The proof that the Independent-Set problem is NP-complete depends on a construct

Question

The proof that the Independent-Set problem is NP-complete depends on a construction given in Theorem 10.18 (p. 460), which reduces 3SAT to Independent Sets. Apply this construction to the 3SAT instance: (u + vv + w)(-v + -w + x)(-u +-x + y)(x + -y + z)(u + -w + -z) Note that - denotes negation, e.g., -v stands for the literal NOT v. Also, remember that the construction involves the creation of nodes denoted [I, j]. The node [I, j] corresponds to the jth literal of the ith clause. For example, [1.2] corresponds to the occurrence of v. After performing the construction, identify from the list below the one pair of nodes that does NOT have an edge between them. [3, 2] and [3, 3] [1, 2] and [5, 3] [2, 3] and [3, 2] [1, 1] and [3, 1]

Explanation / Answer

[3,2] and [3,3] are part of same clause so they will be connected by edge

[1,2] = v, [5,3] = -z, these will not have edge between them as they are not part of same clause neither they are complement of each other

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