# An interesting mathematical result, but is there a simpler way?

I came across this article on the Math Plus Web site, which I recommend for those with an interest in math but who are not necessarily proficient in it. The article gives an introduction to group theory and then explains how the Klein 4 group can be used to give information about final outcomes in peg solitaire.

I thought that was a nice result, but it seems to me that the same conclusions, at least about possible outcomes with one marble left, can be arrived at with symbols 0 and 1 and adding mod 2. I tried it and had a 0 in the center and an even number of 1’s in the other positions, giving a sum of 0. That means that it is not possible to end up with a single marble left in a hole labeled 1. The same conclusion as in the article can be arrived at by using the same type of symmetry arguments.

If I am right, then this could easily be used at a high school level for those who are taught modular arithmetic.

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I’m sorry. It was a lot to read there and I got bored. What’s the problem again. [No I’m not sarcastic and I don’t want to be rude but the font there is small and it’s a lot to read].

Hibernate (9038)

If that’s high school maths then I must be completely retarded. I didn’t understand any of it.

downtide (23456)

If we forget about the article, here is my variation. Imagine we start with a peg solitaire board and number four holes with ones and zeros, for example:
10
01

To get the values for the other holes, just add the values of the adjacent ones mod 2, so we would get:
101
011
110
and so on.

Now here is the trick that the article talks about. Because each hole is the sum of the two adjacent ones, every time there is a jump, the sum of the values of the occupied holes remains unchanged. For example if two occupied holes have 0 and 1, the adjacent hole has value 1. The jump replaces the pegs in 1 and 0 with one in 1. 0 + 1 =1, so the total of all occupied holes remains unchanged. If the occupied holes have values of 1 and 1, the adjacent hole has value 0. The pegs in holes with values of 1 and 1 are replaced with a peg in a hole with value 0, and again the total value is unchanged.

Now suppose that the numbering scheme ends up placing a 0 in the center. You can force this by starting the numbering from the center. What you find is that the remaining holes have an even number of 1’s, so the total is 0 mod 2. The sum of the occupied holes will always add to 0. If we have one peg remaining, it must be in a hole labeled 0, meaning that it is impossible to end up in any of the holes labeled 1.

I hope that clarifies things. I do think this could be explained to those with high school level math.

Yeah—that’s very nice. I think that you can do it with ones and zeroes, as you describe. That strikes me as much simpler.

This problem reminds me a bit of the problem of tiling an mxn board with triominoes (l-shaped or straight), if you’re allowed to cover a single square first. Which single squares can you cover?

The breakthrough in that problem comes similarly, with a choice of coloring. It’s a little less motile, though.

Do you know another one it reminds me of? This one.

finkelitis (1907)

I have seen that solitaire example before. It is due to the mathematician John Conway, who you have probably heard of.