by Dudeney, Henry Ernest, 1857-1930
Available in 215 free installments
Owner:
a, you must consider joins at one end only of the key-line, or you will get repetitions. In other cases you must
consider joins at both ends of the key; but after leaving a in case D, turn always either to right or left?use one
direction only. Figs. 1 and 2 are examples under A; 3 and 4 are examples under B; 5 and 6 come under C;
and 7 is a pretty example of D. Of course, E is a peculiar type, and obviously admits of only one way of Pg212
cutting, for you clearly cannot enter at b or c.
Here is a table of the results:?
a b c Ways.
A = 8 + 17 + 21 = 46
B = 0 + 17 + 21 = 38
C = 15 + 31 + 39 = 85
D = 17 + 29 + 39 = 85
E = 1 + 0 + 0 = 1
41 94 120
255
I have not attempted the task of enumerating the ways of dividing a board 8x8?that is, an ordinary chessboard. Whatever the method adopted, the solution would entail considerable labour.
289.?LIONS AND CROWNS.? solution
Here is the solution. It will be seen that each of the four pieces (after making the cuts along the thick lines) is of exactly the same size and shape, and that each piece contains a lion and a crown. Two of the pieces are shaded so as to make the solution quite clear to the eye.
290.?BOARDS WITH AN ODD NUMBER OF SQUARES.?solution
There are fifteen different ways of cutting the 5x5 board (with the central square removed) into two pieces of the same size and shape. Limitations of space will not allow me to give diagrams of all these, but I will enable the reader to draw them all out for himself without the slightest difficulty. At whatever point on the edge your cut enters, it must always end at a point on the edge, exactly opposite in a line through the centre of the square. Thus, if you enter at point 1 (see Fig. 1) at the top, you must leave at point 1 at the bottom. Now, 1 and 2 are the only two really different points of entry; if we use any others they will simply produce similar solutions. The directions of the cuts in the following fifteen
solutions are indicated by the numbers on the diagram. The duplication of the numbers can lead to no confusion, since every successive number is contiguous to the previous one. But whichever direction you take from the top downwards you must repeat from the bottom upwards, one direction being an exact reflection of the other.
1,4,8.
1,4,3,7,8.
1,4,3,7,10,9.
1,4,3,7,10,6,5,9.
1,4,5,9.
1,4,5,6,10,9.
1,4,5,6,10,7,8.
2,3,4,8.
2,3,4,5,9.
2,3,4,5,6,10,9.
2,3,4,5,6,10,7,8.
2,3,7,8. 2,3,7,10,9. 2,3,7,10,6,5,9. 2,3,7,10,6,5,4,8.
It will be seen that the fourth direction (1, 4, 3, 7, 10, 6, 5, 9) produces the solution shown in Fig. 2. The thirteenth produces the solution given in propounding the puzzle, where the cut entered at the side instead of at the top. The pieces, however, will be of the same shape if turned over, which, as it was stated in the conditions, would not constitute a different solution.
291.?THE GRAND LAMA'S PROBLEM.? solution
The method of dividing the chessboard so that each of the four parts shall be of exactly the same size and shape, and contain one of the gems, is shown in the diagram. The method of shading the squares is adopted to make the shape of the pieces clear to the eye. Two of the pieces are shaded and two left white.
The reader may find it interesting to compare this puzzle with that of the "Weaver" (No. 14, Canterbury Puzzles).
Pg213
292.?THE ABBOT'S WINDOW.? solution
The man who was "learned in strange mysteries" pointed out to Father John that the orders of the Lord Abbot of St. Edmondsbury might be easily carried out by blocking up twelve of the lights in the window as shown by the dark squares in the following sketch:?
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