# Studying equivalence relation on matrices defined by permutation of rows and columns

I'm interested in the equivalence relation $\sim$ on $m\times n$ binary matrices where $(a_{ij})_{ij}=A\sim A'$ if there exists one permutation for the rows and one for the columns of $A$ to make the two identical, i.e. there is $\sigma\in{\cal S}_m,\tau\in{\cal S}_n$ such that $(a_{\sigma(i)\tau(j)})_{ij}=A'$.

My questions:

(i) How can one effectively compute whether two given matrices belong to the same equivalence class?

(ii) How can one, given a finite class ${\cal M}$ of matrices, e.g. all $m\times n$ binary matrices, effectively generate a set of representations of ${\cal M}/\sim$, i.e. a set of matrixes which contains exactly one element in every equivalence class of ${\cal M}/\sim$?

• I would recommend including everything from "My thoughts so far" downward into a self-answer to this question. As it stands you have the question combined with the answer, with the implied question "Is there a better answer?" – Wildcard Jan 5 '17 at 11:04
• Done, thanks for the input. – fweth Jan 5 '17 at 11:06

My thoughts so far:

For any given matrix $A$ there is a function $f_A:{\cal P}([m])\to[n]$ (where $[m]$ denotes $\{1,\dots,m\}$) such that $f_A(X)$ is the size of the set $\{j\in[n]:\forall i\in X.a_{ij}=1\}$, i.e. the number of ones in the bitwise conjunction of the rows $A_i$ for $i\in X$. One can define the equivalence relation $f\sim^* f'$ whenever $f\circ(X\mapsto\{\sigma(i):i\in X\})=f'$ for some $\sigma\in{\cal S}_m$. Then whenever $A\sim A'$ we have $f_A\sim^*f_{A'}$ and my intuition says also vice versa. So (i) would reduce to study the equivalence relation $\sim^*$ and for (ii) one could ask which functions $f:{\cal P}([m])\to[n]$ actually yield a matrix $A$ with $f=f_A$.

Then I had another idea, one can just sort matrices lexicographically, interpreting them as concatenation of all their rows. I wrote a Haskell script to find the minimal $A'$ in $[A]_\sim$ w.r.t. this ordering, I don't know how efficient it is but it should be more efficient than brute forcing through all $\sigma$-$\tau$-combinations. Here foo is some helper function finding the minimal element out of all column only permutations of a given matrix, so getMinEqu is only brute forcing through the row permutations. It works by first ordering the columns such that the first row looks like [f,...,f,t,...,t] and then applying itself to the two submatrices sitting below the fs and below the ts.

import Data.List

f::Bool
f=False

t::Bool
t=True

getInd::[Int]->[a]->[a]
getInd ks xs=[xs!!k|k<-ks]

getMinEqu::[[Bool]]->[[Bool]]
getMinEqu xs=head$sort$map foo$permutations xs where foo::[[Bool]]->[[Bool]] foo []=[] foo(x:xs) |m*n>1 =zipWith(++)((map(const f)falPos) :(foo$map(getInd falPos)$tail xs)) ((map(const t)truPos) :(foo$map(getInd truPos)\$tail xs))
|otherwise=x:xs where
m::Int
m=length(x:xs)
n::Int
n=length x
falPos::[Int]
falPos=elemIndices f x
truPos::[Int]
truPos=elemIndices t x