Chess/Puzzles/Placement/14 Bishops/Solution

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There are several solutions to this puzzle, but they are all quite similar.

Here's a possible one:

a b c d e f g h
8 a8 b8 c8 d8 e8 f8 g8 h8 8
7 a7 b7 c7 d7 e7 f7 g7 h7 7
6 a6 b6 c6 d6 e6 f6 g6 h6 6
5 a5 b5 c5 d5 e5 f5 g5 h5 5
4 a4 b4 c4 d4 e4 f4 g4 h4 4
3 a3 b3 c3 d3 e3 f3 g3 h3 3
2 a2 b2 c2 d2 e2 f2 g2 h2 2
1 a1 b1 c1 d1 e1 f1 g1 h1 1
a b c d e f g h

Proof of maximality[edit]

There are 15 diagonals on the chessboard running from bottom left to top right. They are:

  • a8-a8
  • a7-b8
  • a6-c8
  • a5-d8
  • a4-e8
  • a3-f8
  • a2-g8
  • a1-h8
  • b1-h7
  • c1-h6
  • d1-h5
  • e1-h4
  • f1-h3
  • g1-h2
  • h1-h1

Each of these diagonals can only contain one bishop. Also, the first and last diagonals cannot both contain a bishop, since both are on the diagonal a8-h1. Therefore, we can place at most 13 bishops on the other 13 diagonals, and one bishop on those two diagonals, for a total of 14 bishops. Since 14 bishops is possible, 14 is the maximum number of bishops we can place so no two attack each other.