by Dudeney, Henry Ernest, 1857-1930
Available in 215 free installments
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Father John held that the four corners should also be darkened, but the sage explained that it was desired to obstruct no more light than was absolutely necessary, and he said, anticipating Lord Dundreary, "A single pane can no more be in a line with itself than one bird can go into a corner and flock in solitude. The Abbot's condition was that no diagonal lines should contain an odd number of lights."
Now, when the holy man saw what had been done he was well pleased, and said, "Truly, Father John, thou art a man of deep wisdom, in that thou hast done that which seemed impossible, and yet withal adorned our window with a device of the cross of St. Andrew, whose name I received from my godfathers and godmothers." Thereafter he slept well and arose refreshed. The window might be seen intact to-day in the monastery of St. Edmondsbury, if it existed, which, alas! the window does not.
293.?THE CHINESE CHESSBOARD.? solution
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§18,1
Eighteen is the maximum number of pieces. I give two solutions. The numbered diagram is so cut that the eighteenth piece has the largest area?eight squares?that is possible under the conditions. The second diagram was prepared under the added condition that no piece should contain more than five squares.
No. 74 in The Canterbury Puzzles shows how to cut the board into twelve pieces, all different, each p 9 214 containing five squares, with one square piece of four squares.
294.?THE CHESSBOARD SENTENCE.? solution
The pieces may be fitted together, as shown in the illustration, to form a perfect chessboard.
295.?THE EIGHT ROOKS.?solution
Obviously there must be a rook in every row and every column. Starting with the top row, it is clear that we may put our first rook on any one of eight different squares. Wherever it is placed, we have the option of seven squares for the second rook in the second row. Then we have six squares from which to select the third row, five in the fourth, and so on. Therefore the number of our different ways must be 8x7x6x5x4x3x2 x 1 = 40,320 (that is 8!), which is the correct answer.
How many ways there are if mere reversals and reflections are not counted as different has not yet been determined; it is a difficult problem. But this point, on a smaller square, is considered in the next puzzle.
296.?THE FOUR LIONS.? solution
There are only seven different ways under the conditions. They are as follows: 12 3 4,1243,1324,134 2,1432,2143,2413. Taking the last example, this notation means that we place a lion in the second square of first row, fourth square of second row, first square of third row, and third square of fourth row. The first example is, of course, the one we gave when setting the puzzle.
297.?BISHOPS?UNGUARDED.? solution
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This cannot be done with fewer bishops than eight, and the simplest solution is to place the bishops in line along the fourth or fifth row of the board (see diagram). But it will be noticed that no bishop is here guarded by another, so we consider that point in the next puzzle.
