A side about to lose its move

Not a domination, in that order split the second reduction’s 525 deletions into 448 that leave a survivor — every one a domination in the order on stops — and 77 that leave no option on their side at all. It closed on those:

They are the deletions that empty a side, they are a fifth of the second pass’s work, and no rule on this ladder has looked at one … the question is whether the pairs it happens to can be recognised from the parts: whether a player about to add two reduced forms can tell, from the two forms alone, that one of them is about to lose a move entirely.

They can. The test is four steps, it never builds the sum, and it is right on all 525.

Why no rule had looked at one #

The rung below’s eight candidate rules all named the deleted option by a property of that option — the hottest, the coldest, the widest, the extreme stops. A rule of that shape cannot describe a deletion with no survivor, because the point of naming an option is to distinguish it from the ones that stay, and here nothing stays.

More than that: the rung below scored its candidates on the deletions with a single option removed and a survivor, by construction, so the seventy-seven were outside its population rather than inside it and failing. That is a much better position for a ladder to be in than the alternative — a rule scoring 79 per cent because it is right on four fifths of a mixed population is a rule nobody can improve, and a rule scoring 79 per cent on a population that excludes a known fifth is a rule with a stated gap.

So the object had to change. What is recognisable is not which option goes but whether the side is the kind of side an option can go from, and that is a property of the pair of positions rather than of either option in it.

It is worth saying what the population is, because 525 deletions is a count of events rather than of positions. The pool is the first sixty day-three values with a positive temperature, taken in ordered pairs; a pair enters the census when reducing the two parts and adding them gives something the reduced form of the sum does not — 431 such pairs. On each of those, each side is examined, and a side enters the count when exactly one of the sum’s options there is deleted. That gives 525 side-events, of which 77 leave nothing behind.

So a single pair can contribute a deletion on each side, and most do not contribute on either. The seventy-seven are 15 per cent of the events and a much smaller share of the pairs.

The test #

Take the side in question — Left’s, say. Each part contributes its own Left options, and each of those, taken with the whole of the other part, is a candidate move in the sum. Two reduced forms with one Left option each give two candidates; three at most on this pool.

Delete the candidates that another surviving candidate dominates, in the ordinary partial order on games. The side empties exactly when one candidate survives.

That is right on all 525 deletions: 77 emptied, 448 not, no exceptions in either direction.

Nothing in it builds the sum. The candidates are option-sums, the comparison is the ordinary order, and the recursion beneath the top level — which is what makes a reduction expensive — is never entered.

What the test is saying #

The arithmetic is short and the reading is shorter.

A side empties when the two parts do not offer that player genuinely different moves. If moving in the first part, with the second part left standing, is at least as good as moving in the second part with the first left standing, then there was only ever one move worth having on that side — the sum’s canonical form keeps one option, and the second reduction takes it away.

If the two moves are incomparable, both survive, and the second pass can delete one and leave the other. That is the 448.

So the fourth relation is doing the work again, in the opposite direction from the case that was supposed to be hard: there incomparability made an argument go through, and here it is what keeps a player’s move alive. Two positions offering incomparable moves are two positions offering a genuine choice, and a genuine choice is not something a reduction can throw away.

A test with two readings #

There is a second way to say what the test is doing, and it makes the result look less like a coincidence.

The reduced canonical form of a sum is computed in two stages: build the sum’s canonical form, then apply the extra clause that replaces a position whose stops agree by that number. The first stage is an ordinary reduction and it happens at every level. The top level of that first stage is exactly the test: form the option-sums, delete the dominated ones.

So the test is not a new object at all. It is the first thing the full computation does, extracted and read as an answer rather than as a stage. What makes that legitimate — and what the census establishes — is that whether the side ends up empty is decided at the top level and nowhere below it. The recursion into the options can change what the surviving option is; it cannot change how many there are.

That is worth stating as the general claim, because it is what a reader should take away: the width of a reduced sum’s option set on each side is settled before the recursion starts. The 525 rows are the check on it, and it is the reason a cheap test can answer a question about an expensive object.

What it costs #

The candidates number two on 497 of the 525 deletions and three on the other 28, so the test is two comparisons or at worst six.

Building the sum and reducing it visits the whole of the sum’s tree — an average of 17.5 positions on the pairs sampled here, against the test’s 2.2 comparisons, and the gap widens with the day, because a sum of two day-$n$ values is a day-$(n{+}1)$ value and its tree grows accordingly.

That is the point of a recognition test. The question does this side empty is a question about the reduced sum, and the test answers it without computing the reduced sum. A player at the board can apply it; a player who has to build the sum has already answered the question by other means.

Where it fires and where it does not #

The seventy-seven are worth a paragraph on their own, because what happens to them is larger than a value changing.

A player with a move in the unreduced sum and none in the reduced one has lost the option of playing in that component at all. The reduction that always shrinks is the standing fact that a reduction never adds an option, and this is that fact taken as far as it goes: the side is not merely narrower, it is empty. If every component a player has does that, they are an end — a player with no move anywhere — which is a change to who wins rather than to what the position is worth by a little.

The 448 are where the rung below’s account applies: two incomparable candidates, one deleted by the second pass, and the survivor’s stops dominating the deleted option’s. So the two accounts between them now cover the whole census — 77 recognised in advance by this test, 448 described afterwards by the stop order — with no residue.

What a player would actually do #

Stated as a procedure at the board, the test is short enough to be worth writing out.

