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It is not difficult to determine the number of possible games of tic toe, but what about when rotationally symmetric positions are considered equal? Please do not simply give me the number, I would like the intuition of how it is found. IMPORTANT: I am more talking more about arrangements of x's and o's in a 3X3 grid than actual tictactoe games, thus when somebody "wins" the game it continues.

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    $\begingroup$ tell us how do you determine the total number of possible games, so that we can enhance YOUR answer $\endgroup$ – Thanos Darkadakis Dec 20 '13 at 2:31
  • $\begingroup$ After the first move, of which there are 9 possibilities, there are 8 possibilities for the second move, so 9*8 positions on move 2. There are seven possible turn 3 moves, leading to 9*8*7 positions, and the pattern continues so that eventually there are 9! games. $\endgroup$ – Platonix Dec 20 '13 at 2:38
  • $\begingroup$ not exactly. games can end on 5th, 6th, 7th, 8th or 9th move. $\endgroup$ – Thanos Darkadakis Dec 20 '13 at 2:39
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    $\begingroup$ here is an explanation: se16.info/hgb/tictactoe.htm .There is also a link inside the link... $\endgroup$ – Thanos Darkadakis Dec 20 '13 at 2:42
  • $\begingroup$ Sorry, read my edited question. You are quite correct of course, but I am more interested in the problem of symmetry than the games that "end" $\endgroup$ – Platonix Dec 20 '13 at 2:53
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I'll use Burnside's method: count the number of invariant patterns for each rotation, and average that number over all rotations to get the number of distinguishable patterns.

I assume we're counting patterns of $5$ Xs and $4$ Os. If you're not allowing reflections, there are $4$ rotations of the square in the plane: the identity, $90^o$ degrees either way, or $180^o$.

For the identity rotation, all $\binom94=126$ patterns are invariant.

For a $90^o$ rotation there are $3$ orbits, one of size $1$ (the center) and two of size $4$ (the four corners or the four sides). Since we have $4$ Os, we have to put them in the four corners or the four sides: $2$ invariant patterns.

For the $180^o$ rotation, there is one orbit of size $1$ and there are four orbits of size $2$, consisting of a pair of opposite cells. The Os have to fill two of the size $2$ orbits; the number of invariant patterns is $\binom42=6$.

Thus the number of distinguishable patterns, allowing rotations in the plane but not reflections, is $\dfrac{126+2+2+6}4=34$.

Suppose you also allow the $4$ reflections as symmetries. Each reflection has three orbits of size $1$ (a cell on the axis of reflection) and three orbits of size $2$ (two mirrored cells). The number of invariant patterns for four Os is $\binom30\binom32+\binom32\binom31=12$, and so the number of distinguishable patterns is $\dfrac{126+2+2+6+12+12+12+12}8=23$.

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The following MSE post computes the cycle index for the symmetries of an $N\times N$ board, $N$ odd or even, so that we may apply the Polya Enumeration Theorem, which includes Burnside as a special case. For $N$ odd we get that $$Z(H_N) = \frac{1}{8} \left( a_1^{N^2} + 4 a_1^N a_2^{(N^2-N)/2} + 2 a_1 a_4^{(N^2-1)/4}+a_1 a_2^{(N^2-1)/2}\right).$$ Put $N=3$ to obtain $$Z(H_3) = \frac{1}{8} \left(a_1^9 + 4 a_1^3 a_2^3 + 2 a_1 a_4^2 + a_1 a_2^4\right).$$ The substituted cycle index thus becomes $$Z(H_3)(1+z)= 1/8\, \left( 1+z \right) ^{9} +1/2\, \left( 1+z \right) ^{3} \left( 1+{z}^{2} \right) ^{3} +1/4\, \left( 1+z \right) \left( 1+{z}^{4} \right) ^{2}\\ +1/8\, \left( 1+z \right) \left( 1+{z}^{2} \right) ^{4}$$ which is $$Z(H_3)(1+z) = {z}^{9}+3\,{z}^{8}+8\,{z}^{7}+16\,{z}^{6}+23\,{z}^{5}+23\,{z}^{4} +16\,{z}^{3}+8\,{z}^{2}+3\,z+1,$$ so there are $23$ patterns with five cells marked.

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I'll look at completely-filled boards with five $X$'s and four $O$'s (say).

I'll assume rotations and reflections are not distinct.

I'll count the number of distinct cases where $O$ occupies each possible number of corners. Call this number $O_{corner}$.

If $O_{corner} = 0$, there are $2$ patterns: one of the $O$'s is in the center, or it isn't. There is at most one vacant side space.

If $O_{corner} = 1$, there are $5$ cases. If $O$ is also in the center, one can fill the two adjacent sides, one adjacent and one "far" side, or the two far sides. If $O$ is not in the center, either an adjacent side or a far side can be blank.

If $O_{corner} = 2$, these two can be in opposite corners, or adjacent corners.

Let's take adjacent corners first. For a second, let's number the squares: upper left is $1$; lower right is $9$. Our two corner $O$'s are in $1$ and $3$. If one of the $O$'s is in the center ($5$) the distinct cases are spaces $2, 4,$ and $8$. If $O$ isn't in the center, the distinct cases for the last two are the pairs of spaces $(2,4), (4,6), (2,8),$ and $(6,8)$. We have $7$ cases.

For opposite corners, if $O$ is in the center, any side square is the same, symmetry-wise. If $O$ is not in the center, the other two $O$'s can be both next to one of the corner $O$'s, one next to each corner $O$ on the same side of the diagonal, one next to each corner $O$ on the opposite side of the diagonal. So $4$ more cases.

Total cases for $O_{corner} = 2$ is $11$.

For $O_{corner} = 3$, the fourth $O$ is on the side in between two corner $O$'s, on the side not between two corner $O$'s, or in the center. That's $3$ cases.

For $O_{corner} = 4$, that's all she wrote ($1$).

So, $22$ cases. I think I got them all.

If you don't want to consider any of the symmetries I've done, or if you want to take all different numbers of $X$'s and $O$'s, then you can go through the same exercise and determine as you go along which ones are distinct.

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  • $\begingroup$ I got $23$ patterns using Burnside. Took me a while to figure out which one you missed. With $O_{corner}=1$ and $O$ in the center, there are two ways to fill one adjacent and one far side. $\endgroup$ – bof Dec 20 '13 at 4:57

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