Field

What a dictionary buys and what it costs

A balancing rule is a list of functions, and the list decides two things that pull against each other. What it protects against is decided by parity: an interaction between two odd functions is even, an odd function is orthogonal to an even one at every correlation, and the two things every trial balances — a mean and a median split — are both odd, so their worst case is exactly zero however many of them are held. What it costs is the assignments it leaves, and past a certain thinness those stop being one set: the admissible assignments split into an arrangement and its mirror image, no sequence of admissible single swaps joins them, and the walk that samples them is uniform on half the reference distribution for ever.
The rule is parity, and it runs both ways. At a correlation of 0.5, four combinations of a dictionary and an outcome shape. The joint sign flip (X, Y) → (−X, −Y) leaves the bivariate normal alone at every correlation, so a function that changes sign under it is orthogonal to one that does not. A product of two odd functions is even; a product of an odd and an even one is odd. So an odd dictionary removes exactly none of the first and something of the second, and an even dictionary does the reverse — which it does, to machine precision, in both of the two rows that should be zero. This is one rule where there had been two: that a median split's square is constant, and that a polynomial dictionary contains the products a correlation generates.

A dictionary that is neither

A rule handed two median splits removes none of their interaction; a rule handed two covariates removes none of their product. Those were two results with two explanations, and they are one result with one — and finding it corrected the number underneath both.

What the worst case is worth, one function at a time. The smallest share each dictionary removes, over seven outcome shapes, at a correlation of 0.5. A rule balancing the mean of each covariate has a worst case of exactly zero — against the square, and against both products. Adding a median split to it, which is the second thing every trial balances, leaves the worst case at exactly zero, because a median split is odd and so is a mean. Adding the square instead moves it to 6.8%, and the extra functions after that move it to 7.4%. The worst case is decided by which parities the dictionary contains rather than by how many functions are in it.

What the extra function buys

A rule balancing the mean of each covariate has a worst case of exactly zero. Adding the median split — the other thing every trial balances — leaves it at exactly zero, and one square moves it.

Whichever dial made the set thin, the crossing is at the same thinness. Each curve is one dictionary, swept over eight tolerances at two hundred units; a point above the line is a set thin enough that walking beats hunting. The curves lie nearly on top of one another, which is the answer to whether the crossing is a fact about the tolerance or about the thinness it produces: the crossings sit between one admissible assignment in 176 and one in 268 for dictionaries of 3 to 6 functions. The mechanism is that a hunt costs exactly 1/p and a walk costs almost the same everywhere — between 82 and 394 evaluations per usable draw across the whole table — so the crossing is wherever 1/p reaches a number that does not move.

The set a dictionary leaves

A rule constrained on six functions at a loose tolerance leaves a set as thin as one constrained on three at a tight one. Both sampling methods cross over at the same thinness, and the tolerance where that happens moves by a factor of three.

A thin enough set is not one set. Every admissible set of 14 units this table can enumerate, by how much of the assignment space it admits and how many pieces it falls into under single swaps. A walk is uniform on the piece it starts in and never leaves it. The pieces are not fragments: at 522 admissible assignments the set splits into 3 halves of exactly 520 each, and every assignment's complement is in the other half — no sequence of admissible single swaps takes an assignment to its own mirror image. Two-swap proposals reconnect four of the six disconnected sets here — the two they do not are the thinnest, where a two-unit move rarely lands anywhere admissible either — which makes a bigger proposal a correctness repair rather than the speed dial it was measured as.

The walk that cannot cross

A thin enough admissible set is not one set. It splits into an assignment and its mirror image, no sequence of admissible single swaps joins them, and the walk that samples it is uniform on half the reference distribution for ever.

The best worst cases among twenty-eight pairs. Every pair of two functions from eight — the covariate, its square, cube and fourth power, and cuts at −1, 0, 1 and 2 — held on both covariates, ranked by the smallest share it removes over seven outcome shapes at a correlation of 0.5. The best is the square and a cut at 1 at 7.00%; the mean and the square reach 6.84%. For almost every pair the shape holding the floor is the product of two median splits, which no pair removes more than a small part of.

Two functions a covariate

Every constraint costs admissible assignments at the same rate, so the decision a trial faces is a budget: which two functions of each covariate. Over twenty-eight pairs the worst case never picks the mean. At a correlation of a half it picks the square with a cut at one, beating the mean and the square by sixteen hundredths of a point on a floor held by a shape no pair removes a tenth of — and giving up 41% of a linear outcome to do it. Over the 127 lists of shapes a designer could write, twenty-three different pairs are best somewhere.

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