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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 there is a maximizing solution where all $x_v$'s are integers summing to $0$?

Examples: If $G$ is a path of length $n$, a maximum occurs at $(1,-2,2,-2,\dots,-2,1)$ if $n$ is odd, and $(1,-2,2,\dots,2,-1)$ if $n$ is even. If $G$ is a cycle of length $n$, a maximum occurs at $(1,-1,\dots,1,-1)$ if $n$ is even and $(1,-1,\dots,1,-1,0)$ if $n$ is odd.

The constraints and objective can be written as a linear program by taking out absolute values, and some linear programming facts may be useful.

Alexi
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  • Since you said "... a maximizing solution where all $x_v$'s are integers", is there a maximizing solution for some $G$ that does not satisfy that property ? – VTand Nov 14 '21 at 05:51
  • @VTand Yes, if $G$ is a triangle $abc$ with an edge $cd$, then I believe that $x_a = t$, $x_b = 1-t$, $x_c = -2$, $x_d = 1$ is an optimum for any $t\in[0,1]$. – Alexi Nov 14 '21 at 05:56

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Let $G$ be the graph with vertex set $\{1,2,3,4,5\}$ and all edges except $12$. Here, we have a constraint on every subset $W$ except for $W = \{1,2\}$ and its complement $W = \{3,4,5\}$.

The LP optimum is $3$, achieved for example when $x_1 = 1$, $x_2 = \frac12$, and $x_3 = x_4 = x_5 = -\frac12$.

However, if $x_1,x_2,x_3,x_4,x_5$ are all required to be integers, we can't do better than $2$. First of all, $x_i \in \{-1,0,1\}$ for all $i$. Can we have $x_i = x_j = 1$ for some $i,j$? Only if $\{i,j\} = \{1,2\}$, otherwise the $|x_i + x_j| \le 1$ constraint is violated. But if $x_1 = x_2 = 1$, then $|x_1 + x_2 + x_k| \le 1$ for $k=3,4,5$ forces $x_3 = x_4 = x_5 = -1$, and then $|x_3 + x_4| = 2 > 1$. So we can't set two variables to $1$; similarly, we can't set two variables to $-1$. So the optimal integer solution is to set one variable equal to $1$, another to $-1$, and the rest to $0$.

If we do the same thing with $6$ vertices, it appears that the optimal value is $\frac83$, which is not even an integer (achieved by $x_1 = 1$, $x_2 = \frac13$, and $x_3 = x_4 = x_5 = x_6 = -\frac13$).

Misha Lavrov
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  • Thanks! Still, it looks like the sum of the $x_v$'s is always $0$. Do you see how to prove this? – Alexi Nov 15 '21 at 03:21
  • Not offhand, but at least I have not found any small counterexamples. – Misha Lavrov Nov 15 '21 at 03:40
  • Also - this is not actually a linear program! We can't eliminate absolute values from the objective value of a maximization problem, only from a minimization problem. In general, we can turn a piecewise linear objective function into a linear one if (1) it's convex and we're minimizing, or (2) it's concave and we're maximizing. Here, it's convex, but we're trying to maximize. – Misha Lavrov Nov 15 '21 at 04:59
  • @MishaLavrov Couldn't we minimize $\sum y_i$ where $y_i \geq x_i$ and $y_i \geq -x_i$? – Dániel Garamvölgyi Nov 15 '21 at 19:25
  • @DánielG. That would work great - if we wanted to minimize $\sum |x_i|$, but we want to maximize. – Misha Lavrov Nov 15 '21 at 19:50