Two rows still reward the move

In take-the-ends the coins lie in a row, each player in turn takes a coin from one end, and a coin counts for whoever takes it. Even rows always reward the move proved something that every table of this game had been showing without saying: on a row with an even number of coins the player to move never finishes behind. Number the coins, and the two ends sit on positions of opposite parity. The first player takes whichever end belongs to the richer parity class, which leaves the opponent two ends of the other class to choose between, and the first player then takes the coin that choice exposes. Repeated to the end, the first player collects a whole class, and one of the two classes holds at least half of the total.

That settles Milnor’s incentive condition for every even row, since in a game that treats the two players alike the incentive to move is exactly the statement that the first player does not lose. It says nothing about sums, and sums are what the mean-value theory is about. Two even rows side by side have an even number of coins between them, but the parity strategy does not extend: the opponent can answer a move in one row by moving in the other, and the careful alternation that handed the first player a whole class is broken at the first reply. The essay that proved the theorem named the test and left it — put two even rows side by side, and ask whether the first player can still always avoid losing.

Every pair, from both sides #

The test is small enough to be exhaustive. Take the four coin sets the single-row census used, each with one negative coin and two positive ones, and every row of two or four coins from each: nine rows of two, eighty-one of four. Every pair of rows from the ninety is played out exactly, once with each player moving first.

Not one of the 16,380 sums leaves the first player behind. On 547 in each coin set the two players finish level and on the rest the first player finishes ahead. Because the game treats the players alike, the first player’s margin is the same whoever that player is, so the incentive to move holds on every sum in the census.

The second column is the one that says why. Write $A_1$ and $A_2$ for the margins the first player gets on each row played alone. If each row behaved like a switch — a fight worth $\pm A$ to whoever moves in it first — the sum would be worth $|A_1 - A_2|$ to the first player: take the bigger switch and let the opponent take the smaller. Worth nothing, and worth fighting for is where that arithmetic is set out. The sum is never worth less than that difference, and on 3,147 of the 4,095 pairs in each set it is worth exactly that. On the other 948 it is worth more, and the most any pair gains over the switch arithmetic is both rows’ margins at once.

The four coin sets give identical counts. That is not forced — the margins themselves differ from set to set — and it suggests that which pairs meet the floor depends on the order of the coin values, one negative below two positives, more than on their sizes. It is observed on these four sets and not explained.

Why two rows cannot punish the move #

The floor is not merely observed. There is a strategy that reaches it on every pair, and it uses the single-row theorem at exactly one point.

The strategy. Open the row whose own margin is larger, playing in it as best play on that row alone would. After that, whenever the opponent moves in a row, answer in the same row, again as best play on that row alone would.

Follow what that does to each row. In the row the first player opened, the first player moves only in answer to the opponent, so the moves in that row strictly alternate, first player first — it is played exactly as a row standing alone, and the first player gets at least its margin $A_1$ from it. In a row the opponent opens, the same argument with the roles exchanged gives the first player at least $-A_2$. The two rows add, and the margin is at least $A_1 - A_2$.

There is one place the strategy can be interrupted. When the opponent takes the last coin of the row the first player opened, there is nothing there to answer, and the first player has to move somewhere else. If the other row is untouched, the first player opens it and gets its margin as a bonus. If it is in play, the first player must move in it out of turn — and what is left of it is an even row, because the answers keep that row’s count even between the opponent’s moves. Moving first on an even row is never worse than moving second on it, which is the single-row theorem. So the interruption can only help, and the guarantee stands: on any sum of two even rows the first player gets at least the difference of their margins. That is at least nought, because the richer row was the one opened.

In the drawn pair the opponent takes the obvious reply, opening the second row by taking its 3, and from there each row is played out alone. The first player finishes 6 ahead, exactly $8 - 2$. Most pairs finish there.

A pair worth both of its rows #

The pairs that beat the floor are where the rows stop behaving like switches, and the one that beats it by most shows how.

