seeds

\times

200

problems). A problem’s state is the convention receiving more of its

k{=}4

samples (

A

B

, tie) or unsolved (

U

). The solved set is nearly fixed while nearly half of it changes label,

26\times

more often toward the convention trained last. The tie row carries the

3

transitions

U\to

tie and the

7

tie

\to U

, which is why the prose’s newly-solved and lost counts (

44

and

51

) exceed what the four

U

rows sum to (

41

and

44

). Source cf2b3lr3e5 per-problem arrays.

Panel of Figure 7 — Figure 7: Capability and allocation move on separate clocks, which the range alone hides. The blocked construction’s second stage: a model trained to completion on $A$ and then trained on $B$ , checkpointed every four optimiser steps over that stage’s first $54$ , three seeds averaged. It is not a ladder rung, and its allocation ends at $0.718$ heading toward blocked ’s $0.869$ rather than at the $L{=}54$ rung’s $0.492$ ; (top) allocation $\mathrm{acc}_{B}/\mathrm{TOTAL}$ , (bottom) capability $\mathrm{acc}_{A}+\mathrm{acc}_{B}$ with a band of $\pm$ its floor $0.0270$ around the endpoint. The shaded column is the first four steps. Capability spends $97.4\%$ of its total excursion there, while allocation spends $4.5\%$ of its, so the capability dip precedes the commitment shift rather than accompanying it. The verdict file’s audit fields record that both evaluations use the same $200$ questions with disjoint golds.

Panel of Figure 8 — Figure 8: The averaging wall, measured. Same corpus, volume, budget and seeds; only the data path differs (Definition 1 ). (a) Allocation is flat from $L{=}1$ to $L{=}27$ and across shuf and purerand : $0.407$ to $0.449$ over a $27\times$ change in block length and a change of batch composition. $L{=}54$ is the first rung off that floor at $0.492$ and blocked sits at $0.856$ . (b) Capability over the identical arms spans $0.0183$ , inside the contrast floor: the path relocates commitment without changing what can be solved. Bars are the seed standard deviation, three seeds per point ( shuf and blocked carry eight). Source cf2b3lr3e5 .

arm	cosine	constant	difference	floors
shuf (mixed within batch)	$0.4129$	$0.3630$	${-0.0499}$	$2.6$
$L{=}1$	$0.4491$	$0.4254$	${-0.0237}$	$1.2$
$L{=}3$	$0.4126$	$0.4505$	${+0.0379}$	$2.0$
$L{=}6$	$0.4363$	$0.4592$	${+0.0229}$	$1.2$
$L{=}9$	$0.4140$	$0.4498$	${+0.0358}$	$1.9$
$L{=}18$	$0.4257$	$0.4823$	${+0.0566}$	$3.0$

Table 5: The same ladder under two learning-rate schedules. Ten arms, one corpus, one budget, the identical data path files, and the same three seeds : the intersection, computed rather than assumed, because the published shuf and blocked arms carry eight and a mixed comparison would print most of the eight-against-three difference as a schedule effect. The two families differ in lr_scheduler_type and in nothing else, which we checked against the training arguments and the corpus checksum rather than assuming, and both run to the same step (

324

on the ladder,

162

at the corner). Under a cosine the interior spans

0.0420

, which is

2.20

contrast floors and below anything this design can resolve. Under a constant rate the same interior spans

\mathbf{0.2221}

, which is

11.63

floors, with an intraclass correlation of

0.836

whose bootstrap interval

[0.315,0.926]

lies entirely above zero. The corner does not move : blocked reads

0.8692

and

0.8501

, one contrast floor apart, against the

17.0

floors the rung below it moves. Source: constlr_ladder_table , constant_lr_verdict .

