# Strategic thinking in infinite games, CosmoCaixa Science Museum, Barcelona, March 2023

I am deeply honored to be invited by la Caixa Foundation to give a talk in “The Greats of Science” talk series, to be held 16 March 2023 at the CosmoCaixa Science Museum in Barcelona. This talk series aspires to host “prestigious figures who have contributed towards admirable milestones, studies or discoveries,” who will bring the science to a general audience, aiming to “give viewers the chance to explore the most relevant parts of contemporary sicence through the top scientists of the moment.” Previous speakers include Jane Goodall and nearly a dozen Nobel Prize winners since 2018.

Cosmo Caixa announcement

I hope to rise to those high expectations!

My topic will be: Strategic thinking in infinite games.

Have you time for an infinite game? Many familiar finite games admit natural infinitary analogues, infinite games that may captivate and challenge us with intriguing patterns and sublime complexity. Shall we have a game of infinite chess? Or how about infinite draughts, infinite Hex, infinite Wordle, or infinite Sudoku? In the Chocolatier’s game, the Chocolatier serves up an infinite stream of delicious morsels, while the Glutton aims to eat every one. These games and others illustrate the often subtle strategic aspects of infinite games, and sometimes their downright logical peculiarity. Does every infinite game admit of a winning strategy? Must optimal play be in principle computable? Let us discover the fascinating nature of infinitary strategic thinking.

The theory builds upon the classical finitary result of Zermelo (1913), the fundamental theorem of finite games, which shows that in every finite two-player game of perfect information, one of the players must have a winning strategy or both players have draw-or-better strategies. This result extends to certain infinitary games by means of the ordinal game-value analysis, which assigns transfinite ordinal values $\alpha$ to positions in a game, generalizing the familiar mate-in-$n$ idea of chess to the infinite. Current work realizes high transfinite game values in infinite chess, infinite draughts (checkers), infinite Go, and many other infinite games. The highest-known game value arising in infinite chess is the infinite ordinal $\omega^4$, and every countable ordinal arises in infinite draughts, the optimal result. Games exhibiting high transfinite ordinal game values have a surreal absurd character of play. The winning player will definitely win in finitely many moves, but the doomed losing player controls the process with absurdly long deeply nested patterns of forcing moves that must be answered, as though counting down from the infinite game value—when 0 is reached, the game is over.

# Checkmate is not the same as a forced capture of the enemy king in simplified chess

In my imagination, and perhaps also in historical reality, our current standard rules of chess evolved from a simpler era with a simpler set of rules for the game. Let us call it simplified chess. In simplified chess there was the same 8×8 board with the same pieces as now moving under the same movement rules. But the winning aim was different, and simpler. Namely, the winning goal of simplified chess was simply to capture the enemy king. You won the game by capturing the opposing king, just as you would capture any other piece.

There was therefore no need to mention check or checkmate in the rules. Rather, these words described important situations that might arise during game play. Specifically, you have placed your opponent in check, if you were in a position to capture their king on the next move, unless they did something to prevent it. You placed your opponent in checkmate, if there was indeed nothing that they could do to prevent it.

In particular, in simplified chess there was no rule against leaving your king in check or even moving your king into check. Rather, this was simply an inadvisable move, because your opponent could take it and thereby win. Checkmate was merely the situation that occurred when all your moves were like this.

It is interesting to notice that it is common practice in blitz chess and bullet chess to play with something closer to simplified chess rules—it is often considered winning simply to capture the opposing king, even if there was no checkmate. This is also how the chess variant bughouse is usually played, even in official tournaments. To my way of thinking, there is a certain attractive simplicity to the rules of simplified chess. The modern chess rules might seem to be ridiculous for needlessly codifying into the rules a matter that could simply be played out on the board by actually capturing the king.

Part of what I imagine is that our contemporary rules could easily have evolved from simplified chess from a practice of polite game play. In order to avoid the humiliation of actually capturing and removing the opponent’s king and replacing it with one’s own piece, which indeed might even have been a lowly pawn, a custom developed to declare the game over when this was a foregone conclusion. In other words, I find it very reasonable to suppose that the winning checkmate rule simply arose from a simplified rule set by common practice of respectful game play.

I am not a chess historian, and I don’t really know if chess did indeed evolve from a simpler version of the game like this, but it strikes me as very likely that something like this must have happened. I await comment from the chess historians. Let me add though that I would also find it reasonable to expect that simplified chess might also have had no provision for opening pawns moving two squares instead of just one. Such a rule could arise naturally as an agreed upon compromise to quicken the game and get to the more interesting positions more quickly. But once that rule was adopted, then the en passant rule is a natural corrective to prevent abuse. I speculate that castling may have arisen similarly—perhaps many players in a community customarily took several moves, perhaps in a standard manuver sequence, to accomplish the effect of hiding their kings away toward the corner and also bringing their rooks to the center files; the practice could have simply been codified into a one-move practice.

My main point in this post, however, does not concern these other rules, but rather only the checkmate winning condition rule and to a certain logic-of-chess issue it brings to light.

When teaching chess to beginners, it is common to describe the checkmate winning situation in terms of the necessary possibility of capturing the king. One might say that a checkmate situation means the king is attacked, and there is nothing the opponent can do to prevent the king from being captured.

This explanation suggests a general claim in the logic of chess: a position is in checkmate (in the contemporary rules) if and only if the winning player can necessarily capture the opposing king on the next move (in simplified chess).

This claim is mostly true. In most instances, when you have placed your opponent in checkmate, then you would be able to capture their king on your next move in simplified chess, since all their moves would leave the king in check, and so you could take it straight away.

But I would like to point out something I found interesting about this checkmate logic claim. Namely, it isn’t true in all cases. There is a position, I claim, that is checkmate in the modern rules, but in simplified chess, the winning player would not be able to capture the enemy king on the next move.

My example proceeds from the following (rather silly) position, with black to move. (White pawns move upward.)

Of course, Black should play the winning move: knight to C7, as shown here:

This places the white king in check, and there is nothing to be done about it, and so it is checkmate. Black has won, according to the usual chess rules of today.

But let us consider the position under the rules of simplified chess. Will Black be able to capture the white king? Well, after Black’s nC7, it is now White’s turn. Remember that in simplified chess, it is allowed (though inadvisable) to leave one’s king in check at the end of turn or even to move the king into check. But the trouble with this position is not that White can somehow move to avoid check, but rather simply that White has no moves at all. There are no White moves, not even moves that leave White in check. But therefore, in simplified chess, this position is a stalemate, rather than a Black win. In particular, Black will not actually be able to capture the White king, since White has no moves, and so the game will not proceed to that juncture.

Conclusion: checkmate in contemporary chess is not always the same thing as being necessarily able to capture the opposing king on the next move in simplified chess.

Of course, perhaps in simplified chess, one wouldn’t regard stalemate as a draw, but as a win for the player who placed the opponent in stalemate. That would be fine by me (and it would align with the rules currently used in draughts, where one loses when there is no move available), but it doesn’t negate my point. The position above would still be a Black win under that rule, sure, but still Black wouldn’t be able to capture the White king. That is, my claim would stand that checkmate (in modern rules) is not the same thing as the necessary possibility to capture the opposing king.

I had made a Tweet earlier today about this idea (below), but opted for a fuller explanation in this blog post to explain the idea more carefully.

On Twitter, user Gro-Tsen pointed out a similar situation arises with stalemates. Namely, consider the following position, with White to play:

Perhaps Black had just played qB5, which perhaps was a blunder, since now the White king sits in stalemate, with no move. So in the usual chess rules, this is a drawn position.

But in simplified chess, according to the rules I have described, the White king is not yet explicitly captured, and so he is free still to move around, to A6, A8, B6, or B7 (moving into check), or also capturing the black rook at B8. But in any case, whichever move White makes, Black will be able to capture the White king on the next move. So in the simplified chess rules, this position is White-to-win-in-1, and not a draw.

Furthermore, this position therefore shows once again that checkmate (in normal rules) is not the same as the necessary possibility to capture the king (in simplified rules), since this is a position where Black has a necessary possibility to capture the White king on the next move in simplified chess, but it is not checkmate in ordinary chess, being rather stalemate.

