The follower does the reversing

A gift horse added to a form does not change its value, and — against the grain of everything else the ordinal sum does — it does not change the ordinal sum either. The proof needs both reductions found the shape of the argument for that. Of 10,512 gift horses added to a base’s Left options and carried under four followers, 10,280 are dominated in the sum: some option the base already had is at least as good once the follower is attached. The other 232 are not dominated by anything, and every one of those is reversible: its ordinal sum has a Right option that is at most the whole sum, so canonical form removes it by the other of its two reductions.

That settled that two cases are needed and that two cases are enough. It did not say which Right option does the reversing, and without that the second case is a count rather than a lemma. A reversal has to go through a specific move, and a proof has to name it.

Two places Right can answer #

Write the base as $G$, the gift horse as $h$ and the follower as $f$. The new Left option of the ordinal sum is $h : f$ — the horse with the follower attached — and a Right option of $h : f$ is a Right move in one of two places.

Right may move in the base part, to some $h^R$. In an ordinal sum a move in the base wipes the follower out entirely, so this option is just $h^R$, with nothing underneath.

Or Right may move in the follower, to $h : f^R$, which keeps the whole of $h$ and replaces the follower by one of its own Right options.

For $h : f$ to be reversible there has to be one of these, call it $X$, with $X \le G' : f$ — at most the whole sum, where $G'$ is the base with the horse added. The previous sweep checked that some $X$ exists for all 232 escapes. This one records which, and the answer in the figure is unambiguous: the move inside the follower works every time. The move in the base works most of the time and not always — under $-1$ it reverses 184 of 190, under $\downarrow$ 130 of 132 — which makes it the less useful of the two for a proof, since a lemma that holds most of the time is not a lemma.

The lead left by the earlier essay put it as a guess: if the answer is always the follower’s own contribution — the part of the sum below the base — then the reversibility case has a one-line description. It is always the follower’s own contribution. The guess was right, and it was right more strongly than the guess asked.

Every gift horse, not only the escapes #

The escapes are a biased sample for testing a claim about reversal, because they are defined by domination failing. The stronger test runs the question over every gift horse, dominated or not.

Seven followers this time — the four of the earlier sweep, $\ast$, $1$, $-1$ and $\uparrow$, and three more, $\downarrow$, $\tfrac12$ and $\pm 1$ — and 2,628 gift horses under each. The result is as clean as a sweep can be. Under every follower that gives Right a move, Right’s move inside the follower reverses every gift horse — all 2,628 under each of six followers, 15,768 instances with no exception. Under the one follower that gives Right no move, $1$, there is no such option to use, and there domination covers every gift horse instead.

So the two cases of the proof are not two kinds of gift horse. They are two kinds of follower. Where the follower offers Right a move, every gift horse is reversible through it, including the thousands that are also dominated; where it does not, every gift horse is dominated. The earlier sweep saw a split by instance — 97.8 per cent one way, 2.2 per cent the other — because it asked about domination first and reversal only when domination failed. Asked in the other order, the split is by follower and has no exceptions.

The last column says something too. A Right move in the base part reverses 2,390 of the 2,628 gift horses, and it is the same 2,390 under every follower, including the one with no Right move. That is what a move which wipes the follower out should do: the option $h^R$ does not depend on $f$, and the only place $f$ enters is the whole sum it is compared with. The base-part move is a property of the horse and the base; the follower’s move is a property of the follower. The second is the one that always works.

What the follower’s move leaves #

Once the reversing move is known the lemma it needs can be read off.

For three of the followers the follower’s only Right option is nought, and $h : 0$ is simply $h$ — the ordinal sum with an empty follower is its base. So under $\ast$, $-1$ and $\downarrow$ the reversal says only this: the gift horse is at most the whole sum, $h \le G' : f$. That is very nearly the definition of a gift horse, which is an option at most the base, $h \le G$; the lemma needed is that attaching the follower to the base does not pull it below the horse.

