Selector Width Limit

This overview explains why input width alone is not enough to choose a fast execution path. This matters because a selector can look good on typical cases while making one case much slower than the fixed control. Both rules mention four Boolean inputs. The second repeats A across several branches and offers sharing or factoring opportunities that the first does not present in the same way. Width is one descriptor of the job, not a complete description of its structure. Imagine a router whose feature extraction costs three units and whose chosen exact arm costs nine. A fixed exact method costs ten. The routed total is twelve, so the selected arm's attractive nine-unit execution does not make the dispatcher a winner. With regret defined as routed total divided by the fixed reference, this case has a ratio of one point two. Keep feature extraction and any conversion inside the boundary when the caller must pay them. The ratio convention must be declared. A selector chooses an execution path using information available before it runs. Count the cost of obtaining that information. Here, regret is the selector's total time divided by the fixed control's time. Every case must stay below two. A good median alone can hide a severe losing case. Imagine a toy policy that chooses the packed-word engine above a width threshold. Most cases cost one unit relative to the fixed control, but one costs ten. The median looks harmless while that case takes ten times the control's time. If the frozen rule requires every regret ratio below two, this policy fails. This arithmetic illustrates a gate; it is not a new retained selector experiment. The corrected focused width study did not establish a universal switch rule and reused validation. Later retained q64 selector work also promoted no learned route. Keep those cohorts distinct, and retain the exact fixed fallback rather than silently installing a heuristic suggested by the animation. Sort these four illustrative ratios. The median is the average of zero point nine and one, or zero point nine five. The typical-looking summary can coexist with a tenfold bad case. If the declared acceptance rule requires every case to stay below two, the tenfold case fails it. The good median cannot override that requirement. This is a direct conclusion from the stated gate, not a quiz for the viewer. A selector revised after observing the bad case is a new candidate. It needs an appropriately fresh confirmation set under a frozen protocol. Retesting on the same exposed cases can demonstrate repair there but does not establish unseen-case transfer. In the toy example, case four takes ten times the control's time, so the policy fails the below-two rule. The takeaway is to charge the full selection cost and keep the fixed fallback until a frozen policy passes its validation.