Probabilistic Complexity. Probabilistic Algorithms Def: A probabilistic Turing Machine M is a type of non- deterministic TM, where each non-deterministic.
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Probabilistic Complexity
Probabilistic Algorithms Def: A probabilistic Turing Machine M is a type of
non-deterministic TM, where each non-deterministic step is called a coin-flip step and has 2 legal next moves.
' .
assign a probability to each branch b of
M s computation on input w as follows
Define the probability of branch b to be Pr[b]=2-k, where k is the number of coin-flip steps
that occur on branch b. Define the probability that M accepts w to be
Pr[M accepts w] = Pr[b]
For 0 ½ , we say that M recognize language A with error probability if
wA implies Pr[M accepts w] 1-, and wA implies Pr[M rejects w] 1-
branchaccepting
anisb
BPP Def: BPP is the class of languages that are
recognized by probabilistic polynomial time TM’s with an error probability of 1/3 .
Lemma: [Amplification Lemma]Let 0<<½. Then for any polynomial p(n) a probabilistic poly time TM M1 that operates with error probability has an equivalent probabilistic poly time TM M2 that operates with an error probability of 2-p(n) .
Pf: M1: recognizes a language with error , and a poly
p(n). Construct M2 that recognizes the same language
with an error probability 2-p(n).
M2 =“ On input w
1. Calculate k and repeat the following 2k times2. Simulate M1 on w.
3. If most runs of M1 accept, then accept;
otherwise reject. ”
M1 errs on w with some probability < ½ , /1- <1 . 2
2
2
2
2
Pr[ 2 ]
2(1 )
2( 1) (1 )
( 1)2 (1 ) ( 1)(4 (1 ))
( 1)(4 (1 ))
k i k
i k i
k i k
k k
k k k k
k
M errs exactly i times on k runs
k
i
kk
k
k k
k
(1-)-(1-)<0
Need to show that (k+1)(4(1-))k 2-p(n) .Let t=2p(n), a=1/4(1-), b=max(1,1/loga), c=2logbt, k=bc Claim: (k+1)(1/a)k 1/t
ak = abc abc 2c
2c = 22log(bt) = (bt)2
b1, assume that t9 bt9 bt>2+2log(bt) (bt)2 > bt(2+2logbt) = t(2b+2blogbt) ∵ b 1
i) if 1/loga > 1 b=1/loga abc=a1/loga•C=2 C
ii) if 1/loga < 1 b=1 and a>2 abc=aC>2 C
Hence ak t(2+2blogbt) t(1+2blogbt) t(1+bc) = t(k+1)
(k+1)(4(1- ))k = (k+1)/ak 1/t =2-p(n)
■
Primality: Composite number : 複合數 Prime : 質數
Zp+={1,2,…,p-1}, Zp ={0}Zp
+
Z5 ={0,1,2,3,4}Z6 ={0,1,2,3,4,5}Z6
+={1,2,3,4,5} x mod p is the smallest non-negative y where x y (mod
p) .
Lemma: Suppose that a1,…,at all divide n and (ai,aj)=1 for ij.Then a1…at n.
Pf:By induction. It is clear for t=1Suppose the lemma is true up to t-1,i.e. a1…at-1n.(at,a1…at-1)=1 Exist r and s such that r at + s a1…at-1 =1,ratn + sa1…at-1n=n .
■ by ind. at|n
Thm: (Chinese Remainder theorem)Suppose m=m1m2…mt and (mi,mj)=1 for ij. Let b1,b2,…,bt be integers and consider the system of congruencies :
xb1 (mod m1)xb2 (mod m2)
xbt (mod mt)
.
mofmultiplea
bydiffersolutionstwoany
solutionshassystemThis
Pf:Let ni = m/mi , then (mi,ni)=1. Exist ri and si such that rimi+sini=1.Let ei = sini ei 1 (mod mi)
ei 0 (mod mj), ijLet x0= biei. Then x0 biei (mod mi)
bi (mod mi) x0 is a solution.Suppose x1 is another solution. Then x1–x0 0 (mod mj), for i=1,…,t .That is m1,…,mt divide x1-x0 . m x1-x0
■
t
i 1
The CRT says that a 1-1 correspondence exists between Zm and Zm1×…×Zmt
Thm: (Fermat’s little theorem) If p is a prime number and aZp
+, then ap-1 1 (mod p) .
Pf: 1a,2a,…,(p-1)a1i,jp-1, ia ja (mod p)(ia–ja) 0 (mod p)(i-j)a = k•p p (i-j) i=j . Thus, 1a,2a,…(p-1)a is a permutation of 1,2,…,(p-1).
1a•2a• …•(p-1)a 1•2•…•(p-1) (mod p) (p-1)!ap-1 (p-1)! (mod p) (p-1)!(ap-1 -1) 0 (mod p) (p-1)!(ap-1 -1) k•p
p ap-1 –1, ap-1 1 (mod p) .
■
eg.27-1=26=64, 64 mod 7 =1.
