Randomized Complexity Classes We allow TM to toss coins/throw - - PowerPoint PPT Presentation

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Randomized Complexity Classes We allow TM to toss coins/throw - - PowerPoint PPT Presentation

Feb 24, 2023 405 likes •612 views

Randomized Complexity Classes We allow TM to toss coins/throw dice etc. We write M(x,R) for output of M on input x, coin tosses R Def: L RP <=> poly-time randomized M : x L => Pr R [M(x,R)=1] 1/2 x

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SLIDE 1

Randomized Complexity Classes

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SLIDE 2
  • We allow TM to toss coins/throw dice etc.

We write M(x,R) for output of M on input x, coin tosses R

  • Def: L RP <=> poly-time randomized M :

∈ ∃ x L => Pr ∈

R [M(x,R)=1] ≥ 1/2

x L => Pr ∉

R [M(x,R)=1] = 0

  • Def: L BPP <=> poly-time randomized M :

∈ ∃ x L => Pr ∈

R [M(x,R)=1] ≥ 2/3

x L => Pr ∉

R [M(x,R)=1] ≤ 1/3

  • Exercise: For RP, can replace 1/2 with 1/nc , or

1- 1/2m for m = nc , for any c For BPP, can replace (2/3,1/3) = (1/2 + 1/nc , 1/2-1/nc ) or (1-1/2m , 1/2m ).

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SLIDE 3
  • Exercise: The following are equivalent:

1) L RP ∩ co-RP ∈ 2) There is a randomized poly-time machine M for L : ∀ x, R, M(x,R) {L(x), ?}, ∀ ∈ ∀ x, PrR [M(x,R) = ? ] ≤ 1/2 3) There is a randomized machine M for L : ∀ x, R, M(x,R) = L(x) ∀ the expected running time of M on x is poly(n) This class is known as ZPP.

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SLIDE 4
  • Claim: P ZPP RP BPP

⊆ ⊆ ⊆

  • Proof: By definition.

  • Claim: RP NP

⊆ Proof: ?

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SLIDE 5
  • Claim: P ZPP RP BPP

⊆ ⊆ ⊆

  • Proof: By definition.

  • Claim: RP NP

⊆ Proof: The witness is the random string

  • Big open question, is P = ZPP = RP = BPP?

Surprisingly, this is believed to be the case

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SLIDE 6
  • Claim: BPP P/poly

⊆

  • Proof:

Let L BPP. ∈ Let M(x,R) be a randomized poly-time TM deciding L. Make the error < 2-n. Note that for every x, PrR [ L(x) ≠ M(x,R) ] < 2-n So by the probabilistic method,

?????????????????????????????????????????????????????????

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SLIDE 7
  • Claim: BPP P/poly

⊆

  • Proof:

Let L BPP. ∈ Let M(x,R) be a randomized poly-time TM deciding L. Make the error < 2-n. Note that for every x, PrR [ L(x) ≠ M(x,R) ] < 2-n So by the probabilistic method, there exists some string R* : L(x) = M(x,R*) x. ∀ The circuit corresponding to M(x,R*) is the desired circuit.  Upshot: Randomness is only “useful” for TM, not for circuits.

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SLIDE 8
  • Claim: BPP ∑

⊆

2 P

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SLIDE 9
  • Claim: BPP ∑

⊆

2 P

  • Proof: Let M(x,R) toss |R| = r coins, and have error < 1/r2

Fix x and ask: Can we cover {0,1}r with r shifts of A := { R {0,1} ∈

r : M(x,R) = 1 } ?

For s {0,1} ∈

r , the s-shift is s+A := { s XOR a : a A } {0,1}

∈ ⊆

r

We'll show the answer to this question is equivalent to x L ∈ We then show this question can be asked in ∑2 P

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SLIDE 10
  • Claim: BPP ∑

⊆

2 P

  • Proof: Let M(x,R) toss |R| = r coins, and have error < 1/r2

Fix x and ask: Can we cover {0,1}r with r shifts of A := { R {0,1} ∈

r : M(x,R) = 1 } ?

For s {0,1} ∈

r , the s-shift is s+A := { s XOR a : a A } {0,1}

∈ ⊆

r

  • x L, we show we cannot cover. Note |A| <=

∉ ?

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SLIDE 11
  • Claim: BPP ∑

⊆

2 P

  • Proof: Let M(x,R) toss |R| = r coins, and have error < 1/r2

Fix x and ask: Can we cover {0,1}r with r shifts of A := { R {0,1} ∈

r : M(x,R) = 1 } ?

For s {0,1} ∈

r , the s-shift is s+A := { s XOR a : a A } {0,1}

∈ ⊆

r

  • x ∉ L, we show we cannot cover. Note |A| <= 2r / r2 .

∀ s1, …, sr : |s1+A U s2+A U … U sr+A | ≤ ?

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SLIDE 12
  • Claim: BPP ∑

⊆

2 P

  • Proof: Let M(x,R) toss |R| = r coins, and have error < 1/r2

Fix x and ask: Can we cover {0,1}r with r shifts of A := { R {0,1} ∈

r : M(x,R) = 1 } ?

For s {0,1} ∈

r , the s-shift is s+A := { s XOR a : a A } {0,1}

∈ ⊆

r

  • x ∉ L, we show we cannot cover. Note |A| <= 2r / r2 .

∀ s1, …, sr : |s1+A U s2+A U … U sr+A | ≤ r |A| ≤ ?

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SLIDE 13
  • Claim: BPP ∑

⊆

2 P

  • Proof: Let M(x,R) toss |R| = r coins, and have error < 1/r2

Fix x and ask: Can we cover {0,1}r with r shifts of A := { R {0,1} ∈

r : M(x,R) = 1 } ?