;TLDR Ultimately, my view is that our current rules of chess likely evolved from simplified rules and the idea that checkmate is what occurs when you have a necessary possibility of capturing the enemy king on the next move in those rules. But nevertheless, example chess positions show that these two notions are not quite exactly the same.

# My favorite theorem

What a pleasure it was to be interviewed by Evelyn Lamb and Kevin Knudson for their wonderful podcast series, My Favorite Theorem, available on Apple, Spotify, and any number of other aggregators.

I had a chance to talk about one my most favorite theorems, the fundamental theorem of finite games.

Theorem.(Zermelo 1913) In any two-player finite game of perfect information, one of the players has a winning strategy, or both players have drawing strategies.

Listen to the podcast here: My Favorite Theorem. A transcript is also available.

# Determinacy for proper class games, Seminaire de Logique Lyon-Paris, April 2021

This will be a talk for the Seminaire de Logique Lyon-Paris on 14 April 2021 4pm Paris time (3pm UK). The talk will be held on Zoom at
875 1148 7359
.

Abstract. The principle of open determinacy for class games — two-player games of perfect information with plays of length ω, where the moves are chosen from a possibly proper class, such as games on the ordinals — is not provable in Zermelo-Fraenkel set theory ZFC or Gödel-Bernays set theory GBC, if these theories are consistent, because provably in ZFC there is a definable open proper class game with no definable winning strategy. In fact, the principle of open determinacy and even merely clopen determinacy for class games implies Con(ZFC) and iterated instances Con(Con(ZFC)) and more, because it implies that there is a satisfaction class for first-order truth, and indeed a transfinite tower of truth predicates $\text{Tr}_\alpha$ for iterated truth-about-truth, relative to any class parameter. This is perhaps explained, in light of the Tarskian recursive definition of truth, by the more general fact that the principle of clopen determinacy is exactly equivalent over GBC to the principle of elementary transfinite recursion ETR over well-founded class relations. Meanwhile, the principle of open determinacy for class games is strictly stronger, although it is provable in the stronger theory GBC+$\Pi^1_1$-comprehension, a proper fragment of Kelley-Morse set theory KM.

# The theory of infinite games, including infinite chess, Talk Math With Your Friends, June 2020

This will be accessible online talk about infinite chess and other infinite games for the Talk Math With Your Friends seminar, June 18, 2020 4 pm EST (9 pm UK).  Zoom access information.  Please come talk math with me!

Abstract. I will give an introduction to the theory of infinite games, with examples drawn from infinite chess in order to illustrate various concepts, such as the transfinite game value of a position.

See more of my posts on infinite chess.

# Infinite Sudoku and the Sudoku game

Consider what I call the Sudoku game, recently introduced in the MathOverflow question Who wins two-player Sudoku? posted by Christopher King (user PyRulez). Two players take turns placing numbers on a Sudoku board, obeying the rule that they must never explicitly violate the Sudoku condition: the numbers on any row, column or sub-board square must never repeat. The first player who cannot continue legal play loses. Who wins the game? What is the winning strategy?

The game is not about building a global Sudoku solution, since a move can be legal in this game even when it is not part of any global Sudoku solution, provided only that it doesn’t yet explicitly violate the Sudoku condition. Rather, the Sudoku game is about trying to trap your opponent in a maximal such position, a position which does not yet explicitly violate the Sudoku condition but which cannot be further extended.

In my answer to the question on MathOverflow, I followed an idea suggested to me by my daughter Hypatia, namely that on even-sized boards $n^2\times n^2$ where $n$ is even, then the second player can win with a mirroring strategy: simply copy the opponent’s moves in reflected mirror image through the center of the board. In this way, the second player ensures that the position on the board is always symmetric after her play, and so if the previous move was safe, then her move also will be safe by symmetry. This is therefore a winning strategy for the second player, since any violation of the Sudoku condition will arise on the opponent’s play.

This argument works on even-sized boards precisely because the reflection of every row, column and sub-board square is a totally different row, column and sub-board square, and so any new violation of the Sudoku conditions would reflect to a violation that was already there. The mirror strategy definitely does not work on the odd-sized boards, including the main $9\times 9$ case, since if the opponent plays on the central row, copying directly would immediately introduce a Sudoku violation.

After posting that answer, Orson Peters (user orlp) pointed out that one can modify it to form a winning strategy for the first player on odd-sized boards, including the main $9\times 9$ case. In this case, let the first player begin by playing $5$ in the center square, and then afterwards copy the opponent’s moves, but with the ten’s complement at the reflected location. So if the opponent plays $x$, then the first player plays $10-x$ at the reflected location. In this way, the first player can ensure that the board is ten’s complement symmetric after her moves. The point is that again this is sufficient to know that she will never introduce a violation, since if her $10-x$ appears twice in some row, column or sub-board square, then $x$ must have already appeared twice in the reflected row, column or sub-board square before that move.

This idea is fully general for odd-sized Sudoku boards $n^2\times n^2$, where $n$ is odd. If $n=2k-1$, then the first player starts with $k$ in the very center and afterward plays the $2k$-complement of her opponent’s move at the reflected location.

Conclusion.

1. On even-sized Sudoku boards, the second player wins the Sudoku game by the mirror copying strategy.
2. On odd-sized Sudoku boards, the first players wins the Sudoku game by the complement-mirror copying strategy.

Note that on the even boards, the second player could also play complement mirror copying just as successfully.

What I really want to tell you about, however, is the infinite Sudoku game (following a suggestion of Sam Hopkins). Suppose that we try to play the Sudoku game on a board whose subboard squares are $\mathbb{Z}\times\mathbb{Z}$, so that the full board is a $\mathbb{Z}\times\mathbb{Z}$ array of those squares, making $\mathbb{Z}^2\times\mathbb{Z}^2$ altogether. (Or perhaps you might prefer the board $\mathbb{N}^2\times\mathbb{N}^2$?)

One thing to notice is that on an infinite board, it is no longer possible to get trapped at a finite stage of play, since every finite position can be extended simply by playing a totally new label from the set of labels; such a move would never lead to a new violation of the explicit Sudoku condition.

For this reason, I should like to introduce the Sudoku Solver-Spoiler game variation as follows. There are two players: the Sudoku Solver and the Sudoku Spoiler. The Solver is trying to build a global Sudoku solution on the board, while the Spoiler is trying to prevent this. Both players must obey the Sudoku condition that labels are never to be explicitly repeated in any row, column or sub-board square. On an infinite board, the game proceeds transfinitely, until the board is filled or there are no legal moves. The Solver wins a play of the game, if she successfully builds a global Sudoku solution, which means not only that every location has a label and there are no repetitions in any row, column or sub-board square, but also that every label in fact appears in every row, column and sub-board square. That is, to count as a solution, the labels on any row, column and sub-board square must be a bijection with the set of labels. (On infinite boards, this is a stronger requirement than merely insisting on no repetitions.)

The Solver-Spoiler game makes sense in complete generality on any set $S$, whether finite or infinite. The sub-boards are $S^2=S\times S$, and one has an $S\times S$ array of them, so $S^2\times S^2$ for the whole board. Every row and column has the same size as the sub-board square $S^2$, and the set of labels should also have this size.

Upon reflection, one realizes that what matters about $S$ is just its cardinality, and we really have for every cardinal $\kappa$ the $\kappa$-Sudoku Solver-Spoiler game, whose board is $\kappa^2\times\kappa^2$, a $\kappa\times\kappa$ array of $\kappa\times\kappa$ sub-boards. In particular, the game $\mathbb{Z}^2\times\mathbb{Z}^2$ is actually isomorphic to the game $\mathbb{N}^2\times\mathbb{N}^2$, despite what might feel initially like a very different board geometry.

What I claim is that the Solver has a winning strategy in the all the infinite Sudoku Solver-Spoiler games, in a very general and robust manner.

Theorem. For every infinite cardinal $\kappa$, the Solver has a winning strategy to win the $\kappa$-Sudoku Solver-Spoiler game.

• The strategy will win in $\kappa$ many moves, producing a full Sudoku solution.
• The Solver can win whether she goes first or second, starting from any legal position of size less than $\kappa$.
• The Solver can win even when the Spoiler is allowed to play finitely many labels at once on each turn, or fewer than $\kappa$ many moves (if $\kappa$ is regular), even if the Solver is only allowed one move each turn.
• In the countably infinite Sudoku game, the Solver can win even if the Spoiler is allowed to make infinitely many moves at once, provided only that the resulting position can in principle be extended to a full solution.

