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."
From: Wenhui Chen [view email]
[v1]
Wed, 23 Sep 2026 01:33:30 UTC (146 KB)