The Dutch Association for Logic and Philosophy of the Exact Sciences (VvL) has organized a major annual public online lecture series called LOGIC AT LARGE, where “well-known logicians give public audience talks to a wide audience,” and I am truly honored to have been invited to give this year’s lecture. This will be an online event, the second of the series, scheduled for May 31, 2022 (note change in date!), and further access details will be posted when they become available. Free registration can be made on the VvL Logic at Large web page.

Abstract. Many familiar finite games admit natural infinitary analogues, which often highlight intriguing issues in infinite game theory. Shall we have a game of infinite chess? Or how about infinite draughts, infinite Hex, infinite Go, infinite Wordle, or infinite Sudoku? Let me introduce these games and use them to illustrate various fascinating concepts in the theory of infinite games.

Come enjoy the lecture, and stay for the online socializing event afterwards. Hope to see you there!

Abstract. I consider the natural infinitary variations of the games Wordle and Mastermind, as well as their game-theoretic variations Absurdle and Madstermind, considering these games with infinitely long words and infinite color sequences and allowing transfinite game play. For each game, a secret codeword is hidden, which the codebreaker attempts to discover by making a series of guesses and receiving feedback as to their accuracy. In Wordle with words of any size from a finite alphabet of $n$ letters, including infinite words or even uncountable words, the codebreaker can nevertheless always win in $n$ steps. Meanwhile, the mastermind number đť•ž, defined as the smallest winning set of guesses in infinite Mastermind for sequences of length $\omega$ over a countable set of colors without duplication, is uncountable, but the exact value turns out to be independent of ZFC, for it is provably equal to the eventually different number $\frak{d}({\neq^*})$, which is the same as the covering number of the meager ideal $\text{cov}(\mathcal{M})$. I thus place all the various mastermind numbers, defined for the natural variations of the game, into the hierarchy of cardinal characteristics of the continuum.

This will be an in-person talk for the CUNY Logic Workshop at the Graduate Center of the City University of New York on 11 March 2022.

Abstract. I shall introduce and consider the natural infinitary variations of Wordle, Absurdle, and Mastermind. Infinite Wordle extends the familiar finite game to infinite words and transfinite playâ€”the code-breaker aims to discover a hidden codeword selected from a dictionary $\Delta\subseteq\Sigma^\omega$ of infinite words over a countable alphabet $\Sigma$ by making a sequence of successive guesswords, receiving feedback after each guess concerning its accuracy. For any dictionary using the usual 26-letter alphabet, for example, the code-breaker can win in at most 26 guesses, and more generally in $n$ guesses for alphabets of finite size $n$. Meanwhile, for some dictionaries on an infinite alphabet, infinite play is required, but the code-breaker can always win by stage $\omega$ on a countable alphabet, for any fixed dictionary. Infinite Mastermind, in contrast, is a subtler game than Wordle because only the number and not the position of correct bits is given. When duplication of colors is allowed, nevertheless, the code-breaker can still always win by stage $\omega$, but in the no-duplication variation, no countable number of guesses (even transfinite) is sufficient for the code-breaker to win. I therefore introduce the mastermind number, denoted $\frak{mm}$, defined to be the size of the smallest winning no-duplication Mastermind guessing set, a new cardinal characteristic of the continuum, which I prove is bounded below by the additivity number $\text{add}(\mathcal{M})$ of the meager ideal and bounded above by the covering number $\text{cov}(\mathcal{M})$. In particular, the precise value of the mastermind number is independent of ZFC and can consistently be strictly between $\aleph_1$ and the continuum $2^{\aleph_0}$. In simplified Mastermind, where the feedback given at each stage includes only the numbers of correct and incorrect bits (omitting information about rearrangements), then the corresponding simplified mastermind number is exactly the eventually different number $\frak{d}(\neq^*)$.

I am preparing an article on the topic, which will be available soon.

