213.—­TURKS AND RUSSIANS.

The main point is to discover the smallest possible number of Russians that there could have been.  As the enemy opened fire from all directions, it is clearly necessary to find what is the smallest number of heads that could form sixteen lines with three heads in every line.  Note that I say sixteen, and not thirty-two, because every line taken by a bullet may be also taken by another bullet fired in exactly the opposite direction.  Now, as few as eleven points, or heads, may be arranged to form the required sixteen lines of three, but the discovery of this arrangement is a hard nut.  The diagram at the foot of this page will show exactly how the thing is to be done.

[Illustration]

If, therefore, eleven Russians were in the positions shown by the stars, and the thirty-two Turks in the positions indicated by the black dots, it will be seen, by the lines shown, that each Turk may fire exactly over the heads of three Russians.  But as each bullet kills a man, it is essential that every Turk shall shoot one of his comrades and be shot by him in turn; otherwise we should have to provide extra Russians to be shot, which would be destructive of the correct solution of our problem.  As the firing was simultaneous, this point presents no difficulties.  The answer we thus see is that there were at least eleven Russians amongst whom there was no casualty, and that all the thirty-two Turks were shot by one another.  It was not stated whether the Russians fired any shots, but it will be evident that even if they did their firing could not have been effective:  for if one of their bullets killed a Turk, then we have immediately to provide another man for one of the Turkish bullets to kill; and as the Turks were known to be thirty-two in number, this would necessitate our introducing another Russian soldier and, of course, destroying the solution.  I repeat that the difficulty of the puzzle consists in finding how to arrange eleven points so that they shall form sixteen lines of three.  I am told that the possibility of doing this was first discovered by the Rev. Mr. Wilkinson some twenty years ago.

214.—­THE SIX FROGS.

Move the frogs in the following order:  2, 4, 6, 5, 3, 1 (repeat these moves in the same order twice more), 2, 4, 6.  This is a solution in twenty-one moves—­the fewest possible.

If n, the number of frogs, be even, we require (n squared + n)/2 moves, of which (n squared — n)/2 will be leaps and n simple moves.  If n be odd, we shall need ((n squared + 3n)/2) — 4 moves, of which (n squared — n)/2 will be leaps and 2n — 4 simple moves.

In the even cases write, for the moves, all the even numbers in ascending order and the odd numbers in descending order.  This series must be repeated 1/2n times and followed by the even numbers in ascending order once only.  Thus the solution for 14 frogs will be (2, 4, 6, 8, 10, 12, 14, 13, 11, 9, 7, 5, 3, 1) repeated 7 times and followed by 2, 4, 6, 8, 10, 12, 14 = 105 moves.