The split slips one day deeper

The follower does the reversing replaced a count with a description. The gift-horse theorem for the ordinal sum says that adding a gift horse $h$ to a base $G$ leaves $G : f$ unchanged; its proof needs two cases, and the description said which case applies by looking only at the follower. If $f$ gives Right a move, Right’s move inside the follower reverses the horse’s ordinal sum — on every one of 15,768 instances. If $f$ gives Right no move, an option the base already had dominates it — on every one of 2,628.

A description with no exceptions on the sweep it was fitted to is a conjecture, and the honest thing to do with a conjecture is to test it where it was not fitted. That essay named two such places and made a prediction for one of them in advance. This essay runs both.

The mirror, predicted and run #

The first test is the mirror. Everything in the ordinal sum has a Left–Right mirror image: add a gift horse to a base’s Right options — an option at least the base, which leaves its value unchanged — and the question becomes whether $h : f$ is dominated by an existing Right option, and if not, whether it is reversible through a Left answer at least the whole sum.

The prediction was stated before the sweep was run. The reversing move should be Left’s move inside the follower; the follower with no Left move — $-1$, whose only option is Right’s — should need domination for every horse; and $1$ and $-1$ should exchange the roles they had. What $-1$ did to Left gift horses, $1$ should do to Right ones.

The prediction holds in every detail. Of 2,013 Right gift horses, under each of the six followers that give Left a move, Left’s move inside the follower reverses all 2,013. Under $-1$ there is no such move, and every horse is dominated. The mirror is exact.

It is worth being clear about what that does and does not show. It does not add much evidence for the underlying lemma, because the mirror sweep is the original sweep with signs changed, on the same bases, and a lemma true of one is very nearly guaranteed to be true of the other by symmetry. What it rules out is an accident: a description that held only because gift horses were added on one side, or only because the followers happened to be arranged a certain way relative to the bases. That accident would have shown up here and did not.

Read side by side, the two sweeps also show where the mirror is not a mirror. The escape counts differ — under $\uparrow$ one Left horse escapes domination and 232 Right horses do; under $\tfrac12$ none and 243. That is not a failure of symmetry. The bases are drawn from day three without regard to sign, the horses too, and a base with more Left options than Right has more ways to dominate a Left horse than a Right one. The part that must be symmetric — which followers need domination, and which move reverses — is.

One day deeper #

The second test is harder, and it is the one that could fail for a reason rather than by accident. A day-three base has a form at most three days deep; its options are day-two values and its options’ options are day-one values. A statement about options — which is what “the follower’s move reverses” is — could hold on shallow forms because there is not enough room in them for it to fail.

Day four cannot be enumerated. It can be built, the same way a floor, and not a decline builds its day-four sample: choose one or two day-three values for each side and keep the result when its canonical form is genuinely four days deep. Forty such bases were built, and the horses were the eighty day-three horses of the earlier sweeps together with forty built day-four ones.

The theorem survives: the ordinal sum is unchanged on all 9,709 instances. The two-case proof survives: nothing is neither dominated nor reversible. And the one-line description does not survive. Under $\uparrow$ and under $\tfrac12$, three gift horses are not reversed by the follower’s Right move. Under $\ast$, $-1$, $\downarrow$ and $\pm 1$ the follower’s move reverses every one of the 1,387.

The three exceptions #

Three failures in a thousand are a small number, and the temptation is to call them noise. They are not noise in a deterministic computation, and what they are is worth looking at.

Three pairs of base and horse, and the same three under both followers. Two of the horses are $\ast 2$ and one is $\{0 \mid -1\}$; the bases are wide day-four forms, forty-three to sixty-nine characters written out. Every one of the three is dominated — an option the base already had is at least as good as the horse once the follower is attached — and every one is also reversible through a Right move in the base part. So each is covered twice over by the proof, and neither covering is the follower’s move.

