53

Here is an example of a 12-move game after which White (to move) is stalemated. All 32 units (pieces and pawns) are still on the board. The original version of this concept game was created by Charles Henry Wheeler, and published in Sunny South in 1887, according to Edward Winter's C.N. 3679. Samuel Loyd is often, and wrongly, given credit. [Title ""] [...


31

There are many different aspects of chess which can be formalized mathematically. Since the 19th century at least, chess has been mined as a resource to drive mathematical innovation. So when talking about a mathematical characterization of chess, it's not a single modeling that we are talking about, which grabs every feature, but rather a number of models, ...


17

Disclaimer: This solution is not reachable from the starting position, and is not reachable in a game of Chess960 (thanks Rewan!). [FEN "3bBNRN/2pPpPKQ/2P1P1PR/7P/p7/rp1p1p2/qkpPpP2/nrnbB3 w - - 0 1"] Why does the solution here not work? This is clearly not reachable from the starting position (because of the bishops stuck on the first rank), but the ...


12

I noticed a simplification of @RosieF's solution: Label the squares with two-number labels as follows: 4,2 0,3 1,4 2,0 3,1 1,0 2,1 3,2 4,3 0,4 3,3 4,4 0,0 1,1 2,2 0,1 1,2 2,3 3,4 4,0 2,4 3,0 4,1 0,2 1,3 Then two knights may occupy two squares x,y and X,Y iff those squares' first numbers x and X are different and their second numbers y and Y are ...


11

Thanks for the fun question, and welcome to CSE! For starters, the king is obviously the piece to choose. I will answer this question assuming that the normal rules of chess are used and no special conditions are used other than what the question has. The OP has clarified that both sides need a king and first/last rank pawns can do a double step. ...


9

Label the squares with two-number labels as follows: 4,2 0,3 1,4 2,0 3,1 1,0 2,1 3,2 4,3 0,4 3,3 4,4 0,0 1,1 2,2 0,1 1,2 2,3 3,4 4,0 2,4 3,0 4,1 0,2 1,3 Then two knights may occupy two squares x,y and X,Y iff those squares' first numbers x and X are different and their second numbers y and Y are different. Regard these numbers as modulo 5. Then the two-...


9

This is my answer :) Spoiler alert: Answer is in a comment in the last line of the code block using System; namespace Toroidal5Knights { class Program { static int solution = 0; static bool[,] board = new bool[5, 5]; static int[] dx = { -1, 1, 2, 2, 1, -1, -2, -2 }; static int[] dy = { -2, -2, -1, 1, 2, 2, 1, -1 }; ...


9

We can easily get a reasonably good upper bound on the number of positions. At any point in time, each player has 16 pieces of which the 8 pawns can perhaps be promoted to a knight / bishop / rook / queen. Consider all captured pieces to be off the board but part of the position (this would increase the number of positions but that is fine because we only ...


8

Here are some starting spots in reading up on Game theory which is the mathematical tool that would be most appropriate for making claims about chess. This is a light history of early game theory. Chess is a "perfect information game" and there are some interesting things one can claim about this category of games. See this for example. I would also read ...


7

The question has two parts to it, and Rosie F perfectly answers the first part. Regarding only the second part, the question asks for any possible position where all 32 pieces are stalemated, disregarding legality of such positions. @im_so_meta_even_this_acronym's wonderful answer proves that it indeed possible. However, I want to focus specifically on only ...


7

To answer the second part of your question: Suppose all figures are on the board. Does there exist a transposition of figures such that both of the opponents can't do any move (a stalemate)? No, this is not possible. Pieces are simply too mobile for this, so for a stalemate you need to hem them in (like the white queen in @RosieF's answer) or pin them on ...


7

There are essentially six solutions that are rotationally and translationally unique. Two of them yield five translations and five more rotated translations each, for a total of twenty: NNNNN ..... ..... ..... ..... N.... .N... ..N.. ...N. ....N The remaining four unique solutions can be translated to each of the 25 positions on the board, but are ...


7

(EDIT: substantially revised because I hadn't thought carefully enough about Rosie F's labeling. The result is even cooler!) Thanks for all the great answers. I would like to add my own solution. It's key to label squares in the torus, as Rosie F and Brilliand did: 4,2 0,3 1,4 2,0 3,1 1,0 2,1 3,2 4,3 0,4 3,3 4,4 0,0 1,1 2,2 0,1 1,2 2,3 3,4 4,0 2,4 3,0 4,1 0,...


6

It kind of depends exactly what you mean by artificial cutoffs. There is one explicitly stated artificial cutoff in the FIDE rating regulations. That says that if the rating difference is greater than 400 it will be treated as 400. But there is another hidden cutoff. The actual elo algorithm for calculating expected score or scoring probability involves ...


