[Illustration]

As all the points and lines puzzles that I have given so far, excepting the last, are variations of the case of ten points arranged to form five lines of four, it will be well to consider this particular case generally.  There are six fundamental solutions, and no more, as shown in the six diagrams.  These, for the sake of convenience, I named some years ago the Star, the Dart, the Compasses, the Funnel, the Scissors, and the Nail. (See next page.) Readers will understand that any one of these forms may be distorted in an infinite number of different ways without destroying its real character.

In “The King and the Castles” we have the Star, and its solution gives the Compasses.  In the “Cherries and Plums” solution we find that the Cherries represent the Funnel and the Plums the Dart.  The solution of the “Plantation Puzzle” is an example of the Dart distorted.  Any solution to the “Ten Coins” will represent the Scissors.  Thus examples of all have been given except the Nail.

On a reduced chessboard, 7 by 7, we may place the ten pawns in just three different ways, but they must all represent the Dart.  The “Plantation” shows one way, the Plums show a second way, and the reader may like to find the third way for himself.  On an ordinary chessboard, 8 by 8, we can also get in a beautiful example of the Funnel—­symmetrical in relation to the diagonal of the board.  The smallest board that will take a Star is one 9 by 7.  The Nail requires a board 11 by 7, the Scissors

[Illustration]

11 by 9, and the Compasses 17 by 12.  At least these are the best results recorded in my note-book.  They may be beaten, but I do not think so.  If you divide a chessboard into two parts by a diagonal zigzag line, so that the larger part contains 36 squares and the smaller part 28 squares, you can place three separate schemes on the larger part and one on the smaller part (all Darts) without their conflicting—­that is, they occupy forty different squares.  They can be placed in other ways without a division of the board.  The smallest square board that will contain six different schemes (not fundamentally different), without any line of one scheme crossing the line of another, is 14 by 14; and the smallest board that will contain one scheme entirely enclosed within the lines of a second scheme, without any of the lines of the one, when drawn from point to point, crossing a line of the other, is 14 by 12.

[Illustration:  STAR DART COMPASSES FUNNEL SCISSORS NAIL]

211.—­THE TWELVE MINCE-PIES.

If you ignore the four black pies in our illustration, the remaining twelve are in their original positions.  Now remove the four detached pies to the places occupied by the black ones, and you will have your seven straight rows of four, as shown by the dotted lines.

[Illustration:  The Twelve Mince Pies.]

212.—­THE BURMESE PLANTATION.

The arrangement on the next page is the most symmetrical answer that can probably be found for twenty-one rows, which is, I believe, the greatest number of rows possible.  There are several ways of doing it.