For the other three the follower’s Right option is a game other than nought — $\ast$ for $\uparrow$, $1$ for $\tfrac12$, $-1$ for $\pm1$ — and the reversing option is $h : f^R$, the horse with the follower replaced by the follower’s own Right option. The lemma then compares two ordinal sums, $h : f^R$ against $G' : f$, and has the flavour of a monotonicity statement: a base no larger than $G'$, with the follower moved once in Right’s favour, is at most $G'$ with the follower intact. Note that $f^R$ need not be smaller than $f$ — for $\tfrac12$ it is $1$, which is larger — so the statement is not about the followers being ordered, only about Right having spent a move in one of them. Ordinal sums are not monotone in general — when the nested sum only sees the value is the standing warning that they read forms — so the lemma is a real claim and not a restatement of an obvious one.

One reversal, written out #

The earlier essay ended by saying that its six tables described two reductions and could not show one. A reversal is a small object and is worth seeing whole.

The base is $\uparrow = \{0 \mid \ast\}$. The gift horse is $\ast 2$: it is at most $\uparrow$ — the sum $\uparrow + \ast 2$ is positive, which is the classical fact that up beats every nimber from $\ast 2$ on — so adding it to $\uparrow$’s Left options leaves the value $\uparrow$. The follower is $-1 = \{\,\mid 0\}$.

The base’s only Left option, $0$, becomes $0 : -1 = -1$ in the sum. The horse’s option becomes $\ast 2 : -1$, whose canonical form is $\{0, \ast \mid 0, \ast, \ast 2\}$ — a confused game near nought, and not at most $-1$. So nothing the base already had dominates it; this is an escape.

Right’s move inside the follower takes $-1$ to its Right option $0$ and leaves $\ast 2 : 0 = \ast 2$. The whole sum is $\{0 \mid \uparrow, \ast\}$, and $\ast 2$ is at most it. So the horse’s option is reversible through the follower’s move, it is replaced by the Left options of $\ast 2$ — which are $0$ and $\ast$ — and the ordinal sum is the same as if the horse had never been added. Every line of that is checked by comparison in the figure rather than taken on trust.

What the small example shows is how little the reversal has to do with the horse’s own structure. The option that reverses is the horse itself, unchanged; what makes it work is that the follower $-1$ gives Right a move to nought, and nought as a follower is invisible. The same happens for every escape under $-1$, $\ast$ and $\downarrow$.

The proof, split by follower #

Put together, the measurements say what a written proof of the gift-horse theorem for ordinal sums would look like on this sweep.

Case one: the follower gives Right a move. Then the gift horse’s ordinal sum is reversible through that move, whether or not it is also dominated, and canonical form removes it. The lemma to prove is $h : f^R \le G' : f$ for a gift horse $h$ of $G$, which for a follower whose Right option is nought reduces to $h \le G' : f$.

Case two: the follower gives Right no move — a form with no Right options, whose value is always a whole number of moves for Left, as $1$ is. Then the gift horse is dominated by an option the base already had, and the lemma to prove is that one of $G$’s Left options, with the follower attached, is at least $h : f$.

This is a better shape than the one the earlier essay reached, in two ways. It has no leftover: neither case is “most of the time”. And the cases are chosen by looking at the follower, which is a single small game, rather than by testing domination instance by instance.

The earlier essay also noticed that the number of escapes falls as the follower becomes better for Left — 190 under $-1$, 41 under $\ast$, one under $\uparrow$, none under $1$. With seven followers the ordering holds wherever the followers are comparable: $-1$ gives 190, $\downarrow$ 132, $\ast$ 41, $\uparrow$ one, and $\tfrac12$ and $1$ none; $\pm 1$, which is confused with most of the others, gives 46. That monotone count is a count of how often domination fails, and the new reading explains why it is not the quantity a proof wants: under every follower but $1$, reversal succeeds on all of them, and how often domination also succeeds is beside the point.

Why a proof of the first case is not immediate #

It is worth trying to prove the first case directly, because the attempt shows where the difficulty lives.