The two rows are $[-2\ 3]$ and $[-2\ {-2}\ 3\ {-2}]$. Each alone is worth 5 to whoever moves first: on the short row the first player takes the 3 and the opponent must take the −2, and on the long row the first player takes a −2 from the left, the opponent is forced onto another −2, and the first player then takes the 3. Two switches of size 5 would cancel — whoever takes one, the other player takes the other — and the switch arithmetic says the sum is worth nothing.

It is worth 10. The first player takes the 3 from the short row. The opponent now faces a single −2 on one row and a long row whose ends are both −2: every coin within reach is a −2, and whichever is taken, the first player is the one who can wait. The play runs 3, −2, −2, −2, 3, −2, and the second player ends with three of the four bad coins.

What makes this pair possible is compulsion. A switch is a fight a player can decline: nobody is obliged to move in one, so a switch’s worth is what the player who takes it gains over the one who does not. A coin row has no such freedom at its end, because every coin must be taken by somebody and a player whose turn it is must take one — the rule a pass is not a move examines from the other side, by handing the players a way out of it. Negative coins turn that obligation into a cost, and a pair of rows can pass the cost to one player twice over.

Pairs that beat it with nothing to avoid #

Compulsion is the explanation that suggests itself, and it is not the whole of one. The test is to run the same census on coin sets with no negative coin in them, where no coin is a liability and nobody is ever sorry to take one.

The floor still holds — it is a theorem — and it is still beaten, on 844 of the 4,095 pairs from $\{1, 2, 3\}$, on 980 from $\{1, 2, 5\}$ and on 844 from $\{0, 1, 2\}$. On the set with a negative coin, 932 of the 948 pairs that beat it carry a −2; the other sixteen are pairs of positive coins from the same set, which beat it for the same reason the positive sets do.

The smallest case is two rows of two coins, $[1\ 2]$ and $[2\ 3]$. Each alone is worth 1 to its first mover, who takes the larger coin and leaves the smaller, so as switches they cancel. Together they are worth 2: the first player takes the 3, and whatever the opponent takes next — the lone 2, or either coin of $[1\ 2]$ — the first player takes a 2 and the opponent the 1. The first player ends with 3 and 2 and the opponent with 2 and 1.

The reason is that a row is not one fight. After the first move on $[2\ 3]$ a single coin of 2 is left, and a single coin is itself a fight — worth its value to whoever moves in it. So the row is a fight with another fight behind it, and a sum of such things is not the sum of their first fights. The switch a player is imagining makes the same point about hot positions in general: the summary “worth about m, and worth t to move in” is exact for a plain fight and leaves out whatever stands behind it. On coin rows what stands behind every move is the rest of the row, and the switch arithmetic reads each row as its first fight alone. That is why it is a floor rather than a description, on positive coins as on negative ones.

Three rows and four #

The proof above uses two rows in an essential way. With three, the opponent can answer the first player’s opening by opening a second row, and then a third, and each time the first player’s answer keeps the alternation the opponent chose. The guarantee the argument gives is then the richest row’s margin less the other two together, which is often negative.

So sums of three and four rows are sampled — a thousand sums of three rows of two or four coins from $\{-2, 1, 3\}$, four hundred sums of three four-coin rows of positive coins, and four hundred sums of four two-coin rows from a set of six coins running from −4 to 9. Each is played out exactly, from both sides.

No sum leaves the first player behind. On 225 of the thousand triples, 76 of the four hundred positive triples and 217 of the four-row sums the opener’s argument has nothing to say: the richest row is worth less than all the others together. The incentive holds there anyway.

What does not hold is the switch arithmetic. For several switches the first player’s margin is the alternating sum of their sizes sorted — take the biggest, the opponent takes the next, and so on. With two rows that is the difference of the margins and a proved floor. With three it is violated on 70 of the thousand triples, 30 of the positive ones and 75 of the four-row sums. The picture that explained two rows is wrong about three, and the conclusion it would have explained is right anyway.

Three rows worth less than their switches #

The smallest triple in the sample that falls below the arithmetic is three rows of two coins.

