Five readers (Katsumi Takemura, Seiichi Fusamura, Mitsu-nobu Matsuyama, James Stuart, and Y. Dvir) lowered to 12 the moves required to transfer the 3x3 square of counters diagonally from corner to corner on the order-7 board.

The three-move solution given for the Halma solitaire problem involving the order-4 array on an order-6 board apparently not only is fundamentally unique for ending on one of the central cells but also seems to be the only three-move solution that eliminates all but one counter when this last counter can end anywhere on the board. When the order-4 square is at the center of a standard eight-by-eight chessboard, a pretty four-move solution puts the last counter on a corner of the board. And I found a four-move solution that leaves four counters at the corners of the order-4 square when it is centered on the order-6 field.

Here are some more of my results: The order-5 array (on the order-7 board) has four-move solutions that end on any cell originally occupied; the order-6 formation (on the order-8 board) has six-move solutions to any cell formerly occupied, and a five-move solution to the board's corner, and there are two-move solutions for the order-3 array that place the last counter on any cell of the order-5 board's border.

John W. Harris was the only reader to send results for the order-7 array on the order-9 Japanese chessboard. He found a solution, to the center cell, in seven moves.

If a square array of nine counters are placed at a corner of a 4 x 6 board, it is a pleasant task to shift them to the diagonally opposite corner in ten moves. The small size of the board makes it an attractive puzzle to market. I offer it free to any firm that cares to manufacture it, either with marbles, counters, or pegs in holes. I found a solution in ten moves and proved it to be minimal.

0 0