Lecture on Mar 23, 2004 P vs NP, NPCompleteness

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Lecture on Mar 23, 2004P vs NP, NP-Completeness

• Topics
• The Complexity Classe NP
  
• The P vs NP problem
• NP-Completeness
• Reading Sipser Sections 7.2, 7.3, 7.4

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The Class NP

• Stands for Nondeterministic Polynomial Time,
  Cook 1970.
• Nondeterminism A fundamental concept introduced
  by Rabin Scott, 1959.
• Informally, a nondeterministic algorithm is one
  endowed with the power to make a
  nondeterministic choice in each step.
• Will illustrate but wont treat it formally in
  this course.
• NP traditionally defined using nondeterministic
  TMs.
• Will give an equivalent alternative definition.

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The Class NP

• Def A language L ? NP if ? a polynomial time
  verifier V s.t.
• ? x ? L, ? a proof ?x s.t. V(x, ?x ) ? 1
  
• ?x Proof of membership of x
• ? x ? L, ? ?, V(x, ?) ? 0
• Thus nonmembers have no proof of membership
• Thus NP ? Class of all languages that are
  efficiently verifiable
• ? Class of problems whose
  solutions are efficiently verifiable

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A View of NP

• Envision a language L as consisting of correct
  statements.
• An all-powerful prover P, and a polynomial-time
  verifier V.
• P wants to prove x ? L, writes down a proof ?x
  , sends ?x to V.
• V verifies that ?x is a correct proof of x ?
  L.
• An NP proof system requires that
• Every correct statement has a proof that is
  easily verifiable.
• Every incorrect statement has no proof of
  correctness even an all-powerful malicious
  prover cannot fool you.

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Examples of NP Languages

• Many natural computational problems are in NP.
• Example Is a given integer n a composite
  number?
• COMPOSITES ? ?n? n ? p?q, p, q ? 1
• Claim COMPOSITES ? NP.
• Proof For each composite n ? p?q, (p, q) is a
  proof of compositeness.
• The verifier V, given (n, p, q), simply verifies
  that n ? p?q.
• If n is composite, then n is accepted given (n,
  p, q).
• If n is prime, then n is rejected as ? p, q ? 1,
  n ? p?q.
• In fact, COMPOSITES ? P, as PRIMES ? P.

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Examples of NP Languages

• Theorem P ? NP, i.e. every language in P is in
  NP.
• Intuitively Verifying a solution is easier than
  coming up with one.
• Proof Let L ? P, and A a polynomial-time
  algorithm that decides L.
• Then A is a verifier which can verify without any
  proof.
• There are however problems in NP which are not
  known to be in P.

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CLIQUE

• Def A clique C of a graph G is a complete
  subgraph of G, i.e. every two nodes of C are
  adjacent. A k-clique is a clique with k nodes.
• Does a given graph G has a k-clique?
• CLIQUE ? ?G,k? G has a k-clique

a 4-clique
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CLIQUE

• Claim CLIQUE ? NP.
• Proof For each ?G,k? ? CLIQUE, G has a k-clique
  C.
• The k-clique C is a proof of that G has a
  k-clique.
• The verifier V, given ?G,k,C?, simply verifies
  that C is a k-clique, i.e.
• C is a subgraph of G, C has k nodes, and there is
  an edge between every two nodes of C.
• If ?G,k? ? CLIQUE, then ?G,k? is accepted given
  ?G,k,C?.
• If ?G,k? ? CLIQUE, then ?G,k? is rejected as G
  has no k-clique.
• Do not know whether CLIQUE ? P.
• Cannot try all possible subsets of k nodes of
  such subsets ? (n/k)k , exponentially large in k.
• Still do not know how to do much better than the
  naïve method.

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Hamiltonian Path

• Def A Hamiltonian Path of a graph G is a path
  that visits each node of G exactly once. G is
  Hamiltonian if G has a Hamiltonian path.

Hamiltonian
Not Hamiltonian
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Hamiltonian Path

• Is a given graph G Hamiltonian?
• HAM ? ?G? G is Hamiltonian
• Claim HAM ? NP.
• Proof For each ?G? ? HAM, a Hamiltonian path P
  is a proof of membership.
• The verifier V, given ?G, P?, simply verifies
  that P is a Hamiltonian path,
• P is a path G, P visits each node of G exactly
  once.
• If ?G? ? HAM, then ?G? is accepted given ?G,P?.
• If ?G? ? HAM, then ?G? is rejected as G has no
  Hamiltonian path
• Do not know whether HAM ? P. Cannot try all n!
  possible paths.

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Knapsack (or Subset Sum)

• Given a set S of integers, and a target integer
  k, does S contain a subset that adds up to k?
• KNAPSACK ? ?S,k? S has a subset Y ? y1,,ym ?
  S s.t. ? yi ? k
• Claim KNAPSACK ? NP.
• Proof For each ?S,k? ? KNAPSACK, the subset Y is
  a proof of membership.
• The verifier V, given ?S,k,Y?, simply verifies
  that Y ? S, and elements of Y add up to k.
• Do not know whether KNAPSACK ? P. Cannot try all
  2n possible subsets.

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P vs NP

• Know that P ? NP.
• Grand Challenge Is P ? NP?
• One of the most important open problems of all
  sciences.
• A million-dollar problem prize presented by the
  Clay Mathematics Institute.

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What if P ? NP?

• Many optimization problems can be solved
  efficiently and exactly.
• Much of complexity theory meaningless.
• Approximation algorithms unnecessary.
• Much of cryptography collapses!
• Mathematicians can be replaced by a computer.

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P vs NP

• Common Belief P ? NP
• Philosophical Evidence Verifying a
  proof/solution seems easier than coming up with a
  proof/solution.
• Empirical Evidence Still dont know how to solve
  CLIQUE, HAMILTONIAN PATH, KNAPSACK, much better
  than the naïve method.

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NP-Completeness

• Fundamental concept introduced by Cook 1970,
  Levin 1972
• NP-Complete problems are the hardest problems
  in NP, and 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 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 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) ? (x ? ?y ? ?z) ? (x ? y ? z)
• CNF-SAT ? ??? ? is a satisfiable 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.