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Probabilistic Computation
Lecture 12 Flipping coins, taking chances PP, BPP
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Probabilistic Computation
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Probabilistic Computation Lecture 12 Flipping coins, taking chances - - PowerPoint PPT Presentation
Probabilistic Computation Lecture 12 Flipping coins, taking chances - - PowerPoint PPT Presentation
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
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Probabilistic Computation
Output depends not only on x, but also on random “coin flips”
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Probabilistic Computation
Output depends not only on x, but also on random “coin flips” M,x define a probability distribution over outcomes
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Probabilistic Computation
Output depends not only on x, but also on random “coin flips” M,x define a probability distribution over outcomes If for all x, M(x) equals f(x) with very high probability, could be used as f(x)
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Probabilistic Computation
Output depends not only on x, but also on random “coin flips” M,x define a probability distribution over outcomes If for all x, M(x) equals f(x) with very high probability, could be used as f(x)
x coins M
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Language Decided by a Probabilistic Computation
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Language Decided by a Probabilistic Computation
Different possible definitions of a prob. TM accepting input
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Language Decided by a Probabilistic Computation
Different possible definitions of a prob. TM accepting input M accepts x if pr[M(x)=yes] > 0; rejects if pr[M(x)=yes] = 0
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Language Decided by a Probabilistic Computation
Different possible definitions of a prob. TM accepting input M accepts x if pr[M(x)=yes] > 0; rejects if pr[M(x)=yes] = 0 M accepts x if pr[M(x)=yes] > 1/2; rejects if pr[M(x)=yes] ! 1/2
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Language Decided by a Probabilistic Computation
Different possible definitions of a prob. TM accepting input M accepts x if pr[M(x)=yes] > 0; rejects if pr[M(x)=yes] = 0 M accepts x if pr[M(x)=yes] > 1/2; rejects if pr[M(x)=yes] ! 1/2 M accepts x if pr[M(x)=yes] > 2/3; rejects if pr[M(x)=yes] < 1/3
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Language Decided by a Probabilistic Computation
Different possible definitions of a prob. TM accepting input M accepts x if pr[M(x)=yes] > 0; rejects if pr[M(x)=yes] = 0 M accepts x if pr[M(x)=yes] > 1/2; rejects if pr[M(x)=yes] ! 1/2 M accepts x if pr[M(x)=yes] > 2/3; rejects if pr[M(x)=yes] < 1/3 M accepts x if pr[M(x)=yes] > 2/3; rejects if pr[M(x)=yes] = 0
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Language Decided by a Probabilistic Computation
Different possible definitions of a prob. TM accepting input M accepts x if pr[M(x)=yes] > 0; rejects if pr[M(x)=yes] = 0 M accepts x if pr[M(x)=yes] > 1/2; rejects if pr[M(x)=yes] ! 1/2 M accepts x if pr[M(x)=yes] > 2/3; rejects if pr[M(x)=yes] < 1/3 M accepts x if pr[M(x)=yes] > 2/3; rejects if pr[M(x)=yes] = 0 Last two: If on any x neither, M doesn’ t decide a language!
