Probabilistic Computation Lecture 13 BPP vs. PH 1 Recap 2 Recap - - PowerPoint PPT Presentation

▶
probabilistic computation
SMART_READER_LITE
LIVE PREVIEW

Probabilistic Computation Lecture 13 BPP vs. PH 1 Recap 2 Recap - - PowerPoint PPT Presentation

Dec 21, 2022 475 likes •1.44k views

Probabilistic Computation Lecture 13 BPP vs. PH 1 Recap 2 Recap Probabilistic computation 2 Recap Probabilistic computation NTM (on random certificates) for L: 2 Recap Probabilistic computation NTM (on random certificates)

slide-1
SLIDE 1

Probabilistic Computation

Lecture 13 BPP vs. PH

1

slide-2
SLIDE 2

Recap

2

slide-3
SLIDE 3

Probabilistic computation

Recap

2

slide-4
SLIDE 4

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

Recap

2

slide-5
SLIDE 5

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

Recap

2

slide-6
SLIDE 6

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

Recap

x∉L x∈L

2

slide-7
SLIDE 7

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

Recap

x∉L x∈L x∉L x∈L

2

slide-8
SLIDE 8

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

Recap

x∉L x∈L x∉L x∈L x∉L x∈L

2

slide-9
SLIDE 9

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

Recap

x∉L x∈L x∉L x∈L x∉L x∈L x∉L x∈L

2

slide-10
SLIDE 10

Recap

3

slide-11
SLIDE 11

Recap

PP, RP, co-RP, BPP

3

slide-12
SLIDE 12

Recap

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

3

slide-13
SLIDE 13

Recap

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

3

slide-14
SLIDE 14

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

slide-15
SLIDE 15

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

slide-16
SLIDE 16

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

slide-17
SLIDE 17

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

slide-18
SLIDE 18

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 ⊆ Σ2P ∩ Π2P

3

slide-19
SLIDE 19

BPP vs. NP

4

slide-20
SLIDE 20

BPP vs. NP

Can randomized algorithms efficiently decide all NP problems?

4

slide-21
SLIDE 21

BPP vs. NP

Can randomized algorithms efficiently decide all NP problems? Unlikely: NP ⊆ BPP ⇒ PH = Σ2P

4

slide-22
SLIDE 22

BPP vs. NP

Can randomized algorithms efficiently decide all NP problems? Unlikely: NP ⊆ BPP ⇒ PH = Σ2P Will show BPP ⊆ P/poly

4

slide-23
SLIDE 23

BPP vs. NP

Can randomized algorithms efficiently decide all NP problems? Unlikely: NP ⊆ BPP ⇒ PH = Σ2P Will show BPP ⊆ P/poly Then NP ⊆ BPP ⇒ NP ⊆ P/poly

4

slide-24
SLIDE 24

BPP vs. NP

Can randomized algorithms efficiently decide all NP problems? Unlikely: NP ⊆ BPP ⇒ PH = Σ2P Will show BPP ⊆ P/poly Then NP ⊆ BPP ⇒ NP ⊆ P/poly ⇒ PH = Σ2P

4

slide-25
SLIDE 25

BPP ⊆ P/poly

5

slide-26
SLIDE 26

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

slide-27
SLIDE 27

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 ☑ ☒ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☒ ☒ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ x r

5

slide-28
SLIDE 28

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 Then, can give that random tape as advice ☑ ☒ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☒ ☒ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ x r

5

slide-29
SLIDE 29

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 Then, can give that random tape as advice One such random tape if average (over x) error probability is less than 2-n ☑ ☒ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☒ ☒ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ x r

5

slide-30
SLIDE 30

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 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 ☑ ☒ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☒ ☒ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☑ ☒ ☑ ☑ ☑ ☑ ☑ ☑ x r

5

slide-31
SLIDE 31

BPP vs. PH

6

slide-32
SLIDE 32

BPP vs. PH

BPP ⊆ Σ2P

6

slide-33
SLIDE 33

BPP vs. PH

BPP ⊆ Σ2P So BPP ⊆ Σ2P ∩ Π2P

6

slide-34
SLIDE 34

BPP ⊆ Σ2P

7

slide-35
SLIDE 35

BPP ⊆ Σ2P

x∈L: “for almost all” r, M(x,r)=yes

7

slide-36
SLIDE 36

BPP ⊆ Σ2P

x∈L: “for almost all” r, M(x,r)=yes x∉L: M(x,r)=yes for very few r

7

slide-37
SLIDE 37

BPP ⊆ Σ2P

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

slide-38
SLIDE 38

BPP ⊆ Σ2P

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

slide-39
SLIDE 39

BPP ⊆ Σ2P

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”, ∀r’, for some r “near” r’, M(x,r)=yes }

