There are some chess positions in which one player can force a win. Here’s an extremely simple example:

White just moves their queen up to a8 and checkmates Black. Some winning positions are harder to see than this. Take a look at the following position. Can you find a guaranteed two-move checkmate for White?

And for fun, here’s a harder one, again a guaranteed two-move checkmate, this time by Black:

Notice that in this last one, the opponent had multiple possible moves to choose from. A forced mate does not necessarily mean restricting your opponent to exactly one move on each of their turns. It just means that no matter *what* they do, you can still guarantee a win. **(Edit: with best play from White, this is not actually a forced mate.)** Forced wins can become arbitrarily complicated and difficult to see if you’re looking many moves down the line, as you have to consider all the possible responses your opponent has at each turn. The world record for the longest forced win is the following position:

It’s White’s move, and White does have a strategy for a forced win. It just takes 549 turns to actually do this! (This strategy does violate the 50-move rule, which says that after 50 turns with no pawn moves or capture the game is drawn.) At this link you can watch the entire 549 move game. Most of it is totally incomprehensible to human players, and apparently top chess players that look at this game have reported that the reasoning behind the first 400 moves is opaque to them. Interestingly, White gets a pawn promotion after six moves, and it promotes it to a knight instead of a queen! It turns out that promoting to a queen actually loses for White, and their only way to victory is the knight promotion!

This position is the longest forced win with 7 pieces on the board. There are a few others that are similarly long. All of them represent a glimpse at the perfect play we might expect to see if a hypercomputer could calculate the whole game tree for chess and select the optimal move.

A grandmaster wouldn’t be better at these endgames than someone who had learned chess yesterday. It’s a sort of chess that has nothing to do with chess, a chess that we could never have imagined without computers. The Stiller moves are awesome, almost scary, because you know they are the truth, God’s Algorithm – it’s like being revealed the Meaning of Life, but you don’t understand one word.

Tim Krabbe

With six pieces on the board, the longest mate takes 262 moves (you can play out this position here). For five pieces, it’s 127 moves, for four it’s 43 moves, and the longest 3-man mate takes 28 moves.

But now a natural question arises. We know that a win can be forced in some positions. But how about the opening position? That is, is there a guaranteed win for White (or for Black) starting in this position?

Said more prosaically: Can chess be solved?

Zermelo’s theorem, published in “On an Application of Set Theory to the Theory of the Game of Chess” (1913), was the first formal theorem in game theory. It predated the work of von Neumann (the so-called “father of game theory”) by 15 years. It proves that yes, **it is in fact possible to solve chess**. We don’t know what the solution is, but we know that either White can force a win, or Black can force a win, or the result will be a draw if both play perfectly.

Of course, the guarantee that in principle there is a solution to chess doesn’t tell us much in practice. The exponential blowup in the number of possible games is so enormous that humans will never find this solution. Nonetheless, I still find it fascinating to think that the mystery of chess is ultimately a product of computational limitations, and that in principle, if we had a hypercomputer, we could just find the unique best chess game and watch it play out, either to a win by one side or to a draw. That would be a game that I would *love* to see.

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Here’s another fun thing. There’s an extremely bizarre variant on chess called suicide chess (or anti-chess). The goal of suicide chess is to lose all your pieces. Of course, if all the rules of play were the same, it would be practically impossible to win (since your opponent could always just keep refusing to take a piece that you are offering them). To remedy this, in suicide chess, capturing is mandatory! And if you have multiple possible captures, then you can choose among them.

Suicide chess gameplay is extremely complicated and unusual looking, and evaluating who is winning at any given moment tends to be really difficult, as sudden turnarounds are commonplace compared to ordinary chess. But one simplifying factor is that it tends to be easier to restrict your opponents’ moves. In ordinary chess, you can only restrict your opponents’ moves by blocking off their pieces or threatening their king. But in suicide chess, your opponents’ moves are restricted ANY time you put one of your pieces in their line of fire! This feature of the gameplay makes the exponential blow up in possible games more manageable.

Given this, it probably won’t be much of a surprise that suicide chess is, just like ordinary chess, in principle solvable. But here’s the crazy part. Suicide chess is solved!!

That’s right: it was proven a triple of years ago that White can force a win by moving first with e3!

Here’s the paper. The proof amounts to basically running a program that looks at all possible responses to e3 and expands out the game tree, ultimately showing that all branches can be terminated with White losing all pieces and winning the game.

