there are several ways we can define these sets. we will discuss two
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The Evolution of a Hard Graph Theory Problem – Secure Sets Ron Dutton Computer Science University of Central Florida. Many real world problems involve a collection of entities (e.g., individuals , businesses, or countries) that compete for each others resources . - PowerPoint PPT PresentationTRANSCRIPT
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The Evolution of a Hard Graph Theory Problem – Secure Sets
Ron DuttonComputer Science
University of Central Florida
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• 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.
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• 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.
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• There are several ways we can define these sets. We will discuss two.
Secure Sets and Defensive Alliances.
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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.
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• 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.
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Secure SetsAny 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.
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• 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.
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• 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.
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• 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.
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• 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).
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• 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.
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• 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.
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• 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!!
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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(n2) time at most).
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• That proves Clique is in NP.
• Similarly, Defensive Alliance can be shown to be in NP.
• But, how about SecureSet?
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• 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 2k subsets that must be checked, and k can itself be O(n).
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• 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?
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SecureSet Problem
• Given: A graph G = (V, E) and an integer k.
• Question: Does G have a secure set S V with |S| ≤ k?
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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!!!
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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|?
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• 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.
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• 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.
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• 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?
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• 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.
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• 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.
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• 1) Vertices: V = V' X Y, where– (i) X is an introduced set of n2 vertices partitioned
into sets A, B, and C so that |A| = n2–n–k'–1, B = {b1, b2, …, bn}, and |C| = k'+1, and
– (ii) Y is an introduced set of n2 vertices partitioned into n sets Y1, Y2, …, Yn where Yi = {yi,1, yi,2, …, yi,n}, for 1 ≤ i ≤ n.
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2) Edges: E is defined by– (i) V' induces G', X induces a complete subgraph,
and– (ii) for each vertex vt V', vt is also adjacent to• a) all vertices in A,• b) bt B, and
• c) for 1 ≤ i ≤ n, yi,j Vi, when vj NG'[vt]
3) Finally, let S = X V'.
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• 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| = n2 and, for every x X, |N[x] S| ≥ n2.
• 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'.
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• 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'| ≤ n2,
Equality holds if and only if W dominates G'.
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• Two cases to consider for the set W V'.
• Case 1: W dominates V'. Then, |N[W] S| = |A|+|V'|+|W| = n2–n–k'–1+n+|W| = n2–k'–1+|W|. From the previous comments, |N[W]–S| = n2.
Therefore, |N[W] S| < |N[W]–S| if and only if n2–k'–1+|W| < n2, or |W| ≤ k'.
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• Case 2: W dominates n–t < n vertices of V'. Then, |N[W] S| = |A|+n–t+|W| = n2–k'–1–t+|W|. Again, since |N[W]–S| = n(n–t) = n2–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|.
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• 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.
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• 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.
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• On the other hand, a quite different result holds when P = NP.
• Theorem. If P = NP, there is a deterministic polynomial algorithm for SecureSet.
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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.
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• 4. References/Bibliography• [1] Robert C. Brigham, Ronald D. Dutton, and Stephen T. Hedetniemi, Security in graphs, Discrete Applied
Mathematics, 155 (2007), no. 13, pp. 1708-1714.• [2] R. D. Dutton, R. Lee, and R. C. Brigham, Bounds on a graph's security number, Discrete Applied
Mathematics 156 (2008), no. 5, pp. 695-704.• [3] Ronald D. Dutton, On a graph's security number, Discrete Mathematics, 309 (2009), pp. 4443-4447.• [4] Ronald Dutton, Secure set algorithms and complexity, Congressus Numerantium 180 (2006), pp. 115-121.• [5 ] Rosa I. Enciso and Ronald D. Dutton, Parameterized complexity of secure sets, Congressus Numerantium
189 (2008), pp. 161-168.• [6] M. Garey and D. Johnson, Computers and Intractability: the Theory of NP-Complete, (1979) W H.
Freeman and Co. NY.• [7] Yiu Yu Ho and Ronald Dutton, Rooted secure sets of trees, AKCE, J. Graphs Combin, 6, No. 3 (2009), pp.
373-392. .• [8] K. Kozawa, Y. Otachi, and K. Yamazaki, Security number of grid-like graphs, Discrete Applied Mathematics,
157 (2009), pp 2555-61.• [9] P. Kristiansen, S. M. Hedetniemi, and, S. T. Hedetniemi, Alliances in graphs, J. Combin. Math. Combin.
Comput. 48 (2004), pp. 155-177.• [10] Khurram H. Shafique and R. D. Dutton, On satisfactory partitioning of graphs, Congressus Numerantium
154 (2002), pp. 183-194.