On the most imbalanced orientation of a graph

Abstract

We study the problem of orienting the edges of a graph such that the minimum over all the vertices of the absolute difference between the outdegree and the indegree of a vertex is maximized. We call this minimum the imbalance of the orientation, i.e. the higher it gets, the more imbalanced the orientation is. The studied problem is denoted by ${{\mathrm{\textsc {MaxIm}}}}$. We first characterize graphs for which the optimal objective value of ${{\mathrm{\textsc {MaxIm}}}}$ is zero. Next we show that ${{\mathrm{\textsc {MaxIm}}}}$ is generally NP-hard and cannot be approximated within a ratio of $\frac{1}{2}+\varepsilon$ for any constant $\varepsilon >0$ in polynomial time unless $\texttt {P}=\texttt {NP}$ even if the minimum degree of the graph $\delta$ equals 2. Then we describe a polynomial-time approximation algorithm whose ratio is almost equal to $\frac{1}{2}$. An exact polynomial-time algorithm is also derived for cacti. Finally, two mixed integer linear programming formulations are presented. Several valid inequalities are exhibited with the related separation algorithms. The performance of the strengthened formulations is assessed through several numerical experiments.

Appendix 1: Proof of Lemma 6

Let $\varphi$ be a not-all-equal at most 3-SAT formula with n variables $\{x_1,\ldots ,x_n\}$ and m clauses $\{c_1,\ldots ,c_m\}$ and for all $i\in \llbracket 1,n\rrbracket$, let $k_i\in \mathbb {N}$ be the number of occurences of $x_i$ in $\varphi$. We assume that there is at least one variable $x_i$ that has at least 4 occurences in $\varphi$ (otherwise $\varphi$ is already a not-all-equal at most 3-SAT(3V) formula) and we will build from $\varphi$ a not-all-equal at most 3-SAT(3V) $\varphi '$ such that $\varphi$ and $\varphi '$ are equisatisfiable as follows.

• For all $i\in \llbracket 1,n\rrbracket$, if $k_i\ge 4$ then we introduce $k_i$ new variables $\{x_i^1,\ldots ,x_i^{k_i}\}$ and for $l\in \llbracket 1,k_i\rrbracket$ we replace the l-th occurence of $x_i$ in $\varphi$ with $x_i^l$.

• For all $i\in \llbracket 1,n\rrbracket$, if $k_i\ge 4$ then we add $k_i$ new clauses $\{c_{x_i}^1,\ldots ,c_{x_i}^{k_i}\}$ where for $l\in \llbracket 1,k_i-1\rrbracket$, $c_{x_i}^l=(x_i^l\vee \lnot x_i^{l+1})$ and $c_{x_i}^l=(x_i^l\vee \lnot x_i^1)$.

Suppose there exists an assignment $v:\{x_1,\ldots ,x_n\}\rightarrow \{{{\mathrm{{\text {TRUE}}}}},{{\mathrm{{\text {FALSE}}}}}\}$ of $x_1,\ldots ,x_n$ satisfying $\varphi$. Then

$$\begin{aligned}v':\begin{array}{ll} x_i\mapsto v(x_i)~&{}\quad \forall i\in \llbracket 1,n\rrbracket \text { s.t. }k_i\le 3;\\ x_i^l\mapsto v(x_i)~&{}\quad \forall i\in \llbracket 1,n\rrbracket \text { s.t. }k_i \ge 4\text { and }\forall l\in \llbracket 1,k_i\rrbracket ; \end{array} \end{aligned}$$

is an assignment of the variables $x_i$ and $x_i^l$ satisfying $\varphi '$ for

• $\forall j\in \llbracket 1,m\rrbracket$, the values of the literals of $c_j$ w.r.t. v and $v'$ are piecewise equal so $v'(c_j)=v(c_j)={{\mathrm{{\text {TRUE}}}}}$ and $v'$ is not-all-equal for $c_j$ as well as v is;

• $\forall i\in \llbracket 1,n\rrbracket \text { s.t. }k_i\ge 4$, $\forall l\in \llbracket 1,k_i-1\rrbracket$, $v'(x_i^l)=v'(x_i^{l+1})=v(x_i)$ and $v'(x_i^{k_i})=v'(x_i^{1})=v(x_i)$ so we directly have $\forall l\in \llbracket 1,k_i-1\rrbracket$, $v'(c_{x_i}^l)={{\mathrm{{\text {TRUE}}}}}$ and $v'(c_{x_i}^{k_i})={{\mathrm{{\text {TRUE}}}}}$ and $v'$ is not-all-equal for each of these clauses since they all consist of two literals having opposite values w.r.t. $v'$.