	$1{:}1$ (rr11)		$2{:}1$ (rr21)
	share	capability	share	capability
mean-field prediction	$0.500$	—	$0.333$	—
shuf (measured)	$0.453$	$0.372$	$0.312$	$0.377$
blocked $\text{A}\!\to\!\text{B}$	$0.731$	$0.380$	$0.699$	$0.381$
$B$ only (guard: $\mathrm{acc}_{B}=0.263/0.187>$ floor)	$0.713$	$0.368$	$0.595$	$0.315$

Table 6: The share tracks the mixture weights; the blocked arm does not (natural-conflict ratio corpora,

936

rows, three seeds; read-out rules frozen in the job script the scheduler executed: T1). The

3{:}1

form was excluded by its own learnability guard and redesigned at

2{:}1

; the redesign is labelled, not laundered.

intervention	mean-field prediction	measured	miss	moves the share?
	allocation share of $B$
mixture ratio $1{:}1\to 2{:}1$	$0.500\to 0.333$	$0.453\to 0.312$	$0.047,\;0.021$	yes, as predicted
block length $L{=}1\to 54$	no change predicted		—	no ( $0.407$ – $0.492$ )
batch purity, fixed blocks	no change predicted		—	no ( $3\%$ of the span)
palindromic reordering	no change predicted		—	no (and $4\times$ worse)

Table 7: The one lever the averaging limit leaves free. Proposition 1 ’s mean field is weighted by source proportions , so changing the mixture ratio must move the allocation and changing the path must not. Both halves are measured here on the same corpus family. Tolerance

0.05

was frozen with the prediction; both misses fall on the side of the pretrained prior, which the family measures independently. (T1.)

check	registered bar	measured
S3 order effect of a palindrome	$\mathrm{ASYM}\equiv 0$	$0$ to machine precision	passes
S1 median $\|D_{\text{strang}}\|/\|D_{\text{blocked}}\|$	$<0.5$	$\mathbf{3.65}$ over $300$ draws (IQR $1.88$ – $8.09$ ); $3.7\%$ of draws below the bar	fails
S2 log–log slopes in $\eta$ differ	by $\geq 0.5$	both slopes $0.0$ to four figures across $\eta=0.4\ldots 0.025$	fails
the penalty across a $16\times$ refinement of the step, which is what “no order to improve” means
$\eta=0.4,\,0.2,\,0.1,\,0.05,\,0.025$
blocked		$0.081646,\ 0.081638,\ 0.081634,\ 0.081632,\ 0.081631$

Table 8: The construction splitting theory recommends, and what it did. Registered in the linear model, where every arm is computable exactly. S3 is an implementation guard, not a claim: a palindromic schedule is its own reverse, so its order effect must vanish, and it does. S1 and S2 are the claim and both fail: S1 by a factor of seven against its own bar, and S2 degenerately , since refining the step by

16\times

moves either penalty in the fifth decimal. Source splitting_verdict .

Panel of Figure 9 — Figure 9: Two ways of leaving the averaging wall, one designed and one accidental. (a) The palindromic schedule splitting theory recommends, against the blocked one it was built to beat, in the linear model where the joint arm is available in closed form. It is roughly four times worse , and refining the step by $16\times$ moves either penalty in the fifth decimal: there is no order to improve because the penalty is not a discretisation error. (b) The one intervention that does move the contrast is leaving the lazy regime. Raising the learning rate grows $|D|$ monotonically, $0.0236\to 0.0653$ , through a change of sign; the highest rate also costs capability, $0.268$ against $0.340$ , so it is the system leaving the regime the averaging argument is tight in.

axis swept	on the path?	share, one end to the other	contrast floors
stopping phase, shuf against blocked	yes	$0.4129\to 0.8692$	$23.89$
model scale, 3B to 14B	no	$0.1322\to 0.4732$	$17.85$
mixture ratio, 1:1 to 2:1	no	$0.4531\to 0.3116$	$7.41$
cosine restart, byte-identical path	no	$0.4071\to 0.2912$	$6.07$
block length, $L{=}1$ to $L{=}54$	yes	$0.4071\to 0.4922$	$4.46$
batch purity at fixed block structure	yes	$0.4129\to 0.4253$	$0.65$

Table 9: Every axis this project has swept, on one scale, sorted by how far it moves the allocation. All six are measured on the allocation share of a mixed arm at one model and one budget, so the column is commensurable; the learning-rate sweep is reported on the contrast

D

rather than on the share and is deliberately absent rather than forced onto this axis. The stopping phase is the largest term measured anywhere in the project, which is why it gets a section. The block-length ladder ranks fifth of six, and all three non-path axes move the allocation further than the whole ladder does. Source: axes .