Abstract. We introduce the game of infinite Hex, extending the familiar finite game to natural play on the infinite hexagonal lattice. Whereas the finite game is a win for the first player, we prove in contrast that infinite Hex is a drawâ€”both players have drawing strategies. Meanwhile, the transfinite game-value phenomenon, now abundantly exhibited in infinite chess and infinite draughts, regrettably does not arise in infinite Hex; only finite game values occur. Indeed, every game-valued position in infinite Hex is intrinsically local, meaning that winning play depends only on a fixed finite region of the board. This latter fact is proved under very general hypotheses, establishing the conclusion for all simple stone-placing games.

Abstract. I shall give an introduction to the logic of infinite games, including the theory of transfinite game values, using the case of infinite draughts as a principal illustrative instance. Infinite draughts, also known as infinite checkers, is played like the finite game, but on an infinite checkerboard stretching without end in all four directions. In recent joint work with Davide Leonessi, we proved that every countable ordinal arises as the game value of a position in infinite draughts. Thus, there are positions from which Red has a winning strategy enabling her to win always in finitely many moves, but the length of play can be completely controlled by Black in a manner as though counting down from a given countable ordinal. This result is optimal for games having countably many options at each moveâ€”in short, the omega one of infinite draughts is true omega one.

A joint paper with Davide Leonessi, in which we prove that every countable ordinal arises as the game value of a position in infinite draughts, and this result is optimal for games having countably many options at each move. In short, the omega one of infinite draughts is true omega one.

J. D. Hamkins and D. Leonessi, “Transfinite game values in infinite draughts,” Mathematics arXiv, 2021. [Bibtex]

@ARTICLE{HamkinsLeonessi:Transfinite-game-values-in-infinite-draughts,
author = {Joel David Hamkins and Davide Leonessi},
title = {Transfinite game values in infinite draughts},
journal = {Mathematics arXiv},
year = {2021},
volume = {},
number = {},
pages = {},
month = {},
note = {Under review},
abstract = {},
keywords = {under-review},
source = {},
doi = {},
eprint = {2111.02053},
archivePrefix = {arXiv},
primaryClass = {math.LO},
url = {http://jdh.hamkins.org/transfinite-game-values-in-infinite-draughts},
}

Abstract. Infinite draughts, or checkers, is played just like the finite game, but on an infinite checkerboard extending without bound in all four directions. We prove that every countable ordinal arises as the game value of a position in infinite draughts. Thus, there are positions from which Red has a winning strategy enabling her to win always in finitely many moves, but the length of play can be completely controlled by Black in a manner as though counting down from a given countable ordinal.

Mr. Davide Leonessi successfully defended his dissertation for the Masters of Science degree in Mathematics and Foundations of Computer Science, entitled “Transfinite game values in infinite games,” on 15 September 2021. Davide earned a distinction for his thesis, an outstanding result.

Abstract. The object of this study are countably infinite games with perfect information that allow players to choose among arbitrarily many moves in a turn; in particular, we focus on the generalisations of the finite board games of Hex and Draughts.

In Chapter 1 we develop the theory of transfinite ordinal game values for open infinite games following [Evans-Hamkins 2014], and we focus on the properties of the omega one, that is the supremum of the possible game values, of classes of open games; we moreover design the class of climbing-through-$T$ games as a tool to study the omega one of given game classes.

The original contributions of this research are presented in the following two chapters.

In Chapter 2 we prove classical results about finite Hex and present Infinite Hex, a well-defined infinite generalisation of Hex.

We then introduce the class of stone-placing games, which captures the key features of Infinite Hex and further generalises the class of positional games already studied in the literature within the finite setting of Combinatorial Game Theory.

The main result of this research is the characterization of open stone-placing games in terms of the property of essential locality, which leads to the conclusion that the omega one of any class of open stone-placing games is at most $\omega$. In particular, we obtain that the class of open games of Infinite Hex has the smallest infinite omega one, that is $\omega_1^{\rm Hex}=\omega$.

