Cancelling is not pairing

In the simplest scoring game there is — a row of coins, each player in turn taking a coin from one end and keeping it, the higher total winning — a position does not in general cancel against its negative. Nothing to subtract with counted it: sixty-six of eighty-one rows of four coins, played beside the same row with every coin’s sign changed, leave somebody ahead. The restriction that buys the most then found the fifteen rows that do cancel to be the most useful class of the four it compared — larger than the class everyone expected to be larger, and buying every property asked of it.

Fifteen rows of four is a class small enough to read, and reading it suggests a description. The fifteen are $a\,a\,b\,b$ and $a\,b\,b\,a$ for coins $a$ and $b$: every coin has an equal partner, and the partners nest like brackets. That is a pleasing picture — a row cancels when its coins pair off — and it is what a reader of that table would take away. This essay tests it and finds that it is right exactly where it was noticed and nowhere else.

Pairing off, made precise #

A row pairs off when deleting adjacent equal coins, and repeating, leaves nothing. $3\,1\,1\,3$ pairs off: delete $1\,1$, and $3\,3$ is left, and delete that. $1\,3\,1\,3$ does not: no two adjacent coins are equal, so nothing can be deleted. This is exactly the rule by which a word of brackets reduces, with each coin value playing the part of a kind of bracket, and it captures the picture precisely. A row that pairs off has an even number of coins, and its coins can be matched into equal pairs that never cross.

The census compares that property with cancelling — a row cancels when the row beside its own negative scores nought whichever player moves first — over every row of two to seven coins drawn from three coin sets: $\{-2, 1, 3\}$ and $\{1, 2, 3\}$, which were the families of the earlier essays, and $\{-1, 1, 2\}$, which contains a coin and its negative.

On rows of two and four coins from the first two sets the two properties agree exactly. That is the fifteen rows of four that suggested the picture, and the three of two. From six coins they part. From $\{-2, 1, 3\}$, 87 rows of six pair off and only 67 cancel — every cancelling row of six pairs off, but twenty rows that pair off do not cancel. From $\{1, 2, 3\}$ the failure runs both ways: twenty pairable rows fail to cancel, and eight rows cancel without pairing off at all.

And the third set breaks the picture from the start. With $\{-1, 1, 2\}$ the row $-1\;1$ cancels: its negative is $1\;{-1}$, the same coins reversed, and the row and its negative are then the same game played from opposite ends. On rows of six from that set, 203 rows cancel and only 67 of them pair off.

Pairs that do not cancel #

The twenty pairable rows of six that fail to cancel are the more instructive failure, because they are the ones the picture says should work.

They have a common shape. Each consists of two equal coins, each with a pair beside it on the same side: $1\;1\;2\;1\;1\;2$ is a pair, a coin, a pair, the same coin; $1\;2\;2\;1\;3\;3$ is a coin, a pair, the same coin, a pair. Both pair off — delete the pairs and the two equal coins become adjacent — and in both the first player, in the row plus its negative, finishes ahead by two or four.

Pairing off is a property of the coins; cancelling is a property of the play, and the difference is that the play takes coins from the ends. A pair nested inside a row is only reachable once everything outside it is gone, and in these rows the outer coins can be taken in an order that separates a coin from its partner’s reach. The bracket structure says which coins match. It does not say that either player can take them in matching order.

The rows themselves, before their negatives #

It is worth looking at the rows on their own first, because the row alone and the row beside its negative answer different questions and the difference is the whole subject of this essay.

On its own, $1\;1\;2\;1\;1\;2$ is worth two to whoever moves first: take the $2$ from the right, and the rest splits evenly. $3\;1\;1\;3$ is worth nothing to either — whoever takes a $3$ exposes the other $3$ only after the $1$s are gone, and each player gets one of each. And $1\;2\;2\;1\;3\;3$ is worth nothing to either as well: on its own it has no stake at all, exactly like the row of four that cancels.