Panel of Figure 10 — Figure 10: The stopping phase is a position on a cycle, and the mirror moves it to the other side. (a) The reading of Proposition 3 , labelled post hoc: a driven system oscillates about the mean-field point with amplitude growing in $L$ , and an endpoint measures where the path stopped. It makes one frozen prediction. (b) The measurement (read-out M1): at $L{=}54$ the shift is antisymmetric, $+0.0850$ against $-0.0875$ , with capability moving $0.0025$ . At $L{=}27$ both constructions sit at the floor and the mirror manufactures nothing, as it must.

Panel of Figure 11 — Figure 11: What the key opens, and why what it leaves is a different obstruction. (a) The unkeyed model sits at exactly half the additive ceiling, which is what a coin flip between two mutually exclusive golds scores; the key recovers $74.9\%$ of the gap and leaves $0.0468$ . (b) That residue is coverage , and this panel is the whole of the evidence. Marked $B$ - only , trained on nothing but the spelled convention, answers in numerals at $0.3767$ because numerals are the pretrained default and are therefore written; marked $A$ - only asked for a word scores $0.0025$ , because there is nothing to select. (c) And the residue grows on a family that writes four times less of the minority convention, $+2.92$ floors against a bar of two frozen before the run (V1). It is a necessary-condition test, which V2 or V3 would have refuted and which V1 cannot establish. Sources unlock_verdict , q3unlock_verdict .

read-out	quantity	unkeyed	keyed	verdict
C	arrangement switch $D$	$-0.0985$ ( $5.16$ floors)	$\mathbf{-0.0006}$ ( $0.03$ floors)	C1 : it collapses
L	mean accuracy	$0.1863$	$\mathbf{0.3259}$	L4 : partial unlock
	against ceiling $0.3727$	$50.0\%$	$87.5\%$	$74.9\%$ of the gap
	against L1 target $0.3354$	—	short by $0.50$ floors
the residue, and why it is coverage rather than difficulty
	marked $B$ - only on the numeral evaluation	$0.1283$	$\mathbf{0.3767}$	the key selects

Table 10: What the key opens, and what it cannot. The read-out was frozen in the job script before the runs (C on the switch, L on the level), and the analyzer recovers the three constants it was frozen against from the result base. The residue has a mechanism rather than an excuse: a marked

B

- only model, trained on nothing but the spelled convention, answers in numerals when the key asks for numerals, but a marked

A

- only model asked for a word cannot produce one it never saw. The key selects among conventions already written.

Panel of Figure 12 — Figure 12: Three results asked to travel, and the answers are not uniform. (a) Every quantity whose threshold was frozen before the second family ran, divided by that threshold, so the bar sits at $1$ for all of them and the arrow says which side had to be cleared. Hollow is Qwen2.5-7B, filled is Qwen3-8B-Base, and the line between them is the travel. The guards pass well: the conflict installs better on the new family, so nothing below is an instrument at its floor. The key collapses the switch to $3\%$ of its bar on one family and $66\%$ on the other, and the residue grows as the coverage reading requires. The ladder’s span misses and so does the published row at $1.09$ , which is a fault in the registration rather than in the second family. Rows are ordered by what they decide, not by outcome, and the post-hoc statistic is ruled off. (b) The mirror is the one registration with no scalar bar: its rule is a shape, so it is drawn as one. Sources q3unlock_verdict , q3ladder_verdict , unlock_verdict , rate_verdict .

	quantity	Qwen2.5-7B	Qwen3-8B-Base	read-out
guards	minority convention written, $\mathrm{acc}_{B}(B\ \textsc{only})$	$0.2091$	$0.0520$	—
	on the synthetic conflict	$0.4225$	$\mathbf{0.4813}$	G1 passes
	switch present, $\|D\|$ on that corpus	$11.9$ fl	$\mathbf{15.0}$ fl	G2 passes
the key	$D$ unkeyed	$-0.0985$ ( $5.16$ fl)	$-0.0875$ ( $4.58$ fl)
	$D$ keyed	$-0.0006$ ( $0.03$ fl)	$-0.0125$ ( $0.66$ fl)	K1 collapses
	marked mean against its ceiling	$87.5\%$	$75.5\%$	L4 partial