Fermat test: we say that p passes the Fermat test at a, we mean that ap-1 1 (mod p) . Fermat’s little theorem states that primes pass all Ferm
at tests for aZp+ .
Carmichael numbers:Composite numbers that passes all Fermat tests.
Pseudo-prime = “ On input p:1. Select a1,…,ak randomly in Zp
+ .2. Compute ai
p-1 mod p for each i .3. If all computed values are 1 , accept ; otherwise, rejects . “
Numbers of prime power: N=pk
It is easy to test such type of numbers!
It is clear that (p-1)| (N-1)
Find a number a such that gcd(a, p) =1.
Then aN-1 1 (mod p). Why?
Thus p | gcd ( N, aN-1 -1).
Test prime power
PRIME = “ On input p1. If p is even and p=2 then accept ; else reject .2. Select a1,…,ak randomly in Zp
+ .3. For i=1 to k do4. Compute ai
p-1 mod p and reject if different from 1.5. Let p-1=st where s is odd and t=2h
6. Compute mod p.
7. If some element is not 1, then find the last one that is not 1 and reject if it is not –1.8. All tests have passed at this point, so accept. “
hsi
si
si
si aaaa 2222 ,...,,,
210
Lemma: If p is an odd prime number, then Pr[PRIME accepts p]=1 .
Pf:If p is an odd prime, then it will pass stage 4.If a were a stage 7 witness, some b exists in Zp
+, where b 1 (mod p) and b 1(mod p) b2-1 0 (mod p)(b-1)(b+1) 0 (mod p)
(b-1)(b+1) = cp for some positive integer c. ∵ b 1 (mod p) 0<b+1 , b-1<p . Therefore, p is composite because a prime number
cannot be expressed as a product of numbers that are smaller than it is.
■
Lemma: If p is an odd composite number, thenPr[PRIME accepts p] 2-k .
Pf: Goal: If p is an odd composite number and a is selected ra
ndomly in Zp+, then Pr[a is witness] ½ .
Prove by demonstrating that at least as many witnesses as non-witnesses exist in Zp
+ , i.e. by finding a unique witness for each non-witness.
For every non-witness, the sequence computed in stage 6 is either all 1 or contains –1 at some position followed by 1’s .
1: non-witness of the first kind1,1,1,…,1
-1:non-witness of the second kind-1,1,1,…1
Among all non-witness of 2nd kind, find a non-witness for which the –1 appears in the largest position in the sequence.
Let h be a non-witness.,…,……………..-1,1,…,1
∵ p is composite. We can write p = qr, (q,r)=1, or p is a prime power. We handle former case first.By the CRT, there exists t Zp .
t h (mod q)t 1 (mod r)
j-th0 12 2 2 2, , ...... 1 ( mod p)
j jS S S Sh h h h
r) (mod 1 t
q) (mod 12
2
j
j
s
st
Hence t is a witness because but
p) (mod 12 jst
p) (mod 112 jst
p). (mod 1 ,
r) (mod 1
q) (mod 1
1 p) (mod 1
2
2
22
j
j
jj
s
s
ss
tforsimilarly
t
kpttIf
Next we prove that dt mod p is a unique witness for each non-witness d by 2 observations .
. p) (mod 1(dt) and 1
because witness,a is p mod
)p(mod 1 and )p(mod 1 )1
1
1
22
22
jj
jj
ss
ss
(dt)
dt
dd
1 2
1 212
2) If and are distinct non-witnesses, then
mod p mod p .
The reason is that mod p 1 js
d d
d t d t
t
Thus the number of witnesses must be as large as the number of non-witnesses when p=qr.
2
212
112
1
21
12
p mod p mod
then mod p mod If
1. p mod
11
1
d
dttdttd
ptdtd
tt
jj
j
ss
s
For the case p=qe, where q is a prime and e >1. Let t= 1+ qe-1, which is < p. Thus t p = (1+ q e-1)p = 1 + p qe-1 + (.....) q2(e-1)
= 1 + p(.........) 1 (mod p). Observe that if t p-1 1 (mod p), then t p t !1 (mod p),
which contradicts that t p 1 (mod p). Thus t is a stage 4 witness, since t p-1 ! 1 (mod p). If d is a stage 4 non-witness, then dp-1 1 (mod p), but
then (dt)p-1 ! 1 (mod p), ie, dt is a witness. If d1 and d2 are distinct non-witness, then d1 t mod p d2 t mod p. Otherwise, d1 =d1 tp mod p = d2 tp mod p = d2. Thus the number of stage 4 witnesses must be as large
as the number of non-witnesses.
Thm: PRIMES BPP, actually co-RP.
Def: BPP is the class of all languages L for which there is a non-det poly time TM M, whose computation branches all have the same length, and
when xL Pr[M(x) accepts] 2/3 , when xL Pr[M(x) accepts] < 1/3 .
Def: LRP, if a NTM in poly time. when xL Pr[M(x) accepts] 2/3 , when xL Pr[M(x) accepts] =0 .
RPBPP. ?
ZPP Def: ZPP = RPco-RP .
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