That third row is the sharpest version of the essay’s point. It pairs off, it has temperature nought — the row alone rewards nobody — and beside its negative the first player still wins by two. Having nothing at stake in the row is not enough; what matters is how the row’s play interleaves with the play in its negative, and a row with no stake of its own can create one once it has a partner.

It is tempting to conclude that a row which rewards the first player can never cancel — that whatever the row alone is worth to the opener, the opener in the sum can claim. For the first two coin sets the census agrees: every cancelling row of even length has temperature nought. But the argument does not survive the third set, where thirty-three rows of four cancel and eighteen of them reward the first player on their own. In those rows the negative row offers the second player a way to win the stake back, because a coin and its negative are both in the set and the two rows share coins in a way the first two sets never allow. So the row alone is evidence, not a verdict.

Why copying does not cancel #

Under normal play, a game plus its negative is always a win for the second player, and the reason is a copying strategy: whatever the first player does in one component, the second does the mirror move in the other, and the second player is never the one to run out of moves. Turn the board through a right angle is that argument, and it is the reason every normal-play game has an additive inverse.

In a scoring game the copying strategy is still available — the row and its negative always offer mirror-image moves — and it still guarantees something, but not nought. Suppose Left takes a coin worth $v$ from the row. Left’s score goes up by $v$. Right copies by taking the mirror coin from the negative row, whose value is $-v$; a coin worth $-v$ taken by Right changes Left-minus-Right by $+v$. The two moves together have moved the score by $2v$ in Left’s favour. The copier has not cancelled the first move; the copier has doubled it. That is the precise sense in which a scoring game has nothing to subtract with, and it is why cancelling is a special property of a few rows rather than a theorem about all of them.

What the copier needs is a different correspondence — not coin for mirrored coin, but some pairing of the first player’s possible moves with replies that restores the balance. On rows of four the pairing is available whenever the coins nest, and the bracket picture is a picture of that correspondence. On rows of six it can fail even when the coins nest, which is what the twenty exceptions show, and on the odd row of five it succeeds with no nesting at all.

What cancelling actually asks #

It helps to look at the first move directly.

In $1\;1\;2\;1\;1\;2$ plus its negative, Left has four openings: either end of the row, either end of the negative. Taking the $2$ from the right end of the row leads, with best play from both sides afterwards, to a win by two. The other three openings gain nothing or lose. In $-2\;1\;3\;1\;{-2}$ plus its negative, every opening comes to exactly nought.

That reframes the question, and the census supports the reframing on every row it contains: in a row plus its negative, the player who moves first never finishes behind. Across all 9,828 rows of the census — every length from two to seven, every coin set — the first mover’s score in the row plus its negative is at least nought. So “the row cancels” does not mean that neither player can win; one of them never loses anyway. It means that the first player cannot win either — that moving first in the row plus its negative is worth exactly nothing.

Why the first mover never loses is not proved here. It is not the mirror argument of normal play, which fails in a scoring game in a specific way: copying the opponent’s move in the negative row hands the copier a coin of the same sign the opponent just took, so the two coins double rather than cancel. The observation is that the first mover can always at least break even, and it is recorded as an observation. What it does is turn the search for a description of cancelling rows into a search for rows in which the first mover’s advantage is exactly nought.

A row of five #

The picture’s third failure is a single row, and it is the most surprising.

$-2\;1\;3\;1\;{-2}$ has five coins, so it cannot pair off — an odd number of coins cannot be matched in pairs — and it cancels. The first-move table shows it completely: all four openings come to nought. It is the only cancelling row of five coins from $\{-2, 1, 3\}$, and the row of seven coins has eight such rows, each built around the same core: $-2\;1\;3\;1\;{-2}$ with an extra pair spliced in, as in $-2\;{-2}\;{-2}\;1\;3\;1\;{-2}$.

What makes it cancel is visible if the row is read as a palindrome with a negative coin at each end. Whoever takes an end coin takes a $-2$, which is a cost; the other player can then take the other $-2$ or decline it, and the symmetric middle $1\;3\;1$ is a small game in which moving first is a disadvantage — the opener takes a $1$ and hands the $3$ across — which is the kind of hidden interval a hypothesis has to hold all the way down found breaking Milnor’s bound. The costs balance exactly. That is an explanation of one row; it is not a rule, and the rows of seven built on it are the same explanation repeated.