Table 11: The key, and the ladder, asked to travel. Both were registered before their runs with the branches and thresholds frozen, and both gates were read first. The key replicates and its residue moves as the coverage reading requires. The ladder’s registered branch is W2 , and the row below it is why that verdict is reported together with the threshold it was read against: the published Qwen2.5 ladder does not meet the same bar, which was checkable before the run and was not checked. Sources: q3unlock_verdict , q3ladder_verdict .

registered claim	the rule, frozen first	what came back	verdict
v1 batch composition carries the switch	purerand near blocked $\Rightarrow$ purity; near shuf $\Rightarrow$ order	purerand $0.2242$ against shuf $0.2179$ , blocked $0.4454$ : purity is $3\%$ of the span	dead
v2 optimiser memory cancels contested directions	a response turning on at $L^{\star}\!\approx\!10$ , or at $\tau_{\text{write}}\!=\!13.6$	$M(L)$ flat through both onsets, $\sigma(M)\!\approx\!0.06$ ; only $L{=}54$ rises	dead
Strang a palindromic schedule beats a blocked one	S1 median $\|D_{\text{str}}\|/\|D_{\text{blk}}\|<0.5$ ; S2 slopes differ by $\geq 0.5$	S1 $3.65$ ; S2 both slopes $0$ ; S3 passes	dead
T1 allocation tracks the mixture weights	share within $0.05$ of mean-field $0.500$ and $0.333$	$0.453$ and $0.312$ : misses $0.047$ and $0.021$ , both toward the pretrained prior	held
Ladder every block length lands at the interleaved allocation	$L{=}1\ldots 54$ within noise of shuf ’s own share	$0.407$ – $0.449$ against shuf ’s $0.413$	held
M1 the mirror relocates antisymmetrically at fixed $C$	ending on $A$ moves the same magnitude to the other side	$+0.0850$ against $-0.0875$ , capability moves $0.0025$	held

Table 12: The ledger: every claim this paper registered, and what came back. Read the verdict column as a distribution rather than a score: three mechanisms died, four predictions held, one control was void and one instrument build inverted. The three deaths were derivable in advance from Proposition 1 , which is why they are kept rather than deleted. Rules are quoted from the artefact that froze them; Appendix D gives the commit and the interval for each.

	share
rung	control (one cosine)	treatment (two)	$\Delta$	$t$ (crit $2.776$ )	guard $\|\Delta C\|$
$L{=}27$ primary	$0.4071$	$\mathbf{0.2912}$	$\mathbf{-0.116}$	$\mathbf{-10.10}$	$0.59$ fl
	$0.413,0.419,0.389$	$0.294,0.302,0.278$	$6.07$ fl	SC-3	passes
$L{=}54$ secondary	$0.4922$	$0.2053$	$-0.287$	$-2.35$	$0.26$ fl
	$0.424,0.526,0.527$	$0.024,0.424,0.168$	$15.02$ fl	SC-2	passes
for scale: the entire block-length ladder, $L{=}1$ through $54$ , moves the share $0.085$

Table 13: The schedule control: identical path, one cosine against two. The treatment consumes the control’s own stream in the same order and differs only in that the cosine restarts and the optimiser state resets at the midpoint, which is blocked ’s structure. The guard that voided the single-cosine control is read first and passes on both rungs.

L{=}27

was fixed as primary before either arm ran, because its control dispersion is four times smaller. Its movement is

6.07

contrast floors at

t=-10.1

L{=}54

moves further and does not clear its own noise: its treatment seeds span

0.40

, wider than the effect, and the registration recorded in advance that at this dispersion its silence is not evidence of absence. That is why the mechanical branch label SC-2 is reported with that sentence beside it rather than as “no effect”. Both rungs move the same way, which the registration fixed in advance as the signature of the schedule rather than of a mid-run stopping phase. Source schedctl_verdict .

inference time (CCH)	training time (this paper)
bounded state $\Sigma$ , budget $B$	the parameters, reached only through the update path
the stream: bindings $+$ distractors, kernel $\kappa$	the data path; conflict $\kappa_{A}\neq\kappa_{B}$ between sources
compressive $O(1)$ channel (state mixing)	the averaging limit: only the path’s time-average is written (§ 4.3 )
horizon wall: a fixed window; deferral, not escape	the stopping phase: recency as the one non-average term (§ 5 )
verbatim index channel, cost $\Theta(L)$	a query-time key in the data; selection moves from path to query (§ 5.1 )
write-time code separability (Assumption 1)	the key selects only among conventions already written: $0.3767$ vs $0.0025$ (§ 5.1 )

Table 14: The CCH family: the same objects at inference time and training time. Left column: Chen et al. [2] . Right column: this paper and, for the premise rows, the companion recipe-level paper [ 5 ] .

