NINE JOLLY GAOL BIRDS.
(
Magic Squares Problem.)
Shortly after the episode recorded in the last puzzle occurred, a ninth
prisoner was placed in the vacant cell, and the merry monarch then
offered them all complete liberty on the following strange conditions.
They were required so to rearrange themselves in the cells that their
numbers formed a magic square without their movements causing any two of
them ever to be in the same cell together, except that at the start one
man was allowed to be placed on the shoulders of another man, and thus
add their numbers together, and move as one man. For example, No. 8
might be placed on the shoulders of No. 2, and then they would move
about together as 10. The reader should seek first to solve the puzzle
in the fewest possible moves, and then see that the man who is burdened
has the least possible amount of work to do.
Answer:
There is a pitfall set for the unwary in this little puzzle. At the
start one man is allowed to be placed on the shoulders of another, so as
to give always one empty cell to enable the prisoners to move about
without any two ever being in a cell together. The two united prisoners
are allowed to add their numbers together, and are, of course, permitted
to remain together at the completion of the magic square. But they are
obviously not compelled so to remain together, provided that one of the
pair on his final move does not break the condition of entering a cell
already occupied. After the acute solver has noticed this point, it is
for him to determine which method is the better one--for the two to be
together at the count or to separate. As a matter of fact, the puzzle
can be solved in seventeen moves if the men are to remain together; but
if they separate at the end, they may actually save a move and perform
the feat in sixteen! The trick consists in placing the man in the centre
on the back of one of the corner men, and then working the pair into the
centre before their final separation.
[Illustration:
A B
+---+---+---+ +---+---+---+
| | | | | | | |
| 2 9 4 | | 6 7 2 |
| | | | | | | |
+- -+- -+- -+ +- -+- -+- -+
| | | | | | | |
| 7 5 3 | | 1 5 9 |
| | | | | | | |
+- -+- -+- -+ +- -+- -+- -+
| | | | | | | |
| 6 1 8 | | 8 3 4 |
| | | | | | | |
+---+---+---+ +---+---+---+
]
Here are the moves for getting the men into one or other of the above
two positions. The numbers are those of the men in the order in which
they move into the cell that is for the time being vacant. The pair is
shown in brackets:--
Place 5 on 1. Then, 6, 9, 8, 6, 4, (6), 2, 4, 9, 3, 4, 9, (6), 7, 6, 1.
Place 5 on 9. Then, 4, 1, 2, 4, 6, (14), 8, 6, 1, 7, 6, 1, (14), 3, 4,
9.
Place 5 on 3. Then, 6, (8), 2, 6, 4, 7, 8, 4, 7, 1, 6, 7, (8), 9, 4, 3.
Place 5 on 7. Then, 4, (12), 8, 4, 6, 3, 2, 6, 3, 9, 4, 3, (12), 1, 6,
7.
The first and second solutions produce Diagram A; the second and third
produce Diagram B. There are only sixteen moves in every case. Having
found the fewest moves, we had to consider how we were to make the
burdened man do as little work as possible. It will at once be seen that
as the pair have to go into the centre before separating they must take
at fewest two moves. The labour of the burdened man can only be reduced
by adopting the other method of solution, which, however, forces us to
take another move.
