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CS221: Computational Complexity Prof. Salil Vadhan
Lecture 29: P
#P
⊆ IP
12/4 Scribe: Dave Fetterman
Contents
1 Announcements 1
2 The unsound protocol 1
3 Arithmetization 2
1 Announcements
- Class Friday in MD-125.
- Relevant Readings for this lecture: PCMI lecture notes (Handout), Papadimitriou 12.2.
2 The unsound protocol
Last time, we started proving the following theorem.
Theorem 1 P
#P ⊆ IP
Recall we were giving an interactive proof system for the following problem: E#SAT = {(φ, k) :
φ has exactly k satisfying assignments}.
Last time we saw a protocol exhibiting completeness but not soundness. Here is the old protocol,
which we will subsequently fix.
Observation: φ has exactly k satisfying assignments iff there exist k 0
, k 1
such that:
(x 1
, x 2
, · · · x n
) ≡ φ(0, x 2
,... , x n
) has exactly k 0
satisfying assignments.
(x 1
, x 2
, · · · x n
) ≡ φ(1, x 2
,... , x n
) has exactly k 1
satisfying assignments.
First attempt at an IP for E#SAT:
- φ, k given to both Prover and Verifier
Prover:
- Send k 0 and k 1 to Verifier
Verifier:
- Choose b ∈ { 0 , 1 } at random.
- Send b to Prover.
As in the last lecture, the prover’s strategy is to recursively show that (φb, kb) ∈ E#SAT, and at
the end of n iterations, the verifier can be sure that the trivial formulae will be true or false.
This clearly works correctly if (φ, k) ∈ E#SAT, but we see that the soundness condition can be
violated with probability exponentially near to 1, causing the verifier to accept a false input.
How do we avoid this problem?
- We have the latter two conditions from above, and on false input, we can only guarantee that
one is violated. Verifier has no way of deciding this efficiently.
- Solution: Allow the verifier choices is a larger set (larger than b ∈ { 0 , 1 }).
- Idea: Allow variables to take values in a large set F and guarantee that most values reflect
inconsistency, rather than just half of them.
3 Arithmetization
We will extend φ : { 0 , 1 }
n
→ { 0 , 1 } to
φ : F
n → F, where F is a large finite field, so
φ is a low-degree
polynomial. The benefits of this will stem from the fact that distinct, low-degree polynmials agree
in very few places. Note that we recently saw a very general way of extending boolean functions to
low-degree polys, but that only ensures that
φ can be computed in PSPACE
φ , whereas we need
the verifier to be able to verify
φ itself. Thus we must do this in a very particular way, exploiting
the structure of SAT. What we will achieve is:
φ| { 0 , 1 }
n (^) = φ.
φ) ≤ |φ|
φ(x) can be evaluated in polynomial time given φ and x ∈ F
n .
Now on to the recursive definition, φ 7 →
φ:
→ x i
φ
φ ·
ψ
∈ F randomly
(α n
φ(α 1
,... , α n
- Accept if all of these checks pass, failing otherwise.
Claim 2 This is an interactive proof for E#SAT.
Proof:
- Efficiency: V is poly-time by inspection and our choice of arithmetization.
- Completeness: By construction.
- Soundness: Suppose φ does not have exactly k satisfying assignments. We’ll show that
∀P
∗ , Pr[V accepts : (P
∗ , V )(φ, k)] ≤
nd
|F|
nd
2
d
→ 0. Let p
∗
1
,... , p
∗
n
be polynomials sent by
P
∗
. Let p 1
,... , p n
be polynomials computed according to specified P-strategy. Assume
p 1
(0) + p 1
(1) = # satisfying assignments to φ 6 = k. Then either p
∗
1
(0) + p
∗
1
(1) 6 = k or p
∗
1
= p 1
In the first case, V rejects immediately. Otherwise, w.p. ≥ 1 −
d
|F|
, p
∗
1
(α 1 ) 6 = p 1 (α 1 ) by the
Basic Theorem of Algebra (two distinct polynomials of degree d agree in at most d places).
Suppose p
∗
1
(α 1 ) 6 = p 1 (α 1 ) = p 2 (0) + p 2 (1). Then either
∗
2
(0) + p
∗
2
(1) 6 = p
∗
2
(α 1
) ⇒ V rejects immediately.
∗
2
= p 2 ⇒ w.p. 1 −
d
|F|
over α 2 , p
∗
2
(α 2 ) 6 = p 2 (α 2 )
So by union bound overall, Pr[V rejects or p
∗
n
(αn) 6 = pn(αn)] ≥ 1 − n
d
|F|
2
3
Now that we see interactive proofs are so much more powerful than classical proofs, containing all
of P
#P (and hence the entire PH, by Toda), we’d like to fully characterize their power. This was
done shortly thereafter:
Theorem 3 (Shamir) : IP = PSPACE
You will show on the problem set IP ⊆ PSPACE. The other direction is based on the arithmeti-
zation ideas above, applied to QBF. See the PCMI notes for details. But how are we to interpret
this? Think about PSPACE-complete problems like GO or CHESS. If we have Kasparov (an
“optimal chess oracle”), he can efficiently convince you (a weak, poly-time verifier) that a move is
optimal. Additionally, there emerge many nice philosophical interpretations. For example, it vali-
dates our everyday intuition that one can be taught more through interaction (as in a classroom)
than just by reading.
