solitaire
and Groups
2011–11–07
Do they play nicely?
Do they talk to each other?
And if they do, what do they say to each other?
classical
When it comes to Group Theory and an English board, in standard position, there is, an elegant proof that there are only five single-peg finishes [to which I’ve lost the link].
Let’s have my look at it shall we? The argument runs thusly: say we colour the board like so, using a, b and c as our colours:
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | |
|---|---|---|---|---|---|---|---|
| 0 | - | - | a | b | c | - | - |
| 1 | - | - | b | c | a | - | - |
| 2 | a | b | c | a | b | c | a |
| 3 | b | c | a | b | c | a | b |
| 4 | c | a | b | c | a | b | c |
| 5 | - | - | c | a | b | - | - |
| 6 | - | - | a | b | c | - | - |
Supposing a binary operation, *, on some set {a, b, c} then we have a Group and we can draw up a Caley table for the said same.
| * | e | a | b | c |
|---|---|---|---|---|
| e | e | a | b | c |
| a | a | e | c | b |
| b | b | c | e | a |
| c | c | b | a | e |
This group is Abelian [symmetric about the leading diagonal and hence commutative] which makes things slightly easier for us.
So let’s remove [3, 3], to obtain the standard position, then we have a board that sums to b, an invariant—that is the essential argument of [the proof to which I cannot link.]. But what does invariant, in this sense, mean exactly?
Is it the sum [that could be product of course, you know Group Theory] of the holes what matters? Let’s play a couple of moves from the standard position and have:
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | |
|---|---|---|---|---|---|---|---|
| 0 | - | - | • | • | • | - | - |
| 1 | - | - | • | c | • | - | - |
| 2 | • | • | • | • | b | c | • |
| 3 | • | • | • | • | • | • | • |
| 4 | • | • | • | • | • | • | • |
| 5 | - | - | • | • | • | - | - |
| 6 | - | - | • | • | • | - | - |
For c * b * c = (c * b) * c = a * c = b; invariant, yes? That’s ok then? [This does work.]
But let’s, at least, try it the other way round. Consider the [reverse of table 3] following:
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | |
|---|---|---|---|---|---|---|---|
| 0 | - | - | a | b | c | - | - |
| 1 | - | - | b | • | a | - | - |
| 2 | a | b | c | a | • | • | a |
| 3 | b | c | a | b | c | a | b |
| 4 | c | a | b | c | a | b | c |
| 5 | - | - | c | a | b | - | - |
| 6 | - | - | a | b | c | - | - |
There are eleven a, ten b and nine c; and we have an Abelian Group, hence [diddle this how you will] the result of the sum is e. Which doesn’t say too much, but then we didn’t really expect anything, did we?
But this business of only having five pegs that we can end up with? Why can’t we end up at [1,2]? That’s a b isn’t it?
clump—redux
In my last post [on solitaire] I introduced the concept of a clump. I also hinted that the clump that I’d used had some interesting properties, now it’s time for me to front-up. So here we go:
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | |
|---|---|---|---|---|---|---|---|
| 0 | - | - | a | b | c | - | - |
| 1 | - | - | • | • | • | - | - |
| 2 | a | • | • | • | • | • | a |
| 3 | b | • | • | b | • | • | b |
| 4 | c | • | • | • | • | • | c |
| 5 | - | - | • | • | • | - | - |
| 6 | - | - | a | • | c | - | - |
- [4,3]→[4,0]
- [4,0]→[4,2]
- [6,3]→[4,3]
- [4,4]→[4,6]
- [2,4]→[4,4]
- [5,4]→[3,4]
- [2,5]→[4,5]
- [4,6]→[4,4]
- [2,2]→[0,2]
- [1,4]→[1,2]
- [0,2]→[2,2]
- [3,1]→[3,3]
- [3,3]→[1,3]
- [2,1]→[2,3]
- [5,2]→[3,2]
- [1,3]→[3,3]
- [4,3]→[2,3]
- [4,4]→[2,4]
- [2,4]→[2,2]
- [2,2]→[4,2]
Notice that, considering the holes, we have an e board, what’s going on here? I suggest that you calculate the reverse, and play through the game and think about what’s happening.
Next time we will have a look at that. Also you should ask yourself whether I have made any logical errors in the above. I have big-time, let’s see if you can spot them. Didn’t I say that this might be hard? Sorry, I meant to.