7

slide-40
SLIDE 40

BPP ⊆ Σ2P

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”, ∀r’, for some r “near” r’, M(x,r)=yes } Note: Neighborhood of r is small (polynomially large), so can go through all of them in polynomial time

7

slide-41
SLIDE 41

BPP ⊆ Σ2P

Space of random tapes = {0,1}m Yesx = {r| M(x,r)=yes }

8

slide-42
SLIDE 42

BPP ⊆ Σ2P

x∈L: |Yesx|>(1-2-n)2m Space of random tapes = {0,1}m Yesx = {r| M(x,r)=yes }

8

slide-43
SLIDE 43

BPP ⊆ Σ2P

x∉L: |Yesx|<2-n2m x∈L: |Yesx|>(1-2-n)2m Space of random tapes = {0,1}m Yesx = {r| M(x,r)=yes }

8

slide-44
SLIDE 44

BPP ⊆ Σ2P

x∈L: Will show that there exist a small set of shifts of Yesx that cover all r’ s

x∉L: |Yesx|<2-n2m x∈L: |Yesx|>(1-2-n)2m Space of random tapes = {0,1}m Yesx = {r| M(x,r)=yes }

8

slide-45
SLIDE 45

BPP ⊆ Σ2P

x∈L: Will show that there exist a small set of shifts of Yesx that cover all r’ s x∉L: Yesx very small, so its few shifts cover only a small region

x∉L: |Yesx|<2-n2m x∈L: |Yesx|>(1-2-n)2m Space of random tapes = {0,1}m Yesx = {r| M(x,r)=yes }

8

slide-46
SLIDE 46

BPP ⊆ Σ2P

9

slide-47
SLIDE 47

BPP ⊆ Σ2P

“A small set of shifts”: P = {u1,u2,...,uk}

9

slide-48
SLIDE 48

BPP ⊆ Σ2P

“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

slide-49
SLIDE 49

BPP ⊆ Σ2P

“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)) For each x∈L, does there exist a P s.t. P(Yesx) := ∪r∈Yes(x) P(r) = {0,1}m?

9

slide-50
SLIDE 50

BPP ⊆ Σ2P

“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)) For each x∈L, does there exist a P s.t. P(Yesx) := ∪r∈Yes(x) P(r) = {0,1}m? Yes! For all large S (like Yesx) can indeed find a P s.t. P(S) = {0,1}m

9

slide-51
SLIDE 51

BPP ⊆ Σ2P

“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)) For each x∈L, does there exist a P s.t. P(Yesx) := ∪r∈Yes(x) P(r) = {0,1}m? Yes! For all large S (like Yesx) can indeed find a P s.t. P(S) = {0,1}m In fact, most P work (if k big enough)!

9

slide-52
SLIDE 52

BPP ⊆ Σ2P

10

slide-53
SLIDE 53

BPP ⊆ Σ2P

Probabilistic Method (finding hay in haystack)

10

slide-54
SLIDE 54

BPP ⊆ Σ2P

Probabilistic Method (finding hay in haystack) To prove ∃P with some property

10

slide-55
SLIDE 55

BPP ⊆ Σ2P

Probabilistic Method (finding hay in haystack) To prove ∃P with some property Define a probability distribution over all candidate P’ s and prove that the property holds with positive probability (often even close to one)

10

slide-56
SLIDE 56

BPP ⊆ Σ2P

Probabilistic Method (finding hay in haystack) To prove ∃P with some property Define a probability distribution over all candidate P’ s and prove that the property holds with positive probability (often even close to one) Distribution s.t. easy to prove positive probability of property holding

10

slide-57
SLIDE 57

BPP ⊆ Σ2P

11

slide-58
SLIDE 58

BPP ⊆ Σ2P

Probabilistic method to find P = {u1,u2,...,uk}, s.t. for all large S, P(S) = {0,1}m

