Games induced by the partitioning of a graph

Abstract

The paper aims at generalizing the notion of restricted game on a communication graph, introduced by Myerson. We consider communication graphs with weighted edges, and we define arbitrary ways of partitioning any subset of a graph, which we call correspondences. A particularly useful way to partition a graph is obtained by computing the strength of the graph. The strength of a graph is a measure introduced in graph theory to evaluate the resistance of networks under attacks, and it provides a natural partition of the graph (called the Gusfield correspondence) into resistant components. We perform a general study of the inheritance of superadditivity and convexity for the restricted game associated with a given correspondence. Our main result is to give for cycle-free graphs necessary and sufficient conditions for the inheritance of convexity of the restricted game associated with the Gusfield correspondence.

Appendix:  Proof of Theorem 20

We give a direct proof following the method of van den Nouweland and Borm (1991).

Proof

For all A,B∈N such that A∩B=∅, we have v(A∪B)≥v(A)+v(B). Let us now consider A,B and i∈N such that A⊂B⊆N∖{i}. We have to prove $\tilde{v}(A \cup\lbrace i \rbrace) - \tilde{v}(A) \leq\tilde{v}(B \cup\lbrace i \rbrace) - \tilde{v}(B)$. By definition, we have:

$$ \tilde{v}(A) = \sum_{C \in\mathcal{F}(A), \ C\ \mathrm{maximal}} v(C) $$

(51)

and

$$ \tilde{v}\bigl(A \cup\lbrace i \rbrace\bigr) = \sum _{C \in\mathcal{F}(A \cup \lbrace i \rbrace), \ C \ \mathrm{maximal}} v(C). $$

(52)

Let us denote by C(i) the unique maximal set $C \in\mathcal{F}(A \cup\lbrace i \rbrace)$ such that i∈C. Let us denote by $\mathcal{C}$ the family:

$$ \mathcal{C} := \bigl\lbrace C \in\mathcal{F}(A),\ C \mbox{ maximal in } \mathcal{F}(A) \mbox{ and } C \subset C(i) \bigr\rbrace. $$

(53)

Observe that as i∉A, $C(i) = \lbrace i \rbrace\cup(\bigcup_{C \in\mathcal{C}} C)$. (If $C \in\mathcal{F}$, C⊂A∪{i}, C is maximal in $\mathcal{F}(A \cup\lbrace i \rbrace)$ and i∉C, then C⊂A and C is maximal in $\mathcal{F}(A)$.) Observe also that if $C \in\mathcal{F}(A \cup\lbrace i \rbrace)$, C is maximal and $C \not\subset C(i)$ then C∩C(i)=∅ (partition) and $C \in\mathcal{F}(A)$ with C maximal in $\mathcal{F}(A)$. Hence:

$$ \tilde{v}\bigl(A \cup\lbrace i \rbrace\bigr) - \tilde{v}(A) = v\bigl(C(i)\bigr) - \sum_{C \in\mathcal{C}} v(C). $$

(54)

Analogously, we define D(i) as the maximal set D in $\mathcal{F}(B \cup\lbrace i \rbrace)$ such that i∈D and:

$$ \mathcal{D} := \bigl\lbrace D \in\mathcal{F}(B) ;\ D \mbox{ maximal in } \mathcal{F}(B),\ D \subset D(i)\bigr\rbrace. $$

(55)

Then $D(i) = \lbrace i \rbrace\cup(\bigcup_{D \in\mathcal{D}} D)$ and:

$$ \tilde{v}\bigl(B \cup\lbrace i \rbrace\bigr) - \tilde{v}(B) = v\bigl(D(i)\bigr) - \sum_{D \in\mathcal{D}} v(D). $$

(56)

Hence, it remains to prove that:

$$ v\bigl(C(i)\bigr) - \sum_{C \in\mathcal{C}} v(C) \leq v\bigl(D(i)\bigr) - \sum_{D \in \mathcal{D}} v(D). $$

(57)

We want now to prove that for every $C \in\mathcal{C}$, there exists one and only one $D \in\mathcal{D}$ such that C=D. As A⊂B, A∪{i}⊂B∪{i} and therefore C(i)⊆D(i). Hence, for all $C \in\mathcal{C}$, there exists precisely one $D \in \mathcal{D}$ such that C⊆D. $D \cap C(i) \not= \emptyset$ because $D \cap C(i) \supset C \not= \emptyset$. D⊇C(i) contradicts i∉D. Therefore D⊂C(i). But i∉D and C(i)⊂A∪{i}, then D⊂A. But D is maximal in $\mathcal{F}(B)$, hence D⊂A is maximal in $\mathcal{F}(A)$. As C⊂D⊂A and C and D are maximal in $\mathcal {F}(A)$, we have C=D. We can now number the elements of $\mathcal{C}$ and $\mathcal{D}$ in such a way that $\mathcal{C} = \lbrace C_{1}, C_{2}, \ldots, C_{s} \rbrace$, $\mathcal {D} = \lbrace D_{1}, D_{2}, \ldots, D_{t}\rbrace$ with s≤t and C _r =D _r for all r, 1≤r≤s. Superadditivity of the game (N,v) implies:

$$v \Biggl( \lbrace i \rbrace\cup\bigcup_{r = 1}^{s} D_{r} \cup\bigcup_{r = s+1}^{t} D_{r} \Biggr) \geq v \Biggl( \lbrace i \rbrace\cup\bigcup _{r = 1}^{s} D_{r} \Biggr) + \Biggl( \sum _{r = s+1}^{t} v(D_{r}) \Biggr). $$

Then:

$$v \biggl(\lbrace i \rbrace\cup\bigcup_{D \in\mathcal{D}}D \biggr) - \sum_{D \in\mathcal{D}} v(D) \geq v \Biggl( \lbrace i \rbrace \cup\bigcup_{r = 1}^{s} D_{r} \Biggr) - \Biggl( \sum_{r = 1}^{s} v(D_{r}) \Biggr). $$

As D _r =C _r for all r, 1≤r≤s, we obtain:

That is precisely (57). □