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I have an expression with 500 elements, each of which has a form similar to the next one:

$\frac{Log\Big[\frac{f_{\{x_1,y_1\}}f_{\{x_2,y_2\}}}{f_{\{x_1,x_2\}}f_{\{y_1,y_2\}}}\Big]Log\Big[\frac{f_{\{x_1,y_2\}}f_{\{y_1,x_2\}}}{f_{\{x_1,x_2\}}f_{\{y_1,y_2\}}}\Big]}{Log\Big[\frac{f_{\{x_1,y_1\}}f_{\{x_2,y_2\}}}{f_{\{x_1,y_2\}}f_{\{y_1,x_2\}}}\Big]}$

Where, $f$ is some unknown function (possible to assume it is a Gaussian, but then each $x_i,y_i$ is a 2D vector). There are six variables $x_1,x_2,x_3,y_1,y_2,y_3$, who permute through out the expression in different ways. Most have more Log[]s in the numerator and denominator, some Log[]s have six arguments instead of four.

My aim is to narrow down the number of elements to a more reasonable number so it can be used as a general formula, where I suspect there is some permutational formula out there.

These 500 elements can be divided into six groups, with a certain symmetry between them (for example $1\leftrightarrow 2$). I have tried to further break down these groups using replacement rules but that didn't help reduce their number significantly.

Trying to use Mathematica built-in functions I've found FindSequenceFunction, FindRepeat and FindTransientRepeat unhelpful.

Any suggestions will be gladly accepted!

1st edit: The variables are initially formed in a he following way: $f_{\{x_1,y_1\}}f_{\{x_2,y_2\}}f_{\{x_3,y_3\}}$ and the possible permutations (without changing $x$ and $y$ spots), triplet:

$f_{\{x_1,y_3\}}f_{\{x_2,y_2\}}f_{\{x_3,y_1\}}\\f_{\{x_1,y_1\}}f_{\{x_2,y_3\}}f_{\{x_3,y_2\}}\\f_{\{x_1,y_2\}}f_{\{x_2,y_1\}}f_{\{x_3,y_3\}}$

and doublet:

$f_{\{x_1,y_2\}}f_{\{x_2,y_3\}}f_{\{x_3,y_1\}} \\f_{\{x_1,y_3\}}f_{\{x_2,y_1\}}f_{\{x_3,y_2\}}$

The initial permutation is symmetric under $1\leftrightarrow 2,1\leftrightarrow 3,2\leftrightarrow 3$. The triplet and the doublet are also symmetric under the same permutations, where they transfer to each other. These are the coefficients of the terms, under which it was possible to divide the 500 terms into 6 groups.

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  • $\begingroup$ Just to clarify: Are you sure that your terms are symmetric under some group action, or does it just look like a good place to start to try and reduce the complexity? Are you looking for some sort of generator that you can "symmetrize" to get all the terms? How are these terms in your expression generated? (I ask because it would make it easier for people to answer if they can generate the terms themselves, and not just guess at what they might be like.) $\endgroup$ – aardvark2012 Oct 9 '17 at 9:57
  • $\begingroup$ It is part of a work in process, the terms were computed from a symmetric starting point , so I suspect the symmetry should remain. I am looking for some sort of generator that would "symmetrize" to get all the terms. The expression was generated after some very long computation (I'm working on it for the last couple of months...) $\endgroup$ – Gaby Fleurov Oct 9 '17 at 10:20
  • $\begingroup$ Can you be more precise about the symmetry? I would guess something more complicated that $\mathbf{S}_6$, but knowing the initial symmetries might be useful. Also, is there any "simple" relation between the symmetries and the number of variable pairs that show up in each term? $\endgroup$ – aardvark2012 Oct 9 '17 at 10:33
  • $\begingroup$ I've edited the original post, I hope it clarifies the issue $\endgroup$ – Gaby Fleurov Oct 9 '17 at 11:09

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