– Def: A language L is in BPPc,s ( 0s(n)c(n)1, nN) if there exists a probabilistic poly-time TM M s.t. :

• 1. wL, Pr[M accepts w] c(|w|) ,

• 2. wL, 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), .

2,21, mnmnBPPBPP sc

More on Randomized Class

• Pf:

Given a BPP machine M with c(n), s(n). We construct a BPP machine for the same language with

for any m=O(1).

– 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.

mm nn sandc 2 21

By the definition of BPPc,s we have:

• wL E[Xi] c(n) ,

• wL E[Xi] s(n) .

'1

'1

( ) ( )( ) 1 Pr[ | ]

2

( ) ( )( ) 1 Pr[ | ]

2

k

M iik

M ii

c n s nc n X k w L

c n s nS n X k w L

• 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),

–

–

2

3

12

3

1

Pr[ (1 )]

Pr[ (1 )]

k pk

ii

k pk

ii

X kp e

X kp e

Using Chernoff bounds, choose with

Setting

)1()(2

)()( and )1()(

2

)()( So

)(2

)()( and

)(2

)()(

nsnsnc

ncnsnc

ns

nsnc

nc

nsnc

22 )(

3 and

)(

3

ns

nk

nc

nk

mm

So

■ . )(/1)()(C

2e

]|2

)()(Pr[1)(

21e-1

]|2

)()(Pr[1)(

''

)(3

-

1'

)(3

-

1'

2

2

nQnSn

Lwnsnc

kXnS

Lwnsnc

kXnC

MM

nkns

k

iiM

nknc

k

iiM

m

m

• P/poly and circuit complexity– Def: P/poly={ L | AP, a sequence of strings {Si}iN and a consta

nt k s.t. |Si|=O(ik) and xL (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: LP/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 LP/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 that 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]

2 21 2 ,2

BPPn n

2

2

1n

Pr[R is bad] Pr[R is bad for w]

Therefore, Pr[R is good] = 1-Pr[R is bad] > 0

Thus, there exists a poly size advice string for any input of length n.

■

nw }1,0{

1222

nn

2P

P

NPCO NP

RPCO RP

BPP

• Thm: BPP 2P(Sipser,Lautemann)

• Pf:

Suppose LBPP.

– 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.

• xL y s.t. z M(x,y,z)=1

• xL 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)=ys, sR, yR . Fs(y) is random i

f s is chosen uniformly.• Imagine a new machine, A’(x,y,S), where S is a se

quence random bits (s1,s2,…,sk), yR, x is the input to test if xLA

• A’ is a deterministic TM s.t.:A’(x,y,S)=1 siS, A accepts x with ysi on its random tape.

If xLA, and a specific S is chosen at random, then

I.e. if xLA, then for any SRk, yR s.t. A’(x,y,S)=0 .

3

1]0),,('[Pr

3

2

)(3

)(2

)(3

]1),([Pr

]1),([Pr

]1),( .. [Pr]1),,('[Pr

1

SyxA

nQ

nQ

nQ

k

yxAk

syxA

syxAtsSsSyxA

Ry

Ry

k

ii

Ry

iiRyRy

let k2Q(n)

If xLA, and a specific y is chosen at random

( ){0,1}

Pr [ '( , , ) 0] Pr [ , ( , ) 0]

Pr [ , ( , ) 0]

1

2

Pr [ . . '( , , ) 0] Pr [ '( , , ) 0]

k k

k

k kQ n

i iS R S R

i iS R

k

S R S Ry

A x y S s S A x y s

s S A x s

y s t A x y S A x y S

( )

( )

2 Pr [ '( , , ) 0]

2

2

k

Q n

S R

Q n

k

A x y S

Let k=2Q(n)

I.e. if xLA, then an SRk, s.t. yR, A’(x,y,S)=1

Therefore, xLA SRk, s.t. yR, A’(x,y,S)=1 .

So a 2P machine decides LA by guessing S, guessing all y and checking A’(x,y,S)=1 .

02

21

]0),,('.. ,[Pr1]1),,(' ,[Pr

)(2

)(

nQ

nQ

RSRSSyxAtsySyxARy

kk

• 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) xS, wT, PrhH[h(x)=w]=1/|T|

– 2) xyS, w,zT, PrhH[(h(x)=w)∧(h(y)=z)]=1/|T|2

• eg. Let S={0,1}n, T={0,1}k and for x{0,1}n, let hM,b(x)=Mx+b,where M is a k 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:– 1) For any fixed x{0,1}n- and y{0,1}k

Pr[x+y+1=1]=?

0

,

{0,1}

Pr [ ( ) ] Pr [ ]

Pr [ ] Pr [ ]

2 2 2 2

M bh H

M bkz

k k k k

h x y Mx b y

Mx z b y z

– 2) For xy{0,1}n and w,z{0,1}k .

,

,

,

,

2

Pr [( ( ) ) ( ) )] Pr [ ]

Pr [ ( ) ]

Pr [ ( ) ] Pr [ | ( ) ]

2 Pr [ ]

2 2

2

M bh H

M b

M M b

kM b

k k

k

h x w h y z Mx b w My b z

M x y w z Mx b w

M x y w z Mx b w M x y w z

Mx b w

• Prop: xy {0,1}n- 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), respecti

vely, then it is true.• Since neither x nor y is , 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 .

0

0

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[! sS 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:–

2

1]0 .. ,![Pr

2

12||

]0[Pr]0 ,0 ,[Pr

. 00 :0 )1

0,M

MstsSs

S

MsMssSs

MS

k

sSsM

–

2

2) 0 : , Pr [ 0] 2

Pr [ 0 0] 2 ,

Pr [ 0 | 0] 2 .

kM

kM

kM

S For any fixed s S Ms and

t s S, Mt Ms which implies

Mt Ms

. 2))1|(|21(2

)]0|0[Pr1(2

]0|0,[Pr1(2

]0|0,[Pr]0[Pr

]0,0[Pr

1

,

kkk

stStM

k

Mk

MM

M

S

MsMt

MsMtSst

MsMtstMs

MtstMs

• Successive restrictions:Given a CNF formula on n variables , choose n+1 random vectors and create I for 1 i n+1 as follows:

. 8/12||

]0 ,0[Pr

]0 ,0 ,[Pr

1

k

SsM

M

S

MtstMs

MtSstMss

x

11,..., nvv

)0(...)0()0( 21 ii vxvxvx

• 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 instanc

es are promised to whose either zero or one solution .

• Thm: NPRPUP .• Thm: If UPRP, then NP=RP .