What are the solutions for peg solitaire with the following boards?

competition/games/peg.solitaire

Problem

What are the solutions for peg solitaire with the following boards?

1) What is the solution to peg solitaire on a triangular board (5 on a
side) with the central peg removed?

2) What is the full solution to peg-solitaire on a triangular board for
all possible opening positions?  Bonus: leave the last peg in the
originally empty hole.

3) What is the solution for the following board, where the goal is to
leave one peg in the center?

 **111**
 *11111*
 1111111
 1110111
 1111111
 *11111*
 **111**

Show solution

1) Number the triangle:

      1
     2 3
    4 5 6
   7 8 9 A
  B C D E F.

Remove 5 and make these moves:

  C-5,A-8,2-9,7-2,1-4,3-A,F-6,E-C,4-D,C-E,6-D,E-C,B-D.

It is not possible to start in 5 and end in 5 (or 8 or 9).

Here is a proof.  Label the board as follows:

     A
    A A
   A B A
  A B B A
 A A A A A

Let b(n)=(number of pegs on a square with color B after move number n) modulo 2.

Then b(n)=b(n-1), for all n.

But b(0)=0, so b(13)=0, which is not possible if 1 peg remains in 5.

2) There are four other opening positions: 1,2,4.

The following starting spaces are equivalent:        1, B, F.
The following starting spaces are equivalent:        2, 3, 7, A, C, E.
The following starting spaces are equivalent:        4, 6, D.
The following starting spaces are equivalent:        5, 8, 9.

Here are solutions that start and end in each opening position:

  1:4-1,D-4,7-2,B-D,E-C,6-D,C-E,F-D,1-6,A-3,2-9,D-6,6-1.

  2:7-2,D-4,B-D,E-C,6-D,C-E,F-D,1-6,A-3,3-8,2-7,D-4,7-2.

  4:1-4,7-2,6-1,1-4,D-6,A-3,F-D,C-E,4-D,E-C,B-D,3-8,D-4.

Counting equivalence under rotation and reflection:

If the starting space is 1, there are 14,880 distinct solutions.
If the starting space is 2, there are 14,880 distinct solutions.
If the starting space is 4, there are 42,629 distinct solutions.
If the starting space is 5, there are    775 distinct solutions.

Thus, there are a total of 438,984 solutions, of which 73,164 are distinct.
Many of these solutions consist of the same jumps, but in a different order.

3) There is no solution.

Mark the board with 3 colors like this:

    ABC
   ABCAB
  ABCABCA
  BCABCAB
  CABCABC
   BCABC
    ABC

Let a(n)=(number of pegs on a square with color A after move number n) modulo 2.
Let b(n)=(number of pegs on a square with color B after move number n) modulo 2.
Let c(n)=(number of pegs on a square with color C after move number n) modulo 2.

Then a(n)=1-a(n-1),
     b(n)=1-b(n-1),
     c(n)=1-c(n-1),
for all 0<=n<=35.

But a(0)=b(0)=c(0)=0, so a(35)=b(35)=c(35)=1, which is not possible, as
there is only 1 peg after 35 moves.

There are many other board configurations with known solutions.

The following archive entry contains references to the literature.

see also entry "hi-q"
see also URL http://rocketdownload.com/Details/Stra/pegsol.htm