Lecture on Mar 30, 2004 NPCompleteness

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Lecture on Mar 30, 2004NP-Completeness

• Topics
• NP-Completeness
• Reading Sipser Sections 7.4

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NP-Completeness

• Fundamental concept introduced by Cook 1970,
  Levin 1972
• NP-Complete problems are the hardest problems
  in NP, and they do exist
• An NP-Complete problem represents the entire
  class NP
• To study the P vs NP problem, it suffices to
  study any single NP-Complete problem
• Formal definition of NP-Completeness, by
  polynomial-time reduction, next.

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Polynomial-Time Reduction

• Def Language A is polynomial-time reducible to
  language B, written A ?P B, if ? a
  polynomial-time computable function f ? ? ?
  such that ? w ? ?, w ? A iff f(w) ? B. The
  function f is called a polynomial-time reduction
  from A to B.

A
B
f
f
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NP-Completeness

• Def A language L is NP-Complete if
• L ? NP
• Every language in NP is polynomial-time reducible
  to L

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Simple Facts

• Lemma 1 If A ?P B and B ? P, then A ? P.
• Proof
• Let M be a poly-time algorithm that decides B.
• Let f be a poly-time reduction from A to B.
• To decide A in poly-time On input w,
• Compute f(w)
• Run M on f(w) and output M(f(w))

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Simple Facts

• Lemma 2 If L is NP-Complete and L ? P, then P ?
  NP.
• Proof
• Since L is NP-Complete, by definition every
  language A ? NP is poly-time reducible to L.
• If L ? P, then by Lemma 1, A ? P .
• Thus NP ? P ? P ? NP (as P ? NP)
• Cor 3 P ? NP iff any NP-Complete language is in
  P.

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Simple Facts

• Lemma 3 If A is NP-Complete and A ?P B, then B
  is NP-Complete.
• Proof
• Suppose that A is NPC. Let L be any language in
  NP.
• By definition, ? a poly-time reduction f from L
  to A.
• Let g be a poly-time reduction from A to B.
• Then g?f is a poly-time reduction L to B.
  (Verify!)
• Cor 4 To show that language B is NP-Complete, it
  suffices to show
• B ? NP
• ? an NP-Complete language A s.t. A ?P B

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Satisfiability

• Are there NP-Complete languages? YES!
• Boolean formula An expression involving Boolean
  variables and operations. E.g. ?(x,y,z) ? (?x ?
  y) ? (x ? ?z)
• ? is satisfiable if some assignment of 0/1 to
  variables makes ? evaluate to 1.
• ?(x,y,z) ? (?x ? y) ? (x ? ?z) is satisfied by
  the assignment x ? 0, y ? 1, z ? 0.
• SAT ? ??? ? is a satisfiable Boolean formula

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Satisfiability

• SAT ? NP
• Proof
• For each ??? ? SAT, a satisfying assignment A for
  ? is a proof of membership.
• The verifier V, given ??, A?, simply verifies
  that A satisfies ?.
• If ??? ? SAT, then ??? is accepted given ??, A?.
• If ??? ? SAT, then ??? is rejected given any ??,
  A?, as ? has no satisfying assignment.

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Satisfiability

• Literal A variable or a negated variable. E.g.
  x, ?y
• Clause Literals connected with ?
• E.g. (x ? ?y ? ?z)
• A Boolean formula is in conjunctive normal form,
  or a CNF-formula, if it consists of clauses
  connected by ?
• E.g. ?(x,y,z,w) ? (x ? ?y ? ?z ? w) ? (x ? y ?
  z ? ?w)
• A CNF-formula ? is a 3-CNF-fomula if each clause
  of ? has 3 literals.
• E.g. ?(x,y,z,w) ? (x ? y ? z) ? (x ? ?y ? w)
• CNF-SAT ? ??? ? is a satisfiable CNF-formula
• 3-SAT ? ??? ? is a satisfiable 3-CNF-formula

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The Cook-Levin Theorem

• Theorem (Cook 1970, Levin 1972) SAT is
  NP-Complete. Further, CNF-SAT is also
  NP-Complete.
• Corollary 3-SAT is NP-Complete.
• Proof of Corollary
• Clearly 3-SAT ? NP.
• CNF-SAT ?P 3-SAT.

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CNF-SAT ?P 3-SAT

• Let ? be a given CNF-formula.
  Goal
  Transform ? into a 3-CNF-formula ? s.t. ? is
  satisfiable iff ? is
• Transform each clause of ? into a conjunction of
  clauses of size 3.
• Replace (a1 ? a2 ? a3 ? a4) by (a1 ? a2 ? z) ?
  (?z ? a3 ? a4)
• z a new variable
• In general, replace (a1 ? a2 ? ? ? ak) by
  
  (a1 ? a2 ? z1) ?
  (?z1 ? a3 ? z2) ? (?z2 ? a4 ? z3) ? ? ? (?zk-3 ?
  ak-1 ? ak)
• z1 , ? , zk-3 new variables

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CLIQUE is NP-Complete

• Already showed that CLIQUE ? NP.
• Will show that 3-SAT ?P CLIQUE, i.e. ? a
  poly-time reduction f that
• takes as input a 3-CNF-formula ???
• outputs ?G,k? ? f(???) s.t. ? is satisfiable iff
  G has a k-clique

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Poly-time Reduction from 3-SAT to CLIQUE

• Input A 3-CNF-formula ???. Let k ? of clauses
  in ?.
• Output ?G,k?, where G is constructed as follows
• Nodes in G divided into k triples
• Each triple corresponds to a clause in ?.
• Each node in a triple corresponds to a literal in
  the corresponding clause.
• Edges in G connect all but two types of nodes
• No edge between any two nodes (literals) in the
  same triple (clause)
• No edge between any two nodes (literals) with
  contradictory labels, i.e. one labeled with x and
  the other labeled ?x for some x.

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Poly-time Reduction from 3-SAT to CLIQUE

• Example ? ? (x ? y ? z) ? (?x ? ?y ? ?z) ? (?x
  ? y ? z)

x
y
z
?x
?x
y
?y
z
?z
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Poly-time Reduction from 3-SAT to CLIQUE

• Claim ? satisfiable ? G has a k-clique
• ? satisfiable ? ? an assignment s.t. each of
  the k clauses of ? has at least 1 satisfied
  literal
• Take 1 satisfied literal from each clause. Then
  by construction of G, the corresponding k nodes
  in G form a k-clique in G
• No two of the k literals can be contradictory, as
  all the k literals are satisfied.
• Each of the k literals (nodes) comes from a
  distinct clause (triple).

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Poly-time Reduction from 3-SAT to CLIQUE

• Claim G has a k-clique ? ? satisfiable
• Suppose G has a k-clique C
• Then by construction of G,
• Each of the k nodes (literal) of C comes from a
  distinct triple (clause)
• No two of the k nodes can be contradictory
• Hence we can satisfy all the k corresponding
  literals, thus all the k clauses, and thus ?.