Is it possible to promote all 16 pawns to queens, without capturing any queen in the process, or either side check/stalemating the other?
Without actually working it out, my immediate inclination is: why not? I doubt that this would ever occur in a game that either side was intending to win, but with both sides working toward the goal of achieving it, it should be doable.
The main problem is passing the pawn ranks “through” each other. For each of the 8 columns on the board, one of the pawns must move aside to allow the other past. Since sideways movement of a pawn requires a capture, that will consume 8 of the remaining pieces on the board.
Assume that we split those 50/50 between black and white, leaving 4 non-pawn pieces for each side. One of these will be the king, which we move to the corner, and the remaining three on the three adjacent spots. This prevents checkmate, and since you’re never required to capture a piece, the rest of the pieces (eventually all queens) can dance around the board as they like.
Note that you will eventually statemate based on the “if the same board position repeats three times, it’s a stalemate” rule, but (a) it’ll take a long time if both sides are working to avoid it, and (b) that’s true of any stready-state game. So far as I know, it’s not possible for any chess game to have an infinite number of moves to it.
I would assume the answer is yes. You only need one capture to free up two pawns to promote (the pawn doing the capturing and the pawn opposite it), so that costs you eight of your twelve non-pawn, non-queen, non-king pieces. And you can just build a meat shield out of queens and the remaining four pieces to avoid checkmate and stalemate.
“Threefold repetition.”
Like anyone’s going to argue with that username :- )
I’d think that’s true of any game, period. There are only a finite number of configurations of the board so they have to start repeating eventually. Back-of-the-envelope calc says that a chess match with no captures at all has at most 2*64!/(32!*8!*8!*2^6) + 1 ~= 9.2694534e+42 moves before some configuration repeats twice.
But other games do not necessary have an explicit limit on repeating states. Magic: the Gathering, for example, only declares a stalemate if the loop cannot be broken by player input. It does not say “If you equip Lightning Greaves to the same creature three times, the game ends”.
I think we’re all agreeing, actually. leahcim meant “any chess game”, and was probably referring to my comment about “steady state” games rather than my more general one afterward. Probably most (non-chess) games that have the possibility of repetitive states don’t bother to have a rule against them.
See post #2 here:
I will, and he knows why. But not on this occasion, because of course he is right. I’ve explained this a few times just lately at school: in chess, stalemate is not a synonym for “draw” but a specific category of it, namely that the side to move is not in check but cannot move without exposing himself to check. Example: White Queen on c7, White King on e1, Black King on a8 and Black to move has no legal move.
AFAIK the record number of Queens on a board in a game between masters is still a humble five, and it’s not happened often. But kids with the weaker side at their mercy will often queen as many pawns as they can just for the lulz. Not unoften, they end up stalemating the enemy King just because they can’t keep all their Queens from tripping over each other.
A chess game cannot have an infinite number of moves, but the limit is nowhere near 10[sup]42[/sup] or anything like it; there is another limit that will kick in far sooner than threefold repetition. If fifty consecutive full moves (one by each side being a full move) elapse with no man taken and no pawn moved, the game is a draw. The advancing student is welcome to evaluate the maximum number of pawn moves and captures possible in a game and hence find the upper limit.
By my reckoning, that places the upper limit at 5,950 turns: 50+49 knight moves then move a single pawn forwards a square, then another 50+49 moves and another pawn move, and so on. That repeats 16*6 times for the pawn moves and 30 times for the captures, minus the 8 captures that must be completed with pawns to clear the way.
Each pawn can move 6 times. With 8 pawns on each side, that’s 96 pawn moves. There are 7 other pieces on each side, plus the 8 pawns on each side, so there’s a maximum of 30 captures. (96+30)*50 is 6300 so the longest possible chess game is 6349 moves – and it will include all pawns being promoted, and later being taken.
Sure. This game, for example, works:
[QUOTE=Game]
e2e4 e7e5 g1f3 d7d5 f3d4 e5d4 f2f4 c8f5 e4f5 f8a3 g2g4 g8f6 f1h3 e8g8 e1g1 f8e8 g4g5 e8e3 d2e3 d4d3 g5f6 d5d4 b2a3 b7b5 d1f3 d3d2 f1f2 h7h5 f2g2 b5b4 g2g6 f7g6 f3f1 c7c5 f6f7 g8h7 f5f6 g6g5 h3g4 h5g4 f4f5 d4d3 c2c3 b8c6 c1b2 c6d4 c3d4 b4b3 b1c3 c5c4 c3b5 c4c3 d4d5 c3c2 b2d4 b3b2 d4b6 a7b6 e3e4 g4g3 h2h4 g5g4 f1f4 g7g5 a1f1 b2b1=Q h4h5 b1a1 a3a4 c2c1=Q d5d6 h7h8 a4a5 d8g8 f7f8=Q a8c8 d6d7 c8c2 d7d8=Q d2d1=Q b5c7 b6b5 f8c5 d3d2 a5a6 b5b4 a6a7 b4b3 a7a8=Q b3b2 f6f7 b2b1=Q a2a4 c2a2 f7f8=Q d1b3 d8b8 d2d1=Q f8c8 c1c4 a4a5 c4a4 a5a6 d1d8 f5f6 b3b7 c5c1 g8e6 f1f3 b7c6 g1f1 h8h7 h5h6 h7g6 f1e1 g3g2 f3f1 g2g1=Q h6h7 g1h1 h7h8=Q g4g3 f6f7 g3g2 f7f8=Q g2g1=Q h8h2 g1b6 c7d5 g5g4 a8b7 g4g3 a6a7 g3g2 a7a8=Q g2g1=Q e4e5 e6h3 e5e6 c6c7 e6e7 g6h7 e7e8=Q
[/QUOTE]
(Sorry about the bad notation. Also, I realized after posting that someone linked to a much more legible game above, so…yeah.)
To be a bit pedantic, the game is not automatically a draw under either of these rules. One of the players must claim the draw. It would be perfectly legal for two players to just repeat moves forever, without ending the game.
There are lots of examples of games where a player has missed the fact that they could have claimed such a draw, and they’ve gone on to lose the game.
Having said that, most of the time during these types of discussions (or chess problems where it’s relevant) it’s assumed that the draw will be claimed when it’s available.
After the last capture happens, so the game is down to bare kings, it’s immediately drawn, so you don’t get those last 49 moves.
4 of those captures for each side would need to be done with pawns, making 22 captures in addition to the pawn moves. The first move by a rook or king for either side will also reset the clock so ((96+22)*50)+98=5998?
[edit] No, that’s not right is it?That only applies to 3-fold repetition[/edit]
Yes, changing the castling status doesn’t reset the 50-move counter, although IMO it should, since it is a non-reversible change to the position.
Would a position be considered the same for the purposes of the threefold repetition rule if pieces of the same rank occupy the same squares, but they are not the same pieces, e.g. the original queen is on f1 the first time, a promoted queen is on f1 the second time, and a different promoted queen is on f1 the third time?
Yes. The official rule (9.2; scroll down) specifies “Positions … are considered the same if the same player has the move, pieces of the same kind and colour occupy the same squares, and the possible moves of all the pieces of both players are the same.” See also Wikipedia.
Decades ago I saw a last-round game in the British Championship where the position was repeated six times before they agreed a draw.
Both players needed a win to get a lot more prize money (and probably a shot at the title.)