18.405J/6.841J Advanced Complexity Theory
Lecture 17
Lecturer: Dan Spielman
April 12, 2001
Scribe: Ronnie Misra
Recap
Last time, we discussed the class of \Arthur/Merln games". Recall that:
Denition 1 L 2 AM(k(n)) if 9 a PTIME verier A and a polynomial
p(n) s.t.
x 2 L =) 9 a p(n)-prover P s.t. Pr(A ! P) accepts] > 23
x 2= L =) 8 p(n)-provers P, Pr(A ! P) accepts] < 31
In this lecture, we will show some applications of IP, and discuss the class PCP.
Graph Isomorphism
Although there are no known PTIME algorithms for graph isomorphism, we can show that ISO is
probably not NP-complete:
Theorem 2 If ISO is NP-complete, then 3 = 3 .
Proof In the problem set, we showed that MA AM (proved as NP BP P BP NP P).
P
P
In fact, the AM speedup theorem says that:
AM(k(n)) AM( (2 ) + 1)
=) AM(k) AM 8 constants k
=) MAM AM
We also showed in the problem set that AM ( = BP NP P ) NP/poly.
k n
ISO is NP-complete
=) NISO is coNP-complete
=) coNP MAM
=) coNP AM
=) coNP NP/poly
=) 3 = 3 .
P
P
Probabilistically checkable proofs
We would like to describe a \tough verier" that accepts a proof that can be \spot checked".
Denition 3 A (R(n), Q(n)) verier is a probabilistic, PTIME oracle TM M that
P
gets R(n) random bits
is limited to Q(n) bits from its oracle P
Denition 4 L 2 PCP(R(n), Q(n)) if 9 a (R(n), Q(n)) verier V s.t.
x 2 L =) 9 s.t. PrV � (x) accepts] = 1
x 2= L =) 8 , PrV � (x) accepts] < 12
Here, is the probabilistically checkable proof (PCP). Note that we don't really think of as
an oracle. Instead, we are using OTM notation just as a way to show that V gets access to random
bits of .
Theorem 5 NEXP = PCP(Poly(n), Poly(n))
Even though proofs for languages in NEXP are exponentially long, randomized access allows
such proofs to be veried in PTIME.
Theorem 6 NP = PCP(O(log n), O(1))
As a consequence, we can use PCP to show that some approximation problems are NP-hard.
3SAT approximation is NP-hard
3SAT NP = PCP(O(log n), O(1))
=) 9 a (O(log n), O(1)) verier for 3SAT.
Assume V is non-adaptive, or that it ips R = O(log n) coins, then queries the proof at Q = O(1)
places determined only by these random bits. (If V is adaptive, we can construct a non-adaptive
version this version may use exponentially more queries, but this is still a constant.)
On input , assume V gets random bits r 2 f0 1g . Let q 1 : : : q denote the indices of the
Q bits of that V queries. Thus, if = 1 2 : : : , V reads 1 : : : .
For each r 2 f0 1g , construct a small 3SAT instance on inputs 1 : : : and some
auxilliary bits. Assume has at most S clauses. Since Q = O(1), S = O(1). should be satised
when 1 : : : have values that would cause V to accept.
Let = 01
is polynomial in the size of , and has inputs 1 : : : as well as all the auxilliary bits.
also has some special properties:
2 3SAT =) 2 3SAT, since V will accept for any r 2 f0 1g
2= 3SAT =) Pr V rejects] > 21
01
=) 8 , at most 21 of the clauses are satisable (from aux bits)
=) 8 , 8 aux, there is at least one unsatised clause in at least 12 of the clauses
=) at least a 21 fraction of the clauses of are unsatisable
=) the maximum fraction of satisable clauses of is (1 - 21 ) =) approximating 3SAT
within a factor of 41 is NP-hard, where S is some constant dependant on Q
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