A cancelling pair is a zero

Cancelling is not pairing found that the coin rows which cancel against their own negatives have no short description. They are not the rows whose coins pair off, though they are on rows of four; they are not the rows with nothing at stake, though that is necessary for some coin sets; and they include a row of five coins, $-2\;1\;3\;1\;{-2}$, that no pairing could produce. A class without a description can still have structure, and the structure that matters for a scoring theory is the one the restriction that buys the most asked about and could not measure: is the class closed, and does it behave like a group?

The two questions have direct tests. Closure asks whether two cancelling rows, placed side by side, still cancel against their two negatives. The group question asks something stronger: whether a cancelling row and its negative together are a zero — a position that can be added to any other without changing anything.

What the normal-play version of this looks like #

It helps to have the normal-play theorem in view, because both tests here are that theorem’s two halves asked of a setting where neither is guaranteed.

Under the last-move convention every game $G$ has a negative $-G$, the same game with the players’ roles swapped, and $G + (-G)$ is a second-player win. Turn the board through a right angle is the picture: the second player copies every move in the mirror component and so never runs out of moves first. From that one fact two consequences follow. Sums of zeros are zeros, because a second player can copy in every component at once. And a zero is invisible beside anything, because in $X + G + (-G)$ a player can play their best strategy for $X$ and answer every move in the pair inside the pair. Those two are exactly closure and invisibility, and under normal play they are free.

In a scoring game neither is free, and nothing to subtract with is the account of why: copying in the mirror row doubles a coin rather than cancelling it. So the questions are genuinely open for coin rows, and the answer for the cancelling class is not implied by anything proved about the class. It could have come out otherwise.

Sums of cancelling rows #

Nineteen rows of two, four and five coins from $\{-2, 1, 3\}$ cancel: three of two coins, fifteen of four, and the one of five. Every pair of them — 190 pairs, a row with itself included — was placed side by side and played against the two negatives side by side: four rows at once, each player taking a coin from either end of any row. Every one of the 190 cancels. Nought with Left to move first, nought with Right to move first.

The table groups them by the lengths involved, and the row that matters most is the one with a five in it. The row of five is the one cancelling row with no pairing structure, the one whose cancellation the earlier essay could only explain as a balance of costs at its two ends. It adds like the others. Placed beside any of the eighteen even cancelling rows, or beside itself, the sum still cancels. Whatever property makes a row cancel, it survives addition whether or not the row has the bracket shape.

That is the first property a group needs and it is not automatic. Under normal play every game has an inverse and sums of inverses are inverses of sums as a matter of algebra. In a scoring game the class of games with inverses is a small subset, and nothing about the definition — “this row plus its negative scores nought” — says that the subset is closed. It might have been that two cancelling rows interfere, each giving the other’s negative a tempo it could use. On these 190 pairs they never do.

A zero has to be invisible everywhere #

Closure is a statement about cancelling rows added to each other. A zero has to be invisible when added to anything.

The test takes each cancelling row, puts its negative beside it, and places the pair next to a context row — every row of one, two and three coins from the same set, thirty-nine in all — then plays the three rows together and compares the scores with the context row’s own. If the pair is a zero, the context’s score with Left to move and its score with Right to move must both be exactly what they were.

They are, in all 741 tests. Not one score moves by a single point. A cancelling row with its negative can be dropped beside any short row and the result is that row, from both sides.

The control makes the result mean something. Twenty rows of four that do not cancel, each with its negative, placed beside the same thirty-nine contexts, leave the context unchanged only 328 times in 780 — about two tests in five. In the other 452 both scores move. So invisibility is not a property every row-and-negative pair happens to have because the pair’s coins sum to nothing; it belongs to the cancelling pairs and fails more often than not for the others.

How a pair that is not a zero moves a score #

The first failure of the control is small enough to play by hand, and it shows what a non-zero pair actually contributes.

The context is a single coin worth $-2$. Alone, whoever moves first has to take it: Left moving first scores $-2$, Right moving first gives Left $+2$. Now put $-2\;{-2}\;{-2}\;1$ and its negative $2\;2\;2\;{-1}$ beside it. The pair’s coins total nothing, and yet both scores become nought. The pair has given the player who would have had to take the $-2$ something else to do — a way to spend a turn in the pair without losing ground there — and a spare turn in a scoring game is worth exactly the cost of the move it avoids.

That is the characteristic effect of a non-zero pair, and it is the same object what a pass is worth to a theory studied: a tempo. A pair that can supply a tempo changes the context whenever the context contains a move nobody wants to make. A cancelling pair supplies none. Whatever the first player does in it, the second can answer inside it without loss, so it never offers either player a free move to spend elsewhere. That is not a proof — the tests are what establish invisibility on this range — but it is what the tests are measuring.

What a group buys #

The four-class table of the earlier essay scored each class on three questions: whether the mean-value bound holds on pairs from it, whether comparison by subtraction agrees with comparison in context, and whether every member cancels.

Closure and invisibility explain the third column’s best cell. Comparison by subtraction asks whether $G \ge H$ can be decided by playing $G$ plus the negative of $H$ and checking the score. Under normal play that works because $H$ plus its negative is nought in every context, so adding it changes nothing, and $G - H \ge 0$ is equivalent to $G \ge H + (H - H)$, which is $G \ge H$. The same chain of reasoning goes through in a scoring game exactly when $H$ plus its negative is invisible in every context the comparison uses — which is what the context test measures. So the cancelling class getting all its comparisons right, which the earlier essay reported and did not explain, is a consequence of its members’ pairs being zeros.

