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Given a graph $G=\{V,E\}$, each node $i$ has a value $v_i$. Given budget $k$, we have $k$ chance to add 1 or minus 1 for a node's value, for example, $v'_i=v_i+1$ or $v'_i=v_i-1$. In particular, $v'_i$ denotes the value of node $i$ after $k$ modifications. The target is to maximize absolute value difference between each pair of connected node after $k$ times modifications, which is formulated as: \begin{gather} \max\sum_{e(i,j)\in E}{|v'_i-v'_j|} \\ s.t. \sum_{i\in V}|v'_i-v_i|=k \end{gather}

I want to ask whether this problem is NP-hard or not? I find this function is not submodular or monotone, thus is there any solution with an approximation ratio if it is NP-hard?

Thanks in advance.

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  • $\begingroup$ Is it not the case that an optimal solution keeps picking the same vertex value in the same direction all $k$ times? If it is, then the problem has a pretty straightforward poly-time algorithm (try both possible directions for all possible vertices) and is thus (likely) not NP-hard. $\endgroup$
    – Yonatan N
    Oct 7 '21 at 19:21
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We confirm @YonatanN's conjecture:

Lemma 1. There is always an optimal solution $v'$ such that, for some $i'$, $|v_{i'}' - v_{i'}| = k$, while $v'_i = v_i$ for all $i\ne i'$.

@YonatanN's suggested algorithm (try all possible such solutions and choose the best) will give the following corollary:

Corollary 1. There is a polynomial-time algorithm for the problem in the post.

Proof of Lemma 1. Among optimal solutions, let $v'$ be one with a minimum number of offset indices, that is, $i$ such that $v'_i \ne v_i$. Suppose for contradiction that $v'$ has at least two offset indices $h$ and $j$.

Case 1. First consider the case that $v'_h > v_h$ and $v'_j > v_j$. Fix $\delta = d(h) - d(j)$, where $$d(x) = |\{(x, i) \in E : v'_x > v'_i\}| - |\{(x, i) \in E : v'_x < v'_i\}|.$$ Imagine increasing $v'_h$ while decreasing $v'_j$ at unit rate by an arbitrarily small $\epsilon>0$. This operation (for $\epsilon \le v'_j - v_j$) would preserve $\sum_{i} |v'_i - v_i|$, so the solution would remain feasible, and it would increase the objective value at rate at least $\delta$. We conclude by this thought experiment (and the optimality of $v'$) that $\delta \le 0$.

Now actually decrease $v'_h$ and increase $v'_j$, both at unit rate, until $v'_h = v_h$. This increases the objective at rate at least $-\delta\ge 0$, and preserves feasibility, so it yields another optimal solution with fewer offset indices (as $h$ is no longer offset), contradicting the choice of $v'$.

(We use above that, as we continue to decrease $v'_h$ and increase $v'_j$, the rate of increase in the objective does not decrease. This is because all edges that contribute positively to the rate (e.g. $(i,h)$ with $v'_i > v'_h$) continue to do so, while edges that contribute negatively may start contributing positively.)

Case 2. Next consider the case that $v'_h > v_h$ and $v'_j < v_j$. Consider increasing $v'_h$ and $v'_j$ at unit rate by an arbitrarily small positive amount. This would preserve feasibility while increasing the objective at rate at least $\alpha = d(h) + d(j$). So $\alpha \le 0$. Now decrease both $v'_h$ and $v'_j$ at unit rate until $v'_j = v_j$. This preserves feasibility while increasing the objective at rate at least $-\alpha \ge 0$, yielding another optimal solution with fewer offset indices, contradicting the choice of $v'$.

For each remaining case (e.g. $v'_h < v_h$ and $v'_j < v_j$), a symmetric argument shows that the case cannot happen. It follows that $v'$ has at most one offset index. Let $i'$ be that index, if it exists, and otherwise an arbitrary index. If $|v'_{i'} - v_{i'}| = k$, we are done, so assume $|v'_{i'} - v_{i'}| < k$. Now decreasing $v'_{i'}$ would increase the objective at rate at least $d(i')$, so the optimality of $v'$ implies $d(i') \le 0$. So increase $v'_{i'}$ at unit rate until $|v'_{i'} - v_{i'}| = k$. This preserves feasibility, while increasing the objective at rate at least $-d(i') \ge 0$, giving the desired optimal solution. $~~~\Box$

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