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Probabilistic Complexity

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. Define the probability of branch b to be

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Probabilistic Complexity

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  1. Probabilistic Complexity

  2. 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.

  3. 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 • wA implies Pr[M accepts w]  1-, and • wA implies Pr[M rejects w]  1-

  4. 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) .

  5. 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 times 2. Simulate M1 on w. 3. If most runs of M1 accept, then accept; otherwise reject. ”

  6. M1 errs on w with some probability    < ½ , /1- <1 . (1-)-(1-)<0

  7. 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=abc abc 2c 2c = 22log(bt) = (bt)2 b1, assume that t9  bt9  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

  8. 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) ■

  9. 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) .

  10. Lemma: Suppose that a1,…,at all divide n and (ai,aj)=1 for ij. Then a1…at n. • Pf: By induction. It is clear for t=1 Suppose the lemma is true up to t-1, i.e. a1…at-1n. (at,a1…at-1)=1  Exist r and s such that r  at + s  a1…at-1 =1, ratn + sa1…at-1n=n . ■ at|n by ind.

  11. Thm: (Chinese Remainder theorem) Suppose m=m1m2…mt and (mi,mj)=1 for ij. Let b1,b2,…,bt be integers and consider the system of congruencies : xb1 (mod m1) xb2 (mod m2) xbt (mod mt)

  12. 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), ij Let 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 ■

  13. 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 aZp+, then ap-1  1 (mod p) .

  14. Pf: 1a,2a,…,(p-1)a 1i,jp-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) . ■

  15. 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 Fermat tests for aZp+ .

  16. Carmichael numbers: Composite numbers that passes all Fermat tests. Pseudo-prime = “ On input p: 1. Select a1,…,ak randomly in Zp+ . 2. Compute aip-1 mod p for each i . 3. If all computed values are 1 , accept ; otherwise, rejects . “

  17. Test prime power • 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).

  18. PRIME = “ On input p 1. If p is even and p=2 then accept ; else reject . 2. Select a1,…,ak randomly in Zp+ . 3. For i=1 to k do 4. Compute aip-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. “

  19. 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)

  20. (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. ■

  21. Lemma: If p is an odd composite number, then Pr[PRIME accepts p]  2-k . • Pf: • Goal: If p is an odd composite number and a is selected randomly 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.

  22. 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 kind 1,1,1,…,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.

  23. 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-th

  24. Hence t is a witness because but

  25. Next we prove that dt mod p is a unique witness for each non-witness d by 2 observations .

  26. Thus the number of witnesses must be as large as the number of non-witnesses when p=qr.

  27. 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 =d1tp 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.

  28. 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 xL  Pr[M(x) accepts]  2/3 , • when xL  Pr[M(x) accepts] < 1/3 . • Def: LRP, if  a NTM in poly time. • when xL  Pr[M(x) accepts]  2/3 , • when xL  Pr[M(x) accepts] =0 . • RPBPP. ? • ZPP • Def: ZPP = RPco-RP .

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