Solution to Shatranj Problem, July 2015

Saturday, 8th August 2015

Last month's shatranj problem is the last one for the time being. The solution to it is as follows:

1. Pawn h4-h5 (check), King g6-f7; 2. Knight f5xd6 (check), King f7-e6; 3. Pawn f4-f5 (check), King e6-e5; 4. Knight d6xc4 (check), King e5-d4; 5. Rook h1-d1 (check), King d4xc4; 6. Pawn b2-b3 (check), King c4-b5; 7. Knight b1-c3 (check), King b5-a6; 8. Pawn b4-b5 (check), King a6-a5; 9. Pawn b3-b4 (check), Knight c6xb4; 10. Pawn a3xb4 (check), King a5xb4; 11. Knight b3-d5 (check), King b4-a5; 12. Rook f3-a3 (check), King a5-b5; 13. Rook a3-b3 (check), King b5xc5; 14. Rook b3-c3 (check), King c6-d6; 15. Knight d5-e7 (discovered check), King d6-e5; 16. Knight e7-g6, King e5xe4; 17. Rook c3-c4 (check), King e4xf5; 18. Rook a5-d5 (check), King f5-e4; 19. Knight g6-f4 (check), King e6-f7; 20.Rook d5xd7 (check), King f7-e8; 21. Rook d7-e7 (check), King e7-d7; 22. Knight f4-e6 (checkmate).

This is the classic series of forced moves: note the check on every move. Only a couple of times did white have any choice in how to move; I assume (given my source) that the alternatives would have led to a quicker defeat. The final position is shown in the diagram. This wouldn't be a checkmate in modern chess; the bishop in mediaeval chess had the power of jumping exactly two spaces, so the rook at e7 is defended here.