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Probabilistic Computation Lecture 12 Flipping coins, taking chances - PowerPoint PPT Presentation

May 21, 2023 •535 likes •1.84k views

Probabilistic Computation Lecture 12 Flipping coins, taking chances PP, BPP 1 Probabilistic Computation 2 Probabilistic Computation Output depends not only on x, but also on random coin flips 2 Probabilistic Computation Output

  1. PP, BPP and RP PP = ∪ c>0 PTIME(O(n c )) BPP = ∪ c>0 BPTIME(O(n c )) RP = ∪ c>0 RTIME(O(n c )) 9

  2. PP, BPP and RP PP = ∪ c>0 PTIME(O(n c )) BPP = ∪ c>0 BPTIME(O(n c )) RP = ∪ c>0 RTIME(O(n c )) co-RP 9

  3. co-RTM 10

  4. co-RTM Accept if pr[yes] > 0; reject if pr[yes] = 0 NTM 10

  5. co-RTM Accept if pr[yes] > 0; reject if pr[yes] = 0 NTM Accept if pr[yes] = 1; reject if pr[yes] < 1 co-NTM 10

  6. co-RTM Accept if pr[yes] > 0; reject if pr[yes] = 0 NTM Accept if pr[yes] = 1; reject if pr[yes] < 1 co-NTM Accept if pr[yes] > 1/2; reject if pr[yes] ! 1/2 PTM 10

  7. co-RTM Accept if pr[yes] > 0; reject if pr[yes] = 0 NTM Accept if pr[yes] = 1; reject if pr[yes] < 1 co-NTM Accept if pr[yes] > 1/2; reject if pr[yes] ! 1/2 PTM Accept if pr[yes] " 1/2; reject if pr[yes] < 1/2 co-PTM 10

  8. co-RTM Accept if pr[yes] > 0; reject if pr[yes] = 0 NTM Accept if pr[yes] = 1; reject if pr[yes] < 1 co-NTM Accept if pr[yes] > 1/2; reject if pr[yes] ! 1/2 PTM Accept if pr[yes] " 1/2; reject if pr[yes] < 1/2 co-PTM Accept if pr[yes] > 2/3; reject if pr[yes] < 1/3 BPTM 10

  9. co-RTM Accept if pr[yes] > 0; reject if pr[yes] = 0 NTM Accept if pr[yes] = 1; reject if pr[yes] < 1 co-NTM Accept if pr[yes] > 1/2; reject if pr[yes] ! 1/2 PTM Accept if pr[yes] " 1/2; reject if pr[yes] < 1/2 co-PTM Accept if pr[yes] > 2/3; reject if pr[yes] < 1/3 BPTM Accept if pr[yes] > 2/3; reject if pr[yes] < 1/3 co-BPTM 10

  10. co-RTM Accept if pr[yes] > 0; reject if pr[yes] = 0 NTM Accept if pr[yes] = 1; reject if pr[yes] < 1 co-NTM Accept if pr[yes] > 1/2; reject if pr[yes] ! 1/2 PTM Accept if pr[yes] " 1/2; reject if pr[yes] < 1/2 co-PTM Accept if pr[yes] > 2/3; reject if pr[yes] < 1/3 BPTM Accept if pr[yes] > 2/3; reject if pr[yes] < 1/3 co-BPTM Accept if pr[yes] > 2/3; reject if pr[yes] = 0 RTM 10

  11. co-RTM Accept if pr[yes] > 0; reject if pr[yes] = 0 NTM Accept if pr[yes] = 1; reject if pr[yes] < 1 co-NTM Accept if pr[yes] > 1/2; reject if pr[yes] ! 1/2 PTM Accept if pr[yes] " 1/2; reject if pr[yes] < 1/2 co-PTM Accept if pr[yes] > 2/3; reject if pr[yes] < 1/3 BPTM Accept if pr[yes] > 2/3; reject if pr[yes] < 1/3 co-BPTM Accept if pr[yes] > 2/3; reject if pr[yes] = 0 RTM Accept if pr[yes] = 1; reject if pr[yes] < 1/3 co-RTM 10

  12. RP and co-RP 11

  13. RP and co-RP One sided error (“bounded error” versions of NP and co-NP) 11

  14. RP and co-RP One sided error (“bounded error” versions of NP and co-NP) RP: if yes, may still say no w/p at most 1/3 11

  15. RP and co-RP One sided error (“bounded error” versions of NP and co-NP) RP: if yes, may still say no w/p at most 1/3 i.e., if RTM says no, can be wrong 11

  16. RP and co-RP One sided error (“bounded error” versions of NP and co-NP) RP: if yes, may still say no w/p at most 1/3 i.e., if RTM says no, can be wrong co-RP: if no, may still say yes w/p at most 1/3 11

  17. RP and co-RP One sided error (“bounded error” versions of NP and co-NP) RP: if yes, may still say no w/p at most 1/3 i.e., if RTM says no, can be wrong co-RP: if no, may still say yes w/p at most 1/3 i.e., if co-RTM says yes, can be wrong 11

  18. co-BPP, co-PP 12

  19. co-BPP, co-PP BPP = co-BPP 12

  20. co-BPP, co-PP BPP = co-BPP co-BPTMs are same as BPTMs 12

  21. co-BPP, co-PP BPP = co-BPP co-BPTMs are same as BPTMs In fact PP = co-PP 12

  22. co-BPP, co-PP BPP = co-BPP co-BPTMs are same as BPTMs In fact PP = co-PP PTMs and co-PTMs differ on accepting inputs with Pr [yes]=1/2 12