Proof. Consider first the countably infinite Sudoku game, and assume the initial position is finite and that the Spoiler will make finitely many moves on each turn. Consider what it means for the Solver to win at the limit. It means, first of all, that there are no explicit repetitions in any row, column or sub-board. This requirement will be ensured since it is part of the rules for legal play not to violate it. Next, the Solver wants to ensure that every square has a label on it and that every label appears at least once in every row, every column and every sub-board. If we think of these as individual specific requirements, we have countably many requirements in all, and I claim that we can arrange that the Solver will simply satisfy the $n^{th}$ requirement on her $n^{th}$ play. Given any finite position, she can always find something to place in any given square, using a totally new label if need be. Given any finite position, any row and any particular label $k$, since can always find a place on that row to place the label, which has no conflict with any column or sub-board, since there are infinitely many to choose from and only finitely many conflicts. Similarly with columns and sub-boards. So each of the requirements can always be fulfilled one-at-a-time, and so in $\omega$ many moves she can produce a full solution.

The argument works equally well no matter who goes first or if the Spoiler makes arbitrary finite play, or indeed even infinite play, provided that the play is part of some global solution (perhaps a different one each time), since on each move the Solve can simply meet the requirement by using that solution at that stage.

An essentially similar argument works when $\kappa$ is uncountable, although now the play will proceed for $\kappa$ many steps. Assuming $\kappa^2=\kappa$, a consequence of the axiom of choice, there are $\kappa$ many requirements to meet, and the Solve can meet requirement $\alpha$ on the $\alpha^{th}$ move. If $\kappa$ is regular, we can again allow the Spoiler to make arbitrary size-less-than-$\kappa$ size moves, so that at any stage of play before $\kappa$ the position will still be size less than $\kappa$. (If $\kappa$ is singular, one can allow Spoiler to make finitely many moves at once or indeed even some uniform bounded size $\delta<\kappa$ many moves at once. $\Box$

I find it interesting to draw out the following aspect of the argument:

Observation. Every finite labeling of an infinite Sudoku board that does not yet explicitly violate the Sudoku condition can be extended to a global solution.

Similarly, any size less than $\kappa$ labeling that does not yet explicitly violate the Sudoku condition can be extended to a global solution of the $\kappa$-Sudoku board for any infinite cardinal $\kappa$.

What about asymmetric boards? It has come to my attention that people sometimes look at asymmetric Sudoku boards, whose sub-boards are not square, such as in the six-by-six Sudoku case. In general, one could take Sudoku boards to consist of a $\lambda\times\kappa$ array of sub-boards of size $\kappa\times\lambda$, where $\kappa$ and $\lambda$ are cardinals, not necessarily the same size and not necessarily both infinite or both finite. How does this affect the arguments I’ve given?

In the finite $(n\times m)\times (m\times n)$ case, if one of the numbers is even, then it seems to me that the reflection through the origin strategy works for the second player just as before. And if both are odd, then the first player can again play in the center square and use the mirror-complement strategy to trap the opponent. So that analysis will work fine.

In the case $(\kappa\times\lambda)\times(\lambda\times\kappa)$ where $\lambda\leq\kappa$ and $\kappa=\lambda\kappa$ is infinite, then the proof of the theorem seems to break, since if $\lambda<\kappa$, then with only $\lambda$ many moves, say putting a common symbol in each of the $\lambda$ many rectangles across a row, we can rule out that symbol in a fixed row. So this is a configuration of size less than $\kappa$ that cannot be extended to a full solution. For this reason, it seems likely to me that the Spoiler can win the Sudoko Solver-Spoiler game in the infinite asymmetric case.

Finally, let’s consider the Sudoku Solver-Spoiler game in the purely finite case, which actually is a very natural game, perhaps more natural than what I called the Sudoku game above. It seems to me that the Spoiler should be able to win the Solver-Spoiler game on any nontrivial finite board. But I don’t yet have an argument proving this. I asked a question on MathOverflow: The Sudoku game: Solver-Spoiler variation.

# On the strengths of the class forcing theorem and clopen class game determinacy, Prague set theory seminar, January 2018

This will be a talk for the Prague set theory seminar, January 24, 11:00 am to about 2pm (!).

Abstract. The class forcing theorem is the assertion that every class forcing notion admits corresponding forcing relations. This assertion is not provable in Zermelo-Fraenkel ZFC set theory or Gödel-Bernays GBC set theory, if these theories are consistent, but it is provable in stronger second-order set theories, such as Kelley-Morse KM set theory. In this talk, I shall discuss the exact strength of this theorem, which turns out to be equivalent to the principle of elementary transfinite recursion ETRord for class recursions on the ordinals. The principle of clopen determinacy for class games, in contrast, is strictly stronger, equivalent over GBC to the full principle of ETR for class recursions over arbitrary class well-founded relations. These results and others mark the beginnings of the emerging subject I call the reverse mathematics of second-order set theory.

# Escape!

Here is an interesting game I heard a few days ago from one of my undergraduate students; I’m not sure of the provenance.

The game is played with stones on a grid, which extends indefinitely upward and to the right, like the lattice $\mathbb{N}\times\mathbb{N}$.  The game begins with three stones in the squares nearest the origin at the lower left.  The goal of the game is to vacate all stones from those three squares. At any stage of the game, you may remove a stone and replace it with two stones, one in the square above and one in the square to the right, provided that both of those squares are currently unoccupied.

For example, here is a sample play.

Question. Can you play so as completely to vacate the yellow corner region?

One needs only to move the other stones out of the way so that the corner stones have room to move out. Can you do it? It isn’t so easy, but I encourage you to try.

Here is an online version of the game that I coded up quickly in Scratch: Escape!

My student mentioned the problem to me and some other students in my office on the day of the final exam, and we puzzled over it, but then it was time for the final exam. So I had a chance to think about it while giving the exam and came upon a solution. I’ll post my answer later on, but I’d like to give everyone a chance to think about it first.

Solution. Here is the solution I hit upon, and it seems that many others also found this solution. The main idea is to assign an invariant to the game positions. Let us assign weights to the squares in the lattice according to the following pattern. We give the corner square weight $1/2$, the next diagonal of squares $1/4$ each, and then $1/8$, and so on throughout the whole playing board. Every square should get a corresponding weight according to the indicated pattern.

The weights are specifically arranged so that making a move in the game preserves the total weight of the occupied squares. That is, the total weight of the occupied squares is invariant as play proceeds, because moving a stone with weight $1/2^k$ will create two stones of weight $1/2^{k+1}$, which adds up to the same. Since the original three stones have total weight $\frac 12+\frac14+\frac14=1$, it follows that the total weight remains $1$ after every move in the game.

Meanwhile, let us consider the total weight of all the squares on the board. If you consider the bottom row only, the weights add to $\frac12+\frac14+\frac18+\cdots$, which is the geometric series with sum $1$. The next row has total weight $\frac14+\frac18+\frac1{16}+\cdots$, which adds to $1/2$. And the next adds to $1/4$ and so on. So the total weight of all the squares on the board is $1+\frac12+\frac14+\cdots$, which is $2$.  Since we have $k$ stones with weight $1/2^k$, another way to think about it is that we are essentially establishing the sum $\sum_k\frac k{2^k}=2$.

The subtle conclusion is that after any finite number of moves, only finitely many of those other squares are occupied, and so some of them remain empty. So after only finitely many moves, the total weight of the occupied squares off of the original L-shape is strictly less than $1$. Since the total weight of all the occupied squares is exactly $1$, this means that the L-shape has not been vacated.

So it is impossible to vacate the original L-shape in finitely many moves. $\Box$

Suppose that we relax the one-stone-per-square requirement, and allow you to stack several stones on a single square, provided that you eventually unstack them. In other words, can you play the stacked version of the game, so as to vacate the original three squares, provided that all the piled-up stones eventually are unstacked?

No, it is impossible! And the proof is the same invariant-weight argument as above. The invariance argument does not rely on the one-stone-per-square rule during play, since it is still an invariant if one multiplies the weight of a square by the number of stones resting upon it. So we cannot transform the original stones, with total weight $1$, to any finite number of stones on the rest of the board (with one stone per square in the final position), since those other squares do not have sufficient weight to add up to $1$, even if we allow them to be stacked during intermediate stages of play.

