Summary of Previous Lecture and Plan

1
Summary of Previous Lecture and Plan

• Did last time
• Poly-time verifiers (for Vertex cover)
• P vs NP question
• Implications if PNP
• Plan for today
• Reductions, NP hardness, NP completeness
• Oblivious TM
• SAT is NP complete (Cook-Levin Theorem)
• 3-SAT is NP complete and other reductions

2
Reductions, NP-Hardness, NP-Completeness.

• A?? is polynomial-time (Karp) reducible to B??
  if there is a poly-time computable function f ?
  ?? such that for every x?? holds that
  x?A?f(x)?B. We write A?pB
• B is called NP-hard if every language in NP is
  poly-time reducible to B (?A ?NP? A?pB)
• B is NP-complete if it is NP hard and it is in NP.

3
Facts about reducibility.

• Transitivity A?pB and B?pC, then A?pC (needs
  proof why?)
• a. If A is NP complete and we find a poly- time
  algorithm for A then PNP (why?)
• b. An NP-hard language A is in P if and only if
  PNP
• The following language is NP-complete
  TMSAT(ltMgt,x,1p,1t) ?u?0,1p such that
  M outputs 1 on ltx,ugt within t steps
• (Why?)

4
Not TMSAT - CNFs

• TMSAT defined in terms of Turing Machine and
  there is no hint of how to understand
  NP-completeness in independent terms not much
  of new insight !!!
• Boolean formula ? over n-variables is in CNF if
• All negations are directly preceding variables
  (?xi ). We call a variable or its negation a
  literal
• All disjunctions are inside parenthesis
• C(Vj1J lj), where a term in parentheses
  is called clause, and lj is jth literal of this
  clause
• ? is a conjunction of clauses (?i1m Ci).
• Formula is k-CNF if clauses have at most
  k-literals

5
Expressiveness of CNFs

• Equality of 2 binary strings of length n is
  expressible in CNF
• (x1 v ?y1) ? (?x1 v y1)? ? (xn v ?yn)?(?xnv yn)
• Claim For every Boolean function f0,1n?0,1
  there is an n-variable CNF ? of size n2n such
  that ?(u)f(u) for every u?0,1n.
• Proof ?u?0,1n there exists a clause Cu such
  that Cu(u)0 and Cu(v)1 when u?v. For given f
  let Suf(u)0 and ?f?u?S Cu. Then for every
  u such that f(u)0 holds ?f(u)0 and for every v
  such that f(v)1 holds ?f(v)1

6
SAT

• SAT is a language of all satisfiable CNF formulas
• k-SAT is a language of all satisfiable k-CNF
  formulas
• COOK-LEVIN Theorem SAT is NP-complete.
• Corollary 3-SAT is NP complete (poly-time
  reduction).