Two components, both already reduced, and Left is wondering what happens when they are considered together. Left has one move in the first component and one in the second. The question is whether a move here, with that component untouched is at least as good as a move there, with this one untouched. If it is — either way round — then Left has one move in the sum rather than two, and the second reduction can take it, leaving Left with nothing to do in this pair of components at all.

If neither is at least as good as the other, Left has a real choice and keeps one of the two whatever the reduction does.

That is a question about two positions Left can already see, and it is exactly the kind of question comparing positions is about. What makes it worth asking is that its answer is about a third position — the reduced sum — that Left has not built.

Why the ordinary order works here and not there #

There is a tension worth resolving explicitly, because the rung below’s headline was that none of the 525 deletions is a domination in the ordinary order, and this page’s test is a domination in the ordinary order.

They are dominations between different things. The rung below compared the deleted option with the surviving options of the sum, and found the ordinary order blind to that pair — as it must be, since the reduced form compares modulo infinitesimals and two positions an infinitesimal apart are incomparable.

This page compares the two candidates with each other, before any reduction has run. That comparison is the sum’s own canonicalisation doing its ordinary first step, and it is an ordinary domination because it is an ordinary reduction.

So the two statements are consistent and their combination is the useful one: the ordinary order settles which candidates reach the reduced form, and the stop order settles which of the survivors the second pass then removes. Two orders, two stages, and each blind exactly where the other sees.

The figures, and what they leave out #

Six tables, and the object they are about — a sum of two games with an option missing — is not drawn on any of them.

It could be. A pair of reduced forms, their sum’s canonical form with two options on a side, and the reduced sum with one or none, drawn as three brace expressions in a row, would carry the whole event for one pair. This site’s rule is to show the position beside the notation, and a partizan value born on day three has no position: it is a brace form and nothing else, so the drawing would be brace forms either way.

What the tables do instead is count. That is the right choice for a claim about 525 events and it is a loss for a reader who wants to see one happen. Many forms one value is where a reduction is drawn step by step on a single position, and it is the page to look at if the arithmetic here needs a picture behind it.

What this does not settle #

The test is a test and not an explanation. It says a side empties when the candidates collapse to one, which is nearly a restatement of what the reduced form does at the top level. What it does not say is why the second pass then deletes that last option — the deletion is still a stop-order event and the test does not predict it. What the test predicts is that if a deletion happens on that side, there will be nothing left.

And that is a real gap in what is claimed. The population is the 525 deletions the second pass actually makes; the test is not scored on sides where no deletion happens at all. A player wanting to know whether their side will lose a move needs both this test and a prediction that a deletion occurs, and only the first is here.

Sixty day-three values, and hot ones. The pool is the rung below’s — the first sixty day-three values with a positive temperature, in ordered pairs — because the whole ladder is measured on it. Whether the test survives a wider pool is untested, and the mechanism it rests on is the sum’s canonical form having one option, which is not a fact about a day.

The comparison is the ordinary order and it is not free. Comparing two games is a search over sums, not a look, and comparison is a search is the page about what that costs. Two comparisons is a count of comparisons rather than of work; what makes the test cheap is that it makes two of them on small games instead of building a tree.

And the seventy-seven are one pool’s worth. The test is exact on them and the claim it supports — that emptying is recognisable from the parts — rests on a census in which the largest candidate count is three. A pair of positions with four or five options a side would give a wider candidate list and a domination structure this pool cannot exhibit, and nothing here says the test still reads one survivor correctly there.

The 431 pairs are ordered pairs. A pair and its reverse are counted separately, and the second reduction is not symmetric in them, so the population double-counts unordered pairs that need the second pass in both directions. That is the rung below’s convention and it is inherited here without change; nothing on this page depends on it, and a reader comparing the 431 with a count of unordered pairs elsewhere should know.

Normal play, short games, and the reduced canonical form as defined on this ladder — a position whose two stops agree replaced by that number, and the reduction recursing into the options otherwise.

What the anchor now knows #

Six rungs in, it is worth writing down the state of the account, because the second reduction has gone from a curiosity to a described operation.

That it exists. Adding two reduced forms does not give a reduced form; a second pass is sometimes needed, on 431 pairs of the pool.

What it deletes. One option, on one side, on 525 side-events.

By what relation. Not a domination in the ordinary order — nought of 525 — and a domination in the order on stops on all 448 with a survivor, with the dominator unique on 436 and always from the other part.

And now which sides it can empty, recognisable in advance from the two parts by two comparisons.

What is not known is which pairs need the second pass at all, and that is the one structural question left. Everything above is conditional on a deletion happening; nothing says when one does.

Where the ladder goes next #

The reduced-form anchor has six rungs: what is left when the small change is thrown away, how hot a background has to be, add then reduce again, the option nothing names, not a domination in that order, and now the side that empties.

The rung above is the deletion itself. Both halves of the census are now described — the 77 recognised in advance, the 448 described after the fact — and neither is predicted: nothing on this ladder says which pairs the second pass will act on at all. The rung below counted 431 pairs needing a second reduction out of a much larger population, and the property that separates them from the pairs that reduce in one pass has never been looked for. That is a question about the pair rather than about a side, it is the last unasked one on this anchor, and the population is already built.

Two neighbours are worth the trip. Add then reduce again is where the second pass was found and counted, and it is the page whose 431 everything here is over. And what is left when the small change is thrown away is where the reduced form arrives and where the infinitesimals it discards are named, and it is worth reading beside a page about a reduction removing something much larger than an infinitesimal.