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Given a connected (undirected) graph $G$ with vertex set $V$ of size at least $2$, we are allowed to put a real number $x_v$ on each $v\in V$. The constraint is that, for any $W\subseteq V$ such that the induced subgraphs on both $W$ and $V\setminus W$ are connected, $\displaystyle\left|\sum_{v\in W}x_v\right|\le 1$. We want to maximize $\displaystyle\sum_{v\in V} |x_v|$.

Is it true that for any maximizing solution, the sum of the $x_v$'s is $0$?

It is true for several graphs as seen here, as well as all graphs of size up to $4$. One can observe that if $\{x_v\}_{v\in V}$ is a maximizing solution, then so is $\{-x_v\}_{v\in V}$. But that does not imply that the sum of $x_v$'s must be $0$.

Alexi
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Let $s(W) = \displaystyle \sum_{v\in W} x_v$ be the sum over a subset $W$. $s(V)$ is the total sum of an optimal solution. As we also have an optimal solution by changing all the sign, we can suppose that $s(V)\geq 0$. Suppose by contradiction that we have $s(V)>0$. Lets choose a vertex $v$ with negative value (If all vertices have a positive value and $|V|>1|$, we can change the sign of the smallest value while keeping a valid solution with the same objective value and $s(V)$ still positive). If for every admissible subset $W$, $v\in W$, $s(W)$ is strictly more than $-1$, then we can decrease the value of $v$. So the solution is optimal only if there exists an admissible $W$, $v\in W$ with $s(W) = -1$. However, $V\setminus W$ is also an admissible subset. So we have $s(V) = s(W) + s(V\setminus W)$, and $s(V\setminus w) = s(V) - s(W) > 1$, which is a contradiction. So $s(V) = 0$

caduk
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