Meanwhile, let us consider playing the game on a finite $n\times n$ board, with the rule modified so that stones that would be created in row or column $n+1$ in the infinite game simply do not materialize in the $n\times n$ game. This breaks the proof, since the weight is no longer an invariant for moves on the outer edges. Furthermore, one can win this version of the game. It is easy to see that one can systematically vacate all stones on the upper and outer edges, simply by moving any of them that is available, pushing the remaining stones closer to the outer corner and into oblivion. Similarly, one can vacate the penultimate outer edges, by doing the same thing, which will push stones into the outer edges, which can then be vacated. By reverse induction from the outer edges in, one can vacate every single row and column. Thus, for play on this finite board with the modified rule on the outer edges, one can vacate the entire $n\times n$ board!

Indeed, in the finite $n\times n$ version of the game, there is no way to lose! If one simply continues making legal moves as long as this is possible, then the board will eventually be completely vacated. To see this, notice first that if there are stones on the board, then there is at least one legal move. Suppose that we can make an infinite sequence of legal moves on the $n\times n$ board. Since there are only finitely many squares, some of the squares must have been moved-upon infinitely often. If you consider such a square closest to the origin (or of minimal weight in the scheme of weights above), then since the lower squares are activated only finitely often, it is clear that eventually the given square will replenished for the last time. So it cannot have been activated infinitely often. (Alternatively, argue by induction on squares from the lower left that they are moved-upon at most finitely often.) Indeed, I claim that the number of steps to win, vacating the $n\times n$ board, does not depend on the order of play. One can see this by thinking about the path of a given stone and its clones through the board, ignoring the requirement that a given square carries only one stone. That is, let us make all the moves in parallel time. Since there is no interaction between the stones that would otherwise interfere, it is clear that the number of stones appearing on a given square in total is independent of the order of play. A tenacious person could calculate the sum exactly: each square is becomes occupied by a number of stones that is equal to the number of grid paths to it from one of the original three stones, and one could use this sum to calculate the total length of play on the $n\times n$ board.

# Buckets of fish!

Let me tell you about the game Buckets of fish

This is a two-player game played with finitely many buckets in a line on the beach, each containing a finite number of fish. There is also a large supply of additional fish available nearby, fresh off the boats.

Taking turns, each player selects a bucket and removes exactly one fish from it and then, if desired, adds any finite number of fish from the nearby supply to the buckets to the left.

For example, if we label the buckets from the left as 1, 2, 3 and so on, then a legal move would be to take one fish from bucket 4 and then add ten fish to bucket 1, no fish to bucket 2, and ninety-four fish to bucket 3. The winner is whoever takes the very last fish from the buckets, leaving them empty.

Since huge numbers of fish can often be added to the buckets during play, thereby prolonging the length of play, a skeptical reader may wonder whether the game will necessarily come to an end. Perhaps the players can prolong the game indefinitely? Or must it always come to an end?

Question. Does every play of the game Buckets of fish necessarily come to an end?

The answer is yes, every game must eventually come to a completion. I shall give several arguments.

Theorem. Every play of the game Buckets of fish ends in finitely many moves. All the fish in the buckets, including all the new fish that may have been added during play, will eventually run out by some finite stage during play.

That is, no matter how the players add fish to the buckets during play, even with an endless supply of fish from the boats, they will eventually run out of fish in the buckets and one of the players will take the last fish.

First proof. We prove the claim by (nested) induction on the number of buckets. If there is only one bucket, then there are no buckets to the left of it, and so there is no possibility in this case to add fish to the game. If the one bucket contains $k$ fish, then the game clearly ends in $k$ moves. Assume by induction that all plays using $n$ buckets end in finitely many moves, and suppose that we have a game situation with $n+1$ buckets, with $k$ fish in bucket $n+1$. We now prove by induction on $k$ that all such games terminate. This argument is therefore an instance of nested induction, since we are currently inside our proof by induction on $n$, in the induction step of that proof, and in order to complete it, we are undertaking a separate full induction on $k$. If $k=0$, then there are no fish in bucket $n+1$, and so the game amounts really to a game with only $n$ buckets, which terminates in finitely many steps by our induction hypothesis on $n$. So, let us assume that all plays with $k$ fish in bucket $n+1$ terminate in finitely many moves. Consider a situation where there are $k+1$ many fish in that bucket. I claim that eventually, one of those fish must be taken, since otherwise all the moves will be only on the first $n$ buckets, and all plays on only $n$ buckets terminate in finitely many moves. So at some point, one of the players will take a fish from bucket $n+1$, possibly adding additional fish to the earlier buckets. But this produces a situation with only $k$ fish in bucket $n+1$, which by our induction assumption on $k$ we know will terminate in finitely many steps. So we have proved that no matter how many fish are in bucket $n+1$, the game will end in finitely many moves, and so the original claim is true for $n+1$ buckets. Thus, the theorem is true for any finite number of buckets. QED

A second proof. Let me now give another proof, following an idea arising in a conversation with Miha Habič. We want to prove that there is no infinitely long play of the game Buckets of fish. Suppose toward contradiction that there is a way for the players to conspire to produce an infinite play, starting from some configuration of some finite number $n$ of buckets, each with finitely many fish in them. Fix the particular infinitely long play. Let $m$ be the right-most bucket from which a fish was taken infinitely often during that infinite course of play. It follows, for example, that $m<n$, since the top bucket can be used only finitely often, as it never gets replenished. Since bucket $m$ starts with only finitely many fish in it, and each time it is replenished, it is replenished with only finitely many fish, it follows that in order to have been used infinitely many times, it must also have been replenished infinitely often. But each time it was replenished, it was because there was some bucket further to the right that had been used. Since there are only finitely many buckets to the right of bucket $m$, it follows that one of them must have been used infinitely often. This contradicts the choice of $m$ as the right-most bucket that was used infinitely often. QED

A third proof. Let me now give a third proof, using ordinals. We shall associate with each Buckets-of-fish position a certain ordinal. With the position $$7\quad 2\quad 5\quad 24,$$ for example, we associate the ordinal $$\omega^3\cdot 24+\omega^2\cdot 5+\omega\cdot 2+7.$$ More generally, the number of fish in each bucket of a position becomes the coefficient of the corresponding power of $\omega$, using higher powers for the buckets further to the right. The key observation to make is that these associated ordinals strictly descend for every move of the game, since one is reducing a higher-power coefficient and increasing only lower-power coefficients. Since there is no infinite descending sequence of ordinals, it follows that there is no infinite play in the game Buckets of fish. This idea also shows that the ordinal game values of positions in this game are bounded above by $\omega^\omega$, and every ordinal less than $\omega^\omega$ is realized by some position. QED

OK, fine, so now we know that the game always ends. But how shall we play? What is the winning strategy? Say you are faced with buckets having fish in the amounts: $$4\quad 5\quad 2\quad 0\quad 7\quad 4$$ What is your winning move? Please give it some thought before reading further.

The winning strategy turns out to be simpler than you might have expected.

Theorem. The winning strategy in the game Buckets of fish is to play so as to ensure that every bucket has an even number of fish.

Proof. Notice first, as a warm-up, that in the case that there is only one bucket containing an even number of fish, then the second player will win, since the first player will necessarily make it odd, and then the second player will make it even again, and so on. So it will be the second player who will make it zero, winning the game. So with one bucket, the player who can make the bucket even will be the winner.

Next, notice that if you play so as to give your opponent an even number of fish in every bucket, then whatever move your opponent makes will result in an odd number of fish in the bucket from which he or she takes a fish (and possibly also an odd number of fish in some of the earlier buckets as well, if they happen to add an odd number of fish to some of them). So if you give your opponent an all-even position, then they cannot give you back an all-even position.

Finally, notice that if you are faced with a position that is not all-even, then you can simply take a fish from the right-most odd bucket, thereby making it even, and add fish if necessary to the earlier buckets so as to make them all even. In this way, you can turn any position that is not all-even into an all-even position in one move.