11

slide-59
SLIDE 59

BPP ⊆ Σ2P

Probabilistic method to find P = {u1,u2,...,uk}, s.t. for all large S, P(S) = {0,1}m Distribution over P’ s: randomized experiment to generate P

11

slide-60
SLIDE 60

BPP ⊆ Σ2P

Probabilistic method to find P = {u1,u2,...,uk}, s.t. for all large S, P(S) = {0,1}m Distribution over P’ s: randomized experiment to generate P Pick each ui independently, and uniformly at random from {0,1}m

11

slide-61
SLIDE 61

BPP ⊆ Σ2P

Probabilistic method to find P = {u1,u2,...,uk}, s.t. for all large S, P(S) = {0,1}m Distribution over P’ s: randomized experiment to generate P Pick each ui independently, and uniformly at random from {0,1}m Pr(over P)[P(S) ≠ {0,1}m] = Pr(over P)[∃z z∉P(S)]

11

slide-62
SLIDE 62

BPP ⊆ Σ2P

Probabilistic method to find P = {u1,u2,...,uk}, s.t. for all large S, P(S) = {0,1}m Distribution over P’ s: randomized experiment to generate P Pick each ui independently, and uniformly at random from {0,1}m Pr(over P)[P(S) ≠ {0,1}m] = Pr(over P)[∃z z∉P(S)]

11

slide-63
SLIDE 63

BPP ⊆ Σ2P

Probabilistic method to find P = {u1,u2,...,uk}, s.t. for all large S, P(S) = {0,1}m Distribution over P’ s: randomized experiment to generate P Pick each ui independently, and uniformly at random from {0,1}m Pr(over P)[P(S) ≠ {0,1}m] = Pr(over P)[∃z z∉P(S)]

11

slide-64
SLIDE 64

BPP ⊆ Σ2P

Probabilistic method to find P = {u1,u2,...,uk}, s.t. for all large S, P(S) = {0,1}m Distribution over P’ s: randomized experiment to generate P Pick each ui independently, and uniformly at random from {0,1}m Pr(over P)[P(S) ≠ {0,1}m] = Pr(over P)[∃z z∉P(S)] ≤ Σz Pr(over P)[z∉P(S)]

11

slide-65
SLIDE 65

BPP ⊆ Σ2P

Probabilistic method to find P = {u1,u2,...,uk}, s.t. for all large S, P(S) = {0,1}m Distribution over P’ s: randomized experiment to generate P Pick each ui independently, and uniformly at random from {0,1}m Pr(over P)[P(S) ≠ {0,1}m] = Pr(over P)[∃z z∉P(S)] ≤ Σz Pr(over P)[z∉P(S)] = Σz Pr(over u1..uk)[∀i z⊕ui ∉ S]

11

slide-66
SLIDE 66

BPP ⊆ Σ2P

Probabilistic method to find P = {u1,u2,...,uk}, s.t. for all large S, P(S) = {0,1}m Distribution over P’ s: randomized experiment to generate P Pick each ui independently, and uniformly at random from {0,1}m Pr(over P)[P(S) ≠ {0,1}m] = Pr(over P)[∃z z∉P(S)] ≤ Σz Pr(over P)[z∉P(S)] = Σz Pr(over u1..uk)[∀i z⊕ui ∉ S] = Σz Πi Pr (over ui)[z⊕ui ∉ S]

11

slide-67
SLIDE 67

BPP ⊆ Σ2P

Probabilistic method to find P = {u1,u2,...,uk}, s.t. for all large S, P(S) = {0,1}m Distribution over P’ s: randomized experiment to generate P Pick each ui independently, and uniformly at random from {0,1}m Pr(over P)[P(S) ≠ {0,1}m] = Pr(over P)[∃z z∉P(S)] ≤ Σz Pr(over P)[z∉P(S)] = Σz Pr(over u1..uk)[∀i z⊕ui ∉ S] = Σz Πi Pr (over ui)[z⊕ui ∉ S] = Σz Πi Pr (over ui)[ui ∉ z⊕S]

