## IPSC 2004

## Solution to Problem B – Bishops

This was probably the easiest task. Let's call a diagonal from
top-left to bottom-right a "TLBD" diagonal and a diagonal from
bottom-left to top-right a "BLTR" diagonal.

On a chessboard of size **N × N** there are **2N-1**
"TLBD" diagonals. On each of them we can place at most one
bishop. Therefore the chessboard can contain at most **2N-1**
bishops. In fact, it is impossible to place **2N-1** bishops,
because in such case we would have to place a bishop on every "TLBD"
diagonal, including the top-right and bottom-left corner of the
chessboard. But these two bishops threaten each other. Therefore
we can place at most **2N-2** bishops on the chessboard.

It is indeed possible to place **2N-2** bishops on the
chessboard. For example, one can use all places in the bottom row
and the top row of the chessboard except the two right corners.

There is one special case **N=1**. In that case, the top-right
and bottom-left corner are the same place. The maximal number of
bishops for **N=1** is clearly **1**.