Page:Amusements in mathematics.djvu/205

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, I (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 ${\frac {n^{2}+n}{2}}$ moves, of which will be leaps and $n$ simple moves. If $n$ be odd, we shall need ${\frac {n^{2}+3n}{2}}-4$ moves, of which ${\frac {{n^{2}}-n}{2}}$ 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 ${\frac {1}{2}}n$ 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.

In the odd cases, write the even numbers in ascending order and the odd numbers in descending order, repeat this series ${\frac {1}{2}}(n-1)$ times, follow with the even numbers in ascending order (omitting $n-1$), the odd numbers in descending order (omitting 1), and conclude with all the numbers (odd and even) in their natural order (omitting 1 and $n$). Thus for 11 frogs : (2, 4, 6, 8, 10, 11, 9, 7, 5, 3, 1) repeated 5 times, 2, 4, 6, 8, 11, 9, 7, 5, 3, and 2, 3, 4, 5, 6, 7, 8, 9, 10=73 moves.

This complete general solution is published here for the first time.

215.—THE GRASSHOPPER PUZZLE.

Move the counters in the following order. The moves in brackets are to be made four times in succession. 12, 1, 3, 2, 12, 11, 1, 3, 2 (5, 7, 9, 10, 8, 6, 4), 3, 2, 12, 11, 2, I, 2. The grasshoppers will then be reversed in forty-four moves.

The general solution of this problem is very difficult. Of course it can always be solved by the method given in the solution of the last puzzle, if we have no desire to use the fewest possible moves. But to employ a full economy of moves we have two main points to consider. There are always what I call a lower movement (L) and an upper movement (U). L consists in exchanging certain of the highest numbers, such as 12, 11, 10 in our "Grasshopper Puzzle," with certain of the lower numbers, 1, 2, 3; the former moving in a clockwise direction, the latter in a non-clockwise direction. U consists in reversing the intermediate counters. In the above solution for 12, it will be seen that 12, 11, and 1, 2, 3 are engaged in the L movement, and 4, 5, 6, 7, 8, 9, 10 in the U movement. The L movement needs 16 moves and U 28, making together 44. We might also involve 10 in the L movement, which would result in L 23, U 21, making also together 44 moves. These I call the first and second methods. But any other scheme will entail an increase of moves. You always get these two methods (of equal economy) for odd or even counters, but the point is to determine just how many to involve in L and how many in U. Here is the solution in table form. But first note, in giving values to $n$, that 2, 3, and 4 counters are special cases, requiring respectively 3, 3, and 6 moves, and that 5 and 6 counters do not give a minimum solution by the second method—only by the first.

First Method.

Total No. of Counters.	L Movement.		U Movement.		Total No. of Moves.
	No. of Counters.	No. of Moves.	No. of Counters.	No. of Moves.
$4n$	$n-1$ and $n$	$2(n-1)^{2}+5n-7$	$2n+1$	$2n^{2}+3n+1$	$4(n^{2}+n-1)$
$4n-2$	$n-1$ and„ $n$	$2(n-1)^{2}+5n-7$	$2n-1$	$2(n-1)^{2}+3n-2$	$4n^{2}-5$
$4n+1$	$n$ and„ $n+1$	$2(n^{2}+5n-2$	$2n$	$2n^{2}+3n-4$	$2(2n^{2}+4n-3)$
$4n-1$	$n-1$ and„ $n$	$2(n-1)^{2}+5n-7$	$2n$	Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikisource.org/v1/":): {\displaystyle 2n^2+3n-4}	$4n^{2}+4n-9$

First Method.

Total No. of Counters.	L Movement.		U Movement.		Total No. of Moves.
	No. of Counters.	No. of Moves.	No. of Counters.	No. of Moves.
$4n$	$n$ and $n$	$2n^{2}+3n-4$	$2n$	$2(n-1)^{2}+5n-2$	$4(n^{2}+n-1)$
$4n-2$	$n-1$ and $n-1$	$2(n-1)^{2}+3n-7$	$2n$	$2(n-1)^{2}+5n-2$	$4n^{2}-5$
$4n+1$	$n$ and„ $n$	$2n^{2}+3n-4$	$2n+1$	$2n^{2}+5n-2$	$2(2n^{2}+4n-3)$
$4n-1$	$n$ and„ $n$	$2n^{2}+3n-4$	$2n-1$	$2(n-1)^{2}+5n-7$	$4n^{2}+4n-9$