By following this strategy, a player will ensure that he or she will take the last fish, since the winning move is to make the all-zero position, which is an all-even position, and the opponent cannot produce an all-even position. QED

In the particular position of the game mentioned before the theorem, therefore, the winning move is to take a fish from the bucket with 7 fish and add an odd number of fish to the bucket with 5 fish, thereby producing an all-even position.

Finally, let’s consider a few variations of the game. It is clear that the all-even strategy works in the versions of the game where one is limited to add at most one fish to each of the earlier buckets, and this version of the game is actually playable, since the number of fish does not grow too much. A similar variation arises where one can either or add or remove any number of fish (or just at most one) from any of the earlier buckets, or where one can, say, add either 5 or 6 fish only to each of the earlier buckets. What is important in the argument is simply that one should be able to ensure the all-even nature of the buckets.

For a more interesting variation, consider what I call the Take 3 version of the game, where one can take either one, two or three fish from any bucket and then add any number of fish to the earlier buckets. The game must still eventually end, but what is the winning strategy?

Question. What is your strategy in the Take 3 variation of Buckets of fish?

Please post your answers in the comments, and I’ll post an answer later. One can generalize this to the Take $n$ variation, where on each turn, the player is allowed to take between 1 and $n$ fish from any bucket, and add as many fish as desired to the earlier buckets.

Another puzzling variation is where each player can take any number of fish from a bucket, and then add any number of fish to earlier buckets. Can you find a strategy for this version of the game? Please post in the comments.

# Open and clopen determinacy for proper class games, VCU MAMLS April 2017

This will be a talk for the Mid-Atlantic Mathematical Logic Seminar at Virginia Commonwealth University, a conference to be held April 1-2, 2017.

Abstract. The principle of open determinacy for class games — two-player games of perfect information with plays of length $\omega$, where the moves are chosen from a possibly proper class, such as games on the ordinals — is not provable in Zermelo-Fraenkel set theory ZFC or Gödel-Bernays set theory GBC, if these theories are consistent, because provably in ZFC there is a definable open proper class game with no definable winning strategy. In fact, the principle of open determinacy and even merely clopen determinacy for class games implies Con(ZFC) and iterated instances Con(Con(ZFC)) and more, because it implies that there is a satisfaction class for first-order truth, and indeed a transfinite tower of truth predicates $\text{Tr}_\alpha$ for iterated truth-about-truth, relative to any class parameter. This is perhaps explained, in light of the Tarskian recursive definition of truth, by the more general fact that the principle of clopen determinacy is exactly equivalent over GBC to the principle of elementary transfinite recursion ETR over well-founded class relations. Meanwhile, the principle of open determinacy for class games is provable in the stronger theory GBC+$\Pi^1_1$-comprehension, a proper fragment of Kelley-Morse set theory KM. New work by Hachtman and Sato, respectively has clarified the separation of clopen and open determinacy for class games.

This is joint work with Victoria Gitman. See our article, Open determinacy for class games.

Slides

# The pirate treasure division problem

In my logic course this semester, as a part of the section on the logic of games, we considered the pirate treasure division problem.

Imagine a pirate ship with a crew of fearsome, perfectly logical pirates and a treasure of 100 gold coins to be divided amongst them. How shall they do it? They have long agreed upon the pirate treasure division procedure: The pirates are linearly ordered by rank, with the Captain, the first Lieutenant, the second Lieutenant and so on down the line; but let us simply refer to them as Pirate 1, Pirate 2, Pirate 3 and so on. Pirate 9 is swabbing the decks in preparation. For the division procedure, all the pirates assemble on deck, and the lowest-ranking pirate mounts the plank. Facing the other pirates, she proposes a particular division of the gold — so-and-so many gold pieces to the captain, so-and-so many pieces to Pirate 2 and so on.  The pirates then vote on the plan, including the pirate on the plank, and if a strict majority of the pirates approve of the plan, then it is adopted and that is how the gold is divided. But if the pirate’s plan is not approved by a pirate majority, then regretfully she must walk the plank into the sea (and her death) and the procedure continues with the next-lowest ranking pirate, who of course is now the lowest-ranking pirate.

Suppose that you are pirate 10: what plan do you propose?  Would you think it is a good idea to propose that you get to keep 94 gold pieces for yourself, with the six remaining given to a few of the other pirates? In fact, you can propose just such a thing, and if you do it correctly, your plan will pass!

Before explaining why, let me tell you a little more about the pirates. I mentioned that the pirates are perfectly logical, and not only that, they have the common knowledge that they are all perfectly logical. In particular, in their reasoning they can rely on the fact that the other pirates are logical, and that the other pirates know that they are all logical and that they know that, and so on.

Furthermore, it is common knowledge amongst the pirates that they all share the same pirate value system, with the following strictly ordered list of priorities:

Pirate value system:

1. Stay alive.
2. Get gold.
3. Cause the death of other pirates.
4. Arrange that other’s gold goes to the most senior pirates.

That is, at all costs, each pirate would prefer to avoid death, and if alive, to get as much gold as possible, but having achieved that, would prefer that as many other pirates die as possible (but not so much as to give up even one gold coin for additional deaths), and if all other things are equal, would prefer that whatever gold was not gotten for herself, that it goes as much as possible to the most senior pirates, for the pirates are, in their hearts, conservative people.

So, what plan should you propose as Pirate 10? Well, naturally, the pirates will consider Pirate 10’s plan in light of the alternative, which will be the plan proposed by Pirate 9, which will be compared with the plan of Pirate 8 and so on. Thus, it seems we should propagate our analysis from the bottom, working backwards from what happens with a very small number of pirates.

One pirate. If there is only one pirate, the captain, then she mounts the plank, and clearly she should propose “Pirate 1 gets all the gold”, and she should vote in favor of this plan, and so Pirate 1 gets all the gold, as anyone would have expected.

Two pirates. If there are exactly two pirates, then Pirate 2 will mount the plank, and what will she propose? She needs a majority of the two pirates, which means she must get the captain to vote for her plan. But no matter what plan she proposes, even if it is that all the gold should go to the captain, the captain will vote against the plan, since if Pirate 2 is killed, then the captain will get all the gold anyway, and because of pirate value 3, she would prefer that Pirate 2 is killed off.  So Pirate 2’s plan will not be approved by the captain, and so unfortunately, Pirate 2 will walk the plank.

Three pirates. If there are three pirates, then what will Pirate 3 propose? Well, she needs only two votes, and one of them will be her own. So she must convince either Pirate 1 or Pirate 2 to vote for her plan. But actually, Pirate 2 will have a strong incentive to vote for the plan regardless, since otherwise Pirate 2 will be in the situation of the two-pirate case, which ended with Pirate 2’s death. So Pirate 3 can count on Pirate 2’s vote regardless, and so Pirate 3 will propose:  Pirate 3 gets all the gold! This will be approved by both Pirate 2 and Pirate 3, a majority, and so with three pirates, Pirate 3 gets all the gold.

Four pirates. Pirate 4 needs to have three votes, so she needs to get two of the others to vote for her plan. She notices that if she is to die, then Pirates 1 and 2 will get no gold, and so she realizes that if she offers them each one gold coin, they will prefer that, because of the pirate value system. So Pirate 4 will propose to give one gold coin each to Pirates 1 and 2, and 98 gold coins to herself. This plan will pass with the votes of 1, 2 and 4.

Five pirates. Pirate 5 needs three votes, including her own. She can effectively buy the vote of Pirate 3 with one gold coin, since Pirate 3 will otherwise get nothing in the case of four pirates. And she needs one additional vote, that of Pirate 1 or 2, which she can get by offering two gold coins. Because of pirate value 4, she would prefer that the coins go to the highest ranking pirate, so she offers the plan:  two coins to Pirate 1, nothing to pirate 2, one coin to pirate 3, nothing to Pirate 4 and 97 coins to herself.  This plan will pass with the votes of 1, 3 and 5.

Six pirates. Pirate 6 needs four votes, and she can buy the votes of Pirates 2 and 4 with one gold coin each, and then two gold coins to Pirate 3, which is cheaper than the alternatives. So she proposes:  one coin each to 2 and 4, two coins to 3 and 96 coins for herself, and this passes with the votes of 2, 3, 4 and 6.

Seven pirates. Pirate 7 needs four votes, and she can buy the votes of Pirates 1 and 5 with only one coin each, since they get nothing in the six-pirate case. By offering two coins to Pirate 2, she can also get another vote (and she prefers to give the extra gold to Pirate 2 than to other pirates in light of the pirate values).

