Given universal set $U=X \cup Y = \{x_1, \ldots, x_{n_1} \} \cup \{y_1, \ldots, y_{n_2}\}$ where $X \cap Y = \emptyset$ and sets $\mathcal{S}=\{s_1, \ldots, s_m\}$ such that $s_i \subseteq U$ for all $i=1,\ldots,m$ and integer $k \in \mathbb{Z}^{+}$.

Find $\mathcal{C} \subseteq \mathcal{S}$ (denote $Z = \cup_{s \in \mathcal{C}} s$) that maximizes

$|\{ z \in Z \mid (z \in X \text{ and is covered even times by } \mathcal{C}) \text{ or } (z \in Y \text{ and is covered odd times by } \mathcal{C} )\}|$


  • $|\mathcal{C}| = k$

In other words, we consider an element to be covered if it satisfies the odd/even coverage requirement according to its membership ($X$ or $Y$).

This problem differs from the standard maximum set cover in the definition of "covering an element".

Is this problem NP-hard? Is so, how to prove it?

  • 2
    $\begingroup$ What have you tried so far? Did you try coming up with a reduction or an efficient algorithm? $\endgroup$
    – adrianN
    Commented Jul 14, 2016 at 13:06

1 Answer 1


One can rephrase your question as follows:

Given a matrix $A$ and a vector $b$ over $GF(2)$, find a vector $x$ of weight $k$ such that $|Ax-b|$ is as small as possible.

A very similar problem was shown to be NP-complete by Stadnicki on cstheory:

Given a matrix $A$ and a vector $b$ over $GF(2)$, does there exist a vector $x$ of weight at most $k$ such that $Ax = b$?

This implies that your problem is intractable as well.


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.