# Problem 8. (5 points)

A figure called a "half-bishop" moves one square diagonally. What is the maximum number of non-attacking half-bishops that can be placed on a $7 \times 7$ chessboard?

Answer: 28
Solution:

Consider all diagonals of one direction. On each diagonal of even length, no more than half of the cells can be occupied by semi-bishops, as it can be divided into pairs of cells that cannot be simultaneously occupied.

On each diagonal of odd length $2n+1$, no more than $n+1$ semi-bishops can be placed for the same reasons.

There are a total of 7 odd diagonals, so the maximum possible number of semi-bishops is $\frac{49-7}{2}+7=28$.

An example can be constructed by placing semi-bishops in all cells of rows with odd numbers.

## 2nd Option

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