Correcting the solution for the problem presented on pg 120, ‘A Remarkable Monochromatic,’ from The Chess Mysteries of Sherlock Holmes by Raymond Smullyan
Given this board, we are told that it’s a monochromatic game (meaning that no piece has moved from a square of one colour to a square of a different colour). When Homes is told that it’s White’s move, he declares that a promotion has been made and that a pawn has been captured en passant.
But interestingly, even if it were black’s move, that statement would be true. No matter which colour is to move next, one can demonstrate that a promotion has been made and a pawn has been captured en passant; and unfortunately the presence of white pawns at e2 and g2 in the absence of a white bishop at f1 renders the board impossible as a monochromatic game if it’s White’s move.
It is worth considering some of the surprising restrictions that result in a monochromatic game. Considering that no piece may move from a white square to a black square, or vice versa, we find:
- Knights may never move
- Queen-side castles are not permitted
- The four original rooks may never meet. Each rook can only access 16 squares, which are mutually exclusive to each other. Meaning, each rook can only access every second rank or file. For example, original white rooks can only occupy odd ranks, and original black rooks can only occupy even ranks.
The circumstances of the problem immediately bring our attention to the fact that it is White’s move, so we must consider what Black’s last move could have been. The presence of pawns at d2 and f2 indicate that the Black King couldn’t have arrived at its present position on a2 without first castling. The only two Black pieces that are still on the board that have moved in this game are the king and the pawn on h5. But since the pawn needed to be on h5 for the Black king to leave f1, Black’s last move had to be with its King from b3 (escaping check from the White Queen. But the White’s last move could not have been moving the Queen from c2, d5, or d3 (because it would require that Black ended its second-last turn in check on b3).
The only possible legal possible position for this situation to come from is with a white rook at c2:
White’s last move is to move the Rook to a2 (check), Black takes the rook, and it is White’s move.
But since an original white rook cannot occupy c2 in a monochromatic game, this rook must be the result of a promotion. Smullyan explains that the pawn from a2 or c2 are the only White pawns that could promote on a white square, and this would require 4 captures. But since there are only 3 eligible Black pieces on White squares that can be taken (the Black queen’s bishop, the Black queen’s rook, and the pawn from b7 (the Black King’s Knight was necessarily captured on its home square and so could not be captured by either of the pawns in question)), the promoting White pawn must have captured a Black piece that was on a black square – and the only way this can happen in a monochromatic game is via en passant. So, the pawn moved the 4th rank and then captured a black pawn on b5. Then a black pawn moved a5 or c5 and was captured en passant, and then White promoted on a8 or c8 by capturing the Black queen’s Rook.
And so we reach the solution that because it’s White’s move, there has been a promotion and a pawn has been captured en passant.
Unfortunately the situation presented is not possible if it is indeed White’s move.
We must taken note of the absence of a white bishop on f1. Because there are pawns on e2 and g2, we know the White king’s bishop was captured on its home square, f1. There is no original piece that can accomplish this in a monochromatic game. Black’s knights can’t move, the queen’s rook cannot reach the first rank, and neither the king nor the queen’s bishop can reach f1 with pawns at e1 and g1. The only way the White king’s bishop could have been captured is if Black promoted a pawn to a rook or queen on a white square.
The only available candidate is the pawn from b7, and it would have had to promote on d7. This would require four captures. The only available White pieces on white squares are the pawns from a2 and c2 and the king’s rook. Meaning that Black’s pawn must have captured a piece on a black square (again, the only option for this in a monochromatic game is via en passant). So Black’s pawn must have moved from b2 to b5, captured a White pawn on a4, capturing the pawn from b2 en passant, capturing the pawn on c2, and then capturing the king’s rook on d1 and promoting to a queen or rook; Black’s promoted piece can now capture the White king’s bishop on its home square.
So we see that the situation before us requires two promotions and two captures via en passant. That Black must promote the pawn from b7, while White’s promotion requires the capture of this very same pawn. Further, while White must promote either the pawn from a2 or c2, Black’s promotion requires the capture of both of these pawns.
The situation as presented, is therefore impossible.
This puzzle and its presented solution can be corrected simply by removing the pawn from e2 or
This means that Holmes could have deduced that there had been a promotion and a pawn had been captured en passant as soon as he was informed that it was a monochromatic game. And upon being told that it was White’s move he should have declared that this was an impossiblity, and proceeded to chastise the Lord and Lady Ashley for either dishonesty or improper play.