In Chapter 3 we show a dual result; we define the class of games of Infinite Draughts and explicitly construct open games of arbitrarily high game value with the tools of Chapter 1, concluding that the omega one of the class of open games of Infinite Draughts is as high as possible, that is $\omega_1^{\rm Draughts}=\omega_1$.

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.

I was interviewed 26 August 2021 by mathematician Daniel Rubin on his show, and we had a lively, wideranging discussion spanning mathematics, infinity, and the philosophy of mathematics. Please enjoy!

Abstract. The principle of open determinacy for class games â€” two-player games of perfect information with plays of length Ď‰, where the moves are chosen from a possibly proper class, such as games on the ordinals â€” is not provable in Zermelo-Fraenkel set theory ZFC or GĂ¶del-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.

Welcome to Cantor’s Ice Cream Shoppe! A huge choice of flavorsâ€”pile your cone high with as many scoops as you want!

Have two scoops, or three, four, or more! Why not infinitely many? Would you like $\omega$ many scoops, or $\omega\cdot2+5$ many scoops? You can have any countable ordinal number of scoops on your cone.

And furthermore, after ordering your scoops, you can order more scoops to be placed on topâ€”all I ask is that you let me know how many such extra orders you plan to make. Let’s simply proceed transfinitely. You can announce any countable ordinal $\eta$, which will be the number of successive orders you will make; each order is a countable ordinal number of ice cream scoops to be placed on top of whatever cone is being assembled.

In fact, I’ll even let you change your mind about $\eta$ as we proceed, so as to give you more orders to make a taller cone.

So the process is:

You pick a countable ordinal $\eta$, which is the number of orders you will make.

For each order, you can pick any countable ordinal number of scoops to be added to the top of your ice-cream cone.

After making your order, you can freely increase $\eta$ to any larger countable ordinal, giving you the chance to make as many additional orders as you like.

At each limit stage of the ordering process, the ice cream cone you are assembling has all the scoops you’ve ordered so far, and we set the current $\eta$ value to the supremum of the values you had chosen so far.

If at any stage, you’ve used up your $\eta$ many orders, then the process has completed, and I serve you your ice cream cone. Enjoy!

Question. Can you arrange to achieve uncountably many scoops on your cone?

Although at each stage we place only countably many ice cream scoops onto the cone, nevertheless we can keep giving ourselves extra stages, as many as we want, simply by increasing $\eta$. Can you describe a systematic process of increasing the number of steps that will enable you to make uncountably many orders? This would achieve an unountable ice cream cone.

What is your solution? Give it some thought before proceeding. My solution appears below.

Alas, I claim that at Cantor’s Ice Cream Shoppe you cannot make an ice cream cone with uncountably many scoops. Specifically, I claim that there will inevitably come a countable ordinal stage at which you have used up all your orders.

Suppose that you begin by ordering $\beta_0$ many scoops, and setting a large value $\eta_0$ for the number of orders you will make. You subsequently order $\beta_1$ many additional scoops, and then $\beta_2$ many on top of that, and so on. At each stage, you may also have increased the value of $\eta_0$ to $\eta_1$ and then $\eta_2$ and so on. Probably all of these are enormous countable ordinals, making a huge ice cream cone.

At each stage $\alpha$, provided $\alpha<\eta_\alpha$, then you can make an order of $\beta_\alpha$ many scoops on top of your cone, and increase $\eta_\alpha$ to $\eta_{\alpha+1}$, if desired, or keep it the same.

At a limit stage $\lambda$, your cone has $\sum_{\alpha<\lambda}\beta_\alpha$ many scoops, and we update the $\eta$ value to the supremum of your earlier declarations $\eta_\lambda=\sup_{\alpha<\lambda}\eta_\alpha$.