For s {0,1} ∈

r , the s-shift is s+A := { s XOR a : a A } {0,1}

∈ ⊆

r

  • x ∉ L, we show we cannot cover. Note |A| <= 2r / r2 .

∀ s1, …, sr : |s1+A U s2+A U … U sr+A | ≤ r |A| ≤ r 2r / r2 < 2r

  • x L, we show we can cover.

∈ Idea pick the shifts at random and show Pr[do not cover] < ?

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SLIDE 14
  • Claim: BPP ∑

⊆

2 P

  • Proof: Let M(x,R) toss |R| = r coins, and have error < 1/r2

Fix x and ask: Can we cover {0,1}r with r shifts of A := { R {0,1} ∈

r : M(x,R) = 1 } ?

For s {0,1} ∈

r , the s-shift is s+A := { s XOR a : a A } {0,1}

∈ ⊆

r

  • x ∉ L, we show we cannot cover. Note |A| <= 2r / r2 .

∀ s1, …, sr : |s1+A U s2+A U … U sr+A | ≤ r |A| ≤ r 2r / r2 < 2r

  • x ∈ L, we show we can cover.

Idea pick the shifts at random and show Pr[do not cover] < 1: Prs1, …, sr [ y {0,1} ∃ ∈

r : y U

∉

r sr + A] ≤

?

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SLIDE 15
  • Claim: BPP ∑

⊆

2 P

  • Proof: Let M(x,R) toss |R| = r coins, and have error < 1/r2

Fix x and ask: Can we cover {0,1}r with r shifts of A := { R {0,1} ∈

r : M(x,R) = 1 } ?

For s {0,1} ∈

r , the s-shift is s+A := { s XOR a : a A } {0,1}

∈ ⊆

r

  • x ∉ L, we show we cannot cover. Note |A| <= 2r / r2 .

∀ s1, …, sr : |s1+A U s2+A U … U sr+A | ≤ r |A| ≤ r 2r / r2 < 2r

  • x ∈ L, we show we can cover.

Idea pick the shifts at random and show Pr[do not cover] < 1: Prs1, …, sr [ y {0,1} ∃ ∈

r : y U

∉

r sr + A] ≤

∑y Prs1,…,sr[y U ∉

r sr + A] = ?

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SLIDE 16
  • Claim: BPP ∑

⊆

2 P

  • Proof: Let M(x,R) toss |R| = r coins, and have error < 1/r2

Fix x and ask: Can we cover {0,1}r with r shifts of A := { R {0,1} ∈

r : M(x,R) = 1 } ?

For s {0,1} ∈

r , the s-shift is s+A := { s XOR a : a A } {0,1}

∈ ⊆

r

  • x ∉ L, we show we cannot cover. Note |A| <= 2r / r2 .

∀ s1, …, sr : |s1+A U s2+A U … U sr+A | ≤ r |A| ≤ r 2r / r2 < 2r

  • x ∈ L, we show we can cover.

Idea pick the shifts at random and show Pr[do not cover] < 1: Prs1, …, sr [ y {0,1} ∃ ∈

r : y U

∉

r sr + A] ≤

∑y Prs1,…,sr[y U ∉

r sr + A] = ∑y (Prs[ y s + A])

∉

r ≤ ?

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SLIDE 17
  • Claim: BPP ∑

⊆

2 P

  • Proof: Let M(x,R) toss |R| = r coins, and have error < 1/r2

Fix x and ask: Can we cover {0,1}r with r shifts of A := { R {0,1} ∈

r : M(x,R) = 1 } ?

For s {0,1} ∈

r , the s-shift is s+A := { s XOR a : a A } {0,1}

∈ ⊆

r

  • x ∉ L, we show we cannot cover. Note |A| <= 2r / r2 .

∀ s1, …, sr : |s1+A U s2+A U … U sr+A | ≤ r |A| ≤ r 2r / r2 < 2r

  • x ∈ L, we show we can cover.

Idea pick the shifts at random and show Pr[do not cover] < 1: Prs1, …, sr [ y {0,1} ∃ ∈

r : y U

∉

r sr + A] ≤

∑y Prs1,…,sr[y U ∉

r sr + A] = ∑y (Prs[ y s + A])

∉

r ≤ ∑y (1/r2)r <1

So M(x,R) = 1 <=> ?

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SLIDE 18
  • Claim: BPP ∑

⊆

2 P

  • Proof: Let M(x,R) toss |R| = r coins, and have error < 1/r2

Fix x and ask: Can we cover {0,1}r with r shifts of A := { R {0,1} ∈

r : M(x,R) = 1 } ?

For s {0,1} ∈

r , the s-shift is s+A := { s XOR a : a A } {0,1}

∈ ⊆

r

  • x ∉ L, we show we cannot cover. Note |A| <= 2r / r2 .

∀ s1, …, sr : |s1+A U s2+A U … U sr+A | ≤ r |A| ≤ r 2r / r2 < 2r

  • x ∈ L, we show we can cover.

Idea pick the shifts at random and show Pr[do not cover] < 1: Prs1, …, sr [ y {0,1} ∃ ∈

r : y U

∉

r sr + A] ≤

∑y Prs1,…,sr[y U ∉

r sr + A] = ∑y (Prs[ y s + A])

∉

r ≤ ∑y (1/r2)r <1

So M(x,R) = 1 <=> s ∃

1, …, sr :

y {0,1} ∀ ∈

r , y U

∈

r sr + A

<=> s ∃

1, …, sr :

y {0,1} ∀ ∈

r, Vi=1 r M(x, y + si )=1 

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SLIDE 19
  • Corollary: P = NP => P = BPP.
  • Proof:

?

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SLIDE 20
  • Corollary: P = NP => P = BPP.
  • Proof:

P = NP => P = PH, and so P BPP PH = P ⊆ ⊆ 