Probabilistic Computation Lecture 13 Understanding BPP 1 Recap 2 - PowerPoint PPT Presentation

Probabilistic Computation Lecture 13 Understanding BPP 1

Recap 2

Recap Probabilistic computation 2

Recap Probabilistic computation NTM (on “random certificates”) for L: 2

Recap Probabilistic computation NTM (on “random certificates”) for L: Pr[M(x)=yes]: 2

Recap Probabilistic computation NTM (on “random certificates”) for L: x ∉ L x ∈ L Pr[M(x)=yes]: 2

Recap Probabilistic computation NTM (on “random certificates”) for L: x ∉ L x ∈ L Pr[M(x)=yes]: x ∉ L x ∈ L PTM for L: Pr[yes]: 2

Recap Probabilistic computation NTM (on “random certificates”) for L: x ∉ L x ∈ L Pr[M(x)=yes]: x ∉ L x ∈ L PTM for L: Pr[yes]: x ∉ L x ∈ L BPTM for L: Pr[yes]: 2

Recap Probabilistic computation NTM (on “random certificates”) for L: x ∉ L x ∈ L Pr[M(x)=yes]: x ∉ L x ∈ L PTM for L: Pr[yes]: x ∉ L x ∈ L BPTM for L: Pr[yes]: x ∉ L x ∈ L RTM for L: Pr[yes]: 2

Recap 3

Recap PP, RP, co-RP, BPP 3

Recap PP, RP, co-RP, BPP PP too powerful: NP ⊆ PP 3

Recap PP, RP, co-RP, BPP PP too powerful: NP ⊆ PP RP, BPP, with bounded gap 3

Recap PP, RP, co-RP, BPP PP too powerful: NP ⊆ PP RP, BPP, with bounded gap Gap can be boosted from 1/poly to 1-1/ exp 3

Recap PP, RP, co-RP, BPP PP too powerful: NP ⊆ PP RP, BPP, with bounded gap Gap can be boosted from 1/poly to 1-1/ exp A realistic/useful computational model 3

Recap PP, RP, co-RP, BPP PP too powerful: NP ⊆ PP RP, BPP, with bounded gap Gap can be boosted from 1/poly to 1-1/ exp A realistic/useful computational model Today: 3

Recap PP, RP, co-RP, BPP PP too powerful: NP ⊆ PP RP, BPP, with bounded gap Gap can be boosted from 1/poly to 1-1/ exp A realistic/useful computational model Today: NP ⊈ BPP, unless PH collapses 3

Recap PP, RP, co-RP, BPP PP too powerful: NP ⊆ PP RP, BPP, with bounded gap Gap can be boosted from 1/poly to 1-1/ exp A realistic/useful computational model Today: NP ⊈ BPP, unless PH collapses BPP ⊆ Σ 2P ∩ Π 2P 3

BPP vs. NP 4

BPP vs. NP Can randomized algorithms efficiently decide all NP problems? 4

BPP vs. NP Can randomized algorithms efficiently decide all NP problems? Unlikely: NP ⊆ BPP ⇒ PH = Σ 2P 4

BPP vs. NP Can randomized algorithms efficiently decide all NP problems? Unlikely: NP ⊆ BPP ⇒ PH = Σ 2P Will show BPP ⊆ P/poly 4

BPP vs. NP Can randomized algorithms efficiently decide all NP problems? Unlikely: NP ⊆ BPP ⇒ PH = Σ 2P Will show BPP ⊆ P/poly Then NP ⊆ BPP ⇒ NP ⊆ P/poly 4

BPP vs. NP Can randomized algorithms efficiently decide all NP problems? Unlikely: NP ⊆ BPP ⇒ PH = Σ 2P Will show BPP ⊆ P/poly Then NP ⊆ BPP ⇒ NP ⊆ P/poly ⇒ PH = Σ 2P 4

BPP ⊆ P/poly 5

BPP ⊆ P/poly If error probability is sufficiently small, will show there should be at least one random tape which works for all 2 n inputs of length n 5

BPP ⊆ P/poly If error probability is x r sufficiently small, will show there should be at least one � � � � � � � � � � � � random tape which works for � � � � � � all 2 n inputs of length n � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � 5

BPP ⊆ P/poly If error probability is x r sufficiently small, will show there should be at least one � � � � � � � � � � � � random tape which works for � � � � � � all 2 n inputs of length n � � � � � � Then, can give that � � � � � � random tape as advice � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � 5

BPP ⊆ P/poly If error probability is x r sufficiently small, will show there should be at least one � � � � � � � � � � � � random tape which works for � � � � � � all 2 n inputs of length n � � � � � � Then, can give that � � � � � � random tape as advice � � � � � � � � � � � � One such random tape if � � � � � � average (over x) error � � � � � � probability is less than 2 -n � � � � � � � � � � � � � � � � � � � � � � � � 5

BPP ⊆ P/poly If error probability is x r sufficiently small, will show there should be at least one � � � � � � � � � � � � random tape which works for � � � � � � all 2 n inputs of length n � � � � � � Then, can give that � � � � � � random tape as advice � � � � � � � � � � � � One such random tape if � � � � � � average (over x) error � � � � � � probability is less than 2 -n � � � � � � � � � � � � BPP: can make worst � � � � � � error probability < 2 -n � � � � � � 5