Appendix figures & tables11 assets

Supplementary material from the paper’s appendix.

Appendix

the mechanism, in-model (random non-commuting pairs, lazy regime)
sign reversal of $\mathrm{ASYM}(T)$	across the budget sweep		$\mathbf{72\%}$ of draws
sign reversal of $D(T)$	across the budget sweep		$\mathbf{70\%}$
the same, at $\eta/\eta_{\text{stab}}=0.1$	step size is not the cause		$0.867$ / $0.725$
median rotation at the crossing	sum / difference		$11.2^{\circ}$ / $11.7^{\circ}$
the one-shot proxy, asked to predict where (Spearman)
in-model, $T^{\star}$ from Magnus	against the measured crossing budget	$278$	$+0.031$

Appendix

Table 15: The mechanism survives every test; the one-shot proxy for where it fires survives none. Upper panel: budget-dependent sign reversal is generic in the lazy regime, so the stopping term of Proposition 3 is not an artefact of our corpora. Lower panel: the leading Magnus/commutator score, the quantity a practitioner would compute once at initialisation to predict the crossing: is asked to rank three different things in three different places and fails all three, twice with the wrong sign . Reported together because they are one settled negative about a single object, not three separate disappointments.

Panel of Figure 13 — Figure 13: One object, confirmed as a mechanism and failed as a predictor. The same two halves as Table 15 , drawn, because both halves are claims about a distribution. (a) Over random non-commuting pairs in the lazy regime, $\mathrm{ASYM}(T)$ reverses sign in $72\%$ of draws and $D(T)$ in $70\%$ , and the rate stays above chance across a $19\times$ sweep of the step size, so the reversal is something the finite-budget theory produces rather than a property of our corpora. The dotted series is the fraction of budgets the sweep can still represent, which falls above $\eta/\eta_{\text{stab}}=1.3$ and is why the $D$ series falls with it; it is drawn rather than left out. (b) Every ranking the one-shot commutator score is asked for, on one axis. It is uncorrelated with the in-model crossing budget over $278$ pairs and inverted against the measured order effect at both budgets. Sources asym_budget_verdict , commutator_verdict .

learning rate	$\bar{D}$	per-seed $D$	$\overline{\mathrm{acc}}_{\textsc{only}}$
$1\times 10^{-5}$	$-0.0236$	$-0.0125,\;-0.0188,\;-0.0396$	$0.3014$
$3\times 10^{-5}$	$+0.0507$	$+0.0229,\;+0.0833,\;+0.0459$	$0.3396$
$1\times 10^{-4}$	$+0.0653$	$+0.0729,\;+0.0604,\;+0.0625$	$0.2680$

Appendix

Table 16: The switch grows with the learning rate, monotonically, and the arm that is supposed to be inert does not move with it. Three learning rates at three epochs on the natural conflict, three seeds each;

\bar{D}

is the mean allocation contrast and

\overline{\mathrm{acc}}_{\textsc{only}}

the mean single-convention accuracy, which is what a capacity explanation would have to move and does not. Raising the rate is the one intervention that takes the system further from the lazy regime the theory is derived in, and

|D|

grows with it: which is the direction the suppression argument predicts.

