THE GRASSHOPPER PUZZLE.
(
Moving Counter Problem)
It has been suggested that this puzzle was a great favourite among the
young apprentices of the City of London in the sixteenth and seventeenth
centuries. Readers will have noticed the curious brass grasshopper on
the Royal Exchange. This long-lived creature escaped the fires of 1666
and 1838. The grasshopper, after his kind, was the crest of Sir Thomas
Gresham, merchant grocer, who died in 1579, and from this cause it has
been used as a sign by grocers in general. Unfortunately for the legend
as to its origin, the puzzle was only produced by myself so late as the
year 1900. On twelve of the thirteen black discs are placed numbered
counters or grasshoppers. The puzzle is to reverse their order, so that
they shall read, 1, 2, 3, 4, etc., in the opposite direction, with the
vacant disc left in the same position as at present. Move one at a time
in any order, either to the adjoining vacant disc or by jumping over one
grasshopper, like the moves in draughts. The moves or leaps may be made
in either direction that is at any time possible. What are the fewest
possible moves in which it can be done?
Answer:
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, 1, 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.| L MOVEMENT. | U MOVEMENT. | |
| of +-------------+-------------+----------+------------+ Total No. |
| Counters.| No. of | No. of | No. of | No. of | of Moves. |
| | Counters. | Moves. |Counters. | Moves. | |
+----------+-------------+-------------+----------+------------+-----------+
| 4n | n-1 and n |2(n-1) squared+5n-7 | 2n+1 |2n squared+3n+1 |4(n squared+n-1) |
| 4n-2 | n-1 " n |2(n-1) squared+5n-7 | 2n-1 |2(n-1) squared+3n-2|4n squared-5 |
| 4n+1 | n " n+1 |2n squared+5n-2 | 2n |2n squared+3n-4 |2(2n squared+4n-3)|
| 4n-1 | n-1 " n |2(n-1) squared+5n-7 | 2n |2n squared+3n-4 |4n squared+4n-9 |
+----------+-------------+-------------+----------+------------+-----------+
SECOND METHOD.
+---------+--------------------------+-------------------------+-----------+
|Total No.| L MOVEMENT. | U MOVEMENT. | |
| of +-------------+------------+----------+--------------+ Total No. |
|Counters.| No. of | No. of | No. of | No. of | of Moves. |
| | Counters. | Moves. | Counters.| Moves. | |
+---------+-------------+------------+----------+--------------+-----------+
| 4n | n and n |2n squared+3n-4 | 2n | 2(n-1) squared+5n-2 |4(n squared+n-1) |
| 4n-2 | n-1 " n-1 |2(n-1) squared+3n-7| 2n | 2(n-1) squared+5n-2 |4n squared-5 |
| 4n+1 | n " n |2n squared+3n-4 | 2n+1 | 2n squared+5n-2 |2(2n squared+4n-3)|
| 4n-1 | n " n |2n squared+3n-4 | 2n-1 | 2(n-1) squared+5n-7 |4n squared+4n-9 |
+---------+-------------+------------+----------+--------------+-----------+
More generally we may say that with m counters, where m is even and
greater than 4, we require (m squared + 4m - 16)/4 moves; and where m is odd
and greater than 3, (m squared + 6m - 31)/4 moves. I have thus shown the
reader how to find the minimum number of moves for any case, and the
character and direction of the moves. I will leave him to discover for
himself how the actual order of moves is to be determined. This is a
hard nut, and requires careful adjustment of the L and the U
movements, so that they may be mutually accommodating.
