The Canterbury Puzzles Part 30

123 87 4

56 -------- = 5 moves.

E312 87 4

E -------- = 9 moves.

87654321

Twenty-six moves in all.

88.--_The Eccentric Market-woman._

The smallest possible number of eggs that Mrs. Covey could have taken to market is 719. After selling half the number and giving half an egg over she would have 359 left; after the second transaction she would have 239 left; after the third deal, 179; and after the fourth, 143. This last number she could divide equally among her thirteen friends, giving each 11, and she would not have broken an egg.

89.--_The Primrose Puzzle._

The two words that solve this puzzle are BLUEBELL and PEARTREE. Place the letters as follows: B 3-1, L 6-8, U 5-3, E 4-6, B 7-5, E 2-4, L 9-7, L 9-2. This means that you take B, jump from 3 to 1, and write it down on 1; and so on. The second word can be inserted in the same order. The solution depends on finding those words in which the second and eighth letters are the same, and also the fourth and sixth the same, because these letters interchange without destroying the words. MARITIMA (or sea-pink) would also solve the puzzle if it were an English word.

Compare with No. 226 in _A. in M._

90.--_The Round Table._

Here is the way of arranging the seven men:--

A B C D E F G A C D B G E F A D B C F G E A G B F E C D A F C E G D B A E D G F B C A C E B G F D A D G C F E B A B F D E G C A E F D C G B A G E B D F C A F G C B E D A E B F C D G A G C E D B F A F D G B C E

Of course, at a circular table, A will be next to the man at the end of the line.

I first gave this problem for six persons on ten days, in the _Daily Mail_ for the 13th and 16th October 1905, and it has since been discussed in various periodicals by mathematicians. Of course, it is easily seen that the maximum number of sittings for _n_ persons is (_n_ - 1)(_n_ -2)/2 ways. The comparatively easy method for solving all cases where _n_ is a prime+1 was first discovered by Ernest Bergholt. I then pointed out the form and construction of a solution that I had obtained for 10 persons, from which E. D. Bewley found a general method for all even numbers. The odd numbers, however, are extremely difficult, and for a long time no progress could be made with their solution, the only numbers that could be worked being 7 (given above) and 5, 9, 17, and 33, these last four being all powers of 2+1. At last, however (though not without much difficulty), I discovered a subtle method for solving all cases, and have written out schedules for every number up to 25 inclusive. The case of 11 has been solved also by W. Nash. Perhaps the reader will like to try his hand at 13. He will find it an extraordinarily hard nut.

The solutions for all cases up to 12 inclusive are given in _A. in M._, pp. 205, 206.

91.--_The Five Tea Tins._

There are twelve ways of arranging the boxes without considering the pictures. If the thirty pictures were all different the answer would be 93,312. But the necessary deductions for cases where changes of boxes may be made without affecting the order of pictures amount to 1,728, and the boxes may therefore be arranged, in accordance with the conditions, in 91,584 different ways. I will leave my readers to discover for themselves how the figures are to be arrived at.

92.--_The Four Porkers._

The number of ways in which the four pigs may be placed in the thirty-six sties in accordance with the conditions is seventeen, including the example that I gave, not counting the reversals and reflections of these arrangements as different. Jaenisch, in his _a.n.a.lyse Mathematique au jeu des echecs_ (1862), quotes the statement that there are just twenty-one solutions to the little problem on which this puzzle is based. As I had myself only recorded seventeen, I examined the matter again, and found that he was in error, and, doubtless, had mistaken reversals for different arrangements.

Here are the seventeen answers. The figures indicate the rows, and their positions show the columns. Thus, 104603 means that we place a pig in the first row of the _first_ column, in no row of the _second_ column, in the fourth row of the _third_ column, in the sixth row of the _fourth_ column, in no row of the _fifth_ column, and in the third row of the _sixth_ column. The arrangement E is that which I gave in diagram form:--

A. 104603 B. 136002 C. 140502 D. 140520 E. 160025 F. 160304 G. 201405 H. 201605 I. 205104 J. 206104 K. 241005 L. 250014 M. 250630 N. 260015 O. 261005 P. 261040 Q. 306104

It will be found that forms N and Q are semi-symmetrical with regard to the centre, and therefore give only two arrangements each by reversal and reflection; that form H is quarter-symmetrical, and gives only four arrangements; while all the fourteen others yield by reversal and reflection eight arrangements each. Therefore the pigs may be placed in (2 2) + (4 1) + (8 14) = 120 different ways by reversing and reflecting all the seventeen forms.

Three pigs alone may be placed so that every sty is in line with a pig, provided that the pigs are not forbidden to be in line with one another; but there is only one way of doing it (if we do not count reversals as different), as follows: 105030.

93.--_The Number Blocks._

Arrange the blocks so as to form the two multiplication sums 915 64 and 732 80, and the product in both cases will be the same: 58,560.

94.--_Foxes and Geese._

The smallest possible number of moves is twenty-two--that is, eleven for the foxes and eleven for the geese. Here is one way of solving the puzzle:

10--5 11--6 12--7 5--12 6--1 7--6 ---- ---- ---- ---- ---- ---- 1--8 2--9 3--4 8--3 9--10 4--9

12--7 1--8 6--1 7--2 8--3 ---- ---- ---- ---- ---- 3--4 10--5 9--10 4--11 5--12

Of course, the reader will play the first move in the top line, then the first move in the second line, then the second move in the top line, and so on alternately.

[Ill.u.s.tration]

In _A. in M._, p. 230, I have explained fully my "b.u.t.tons and string"

method of solving puzzles on chequered boards. In Diagram A is shown the puzzle in the form in which it may be presented on a portion of the chessboard with six knights. A comparison with the ill.u.s.tration on page 141 will show that I have there dispensed with the necessity of explaining the knight's move to the uninstructed reader by lines that indicate those moves. The two puzzles are the same thing in different dress. Now compare page 141 with Diagram B, and it will be seen that by disentangling the strings I have obtained a simplified diagram without altering the essential relations between the b.u.t.tons or discs. The reader will now satisfy himself without any difficulty that the puzzle requires eleven moves for the foxes and eleven for the geese. He will see that a goose on 1 or 3 must go to 8, to avoid being one move from a fox and to enable the fox on 11 to come on to the ring. If we play 1--8, then it is clearly best to play 10--5 and not 12--5 for the foxes. When they are all on the circle, then they simply promenade round it in a clockwise direction, taking care to reserve 8--3 and 5--12 for the final moves. It is thus rendered ridiculously easy by this method. See also notes on solutions to Nos. 13 and 85.

95.--_Robinson Crusoe's Table._

The diagram shows how the piece of wood should be cut in two pieces to form the square table-top. A, B, C, D are the corners of the table. The way in which the piece E fits into the piece F will be obvious to the eye of the reader. The shaded part is the wood that is discarded.