Panel of Figure 14 — Figure 14: The commutator proxy fails twice, and the two failures are different. (a) The score a practitioner would compute once at $\theta_{0}$ ranks the pairs backwards , at both budgets (Spearman $-0.543$ at one epoch, $-0.600$ at three). A weak predictor scatters about zero; this one is consistently inverted, which is stronger than uninformative. The dashed line is the contrast floor. (b) The deeper failure: four of six pairs change the sign of their order effect between budgets, so there is no fixed ordering for any once-computed quantity to predict. Source commutator_verdict .

pair	$\\|[A,B]\\|$	$T^{\star}$	$\mathrm{ASYM}$ ( $1$ ep)	$\mathrm{ASYM}$ ( $3$ ep)	sign flips?
algebra–calculus	$89{,}002$	$0.381$	$+0.0743$	$+0.0020$
algebra–geometry	$81{,}524$	$0.459$	$+0.0320$	$-0.0333$	yes
algebra–combinatorics	$70{,}810$	$0.281$	$+0.0779$	$-0.0619$	yes
calculus–geometry	$131{,}642$	$0.319$	$-0.0396$	$+0.0041$	yes
calculus–combinatorics	$112{,}524$	$0.355$	$+0.0104$	$-0.0125$	yes
geometry–combinatorics	$95{,}175$	$0.361$	$+0.0506$	$+0.0091$

Appendix

Table 17: The six pairs the proxy is asked to rank, and the ranking it produces. The commutator norm is evaluated once at

\theta_{0}

on Qwen2.5-7B activations at layer fraction

0.75

;

T^{\star}

is the leading Magnus budget built from it;

\mathrm{ASYM}

is the measured order effect, three seeds, contrast floor

0.0191

. Two things are fatal to a static predictor and both are visible by eye: the commutator ranks

|\mathrm{ASYM}|

with the wrong sign at both budgets, and four of six pairs change that sign between budgets, so the quantity being ranked is not a property of the pair alone. Source commutator_verdict .

	$\Rightarrow$ the $+1$ convention lost to the pretrained prior; $D$ measured at the instrument’s floor
build	arm	against the shifted gold	against the unshifted gold
v1 (algebra pool, $1$ epoch)	conflict arms	$\approx 0.02$	$0.39$ – $0.50$
the read-out that followed, unchanged in rule
v1	control $D$	$+0.0461$ ( $2.41\sigma$ ) — the arm that should have been inert moved
v1	conflict $D$	$+0.0020$ ( $0.10\sigma$ ) — the arm that should have moved did not

Appendix

Table 18: Why v1 read at the floor: the treatment was never installed. Every conflict arm is scored twice, against the shifted golds the

+1

convention defines and against the unshifted ones the pretrained model already produces. Had the conflicting convention been learned, the first column would carry the mass. It does not, so

D

in v1 was a difference between two arms neither of which had received the treatment, and the control’s

2.41\sigma

was the only variance in the experiment. Source the conflict_switch family, three seeds,

238

rows per condition at one epoch.

Panel of Figure 15 — Figure 15: An instrument reporting a null because its treatment was absent. (a) v1’s conflict arms scored against both golds. The $+1$ convention the arms were trained on is worth ${\approx}0.02$ ; the unshifted convention the base already writes is worth $0.39$ – $0.50$ . Had the treatment installed, the first bar would carry the mass. (b) What the read-out rule then returned, in units of the seed standard error, and what the same rule returned after the redesign. In v1 the arm registered to stay inert moved $2.41$ and the arm registered to move returned $0.10$ : the control is the only thing in the experiment with variance to report. The v3 pair is the three-seed read-out, matched to v1’s three seeds rather than the eight-seed pair Table 12 prints. Nothing about the rule changed between them. Sources conflict_switch_verdict , conflict_switch2b3lr3e5_verdict .

resolution	$0.5\hat{\sigma}$	$1\hat{\sigma}$	$1.5\hat{\sigma}$	$\mathbf{2}\hat{\sigma}$	$3\hat{\sigma}$	$4\hat{\sigma}$
Qwen2.5-7B, clusters	$4$	$3$	$3$	$\mathbf{2}$	$2$	$2$
$R_{\mathrm{real}}$ (bits)	$2.00$	$1.58$	$1.58$	$\mathbf{1.00}$	$1.00$	$1.00$
$R_{\mathrm{cap}}$ (bits)	$5.30$	$4.30$	$3.72$	$\mathbf{3.30}$	$2.72$	$2.30$
Qwen3-8B-Base, clusters	$4$	$2$	$2$	$\mathbf{2}$	$2$	$2$
$R_{\mathrm{real}}$ (bits)	$2.00$	$1.00$	$1.00$	$\mathbf{1.00}$	$1.00$	$1.00$

Appendix

Table 19: The headline pair, swept over the resolution constant it is defined at.

