I've always been interested in puzzles. Cryptic crosswords, Sudoku, Cats mice and cheese crossing a river. For example the old question of dominoes and a chessboard - If a domino covers two chessboard squares, can you place the dominoes in such a way that when all other squares are covered, two opposite corners are visible? If so, how? If not, why not? *
I was thinking the other day about pawns and Queens. The pawn is arguably the most powerful piece on the chessboard next to the Queen, because every pawn is itself a potential Queen. Once it attains the eighth rank it can blossom into any major piece, and although sometimes it may choose to become a lesser piece, the Queen is usually the promotion of choice.
I wondered how many Queens were possible in a game of chess. (Forgive me if this is already on "Google" but I couldn't find it) There are two Queens at the start of each game of course, and sixteen pawns, so one might think that the total possible would be eighteen. But wait, if there are two rows of pawns facing each other, the only way that they could pass each other to become Queens is if some pawns are removed from the board to make passage for others.
For example, the pawn at B4 could take the pawn at A5, allowing the white pawns at A4 and A5 and the black pawn at B5 to advance to Queen. If the same pattern is repeated across the board this would result in another nine pawns advancing to Queen status, a total of fourteen Queens in all, including the original two.
This all assumes, of course, that both White and Black cooperate in this venture and that the Queens do not capture each other - an assumption that is not exactly aligned with the priciples of chess!
Is this analysis correct or can more Queens be present at any one time?
And what is the maximum number of Queens ever recorded on a chessboard in a real game? (ie a game where the purpose was winning rather than having fun).
I'd be really interested to know.
* PS: The answer to the dominoes question is quite simple. If you can't work it out, send me a message and I'll e-mail you the answer. :-)