What I claim now is that there will inevitably come a countable stage $\lambda$ for which $\lambda=\eta_\lambda$, meaning that you have used up all your orders with no possibility to further increase $\eta$. To see this, consider the sequence $$\eta_0\leq \eta_{\eta_0}\leq \eta_{\eta_{\eta_0}}\leq\cdots$$ We can define the sequence recursively by $\lambda_0=\eta_0$ and $\lambda_{n+1}=\eta_{\lambda_n}$. Let $\lambda=\sup_{n<\omega}\lambda_n$, the limit of this sequence. This is a countable supremum of countable ordinals and hence countable. But notice that $$\eta_\lambda=\sup_{n<\omega}\eta_{\lambda_n}=\sup_{n<\omega}\lambda_{n+1}=\lambda.$$ That is, $\eta_\lambda=\lambda$ itself, and so your orders have run out at $\lambda$, with no possibility to add more scoops or to increase $\eta$. So your order process completed at a countable stage, and you have therefore altogether only a countable ordinal number of scoops of ice cream. I’m truly very sorry at your pitiable impoverishment.

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.

I saw the following image on Twitter and Reddit, an image suggesting an entire class of infinitary analogues of the game Connect-Four. What fun! Let’s figure it out!

I’m not sure to whom the image or the idea is due. Please comment if you have information. (See comments below for current information.)

The rules will naturally generalize those in Connect-Four. Namely, starting from an empty board, the players take turns placing their coins into the $\omega\times 4$ grid. When a coin is placed in a column, it falls down to occupy the lowest available cell. Let us assume for now that the game proceeds for $\omega$ many moves, whether or not the board becomes full, and the goal is to make a connected sequence in a row of $\omega$ many coins of your color (you don’t have to fill the whole row, but rather a connected infinite segment of it suffices). A draw occurs when neither or both players achieve their goals.

In the $\omega\times 6$ version of the game that is shown, and indeed in the $\omega\times n$ version for any finite $n$, I claim that neither player can force a win; both players have drawing strategies.

Theorem. In the game Connect-$\omega$ on a board of size $\omega\times n$, where $n$ is finite, neither player has a winning strategy; both players have drawing strategies.

Proof. For a concrete way to see this, observe that either player can ensure that there are infinitely many of their coins on the bottom row: they simply place a coin into some far-out empty column. This blocks a win for the opponent on the bottom row. Next, observe that neither player can afford to follow the strategy of always answering those moves on top, since this would lead to a draw, with a mostly empty board. Thus, it must happen that infinitely often we are able to place a coin onto the second row. This blocks a win for the opponent on the second row. And so on. In this way, either players can achieve infinitely many of their coins on each row, thereby blocking any row as a win for their opponent. So both players have drawing strategies. $\Box$

Let me point out that on a board of size $\omega\times n$, where $n$ is odd, we can also make this conclusion by a strategy-stealing argument. Specifically, I argue first that the first player can have no winning strategy. Suppose $\sigma$ is a winning strategy for the first player on the $\omega\times n$ board, with $n$ odd, and let us describe a strategy for the second player. After the first move, the second player mentally ignores a finite left-initial segment of the playing board, which includes that first move and with a odd number of cells altogether in it (and hence an even number of empty cells remaining); the second player will now aim to win on the now-empty right-side of the board, by playing as though playing first in a new game, using strategy $\sigma$. If the first player should ever happen to play on the ignored left side of the board, then the second player can answer somewhere there (it doesn’t matter where). In this way, the second player plays with $\sigma$ as though he is the first player, and so $\sigma$ cannot be winning for the first player, since in this way the second player would win in this stolen manner.

Similarly, let us argue by strategy-stealing that the second player cannot have a winning strategy on the board $\omega\times n$ for odd finite $n$. Suppose that $\tau$ is a winning strategy for the second player on such a board. Let the first player always play at first in the left-most column. Because $n$ is odd, the second player will eventually have to play first in the second or later columns, leaving an even number of empty cells in the first column (perhaps zero). At this point, the first player can play as though he was the second player on the right-side board containing only that fresh move. If the opponent plays again to the left, then our player can also play in that region (since there were an even number of empty cells). Thus, the first player can steal the strategy $\tau$, and so it cannot be winning for the second player.