The rows are $[1\ {-2}]$, $[1\ 3]$ and $[3\ {-2}]$. Alone they are worth 3, 2 and 5 to their first movers: on each row with a −2 the first mover takes the good coin and the opponent is forced onto the bad one. Read as switches of sizes 5, 3 and 2, the sum should be worth $5 - 3 + 2 = 4$ to the first player.

It is worth nothing. The two −2 coins are the reason, though not in a way one sentence captures. The switch arithmetic counts each of them as a coin the opponent is forced to take, because on its own row the second mover is. On the board together a player can decline a −2 only while some better coin remains, and with three rows the better coins run out in an order that leaves one −2 to each side. Played out, each player takes one −2, one 1 and one 3, and the sum is level.

That is the same compulsion that made the pair worth both of its margins, working the other way. A negative coin is a liability that somebody must take, and the switch arithmetic assigns every liability to the player who would carry it on its own row. In a sum the liabilities are shared by whoever runs out of better moves, and that can favour either side.

What the sums cannot show #

The two-row result is a theorem; the rest is a census. The opener’s strategy proves that every sum of two even rows is worth at least the difference of their margins, for any coin values. Sums of three and four rows are sampled — 1,800 sums in all — and none is below nought, and nothing proves that none ever is. The obvious attempts to extend the proof all use the switch arithmetic, and the switch arithmetic is exactly what fails at three.

The incentive at the start is not the hereditary condition. Milnor’s bound needs the incentive at every position the play can reach, and a hypothesis has to hold all the way down found that condition failing on odd intervals of a single row. A sum of even rows becomes a sum containing an odd row after one move, and the census here tests only the positions where every row is even. It extends the class of positions with the incentive; it does not show that sums of even rows satisfy Milnor’s hypothesis.

And the rows are short. Two and four coins in the exhaustive census, two, four and six in the samples, from sets of three to six coin values. The proof covers any length; the census covers what it covers.

The convention the sums are scored under #

Take-the-ends on a disjunctive sum of rows: the player to move takes one coin from either end of any non-empty row, a coin counts for the player who takes it, and nobody may pass while a coin remains. The score is the first player’s total less the second’s. Because the rules treat the two players alike, the margin with Left to move first equals the margin with Right to move first, so the incentive to move — Left does at least as well moving first as moving second — is the statement that the first player’s margin is at least nought. A row’s own margin $A$ is the first player’s margin on that row alone, and the switch arithmetic reads a row as a switch of size $A$.

The surprise: the floor is exact on three pairs in four and still the wrong picture #

The switch picture is how temperature theory thinks about hot games: every component is a fight of some size, the first player takes the biggest, and the sum is an alternating total. It is right on three pairs of rows in four, and for two rows it is even a theorem that it is never too generous. What it misses on the rest is two things, neither of them subtle. A row is a sequence of fights and not one, so what is left after a move is still worth taking. And every coin must be taken, so a player out of good moves can be made to take a bad one.

Both are invisible on a single row, where the parity strategy controls the whole order of play. Both appear the moment a player has a choice of rows. The pair of positive rows worth 2 is the first at work; the pair worth 10 and the triple worth nothing are the second, benefiting first one player and then the other. Outcomes do not add is the normal-play version of the same lesson, that what a part is worth alone is not what it contributes to a sum. Here the part is worth its margin alone, the margin is a floor for two parts and nothing for three, and the incentive to move — the thing the margins were supposed to explain — holds on every sum tried regardless.

Still open: a proof that reaches three rows #

The census says that no sum of even rows punishes the move, and the only proof in hand stops at two. A proof for three would have to do without the switch arithmetic, since three rows can be worth less than their switches. The natural candidate is a strategy that keeps the parity argument alive across rows — choosing, in each row, the class the first player means to take, and answering the opponent so that no row ever hands the opponent two coins of the first player’s class in a row. Whether such a strategy exists is the question, and the measurement that would test a candidate is the same census with the strategy playing one side: if it is never behind on any sampled sum, it is worth trying to prove; if it is, the triple where it fails is where to look. Counting at the end is where the parity strategy was first drawn, and what a pass is worth to a theory is the other place in these essays where the obligation to move turned out to be doing the work a theory attributed to something else.