And the failure of subtraction outside the class is the same fact read backwards. A non-cancelling $H$ plus its negative supplies a tempo in some contexts, so $G - H$ is not a faithful stand-in for “$G$ against $H$”, and the difference test goes wrong in exactly the places the tempo is worth something.

What a solver could do with it #

Invisibility is a statement about positions, and it has a practical reading. A position containing a row and its negative, where the row cancels, can have both deleted before any search begins, and every score of the position is unchanged. That is a simplification rule — the scoring analogue of cancelling $G$ against $-G$ in a sum of normal-play games — and it is licensed exactly for the cancelling rows. For any other row, deleting the row and its negative from a position changes the answer more often than not, as the control shows.

That licence is worth comparing with the ones equal in this company and the company that is closed measured for normal play, where a restricted notion of equality licenses substitution only inside a company closed under addition. Here the company is every short coin row, the equality is “same score from both sides”, and the licence is unconditional within the range tested: any context, any cancelling pair. A class that licenses deletion in every context tried is the strongest kind of licence a scoring theory has so far produced, and it covers nineteen rows.

It is also the only such licence available. Nothing else about coin rows has an inverse: a non-cancelling row plus its negative is a position with a tempo in it, and a solver that deleted it would be deleting a move somebody needed. Counting at the end changes everything set out the cost of the scoring convention as the loss of the group structure; the cancelling rows are the part of that structure that survives.

Where the argument has gaps #

The two tests are exhaustive over their ranges and silent beyond them, and three gaps are worth naming.

The contexts are short. Every context is a single row of at most three coins. A zero must be invisible beside positions of any size, including sums of several rows, and a tempo that only matters in a long position would not show up here. The closure test covers some of that ground — the sums of two cancelling rows are themselves contexts of up to ten coins for each other — but not all of it.

The class is small. Nineteen rows is every cancelling row of up to five coins from one coin set. The rows of six and seven that cancel were not included, because four rows of six coins each is a position with millions of states and 190 of them would take hours rather than minutes. The odd row of five is the only member without pairing structure, so the claim that non-pairable members add like the others rests on one row.

Invisibility is measured, not derived. The explanation offered above — a cancelling pair supplies no tempo — is a reading of the results, not an argument from the definition. A proof would show that in any context, the second player can answer every move in the pair inside the pair without loss; the tests show that the scores never move, which is what such a strategy would produce, but they do not exhibit the strategy.

How the positions were played, and what they cost #

Every number here is an exact score, found by playing the position out with best play for both sides, and the way that is done sets the limits of the study.

A position of several rows is described by which interval of each row is still on the table and whose turn it is. The score from a position is the best the player to move can do over their possible moves — a coin from either end of any non-empty row — given best play afterwards, and it is computed once per position and remembered. A row of $n$ coins has about $n^2/2$ intervals, so a position of four rows of four and five coins has at most a few hundred thousand states, and the 190 closure tests together take about a minute. The context tests are cheaper, three rows at a time.

The same method on four rows of six coins would need tens of millions of states per test, and the cancelling class at six coins has 67 members from this coin set alone, giving over two thousand pairs. That is the reason the rows of six and seven are absent, and it is the first thing a larger study would change: the question of whether the non-pairable cancelling rows add like the pairable ones deserves more than one row of five as evidence, and the rows of seven are where the next non-pairable members live.

The one structural shortcut taken is that two copies of the same row in a position are interchangeable, so a position is recorded by the multiset of intervals of each distinct row rather than the list. That is what lets a row be tested against itself — the pair with $i = j$ in the closure table — without doubling the work.

The convention named #

Take-the-ends throughout: players alternately take a coin from either end of any row and keep it, the score is Left’s total minus Right’s, and a sum of rows is played by choosing any end of any row. The negative of a row changes every coin’s sign. Two positions are equal in a context when their scores with Left moving first and with Right moving first are both the same, and a pair is invisible in a context when adding it leaves both scores unchanged. Nothing here uses the last-move convention; every number is a final score.

What the tables cannot show #

The tables show that the cancelling rows behave like a group on the positions tested. They cannot show that the class is a group in the algebraic sense, because a scoring game’s sum of rows is not a row, and the class as defined contains only rows. The honest object is the set of all positions — sums of any number of rows — whose sum with their negative cancels, and closure says the cancelling rows generate part of it. How large that set is, and whether every member of it is a sum of cancelling rows, is a different question.

Nor can the tables say anything about the mean-value bound for sums of cancelling rows, which was the first of the three properties the earlier table scored. The even cancelling rows have temperature nought for this coin set, so the bound is trivial on them; the row of five does not, and whether sums of cancelling rows keep a temperature of nought, or keep the bound, was not checked here.

Still open: what the other classes exclude #

The cancelling class has now been shown to add and to vanish. The other two restricted classes of the earlier table — rows with an incentive to move at the top, and rows with that incentive at every subposition — were measured in the same table and never examined as classes. The first of them held every row of four in that table: all eighty-one rows of four coins had an incentive to move, which means that on rows of four the condition excluded nothing at all. A condition that excludes nothing on one length is either a coincidence of that length or a theorem about it, and which of the two it is decides what Milnor’s hypothesis — the condition the mean-value theory rests on — actually says about coin rows.

There is a reason to expect the answer to be a theorem rather than a coincidence, and it is the same parity that runs through every coin-row result so far. A row of four has two ends of opposite parity, and a player who moves first can always choose which parity class of coins to collect. If that choice is always enough to avoid falling behind, then every even row rewards the move at least a little, the incentive condition holds on every one of them, and the class defined by it on rows of four is not a restriction at all. The measurement that settles it is the same census run on every length and several coin sets, and the argument, if there is one, is a single paragraph about which coins the first player can reach.