I am unsure about how to implement the strategy stealing arguments when $n$ is even. I shall give this more thought. In any case, the theorem for this case was already proved directly by the initial concrete argument, and in this sense we do not actually need the strategy stealing argument for this case.

Meanwhile, it is natural also to consider the $n\times\omega$ version of the game, which has only finitely many columns, each infinite. The players aim to get a sequence of $\omega$-many coins in a column. This is clearly impossible, as the opponent can prevent a win by always playing atop the most recent move. Thus:

Theorem. In the game Connect-$\omega$ on a board of size $n\times\omega$, where $n$ is finite, neither player has a winning strategy; both players have drawing strategies.

Perhaps the most natural variation of the game, however, occurs with a board of size $\omega\times\omega$. In this version, like the original Connect-Four, a player can win by either making a connected row of $\omega$ many coins, or a connected column or a connected diagonal of $\omega$ many coins. Note that we orient the $\omega$ size column upwards, so that there is no top cell, but rather, one plays by selecting a not-yet-filled column and then occupying the lowest available cell in that column.

Theorem. In the game Connect-$\omega$ on a board of size $\omega\times\omega$, neither player has a winning strategy. Both players have drawing strategies.

Proof. Consider the strategy-stealing arguments. If the first player has a winning strategy $\sigma$, then let us describe a strategy for the second player. After the first move, the second player will ignore finitely many columns at the left, including that first actual move, aiming to play on the empty right-side of the board as though the first player using stolen strategy $\sigma$ (but with colors swapped). This will work fine, as long as the first player also plays on that part of the board. Whenever the first player plays on the ignored left-most part, simply respond by playing atop. This prevents a win in that part of the part, and so the second player will win on the right-side by pretending to be first there. So there can be no such winning strategy $\sigma$ for the first player.

If the second player has a winning strategy $\tau$, then as before let the first player always play in the first column, until $\tau$ directs the second player to play in another column, which must eventually happen if $\tau$ is winning. At that moment, the first player can pretend to be second on the subboard omitting the first column. So $\tau$ cannot have been winning after all for the second player. $\Box$

In the analysis above, I was considering the game that proceeded in time $\omega$, with $\omega$ many moves. But such a play of the game may not actually have filled the board completely. So it is natural to consider a version of the game where the players continue to play transfinitely, if the board is not yet full.

So let us consider now the transfinite-play version of the game, where play proceeds transfinitely through the ordinals, until either the board is filled or one of the players has achieved the winning goal. Let us assume that the first player also plays first at limit stages, at $\omega$ and $\omega\cdot 2$ and $\omega^2$, and so on, if game play should happen to proceed for that long.

The concrete arguments that I gave above continue to work for the transfinite-play game on the boards of size $\omega\times n$ and $n\times\omega$.

Theorem. In the transfinite-play version of Connect-$\omega$ on boards of size $\omega\times n$ or $n\times\omega$, where $n$ is finite, neither player can have a winning strategy. Indeed, both players can force a draw while also filling the board in $\omega$ moves.

Proof. It is clear that on the $n\times\omega$ board, either player can force each column to have infinitely many coins of their color, and this fills the board, while also preventing a win for the opponent, as desired.

On the $\omega\times n$ board, consider a variation of the strategy I described above. I shall simply always play in the first available empty column, thereby placing my coin on the bottom row, until the opponent also plays in a fresh column. At that moment, I shall play atop his coin, thereby placing another coin in the second row; immediately after this, I also play subsequently in the left-most available column (so as to force the board to be filled). I then continue playing in the bottom row, until the opponent also does, which she must, and then I can add another coin to the second row and so on. By always playing the first-available second-row slot with all-empty second rows to the right, I can ensure that the opponent will eventually also make a second-row play (since otherwise I will have a winning condition on the second row), and at such a moment, I can also make a third-row play. By continuing in this way, I am able to place infinitely many coins on each row, while also forcing that the board becomes filled. $\Box$

Unfortunately, the transfinite-play game seems to break the strategy-stealing arguments, since the play is not symmetric for the players, as the first player plays first at limit stages.