Eight pirates. Pirate 8 needs five votes, and she can buy the votes of Pirates 3, 4 and 6 with one coin each, and ensure another vote by giving two coins to Pirate 1, keeping the other 95 coins for herself. With her own vote, this plan will pass.

Nine pirates. Pirate 9 needs five votes, and she can buy the votes of Pirates 2, 5 and 7 with one coin each, with two coins to Pirate 3 and her own vote, the plan will pass.

Ten pirates. In light of the division offered by Pirate 9, we can now see that Pirate 10 can ensure six votes by proposing to give one coin each to Pirates 1, 4, 6 and 8, two coins to Pirate 2, and the remaining 94 coins for herself. This plan will pass with those pirates voting in favor (and herself), because they each get more gold this way than they would under the plan of Pirate 9.

We can summarize the various proposals in a table, where the $n^{\rm th}$ row corresponds to the proposal of Pirate $n$.

1 2 3 4 5 6 7 8 9 10
One pirate 100
Two pirates * X
Three pirates 0 0 100
Four pirates 1 1 0 98
Five pirates 2 0 1 0 97
Six pirates 0 1 2 1 0 96
Seven pirates 1 2 0 0 1 0 96
Eight pirates 2 0 1 1 0 1 0 95
Nine pirates 0 1 2 0 1 0 1 0 95
Ten pirates 1 2 0 1 0 1 0 1 0 94

There are a few things to notice, which we can use to deduce how the pattern will continue. Notice that in each row beyond the third row, the number of pirates that get no coins is almost half (the largest integer strictly less than half), exactly one pirate gets two coins, and the remainder get one coin, except for the proposer herself, who gets all the rest. This pattern is sustainable for as long as there is enough gold to implement it, because each pirate can effectively buy the votes of the pirates getting $0$ under the alternative plan with one fewer pirate, and this will be at most one less than half of the previous number; then, she can buy one more vote by giving two coins to one of the pirates who got only one coin in the alternative plan; and with her own vote this will be half plus one, which is a majority. We can furthermore observe that by the pirate value system, the two coins will always go to either Pirate 1, 2 or 3, since one of these will always be the top-ranked pirate having one coin on the previous round. They each cycle with the pattern of 0 coins, one coin, two coins in the various proposals. At least until the gold becomes limited, all the other pirates from Pirate 4 onwards will alternate between zero coins and one coin with each subsequent proposal, and Pirate $n-1$ will always get zero from Pirate $n$.

For this reason, we can see that the pattern continues upward until at least Pirate 199, whose proposal will follow the pattern:

199 Pirates: 1 2 0 0 1 0 1 0 1 0 1 0 1 $\dots$ 1 0 1 0 0

It is with Pirate 199, specifically, that for the first time it takes all one hundred coins to buy the other votes, since she must give ninety-eight pirates one coin each, and two coins to Pirate 2 in order to have one hundred votes altogether, including her own, leaving no coins left over for herself.

For this reason, Pirate 200 will have a successful proposal, since she no longer needs to spend two coins for one vote, as the proposal of Pirate 199 has one hundred pirates getting zero. So Pirate 200 can get 100 votes by proposing one coin to everyone who would get zero from 199, plus her own vote, for a majority of 101 votes.

200 pirates: 0 0 1 1 0 1 0 1 0 1 0 $\dots$ 0 1 0 1 1 0

Pirate 201 also needs 101 votes, which she can get by giving all the zeros of the 200 case one coin each, plus her own vote. The unfortunate Pirate 202, however, needs 102 votes, and this will not be possible, since she has only 100 coins, and so Pirate 202 will die. The interesting thing to notice next is that Pirate 203 will therefore be able to count on the vote of Pirate 202 without paying any gold for it, and so since she needs only 100 additional votes (after her own vote and Pirate 202’s vote), she will be able to buy 100 votes for one coin each. Pirate 204 will again be one coin short, and so she will die. Although Pirate 205 will be able to count on that one additional free vote, this will be insufficient to gain a passing proposal, because she will be able to buy one hundred votes with the coins, plus her own vote and the free vote of Pirate 204, making 102 votes altogether, which is not a majority. Similarly, Pirate 206 will fall short, because even with her vote and the free votes of 204 and 205, she will be able to get at most 103 votes, which is not a majority. Thus, Pirate 207 will be able to count on the votes of Pirates 204, 205, and 206, which with her own vote and 100 more votes gotten by giving one coin each to the pirates who would otherwise get nothing, we can obtain 104 votes, which is a majority.

The reader is encouraged to investigate further to see how the pattern continues. It is a fun problem to work out! What emerges is the phenomenon by which longer and longer sequences of pirates in a row find themselves unable to make a winning proposal, and then suddenly a pirate is able to survive by counting on their votes.

It is very interesting also to work out what happens when there is a very small number of coins. For example, if there is only one gold coin, then already Pirate 4 is unable to make a passing proposal, since she can buy only one other vote, and with her own this will make only two votes, falling short of a majority. With only one coin, Pirate 5 will survive by buying a vote from Pirate 1 and counting on the vote of Pirate 4 and her own vote, for a majority.

Even the case of zero coins is interesting to think about! In this case, there is no gold to distribute, and so the voting is really just about whether the pirate should walk the plank or not. If only one pirate, she will live. Pirate 2 will die, since Pirate 1 will vote against. But for that reason, Pirate 2 will vote in favor of Pirate 3, who will live. The pattern that emerges is:

lives, dies, lives, dies, dies, dies, lives, dies, dies, dies, dies, dies, dies, dies, lives, ….

After each successful proposal, where the pirates lives, for subsequently larger numbers of pirates, there must be many deaths in a row in order for the proposal to count on enough votes. So after each “lives” in the pattern, you have to double the length with many “dies” in a row, before there will be enough votes to support the next pirate who lives.

See also the Pirate Game entry on Wikipedia, which is a slightly different formulation of the puzzle, since tie-votes are effectively counted as success in that variation. For this reason, the outcomes are different in that variation. I prefer the strict-majority variation, since I find it interesting that one must sometimes use two gold coins to gain the majority, and also because the death of Pirate 2 arrives right away in an interesting way, rather than having to wait for 200 or more pirates as with the plurality version.

Another (inessential) difference in presentation is that in the other version of the puzzle, they have the captain on the plank first, and then always the highest-ranking pirate making the proposal, rather than the lowest-ranking pirate. This corresponds simply to inverting the ranking, and so it doesn’t change the results.

The puzzle appears to have been around for some time, but I am unsure of the exact provenance. Ian Stewart wrote a popular 1998 article for Scientific American analyzing the patterns that arise when the number of pirates is large in comparison with the number of gold pieces.

# Open determinacy for games on the ordinals is stronger than ZFC, CUNY Logic Workshop, October 2015

This will be a talk for the CUNY Logic Workshop on October 2, 2015.

Abstract. The principle of open determinacy for class games — two-player games of perfect information with plays of length $\omega$, where the moves are chosen from a possibly proper class, such as games on the ordinals — is not provable in Zermelo-Fraenkel set theory ZFC or Gödel-Bernays set theory GBC, if these theories are consistent, because provably in ZFC there is a definable open proper class game with no definable winning strategy. In fact, the principle of open determinacy and even merely clopen determinacy for class games implies Con(ZFC) and iterated instances Con(Con(ZFC)) and more, because it implies that there is a satisfaction class for first-order truth, and indeed a transfinite tower of truth predicates $\text{Tr}_\alpha$ for iterated truth-about-truth, relative to any class parameter. This is perhaps explained, in light of the Tarskian recursive definition of truth, by the more general fact that the principle of clopen determinacy is exactly equivalent over GBC to the principle of elementary transfinite recursion ETR over well-founded class relations. Meanwhile, the principle of open determinacy for class games is provable in the stronger theory GBC+$\Pi^1_1$-comprehension, a proper fragment of Kelley-Morse set theory KM.

This is joint work with Victoria Gitman, with the helpful participation of Thomas Johnstone.

