Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $k$ be a field with $p$ elements. Consider the following computational problem

Input: a natural number $n$, $n^2$ linear forms $M_{ij}$, $i,j=1,\ldots n$ in $n^2$ variables $X_{11}, \ldots X_{nn}$.

Problem: Is there an assignement of values to the variables $X_{ij}$ so that the matrix $M_{ij}$ is invertible?

${}$

Question: What is known about algorithms for this problem?

As usually, let's assume the addition and multiplication in the field to have computational cost $1$.

The naive algorithm of checking each assignment of the variables $X_{ij}$ takes time bounded by a polynomial in $p^{n^2}$. I'd be interested to know if there is an improvement to polynomial in $p^n$ (or better).

EDIT: Below Emil Jeřábek shows that the problem is NP-complete, but the reduction from 3-SAT is done in such a way that it still could be that there is an improvement to $p^n$ without proving anything unexpected about 3-SAT.

EDIT: The special case when each $M_{ij}$ is equal either to $0$ or to $X_{ij}$ is solved below by Emil Jeřábek.

EDIT: I've decided to ask a more specific follow-up question.

share|improve this question
    
I've posted it also to cstheory.stackexchange.com/questions/9316/… –  Łukasz Grabowski Dec 6 '11 at 17:52
3  
As an experienced user, you are aware that simultaneous cross-posting is discouraged on both these sites? –  Emil Jeřábek Dec 6 '11 at 18:42
1  
no, thanks for letting me know, i'll keep it mind –  Łukasz Grabowski Dec 7 '11 at 0:09
    
Trying to solve $M_{ij}=c_{ij}$ is a linear system in n^2 unknowns and n^2 equations. If it has a solution for suitable $c_{ij}$ you are done, but the matrix might not be full rank. –  joro Dec 7 '11 at 16:08
    
@joro: this is just another naive algorithm which takes time bounded by a polynomial in $p^{n^2}$, as this is how many possibilities for $c_{ij}$ there are, and it's not difficult to find examples where only $p-1$ choices for $c_{ij}$ lead to an invertible matrix. –  Łukasz Grabowski Dec 7 '11 at 16:31
add comment

1 Answer 1

up vote 5 down vote accepted

The determinant of $M$, considered as a matrix over the polynomial ring $R=k[X_{11},\dots,X_{nn}]$, is a polynomial $f\in R$, and your problem is to determine whether $f$ defines the constant $0$ function over $k$.

There are several division-free algorithms for computation of determinant in any commutative ring using polynomially many ring operations. In principle, you can use such an algorithm to compute $f$, but this may result in a long expression, since $f$ may have exponentially many terms. It is better to combine this with a polynomial identity testing algorithm: since $f$ has degree $n$, the Schwartz–Zippel lemma tells you that $f(a_{11},\dots,a_{nn})\ne0$ for randomly chosen $a_{ij}\in k$ with probability at least $1-n/p$, as long as $f\ne0$ and $p>n$. Thus, if (say) $p>2n$, you don’t actually have to evaluate $f$, you have a simple probabilistic polynomial-time algorithm: choose random assignment to your variables, and test whether the resulting matrix is nonsingular.

In the special case where each $M_{ij}$ is either $0$ or $X_{ij}$, things are much simpler: it is easy to see that there is an assignment making the matrix nonsingular if and only if there is a permutation $\pi$ such that $M_{i\pi(i)}\ne0$ for each $i$, in other words, if and only if the bipartite graph whose biadjacency matrix is defined from $M$ by replacing every $X_{ij}$ with $1$ has a perfect matching. There are various efficient algorithm for finding perfect matchings and/or checking their existence, see e.g. http://en.wikipedia.org/wiki/Perfect_matching#Algorithms_and_computational_complexity .

EDIT: For a fixed $p$, the full problem is NP-complete. The reduction is simplest for $\mathbb F_2$. Assume we are given a $3$-CNF $C(x_1,\dots,x_n)=\bigwedge_{i< m}\bigvee_{j< 3}l_{ij}$, where each $l_{ij}$ is either some $x_k$ or $\neg x_k$. We identify $\neg x_k$ with the linear polynomial $1+x_k$. Then $C$ is satisfiable if and only if the polynomial $f(x_1,\dots,x_n)=\prod_i(1+\prod_jl_{ij})$ assumes a nonzero value for some assignment over $\mathbb F_2$. Let $M$ be the block diagonal matrix consisting of $m$ blocks $M_i$, where $$M_i=\begin{pmatrix}1&l_{i0}&0\\\0&1&l_{i1}\\\\l_{i2}&0&1\end{pmatrix}.$$ Then $\det(M)=f$, thus there is an assignment in $\mathbb F_2$ making $M$ invertible iff $C$ is satisfiable. One can get rid of the constants $1$ in the matrix as follows: we include in $M$ the identity matrix as yet another block, and then we replace $1$ with a new variable $x_0$ everywhere (i.e., in the $1$ entries as well as in the entries of the form $1+x_k$). Any assignment making the new matrix invertible must have $x_0=1$, and then it works as before. The new matrix has only entries of the form $0$, $x_k$, $x_0+x_k$.

share|improve this answer
    
Thanks, this solves the special problem. I'll reformulate the question to make clear that I mean $p$ to be fixed; in particular the probabilistic polynomial-time algorithm you suggest is of not much use. –  Łukasz Grabowski Dec 7 '11 at 0:11
    
After thinking about it, I don't understand your solution to the special problem. I agree that it's equivalen to checking if there is a permutaion $\pi$ such that $M_{i\pi(i)}\neq 0$ for all $i$. But what doe it have to do with perfect matchings? Could you give a reference or expand the argument? –  Łukasz Grabowski Dec 7 '11 at 13:03
    
Also, if you conider the case when $M_{ij}$ is non-zero for $(i,j)= (1,2), (2,3), (3,1)$, there's no perfect matching. Perhaps you meant that vertiices lie in a union of disjoint cycles? (mathings would be then "cycles of length 2", but then the problem dangerously reminds me of finding a hamiltonian cycle) –  Łukasz Grabowski Dec 7 '11 at 13:06
    
I’m talking about bipartite graphs. Your matrix corresponds to the graph with vertex set $U\cup V$, $U=\{u_1,u_2,u_3\}$, $V=\{v_1,v_2,v_3\}$, and edge set $E=\{(u_1,v_2),(u_2,v_3),(u_3,v_1)\}$. This graph has a perfect matching, in fact, it is a perfect matching. –  Emil Jeřábek Dec 7 '11 at 13:12
1  
It's useful to know it's NP complete, but note that you encode a 3-sat instance of $m$ clauses with $n$ variables into a $3m \times 3m$ matrix. I believe this can be solved in a number of steps which is polynomial in $2^m$, so this doesn't seem to directly answer my question whether you can go from $p^{n^2}$ to $p^n$. –  Łukasz Grabowski Dec 7 '11 at 14:14
show 7 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.