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Interactive Proofs
Lecture 18 AM
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Interactive Proofs
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Interactive Proofs Lecture 18 AM 1 Interactive Proofs 2 - - PowerPoint PPT Presentation
Interactive Proofs Lecture 18 AM 1 Interactive Proofs 2 - - PowerPoint PPT Presentation
Interactive Proofs Lecture 18 AM 1 Interactive Proofs 2 Interactive Proofs IP[k] 2 Interactive Proofs IP[k] IP[poly] = PSPACE 2 Interactive Proofs IP[k] IP[poly] = PSPACE IP protocol for TQBF using arithmetization 2 Interactive
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Interactive Proofs
IP[k]
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Interactive Proofs
IP[k] IP[poly] = PSPACE
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Interactive Proofs
IP[k] IP[poly] = PSPACE IP protocol for TQBF using arithmetization
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Interactive Proofs
IP[k] IP[poly] = PSPACE IP protocol for TQBF using arithmetization We saw IP protocol for sum-check
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Interactive Proofs
IP[k] IP[poly] = PSPACE IP protocol for TQBF using arithmetization We saw IP protocol for sum-check IP[const] = AM[const]
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Interactive Proofs
IP[k] IP[poly] = PSPACE IP protocol for TQBF using arithmetization We saw IP protocol for sum-check IP[const] = AM[const] We saw public coin protocol for Graph Non-Isomorphism
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Interactive Proofs
IP[k] IP[poly] = PSPACE IP protocol for TQBF using arithmetization We saw IP protocol for sum-check IP[const] = AM[const] We saw public coin protocol for Graph Non-Isomorphism Using 2-universal hash functions
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Interactive Proofs
IP[k] IP[poly] = PSPACE IP protocol for TQBF using arithmetization We saw IP protocol for sum-check IP[const] = AM[const] We saw public coin protocol for Graph Non-Isomorphism Using 2-universal hash functions Today: Collapse of the AM hierarchy
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Interactive Proofs
IP[k] IP[poly] = PSPACE IP protocol for TQBF using arithmetization We saw IP protocol for sum-check IP[const] = AM[const] We saw public coin protocol for Graph Non-Isomorphism Using 2-universal hash functions Today: Collapse of the AM hierarchy AM[const] = AM[2]
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Recall AM
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Recall AM
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Recall AM
AM[2] (or simply AM)
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Recall AM
AM[2] (or simply AM) Input x
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Recall AM
AM[2] (or simply AM) Input x Random coins r come from a beacon
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Recall AM
AM[2] (or simply AM) Input x Random coins r come from a beacon
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Recall AM
AM[2] (or simply AM) Input x Random coins r come from a beacon Unbounded prover Merlin sends a “proof” message a
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Recall AM
AM[2] (or simply AM) Input x Random coins r come from a beacon Unbounded prover Merlin sends a “proof” message a Polynomial time verifier Arthur runs a deterministic verification procedure R(x;r,a), and outputs Yes or No
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Recall AM
AM[2] (or simply AM) Input x Random coins r come from a beacon Unbounded prover Merlin sends a “proof” message a Polynomial time verifier Arthur runs a deterministic verification procedure R(x;r,a), and outputs Yes or No L is said to have an AM protocol if
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Recall AM
AM[2] (or simply AM) Input x Random coins r come from a beacon Unbounded prover Merlin sends a “proof” message a Polynomial time verifier Arthur runs a deterministic verification procedure R(x;r,a), and outputs Yes or No L is said to have an AM protocol if x∈L ⇔ max Pr[Yes] > 2/3
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Recall AM
AM[2] (or simply AM) Input x Random coins r come from a beacon Unbounded prover Merlin sends a “proof” message a Polynomial time verifier Arthur runs a deterministic verification procedure R(x;r,a), and outputs Yes or No L is said to have an AM protocol if x∈L ⇔ max Pr[Yes] > 2/3 x∉L ⇔ max Pr[Yes] < 1/3
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max Pr[Yes]
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max Pr[Yes]
Quantity of interest
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max Pr[Yes]
Quantity of interest Maximum (over prover strategies) probability (over coins from the beacon)
- f Arthur saying yes
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max Pr[Yes]
Quantity of interest Maximum (over prover strategies) probability (over coins from the beacon)
- f Arthur saying yes
Evaluate the “Avg-Max tree”
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max Pr[Yes]
Max
Avg
R
Max
R R
Max
... ... ... ...
Quantity of interest Maximum (over prover strategies) probability (over coins from the beacon)
- f Arthur saying yes
Evaluate the “Avg-Max tree”
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max Pr[Yes]
Max
Avg
R
Max
R R
Max
... ... ... ...
Quantity of interest Maximum (over prover strategies) probability (over coins from the beacon)
- f Arthur saying yes
Evaluate the “Avg-Max tree” Leaves: Pr[yes] = 0 or 1, as determined by Arthur’ s program
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max Pr[Yes]
Max
Avg
R
Max
R R
Max
... ... ... ...
Quantity of interest Maximum (over prover strategies) probability (over coins from the beacon)
- f Arthur saying yes
Evaluate the “Avg-Max tree” Leaves: Pr[yes] = 0 or 1, as determined by Arthur’ s program Max nodes: maximum of children
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max Pr[Yes]
Max
Avg
R
Max
R R
Max
... ... ... ...
