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More on Randomized Class
– Def: A language L is in BPPc,s ( 0s(n)c(n)1, nN) if there
exists a probabilistic poly-time TM M s.t. :
• 1. wL, Pr[M accepts w]  c(|w|) ,
• 2. wL, Pr[M(x) accepts]  s(|w|) .
• Thm: (Amplification of BPP)
For all choices of poly. computable functions c(n) and s(n) : {0,1}n
{0,1}, such that there exists a poly. Q(n) s.t. n c(n)-s(n)  1/Q(n)
and m=O(1),
BPP  BPP m m .
c,s
1 2 n , 2 n
• Pf:
Given a BPP machine M with c(n), s(n). We construct a
BPP machine for the same language with c  1  2 n and s  2 n
for any m=O(1).
m
– Define M’:
1. Run M on k times independently.
2. Accept if the number of time M accepted is  k‧(c(n)+s(n))/2 .
Xi: indicator random variable for the event that M accepts w.
m
By the definition of BPPc,s we have:
• wL  E[Xi]  c(n) ,
• wL  E[Xi]  s(n) .
k
cM ' (n )  1  Pr[  X i
i 1
k
S M ' (n )  1  Pr[  X i
i 1
 c( n )  s ( n ) 
 k 
 | w  L]
2


 c( n )  s ( n ) 
 k 
 | w  L]
2


• Chernoff bound:
For any k independent identically distributed random variable X1,… Xk
with values in {0,1}, and with expected values E[Xi]=p, for any
(0,1),
–
k
Pr[  X i  kp(1   )]  e

2
3
pk
i 1
–
k
Pr[  X i  kp(1   )]  e
i 1

2
3
pk
Using Chernoff bounds, choose  with

c ( n)  s ( n)
c ( n)  s ( n)
and  
2c(n)
2 s ( n)
c ( n)  s ( n)
 c ( n)  s ( n)

So

c
(
n
)

(
1


)
and

s
(
n
)

(
1


)


2
2


Setting
3n m
3n m
k
and k 
2
c ( n)  
s ( n)   2
So
 c ( n)  s ( n) 
CM ' (n)  1  Pr[  X i  k  
 | w  L]
2


i 1
k
 1- e
-
2
3
c ( n )k
 1  2n
m
 c ( n)  s ( n) 
S M ' (n)  1  Pr[  X i  k  
 | w  L]
2


i 1
k
-
2
 s ( n ) k
nm
e
2
C M ' ( n)  S M ' ( n)  1 / Q ( n) .
3
■
• P/poly and circuit complexity
– Def: P/poly={ L | AP, a sequence of strings {Si}iN and a
constant k s.t. |Si|=O(ik) and xL  (x,S|x|)A }
– Def: A language L has poly circuit complexity if there exists a
constant k such that for all n, the function fn that is 1 iff its input
(of length n) is in L, has circuit complexity O(nk) .
• Prop: LP/poly iff L has poly. circuit complexity.
• Pf:
 : If L has poly circuit complexity, then for each n, there is a circuit
of size poly in n that decides membership in L for all words of length n.
Encode this circuit on a string, Sn ~ poly size.
Construct a poly time TM taking x and S|x| and simulate S|x| on input x.
 : Assume LP/poly.
If M decides L in TIME(O(nk)), then we can construct a circuit ck of
size O(n2k) that simulates M running on input strings of length n .
Hardwired in each machine will be the advice strings Sn, which is
constant for each input size n and which grows polynomial in n .
■
• Thm: BPP  P/poly.
• Pf:
Let L be an arbitrary language in BPP.
– By amplification of BPP, we have a TM M BPP  n2  n2
1 2 ,2
decides L .
Classify all possible random string R as follows:
• R is bad for an input x if M(x,R) is wrong.
• R is bad if there exists an input w for which R is bad.
• R is good otherwise.
Fix w, Pr[R is bad for w] 
1
2
n2
that
Pr[R is bad]  
n 2
2 2 1
Pr[R is bad for w] 
Therefore, Pr[R is good] = 1-Pr[R is bad] > 0
Therefore, there exists a poly size advice string for any input of length
n.
■
n
w{0,1}n

2
P
CO  NP
NP
CO  RP
RP
P
BPP
• Thm: BPP  2P (Sipser,Lautemann)
• Pf:
Suppose LBPP.
– Goal: Show that there is a 2P Machine that decides L.
– I.e. show that  a deterministic poly time TM M(x,y,z)
s.t.
• xL  y s.t. z M(x,y,z)=1
• xL  y z s.t. M(x,y,z)=0 .
Let A be a BPP machine that uses Q(n) random bits with c(n)= ½ and
s(n)=1/3Q(n) where n is the input length and Q(n) is poly.Let R be the
set of all random string of length Q(n) used on A’s random
tape.|R|=2Q(n) .
– Define Fs(y)=ys, sR, yR . Fs(y) is random if s is chosen
uniformly.
Imagine a new machine, A’(x,y,S), where S is a sequence random bits
(s1,s2,…,sk), yR, x is the input to test
if xLA
A’ is a deterministic TM s.t.:
A’(x,y,S)=1  siS, A accepts x with ysi on its random tape.
If xLA, and a specific S is chosen at random, then
Pr [ A' ( x, y, S )  1]  Pr [si  S s.t. A( x, y  si )  1]
yR
yR
k
  Pr [ A( x, y  si )  1]
i 1
yR
 k  Pr [ A( x, y )  1]
yR
k
2Q (n) 2


3Q (n) 3Q(n) 3
1
let k2Q(n)
 Pr [ A' ( x, y, S )  0] 
yR
3

I.e. if xLA, then for any SRk, yR s.t. A’(x,y,S)=0 .
If xLA, and a specific y is chosen at random
Pr [ A '( x, y, S )  0]  Prk [si  S , A( x, y  si )  0]
SR k
SR
 Prk [si  S , A( x, si )  0]
S R
k
1
 