Nevertheless, following some ideas of Timothy Gowers in the comments below, let me show that the second player has a drawing strategy.

Theorem. In the transfinite-play version of Connect-$\omega$ on a board of size $\omega\times\omega$, the second player has a drawing strategy.

Proof. We shall arrange that the second player will block all possible winning configurations for the first player, or to have column wins for each player. To block all row wins, the second player will arrange to occupy infinitely many cells in each row; to block all diagonal wins, the second player will aim to occupy infinitely many cells on each possible diagonal; and to block the column wins, the second player will aim either to have infinitely many cells on each column or to copy a winning column of the opponent on another column.

To achieve these things, we simply play as follows. Take the columns in successive groups of three. On the first column in each block of three, that is on the columns indexed $3m$, the second player will always answer a move by the first player on this column. In this way, the second player occupies every other cell on these columns—all at the same height. This will block all diagonal wins, because every diagonal winning configuration will need to go through such a cell.

On the remaining two columns in each group of three, columns $3m+1$ and $3m+2$, let the second player simply copy moves of the opponent on one of these columns by playing on the other. These moves will therefore be opposite colors, but at the same height. In this way, the second player ensures that he has infinitely many coins on each row, blocking the row wins. And also, this strategy ensures that in these two columns, at any limit stage, either neither player has achieved a winning configuration or both have.

Thus, we have produced a drawing strategy for the second player. $\Box$

Thus, there is no advantage to going first. What remains is to determine if the first player also has a drawing strategy, or whether the second player can actually force a win.

Gowers explains in the comments below also how to achieve such a copying mechanism to work on a diagonal, instead of just on a column.

I find it also fascinating to consider the natural generalizations of the game to larger ordinals. We may consider the game of Connect-$\alpha$ on a board of size $\kappa\times\lambda$, for any ordinals $\alpha,\kappa,\lambda$, with transfinite play, proceeding until the board is filled or the winning conditions are achieved. Clearly, there are some instances of this game where a player has a winning strategy, such as the original Connect-Four configuration, where the first player wins, and presumably many other instances.

Question. In which instances of Connect-$\alpha$ on a board of size $\kappa\times\lambda$ does one of the players have a winning strategy?

It seems to me that the groups-of-three-columns strategy described above generalizes to show that the second player has at least a drawing strategy in Connect-$\alpha$ on board $\kappa\times\lambda$, whenever $\alpha$ is infinite.

Consider what I call the Sudoku game, recently introduced in the MathOverflow question Who wins two-player Sudoku?Â 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.

InÂ 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.

On even-sized Sudoku boards, the second player wins the Sudoku game by the mirror copying strategy.

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 SudokuÂ 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 theÂ 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.

This will be a talk for the CUNY Set Theory Seminar, April 27, 2018, GC Room 6417, 10-11:45am (please note corrected date).

Abstract.Â Open class determinacy is the principle of second order set theory asserting of every two-player game of perfect information, with plays coming from a (possibly proper) class $X$ and the winning condition determined by an open subclass of $X^\omega$, that one of the players has a winning strategy. This principle finds itself about midway up the hierarchy of second-order set theories between GĂ¶del-Bernays set theory and Kelley-Morse, a bit stronger than the principle of elementary transfinite recursion ETR, which is equivalent to clopen determinacy, but weaker than GBC+$\Pi^1_1$-comprehension. In this talk, I’ll given an account of my recent joint work with W. Hugh Woodin, proving that open class determinacy is preserved by forcing. A central part of the proof is to show that in any forcing extension of a model of open class determinacy, every well-founded class relation in the extension is ranked by a ground model well-order relation. This work therefore fits into the emerging focus in set theory on the interaction of fundamental principles of second-order set theory with fundamental set theoretic tools, such as forcing. It remains open whether clopen determinacy or equivalently ETR is preserved by set forcing, even in the case of the forcing merely to add a Cohen real.