Related article and posts:

# Transfinite Nim

Shall we have a game of transfinite Nim? One of us sets up finitely many piles of wooden blocks, each pile having some ordinal height, possibly transfinite, and the other of us decides who shall make the first move. Taking turns, we each successively remove a top part of any one pile of our choosing, making it strictly shorter. Whoever takes the very last block wins. (It is fine to remove an entire pile on a turn or to remove blocks from a different pile on a later turn.)

In my challenge problem last week, for example, I set up six piles with heights:
$$1\qquad \omega+3\qquad \omega^\omega+5 \qquad \omega^{\omega+3}+\omega^\omega\cdot3+\omega\cdot 5+7\qquad \epsilon_0\qquad \omega_1$$Would you want to go first or second? What is the best move? In general, we can start with any finite number of piles of arbitrary ordinal heights — what is the general winning strategy?

Before proceeding with the transfinite case, however, let’s review the winning strategy in ordinary finite Nim, which I explained in my post last week concerning my visit to the 7th/8th grade Math Team at my son’s school. To say it quickly again, a finite Nim position is balanced, if when you consider the binary representations of the pile heights, there are an even number of ones in each binary place position. Another way to say this, and this is how I explained it to the school kids, is that if you think of each pile height as a sum of distinct powers of two, then any power of two that arises in any pile does so an even number of times overall for all the piles. The mathematical facts to establish are that (1) any move on a balanced position will unbalance it; and (2) any unbalanced position admits a balancing move. Since the winning move of taking the very last block is a balancing move, it follows that the winning strategy is to balance whatever position with which you are faced. At the start, if the position is unbalanced, then you should go first and balance it; if it is already balanced, then you should go second and adopt the balancing strategy. It may be interesting to note that this winning strategy is unique in the sense that any move that does not balance the position is a losing move, since the opposing player can adopt the balancing strategy from that point on. But of course there is often a choice of balancing moves.

Does this balancing strategy idea continue to apply to transfinite Nim? Yes! All we need to do is to develop a little of the theory of transfinite binary representation. Let me assume that you are all familiar with the usual ordinal arithmetic, for which $\alpha+\beta$ is the ordinal whose order type is isomorphic to a copy of $\alpha$ followed by a copy of $\beta$, and $\alpha\cdot\beta$ is the ordinal whose order type is isomorphic to $\beta$ many copies of $\alpha$. Consider now ordinal exponentiation, which can be defined recursively as follows:
$$\alpha^0=1$$ $$\alpha^{\beta+1}=\alpha^\beta\cdot\alpha$$ $$\alpha^\lambda=\sup_{\beta<\lambda} \alpha^\beta\qquad\lambda\text{ limit}$$ It turns out that $\alpha^\beta$ is the order-type of the finite-support functions from $\beta$ to $\alpha$, under the suitable lexical order. Ordinal exponentiation should not be confused with cardinal exponentiation, since they are very different. For example, with ordinal exponentiation, one has $$2^\omega=\sup_{n<\omega}2^n=\omega,$$which of course is not the case with cardinal exponentiation. In this post, I use only ordinal exponentiation.

Theorem. Every ordinal $\beta$ has a unique representation as a decreasing finite sum of ordinal powers of two. $$\beta=2^{\beta_n}+\cdots+2^{\beta_0}, \qquad \beta_n>\cdots>\beta_0$$

The proof is easy! We simply prove it by transfinite induction on $\beta$. If the theorem holds below an ordinal $\beta$, first let $2^\alpha$ be the largest power of two that is at most $\beta$, so that $\beta=2^\alpha+\gamma$ for some ordinal $\gamma$. It follows that $\gamma<2^\alpha$, for otherwise we could have made $2^{\alpha+1}\leq\beta$. Thus, by induction, $\gamma$ has a representation with powers of two, and so we may simply add $2^\alpha$ at the front to represent $\beta$. To see that the representations are unique, first establish that any power of two is equal to or more than the supremum of the finite decreasing sums of any strictly smaller powers of two. From this, it follows that any representation of $\beta$ as above must have used $2^\alpha$ just as we did for the first term, because otherwise it couldn’t be large enough, and then the representation of the remaining part $\gamma$ is unique by induction, and so we get uniqueness for the representation of $\beta$. QED

Thus, the theorem shows that every ordinal has a unique binary representation in the ordinals, with finitely many nonzero bits. Suppose that we are given a position in transfinite Nim with piles of ordinal heights $\eta_0,\ldots,\eta_n$. We define that such a position is balanced, if every power of two appearing in the representation of any of the piles appears an even number of times overall for all the piles.

The mathematical facts to establish are (1) any move on a balanced position will unbalance it; and (2) every unbalanced position has a balancing move. These facts can be proved in the transfinite case in essentially the same manner as the finite case. Namely, if a position is balanced, then any move affects only one pile, changing the ordinal powers of two that appear in it, and thereby destroy the balanced parity of whichever powers of two are affected. And if a position is unbalanced, then look at the largest unbalanced ordinal power of two appearing, and make a move on any pile having such a power of two in its representation, reducing it so as exactly to balance all the smaller powers of two appearing in the position.

Finally, those two facts again imply that the balancing strategy is a winning strategy, since the winning move of taking the last block or blocks is a balancing move, down to the all-zero position, which is balanced.

In the case of my challenge problem above, we may represent the ordinals in binary. We know how to do that in the case of 1, 3, 5 and 7, and actually those numbers are balanced. Here are some other useful binary representations:

$\omega+3=2^\omega+2+1$

$\omega^\omega+5 = (2^\omega)^\omega+5=2^{\omega^2}+4+1$

$\omega^{\omega+3}=(2^\omega)^{\omega+3}=2^{\omega^2+\omega\cdot 3}$

$\omega^\omega\cdot3=(2^\omega)^\omega\cdot 3=2^{\omega^2}\cdot 2+2^{\omega^2}=2^{\omega^2+1}+2^{\omega^2}$

$\omega\cdot 5+7 =2^{\omega}\cdot 2^2+2^\omega+7=2^{\omega+2}+2^\omega+4+2+1$

$\epsilon_0 = 2^{\epsilon_0}$

$\omega_1=2^{\omega_1}$

I emphasize again that this is ordinal exponentiation. The Nim position of the challenge problem above is easily seen to be unbalanced in several ways. For example, the $\omega_1$ term among others appears only once. Thus, we definitely want to go first in this position. And since $\omega_1$ is the largest unbalanced power of two and it appears only once, we know that we must play on the $\omega_1$ pile. Once one represents all the ordinals in terms of their powers of two representation, one sees that the unique winning move is to reduce the $\omega_1$ pile to have ordinal height
$$\epsilon_0+\omega^{\omega+3}+\omega^\omega\cdot 2+\omega\cdot 4.$$This will exactly balance all the smaller powers of two in the other piles and therefore leaves a balanced position overall. In general, the winning strategy in transfinite Nim, just as for finite Nim, is always to leave a balanced position.

Special honors to Pedro Sánchez Terraf for being the only one to post the winning move in the comments on the other post!

# Win at Nim! The secret mathematical strategy for kids (with challange problems in transfinite Nim for the rest of us)

Welcome to my latest instance of Math for Kids!

Today I had the pleasure to make an interactive mathematical presentation at my son’s school to the 7th / 8th grade Math Team, about 30 math-enthusiastic kids (twelve and thirteen years old) along with their math teachers and the chair of the school math department.

The topic was the game of Nim! This game has a secret mathematical strategy enabling anyone with that secret knowledge to win against those without it. It is a great game for kids, because with the strategy they can realistically expect to beat their parents, friends, siblings and parent’s friends almost every single time!

To play Nim, one player sets up a number of piles of blocks, and the opponent chooses whether to go first or second. The players take turns removing blocks — each player may remove any number of blocks (at least one) from any one pile, and it is fine to take a whole pile — whichever player takes the last block wins.

For the math team, we played a few demonstration games, in which I was able to beat all the brave challengers, and then the kids paired off to play each other and gain familiarity with the game. Then, it was time for the first strategy discussion.

What could the secret winning strategy be? I explained to the kids a trick that mathematicians often use when approaching a difficult problem, namely, to consider in detail some very simple special cases or boundary instances of the problem. It often happens that these special cases reveal a way of thinking about the problem that applies much more generally.

Perhaps one of the easiest special cases of Nim occurs when there is only one pile. If there is only one pile, then clearly one wants to go first, in order to make the winning move: take the entire pile!

