1. Probabilistic Algorithms Michael Sipser Presented by: Brian Lawnichak

2. 2 Introduction Probabilistic Algorithm –uses the result of a random process –“flips a coin” to decide next execution Purpose –saves on calculating the actual best choice –avoids introducing a bias –e.g. query individuals in a large population

3. 3 Probabilistic Turing Machine Definition 10.3 Nondeterministic Turing Machine M –each nondeterministic step is a coin flip –two legal next moves –probability is given to each branch b of M Pr[b] = 2 -k –where k is the number of coin flips on b

4. 4 M on input w Probability that M accepts input w Pr[M accepts w] =  Pr[b] Probability that M rejects input w Pr[M rejects w] = 1 – Pr[M accepts w] What if there is a bad coin flip? –is this algorithm 100% correct? –errors should be accounted for

5. 5 Error Probability  Allow the Turing machine an error probability  where 0    ½ M recognizes language L with error probability  if –w  L implies Pr[M accepts w]  1 –  –w  L implies Pr[M rejects w]  1 –  We say that TM M is bounded by 

6. 6 The Class BPP Bounded Probabilistic Polynomial time Turing machine M 1 Time and Space complexity same as a nondeterministic TM Definition 10.4 –BPP is the class of languages recognized by M 1 with error probability  = 1/3

7. 7 Amplification An error probability of 33% is lousy Could we improve upon this? Amplification lemma –uses any error 0    ½ –allows us to make the error probability exponentially small

8. 8 Lemma 10.5 Given fixed  and polynomial poly(n) M 1 operates with error probability   an equivalent BPP TM M 2 that operates with error probability 2 -poly(n) –M 2 simulates M 1 by running it a polynomial number of times and taking the majority decision vote

9. 9 Will M 2 Really Work? Box has 2/3 green and 1/3 red balls –M 1 samples one ball at random to decide –errs with probability  M 2 runs M 1 poly(n) times to decide –each run of M 1 gives us err of  –running multiple times gives us exponentially small probability

10. 10 Proof Using M 1 show that M 2 recognizes the same language with error 2 -poly(n) t = 2 poly(n) a = 1/(4  (1-  ) b = max(1,1/log(a)) c = 2 log(bt) k =  bc 

11. 11 Proof [cont.] M 2 = “On input w –find k and repeat the following 2k times – simulate M 1 on input w –if most runs of M 1 accept then accept; otherwise reject

12. 12 Verification We have built M 2 but must now verify that M 2 is equivalent to M 1 Assumptions –t  9 while conserving generality –M 1 errs on w with     ½ –M 2 obtains at least k erroneous results on 2k runs of M 1

13. 13 Verification [cont.] Probability of M2 obtaining more than k erroneous on 2k runs is We can allow i = k because  /(1 -  )  1 when   ½

14. 14 Verification [cont.] With i = k we have an upper bound We bound the combination by 2 2k, the number of all subsets giving us  (k+1)2 2k  k (1-  ) k  (k+1)(4  (1-  )) k

15. 15 Verification [cont.] We can allow  (1 -  )   (1 -  ) because     ½ giving us  (k+1)(4  (1-  )) k  (k+1)(1/a) k To show that this is at most 2 -poly(n), we show that a k  (k+1)t

16. 16 Verification [cont.] We use a series of inequalities a k = a  bc   a bc  2 c = 2 2log(bt) = (bt) 2 b  1 and t  9 so that bt  9; therefore (bt) 2  bt(2+2log(bt)) = t(2b+2blog(bt)) Since b  1, we come up with a k  t(2+2blog(bt))  t(1+  2blog(bt)  )  t(1+  bc  ) = (k+1)t

17. 17 Applications Known: –BPP = co-BPP Unknown: –NP  BPP –BPP  NP Existence of certain strong pseudo- random number generators implies that P = BPP above information was gleaned from http://encyclopedia.thefreedictionary.com/BPP

18. 18 Applications [cont.] Security –Chinese Remainder Theorem simplifies modular arithmetic –increases the efficiency of decryption –given pairwise relatively prime moduli {p1,...,pn} and arbitrary {a1,...,an} –there exists a unique key above information was gleaned from: “FAQ on Public-Key Crypt” on the Google sci.crypt newsgroup

19. 19 Applications [cont.] Primality –don’t bother testing all positive integers less than x to show that x is prime –select a subset of numbers {1, …, x-1} chosen randomly and test for those

20. 20 Conclusions Using random processes for nondeter- minism saves time and avoids bias Error should be accounted for Bounded by exponentially small error