R_{\mathrm{real}}

counts the endpoint states the ten arms occupy under single linkage;

R_{\mathrm{cap}}

is the logarithm of the range in units of the resolution and therefore gains exactly one bit per halving, which

R_{\mathrm{real}}

does not. Two clusters is what both families give from

2\hat{\sigma}

upward , and the second family gives it from

1\hat{\sigma}

upward. Below

2\hat{\sigma}

the first family resolves a third state, so the paper’s constant sits at the lower edge of the interval where the two families agree rather than in its middle, rather than presenting

1

bit as resolution-free. The interval is wide above the choice and the between-cluster gap exceeds the largest within-cluster gap by

8.7\times

on the first family and

20.7\times

on the second, which is why the count is stable there. Source: rate_sensitivity .

Panel of Figure 16 — Figure 16: The count, and the two reasons it is a count rather than a rate. (a) The ten arm endpoints of the cosine family on the share axis, with the $2\hat{\sigma}$ window that decides whether two of them are one state drawn at the size it actually is. Nine arms fall inside a span whose largest internal gap is $0.0431$ ; the tenth sits $0.377$ away, $8.7$ times that gap. The separation is not a knife-edge and does not need the constant to be exactly two. (b) The constant swept. The number of resolvable states is flat from $2\hat{\sigma}$ to $4\hat{\sigma}$ , while $R_{\mathrm{cap}}$ , which is $\log_{2}$ of a volume ratio rather than a count, falls at every step of the sweep. Sources rate_verdict , rate_sensitivity .

claim	outcome	artefact carrying the threshold	registered	result (verdict file, mtime)	gap
instrument, the positive control	confirmed	slurm_natconflict.sbatch	65acb7a 08-01 12:52	natconflict 08-03 08:23	$+1$ d 19h
v1 batch composition	confirmed	slurm_batchcomp.sbatch , decision rule	2e5f5a9 08-02 12:42	batchcomp 08-02 15:32	$+2$ h 50m
v2 and the ladder	no branch fired	slurm_pathspec.sbatch and PLAN.md : predicts $L^{\star}\!\approx\!10$	cd53c09 08-02 16:02	pathspec 08-03 02:33	$+10$ h 30m
M1 the mirror	M1 confirmed	slurm_dpd_mirror.sbatch : an asymmetry falsifies the limit-cycle picture	a85c247 08-03 15:32	dpd_mirror 08-06 07:44	$+2$ d 16h
C , L the key	C1 , L4 confirmed	slurm_disambig.sbatch , C on the switch and L on the level	4830e37 08-04 06:46	disambig 08-04 10:49	$+4$ h 03m
T1 the mixture ratio	T1 confirmed	slurm_ratio2.sbatch , tolerance $0.05$ frozen with the prediction	ff94a95 08-05 06:22	ratio2 08-05 16:20	$+9$ h 58m

Appendix

Table 20: Seventeen registrations, what each returned, and the interval each one bought. Seven confirmed at their primary branch, two decided at another, two came back partial, three void on a guard, and three returned no branch at all. Fifteen were frozen from about an hour to nearly three days ahead of the result they decided. Two are not, and both are stated rather than averaged away: the palindromic check is a linear-model computation of a few seconds whose thresholds and whose answer entered the repository in the same commit, so it can assert an ordering it cannot demonstrate; and the amplitude-onset registration was withdrawn before any training because the rungs it named cannot be built at this budget, which is a fact about the ladder rather than about either family. The second-family replications carry positive intervals and are nonetheless printed here together with their answers rather than announced in an earlier version, which is the weaker of the two guarantees a registration can offer. The last two rows’ intervals are measured against a file timestamp on the shared volume rather than against a commit, which is weaker again and is why those cells say no commit instead of a hash. Of the two only NI is still recomputable from this tree: CM ’s registration gained an outcome section after its verdict, and writing that section overwrote the timestamp its interval was measured from. NI ’s was left byte-frozen for that reason and its outcome is recorded here and in its verdict file instead.

