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# The triangle puzzle

## Question

You've probably seen this puzzle. There are 15 holes in triangular array. (See sketch below.) The game begins with pegs in 14 of the holes. The play is to jump pegs over adjacent pegs, removing the "jumped" pegs afterward, as in checkers. The jump is made in a straight line. To make a jump, you need a contiguous group consisting of { peg1, peg2, hole } in a straight line. Peg1 ends up in the hole, and peg2 is removed. The object is to make 13 legal jumps and end up with a single peg.

This happens about 6% of the time. That is, about 94% of the time you get a configuration, with more than one peg remaining, that permits no further legal jumps. In some games the peg must end up in the original empty hole, and that happens only about 3% of the time. So, it's not a trivial puzzle.

This puzzle asks for something different, and easier: Lose as badly as possible. That is, select a location for the empty hole, and then find a sequence of moves that leaves the greatest number of pegs on the board where there are no more legal jumps.

It's simple enough to play, even without the game, by marking hole locations on a sheet of paper and using pennies.

Spoiler

Hint: you might first want to sketch patterns of pegs that have no further legal jumps. Then decide where to place the initial empty hole and how to make the jumps to get to the desired configuration.

As already stated, there are 15 holes. There are also 36 possible jumps. For convenience in writing sequences of jumps, they can be numbered, as follows:

Number the jumps like this:         and the holes
--------------------------->  o     -------------     1
So Jump #1 means the         / \      like this:
peg in hole #1 jumps        1   2     ---------->   2   3
over the peg in hole
#2 into the empty                                 4   5   6
hole #4.                 o         o
/ \       / \           7   8   9  10
Jump #18 is peg 7      3   4     5   6
over peg 8, into      7              13       11  12  13  14  15
hole 9.              /                 \
o-8       o      14-o
Holes 4, 6, 13     / \       / \       / \
begin 4 jumps;    9  10    11  12    15  16
the others      17        19    21        23
begin two.      /         /       \         \
o-18      o-20   22-o      24-o
There are
36 jumps.
25       27    29  30   33        35
/         /       \ /      \         \
o-26      o-28   31-o-32  34-o      36-o

With symmetries taken into account, the holes have four equivalence classes:

• Corners (1, 11, 15)
• Adjacent to corners (2, 3, 7, 10, 12, 14)
• Edge centers (4, 6, 13)
• Centers (5, 8, 9)

This means that there are just four distinct places for the empty hole to start a game: { 1 2 4 5 }. All other holes are symmetrically equivalent to one of these.

Just to be sure the numbering above is understood, here is a winning game of the normal type. Start with pegs in every hole except #1. (The top hole is empty.) Then make these jumps: { 7 14 2 17 23 27 34 26 30 6 35 14 7 }. If done correctly, the original hole #1 contains the final peg.

Enjoy.

## Recommended Posts

• 2

The two moves earlier solution:

Spoiler

Let the initial empty peg-hole be #4.
Jumps: {33,  4, 27,  21}
Remaining pegs: ten at {1, 2, 3, 4, 6, 7, 10, 11, 13, 15}.

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Spoiler

One of the solutions (due to symmetry), with the original empty peg-hole at #1 is with the jumps {7, 29, 24, 18, 15, 2}. This pattern of jumps will leave 8 pegs -- 3 in the third and 5 in the 5th (bottom) row of the triangle.

Edited by DejMar
correction of row identification
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On 5/31/2018 at 9:10 AM, DejMar said:
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One of the solutions (due to symmetry), with the original empty peg-hole at #1 is with the jumps {7, 29, 24, 18, 15, 2}. This pattern of jumps will leave 8 pegs -- 3 in the third and 5 in the 5th (bottom) row of the triangle.

Very nice!

There is another solution that ends two moves earlier, leaving pegs only on the edges.

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6 hours ago, DejMar said:

The two moves earlier solution:

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Yup.

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