A condition that is necessary, and only sometimes #

One more candidate description is available and is nearly right: a cancelling row is one with nothing at stake in the row itself — it scores the same whoever moves first, so its temperature is nought.

With $\{-2, 1, 3\}$ and $\{1, 2, 3\}$ it is a necessary condition on even rows: every cancelling row of four and of six has temperature nought. It is far from sufficient — seventy-nine rows of six from the first set have temperature nought and sixty-seven cancel — and it fails outright on odd rows, where the one cancelling row of five and the eight of seven all have a temperature. With $\{-1, 1, 2\}$, which holds a coin and its negative, even the necessary half fails: thirty-three rows of four cancel and only fifteen have temperature nought.

So temperature, which is the first thing a reader of Milnor’s theory would reach for, is the wrong quantity. It says how much the row alone rewards moving first, and cancelling is about the row beside its negative, where the rewards can offset in ways the row alone does not record.

What the class is, then #

Put together, the census says that the cancelling rows are not described by any of the short properties tried. They are not the pairable rows, though the two coincide on rows of four for coin sets without a coin and its negative. They are not the rows of temperature nought, though for those coin sets every even cancelling row has temperature nought. They include odd rows, built around a palindrome with costly ends, that no pairing could produce. And the one thing true of every row swept — the first mover in the row plus its negative never finishes behind — describes all rows, not the cancelling ones.

That matters for how the class was used. The restriction that buys the most treated the cancelling rows as a class and found it bought everything asked of it on rows of four. A class defined by a computation — play the row against its negative and see — is a perfectly good class, and nothing here takes that result back. What it takes back is the picture that would have let a reader recognise a member without the computation. On rows of four the picture works; beyond them there is no substitute for playing the sum.

What the earlier census counted #

The rows of four were counted once before, under a different question, and the earlier count reads differently now.

That essay’s point was the sixty-six rows that fail: a scoring game has no inverses in general. Its fifteen successes were incidental, and fifteen rows of four is exactly the range in which the bracket picture is true. A reader who went from that table to rows of six would have been right to expect the picture to carry on and wrong to rely on it. The census here is the check that could only be made one length further on — the same lesson what a pass is worth to a theory draws about which rows a theory is tested on.

The convention named #

The game is take-the-ends: two players alternately take a coin from either end of any row, keep it, and the score is Left’s total minus Right’s. A row’s negative is the same row with every coin’s sign changed, so a coin worth $v$ to whoever takes it becomes a coin worth $-v$. A row cancels when the row plus its negative scores nought with Left to move first and nought with Right to move first. All scores are exact, computed by playing every interval of every row in every combination.

The coin sets are small by design. Three values per set and rows of two to seven coins give 3,276 rows per coin set, few enough to play every one against its negative in full; larger sets would change the counts and might change the shapes.

What the census cannot show #

The census cannot say what the cancelling rows are, only what they are not. It rules out pairing, zero temperature, and — by its silence — any description a reader would guess from rows of four. A positive description might exist in terms of the play itself: a row cancels when some strategy for the second player in the row plus its negative always restores a balance. No such strategy is exhibited here.

Nor does the census explain why the first mover never finishes behind in a row plus its negative. That fact is true of all 9,828 rows swept, and it is exactly the fact a proof of any description of cancelling rows would have to start from.

Still open: whether the cancelling rows are a group #

A class with no short description can still have the structure that matters. The restriction that buys the most asked whether any of its four classes is closed under addition, and for the cancelling rows that question is now the natural next one. Two cancelling rows, placed side by side, make a two-row position; its negative is the two negatives side by side; and the question is whether the four rows together score nought. If they always do, the class is closed, and a stronger test is available: whether a cancelling row and its negative, placed beside any other position, leave that position’s scores exactly where they were. A pair that does that is a zero in the only sense a scoring game can have one, and a class whose members all make zeros with their negatives is as close to a group as scoring games get.