BPP vs. PH 6

BPP vs. PH BPP ⊆ Σ 2P 6

BPP vs. PH BPP ⊆ Σ 2P So BPP ⊆ Σ 2P ∩ Π 2P 6

BPP ⊆ Σ 2P 7

BPP ⊆ Σ 2P x ∈ L: “for almost all” r, M(x,r)=yes 7

BPP ⊆ Σ 2P x ∈ L: “for almost all” r, M(x,r)=yes x ∉ L: M(x,r)=yes for very few r 7

BPP ⊆ Σ 2P x ∈ L: “for almost all” r, M(x,r)=yes x ∉ L: M(x,r)=yes for very few r L = { x| for almost all r, M(x,r)=yes } 7

BPP ⊆ Σ 2P x ∈ L: “for almost all” r, M(x,r)=yes x ∉ L: M(x,r)=yes for very few r L = { x| for almost all r, M(x,r)=yes } If it were “for all”, in coNP 7

BPP ⊆ Σ 2P x ∈ L: “for almost all” r, M(x,r)=yes x ∉ L: M(x,r)=yes for very few r L = { x| for almost all r, M(x,r)=yes } If it were “for all”, in coNP L = { x| ∃ a small “neighborhood”, ∀ z, for some r “near” z, M(x,r)=yes } 7

BPP ⊆ Σ 2P x ∈ L: “for almost all” r, M(x,r)=yes x ∉ L: M(x,r)=yes for very few r L = { x| for almost all r, M(x,r)=yes } If it were “for all”, in coNP L = { x| ∃ a small “neighborhood”, ∀ z, for some r “near” z, M(x,r)=yes } Note: Neighborhood of z is small (polynomially large), so can go through all of them in polynomial time 7

BPP ⊆ Σ 2P Space of random tapes = {0,1} m Yes x = {r| M(x,r)=yes } 8

BPP ⊆ Σ 2P x ∈ L: |Yes x |>(1-2 -n )2 m Space of random tapes = {0,1} m Yes x = {r| M(x,r)=yes } 8

BPP ⊆ Σ 2P x ∉ L: |Yes x |<2 -n 2 m x ∈ L: |Yes x |>(1-2 -n )2 m Space of random tapes = {0,1} m Yes x = {r| M(x,r)=yes } 8

BPP ⊆ Σ 2P x ∉ L: |Yes x |<2 -n 2 m x ∈ L: |Yes x |>(1-2 -n )2 m Space of random tapes = {0,1} m Yes x = {r| M(x,r)=yes } x ∈ L: Will show that there exist a small set of shifts of Yes x that cover all z 8

BPP ⊆ Σ 2P x ∉ L: |Yes x |<2 -n 2 m x ∈ L: |Yes x |>(1-2 -n )2 m Space of random tapes = {0,1} m Yes x = {r| M(x,r)=yes } x ∈ L: Will show that there exist a small set of shifts of Yes x that cover all z If z is a shift of r ∈ Yes x , r is in the neighborhood of z 8

BPP ⊆ Σ 2P x ∉ L: |Yes x |<2 -n 2 m x ∈ L: |Yes x |>(1-2 -n )2 m Space of random tapes = {0,1} m Yes x = {r| M(x,r)=yes } x ∈ L: Will show that there exist a small set of shifts of Yes x that cover all z If z is a shift of r ∈ Yes x , r is in the neighborhood of z x ∉ L: Yes x very small, so its few shifts cover only a small region 8

BPP ⊆ Σ 2P 9

BPP ⊆ Σ 2P “A small set of shifts”: P = {u 1 ,u 2 ,...,u k } 9

BPP ⊆ Σ 2P “A small set of shifts”: P = {u 1 ,u 2 ,...,u k } P(r)={ r ⊕ u 1 ,r ⊕ u 2 ,...,r ⊕ u k } where r, u i are m-bit strings, and k is “small” (poly(n)) 9

BPP ⊆ Σ 2P “A small set of shifts”: P = {u 1 ,u 2 ,...,u k } P(r)={ r ⊕ u 1 ,r ⊕ u 2 ,...,r ⊕ u k } where r, u i are m-bit strings, and k is “small” (poly(n)) For each x ∈ L, does there exist a P s.t. P(Yes x ) := ∪ r ∈ Yes(x) P(r) = {0,1} m ? 9

BPP ⊆ Σ 2P “A small set of shifts”: P = {u 1 ,u 2 ,...,u k } P(r)={ r ⊕ u 1 ,r ⊕ u 2 ,...,r ⊕ u k } where r, u i are m-bit strings, and k is “small” (poly(n)) For each x ∈ L, does there exist a P s.t. P(Yes x ) := ∪ r ∈ Yes(x) P(r) = {0,1} m ? Yes! For all large S (like Yes x ) can indeed find a P s.t. P(S) = {0,1} m 9

BPP ⊆ Σ 2P “A small set of shifts”: P = {u 1 ,u 2 ,...,u k } P(r)={ r ⊕ u 1 ,r ⊕ u 2 ,...,r ⊕ u k } where r, u i are m-bit strings, and k is “small” (poly(n)) For each x ∈ L, does there exist a P s.t. P(Yes x ) := ∪ r ∈ Yes(x) P(r) = {0,1} m ? Yes! For all large S (like Yes x ) can indeed find a P s.t. P(S) = {0,1} m In fact, most P work (if k big enough)! 9

BPP ⊆ Σ 2P 10