The Canterbury Puzzles Part 25

48.--_The Riddle of the Frogs' Ring._

The fewest possible moves in which this puzzle can be solved are 118. I will give the complete solution. The black figures on white discs move in the directions of the hands of a clock, and the white figures on black discs the other way. The following are the numbers in the order in which they move. Whether you have to make a simple move or a leaping move will be clear from the position, as you never can have an alternative. The moves enclosed in brackets are to be played five times over: 6, 7, 8, 6, 5, 4, 7, 8, 9, 10, 6, 5, 4, 3, 2, 7, 8, 9, 10, 11 (6, 5, 4, 3, 2, 1), 6, 5, 4, 3, 2, 12, (7, 8, 9, 10, 11, 12), 7, 8, 9, 10, 11, 1, 6, 5, 4, 3, 2, 12, 7, 8, 9, 10, 11, 6, 5, 4, 3, 2, 8, 9, 10, 11, 4, 3, 2, 10, 11, 2. We thus have made 118 moves within the conditions, the black frogs have changed places with the white ones, and 1 and 12 are side by side in the positions stipulated.

The general solution in the case of this puzzle is 3_n_^{2} + 2_n_ - 2 moves, where the number of frogs of each colour is _n_. The law governing the sequence of moves is easily discovered by an examination of the simpler cases, where _n_ = 2, 3, and 4.

If, instead of 11 and 12 changing places, the 6 and 7 must interchange, the expression is _n_^{2} + 4_n_ + 2 moves. If we give _n_ the value 6, as in the example of the Frogs' Ring, the number of moves would be 62.

For a general solution of the case where frogs of one colour reverse their order, leaving the blank s.p.a.ce in the same position, and each frog is allowed to be moved in either direction (leaping, of course, over his own colour), see "The Gra.s.shopper Puzzle" in _A. in M._, p. 193.

THE STRANGE ESCAPE OF THE KING'S JESTER

Although the king's jester promised that he would "thereafter make the manner thereof plain to all," there is no record of his having ever done so. I will therefore submit to the reader my own views as to the probable solutions to the mysteries involved.

49.--_The Mysterious Rope._

When the jester "divided his rope in half," it does not follow that he cut it into two parts, each half the original length of the rope. No doubt he simply untwisted the strands, and so divided it into two ropes, each of the original length, but one-half the thickness. He would thus be able to tie the two together and make a rope nearly twice the original length, with which it is quite conceivable that he made good his escape from the dungeon.

50.--_The Underground Maze._

How did the jester find his way out of the maze in the dark? He had simply to grope his way to a wall and then keep on walking without once removing his left hand (or right hand) from the wall. Starting from A, the dotted line will make the route clear when he goes to the left. If the reader tries the route to the right in the same way he will be equally successful; in fact, the two routes unite and cover every part of the walls of the maze except those two detached parts on the left-hand side--one piece like a U, and the other like a distorted E. This rule will apply to the majority of mazes and puzzle gardens; but if the centre were enclosed by an isolated wall in the form of a split ring, the jester would simply have gone round and round this ring.

See the article, "Mazes, and How to Thread Them," _in A. in M._

51.--_The Secret Lock._

This puzzle entailed the finding of an English word of three letters, each letter being found on a different dial. Now, there is no English word composed of consonants alone, and the only vowel appearing anywhere on the dials is Y. No English word begins with Y and has the two other letters consonants, and all the words of three letters ending in Y (with two consonants) either begin with an S or have H, L, or R as their second letter. But these four consonants do not appear. Therefore Y must occur in the middle, and the only word that I can find is "PYX," and there can be little doubt that this was the word. At any rate, it solves our puzzle.

52.--_Crossing the Moat._

No doubt some of my readers will smile at the statement that a man in a boat on smooth water can pull himself across with the tiller rope! But it is a fact. If the jester had fastened the end of his rope to the stern of the boat and then, while standing in the bows, had given a series of violent jerks, the boat would have been propelled forward. This has often been put to a practical test, and it is said that a speed of two or three miles an hour may be attained. See W. W. Rouse Ball's _Mathematical Recreations_.

53.--_The Royal Gardens._

[Ill.u.s.tration]

This puzzle must have struck many readers as being absolutely impossible.

The jester said: "I had, of a truth, entered every one of the sixteen gardens once, and never more than once." If we follow the route shown in the accompanying diagram, we find that there is no difficulty in once entering all the gardens but one before reaching the last garden containing the exit B. The difficulty is to get into the garden with a star, because if we leave the B garden we are compelled to enter it a second time before escaping, and no garden may be entered twice. The trick consists in the fact that you may enter that starred garden without necessarily leaving the other. If, when the jester got to the gateway where the dotted line makes a sharp bend, his intention had been to hide in the starred garden, but after he had put one foot through the doorway, upon the star, he discovered it was a false alarm and withdrew, he could truly say: "I entered the starred garden, because I put my foot and part of my body in it; and I did not enter the other garden twice, because, after once going in I never left it until I made my exit at B." This is the only answer possible, and it was doubtless that which the jester intended.

See "The Languis.h.i.+ng Maiden," in _A. in M._

54.--_Bridging the Ditch._

[Ill.u.s.tration]

The solution to this puzzle is best explained by the ill.u.s.tration. If he had placed his eight planks, in the manner shown, across the angle of the ditch, he would have been able to cross without much trouble. The king's jester might thus have well overcome all his difficulties and got safely away, as he has told us that he succeeded in doing.

THE SQUIRE'S CHRISTMAS PUZZLE PARTY

_HOW THE VARIOUS TRICKS WERE DONE_

The record of one of Squire Davidge's annual "Puzzle Parties," made by the old gentleman's young lady relative, who had often spent a merry Christmas at Stoke Courcy Hall, does not contain the solutions of the mysteries. So I will give my own answers to the puzzles and try to make them as clear as possible to those who may be more or less novices in such matters.

55.--_The Three Teacups._

[Ill.u.s.tration]

Miss Charity Lockyer clearly must have had a trick up her sleeve, and I think it highly probable that it was conceived on the following lines.

She proposed that ten lumps of sugar should be placed in three teacups, so that there should be an odd number of lumps in every cup. The ill.u.s.tration perhaps shows Miss Charity's answer, and the figures on the cups indicate the number of lumps that have been separately placed in them. By placing the cup that holds one lump inside the one that holds two lumps, it can be correctly stated that every cup contains an odd number of lumps. One cup holds seven lumps, another holds one lump, while the third cup holds three lumps. It is evident that if a cup contains another cup it also contains the contents of that second cup.

There are in all fifteen different solutions to this puzzle. Here they are:--

1 0 9 1 4 5 9 0 1 3 0 7 7 0 3 7 2 1 1 2 7 5 2 3 5 4 1 5 0 5 3 4 3 3 6 1 3 2 5 1 6 3 1 8 1

The first two numbers in a triplet represent respectively the number of lumps to be placed in the inner and outer of the two cups that are placed one inside the other. It will be noted that the outer cup of the pair may itself be empty.