18.405J/6.841J Advanced Complexity Theory
March 1, 2001
Lecture 7
Lecturer: Dan Spielman
Scribe: Abhinav Kumar
Today we will talk about Randomized Complexity classes.
Last time we showed BPP P/poly. Today we will show BPP
BPP error amplication
2 P \ 2 P
Error amplication means decreasing the probability of error. The key tool in this regard is the use
of Cherno bounds.
Theorem 1 Let X1 Xn2 ::: Xn be independent, identically distributed random variables taking valXi , = EY].
Then PrjY ; j > ] < e; 2 =3 , for 2 0 1].
Proof: See handout on probability.
ues in 0,1]. Let Y =
P
i�0
Amplication Lemma: Let L be a language such that there is a randomized polynomial-time TM
M such that,
x 2 L ) PrM (x)accepts] C (jxj)
x 2= L ) PrM (x)accepts S (jxj)
where C (n) ; S (n) p(1n) for some poly p()
Then 8b > 0 9 a random ptime TM M' such that
x 2 L ) PrM 0 (x)accepts] 1 ; 2;jxjbb
x 2= L ) PrM 0 (x)accepts 2;jxj
(Remark : This is reasonably tight)
Proof: M' runs M k = 12(p(n)3 + nb times, n = jxj. Accepts if M accepted more than k s(n)+2 c(n)
times. Applying Cherno's bound, we get the desired behaviour of M'.
Theorem 2 (Sipser) BPP 2P .
Note: This implies co-BPP 2 P , but co-BPP = BPP since denition of BPP is symmetric.
Proof:
Let L be a language in BPP. We know 9A 2P and a function f (n) = nO(1) such that
If w 2 L then r:jrPr
(r w) 2 A] > 1 ; 2;n , and
j�f (n)
if w 2= L then r:jrPr
(r w) 2 A] < 2;n .
j�f (n)
where n = jwj.
Def. Dene Rw = fr : jrj = f (n) such that (r w) 2 Ag. Correspondingly,
If w 2 L then jRw j > (1 ; 2;n )2f (n) and
if w 2= L then jRw j < 2;n 2f (n) .
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The idea here is to take a number of translations of Rw , and see if they cover the entire space
f0 1gf (n) . Each translation of Rw has the same size as Rw , and if Rw is most of the space (ie,
w 2 L) then this collection of translations would be likely to cover the space. However, if Rw is very
small (ie, w 2= L) then this collection could never cover the space. More formally,
Def. (translation) Let S f0 1gf (n) . For t 2 f0 1gf (n) let the translation S t be dened as
fx : x t 2 S g
where x t is dened as the XOR of the two strings (or, the bitwise sum modulo 2).
Claim. (1) If jS j > (1 ; 2;n )2f (n) then 9 = ft1 : : : tf (n) g such that
f (n)
(S ti ) = f0 1gf (n)
i�1
(2) If jS j < 2;n 2f (n) then 8 = ft1 : : : tf (n) g,
f (n)
(S ti ) =6 f0 1gf (n)
i�1
First, we show that the claim proves the theorem. If the claim is true, we can design a 2 P
machine M to solve L as follows:
1. Use 9 states to generate .
2. Use 8 states to generate r 2 f0 1gf (n) .
3. Check if r 2 i Rx ti and accept if so, otherwise, reject.
This is polynomial time, since we can check whether r 2 Rx in polynomial time, and f (n) is
polynomial in n. By the claim, if x 2 L then on any correct we accept. If x 2= L we reject, since
there is no such . Therefore, we only have to prove the claim.
First, we prove part (2) of the claim. If jS j < 2;n 2f (n) then
i (S ti ) f (n)2;n2f (n) :
Since f (n) = nO(1) f (n)2n < 1 for suciently large n. Therefore, this union doesn't cover
f0 1gf (n) .
Note that the "suciently large n" clause here doesn't cause a problem. If we take this into
account, we need only hard-code the correct answer for all words smaller than this bound into our
2 P machine.
Next, we prove part (1). Let
f (n)
p = Pr
8 r 2 i�1
(ti S )]
jrj�f (n)
f (n)
= 1 ; Pr
9 r 2= i�1
(ti S )]
jrj�f (n)
f (n)
1 ; jrj�f (n)Pr
r 2= i�1
(ti S )]:
We choose the ti 's independently, so we can consider them independently. Therefore,
f (n)
p 1 ; jrj�f (n) i
Prti r 2= ti S )]
�1
f (n) ;n
= 1 ; jrj�f (n) i
2
�1
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since r 2 ti S if and only if ti 2 r S ,
= 1 ; 2f (n) (2;n )f (n) = 1 ; 2;f (n)(n;1) > 0:
Since this probability is nonzero, there must be at least some for which the union of the translations determined by covers f0 1gf (n) . This completes the proof.
Verifying Polynomial identities
Let p be a given polynomial in k variables, q1 ::: qk given polnomials in m variable. The equation
p(q1(y1 ::: ym ) ::: qk (y1 ::: ym )) = 0
can be dicult to check deterministically. Expanding the polynomial out would give us an exponential number of terms. However, if we use randomization there is an easy test. Choose x1 ::: xm
at random and see if we get zero. If we choose x0i s in a large enough range, the probability that the
test is passed but p is not identically zero becomes exponentially small. So we can check polynomial
identities in co-RP.
Lemma 3 (Schwartz's Lemma) Let P (x1 x2 : : : xn) be a polynomial of degree d. Then if P 6
0, then
Prx1 x2 :::xn 2S P (x1 x2 : : : xn ) = 0] jdn
Sj :
Proof: We use induction on n.
Base case (n = 1) : If P 6 0, then there are at most d zeroes in p. At most d = d 1 of them are in
S.
Inductive step: Write
d1
X
P (x1 x2 : : : xn ) = xi1 Pi(x2 : : : xn ):
By hypothesis,
i�0
Prx2 ::: xn Pd1 (x2 : : : xn ) = 0] (n j;S j1)d :
If Pd1 (x2 : : : xn ) =
6 0, then Prx1 P (x1 : : : xn )] jdS1j .
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