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Example Question on Linear Program, Dual and NP-Complete Proof . COT5405 Spring 11. Question. Given an undirected connected graph G = ( V,E ) and a positive integer k ≤ | V|. T wo vertices u and v are connected if and only if there exists at least one path from u to v.
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Example Question on Linear Program, Dual and NP-Complete Proof COT5405 Spring 11
Question • Given an undirected connected graph G = (V,E) and a positive integer k ≤ |V|. • Two vertices u and v are connected if and only if there exists at least one path from u to v. • For all the possible vertex pairs, we want to remove k vertices from G, so that • the number of connected vertex pairs in the resulting graph is minimized. • We call it k-CNP (critical node problem)
Integer Program • Variables: • uij = 1 if vertex i and j are connected in the resulted graph, otherwise uij = 0. Note uii = 1. • vi = 1 if vertex i is removed, otherwise vi = 0. • The objective function and 2 constraints
Leftover Connectivity • Consider node pairs i and j, where (i,j) is NOT an edge. h i j h i j
|E| rows 1 row
Ready to write dual • How many constraints in Primal – • How many constraints in Dual – • Dual variable • For the constraint on (i,j) belonging to E, define xij • For the constraint oni,j,hbelonging to V, define yijh • For the constraint on the aggregate vi, define z
NP-Complete Proof • Decision Version • Given an undirected connected graph G and positive integer k • a value L<n(n-1)/2 • is there a set of k vertices, whose removal makes the number of connected vertex pairs in the resulting graph is at most L?
In NP • Given such a set of k vertices, • Remove them from the graph, • Calculate the number of connected pairs using DFS or BFS in polynomial time, • Compare with L – Give answer: Yes or No
Is NP-hard Reduction from Vertex Cover (VC) • Instance of VC: given a graph G = (V,E) where |V|= n, is there a vertex cover of size at most k? • Instance of k-CNP: on the graph G, is there a set of k vertices whose removal makes the # of connected pairs 0?
Is NP-hard • Forward: If we can have a VC of size k ---> delete those k nodes ---> connectivity = 0 • Backward: If we can delete k nodes to make # connections 0 --> no edges left -> vertex cover of size k
NP-Completeness • In NP • NP-hard • For an alternative proof, please refer to A. Arulselvan et al, ``Detecting Critical Nodes In Sparse Graphs’’, J. Computers and Operations Research, 2009. http://plaza.ufl.edu/clayton8/cnp.pdf
Thank You Q & A