2
Prk [y s.t. A '( x, y, S )  0] 
SR

y{0,1}Q ( n )
Prk [ A '( x, y, S )  0]
SR
 2Q ( n )  Prk [ A '( x, y, S )  0]
S R
 2Q ( n ) 
 k 
 2 
Let k=2Q(n)
Prk [y  R, A' ( x, y, S )  1]  1  Prk [y, s.t. A' ( x, y, S )  0]
SR
SR
2Q ( n )
 1  2Q ( n )  0
2
I.e. if xLA, then  an SRk, s.t. yR, A’(x,y,S)=1
Therefore, xLA  SRk, s.t. yR, A’(x,y,S)=1 .
So a 2P machine decides LA by guessing S, guessing all y and
checking A’(x,y,S)=1 .
■
• USAT:  is USAT if  is satisfied by exactly one truth assignment.
– Suppose  is satisfiable by at most one truth assignment. We want
to decide if USAT .
It turns out to decide USAT is as difficult as to decide SAT.
• Randomized reduction from SAT to USAT.
M: randomized poly-time TM M s.t.
– SAT  M()SAT (USAT)
– SAT  Prob[M()USAT]  1/8 .
• Universal Hashing:
Given sets S and T, a family H of functions from S to T is a
universal family of hash functions from S to T if
– 1) xS, wT, PrhH[h(x)=w]=1/|T|
– 2) xyS, w,zT, PrhH[(h(x)=w)∧(h(y)=z)]=1/|T|2
• eg.
Let S={0,1}n, T={0,1}k
for x{0,1}n, let hM,b(x)=Mx+b
where M is a k x n Boolean matrix, b is a column vector is {0,1}k .
– H={ hM,b: for all possible M and b } .
• Prop: The above H is a family of universal hash functions from
{0,1}n to {0,1}k .
• Pf:

n
– 1) For any fixed x{0,1} - 0 and y{0,1}k
Pr [h( x )  y ]  PrM ,b [ Mx  b  y ]
hH


z{0,1}k
PrM [ Mx  z ]  Prb [b  y  z ]
 2k  2  k  2  k  2  k
Pr[x+y+1=1]=?
– 2) For xy{0,1}n and w,z{0,1}k .
Pr [( h( x )  w)  h( y )  z )]  PrM ,b [ Mx  b  w  My  b  z ]
hH
 PrM ,b [ M ( x  y )  w  z  Mx  b  w]
 PrM [ M ( x  y )  w  z ]  PrM ,b [ Mx  b  w | M ( x  y )  w  z ]
 2 k  PrM ,b [ Mx  b  w]
 2 k  2 k
 22k
■
• Prop: xy
{0,1}n-

0 and w,z{0,1}k, we have
PrM[Mx=w∧My=z]=1/22k .
• Pf:
If x and y are e1=(1,0,…,0) and e2=(0,1,0,…,0), respectively, then it is
true.

Since neither x nor y is 0 , they’re linear independent.
Thus, there exists rank n matrix A s.t. Ax=e1, and Ay=e2 .
∵ rank(A)=n, MA is random if M is chosen randomly.
So, the truth of the proposition is clear .
■
M
x
(a1,…,an)  (b1,…,bn)=c
fixed
Pr[a1b1+a2b2+…+anbn=c]=?
a1,a2,…,an{0,1} are selected randomly, c{0,1} .
• Prop: Let S{0,1}n, with 2k-2|S|2k-1 ,Then
Pr[! sS s.t. Ms= 0 ]1/8, where the probability is taken over the
uniform choice of M from the set of all k x n Boolean matrices.
• Pf:
–

 
1) 0  S : M 0  0 .


Pr M [s  S , s  0, Ms  0] 

[ Ms  0]
 Pr
 M
sS , s  0
 | S | 2  k 
 1
 Pr[! s  S , s.t. Ms  0] 
2
1
2
–
2) 0  S : For any fixed s  S , PrM [ Ms  0]  2  k and
t  s  S, PrM [ Mt  0  Ms  0]  2 2k , which implies
PrM [ Mt  0 | Ms  0]  2  k .


PrM [ Ms  0  t  s, Mt  0]



 PrM [ Ms  0]  PrM [t  s, Mt  0 | Ms  0]


k
 2 (1  PrM [t  s  S , Mt  0 | Ms  0]


k
 2 (1   PrM [ Mt  0 | Ms  0])
tS ,t  s
 2 k (1  2  k (| S | 1))  2  k 1 .


PrM [s, Ms  0  t  s  S , Mt  0]


  PrM [ Ms  0  t  s, Mt  0]
sS
| S | 2  k 1  1 / 8 .
• Successive restrictions:

x
Given a CNF formula  on n variables , choose n+1 random vectors


v1 ,..., vn1 and create I for 1in+1 as follows:
 
 
 
i    ( x  v1  0)  ( x  v2  0)  ...  ( x  vi  0)
• Lemma:
If  is not satisfiable, then none of the i’s are satisfiable .
• Lemma:
If  is satisfiable, then with probability at least 1/8, at least one of the
i’s has a unique satisfying assignment .
• Pf:
Let S be the set of satisfying assignments of : by hypothesis |S|1.
Let k be such that 2k-2|S|<2k-1 .
By the previous prop., k has a probability  1/8 of having exactly one
satisfying assignment .
■
Thus, detecting unique solutions is an hard as NP .
– UP:the class of promise problems where instances are promised to
whose either zero or one solution .
• Thm: NPRPUP .
• Thm: If UPRP, then NP=RP .