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CS221: Computational Complexity Prof. Salil Vadhan
Lecture 17: Randomized Complexity
and Identity Testing
10/30 Scribe: Grant Schoenebeck
Contents
1 Identity Testing 1
2 Randomized Complexity Classes 3
1 Identity Testing
Given algebraic expressions p(x 1 , x 2 ,... , xn), q(x 1 , x 2 ,... , xn) decide is p ≡ q?
We can interpret equivalence in two different ways: as formal polynomials or as functions from Zn^ → Z. For polynomials over Z, these two interpretations are the same.
We gave the following randomized algorithm to solve the problem:
- Let m = max{|p|, |q|}, M = 2m, S = { 1 ,... , M }.
- Randomly choose α 1 , α 2 ,... , αn ⇐ S.
- Accept if p(x 1 , x 2 ,... , xn) = q(x 1 , x 2 ,... , xn).
Proposition 1
if p ≡ q then algorithm always accepts.
if p 6 ≡ q then algorithm accepts with probability ≤ (^) Mm = 2−Ω(m)^ (over its coin tosses).
is obvious.
follows from the 2 lemmas below.
Definition 2 The total degree of a polynomial is the maximum of the sums of coefficients in each term (when written in canonical form
i 1 ,...,in∈N ci^1 ,...,in^ x
i 1 1 x
i 2 2 · · ·^ x
in n )
Example 3 The total degree of f (x, y) = x^2 y^3 + x^4 is 5.
Lemma 4 The total degree of expression p(x 1 , x 2 ,... , xn) is at most |p|.
Proof:
(by induction on |p|)
deg(p 1 · p 2 ) = deg(p 1 ) + deg(p 2 ) ≤ (by inductive hypothesis)|p 1 | + |p 2 | ≤ |p 1 · p 2 |. A similar argument shows that deg(p 1 + p 2 ) ≤ |p 1 + p 2 |.
Lemma 5 (Schwartz–Zippel) If r is a nonzero polynomial,
Pr (α 1 ,α 2 ,...,αn)←S
[r(α 1 , α 2 ,... , αn) = 0] ≤
deg r |S|
where deg r is the total degree of r.
Proof: (By induction on n, the number of variables)
Base case: n = 1. Prα←S [r(α) = 0] ≤ deg |S|^ r. This follows from a fundamental theorem of algebra that states the the number of solutions to a non-zero polynomial is at most the degree of the polynomial.
Inductive step. Observe that
r(x 1 ,... , xn) =
∑^ k
i=
ri(x 1 , x 2 ,... , xn− 1 )xin, (1)
where rk 6 = 0. Then
Pr[r(¯α) = 0]
≤ Pr[rk(α 1 , α 2 ,... , αn− 1 ) = 0] + Pr[r(α 1 , α 2 ,... , αn) = 0|rk(α 1 , α 2 ,... , αn− 1 ) 6 = 0]
≤
deg rk |S|
k |S|
The first term follows from the inductive hypothesis, while the second term is a result of the same algebraic property we used in the base case. Note now that deg rk + k ≤ deg r, by Equation (1).
- If x ∈ L ⇒ Prcoin flips [M (x) accepts ] ≥ 12.
- If x 6 ∈ L ⇒ Prcoin flips [M (x) accepts ] = 0
Thus, the algorithm can make errors only on yes instances. We stress that the above holds for all inputs; the randomness is only over the algorithm’s coin tosses. We have shown:
Theorem 13 IdentityTest is in co-RP.
Lemma 14 (RP Amplification) If L ∈ RP ⇒ ∀ polynomial p ∃ an RP algorithm for L with error probability ≤ 2 −p(n)
Proof: Given error 12 algorithm M for L, run M (x) p(n) times and accept if M ever accepts, otherwise reject.
The probability that M accepts if x 6 ∈ L is still 0. But if x ∈ L the probability that it rejects is at most ( 12 )p(n).
What about 2-sided error?
Definition 15 (Bounded-error probabilistic polynomial time) L ∈ BPP if ∃ a polynomial time probabilistic TM M such that
- If x ∈ L ⇒ Prcoin flips [M (x) accepts ] ≥ 23.
- If x 6 ∈ L ⇒ Prcoin flips [M (x) accepts ] ≤ 13.
Lemma 16 (BPP Amplification) If L ∈ BPP ⇒ ∀ polynomial p∃ a probabilistic polynomial-time algorithm for L with error probability ≤ 2 −p(n)
Proof: Given a BPP algorithm M and input x, run M on x k times and decide by majority vote.
Fact: The error decreases as 2−Ω(k). This is proved using a fact from proba- bility known as the Chernoff Bound (Lemma 11.9 in Pap.). So by picking an appropriate k (polynomial in the length of x) we get the desired bound.
Does BPP = P? We don’t yet know but there is a lot of evidence that if they are not equal they are very close.
We do know that RP ⊆ NP because the certificate could just be the flips that give you an accepting computation. For the same reason co-RP ⊆ co-NP. We will prove that BPP ⊂ Σ 2 P next time.
Σ 2 P Π 2 P
co − N P BP P N P
co − RP RP
P