11

slide-68
SLIDE 68

BPP ⊆ Σ2P

Probabilistic method to find P = {u1,u2,...,uk}, s.t. for all large S, P(S) = {0,1}m Distribution over P’ s: randomized experiment to generate P Pick each ui independently, and uniformly at random from {0,1}m Pr(over P)[P(S) ≠ {0,1}m] = Pr(over P)[∃z z∉P(S)] ≤ Σz Pr(over P)[z∉P(S)] = Σz Pr(over u1..uk)[∀i z⊕ui ∉ S] = Σz Πi Pr (over ui)[z⊕ui ∉ S] = Σz Πi (|Sc|/2m) = Σz Πi Pr (over ui)[ui ∉ z⊕S]

11

slide-69
SLIDE 69

BPP ⊆ Σ2P

Probabilistic method to find P = {u1,u2,...,uk}, s.t. for all large S, P(S) = {0,1}m Distribution over P’ s: randomized experiment to generate P Pick each ui independently, and uniformly at random from {0,1}m Pr(over P)[P(S) ≠ {0,1}m] = Pr(over P)[∃z z∉P(S)] < Σz Πi 2-n ≤ Σz Pr(over P)[z∉P(S)] = Σz Pr(over u1..uk)[∀i z⊕ui ∉ S] = Σz Πi Pr (over ui)[z⊕ui ∉ S] = Σz Πi (|Sc|/2m) = Σz Πi Pr (over ui)[ui ∉ z⊕S]

11

slide-70
SLIDE 70

BPP ⊆ Σ2P

Probabilistic method to find P = {u1,u2,...,uk}, s.t. for all large S, P(S) = {0,1}m Distribution over P’ s: randomized experiment to generate P Pick each ui independently, and uniformly at random from {0,1}m Pr(over P)[P(S) ≠ {0,1}m] = Pr(over P)[∃z z∉P(S)] < Σz Πi 2-n ≤ Σz Pr(over P)[z∉P(S)] = Σz Pr(over u1..uk)[∀i z⊕ui ∉ S] = Σz Πi Pr (over ui)[z⊕ui ∉ S] = Σz Πi (|Sc|/2m) = 2m.(2-n)k = 1 = Σz Πi Pr (over ui)[ui ∉ z⊕S]

11

slide-71
SLIDE 71

BPP ⊆ Σ2P

Probabilistic method to find P = {u1,u2,...,uk}, s.t. for all large S, P(S) = {0,1}m Distribution over P’ s: randomized experiment to generate P Pick each ui independently, and uniformly at random from {0,1}m Pr(over P)[P(S) ≠ {0,1}m] = Pr(over P)[∃z z∉P(S)] So (with |S|>(1-2-n)2m and k=m/n), ∃P, P(S) = {0,1}m < Σz Πi 2-n ≤ Σz Pr(over P)[z∉P(S)] = Σz Pr(over u1..uk)[∀i z⊕ui ∉ S] = Σz Πi Pr (over ui)[z⊕ui ∉ S] = Σz Πi (|Sc|/2m) = 2m.(2-n)k = 1 = Σz Πi Pr (over ui)[ui ∉ z⊕S]

11

slide-72
SLIDE 72

BPP ⊆ Σ2P

x∉L: |Yesx|<2-n2m x∈L: |Yesx|>(1-2-n)2m Space of random strings = {0,1}m Yesx = {r| M(x,r)=yes }

12

slide-73
SLIDE 73

For each x∈L, ∃P (of size k=m/n) s.t. P(Yesx)={0,1}m

BPP ⊆ Σ2P

x∉L: |Yesx|<2-n2m x∈L: |Yesx|>(1-2-n)2m Space of random strings = {0,1}m Yesx = {r| M(x,r)=yes }

12

slide-74
SLIDE 74

For each x∈L, ∃P (of size k=m/n) s.t. P(Yesx)={0,1}m For each x∉L, P(Yesx) ⊊ {0,1}m

BPP ⊆ Σ2P

x∉L: |Yesx|<2-n2m x∈L: |Yesx|>(1-2-n)2m Space of random strings = {0,1}m Yesx = {r| M(x,r)=yes }

12

slide-75
SLIDE 75

For each x∈L, ∃P (of size k=m/n) s.t. P(Yesx)={0,1}m For each x∉L, P(Yesx) ⊊ {0,1}m | P(Yesx) | ≤ k| Yesx | = (m/n) 2-n2m < 2m

BPP ⊆ Σ2P

x∉L: |Yesx|<2-n2m x∈L: |Yesx|>(1-2-n)2m Space of random strings = {0,1}m Yesx = {r| M(x,r)=yes }

