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Interactive proof systems, a type of proof system that adds randomization and interaction to classical proof systems. The document focuses on the application of interactive proof systems to the problem of graph nonisomorphism. Interactive proof systems are philosophically interesting, potentially allowing for the proof of a larger class of assertions, and have applications in cryptography and efficiency in verification. An example of an interactive proof system for graph nonisomorphism and proves its completeness and soundness.
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CS221: Computational Complexity Prof. Salil Vadhan
12/02 Scribe: Ruggiero Cavallo
1 Announcements 1
2 Agenda 1
3 Classical Proof Systems 1
4 Interactive Proof Systems 2
1 if number of n inputs equal to 1 ≥ n/ 2 0 otherwise
In classical proof systems, we write down mathematical assertions as strings. Then, we have some language L that equals the set of “true” assertions. For example,
Sat ↔ true statements of form “φ is satisfiable.”
A proof system for a language L is a verification algorithm V that has the following properties:
Note that NP = languages with classical proof systems (think of V as checking the relation of the NP language).
For interactive proof systems, we add two new ingredients:
Why introduce these two ingredients?
Example: interactive proof system for Graph Nonisomorphism.
G = ([n], E), permutation π : [n] → [n] (where [n] = {1,... , n})
π(G) , ([n], {(π(u), π(v)) (u, v) ∈ E})
G and H are isomorphic (written G ∼= H) if ∃π such that π(G) = H
Graph Isomorphism is the language GI = {(G 0 , G 1 ) : G 0 ∼= G 1 }.
Note that you can make 2/3 in ( 1 ) and 1/3 in ( 2 ) arbitrarily close to 1, 0, respectively, by the usual method of repeating many times and having the verifier rule by majority. Also note that there is no constraint on P ’s efficiency.
We define IP to be the class of languages with interactive proof systems.
Theorem 4 GNI ∈ IP
How big is IP?
Theorem 5 (Lund, Fortnow, Karloff, Nisan ’92) co-NP ⊆ IP, in fact P#P^ ⊆ IP.
Proof Sketch: Convince poly-time verifier that a formula is UNsatisfiable.
E#SAT = {(φ, k) : φ has exactly k satisfying assignments}
Observation: φ has exactly k satisfying assignments iff ∃k 0 , k 1 such that:
φ and k are given to both the verifier V and the prover P.
P:
The prover will recursively prove that (φb, kb) ∈ E#SAT. Each time we’re eliminating one variable; at the end the verifier can just check for itself.
But actually, if (φ, k) ∈/ E#SAT, V might accept with probability exponentially close to 1 (i.e., 1 − 2 −n). We’ll fix this next time.
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