–def: a language l is in bpp c,s ( 0 s(n) c(n) 1, n n) if there exists a probabilistic...
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– 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
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2
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nsnsnc
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22 )(
3 and
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3
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nk
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So
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nQnSn
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kXnS
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• 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 .
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