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Language Decided by a Probabilistic Computation
Different possible definitions of a prob. TM accepting input M accepts x if pr[M(x)=yes] > 0; rejects if pr[M(x)=yes] = 0 M accepts x if pr[M(x)=yes] > 1/2; rejects if pr[M(x)=yes] ! 1/2 M accepts x if pr[M(x)=yes] > 2/3; rejects if pr[M(x)=yes] < 1/3 M accepts x if pr[M(x)=yes] > 2/3; rejects if pr[M(x)=yes] = 0 Last two: If on any x neither, M doesn’ t decide a language! When M does decide, much better than random guess
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Probabilistic TM
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Probabilistic TM
Like an NTM, but the two possible transitions are considered to be taken with equal probability
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Probabilistic TM
Like an NTM, but the two possible transitions are considered to be taken with equal probability
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Probabilistic TM
Like an NTM, but the two possible transitions are considered to be taken with equal probability
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Probabilistic TM
Like an NTM, but the two possible transitions are considered to be taken with equal probability
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Probabilistic TM
Like an NTM, but the two possible transitions are considered to be taken with equal probability
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Probabilistic TM
Like an NTM, but the two possible transitions are considered to be taken with equal probability Defines a probability with which an input is accepted
- r rejected
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Probabilistic TM
Like an NTM, but the two possible transitions are considered to be taken with equal probability Defines a probability with which an input is accepted
- r rejected
1/8 1/8 1/ 4 1/ 4 1/ 4
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Random Tape
x coins
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Random Tape
Random choice: flipping a fair coin
x coins
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Random Tape
Random choice: flipping a fair coin Coin flip is written on a read-once “random tape’’
x coins
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Random Tape
Random choice: flipping a fair coin Coin flip is written on a read-once “random tape’’ Enough coin flips made and written
- n the tape first, then start
execution
x coins
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Random Tape
Random choice: flipping a fair coin Coin flip is written on a read-once “random tape’’ Enough coin flips made and written
- n the tape first, then start
execution When considering bounded time TMs length of random tape (max coins used) also bounded
x coins
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Random Tape
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Random Tape
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Random Tape
1/8
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Random Tape
1/8
1
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Random Tape
1/8 1/8
1
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Random Tape
1/8 1/8
1 1
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Random Tape
1/8 1/8 1/ 4
1 1
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Random Tape
1/8 1/8 1/ 4
1 1 1
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Random Tape
1/8 1/8 1/ 4 1/ 4
1 1 1
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Random Tape
1/8 1/8 1/ 4 1/ 4
1 1 1 1 1
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Random Tape
1/8 1/8 1/ 4 1/ 4 1/ 4
1 1 1 1 1
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Random Tape
1/8 1/8 1/ 4 1/ 4 1/ 4
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
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Language Decided by a Probabilistic Computation
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Language Decided by a Probabilistic Computation
Different possible definitions of accepting input
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Language Decided by a Probabilistic Computation
Different possible definitions of accepting input Accept if pr[yes] > 0; reject if pr[yes] = 0
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Language Decided by a Probabilistic Computation
Different possible definitions of accepting input Accept if pr[yes] > 0; reject if pr[yes] = 0 NTM
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Language Decided by a Probabilistic Computation
Different possible definitions of accepting input Accept if pr[yes] > 0; reject if pr[yes] = 0 Accept if pr[yes] > 1/2; reject if pr[yes] ! 1/2
NTM
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Language Decided by a Probabilistic Computation
Different possible definitions of accepting input Accept if pr[yes] > 0; reject if pr[yes] = 0 Accept if pr[yes] > 1/2; reject if pr[yes] ! 1/2
NTM PTM
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Language Decided by a Probabilistic Computation
Different possible definitions of accepting input Accept if pr[yes] > 0; reject if pr[yes] = 0 Accept if pr[yes] > 1/2; reject if pr[yes] ! 1/2 Accept if pr[yes] > 2/3; reject if pr[yes] < 1/3
NTM PTM
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Language Decided by a Probabilistic Computation
Different possible definitions of accepting input Accept if pr[yes] > 0; reject if pr[yes] = 0 Accept if pr[yes] > 1/2; reject if pr[yes] ! 1/2 Accept if pr[yes] > 2/3; reject if pr[yes] < 1/3
NTM PTM BPTM
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Language Decided by a Probabilistic Computation
Different possible definitions of accepting input Accept if pr[yes] > 0; reject if pr[yes] = 0 Accept if pr[yes] > 1/2; reject if pr[yes] ! 1/2 Accept if pr[yes] > 2/3; reject if pr[yes] < 1/3 Accept if pr[yes] > 2/3; reject if pr[yes] = 0
NTM PTM BPTM
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Language Decided by a Probabilistic Computation
Different possible definitions of accepting input Accept if pr[yes] > 0; reject if pr[yes] = 0 Accept if pr[yes] > 1/2; reject if pr[yes] ! 1/2 Accept if pr[yes] > 2/3; reject if pr[yes] < 1/3 Accept if pr[yes] > 2/3; reject if pr[yes] = 0
NTM PTM BPTM RTM
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Language Decided by a Probabilistic Computation
Different possible definitions of accepting input Accept if pr[yes] > 0; reject if pr[yes] = 0 Accept if pr[yes] > 1/2; reject if pr[yes] ! 1/2 Accept if pr[yes] > 2/3; reject if pr[yes] < 1/3 Accept if pr[yes] > 2/3; reject if pr[yes] = 0 (Not standard nomenclature!)