The claim is $h : f^R \le G' : f$, and a comparison is decided by the difference game: the claim holds when Left, moving second, wins $G' : f - h : f^R$. The natural strategy for Left is to copy — to answer each move in one component by the corresponding move in the other, the way every mirror argument on equal games goes. Two things stop the copy from working as it stands. The two sides have different bases, $G'$ and $h$, and the only relation between them is $h \le G$, which is a statement about the difference game $G - h$ and not a move-by-move correspondence. And the two sides have different followers, $f$ and $f^R$, so copying in the followers runs one move out of step.

In a disjunctive sum those obstacles are routine: Left plays the winning strategy for $G - h$ in the bases and copies in the followers, and the two strategies interleave because the components are independent. In an ordinal sum they are not independent. A move in a base destroys that component’s follower, so a strategy for the bases can wipe out the very position the follower strategy was copying against. That is the same property that makes the ordinal sum read forms rather than values, and it is why canonical form cannot be applied to the base before taking the ordinal sum. A proof of case one has to handle a base move that ends the follower game in the middle of a copy — and on every instance measured, the answer is that the move Right would like to make in the follower is already enough to make the horse redundant.

Why this matters beyond the theorem #

The ordinal sum is not a curiosity. A Hackenbush stalk is an ordinal sum of its edges, each edge sitting on the one below it, and Hackenbush is a numeral reads binary expansions off stalks for exactly that reason. The gift-horse theorem says which ways of rewriting the bottom of a stalk leave everything above it unchanged, and no fifth value found that the forms the ordinal sum can tell apart are, apart from gift horses, confined to the four values born on the first day. A proof of the theorem would turn that measurement into a description of when a stalk’s lower part may be simplified without touching the upper part.

The finding here is a step towards that proof, and it is the kind of step that is only visible by recording more than was asked. The earlier sweep recorded whether each escape was reversible; this one recorded which move reversed it, and ran the question over the instances that did not need it. The first record gave a count. The second gave a case split with no leftovers.

The convention named #

Everything here is the ordinal sum under normal play: the follower is attached below the base, a move in the base destroys the follower, and values are compared by the difference game in which the player unable to move loses. Both reductions of canonical form are normal-play reductions and preserve the value under the disjunctive sum; they are being used here on an operation that does not preserve values, which is the reason the theorem needs a proof at all.

The gift horses are added to Left options only. The follower’s Right move is the relevant one because the reversal a Left option needs is a Right answer. The mirror — horses added to Right options, reversed through the follower’s Left move — is a prediction of this essay and not a result of it.

What the tables cannot show #

The census says the lemma holds on 15,768 instances and says nothing about why. Neither case is proved here, and the harder one is the first: $h : f^R \le G' : f$ compares two ordinal sums, and the tools for comparing ordinal sums are exactly the tools six essays on the ordinal sum have found unreliable.

The sweep is also bounded in a way that matters for a lemma. It is one hundred day-three bases and eighty day-three horses, drawn with one seed. A lemma that fails on a wider base or a deeper horse would not show up, and nothing about 15,768 successes rules that out. The description is a conjecture fitted to a sweep, with an unusually clean fit.

Still open: the mirror, and one day deeper #

Two tests would turn the fit into evidence, and neither was used to find it.

The mirror is the first. Add gift horses to Right options instead, and the prediction is exact: the reversing move should be Left’s move inside the follower; the follower with no Left move, $-1$, should need domination throughout; and $1$ and $-1$ should exchange roles. If any of that fails, the description is an accident of adding horses on one side.

One day deeper is the second. The day-three sweep cannot contain a base whose form is four days deep, and forms get wider as they get deeper — which is exactly where a statement about options could stop holding. A sample of day-four bases, built rather than enumerated as a floor, and not a decline builds its sample, with the same horses and some deeper ones, would test whether the follower’s move still reverses everything once the base has more room for the ordinal sum to see. An option nobody would take is where gift horses were first introduced, and it is worth remembering that the whole reason they need a proof in an ordinal sum is that everything else about equal forms fails there.