+...+...+...+...+.......+.......+
: ::::: ::::: ::::: :::::
+...+...+...+...+...+...+...+...+
::::: ::::: ::::: ::::: :
+...+...+...+...+...+...+...+...+
: ::::: ::::: ::::: :::::
+...+...+...+...+...+...+.......+
::::: B ::B:: B ::::: B ::B:: :
+...........+...+...+...+...+...+
: ::B:: B ::B:: ::B:: B :::::
+...+...+...+...+...+...+...+...+
::::: ::::: ::::: ::::: :
+...+...+...+...+...+...+...+...+
: ::::: ::::: ::::: :::::
+...+...+...+...+.......+...+...+
::::: ::::: ::::: ::::: :
+...+...+...+...+.......+...+...+
This puzzle is quite easy if you first of all give it a little thought.
You need only consider squares of one colour, for whatever can be done
in the case of the white squares can always be repeated on the black,
and they are here quite independent of one another. This equality, of
course, is in consequence of the fact that the number of squares on an
ordinary chessboard, sixty-four, is an even number. If a square
chequered board has an odd number of squares, then there will always be
one more square of one colour than of the other.
Ten bishops are necessary in order that every square shall be attacked
and every bishop guarded by another bishop. I give one way of arranging
them in the diagram. It will be noticed that the two central bishops in
the group of six on the left-hand side of the board serve no purpose,
except to protect those bishops that are on adjoining squares. Another
solution would therefore be obtained by simply raising the upper one of
these one square and placing the other a square lower down.