Panel of Figure 17 — Figure 17: Seventeen registrations, and the interval each one bought. Table 20 in one axis: the distance between the commit that froze a threshold and the file that answered it. Fifteen run from about an hour to nearly three days ahead of their result. The two that are not intervals are drawn on the same axis rather than dropped from the count: the palindromic check put its thresholds and its answer in one commit, so it can assert an ordering it cannot demonstrate, and the amplitude-onset registration was withdrawn before any training because the rungs it named cannot be built at this budget. Source tab_registry , whose gap column is computed from the commit and the verdict file’s mtime.

Learning Rate Scheduling

Learning rate scheduling studies how to vary optimizer step sizes during training to improve stability, convergence, efficiency, and generalization.

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Mar 24, 2026

The Order Is The Message

In a controlled experiment on modular arithmetic (

p = 9973

), varying only example ordering while holding all else constant, two fixed-ordering strategies achieve 99.5% test accuracy by epochs 487 and 659 respectively from a training set comprising 0.3% of the input space, well below established sample complexity lower bounds for this task under IID ordering. The IID baseline achieves 0.30% after 5{,}000 epochs from identical data. An adversarially structured ordering suppresses learning entirely. The generalizing model reliably constructs a Fourier representation whose fundamental frequency is the Fourier dual of the ordering structure, encoding information present in no individual training example, with the same fundamental emerging across all seeds tested regardless of initialization or training set composition. We discuss implications for training efficiency, the reinterpretation of grokking, and the safety risks of a channel that evades all content-level auditing.

Jordan LeDoux

Independent Researcher

Data-Efficient Learning Training Data Curation Frequency-Domain Feature Learning Grokking

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Jun 28, 2026

Optimizer Memory Makes Shuffle Order a First-Order Source of Fine-Tuning Noise

Shuffle order can be a larger source of fine-tuning noise than a memoryless analysis predicts: fixed-clock optimizer memory makes local equal-multiset contrasts first order in the learning rate rather than second order, and the resulting order channel can be large enough for a single seed to flip a close A/B comparison. We isolate this mechanism and derive a fit-free way to size the noise it produces. For a memoryless optimizer, reordering an equal multiset has no first-order endpoint term; the leading local contrast is the

O(η^2)

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τ=ηk

, so the same gradient can receive a position-dependent endpoint weight. For any fixed finite measurement window, a lifted-state expansion gives an

O(η)

equal-multiset contrast whenever the first-order replay coefficient is nonzero, while regular and clock-matched controls remain

O(η^2)

; a bare fixed-

β

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β

momentum, and 4.00 for SGD; matching the memory clock to

τ

restores the regular exponent. For AdamW with a frozen preconditioner, the same impulse-weight kernel gives a closed-form asymptotic order-variance floor after the local potentials are measured, with no fitted coefficients. The result is local to the measurement window (independent reshuffling can average the channel across windows), but it yields order-noise error bars, positional attribution weights, and a seed-budget criterion for fine-tuning comparisons.

John Sweeney

Stochastic Optimization Fine-Tuning Adaptive Gradient Methods

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Oct 31, 2025

Training LLMs Beyond Next Token Prediction - Filling the Mutual Information Gap

Autoregressive language models are usually trained with next-token prediction under a fixed left-to-right target order. This paper studies target-token organization as a training-time design choice across arithmetic reasoning, multi-label text classification, and text generation. Controlled order studies show that the default NTP order can be far from the best-performing tested organization, while fixed hand-crafted orders such as reverse order do not generalize across tasks. To address this, Max(MI(S;t)) is introduced as a deterministic mutual-information-based heuristic that prioritizes information-rich target tokens. For fixed-length outputs, it constructs task-specific prediction orders; for open-vocabulary text generation, the same principle is applied through lightweight prefix augmentation. Experiments with GPT-style models and pretrained Llama/Qwen variants show that Max(MI(S;t)) often improves over standard NTP and simple baselines such as reverse order, TF-IDF, and conditional-probability selection, without changing model architecture or autoregressive decoding.

Chun-Hao Yang, Bo-Han Feng, Tzu-Yuan Lai +3

National Taiwan University

Autoregressive Language Modeling Sequence Prediction Autoregressive Transformer

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Abstract

Learning Rate Scheduling

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