Five new solitaire problems and their solutions

To save referring back to the statements of the problems, the five Soltaire problems are restated and followed by their solutions. The notes are by our guest editor John Beasley.

A hole initially full

A hole initially empty

A marked peg

A hole to be left full at the end

	a	b	c	d	e	f	g
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2
3
4
5
6
7

I am reluctant to make an unqualified claim to this, because “vacate d4, finish at a4 and g4” is a natural problem to try on the 37-hole board and it must have occurred to somebody to see if it could be done interchanging the pegs originally in these holes, but I haven't seen it anywhere else.

d2-d4, b2-d2, d1-d3, f2-d2, c4-c2, e3-c3, c2-c4, a3-c3, c4-c2, c1-c3, g3-e3, e4-e2, e1-e3, c6-c4, a5-c5, c4-c6, c7-c5, d5-b5, e6-e4, g5-e5, e4-e6, e7-e5, d7-d5-f5, after which we have

	a	b	c	d	e	f	g
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2
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6
7

and the rest is easy.

	a	b	c	d	e	f	g
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2
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7
8
9

This was discovered in the course of the investigation described above [in issue 28]. George's computer originally threw out a solution in 24 moves, my solution by hand took 23; a subsequent analysis by computer to find the shortest possible solution got the number down to 21.

I originally played d3-d1, d5-d3, b5-d5, c3-c5, c1-c3, c6-c4-c2, e5-c5, a6-c6, d6-b6, a4-a6-c6-c4, e1-c1-c3-c5, c8-c6-c4, b4-d4-d2, e7-e5, e9-e7, e4-e6-e8, g6-e6, d8-d6-f6, g4-g6-e6, c9-e9-e7-e5, e2-e4, f4-d4, f5-d5-d3-d1. This was the result of a detailed analysis of debts and surpluses using pencil and paper, and had George not told me that the problem was solvable I would have assumed it wasn't; indeed, at one point I was sure I had proved it. George's computer subsequently reduced the number of moves to 21 by playing d3-d1, d5-d3, f4-d4-d2, b5-d5, e6-e4, e3-e5 (6), c7-c5, c9-c7, b4-d4, e1-e3, c2-c4-c6-c8 (11), a6-c6, g6-e6, d6-f6, d8-d6-b6 (15), a4-a6-c6, e8-e6-e4-e2, e9-c9-c7-c5-e5 (18), g4-g6-e6-e4, c1-e1-e3-e5, f5-d5-d3-d1.

	a	b	c	d	e	f	g	h
1
2
3
4
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8

“Here is something I found with 63 poker chips and a chessboard.” John does it in 25 moves, only one more than the number immediately established as necessary by the 8×8 version of Merson's ‘region’ analysis.

f6-d6, c6-e6, f8-f6-d6, c8-c6-e6, a8-c8 (5), d8-b8, h8-f8-d8-d6-f6, g6-e6-e8, g4-g6-g8, a6-a8-c8 (10), e4-g4, h4-f4, c4-e4-g4, d2-d4, a4-a6-c6-c4-e4-e6 (15), b3-d3, c1-c3, a2-a4-c4-c2, a1-c1, d1-b1-b3 (20), f2-f4-f6-d6-d4-d2, f1-d1-d3-f3, h2-h4-f4-f2, h1-f1-f3-h3, h6-h8-f8-d8-b8-b6-b4-b2-d2-f2-h2-h4-h6. Move 8 (g6-e6-e8) is the one that is not an initial exit from one of the Merson regions.

	a	b	c	d	e	f	g	h	i
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9

“Can the 41 cell board be cleared in less than 12 moves? Probably. Is a longer sweep possible on this board? Don't know, it is possible to set up a 26 peg sweep, but not if you start with a single vacancy.”

e5-c7, c3-e5, f4-d6, g7-e5, h4-f6-d4 (5), c7-e5-g5, e9-c7, b6-d8, i5-g7-e9-c7, a5-c3-e5 (10), e1-c3, and we are set up for the sweep:

	a	b	c	d	e	f	g	h	i
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2
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4
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7
8
9

f2-h4-f4-f2-d2-f4-d4-d2-b4-b6-d4-d6-b6-d8-d6-f6-d8-f8-f6-h6-h4-f6-d4-b4.

	a	b	c	d	e	f
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2
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6

John's proof that 13 moves are required: each of the 12 Merson regions around the edge requires a first escape, and the first jump has to be by a centre peg. “It is so simple, maybe even a computer could do it! There could be a 16 peg sweep, 12 move game by starting with the vacancy somewhere else, but it is unlikely to be symmetrical.”

d4-b2, a1-c3 (2), b6-d4-b2, a3-a1-c3, a5-a3-c5, d6-b6, a6-c6 (7), f2-d4-b2, c1-a1-c3, e1-c1-e3, f4-f2, f1-f3 (12) and we are set up for the sweep:

	a	b	c	d	e	f
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6

f6-d6-b6-b4-d6-d4-b4-b2-d4-f6-f4-f2-d2-f4-d4-d2-b2. John uses a binumeric notation in order to bring out the symmetry.