The table records more than the verdict, and the extra columns are where the three agree. Every failure is a confusion. The follower’s option $h : f^R$ is never above the whole sum; it is confused with it, neither at most nor at least, which is the fourth relation confused is not the same as unknown is about. And every failing base has a single Left option, which is the option that dominates the horse.

Both features have a reading. A single Left option is the narrowest a base can be on that side, and a narrow Left side is exactly where one existing option is likely to be strong enough to dominate the horse outright — so these are bases on which the first case of the proof is doing all the work, and the second case has no need to succeed. And a confusion is the characteristic failure of a comparison between two ordinal sums with different followers: $h : \ast$ against $G' : \uparrow$, say, where the base parts differ and the followers differ and the two differences point in incompatible directions.

Take the first row. The horse is $\ast 2$ and the follower $\uparrow = \{0 \mid \ast\}$, whose Right move is to $\ast$. The follower’s option is $\ast 2 : \ast$, which is $\ast 3$ — a nimber, and a nimber is confused with any game whose Left and Right stops straddle nought in the right way. The whole sum here is a hot game built on a base with Left option $\{2 \mid \ldots\}$, so its stops are far apart and a nimber sits inside the gap: neither player can guarantee to do better with $\ast 3$ than with the whole sum. That is a reading of one row and not a checked account of all three: the stops of the three whole sums were not computed, and the explanation would be confirmed by computing them and seeing the nimber fall between the two stops each time.

Which followers they fail under is the part to notice. $\uparrow$ and $\tfrac12$ are exactly the followers whose Right move goes to a game other than nought — to $\ast$ and to $1$ — while every follower whose Right move goes to nought, where the reversing option is the horse itself, reverses everything. The one follower that breaks the pattern the other way is $\pm1$, whose Right move goes to $-1$ and still reverses every horse. So the failures sit among the followers where the lemma is a genuine comparison of two ordinal sums, $h : f^R$ against $G' : f$, and never among the followers where it reduces to $h \le G' : f$.

That is a reasonable place for a lemma to fail. When the follower’s Right option is nought, the reversal needs only that the horse is at most the whole sum — close to the definition of a gift horse. When it is not nought, the reversal needs a comparison between two ordinal sums with different followers, and ordinal sums with different followers are exactly what the ordinal sum’s form-reading makes unpredictable.

The deeper mirror, run as well #

Having found failures at day four on the Left, the mirror of the day-four sweep is the obvious check that the failures are about depth and not about side. It was run on the same forty bases with the gift horses added to Right options instead.

It fails too, and in the mirror-image place. Under $\downarrow$ — the mirror of $\uparrow$ — three Right gift horses are not reversed by Left’s move inside the follower, and under every other follower with a Left move every horse is reversed. The three have the same shape as the Left ones: the horse is $\ast 2$, the follower’s option is $\ast 3$, it is confused with the whole sum, every base has a single Right option, and that option dominates. The follower $\tfrac12$ has no mirror image among the seven followers used, so its row has no counterpart to compare.

So the failure is not an artefact of one side. It is a property of depth, it respects the symmetry exactly, and it has the same structure wherever it occurs: a nimber option confused with a hot sum over a base with one option on the relevant side. That is a good deal more specific than three unexplained exceptions, and it is the specificity a proof of the combined case could use.

How the sample was built, and what it can miss #

The day-four bases are forty of them, and how they were built decides what the sweep can see. Each is one or two day-three values chosen for each side, kept only if its canonical form is genuinely four days deep. That favours forms with few options, since a random pair of day-three values often collapses under domination or reversal to something shallower. The average base carries about thirty-five gift horses from the pool against about twenty-six for a day-three base, so the day-four sweep tests the description more densely per base even though it has fewer bases.

A sample built this way can miss whole kinds of base. Wide day-four forms — four or five options a side — are underrepresented, and they are precisely the forms where a statement about options has most room to fail. So the three exceptions are a floor on how often the description fails at day four, not an estimate. The same caution attaches to the four followers under which the description held: they held on this sample, and the sample leans towards bases where holding is easier.

