Part 27

27.--The Dyer's Puzzle.

The correct answer is 18,816 different ways. The general formula for six fleurs-de-lys for all squares greater than 2^{2} is simply this: Six times the square of the number of combinations of n things, taken three at a time, where n represents the number of fleurs-de-lys in the side of the square. Of course where n is even the remainders in rows and columns will be even, and where n is odd the remainders will be odd.

For further solution, see No. 358 in A. in M.

28.--The Great Dispute between the Friar and the Sompnour.

In this little problem we attempted to show how, by sophistical reasoning, it may apparently be proved that the diagonal of a square is of precisely the same length as two of the sides. The puzzle was to discover the fallacy, because it is a very obvious fallacy if we admit that the shortest distance between two points is a straight line. But where does the error come in?

Well, it is perfectly true that so long as our zigzag path is formed of "steps" parallel to the sides of the square that path must be of the same length as the two sides. It does not matter if you have to use the most powerful microscope obtainable; the rule is always true if the path is made up of steps in that way. But the error lies in the assumption that such a zigzag path can ever become a straight line. You may go on increasing the number of steps infinitely--that is, there is no limit whatever theoretically to the number of steps that can be made--but you can never reach a straight line by such a method. In fact it is just as much a "jump" to a straight line if you have a billion steps as it is at the very outset to pass from the two sides to the diagonal. It would be just as absurd to say we might go on dropping marbles into a basket until they become sovereigns as to say we can increase the number of our steps until they become a straight line. There is the whole thing in a nutshell.

29.--Chaucer's Puzzle.

The surface of water or other liquid is always spherical, and the greater any sphere is the less is its convexity. Hence the top diameter of any vessel at the summit of a mountain will form the base of the segment of a greater sphere than it would at the bottom. This sphere, being greater, must (from what has been already said) be less convex; or, in other words, the spherical surface of the water must be less above the brim of the vessel, and consequently it will hold less at the top of a mountain than at the bottom. The reader is therefore free to select any mountain he likes in Italy--or elsewhere!

30.--The Puzzle of the Canon's Yeoman.

The number of different ways is 63,504. The general formula for such arrangements, when the number of letters in the sentence is 2n + 1, and it is a palindrome without diagonal readings, is [4(2^n - 1)]^2.

I think it will be well to give here a formula for the general solution of each of the four most common forms of the diamond-letter puzzle. By the word "line" I mean the complete diagonal. Thus in A, B, C, and D, the lines respectively contain 5, 5, 7, and 9 letters. A has a non-palindrome line (the word being BOY), and the general solution for such cases, where the line contains 2n + 1 letters, is 4(2^n - 1). Where the line is a single palindrome, with its middle letter in the centre, as in B, the general formula is [4(2^n - 1)]^{2}. This is the form of the Rat-catcher's Puzzle, and therefore the expression that I have given above. In cases C and D we have double palindromes, but these two represent very different types. In C, where the line contains 4^n-1 letters, the general expression is 4^(2^{2n}-2). But D is by far the most difficult case of all.

I had better here state that in the diamonds under consideration (i.) no diagonal readings are allowed--these have to be dealt with specially in cases where they are possible and admitted; (ii.) readings may start anywhere; (iii.) readings may go backwards and forwards, using letters more than once in a single reading, but not the same letter twice in immediate succession. This last condition will be understood if the reader glances at C, where it is impossible to go forwards and backwards in a reading without repeating the first O touched--a proceeding which I have said is not allowed. In the case D it is very different, and this is what accounts for its greater difficulty. The formula for D is this:

(n+5)×2^{2n+2} + (2^{n+2}×(1×3×5×7 . . . . . (2n-1)) / n) - 2^{n+4} - 8 ]

where the number of letters in the line is 4n+1. In the example given there are therefore 400 readings for n = 2.

See also Nos. 256, 257, and 258 in A. in M.

Y YOY YOBOY YOY Y

B L LEL LEVEL LEL L

N NON NOOON NOONOON NOOON NON N

L LEL LEVEL LEVEVEL LEVELEVEL LEVEVEL LEVEL LEL L ]

31.--The Manciple's Puzzle.

The simple Ploughman, who was so ridiculed for his opinion, was perfectly correct: the Miller should receive seven pieces of money, and the Weaver only one. As all three ate equal shares of the bread, it should be evident that each ate 8/3 of a loaf. Therefore, as the Miller provided 15/3 and ate 8/3, he contributed 7/3 to the Manciple's meal; whereas the Weaver provided 9/3, ate 8/3, and contributed only 1/3. Therefore, since they contributed to the Manciple in the proportion of 7 to 1, they must divide the eight pieces of money in the same proportion.

PUZZLING TIMES AT SOLVAMHALL CASTLE

SIR HUGH EXPLAINS HIS PROBLEMS

The friends of Sir Hugh de Fortibus were so perplexed over many of his strange puzzles that at a gathering of his kinsmen and retainers he undertook to explain his posers.

"Of a truth," said he, "some of the riddles that I have put forth would greatly tax the wit of the unlettered knave to rede; yet will I try to show the manner thereof in such way that all may have understanding. For many there be who cannot of themselves do all these things, but will yet study them to their gain when they be given the answers, and will take pleasure therein."

32.--The Game of Bandy-Ball.

Sir Hugh explained, in answer to this puzzle, that as the nine holes were 300, 250, 200, 325, 275, 350, 225, 375, and 400 yards apart, if a man could always strike the ball in a perfectly straight line and send it at will a distance of either 125 yards or 100 yards, he might go round the whole course in 26 strokes. This is clearly correct, for if we call the 125 stroke the "drive" and the 100 stroke the "approach," he could play as follows:--The first hole could be reached in 3 approaches, the second in 2 drives, the third in 2 approaches, the fourth in 2 approaches and 1 drive, the fifth in 3 drives and 1 backward approach, the sixth in 2 drives and 1 approach, the seventh in 1 drive and 1 approach, the eighth in 3 drives, and the ninth hole in 4 approaches. There are thus 26 strokes in all, and the feat cannot be performed in fewer.

33.--Tilting at the Ring.