Counting the variety of means to damage the proportion of an N x N grid by positioning one X and also one O at once

I have actually lately been exercising the complete video game tree for tic - tac - toe, simply for enjoyable. I am making use of the popular equivalence relationship of rotations/reflections to streamline this tree in the typical means (which starts by keeping in mind that there are just 3 opening up actions : side, edge, and also facility). I in fact can not locate a photo of the complete video game tree for tic - tac - toe, so if any person can give a link I would certainly value that.

This job encourages the adhering to ;

Let there be offered an N x N grid ; and also allow $m$ be an all-natural number.

Allow $I$ be the set of all feasible means to position $m$ duplicates of the letter $X$ in the grid, and also $m$ duplicates of the letter $O$ in the grid (with the constraint that we can just position one letter per room of the grid ; to put it simply, simply visualize playing a video game of tic - tac - toe on an N x N board for an also variety of actions).

Trouble:

How several means exist to entirely damage the rotational/reflection proportion of an N x N grid by positioning $m$ duplicates of $X$ and also $m$ duplicates of $O$?

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2019-05-13 03:37:16
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Answers: 1

I assume you are attempting to ask: the amount of such grids confess a non - unimportant proportion (in the dihedral team)?

For dealt with m 1, asymptotically mostly all (0,1, - 1) - matrices with specifically m 1 is and also m - 1 is do not confess a non - unimportant proportion (rotation/reflection). For these matrices, we can take into consideration the X is as the 1 is and also the (letter) O is as the - 1's. The (number) 0 is stand for the vacant cells.

Given that m is dealt with, there are ${{N^2} \choose {m,m,N^2-2m}}=N^{4m}/m!^2+o(N^{4m})$ (0,1, - 1) - matrices in complete with specifically m 1 is and also m - 1's.

The variety of such matrices that are maintained under transposition is \ [\ amount _ i, j N \ pick i, j, n - i - j N \ pick 2 \ pick (m - i)/ 2 N \ pick 2 - (m - i)/ 2 \ pick (m - j)/ 2 \ ] where i is the variety of 1 gets on the major angled, j is the variety of - 1 gets on the major diagonal. Consequently we call for 0i+jm and also m - i and also m - j also. An unrefined upper bound to the above summand is $\mathrm{const} \cdot N^{i+j+2(m-i)/2+2(m-j)/2} \leq N^{3m}=o(N^{4m})$. Given that m is dealt with, there is just a limited variety of sets i, j which is summed over, so the total outcome is $o(N^{4m})$.

The variety of such matrices that confess any one of the various other non - unimportant proportions is bounded over by the above formula additionally. So asymptotically mostly all (0,1, - 1) - matrices with specifically m 1 is and also m - 1 is do not confess a non - unimportant proportion.

In fact, I believe that this outcome would certainly hold true without the dealt with m problem, yet I can not consider just how to confirm it off - hand.

So in response to the inquiry, unless you are managing a really tiny instance, and also unless you are purposely attempting to create a proportion, you are not likely to create a matrix that has a non - unimportant proportion. For the search tree, a lot of the moment you will certainly have the ability to recognize 8 nodes representing equal video games of naughts and also crosses.

[PS: it would certainly be unpleasant (although feasible) to locate a specific formula along these lines for the variety of such matrices that confess a non - unimportant proportion. ]

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2019-05-22 22:45:03
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