Summary of previous class

1
Summary of previous class

• Did reductions
• 0/1-LP
• 0/1-LPE
• Integer LP
• Poly Eq over 0,1
• Defined class coNP L ?L?NP
• Is NP?coNP empty?

2
Main Question

• Of course NP ? co-NP
• Most think not hard to believe that there is a
  short certificate for tautology (that all
  assignments satisfy formula)
• If PNP then co-NPNP (prove!)
• So if we prove co-NP?NP then P?NP must hold.

3
Co-NP completeness

• Definition A language L is said to be
  coNP-complete if it is in co-NP and every co-NP
  language is polynomial time Karp reducible to it.
• Tautologies are true Boolean formulas
• TAUT ? ? is satisfied by every assignment
• co-SAT? ? unsatisfiable formula in CNF

4
Co-NP completeness (continued)

• Theorem TAUT and coSAT are coNP-complete.
• Proof Consider L?co-NP. Then for x??L using
  Cook-Levin reduction we have satisfiable ?x. Then
  for y?L ( ? ?L) the corresponding ?y must be
  unsatisfiable so co-SAT is coNP-complete.
  However, for this reason ??y must be a a true
  formula (a tautology), so TAUT is coNP-complete.

5
Co-SAT and proof systems

• Definition A proof system P for co-SAT is a
  2-tape TM P(x,y) such that
• P runs in time polynomial both in x and in y
• CNF ? is in co-SAT if and only if there exists a
  string ? (proof) such that P(?,?) returns yes.
• Definition A proof system P is a super proof
  system if for any CNF ? there exists a proof ?
  that is polynomial with degree k in ? and P(?,?)
  returns yes
• (i.e. ?k???? (???K) ? (P(?,?) true)

6
Co-NP?NP and super proof system

• Theorem (Cook-Rehow). There exists a super proof
  system (SPS) if and only if NPcoNP
• Proof (?)Let P be such SPS. So for any
  unsatisfiable CNF ? there is a proof ? s.t.
  ??k and P(?,?)1, i.e. co-SAT?NP, so
  co-NP?NP. But then SAT is in co-NP too (co-SAT is
  in NP). Hence NP?co-NP.
• (?) Let coNPNP. Then co-SAT?NP so for every
  unsatisfiable ? there is a polynomial proof
  certificate c such that V(?,c) where V is a
  verifier for co-SAT that runs in polytime in
  first input.

7
Finding proof resolution

• Resolution proof is allowed to use the following
  inference rules
• Resolution
• Weakening
• .Soundness if the premise(s) of the rule is
  (are) satisfiable then so are the conclusion of
  the rule.

8
Resolution proofs

• Let ??i1m Ci be a CNF formula with m clauses
  over n variables. Clauses Ci are also called
  axioms
• A resolution proof of length s for ? is a
  sequence ? (L1,L2,,Ls) is of lines such that
• Each line is a clause a clause can be either an
  axiom or can be derived from previous lines using
  resolution and/or weakening rules
• Last line is an empty clause ?

9
Completeness of resolution

• Theorem A CNF formula is unsatisfiable if it has
  a resolution proof.
• (?) follows from soundness (we didnt loose any
  satsisfying assignment)
• (?)By induction on a number of variables. Basis
  n0, proof consists only of ?
• Hypothesis all unsatisfiable CNFs over n-1
  variables have resolution proofs.
• Step From ? over n variables construct ? over
  n-1 variables s.t. ? is unsatisfiable and is
  derived from ? using resolution and weakening.

10
Completeness of resolution (continued)

• Fix ??0,1n-1 an assignment to n-1 variables.
  Since ? is unsatisfiable it must have C0 and C1
  clauses such that C0 is falsified when we further
  assign xn0 (i.e. on ?,0) and C1 is falsified
  when we further assign xn1 (i.e. on ?,1).
• If C0 does not contain literal xn (since is false
  under ?,0 cannot contain ?xn), let C?C0
• If C1 does not contain literal ?xn (since is
  false under ?,1 cannot contain xn), let C?C1
• Otherwise C0D v xn and C1E v ?xn for some D and
  E clauses on n-1 variables. Then let C?DvE
  derived by one application of resolution rule
• Let ?? C?. It is false by soundness, so it
  has a
• resolution proof by inductive hypothesis.