NTM PTM BPTM RTM
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PTIME, BPTIME and RTIME
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PTIME, BPTIME and RTIME
T-time probabilistic TM
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PTIME, BPTIME and RTIME
T-time probabilistic TM On all inputs x, on any random tape, terminates in T(|x|) time and outputs “yes” or “no. ”
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PTIME, BPTIME and RTIME
T-time probabilistic TM On all inputs x, on any random tape, terminates in T(|x|) time and outputs “yes” or “no. ” Just like NTIME(T)
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PTIME, BPTIME and RTIME
T-time probabilistic TM On all inputs x, on any random tape, terminates in T(|x|) time and outputs “yes” or “no. ” Just like NTIME(T) BPTIME(T) = class of languages decided by BPTMs in time T
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PTIME, BPTIME and RTIME
T-time probabilistic TM On all inputs x, on any random tape, terminates in T(|x|) time and outputs “yes” or “no. ” Just like NTIME(T) BPTIME(T) = class of languages decided by BPTMs in time T Similarly PTIME(T) and RTIME(T)
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PP, BPP and RP
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PP, BPP and RP
PP = ∪c>0 PTIME(O(nc))
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PP, BPP and RP
PP = ∪c>0 PTIME(O(nc)) BPP = ∪c>0 BPTIME(O(nc))
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PP, BPP and RP
PP = ∪c>0 PTIME(O(nc)) BPP = ∪c>0 BPTIME(O(nc)) RP = ∪c>0 RTIME(O(nc))
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PP, BPP and RP
PP = ∪c>0 PTIME(O(nc)) BPP = ∪c>0 BPTIME(O(nc)) RP = ∪c>0 RTIME(O(nc)) co-RP
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co-RTM
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co-RTM
Accept if pr[yes] > 0; reject if pr[yes] = 0 NTM
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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
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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
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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
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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
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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
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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
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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
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RP and co-RP
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RP and co-RP
One sided error (“bounded error” versions of NP and co-NP)
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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
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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
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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
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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
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co-BPP, co-PP
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co-BPP, co-PP
BPP = co-BPP
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co-BPP, co-PP
BPP = co-BPP co-BPTMs are same as BPTMs
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co-BPP, co-PP
BPP = co-BPP co-BPTMs are same as BPTMs In fact PP = co-PP
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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
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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
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PP = co-PP
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PP = co-PP
Modifying a PTM M to an equivalent PTM M’, so that for all x Pr[M’(x)=yes] # 1/2
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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
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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
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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
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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
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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
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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 then M and M’ equivalent M
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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 then M and M’ equivalent What is such an ε? M
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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 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
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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 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 M
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PP and NP
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NP ⊆ PP
PP and NP
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NP ⊆ PP Use random-tape as non- deterministic choices of NTM M
PP and NP
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NP ⊆ PP Use random-tape as non- deterministic choices of NTM M
PP and NP
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NP ⊆ PP Use random-tape as non- deterministic choices of NTM M If M rejects, accept with 1/2 prob., else accept
PP and NP
