On the Maximum Connectivity Improvement Problem

Abstract

In this paper, we define a new problem called the Maximum Connectivity Improvement (MCI) problem: given a directed graph $G = (V,E)$, a weight function $w:V \rightarrow \mathbb {N}_{\ge 0}$, a profit function $p:V \rightarrow \mathbb {N}_{\ge 0}$, and an integer B, find a set S of at most B edges not in E that maximises $f(S)=\sum _{v\in V}w_v\cdot p(R(v,S))$, where p(R(v, S)) is the sum of the profits of the nodes reachable from node v when the edges in S are added to G. We first show that we can focus on Directed Acyclic Graphs (DAG) without loss of generality. We prove that the MCI problem on DAG is $NP$-Hard to approximate to within a factor greater than $1-1/e$ even if we restrict to graphs with a single source or a single sink, and MCI remains $NP$-Complete if we further restrict to unitary weights. We devise a polynomial time algorithm based on dynamic programming to solve the MCI problem on trees with a single source. We propose a polynomial time greedy algorithm that guarantees $(1-1/e)$-approximation ratio on DAGs with a single source or a single sink.

Appendices

Appendix 1.A Omitted Proofs

Lemma 3

Given a graph G and its condensation $G'$, it yields: $f(G')=f(G)$.

Proof

First, consider two nodes u and v that belong to the same strongly connected component $C_{v'}$ in $G'$. Clearly, $R(u,G)=R(v,G)$.

Moreover, it holds $p(R(v,G))=p(R(v',G'))$ because $R(v',G')$ contains one node for each different strongly connected component in R(u, G) and thus:

$$p(R(v',G'))=\sum _{u' \in R(v', G')} p_{u'}=\sum _{u' \in R(v', G')} \sum _{u \in C_{u'}} p(u)= \sum _{u \in R(v,G)} p(u)=p(R(v,G))$$

Denoted $C_{v'}$ the strongly connected component represented by $v'$, we have:

$$\begin{aligned} f(G')&= \sum _{\mathbf{v'} \in V'} w_\mathbf{{v'}} p(R(\mathbf{v'},G'))\\&= \sum _{\mathbf{v'} \in V'} w_{\mathbf{v'}} \left( \sum _{u' \in R(\mathbf{v'}, G')} \sum _{u \in C_{u'}} p(u) \right) \\&= \sum _{\mathbf{v'} : C_{\mathbf{v'}} \in \mathcal {C}} w_{\mathbf{v'}} \left( \sum _{u' \in R(\mathbf{v'}, G')} \sum _{u \in C_{u'}} p(u) \right) \\&= \sum _{\mathbf{v'} : C_{\mathbf{v'}} \in \mathcal {C}} \left( \sum _{v \in C_{\mathbf{v'}}} w_v \right) \left( \sum _{u' \in R(\mathbf{v'}, G')} \sum _{u \in C_{u'}} p(u) \right) \\&= \sum _{\mathbf{v'} : C_{v'} \in \mathcal {C}} \sum _{v \in C_{\mathbf{v'}}} \left( w_v \left( \sum _{u' \in R(\mathbf{v'}, G')} \sum _{u \in C_{u'}} p(u) \right) \right) \\&= \sum _{\mathbf{v'} : C_{\mathbf{v'}} \in \mathcal {C}} \sum _{v \in C_{\mathbf{v'}}} \left( w_v \sum _{u \in R(v,G)} p(u) \right) \\&= \sum _{\mathbf{v'} : C_{\mathbf{v'}} \in \mathcal {C}} \sum _{v \in C_{\mathbf{v'}}} w_v p(R(v,G))\\&= \sum _{v \in V} w_v p(R(v, G)) = f(G) \end{aligned}$$

   $\square$

Appendix 1.B Omitted Images

Fig. 2.

Example of Algorithm 1. Consider the node c with $w_c=2$, $p_v=1\;\forall v\in V$ and $B=2$. We have: $g(d, 2) = 19$, $g(d, 1) = 12$, $g(e, 1) = 7+2(6)=19$. Therefore $g(c, 2) = g(e, 1) + g(d, 1) + w_c \cdot (p(T(r)) - p(T(v)))= 35$.

Appendix 1.C Generic Trees Algorithm

Given a generic rooted tree $T=(V,E)$, let us transform it into a rooted binary tree $T'=(V\cup U, E')$ with weights $w', p'$ by adding dummy nodes U as follows:

• 1. Let the root r of T be the root of $T'$.

• 2. For each non-leaf node v, let $v_1, v_2, \ldots v_{l}$ be the children of v:

  • (a) Add edge $(v,v_1)$ to $E'$;

  • (b) If $l=2$ add $(v,v_2)$ to $E'$;

  • (c) If $l>2$, add $l-2$ dummy nodes $u_{v_2}, u_{v_3}, \ldots , u_{v_{l-2}}, u_{v_{l-1}}$

  • (d) Add edge $(v,u_{v_2})$ and edges $(u_{v_i}, u_{v_{i+1}})$ to $E'$, for each $2 \le i \le l-2$;

  • (e) Add edge $(u_{v_i},v_i)$ to $E'$, for each $2 \le i \le l-1$;

  • (f) If $l>2$, add edge $(u_{i_{l-1}},v_l)$ to $E'$.

• 3. If $v\in V$, then $w'_v=w_v$, otherwise $w'_v=0$ and $p'_v=p_v$, otherwise $p'_v=0$.

See Fig. 3 for an example of the transformation.

Fig. 3.

Example of transformation from general tree to binary tree