Title: CSci 5403 Lecture 15
1CSci 5403
COMPLEXITY THEORY
LECTURE XVII INTERACTIVE PROOFS I
2WHAT IS A PROOF?
G ? HAM v1,v2,,vn
Prover
Verifier
A traditional proof is a static sequence of
symbols that establishes that x?L.
NP languages with short proofs
3INTERACTIVE PROOFS
x?L
?
?
Unbounded Prover
Efficient Verifier
almost certainly so.
4EXAMPLE
Recall that graphs H and G are isomorphic, written
HG, if there is a 1-1, onto p G ? H such that
(u,v) ? G iff (p(u),p(v)) ? H.
Then ISO (H,G) HG ? NP the proof is p.
Thus NONISO (H,G) H ? G ? coNP.
ISO is neither known to be in P nor NP-complete.
Is NONISO ? NP? Seems hard
5INTERACTIVE PROOFS
G0 ? G1
The proof is too long for you to read.
I am crushed.
Intriguing.
Unbounded Prover
Traditional Prover
Efficient Verifier
6INTERACTIVE PROOFS
Definition. An interactive proof system has two
players
?
The conversation (V?P)(xr) is a1,a2,,ak1.
(V?P)(xr) accepts if ak1 accept
We say that (V?P)(xr) has k rounds.
7INTERACTIVE PROOFS
Definition. Pr(P?V)(x) accepts
Prr(P?V)(xr) accepts PrV accepts x maxP
Pr(P?V)(x) accepts
- Definition. (P,V) is an interactive proof system
for L - x?L ? PrV accepts x 2/3
- x ? L ? PrV accepts x 1/3
(Completeness)
(Soundness)
Definition. L?IPk if ?k(n)-round interactive
proof system (P,V) for L and constant c such
that (P?V)(xr) xc
Definition. IP ?c IPnc
8DEFINITION ISSUES
Theorem. If L has an interactive proof system
with deterministic verifier, then L ? NP.
Theorem. IP ? PSPACE.
Proof. Given V, x we can find the set of
responses that maximize PrV accepts x in
polynomial space.
Theorem. Let pin(n) minx?Ln0,1n PrV accepts
x, pout(n) max x??n0,1n PrV accepts
x. If pin(n) pout(n) 1/poly(n), then L?IP.
Proof. Iterate O(1/(pin(n) pout(n))2) times.
9ARTHUR-MERLIN PROOFS
An Arthur-Merlin verifier has no private
randomness
x ? L
a1 random r1
a2 P(x,r1)
a3 random r2
?
ak P(x,a1,,ak-1)
Arthur
Merlin
V(x,a1,,ak) accept/reject
Definition. L ? AMk if there exists
k(n)-round Arthur-Merlin interactive proof system
for L.
Definition. AM AM2. MA MA2
10NONISO ?AMk?
b
G0 ? G1
Unbounded Prover
Efficient Verifier
Theorem. For every N ? N, IP(n) ?
AM(n)2.
We will show a simpler result NONISO ? AM.
11Consider the set S H H G0 or H G1 .
Assume neither Gb has nontrivial automorphisms.
Then G0G1 ? S n!, but G0?G1 ? S2n!
Idea use hashing to prove that S2n!
Lemma. Let H X ? T be a pairwise
independent hash family, and let W ? X . Then
W/T - ½ (W/T)2 Prh,t?w ? W. h(w)t
W/T.
Proof. Apply Inclusion-Exclusion, using ?w,h .
Prth(w)t 1/T and ?w1,w2,t.
Prhh(w1)h(w2)t 1/T2.
12NONISO ? AM2
Let H 0,1n² ? 1,2,,4n! be a family of
pairwise-independent hash functions.
h ?R H, t ?R 4n!
G, b, p
accept if h(G)t and G p(Gb)
Arthur
Merlin
PrV accepts Pr?G ? S. h(G)t.
G0G1 ? Sn! ? PrV accepts n!/4n! ¼.
G0?G1 ? PrV accepts n!/2n! ½(n!/2n!)2
3/8.
13Theorem. If L has a k-round AM proof system
with soundness error 1/3, it has a k-round AM
proof system with soundness error 2-n.
Theorem. If L has a k-round AM proof system with
completeness 2/3, then L has a k1 round AM
proof system with completeness 1.
Proof. See homework.
Theorem. AMO(1) AM.
14Theorem. If ISO is NP complete, then PHS2.
Proof. If ISO is NP complete, then NONISO
is coNP complete.
Then there exists such that (?) ? NONISO iff
?y.?(y).
Because NONISO ? AM, there exists V such that
maxa PrV(x,r,a) 1 1 when x ? NONISO and
maxa PrV(x,r,a) 1 lt 1 otherwise.
Thus ?r?x,a V((?(x,)),r,a) iff ?x?y ?(x,y).
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