The Dyer's Puzzle
(
CANTERBURY PUZZLES)
One of the pilgrims was a Dyer, but Chaucer tells us nothing about him, the Tales being incomplete. Time after time the company had pressed this individual to produce a puzzle of some kind, but without effect. The poor fellow tried his best to follow the examples of his friends the Tapiser, the Weaver, and the Haberdasher; but the necessary idea would not come, rack his brains as he would. All things, however, come to those who wait—and persevere—and one morning he announced, in a state of considerable excitement, that he had a poser to set before them. He brought out a square piece of silk on which were embroidered a number of fleurs-de-lys in rows, as shown in our illustration.
"Lordings," said the Dyer, "hearken anon unto my riddle. Since I was awakened at dawn by the crowing of cocks—for which din may our host never thrive—I have sought an answer thereto, but by St. Bernard I have found it not. There be sixty-and-four flowers-de-luce, and the riddle is to show how I may remove six of these so that there may yet be an even number of the flowers in every row and every column."
The Dyer was abashed when every one of the company showed without any difficulty whatever, and each in a different way, how this might be done. But the good Clerk of Oxenford was seen to whisper something to the Dyer, who added, "Hold, my masters! What I have said is not all. Ye must find in how many different ways it may be done!" All agreed that this was quite another matter. And only a few of the company got the right answer.
Answer:
The correct answer is 18,816 different ways. The general formula for six fleurs-de-lys for all squares greater than 22 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.
