I have a system of linear equations over a finite field $\mathbb F_p \cong \mathbb Z_p$, and I'm interested in the decision problem of whether there exists a solution where all of the variables $x_i$ are in the set $\{0, 1\} \subset \mathbb F$. In particular, I'm trying to determine whether this problem is NP-hard.


One system of equations over $\mathbb F_3$ is: $$ \begin{alignat*}{2} &x_1\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} + x_2\begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix} + x_3\begin{bmatrix}0 \\ 2 \\ 0\end{bmatrix} + x_4\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix} \\ + \, &x_5\begin{bmatrix}2 \\ 1 \\ 0\end{bmatrix} + x_6\begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix} + x_7\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix} + x_8\begin{bmatrix}1 \\ 2 \\ 0\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 2\end{bmatrix}. \end{alignat*} $$ This system of equations is satisfiable with entries in $\{0,1\}^8 \subset \mathbb F^8$, namely $$ \begin{align*} (x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8) &= (0,0,0,1,0,1,1,1) \hspace{1em}\text{or}\\ (x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8) &= (0,1,0,1,0,1,1,1). \end{align*} $$

An unhelpful (?) reduction

One suggestion that was given to me was turning this into a system of quadratic equations in the following way: define auxiliary functions coordinatewise $$ \begin{align*} w_1 &= x_1 + x_4 + 2x_5 + x_7 + x_8 \\ w_2 &= x_1 + 2x_3 + x_5 + x_6 + 2x_8 \\ w_3 &= x_4 + x_6 - 2, \end{align*} $$ and use these to solve the system of quadratic and linear equations $$ w_1 = w_2 = w_3 = x_1^2 - x_1 = \cdots = x_8^2 - x_8 = 0. $$

However, the MQ-problem (Multivariate Quadratic equations over a finite field) is NP-hard, so this reduction doesn't help. However, this set-up is a quite special case, so I'm holding out some hope that the original problem might still be in P.

Is there a polynomial-time algorithm for determining the existence of a solution of linear equations over a finite field with restricted variables? Or is it known if this problem is NP-hard like the MQ-problem?


1 Answer 1


You can reduce 1-IN-3 SAT to your problem (an instance is a 3CNF, and we want to find a satisfying assignment having exactly one satisfied literal per clause), assuming $p \geq 3$.

A clause $x \lor y \lor z$ is encoded as the constraint $x+y+z=1$.

A clause $\bar x \lor y \lor z$ is encoded as the constraint $1-x+y+z = 1$; and so on.

When $p = 2$, your problem becomes easy.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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