12

slide-76
SLIDE 76

For each x∈L, ∃P (of size k=m/n) s.t. P(Yesx)={0,1}m For each x∉L, P(Yesx) ⊊ {0,1}m | P(Yesx) | ≤ k| Yesx | = (m/n) 2-n2m < 2m L = { x| ∃P ∀r’ for some r∈P-1(r’) M(x,r)=yes }

BPP ⊆ Σ2P

x∉L: |Yesx|<2-n2m x∈L: |Yesx|>(1-2-n)2m Space of random strings = {0,1}m Yesx = {r| M(x,r)=yes }

12

slide-77
SLIDE 77

BPP-Complete Problem?

13

slide-78
SLIDE 78

BPP-Complete Problem?

Not known!

13

slide-79
SLIDE 79

BPP-Complete Problem?

Not known! Usual attempt: L = { (M,x,1t) | M(x)=yes in time t with probability > 2/3}

13

slide-80
SLIDE 80

BPP-Complete Problem?

Not known! Usual attempt: L = { (M,x,1t) | M(x)=yes in time t with probability > 2/3} Is indeed BPP-Hard

13

slide-81
SLIDE 81

BPP-Complete Problem?

Not known! Usual attempt: L = { (M,x,1t) | M(x)=yes in time t with probability > 2/3} Is indeed BPP-Hard But in BPP?

13

slide-82
SLIDE 82

BPP-Complete Problem?

Not known! Usual attempt: L = { (M,x,1t) | M(x)=yes in time t with probability > 2/3} Is indeed BPP-Hard But in BPP? Just run M(x) for t steps and accept if it accepts?

13

slide-83
SLIDE 83

BPP-Complete Problem?

Not known! Usual attempt: L = { (M,x,1t) | M(x)=yes in time t with probability > 2/3} Is indeed BPP-Hard But in BPP? Just run M(x) for t steps and accept if it accepts? If (M.x.1t) in L, we will indeed accept with prob. > 2/3

13

slide-84
SLIDE 84

BPP-Complete Problem?

Not known! Usual attempt: L = { (M,x,1t) | M(x)=yes in time t with probability > 2/3} Is indeed BPP-Hard But in BPP? Just run M(x) for t steps and accept if it accepts? If (M.x.1t) in L, we will indeed accept with prob. > 2/3 But M may not have a bounded gap. Then, if (M,x,1t) not in L, we may accept with prob. very close to 2/3.

13

slide-85
SLIDE 85

BPTIME-Hierarchy Theorem?

14

slide-86
SLIDE 86

BPTIME-Hierarchy Theorem?

BPTIME(n) ⊊ BPTIME(n100)?

14

slide-87
SLIDE 87

BPTIME-Hierarchy Theorem?

BPTIME(n) ⊊ BPTIME(n100)? Not known!

14

slide-88
SLIDE 88

BPTIME-Hierarchy Theorem?

BPTIME(n) ⊊ BPTIME(n100)? Not known! But is true for BPTIME(T)/1

14

slide-89
SLIDE 89

Today

15

slide-90
SLIDE 90

Today

Probabilistic computation

15

slide-91
SLIDE 91

Today

Probabilistic computation BPP ⊆ P/poly (so if NP ⊆ BPP, then PH=Σ2P)

15

slide-92
SLIDE 92

Today

Probabilistic computation BPP ⊆ P/poly (so if NP ⊆ BPP, then PH=Σ2P) BPP ⊆ Σ2P ∩ Π2P

15

slide-93
SLIDE 93

Today

Probabilistic computation BPP ⊆ P/poly (so if NP ⊆ BPP, then PH=Σ2P) BPP ⊆ Σ2P ∩ Π2P Coming up

15

slide-94
SLIDE 94

Today

Probabilistic computation BPP ⊆ P/poly (so if NP ⊆ BPP, then PH=Σ2P) BPP ⊆ Σ2P ∩ Π2P Coming up Basic randomized algorithmic techniques

15

slide-95
SLIDE 95

Today

Probabilistic computation BPP ⊆ P/poly (so if NP ⊆ BPP, then PH=Σ2P) BPP ⊆ Σ2P ∩ Π2P Coming up Basic randomized algorithmic techniques Saving on randomness

15