The Riddle Of The Frogs' Ring
(
THE STRANGE ESCAPE OF THE KING'S JESTER)
One Christmas the Abbot offered a prize of a large black jack mounted in silver, to be engraved with the name of the monk who should put forth the best new riddle. This tournament of wit was won by Brother Benedict, who, curiously enough, never before or after gave out anything that did not excite the ridicule of his brethren. It was called the "Frogs' Ring."
A ring was made with chalk on the floor of the hall, and divided into thirteen compartments, in which twelve discs of wood (called "frogs") were placed in the order shown in our illustration, one place being left vacant. The numbers 1 to 6 were painted white and the numbers 7 to 12 black. The puzzle was to get all the white numbers where the black ones were, and vice versa. The white frogs move round in one direction, and the black ones the opposite way. They may move in any order one step at a time, or jumping over one of the opposite colour to the place beyond, just as we play draughts to-day. The only other condition is that when all the frogs have changed sides, the 1 must be where the 12 now is and the 12 in the place now occupied by 1. The puzzle was to perform the feat in as few moves as possible. How many moves are necessary?
I will conclude in the words of the old writer: "These be some of the riddles which the monks of Riddlewell did set forth and expound each to the others in the merry days of the good Abbot David."
Answer:
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 3n2 + 2n - 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 n2 + 4n + 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 space in the same position, and each frog is allowed to be moved in either direction (leaping, of course, over his own colour), see "The Grasshopper Puzzle" in A. in M., p. 193.
Informational
