The Evolution of a Hard Graph Theory Problem – Secure Sets Ron Dutton Computer Science University of Central Florida 1

Many real world problems involve a collection of entities (e.g., individuals, businesses, or countries) that compete for each others resources. A group of these entities can reach mutual agreements, or pacts, to defend themselves against a common adversary. Often, we can model such problems as a graph to more easily study properties and algorithms related to these problems. 2

Throughout, assume G = (V, E) is a connected graph. For a set S  V, a vertex in N[S]–S  V–S can "attack" any one of it's neighbors in S, while a vertex in S can "defend" itself or any one of its neighbors in S. N[S]–S is sometimes called the "boundary" of S. 3

There are several ways we can define these sets. We will discuss two. Secure Sets and Defensive Alliances. 4

Defensive Alliances A set S  V is a defensive alliance when every vertex in S has, with itself, as many neighbors in the set as outside the set – Every vertex in the alliance can be "protected." Formally, Definition: S  V is a defensive alliance when |N[x]  S| ≥ |N[x]–S|, for every x  S. 5

Defensive alliances have been used to model a variety of real world applications and requires only that each individual vertex in a set S be defendable. The defensive alliance number, da(G), satisfies  (  (G)+1)/2  ≤ da(G) ≤  n/2 . 6

Secure Sets Any and all vertices in S can be attacked simultaneously. An attack on S is an assignment for each vertex in N[S]–S to attack exactly one of its neighbors in S. For a given attack, a defense by S is an assignment for each vertex in S to defend either itself or a neighbor in S. 7

For a given attack and defense, a single vertex v  S is defended if the number of defenders assigned to v is at least that of the number of attackers assigned to v. An attack is defendable if there is a defense in which all vertices of S are simultaneously defended. 8

Informally, a group (set) is secure when it can successfully defend itself against any possible take-over or attack. Definition: S  V is secure when all possible attacks by N[S]–S are defendable. 9

S = V is always secure. So, there is always a smallest secure set. The security number of G, s(G), is the number of vertices in a smallest secure set in G. 10

Every secure set is a defensive alliance. Therefore, da(G) ≤ s(G). But, not all defensive alliances are secure. There are graphs for which s(G) =  (n+1)/2  >  n/2  (one of the classes of Kneser Graphs)  (n+1)/2  has been conjectured to be an upper bound for s(G). 11

A graph theoretic characterization of secure sets: Theorem. S  V is secure if and only if |N[X]  S| ≥ |N[X]–S| for all X  S. 12

Determining whether a graph G has a defensive alliance with no more than k vertices has been shown to be NP–Complete. The similar problem for secure sets remains open and may not even be in the set NP. 13

To see what that means: The set NP is the set of Decision Problems whose Yes instances can – with a little help – be verified to be Yes instances in polynomial time. "Little Help"? Actually, a lot of help!! 14

Example: Clique Given a graph G and an integer k. Does G have a clique with at least k vertices? If the answer is Yes, then there is a set of k vertices that induce a complete subgraph. Suppose someone (??) GAVE you that set. Then we could verify it's correctness very easily (O(n 2 ) time at most). 15

That proves Clique is in NP. Similarly, Defensive Alliance can be shown to be in NP. But, how about SecureSet? 16

Suppose you have a YES instance of SecureSet and someone gave you the correct set of vertices. How can you, in polynomial time, check that they really are a Secure Set? We don't know!! The characterization theorem suggests that we must check all subsets of S. There are 2 k subsets that must be checked, and k can itself be O(n). 17

That does not mean SecureSet is NOT in NP. We may just not know what "help" is needed. But, it's pretty good evidence. So, what now? 18

SecureSet Problem Given: A graph G = (V, E) and an integer k. Question: Does G have a secure set S  V with |S| ≤ k? 19

IsSecure Given: A graph G = (V, E) and a set S  V. Question: Is S a secure set? {That is,  X  S is |N[X]  S| ≥ |N[X]–S|?} This problem doesn't seem to be in NP either!!! 20

IsNotSecure (Co–IsSecure) Given: A graph G = (V, E) and a set S  V. Question: Does there exist W  S such that |N[W]  S| < |N[W]–S|? 21

