# Solution

(related to Problem: The Victoria Cross Puzzle)

In solving this puzzle there were two things to be achieved: first, so to manipulate the counters that the word VICTORIA should read round the cross in the same direction, only with the $V$ on one of the dark arms; and secondly, to perform the feat in the fewest possible moves. Now, as a matter of fact, it would be impossible to perform the first part in any way whatever if all the letters of the word were different; but as there are two $I$'s, it can be done by making these letters change places — that is, the first $I$ changes from the 2nd place to the 7th, and the second $I$ from the 7th place to the 2nd. But the point $I$ referred to, when introducing the puzzle, as a little remarkable is this: that a solution in twenty-two moves is obtainable by moving the letters in the order of the following words: "A VICTOR! A VICTOR! A VICTOR I!"

There are, however, just six solutions in eighteen moves, and the following is one of them: $I (1),$ $V, A,$ $I (2),$ $R, O,$ $T, I$ $(1), I$ $(2),$ $A, V,$ $I (2),$ $I (1),$ $C, I$ $(2), V,$ $A, I (1).$ The first and second I in the word are distinguished by the numbers $1$ and $2.$

It will be noticed that in the first solution given above one of the $I$'s never moves, though the movements of the other letters cause it to change its relative position. There is another peculiarity I may point out — that there is a solution in twenty-eight moves requiring no letter to move to the central division except the $I$'s. I may also mention that, in each of the solutions in eighteen moves, the letters $C, T, O, R$ move once only, while the second $I$ always moves four times, the $V$ always being transferred to the right arm of the cross.

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### References

#### Project Gutenberg

1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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