Quantity of interest Maximum (over prover strategies) probability (over coins from the beacon)
- f Arthur saying yes
Evaluate the “Avg-Max tree” Leaves: Pr[yes] = 0 or 1, as determined by Arthur’ s program Max nodes: maximum of children Avg node: average of children
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max Pr[Yes]
Max
Avg
R
Max
R R
Max
... ... ... ...
Quantity of interest Maximum (over prover strategies) probability (over coins from the beacon)
- f Arthur saying yes
Evaluate the “Avg-Max tree” Leaves: Pr[yes] = 0 or 1, as determined by Arthur’ s program Max nodes: maximum of children Avg node: average of children Extends to AM[k], with k alternating levels
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Soundness Amplification
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Soundness Amplification
Recall error reduction in BPP algorithms
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Soundness Amplification
Recall error reduction in BPP algorithms By repeating and taking majority
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Soundness Amplification
Recall error reduction in BPP algorithms By repeating and taking majority Exponential error reduction (by Chernoff bound)
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Soundness Amplification
Recall error reduction in BPP algorithms By repeating and taking majority Exponential error reduction (by Chernoff bound) Extends to MA
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Soundness Amplification
Recall error reduction in BPP algorithms By repeating and taking majority Exponential error reduction (by Chernoff bound) Extends to MA Given input and any answer from Merlin, to determine Pr[Yes]
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Soundness Amplification
Recall error reduction in BPP algorithms By repeating and taking majority Exponential error reduction (by Chernoff bound) Extends to MA Given input and any answer from Merlin, to determine Pr[Yes] Run many independent verifications (using independent random strings from the beacon). Chernoff bound holds.
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Soundness Amplification
Recall error reduction in BPP algorithms By repeating and taking majority Exponential error reduction (by Chernoff bound) Extends to MA Given input and any answer from Merlin, to determine Pr[Yes] Run many independent verifications (using independent random strings from the beacon). Chernoff bound holds. Increased the length of the second message
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Parallel Repetition for AM[k]
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Parallel Repetition for AM[k]
Soundness amplification by sequential repetition/majority
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Parallel Repetition for AM[k]
Soundness amplification by sequential repetition/majority Exponential amplification, just like in MA. But be careful! Not independent executions (Merlin can adapt strategy over the repetitions.) But not a problem!
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Parallel Repetition for AM[k]
Soundness amplification by sequential repetition/majority Exponential amplification, just like in MA. But be careful! Not independent executions (Merlin can adapt strategy over the repetitions.) But not a problem! But increases rounds
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Parallel Repetition for AM[k]
Soundness amplification by sequential repetition/majority Exponential amplification, just like in MA. But be careful! Not independent executions (Merlin can adapt strategy over the repetitions.) But not a problem! But increases rounds Soundness amplification without increasing rounds
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Parallel Repetition for AM[k]
Soundness amplification by sequential repetition/majority Exponential amplification, just like in MA. But be careful! Not independent executions (Merlin can adapt strategy over the repetitions.) But not a problem! But increases rounds Soundness amplification without increasing rounds Parallel repetition
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Parallel Repetition for AM[k]
Soundness amplification by sequential repetition/majority Exponential amplification, just like in MA. But be careful! Not independent executions (Merlin can adapt strategy over the repetitions.) But not a problem! But increases rounds Soundness amplification without increasing rounds Parallel repetition More careful! Merlin’ s answers (and probability of proof being rejected) in the parallel sessions could be correlated
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Parallel Repetition for AM[k]
Soundness amplification by sequential repetition/majority Exponential amplification, just like in MA. But be careful! Not independent executions (Merlin can adapt strategy over the repetitions.) But not a problem! But increases rounds Soundness amplification without increasing rounds Parallel repetition More careful! Merlin’ s answers (and probability of proof being rejected) in the parallel sessions could be correlated Still turns out to give exponential amplification
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MA ⊆ AM
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MA ⊆ AM
Publishing random test before receiving proof
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MA ⊆ AM
Publishing random test before receiving proof Completeness is no worse
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MA ⊆ AM
Publishing random test before receiving proof Completeness is no worse If MA soundness error is sufficiently small, can use union bound over all Merlin messages to get that the AM soundness error is still small
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MA ⊆ AM
Publishing random test before receiving proof Completeness is no worse If MA soundness error is sufficiently small, can use union bound over all Merlin messages to get that the AM soundness error is still small If MA soundness error < 1/2m+2, where m is the length
- f Merlin’
s message, AM soundness error < 1/ 4
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MA ⊆ AM
Publishing random test before receiving proof Completeness is no worse If MA soundness error is sufficiently small, can use union bound over all Merlin messages to get that the AM soundness error is still small If MA soundness error < 1/2m+2, where m is the length
- f Merlin’
s message, AM soundness error < 1/ 4 Note: Argument similar to why BPP ⊆ P/poly
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MA ⊆ AM
Publishing random test before receiving proof Completeness is no worse If MA soundness error is sufficiently small, can use union bound over all Merlin messages to get that the AM soundness error is still small If MA soundness error < 1/2m+2, where m is the length
- f Merlin’
s message, AM soundness error < 1/ 4 Note: Argument similar to why BPP ⊆ P/poly Extends to MAM ⊆ AM
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MA ⊆ AM
Publishing random test before receiving proof Completeness is no worse If MA soundness error is sufficiently small, can use union bound over all Merlin messages to get that the AM soundness error is still small If MA soundness error < 1/2m+2, where m is the length
- f Merlin’
s message, AM soundness error < 1/ 4 Note: Argument similar to why BPP ⊆ P/poly Extends to MAM ