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CS 461 – Nov. 30

CS 461 – Nov. 30. Section 7.5 How to show a problem is NP-complete Show it’s in NP. Show that it corresponds to another problem already known to be NP-complete. Example NP-complete problems 3SAT, special case of SAT Clique Subset sum. Review:. Other problems.

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CS 461 – Nov. 30

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  1. CS 461 – Nov. 30 Section 7.5 • How to show a problem is NP-complete • Show it’s in NP. • Show that it corresponds to another problem already known to be NP-complete. • Example NP-complete problems • 3SAT, special case of SAT • Clique • Subset sum

  2. Review: Other problems • We now have 1 problem, “SAT”, proved to be NP-complete by definition. • To show that some other problem is NP-complete, all we need to do is create a polynomial-time correspondence with SAT, or other problem already proved to be NP-complete. • Not necessary to start over with arbitrary NTM. • Examples: • 3SAT: the satisfiability question assuming the boolean formula is of the form (_  _  _)  (_  _  _)  (_  _  _)  … i.e. CNF in which each factor has exactly 3 terms. • Clique, subset-sum, isomorphic, Hamiltonian path, etc.

  3. 3SAT • Want to show this is NP-complete. • Note that we can’t simply say “3SAT is a special case of SAT, therefore it’s NP-complete.” • Consider this special case of SAT: all boolean formulas containing just one variable and no operators. Now the problem is trivially in P. • It’s possible that a “special case” of an NP-problem could become simple enough to be in P. But in the case of 3SAT we show it’s still NP-complete. • Simple approach: re-do our proof of SAT, but this time make sure each  formula we write is expressed: • in CNF (conjunctive normal form) • with exactly 3 terms per factor

  4. 3SAT in CNF • The  we created for SAT is almost already in CNF. • CNF means an AND of OR’s. •  has 4 factors. See book for precise definitions. • 3 of the 4 factors are in CNF. Only 4th factor move needs modifying. • move is more complex: an AND of OR’s of AND’s. We can use DeMorgan’s Law to convert OR’s of AND’s into AND’s of OR’s. Now we have CNF.

  5. continued • We also need exactly 3 terms per factor (_  _  _) . • If fewer than 3, just repeat a term. e.g. (p  q  q) • To reduce 4 terms to 3, we would do this: (x1  x2  x3  x4) = (x1  x2  d)  (d’  x3  x4) where “d” is a new dummy variable. See why it works? Consider all xi false. Consider one xi true. • For more than 4 terms the technique generalizes: (x1  x2  x3  … xn) = (x1  x2  d1)  (d1’  x3  d2)  (d2’  x4  d3)…  (dn-3’  xn-1  xn)

  6. Clique problem • Given a graph G and a number k, does G contain a complete subgraph of k vertices? • Alternatively, given a graph G, find the size (k) of the largest complete subgraph. • This problem is in NP, because we could be given a subgraph to consider, and we can verify in O(k2) time that it’s complete by checking that all required edges are present. • To complete the proof that “Clique” is NP-complete, we will give a polynomial-time reduction from 3SAT to clique. • We already know that 3SAT is NP-complete. • We must convert a boolean formula  into a graph G, such that the  is true iff G has a clique.

  7. 3SAT  Clique • We’re given , which has k factors, 3 terms per factor. • Create the graph G as follows. • The vertices are arranged in groups of 3. • Each “triple” of vertices corresponds to a factor in  • Label each vertex with corresponding term (e.g. x1’). Note that the same label may appear in other triples. • Edges connect all vertices except: • Vertices within the same triple • Vertices that are opposites of each other (as x1 and x1’) • Why does this construction work? • We’ll show:  is “true”  G has a clique with k vertices.

  8.  True  clique • Suppose  = true. • One of the terms in each triple must be true. • Looking at G, the vertices are grouped in k triples. • There are lots of edges between triples. The only type of edge between triples we don’t have is between contradictory terms like x1 and x1’. • A “true term” in a triple can always be connected to a “true term” in another triple. • Therefore, there is a clique of size k. • Suppose G has a clique of size k. • Each vertex in the clique must be in a different triple. • Let’s assign truth values to each vertex to make each clique vertex true. There will be no contradictions. • Therefore, each triple will have one term that is true, so  is true.

  9. Summary • Thus far we have shown: • SAT is NP-complete • Therefore, 3SAT is NP-complete • Therefore, Clique is NP-complete • Please watch this video lecture by Ron Graham: RonaldLG1988.mpeg on the class Web site • Next: • Wrap up section 7.5 • Review

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