What a proof should now be #

The two-case proof still stands, and the case split has to be re-drawn. On the evidence of four sweeps:

When the follower’s reversing move goes to nought — Right’s move in $\ast$, $-1$ and $\downarrow$ for Left horses, Left’s move in $\ast$, $1$ and $\uparrow$ for Right ones — the reversal through the follower’s move has held on every instance of all four sweeps. Every failure found is under a follower whose relevant move goes somewhere other than nought. A lemma of the form “a gift horse of $G$ is at most $G' : f$ when $f^R = 0$” is the strongest candidate the measurements support.

When the follower gives Right no move — $1$ on the Left side, $-1$ on the Right — domination has held on every instance.

When the follower’s Right option is some other game, the follower’s move reverses almost every horse and not all of them, and the remainder is covered by domination. A proof of this case cannot be a single lemma about the follower’s move; it has to combine the two reductions, as the original two-case proof did.

This is less tidy than the description the day-three sweep suggested, and more honest. It also sharpens where effort should go. The case where the follower’s Right option is nought is the simplest candidate for a proof and has not failed; it is the one to try first. The case where it is some other game has a small, specific set of counterexamples to the simple statement, and those three examples are exactly the objects a proof of the combined case would have to handle.

Why the ordinal sum keeps producing this shape #

This is the fourth time a clean statement about the ordinal sum has been fitted, tested one step further, and found to need a qualification. What the colon respects found that the operation reads value everywhere except at four forms; no fifth value found those four stable under depth; the proof of the gift-horse theorem needed a second reduction; and now the second reduction’s clean description needs a second case of its own.

The pattern has a cause. The ordinal sum is the one operation in this part of the subject that reads the form of its argument rather than its value, and forms are where almost-true statements live: two forms of one value agree on everything that can be computed from the value and may disagree on anything else. A statement fitted on shallow forms is a statement about what those forms have in common, and deeper forms have more room to differ. Hackenbush is a numeral is where this matters in play, since a stalk is an ordinal sum of its edges and every rewriting of its lower part is a claim of exactly this kind.

The convention named #

Normal play throughout: the player unable to move loses, and comparisons are decided by the difference game. The ordinal sum is Conway’s — a move in the base destroys the follower — and the gift horses are defined against the value of the base, not against its form. The day-four sample is built rather than enumerated, biased towards narrow forms in the way every built sample on this subject is, and its forty bases are a test of a description, not a census of day four.

What the counts cannot show #

A counterexample to a description is a fact; a pattern among three counterexamples is a guess. The observation that the failures sit only among followers with a non-zero Right option rests on three base-and-horse pairs and four followers without failures, and a larger day-four sample could produce a failure under $\ast$ or $-1$ and end the guess. What would not change is that the one-line description is false: three instances are enough for that.

Nor do the counts fully show why the three fail. The table records that each failure is a confusion on a base with one Left option, and the worked row above gives a reason in terms of stops; whether that reason covers the other two rows, or whether each confusion arises differently, is a question the stops of the three sums would answer and this sweep does not record.

Still open: the nought case, proved #

The measurements have narrowed the target twice. The first narrowing found that the reversal case is carried by the follower’s move; the second found that this is reliable only when the follower’s Right move goes to nought. The natural next step is to prove that case outright — that for a gift horse $h$ of $G$ and a follower $f$ with Right option $0$, $h \le G' : f$ — rather than to measure it further, since the proof needs both reductions has now been followed by two essays of measurement and each has made the argument’s shape clearer without supplying it.

The statement has one feature that makes it tractable. Its left side is just $h$, with no ordinal sum at all, so the comparison is between an ordinary game and an ordinal sum, and the difference game $G' : f - h$ has one component that behaves like any disjunctive summand. The other sum, the one that nests is where the ordinal sum was introduced, and canonical form is where both reductions are defined; a proof of the nought case would need little beyond those two.