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. Lemma: [Amplification Lemma]
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 . ■
by ind.
at-1n

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. Thm: (Fermat’s little theorem)
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. Fermat test: 27-1=26=64, 64 mod 7 =1.
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. Pseudo-prime = “ On input p:
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. ∵ p is composite.  We can write p = qr, (q,r)=1 .
Let h be a non-witness.
,…,……………..-1,1,…,1
∵ p is composite.  We can write p = qr, (q,r)=1 .
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.■
Thm: PRIMESBPP .

27. 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] =0 .
RPBPP. ?
ZPP
Def: ZPP = RPco-RP .