-+---+---+---+---+---+---+
| V | I | | | | | E | L |
+---+---+---+---+---+---+---+---+
| E | L | | | | | V | I |
+---+---+---+---+---+---+---+---+
| | | E | V | L | I | | |
+---+---+---+---+---+---+---+---+
| | | L | I | E | V | | |
+---+---+---+---+---+---+---+---+

]

If the reader will examine the above diagram, he will see that I have so
placed eight V's, eight E's, eight I's, and eight L's in the diagram
that no letter is in line with a similar one horizontally, vertically,
or diagonally. Thus, no V is in line with another V, no E with another
E, and so on. There are a great many different ways of arranging the
letters under this condition. The puzzle is to find an arrangement that
produces the greatest possible number of four-letter words, reading
upwards and downwards, backwards and forwards, or diagonally. All
repetitions count as different words, and the five variations that may
be used are: VEIL, VILE, LEVI, LIVE, and EVIL.

This will be made perfectly clear when I say that the above arrangement
scores eight, because the top and bottom row both give VEIL; the second
and seventh columns both give VEIL; and the two diagonals, starting from
the L in the 5th row and E in the 8th row, both give LIVE and EVIL.
There are therefore eight different readings of the words in all.

This difficult word puzzle is given as an example of the use of
chessboard analysis in solving such things. Only a person who is
familiar with the "Eight Queens" problem could hope to solve it.

304.--BACHET'S SQUARE.

One of the oldest card puzzles is by Claude Caspar Bachet de Meziriac,
first published, I believe, in the 1624 edition of his work. Rearrange
the sixteen court cards (including the aces) in a square so that in no
row of four cards, horizontal, vertical, or diagonal, shall be found two
cards of the same suit or the same value. This in itself is easy enough,
but a point of the puzzle is to find in how many different ways this may
be done. The eminent French mathematician A. Labosne, in his modern
edition of Bachet, gives the answer incorrectly. And yet the puzzle is
really quite easy. Any arrangement produces seven more by turning the
square round and reflecting it in a mirror. These are counted as
different by Bachet.

Note "row of four cards," so that the only diagonals we have here to
consider are the two long ones.

305.--THE THIRTY-SIX LETTER-BLOC