Two balanced piles

A slightly less trivial and probably more informative case arises when there are exactly two piles. If the stacks have the same height, then the kids realized that the second player could make copying moves so as to preserve this balanced situation. The key insight now is that this copying strategy is a winning strategy, because if one can always copy, then in particular one will have a move whenever the opponent did, and so the opponent will never take the last block. With two piles, therefore, one wants always to make them balanced. If they are initially unbalanced, then choose to go first and follow the balancing strategy. If they are initially balanced, then choose to go second, and copy whatever moves your opponent makes to rebalance them.

A balanced position

With that insight, it is not difficult to see that it is winning to leave a position with any number of pairs of balanced piles. One can in effect play on each pair separately, because whenever the opponent makes a move on one of the piles, one can copy the move with the corresponding partner pile. In this way, we may count such a position overall as balanced. The more fundamental game-theoretic observation to make is that balanced piles in effect cancel each other out in any position, and one can ignore them when analyzing a position. When two balanced piles are present in a possibly more complicated position, one can pretend that they aren’t there, precisely because whenever your opponent plays on one of them, you can copy the move on the other, and so any winning strategy for the position in which those piles are absent can be converted into a winning strategy in which the balanced piles are present.

This idea now provides a complete winning strategy in the case that all piles have height one or two at most. One wants to leave a position with an even number of piles of each height. If only one height has an odd number of piles, then take a whole pile of that height. And if there are odd numbers of piles both of height one and two, then turn a height-two pile into a pile of height one, and this will make them both even. So any unbalanced position can be balanced, and any move on a balanced position will unbalance it.

1+2+3 counts as balanced

Let’s now consider that there may be piles of height three. For example, consider the basic position with piles of height one, two and three. The observation to make here is that any move on this position can be replied to with a move that leaves it balanced (check it yourself to be sure!). It follows that this position is winning to leave for the other player (and so one should go second on $1+2+3$). It would be nice if we could consider this position itself as already balanced in some sense. Indeed, we may incorporate this situation into the balancing idea if we think of the pile of height three as really consisting of two subpiles, one of height two and one of height one. In this way, the Nim position 1+2+3 counts as balanced, since the 3 counts as 2+1, which balances the other stacks.  The 1+2+3 position has two stacks of height two and two of height one, when one regards the stack of height three as having a substack of height two and a substack of height one.

This way of thinking produces a complete winning strategy for Nim positions involving piles of height at most three. (And this is a strategy that can be mastered even by very young children — a few years ago I had talked about Nim with much younger children, Math for six-year-olds: Win at Nim!, first-graders at my daughter’s school, and at that time we concentrated on posititions with piles of height at most three. Older kids, however, can handle the full strategy.) Namely, the winning strategy in this case is to strive to balance the position, to make an even number overall of piles of height one and two, where we count piles of height three as one each of one and two. If you always give your opponent a balanced position, then  you will win!  Faced with an unbalanced position, it is a fact that you can always find a balancing move, and any move on an balanced position will unbalance it.  If the game is just starting, and you are deciding whether to go first or second, you should determine whether it is balanced yet or not.  If it unbalanced, then you should go first and make the balancing move; if it is already balanced, then you should go second and adopt the copying strategy, in which you re-balance the position with each move.

The general winning strategy, of course, goes beyond three. The key idea is to realize that what is really going on when we represent $3$ as $2+1$ is that we are using the binary representation of the number $3$. To explain, I wrote the following numbers on the chalkboard $$1,\ 2,\ 4,\ 8,\ 16,\ 32,\ 64,\ \cdots$$ and was very pleased when the kids immediately shouted out, “The powers of two!” I explained that any natural number can be expressed uniquely as a sum of distinct powers of two. Asked for a favorite number less than one hundred, one student suggested $88$, and together we calculated $$88=64+16+8,$$ which means that the binary representation of $88$ is $1011000$, which I read off as, “one $64$, no $32$s, one $16$, one $8$, no $4$s, no $2$s and no $1$s. This is just the same as thinking of $9572$ as 9 thousands, 5 hundreds, 7 tens and 2 ones, using the powers of ten. It is interesting to learn that one may easily count very high on one hand using binary, up to 1023 on two hands!

The general strategy is to view every Nim pile as consisting of subpiles whose height is a power of two, and to make sure that one leaves a position that is balanced in the sense that every power of two has an even number of such instances in the position. So we think of $3$ as really $2+1$ for the purposes of balancing; $4$ counts as itself because it is a power of two, but $5$ counts as $4+1$ and $6$ counts as $4+2$ and $7$ as $4+2+1$. Another way to describe the strategy is that we express all the pile heights in binary, and we want an even number of $1$s in each binary place position.

The mathematical facts to verify are (1) any move on a balanced position in this powers-of-two sense will cause it to become unbalanced, and (2) any unbalanced position can be balanced in one move. It follows that leaving balanced positions is a winning strategy, because the winning move of taking the last block is a balancing move rather than an unbalancing move.

One can prove statement (1) by realizing that when you move a single stack, the binary representation changes, and so whichever binary digits changed will now become unbalanced.  For statement (2), consider the largest unbalanced power of two $2^k$ and move on any stack that contains a $2^k$ size substack. Since $2^k-1=111\cdots11$ in binary, one can attain any binary pattern for the smaller height stacks by removing between $1$ and $2^k$ many blocks. So one can balance the position.

As a practical matter, the proof of (2) also shows how one can find a (winning) balancing move, which can otherwise be difficult in some cases: look for the largest unbalanced power of two, and move on any pile containing such a subpile, making sure to leave a balanced position.

In most actual instances of Nim, the pile heights are rarely very tall, and so one is usually considering just $1$, $2$ and $4$ as the powers of two that arise.  A traditional starting configuration has piles of height 1, 3, 5, and 7, and this position is balanced, because one may view it as: $1, 2+1, 4+1, 4+2+1$, and there are an even number of 1s, 2s and 4s.

It is interesting to consider also the Misère form of Nim, where one wants NOT to take the last block. This version of the game also has a secret mathematical strategy, which I shall reveal later on.

Challenge 1.   What is the winning strategy in Misère Nim?

If you figure it out, please post a comment! I’ll post the solution later. One might naively expect that the winning strategy of Misère Nim is somehow totally opposite to the winning strategy of regular Nim, but in fact, the positions $1,2,3$ and $1,3,5,7$ are winning for the second player both in Nim and also in Misère Nim. Indeed, I claim that all nontrivial Nim positions that are winning for regular Nim (with a suitable meaning of “nontrivial”) are also winning for Misère Nim. Can you prove it?

Another interesting generalization, for the set-theorists, is to consider transfinite Nim, where the piles can have transfinite ordinal height. So we have finitely many piles of ordinal height, perhaps infinite, and a move consists of making any one pile strictly shorter. Since there are no infinite descending sequence of ordinals, the game will terminate in finitely many moves, and the winner is whowever removes the last block.

Challenge 2.  Who wins the transfinite Nim game with piles of heights: $$1\qquad \omega+3\qquad \omega^\omega+5\qquad \omega^{\omega+3}+\omega^\omega\cdot3+\omega\cdot 5+7\qquad \epsilon_0\qquad \omega_1$$ and what are the winning moves? What is the general winning strategy for transfinite Nim?

Post your solutions! You can also see my solution and further discussion.

# An introduction to the theory of infinite games, with examples from infinite chess, University of Connecticut, December 2014

This will be a talk for the interdisciplinary Group in Philosophical and Mathematical Logic at the University of Connecticut in Storrs, on December 5, 2014.

Abstract. I shall give a general introduction to the theory of infinite games, with a focus on the theory of transfinite ordinal game values. These ordinal game values can be used to show that every open game — a game that, when won for a particular player, is won after finitely many moves — has a winning strategy for one of the players. By means of various example games, I hope to convey the extremely concrete game-theoretic meaning of these game values for various particular small infinite ordinals. Some of the examples will be drawn from infinite chess, which is chess played on a chessboard stretching infinitely without boundary in every direction, and the talk will include animations of infinite chess positions having large numbers of pieces (or infinitely many) with hundreds of pieces making coordinated attacks on the chessboard. Meanwhile, the exact value of the omega one of chess, denoted $\omega_1^{\mathfrak{Ch}_{\!\!\!\!\sim}}$, is not currently known.