Dripread - Book

board. Of course, where n is even the unoccupied squares in the rows and columns will be even, and where n is odd the number of squares will be odd. Here n is 8, so the answer is 18,816 different ways. This is "The Dyer's Puzzle" (Canterbury Puzzles, No. 27) in another form. I repeat it here in order to explain a method of solving that will be readily grasped by the novice. First of all, it is evident that if we put a pawn on any line, we must put a second one in that line in order that the remainder may be even in number. We cannot put four or six in any row without making it impossible to get an even number in all the columns interfered with. We have, therefore, to put two pawns in each of three rows and in each of three columns. Now, there are just six schemes or arrangements that fulfil these conditions, and these are shown in Diagrams A to F, inclusive, on next page.

I will just remark in passing that A and B are the only distinctive arrangements, because, if you give A a quarter-turn, you get F; and if you give B three quarter-turns in the direction that a clock hand moves, you will get successively C, D, and E. No matter how you may place your six pawns, if you have complied with the conditions of the puzzle they will fall under one of these arrangements. Of course it will be understood that mere expansions do not destroy the essential character of the arrangements. Thus G is only an expansion of form A. The solution therefore consists in finding the number of these expansions. Supposing we confine our operations to the first three rows, as in G, then with the pairs a and b placed in the first and second columns the pair c may be disposed in any one of the remaining six columns, and so give six solutions. Now slide pair b into the third column, and there are five possible positions for c. Slide b into the fourth column, and c may produce four new solutions. And so on, until (still leaving a in the first column) you haveb in the seventh column, and there is only one place for c ?in the eighth column. Then you may put a in the second column, b in the third, and c in the fourth, and start sliding c and b as before for another series of solutions.

We find thus that, by using form A alone and confining our operations to the three top rows, we get as many answers as there are combinations of 8 things taken 3 at a time. This is ( 8 x 7 x 6 V(i x 2 x 3) = 56. And it will at once strike the reader that if there are 56 different ways of electing the columns, there must be for each of these ways just 56 ways of selecting the rows, for we may simultaneously work that "sliding" process downwards to the very bottom in exactly the same way as we have worked from left to right. Therefore the total number of ways in which form A may be applied is 56 x 6 = 3,136. But there are, as we have seen, six arrangements, and we have only dealt with one of these, A. We must, therefore, multiply this result by 6, which gives us 3,136 * 6 = 18,816, which is the total number of ways, as we have already stated.

359.?COUNTER SOLITAIRE.? solution

Play as follows: 3?11, 9?10, 1?2, 7?15, 8?16, 8?7, 5?13, 1?4, 8?5, 6?14, 3?8, 6?3, 6?12, 1 ?6, 1?9, and all the counters will have been removed, with the exception of No. 1, as required by the conditions.

360.?CHESSBOARD SOLITAIRE.? solution

Play as follows: 7?15, 8?16, 8?7, 2?10, 1?9, 1?2, 5?13, 3?4, 6?3, 11?1, 14?8, 6?12, 5?6, 5 ?11, 31?23, 32?24, 32?31, 26?18, 25?17, 25?26, 22?32, 14?22, 29?21, 14?29, 27?28, 30 ?27, 25?14, 30?20, 25?30, 25?5. The two counters left on the board are 25 and 19?both belonging to the same group, as stipulated?and 19 has never been moved from its original place.

I do not think any solution is possible in which only one counter is left on the board.

Pg234

361.?THE MONSTROSITY? solution

4. PtoB5 KtoB2

5. QtoKsq KtoKt3

6. QtoKt3 KttoQR3

7. QtoKt8 PtoKR4

8. KttoKB3 RtoR3

9. KttoK5 RtoKt3

10. Q takes B RtoKt6, ch

11. P takes R KtoKt4

12. RtoR4 PtoB3

13. RtoQ4 P takes Kt

14. PtoQKt4 P takes R, ch

15. KtoB4 PtoR5

16. QtoK8 PtoR6

17. KttoB3, ch P takes Kt

18. BtoR3 PtoR7

19. RtoKtsq PtoR8(Q)

20. RtoKt2 P takes R

21. KtoKt5 QtoKKt8

22. QtoR5 KtoR5

23. PtoKt5 RtoBsq

24. PtoKt6 RtoB2

25. P takes R PtoKt8(B)

26. PtoB8(R) QtoB2

27. BtoQ6 KttoKt5

28. KtoKt6 KtoR6

29. RtoR8 KtoKt7

Amusements in Mathematics (PDF)