Henry Ernest Dudeney/Modern Puzzles/Moving Counter Problems

Henry Ernest Dudeney: Modern Puzzles: Moving Counter Problems

The puzzle is to remove all but one counter by a succession of leaps.

A counter can leap over another adjoining it to the next square beyond, if vacant,

and in making the leap you remove the one jumped over.

But no leap may be made in a diagonal direction.

The following is a solution in eight moves:

$5 - 13$, $(6 - 14, 6 - 5)$, $16 - 15$, $(3 - 13, 3 - 6)$, $2 - 10$, $(8 - 7, 8 - 16, 8 - 3)$, $(1 - 9, 1 - 2, 1 - 8)$, $(4 - 12, 4 - 1)$

This means that $5$ leaps over $13$ and $13$ is removed, then $6$ leaps over $14$ and $14$ is removed, and so on.

The leaps within a bracket count as one move, because the leaps are made with the same counter in succession.

It will be seen that No. $4$ makes the last leap.

Now try to find a solution, in seven moves, in which No. $1$ makes the last leap.

There are forty-nine fishing-boats in the North Sea.

How could an enemy ram and sink the lot in twelve straight courses,

starting at $A$ and finishing up at the same place?

Make a rough board, as shown, and place seventeen counters on the squares indicated.

The puzzle is to remove all but one by a series of leaping moves, as in draughts or solitaire.

A counter can be made to leap over another to the next square beyond, if vacant, and you then remove the one jumped over.

It will be seen that the first leap must be made by the central counter, No. $9$, and one has the choice of $8$ directions.

A continuous series of leaps with the same counter will count as a single move.

It is required to take off $26$ counters in $4$ moves, leaving the No. $9$ on its original centre square.

Every play must be a leap.

Divide a sheet of paper into six compartments, as shown in the diagram,

and place a pile of $15$ counters, numbered consecutively $1$, $2$, $3$, $\ldots$, $15$ downwards, in compartment $A$.

The puzzle is to transfer the complete pile, in the fewest possible moves, to compartment $F$.

You can move the counters one at a time to any compartment,

but may never place a counter on one that bears a smaller number than itself.

Thus, if you place $1$ on $B$ and $2$ on $C$, you can then place $1$ on $2$, but not $2$ on $1$.