Not only do we know that by starting with e3, White is guaranteed a win, we also know that Black can force a win if White starts with any of the following moves: a3, b4, c3, d3, d4, e4, f3, f4, h3, h4, Nc3, Nf3. As far as I was able to tell, there are only six opening moves remaining for which we don’t know if White wins, Black wins, or they draw: a4, b3, c4, e3, g3, and g4.

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Alright, final chess variant trivia. Infinite chess is just chess, but played on an infinite board.

There’s a mind-blowing connection between infinite chess and mathematical logic. As a refresher, a little while back I discussed the first-order theory of Peano arithmetic. This is the theory of natural numbers with addition and multiplication. If you recall, we found that Peano arithmetic was incomplete (in that not all first-order sentences about the natural numbers can be proven from its axioms). First order PA is also *undecidable*, in that there exists no algorithm that takes in a first order sentence and returns whether it is provable from the axioms. (In fact, first order logic *in general* is undecidable! To get decidability, you have to go to a weaker fragment of first order logic known as monadic predicate calculus, in which predicates take only one argument and there are no functions. As soon as you introduce a single binary predicate, you lose decidability.)

Okay, so first order PA (the theory of natural numbers with addition and multiplication) is incomplete and undecidable. But there are weaker fragments of first order PA that *are* decidable! Take away multiplication, and you have *Presburger arithmetic*, the theory of natural numbers with addition. Take away addition, and you have *Skolem arithmetic*, the theory of natural numbers with multiplication. Both of these fragments are *significantly* weaker than Peano arithmetic (each is unable to prove general statements about the missing operation, like that multiplication is commutative for Presburger arithmetic). But in exchange for this weakness, you get both decidability and completeness!

How does all this relate to infinite chess? Well, consider the problem of determining whether there exists a checkmate in *n* turns from a given starting position. This seems like a *really hard *problem, because unlike in ordinary chess, now it’s possible for there to be literally infinite possible moves for a given player from a position. (For instance, a queen on an empty diagonal can move to any of the infinite locations on this diagonal.) So apparently, the game tree for infinite chess, in general, branches infinitely. Given this, we might expect that this problem is not decidable.

Well, it turns out that any instance of this problem (any particular board setup, with the question of whether there’s a mate-in-*n* for one of the players) can be translated into a sentence in Presburger arithmetic. You do this by translating a position into a fixed length sequence of natural numbers, where each piece is given a sequence of numbers indicating its type and location. The possibility of attacks can be represented as equations about these numbers. And since the distance pieces (bishops, rooks, and queens – those that have in general an infinite number of available moves) all move in straight lines, there are simple equations expressible in Presburger arithmetic that describe whether these pieces can attack other pieces! From the attack relations, you can build up more complicated relations, including the mate-in-*n* relation.

So we have a translation from the mate-in-*n* problem to a sentence in Presburger arithmetic. But Presburger arithmetic is decidable! So there must also be a decision procedure for the mate-in-*n* problem in infinite chess. And not only is there a decision procedure for the mate-in-*n* problem, but there’s an algorithm that gives the precise strategy that achieves the win in the fewest number of moves!

Here’s the paper in which all of this is proven. It’s pretty wild. Many other infinite chess problems can be proven to be decidable by the same method (demonstrating interpretability of the problem in Presburger arithmetic). But interestingly, not all of them! This has a lot to do with the limitations of first-order logic. The question of whether, in general, there is a forced win from a given position can* not* be shown to be decidable in this way. (This relates to the general impossibility in first-order logic of expressing infinitely long statements. Determining whether a given position is a winning position for a given player requires looking at the mate-in-*n* problem, but without any upper bound on what this *n* is – on how many moves the win may take.) It’s not even clear whether the winning-position problem can be phrased in first-order arithmetic, or whether it requires going to second-order!

The paper takes this one step further. This proof of the decidability of the mate-in-*n* problem for infinite chess doesn’t crucially rest upon the two-dimensionality of the chess board. We could easily translate the proof to a three-dimensional board, just by changing the way we code positions! So in fact, we have a proof that the mate-in-*n* problem for k-dimensional infinite chess is decidable!

I’ll leave you with this infinite chess puzzle:

It’s White’s turn. Can they guarantee a checkmate in 12 moves or less?

> And for fun, here’s a harder one, again a guaranteed two-move checkmate, this time by Black:

I don’t see a forced mate here. Black can win the queen but there isn’t an immediate mate

Huh, you’re totally right. Stockfish agrees that there’s no forced mate here (also no guaranteed way to win the queen). I’m not sure what I was thinking when I wrote this! Thanks for the fact check, will update the post.