6

The FEN consists of six parts (see definition of FEN here): piece placement: at most 64 characters, one for each square plus 7 times "/" to delimit ranks = 71 characters active color: one character ("w" or "b") castling: at most 4 characters en passant: one or two characters half move clock full move number In addition there are 5 space characters between ...


4

It's possible to attack all squares. Q - queen N - nightrider Yellow background - attacked only by a queen Red background - attacked only by a nightrider Orange background - attacked by pieces of both types Queens are placed such that 5 longest diagonals in each direction are attacked. This can be done with just 6 queens, so 2 are redundant and result ...


4

The variability in number of legal moves is about the same for black & white... except for that odd little bump for white on move 7. This phenomenon demands further research!


4

The best example I could find in PDB is this one: [Title "Theodor Steudel: Problemkiste 66, Dec 1989, no. 2339. Za6 in 12"] [StartFlipped "0"] [fen "8/1p1p1p2/1P1P1P1P/8/p7/p1p1p1p1/2P1P1P1/k1K w K - 0 1"] 1.h7 a2 2.h8=N a3 3.Ng6 fxg6 4.f7 g5 5.f8=N g4 6.Ne6 dxe6 7.d7 e5 8.d8=N e4 9.Nc6 bxc6 10.b7 c5 11.b8=N c4 12.Na6 (Far greater lengths are possible in ...


4

Using the definition of FEN it is relatively straightforward to compute an upper bound: piece placement: 64 (pieces/squares) + 7 (slashes) color: 1 (w/b) castling: 4 (KQkq) en passant: 2 (e3) halfmove clock: 3 (100) fullmove number: 4 (considering that maximum game length < 9999 due to the 50-move rule) spaces between fields: 5 So in total we get as an ...


3

Indeed the limit of this scenario would be infinity for the perpetually winning player. This is because FIDE has a rule saying that if one player is rated more than 400 points higher than his opponent, their difference should be set to 400 when calculating rating gain/loss. That means no matter how weak the opposition is, our winning player will receive ...


3

Odd if you count the initial position and Even if not. Mirrored positions Almost any position, including checkmates and stalemates can be flipped by switching colors. Mirroring the chess board horizontally between 4 and 5 rows, you cannot find a position that colors can be switched and produce an illegal position. So for each position there exists a ...


3

I'd hate to say it, but mathematically speaking chess as we play it is very boring. It is a perfect information game, without any non-determinism, for two alternating players. This means that chess is either a win for white, a win for black, or a draw. The optimal strategy for chess is trivial and known: minimax. As far as maths is concerned, chess is solved....


3

Not really. At least not serious mathematicians. It is covered somewhat in Game Theory but other approaches like Group Theory do not seem to fit; although there might some higher dimensional non-linearish type description that could be found. So Algebra which includes Groups, and Topology too, seem to be out but perhaps it could fit in Analysis if ...


3

If we work solely from the standard mathematical formula for Elo ratings, it's possible to prove that the winning player can get an arbitrarily high rating, but that it'll probably take exponentially long (i.e., the winning player's rating increases logarithmically). I'll suppose that the initial Elo rating of the two players is 0, not 1500, because it ...


3

Your estimate is a bit high. In particular you overestimate the number of king moves. At most a king has 8 moves (not including castling which does not apply here), however here, both the black and the white king have a lot less legal moves than your estimated 10. The white king because it is on the edge of the board and the black king because it is not ...


3

To classify pawn formations, a good source is chess master books that explore that topic. The book by Boris Persits comes to mind, for example. I add that it seems impractical as you have been told, to make an exhaustive index of all the combinations that pawn chains can take. However, it is useful to connect families of openings (classified by ECO, or by ...


2

EDIT: Final calculation matches actual proof game exactly. That’s a nice new chess-math question, which deserves wider circulation. Begin by promoting a White pawn as quickly as possible, because until that happens, White is just haemorrhaging pawn moves. Now can promote a second pawn more slowly. This promoted piece is captured by Black pawn to open a ...


2

The longest game I've ever found in tournament play is Nikolic - Arsovic, played in Belgrade in 1989. You can watch it here. It ended in a draw after 269 moves in a rook+bishop-vs-rook endgame


2

This is my best attempt thus far, which I believe attacks 236 (all but 20) squares. I used the staircase solution for 8 queens on the central 8x8 region in order to consequentially fully cover the 4 adjacent 8x4 side regions while putting many diagonals through the corner 4x4 regions. Then I placed the nightriders so as to not obstruct any of the queens' ...


2

It depends on the exact position. If white only has the king left, and in your position black only their rook, bishop and king, then it would take forever. Or at least untill the 50 move rule kicks in. Give black more pawns and each one of their moves could add another 50 to the duration. With more pieces and or pawns it might take more or less time. ...


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