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NP ⊆ PP Use random-tape as non- deterministic choices of NTM M If M rejects, accept with 1/2 prob., else accept
PP and NP
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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
PP and NP
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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 Accepting gap can be exponentially small
PP and NP
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Bounding Probability Gap
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Bounding Probability Gap
Gap
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Bounding Probability Gap
Gap Minx∈L Pr[M(x)=yes] - Maxx∉L Pr[M(x)=yes]
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Bounding Probability Gap
Gap Minx∈L Pr[M(x)=yes] - Maxx∉L Pr[M(x)=yes] BPP, RP, coRP require M to have gap some constant (1/3, 2/3)
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Bounding Probability Gap
Gap Minx∈L Pr[M(x)=yes] - Maxx∉L Pr[M(x)=yes] BPP, RP, coRP require M to have gap some constant (1/3, 2/3) Setting gap = 1/nc is enough
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Bounding Probability Gap
Gap Minx∈L Pr[M(x)=yes] - Maxx∉L Pr[M(x)=yes] BPP, RP, coRP require M to have gap some constant (1/3, 2/3) Setting gap = 1/nc is enough Can be boosted to gap = 1 - 1/2n^d in polynomial time
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Soundness Amplification for RP
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Soundness Amplification for RP
M’(x): Repeat M(x) t times and if any yes, say yes
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Soundness Amplification for RP
M’(x): Repeat M(x) t times and if any yes, say yes If x∉L: Pr[M(x)=no] = 1. So Pr[M’(x)=no] = 1
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Soundness Amplification for RP
M’(x): Repeat M(x) t times and if any yes, say yes If x∉L: Pr[M(x)=no] = 1. So Pr[M’(x)=no] = 1 If x∈L: Pr[M(x)=no] ! 1-δ (when gap = δ). Then Pr[M’(x)=no] ! (1-δ)t
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Soundness Amplification for RP
M’(x): Repeat M(x) t times and if any yes, say yes If x∉L: Pr[M(x)=no] = 1. So Pr[M’(x)=no] = 1 If x∈L: Pr[M(x)=no] ! 1-δ (when gap = δ). Then Pr[M’(x)=no] ! (1-δ)t With t = nd/δ, Pr[M’(x)=no] < e-(n^d) (1-δ < e-δ)
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Soundness Amplification for RP
M’(x): Repeat M(x) t times and if any yes, say yes If x∉L: Pr[M(x)=no] = 1. So Pr[M’(x)=no] = 1 If x∈L: Pr[M(x)=no] ! 1-δ (when gap = δ). Then Pr[M’(x)=no] ! (1-δ)t With t = nd/δ, Pr[M’(x)=no] < e-(n^d) (1-δ < e-δ) For δ = n-c, t=nd+c is polynomial
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Soundness Amplification for BPP
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Soundness Amplification for BPP
Repeat M(x) t times and take majority
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Soundness Amplification for BPP
Repeat M(x) t times and take majority i.e. estimate Pr[M(x)=yes] and check if it is > 1/2
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Soundness Amplification for BPP
Repeat M(x) t times and take majority i.e. estimate Pr[M(x)=yes] and check if it is > 1/2 Error only if |estimate - real| " gap/2
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Soundness Amplification for BPP
Repeat M(x) t times and take majority i.e. estimate Pr[M(x)=yes] and check if it is > 1/2 Error only if |estimate - real| " gap/2 Estimation error goes down exponentially with t: Chernoff bound
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Soundness Amplification for BPP
Repeat M(x) t times and take majority i.e. estimate Pr[M(x)=yes] and check if it is > 1/2 Error only if |estimate - real| " gap/2 Estimation error goes down exponentially with t: Chernoff bound Pr[ |estimate - real| " δ/2 ] ! 2-!(t.δ^2)
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Soundness Amplification for BPP
Repeat M(x) t times and take majority i.e. estimate Pr[M(x)=yes] and check if it is > 1/2 Error only if |estimate - real| " gap/2 Estimation error goes down exponentially with t: Chernoff bound Pr[ |estimate - real| " δ/2 ] ! 2-!(t.δ^2) t = O(nd/δ2) enough for Pr[error] ! 2-n^d
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Today
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Today
Probabilistic computation
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Today
Probabilistic computation PP, RP, co-RP, BPP
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Today
Probabilistic computation PP, RP, co-RP, BPP PP too powerful: NP ⊆ PP
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Today
Probabilistic computation PP, RP, co-RP, BPP PP too powerful: NP ⊆ PP Constant gap: BPP, RP, co-RP
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Today
Probabilistic computation PP, RP, co-RP, BPP PP too powerful: NP ⊆ PP Constant gap: BPP, RP, co-RP RP, co-RP one-sided error
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SLIDE 127
Today
Probabilistic computation PP, RP, co-RP, BPP PP too powerful: NP ⊆ PP Constant gap: BPP, RP, co-RP RP, co-RP one-sided error Soundness Amplification: for RP, for BPP
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SLIDE 128
Today
Probabilistic computation PP, RP, co-RP, BPP PP too powerful: NP ⊆ PP Constant gap: BPP, RP, co-RP RP, co-RP one-sided error Soundness Amplification: for RP, for BPP From gap 1/poly to 1-1/ exp
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SLIDE 129
Today
Probabilistic computation PP, RP, co-RP, BPP PP too powerful: NP ⊆ PP Constant gap: BPP, RP, co-RP RP, co-RP one-sided error Soundness Amplification: for RP, for BPP From gap 1/poly to 1-1/ exp Next: more on BPP and relatives
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