As an example, for a formula

$$\begin{aligned} \varphi= & {} (x_1\vee \lnot x_2\vee x_3)\wedge (\lnot x_1\vee \lnot x_3\vee x_4)\wedge (x_1\vee \lnot x_2)\wedge (\lnot x_1\vee \lnot x_3\vee \lnot x_4)\\&\wedge (x_1\vee x_3), \end{aligned}$$

where $x_1$ occurs five times and $x_3$ four so we add nine new variables $x_1^1$, $x_1^2$, $x_1^3$, $x_1^4$, $x_1^5$, $x_3^1$, $x_3^2$, $x_3^3$ and $x_3^4$ and nine new clauses:

$$\begin{aligned} \varphi '=&(x_1^1\vee \lnot x_2\vee x_3^1)\wedge (\lnot x_1^2\vee \lnot x_3^2\vee x_4)\wedge (x_1^3\vee \lnot x_2)\wedge (\lnot x_1^4\vee \lnot x_3^3\vee \lnot x_4)\\&\wedge (x_1^5\vee x_3^4)\wedge (x_1^1\vee \lnot x_1^2)\wedge (x_1^2\vee \lnot x_1^3)\wedge (x_1^3\vee \lnot x_1^4)\wedge (x_1^4\vee \lnot x_1^5)\\&\wedge (x_1^5\vee \lnot x_1^1)\wedge (x_3^1\vee \lnot x_3^2)\wedge (x_3^2\vee \lnot x_3^3)\wedge (x_3^3\vee \lnot x_3^4)\wedge (x_3^4\vee \lnot x_3^1). \end{aligned}$$

Now suppose there exists an assignment $v'$ of the $x_i$ and $x_i^l$ satisfying $\varphi '$ and let $i\in \llbracket 1,n\rrbracket$ such that $k_i \ge 4$. If we take a look at the clauses $c_{x_i}^1,\ldots ,c_{x_i}^{k_i}$, we notice that if $v'(x_i^1)={{\mathrm{{\text {FALSE}}}}}$ then for $c_{x_i}^1$ to be satisfied, $v'(\lnot x_i^2)={{\mathrm{{\text {TRUE}}}}}$, i.e. $v'(x_i^2)={{\mathrm{{\text {FALSE}}}}}$, then for $c_{x_i}^2$ to be satisfied, $v'(\lnot x_i^3)={{\mathrm{{\text {TRUE}}}}}$ ...etc. Repeating this argument, we obtain that if $v'(x_i^1)={{\mathrm{{\text {FALSE}}}}}$ then $v'(x_i^1)=v'(x_i^2)=\cdots =v'(x_i^{k_i})={{\mathrm{{\text {FALSE}}}}}$. Similarly, if $v'(x_i^{k_i})={{\mathrm{{\text {TRUE}}}}}$ then for $c_{x_i}^{k_i}$ to be satisfied, $v'(\lnot x_i^{k_i-1})={{\mathrm{{\text {FALSE}}}}}$, i.e. $v'(x_i^{k_i-1})={{\mathrm{{\text {TRUE}}}}}$, then for $c_{x_i}^{k_i-1}$ to be satisfied, $v'(\lnot x_i^{k_i-2})={{\mathrm{{\text {FALSE}}}}}$ ...etc. Hence if $v'(x_i^{k_i})={{\mathrm{{\text {TRUE}}}}}$ then $v'(x_i^{k_i})=v'(x_i^{k_i-1})=\cdots =v'(x_i^1)={{\mathrm{{\text {TRUE}}}}}$. This yields that

$$\begin{aligned} \forall i\in \llbracket 1,n\rrbracket \text { s.t. }k_i \ge 4,~v'(x_i^1)=v'(x_i^2)=\cdots =v'(x_i^{k_i}). \end{aligned}$$

Hence for all $i\in \llbracket 1,n\rrbracket$ such that $k_i\ge 4$, we can replace $x_i^1,\ldots ,x_i^{k_i}$ by a unique variable $x_i$ and doing so the clauses $c_{x_i}^1,\ldots ,c_{x_i}^{k_i}$ become trivial and can be removed and only $\varphi$ remains. So the following assignment of $x_1,\ldots ,x_n$:

$$\begin{aligned} v:\begin{array}{ll} x_i\mapsto v'(x_i)~&{}\quad \forall i\in \llbracket 1,n\rrbracket \text { s.t. }k_i\le 3;\\ x_i\mapsto v'(x_i^1)~&{}\quad \forall i\in \llbracket 1,n\rrbracket \text { s.t. }k_i \ge 4; \end{array} \end{aligned}$$

satisfies $\varphi$. We have just shown that $\varphi$ and $\varphi '$ are equisatisfiable.