18.405J/6.841J Advanced Complexity Theory
May 10, 2001
Lecture 22
Lecturer: Daniel A. Spielman
1 Derandomizing Logspace Computations (Cont.)
In today's lecture we wil nish the derandomization of Logspace-bounded computations. Our algorithm runs in space S , time 2 , using only O(S 3 ) random bits. In the last lecture we proved
S
Lemma 1 There exists a constant k > 0 such that for all r c, there exists a polynomial time
computable c;generator g : f0 1g + ! f0 1g f0 1g .
such that 8A1 and A2
r
kc
r
r
A1 : f0 1g ! f0 1g
A2 : f0 1g f0 1g ! f0 1g
r
8b2 2 f0 1g
c
�
�
�
�
c
r
c
A2 (A1 (x1 ) x2 ) = b2 ] ; Probr+kc A2 (A1 (g (z )) g (z )) = b
�
Prob
x1 x2
r
2f0 1gr
z
l
2f0 1g
r
�
��
�
2 �
< 2;
c
We are going to use Lemma 1 recursively, see Figure 1
Denition 2 c = 2S 2
g : f0 1g2( +1) 2 ! f0 1g2 2 f0 1g2 2 8i = 1 : : : S
Dene the functions G : f0 1g2( +1) 2 ! f0 1g2 2 inductively as follows:
G1 (z ) = g1 (z ) g1 (z ))
G (z ) = G ;1 (g (z )) G ;1 (g (z )) 8i = 2 : : : S
Claim 3 G is an -generator for space S and time S 2 2 where
( +1)
= 22 +1 ;;11 2;2 2
Note: In the case of i = S , the running time is S 2 2 . G takes O(S 3 ) random bits. 2;S 2,
i
i
kS
i
i
ikS
ikS
i kS
kS
l
i
i
l
i
i
r
i
i
i
i
S
i
S
i
S
S
S
which is pretty small.
Proof We proof by induction.
The base case is G1 (z ) 2= g1 (z ) g1 (z )) : f0 1g4
c = 2S 2 > S and 1 = 2;2 .
Inductive part (Figure 2): we assume that
l
r
kS
2
S
8b1
8b1 b2
�
�
�
�
�x2
r
�
�
�
�
�x1
Prob
2f0 1g2i;1 �S2
Prob
2f0 1g2i;1 �S2
S
! f0 1g2
kS
2
f0 1g2 2 . Thus r = 2kS 2,
kS
�
�
�
1 �
�
�
�
�
2 �
�
A1 (x1 ) = b1 ] ; Prob
A1 (G ;1 (y1 )) = b ] < ;1
1
y
i
i
A2 (b1 x2 ) = b2 ] ; Prob A2 (b1 G ;1 (y2 )) = b ] < ;1
y2
23-1
i
i
Figure 1: Recursive Pattern. time = 2 , S = O(log n)
s
23-2
A1
s
A2
s
Figure 2: Inductive Picture
Then we sum over b1 (jb1 j = 2 )and apply the previous two inequalities to get
S
�
�
�
�x
8b2
�
�
2 ��
Prob A2 (A1 (x1 ) x2 ) = b2 ] ; Prob A2 (G ;1 (y1 )) G ;1 (y2 )) = b ] < 2 2 ;1
S
1
x2
i
y1 y2
i
i
Since g is a 2 S 2 -generator, we have
i
8b2
�
�
�
�y
Prob A2 (G ;1 (y1 )) G ;1 (y2 )) = b2 ] ; Prob A2 (A1 (G ;1 (g (z ))) G ;1 (g (z ))) = b
�
1
y2
i
i
l
i
i
z
i
r
i
�
��
�
2 �
< 2;2
S
2
By triangular inequality,
8b2
�
�
�
�x
�
��
Prob A2 (A1 (x1 ) x2 ) = b2 ] ; Prob A2 (A1 (G ;1 (g (z ))) G ;1 (g (z ))) = b2 �� < 2 2 ;1 + 2;2 = �
1
x2
i
z
l
i
i
r
i
S
i
Another approach is the recurtion using hash functions. We can achieve S 3 radom bits, and
more eciently only S 2 bits by randomly choose hash functions.
Constructs hash functions:
h1 ::: h = f0 1g ! f0 1g
s
s
23-3
s
S
2
i
And denes
G1 (x) = x h1 (x)
G2 (x) = G1 (x) G1 (h2 (x))
:::
G (x) = G ;1 (x) G ;1 (h (x))
i
i
i
i
2 Hardness vs. Randomness by Nissan Wigderson
Denition 4 for any f : f0 1g ! f0 1g, hardness is h(n) if 8 circuit C of size h(n)
n
1+ 1
Prob
f
(
x
)
=
C
(
x
)]
<
n
x2f0 1g
2 h(n)
Assume a function of hardness nlog is computable in time 2 k for some k, we can prove
n
n
BPP DTIME(2 ) 8 > 0
n
hardness h(n) = 2; nc computable in EXP ) BPP = P:
We will continue this topic in the next lecture.
23-4