Solomon's Stone's is a simple abstract game from Solbenk. They sent me a copy to evaluate and review, which I did. My take was that it looked like a nice game for casual players and non-gamers, but held more interest as a puzzle for gamers.
On your turn, take any number of stones from a single row or column. Last person to take a stone loses.
Our first idea was to look at smaller triangles of size N. I.e.
(N = 1)
(N = 2)
(N = 3)
and so on, trying to solve the simpler patterns. Clearly, if a game of size N is a win for the second player, then a game of size N+1 is a win for first player, who removes the largest row or column, thereby reducing the game to N.
Game N = 1 is a win for second player.
Game N = 2 is a win for first player, who reduces the game to N = 1.
Game N = 3 is a win for first player, who takes stones A1 + C1 (top and bottom left on above diagram).
This didn't seem to be forming any patterns; N = 4 was not immediately obvious. One of my group members claims to have solved N = 4 and N = 5 as first player wins, but I didn't hear any more on the subject. That was a month ago.
The publisher wrote to me that some people tried to solve the game using symmetrical moves, but that this wasn't ultimately successful.
I thought a bit more about it this weekend.
Let's start with some new terminology. A(N) is an arrangement of N pieces on a Solomon's Stone board such that all N pieces are on different rows and columns. E.g. for A(7) there is only a single possible arrangement:
The stone in row 1 must be in column 1. The stone in row 2 must be in any column that the stone in row 1 doesn't occupy, which must be column 2. And so on.
For A(6), there are 127 possible arrangements:
I. There is one solution for A(6) if the stones occupy the first six rows, and it looks similar to the above diagram.
II. There are two solutions for A(6) if the stones occupy rows 1 through 5 and row 7. Row 7's stone can be placed either in column 6 or column 7.
III. There are four possible solutions for A(6) if the stones occupy rows 1 through 4 and rows 6 and 7. If row 6's stone is place in column 5, row 7's stone can be placed in column 6 or 7. If row 6's stone is placed in column 6, row 7's stone can be placed in column 5 or 7.
The pattern is now obvious. For the shape:
The first row has two possible choices. For either choice, the next row has two possible choices. And so on, leading to 2^N possible arrangements.
Continuing from above,
1 + 2 + 4 + 8 + 16 + 32 + 64 = 127 (or simply 2^(N+1)-1)
First player loses for A(odd), while second player loses for A(even). One winning strategy is therefore to force your opponent to play an A(odd) position. If you see a position which can be reduced to A(odd), you win. Let's call this position R(1). R(1) is a position that can be reduced in one play to A(odd). All A(even) positions are R(1).
You should not leave your opponent in R(1). In fact, you want to try to force your opponent to leave you in R(1).
Unfortunately, I believe that a savvy opponent should always be able to, rather than leave you in R(1), remove a necessary stone for your A(odd) configuration. Seeing as there are 127 possible A(6) configurations, not to mention A(4) and A(2) configurations, you may be able to set up a fork situation, so that your opponent cannot block both A(even) positions at the same time.
At least, this is where I got to in my thinking over the weekend. With a board as small as Solomon's Stones, it's possible that I'm overlooking a simpler solution.