  23. co-BPP, co-PP BPP = co-BPP co-BPTMs are same as BPTMs In fact PP = co-PP PTMs and co-PTMs differ on accepting inputs with Pr [yes]=1/2 But can modify a PTM so that Pr[M(x)=yes] # 1/2 for all x, without changing language accepted 12

  24. PP = co-PP 13

  25. PP = co-PP Modifying a PTM M to an equivalent PTM M’, so that for all x Pr[M’(x)=yes] # 1/2 13

  26. PP = co-PP Modifying a PTM M to an equivalent PTM M’, so that for all x Pr[M’(x)=yes] # 1/2 Consider M’(x): w.p. 1/2 run M(x); w.p. 1/2, ignore input and say yes w.p. 1/2 - ε , and say no w.p. 1/2 + ε 13

  27. PP = co-PP Modifying a PTM M to an equivalent PTM M’, so that for all x Pr[M’(x)=yes] # 1/2 Consider M’(x): w.p. 1/2 run M(x); w.p. 1/2, ignore input and say yes w.p. 1/2 - ε , and say no w.p. 1/2 + ε 13

  28. PP = co-PP Modifying a PTM M to an equivalent PTM M’, so that for all x Pr[M’(x)=yes] # 1/2 Consider M’(x): w.p. 1/2 run M(x); w.p. 1/2, ignore input and say yes w.p. 1/2 - ε , and say no w.p. 1/2 + ε M 13

  29. PP = co-PP Modifying a PTM M to an equivalent PTM M’, so that for all x Pr[M’(x)=yes] # 1/2 Consider M’(x): w.p. 1/2 run M(x); w.p. 1/2, ignore input and say yes w.p. 1/2 - ε , and say no w.p. 1/2 + ε M 13

  30. PP = co-PP Modifying a PTM M to an equivalent PTM M’, so that for all x Pr[M’(x)=yes] # 1/2 Consider M’(x): w.p. 1/2 run M(x); w.p. 1/2, ignore input and say yes w.p. 1/2 - ε , and say no w.p. 1/2 + ε pr[M’(x)=yes] = pr[M(x)=yes]/2 + (1/2 - ε )/2 M 13

  31. PP = co-PP Modifying a PTM M to an equivalent PTM M’, so that for all x Pr[M’(x)=yes] # 1/2 Consider M’(x): w.p. 1/2 run M(x); w.p. 1/2, ignore input and say yes w.p. 1/2 - ε , and say no w.p. 1/2 + ε pr[M’(x)=yes] = pr[M(x)=yes]/2 + (1/2 - ε )/2 If pr[M(x)=yes] > 1/2 ⇒ pr[M(x)=yes] > 1/2 + ε M then M and M’ equivalent 13

  32. PP = co-PP Modifying a PTM M to an equivalent PTM M’, so that for all x Pr[M’(x)=yes] # 1/2 Consider M’(x): w.p. 1/2 run M(x); w.p. 1/2, ignore input and say yes w.p. 1/2 - ε , and say no w.p. 1/2 + ε pr[M’(x)=yes] = pr[M(x)=yes]/2 + (1/2 - ε )/2 If pr[M(x)=yes] > 1/2 ⇒ pr[M(x)=yes] > 1/2 + ε M then M and M’ equivalent What is such an ε ? 13

  33. PP = co-PP Modifying a PTM M to an equivalent PTM M’, so that for all x Pr[M’(x)=yes] # 1/2 Consider M’(x): w.p. 1/2 run M(x); w.p. 1/2, ignore input and say yes w.p. 1/2 - ε , and say no w.p. 1/2 + ε pr[M’(x)=yes] = pr[M(x)=yes]/2 + (1/2 - ε )/2 If pr[M(x)=yes] > 1/2 ⇒ pr[M(x)=yes] > 1/2 + ε M then M and M’ equivalent What is such an ε ? 2 -(m+1) where no. of coins used by M is at most m 13

  34. PP = co-PP Modifying a PTM M to an equivalent PTM M’, so that for all x Pr[M’(x)=yes] # 1/2 Consider M’(x): w.p. 1/2 run M(x); w.p. 1/2, ignore input and say yes w.p. 1/2 - ε , and say no w.p. 1/2 + ε pr[M’(x)=yes] = pr[M(x)=yes]/2 + (1/2 - ε )/2 If pr[M(x)=yes] > 1/2 ⇒ pr[M(x)=yes] > 1/2 + ε M then M and M’ equivalent What is such an ε ? 2 -(m+1) where no. of coins used by M is at most m M’ tosses at most m+2 coins 13

  35. PP and NP 14

  36. PP and NP NP ⊆ PP 14

  37. PP and NP NP ⊆ PP Use random-tape as non- deterministic choices of NTM M 14

  38. PP and NP NP ⊆ PP Use random-tape as non- deterministic choices of NTM M 14

  39. PP and NP NP ⊆ PP Use random-tape as non- deterministic choices of NTM M If M rejects, accept with 1/2 prob., else accept 14

  40. PP and NP NP ⊆ PP Use random-tape as non- deterministic choices of NTM M If M rejects, accept with 1/2 prob., else accept 14

  41. PP and NP NP ⊆ PP Use random-tape as non- deterministic choices of NTM M If M rejects, accept with 1/2 prob., else accept If even one thread of M(x) accepts, pr[M(x)=yes] > 1/2 14

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