11
Measures for resolution.

• LengthR(?) minimum length of resolution proof
  for ? (by convention ? if ? satisfiable).
• WidthR(?) minimum (among all proofs) of a
  maximum size of the clause in a resolution proof
  for ? (by convention ? if ? satisfiable).

12
Hard formulas for resolution.

• Pigeonhole principle PHPn one can put n1
  pigeons into n holes that can fit only one
  pigeon
• Variables xij ith pigeon sits in jth hole
• Clauses
• pigeon I occupies at least one hole Pi Vj1n
  xij
• Pigeons i and j cannot both occupy hole k
  Hi,j,k?xik v?xjk

Theorem (Haken, 1989) LengthR(PHPn)2?(n)
13
Another important result

• Not only resolution is not a super proof system
  but also it cannot show that a given function is
  not in P. More exactly
• Theorem (Raz 2002) Resolution cannot efficiently
  prove statements of the type function f is not
  in P/poly.
• P/poly is a class of algorithms that run in
  poly-time using subroutines that work in
  poly-time but can use different subroutine for
  each input length. Of course P?P/poly.

14
Randomized computations

• Should have a model that uses random numbers
• Would be natural to have it by extending models
  for deterministic and non-deterministic
  computations

15
Probabilistic TM

• Probabilistic TM (PTM) is a TM that has two
  associated transition functions ?0 and ?1. In
  addition it has a read-only service tape.
• Execution of PTM on an instance of a problem
  begins with a request to an oracle to write a
  random number (infinite sequence of randomly
  generated 0s and 1s) on a service tape.
• At any further step of an execution the machine
  reads a symbol from a service tape, moves service
  head right and if the symbol was 0 then PTM
  applies ?0, otherwise it applies ?1.

16
PTM-continued

• On an input x a random number r defines a trace
  Tr of a PTM M that uses kkM(r,x) random bits
  from r
• We define a probability of the trace Tr to be
• Pr(Tr)2-k
• The set S of all traces of M on x with
  probability distribution Pr is a probability
  space, i.e. ?t?SPr(t)1
• Let ACC(x) be a subset of S containing traces
  that accept x, then Pr(M(x)1)?t?Acc(x)Pr(t)

17
Runtime of the PTM

• Worst case runtime of PTM M on input x
• TimeM(x) max time(t) t?S,
• Worst case runtime of PTM M
• TimeM(n) max TimeM(x) x is an input, xn
• Expected runtime of PTM M on input x
• ETimeM(x) ?t?SPr(t)time(t)
• Expected runtime of PTM M
• ETimeM(n)maxETimeM(x) x is an input, xn

18
Randomized Algorithms

• Definition of randomized complexity classes in
  terms probabilistic Turing machines.
• Las Vegas algorithms Answers are always correct,
  but runtime depends on random choices. Example
  Randomized Quicksort.
• Monte Carlo algorithms Runtime does not depend
  on random choices, but answers may be incorrect
  (with small probability).

19
Las Vegas algorithms

• We define class ZTIME(T(n)) of PTMs that never
  errs and runs in expected time T(n). That is PTM
  M ?ZTIME(T(n)) if the following conditions hold
• x?L ? Pr(M accepts x)1
• x?L ? Pr(M holds without accepting x)1
• ETimeM(n) O(T(n))
• Las Vegas algorithms are algorithms in the class
  ZPPUcgt0ZTIME(nc).

20
Las Vegas algorithms - Choicen

• Can output ? I cant solve the problem in one
  random attempt if x?L
• Problem decide Choicen 0,1n?1,,n?0,1
  using n/21 bits.
• Alice and Bob need to find a solution for the
  following problem
• Alice will be given n bit binary number, Bob will
  be given a number j between 1 and n. They need to
  find out what is the value of jth in Alices
  number. Alice is allowed to show Bob less than n
  bits.

21
Choicen - solution

• Before starting with the problem Alice and Bob
  agree that if the first bit that Alice shows is 0
  then the rest are first-half bits, if she shows 1
  then the rest are second-half bits.
• Alice chooses a bit at random and shows to Bob
  this bit first and then corresponding half.
• If Bob see the needed bit he gives an answer.
  Otherwise he replies do not know. Alice shows the
  other half
• E(time(n))1/211/221.5 lt2!