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 To d a y: 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 To d a y: 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 To d a y: NP ⊈ BPP, unless PH collapses P P BPP ⊆ Σ2 ∩ Π2 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? P Unlikely: NP ⊆ BPP ⇒ PH = Σ2 4 BPP vs. NP Can randomized algorithms efficiently decide all NP problems? P Unlikely: NP ⊆ BPP ⇒ PH = Σ2 Will show BPP ⊆ P/poly 4 BPP vs. NP Can randomized algorithms efficiently decide all NP problems? P Unlikely: NP ⊆ BPP ⇒ PH = Σ2 Will show BPP ⊆ P/poly Then NP ⊆ BPP ⇒ NP ⊆ P/poly 4 BPP vs. NP Can randomized algorithms efficiently decide all NP problems? P Unlikely: NP ⊆ BPP ⇒ PH = Σ2 Will show BPP ⊆ P/poly Then NP ⊆ BPP ⇒ NP ⊆ P/poly P ⇒ PH = Σ2 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 2n inputs of length n 5 BPP ⊆ P/poly If error probability is sufficiently small, will show r x there should be at least one ჲ ჳ ჳ ჲ ჲ ჲ random tape which works for ჲ ჲ ჲ ჲ ჲ ჳ all 2n inputs of length n ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჳ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჳ ჲ ჳ ჳ ჲ ჲ ჲ ჲ ჳ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჳ ჲ ჲ ჲ ჲ ჲ ჲ ჳ ჲ ჲ ჲ ჲ ჳ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჳ ჲ ჲ ჲ ჲ ჲ ჲ 5 BPP ⊆ P/poly If error probability is sufficiently small, will show r x there should be at least one ჲ ჳ ჳ ჲ ჲ ჲ random tape which works for ჲ ჲ ჲ ჲ ჲ ჳ all 2n inputs of length n ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჳ ჲ ჲ ჲ Then, can give that ჲ ჲ ჲ ჲ ჳ ჲ random tape as advice ჳ ჳ ჲ ჲ ჲ ჲ ჳ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჳ ჲ ჲ ჲ ჲ ჲ ჲ ჳ ჲ ჲ ჲ ჲ ჳ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჳ ჲ ჲ ჲ ჲ ჲ ჲ 5 BPP ⊆ P/poly If error probability is sufficiently small, will show r x there should be at least one ჲ ჳ ჳ ჲ ჲ ჲ random tape which works for ჲ ჲ ჲ ჲ ჲ ჳ all 2n inputs of length n ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჳ ჲ ჲ ჲ Then, can give that ჲ ჲ ჲ ჲ ჳ ჲ random tape as advice ჳ ჳ ჲ ჲ ჲ ჲ ჳ ჲ ჲ ჲ ჲ ჲ One such random tape if ჲ ჲ ჲ ჳ ჲ ჲ average (over x) error ჲ ჲ ჲ ჲ ჳ ჲ -n probability is less than 2 ჲ ჲ ჲ ჳ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჳ ჲ ჲ ჲ ჲ ჲ ჲ 5 BPP ⊆ P/poly If error probability is sufficiently small, will show r x there should be at least one ჲ ჳ ჳ ჲ ჲ ჲ random tape which works for ჲ ჲ ჲ ჲ ჲ ჳ all 2n inputs of length n ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჳ ჲ ჲ ჲ Then, can give that ჲ ჲ ჲ ჲ ჳ ჲ random tape as advice ჳ ჳ ჲ ჲ ჲ ჲ ჳ ჲ ჲ ჲ ჲ ჲ One such random tape if ჲ ჲ ჲ ჳ ჲ ჲ average (over x) error ჲ ჲ ჲ ჲ ჳ ჲ -n probability is less than 2 ჲ ჲ ჲ ჳ ჲ ჲ ჲ ჲ ჲ ჲ ჲ ჲ BPP: can make worst ჲ ჲ ჲ ჲ ჲ ჳ error probability < 2-n ჲ ჲ ჲ ჲ ჲ ჲ 5 BPP vs. PH 6 BPP vs. PH P BPP ⊆ Σ2 6 BPP vs. PH P BPP ⊆ Σ2 P P So BPP ⊆ Σ2 ∩ Π2 6 P BPP ⊆ Σ2 7 P BPP ⊆ Σ2 x∈L: “for almost all” r, M(x,r)=yes 7 P BPP ⊆ Σ2 x∈L: “for almost all” r, M(x,r)=yes x∉L: M(x,r)=yes for very few r 7 P BPP ⊆ Σ2 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 P BPP ⊆ Σ2 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 P BPP ⊆ Σ2 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 P BPP ⊆ Σ2 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 P BPP ⊆ Σ2 Space of random tapes = {0,1}m Yes x = {r| M(x,r)=yes } 8 P BPP ⊆ Σ2 -n m x∈L: |Yesx|>(1-2 )2 Space of random tapes = {0,1}m Yes x = {r| M(x,r)=yes } 8 P BPP ⊆ Σ2 -n m -n m x∈L: |Yesx|>(1-2 )2 x∉L: |Yesx|<2 2 Space of random tapes = {0,1}m Yes x = {r| M(x,r)=yes } 8 P BPP ⊆ Σ2 -n m -n m x∈L: |Yesx|>(1-2 )2 x∉L: |Yesx|<2 2 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 P BPP ⊆ Σ2 -n m -n m x∈L: |Yesx|>(1-2 )2 x∉L: |Yesx|<2 2 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 ∈ Yesx, r is in the neighborhood of z 8 P BPP ⊆ Σ2 -n m -n m x∈L: |Yesx|>(1-2 )2 x∉L: |Yesx|<2 2 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 ∈ Yesx, r is in the neighborhood of z x∉L: Yesx very small, so its few shifts cover only a small region 8 P BPP ⊆ Σ2 9 P BPP ⊆ Σ2 “A small set of shifts”: P = {u1,u2,...,uk} 9 P BPP ⊆ Σ2 “A small set of shifts”: P = {u1,u2,...,uk} P(r)={ r⊕u1,r⊕u2,...,r⊕uk} where r, ui are m-bit strings, and k is “small” (poly(n)) 9 P BPP ⊆ Σ2 “A small set of shifts”: P = {u1,u2,...,uk} P(r)={ r⊕u1,r⊕u2,...,r⊕uk} where r, ui are m-bit strings, and k is “small” (poly(n)) m For each x∈L, does there exist a P s.t.