Theorem. IsNotSecure is in the set NP. Given a Yes instance and a set W  S that makes it a Yes instance, we can verify W is not a secure subset in polynomial time. That's enough, and proves the claim. 22

We now show IsNotSecure is NP–Complete We need to show IsNotSecure is as "hard" as some (any) known NP–Complete problem. That is, for some problem X in NP–Complete, design a polynomial time "conversion" algorithm that transforms any instance of X into an instance of IsNotSecure, which "preserves" Yes/No answers. 23

A known NP–Complete problem: Domination Given: A graph G = (V, E) and an integer k. Question: Is there a set D  V such that |D| ≤ k and N[D] = V? 24

Theorem. IsNotSecure is NP–Complete. Proof: Let G' = (V', E') and an integer k' be an arbitrary instance of Domination. We may assume 1 ≤ k' < n = |V'|, otherwise the conclusion is immediate. 25

Our task: Provide a way to convert an arbitrary instance I of Domination into an instance, p(I), of IsNotSecure in such a way that (1) I is a Yes instance (in Domination) if and only if p(I) is a Yes instance (in IsNotSecure), and (2) Do it in polynomial time. 26

1) Vertices: V = V'  X  Y, where – (i) X is an introduced set of n 2 vertices partitioned into sets A, B, and C so that |A| = n 2 –n–k'–1, B = {b 1, b 2, …, b n }, and |C| = k'+1, and – (ii) Y is an introduced set of n 2 vertices partitioned into n sets Y 1, Y 2, …, Y n where Y i = {y i,1, y i,2, …, y i,n }, for 1 ≤ i ≤ n. 27

2) Edges: E is defined by – (i) V' induces G', X induces a complete subgraph, and – (ii) for each vertex v t  V', v t is also adjacent to a) all vertices in A, b) b t  B, and c) for 1 ≤ i ≤ n, y i,j  V i, when v j  N G' [v t ] 3) Finally, let S = X  V'. 28

Vertices in A and C are unlabeled and there are no edges between vertices of Y. For any W  S, |N[W] – S| ≤ |V – S| = n 2 and, for every x  X, |N[x]  S| ≥ n 2. Therefore, if there exists a set W  S for which |N[W]  S| < |N[W]–S|, W cannot contain any vertex in X. Hence, candidate sets W must be subsets of V'. 29

When W  V', the number of vertices in the closed neighborhood of W in V – S is exactly |N[W]–S| = n|N[W]  V'| ≤ n 2, Equality holds if and only if W dominates G'. 30

Two cases to consider for the set W  V'. Case 1: W dominates V'. Then, |N[W]  S| = |A|+|V'|+|W| = n 2 –n–k'– 1+n+|W| = n 2 –k'–1+|W|. From the previous comments, |N[W]–S| = n 2. Therefore, |N[W]  S| < |N[W]–S| if and only if n 2 –k'–1+|W| < n 2, or |W| ≤ k'. 31

Case 2: W dominates n–t < n vertices of V'. Then, |N[W]  S| = |A|+n–t+|W| = n 2 –k'–1–t+|W|. Again, since |N[W]–S| = n(n–t) = n 2 –nt, |N[W]  S| < |N[W]–S| if and only if (n–1)t < k'+1–|W|. Since t ≥ 1 and |W| ≥ 1, we must have k' ≥ n, contradicting the assumption that k' < n. Thus, in this case, N[W]  S| ≥ |N[W]–S|. 32

Therefore, W  V' is a dominating set of G' and |W| ≤ k' if and only if, in the constructed graph G, |N[W]  S| < |N[W]–S|. Thus, IsNotSecure is NP–Complete. 33

It follows that IsSecure (the complement problem of IsNotSecure) is in Co–NP Complete. Therefore, if P ≠ NP, to verify a "yes" instance of SecureSet would seem to require an oracle to first determine the set S, and then a nondeterministic algorithm to verify that S is secure. 34

On the other hand, a quite different result holds when P = NP. Theorem. If P = NP, there is a deterministic polynomial algorithm for SecureSet. 35

Proof: When P = NP, then IsNotSecure and IsSecure are both in P. Thus, there would be a deterministic polynomial verifier for SecureSet and, hence, SecureSet would be in NP. But, if P = NP, and SecureSet is in NP, then SecureSet is in P. 36

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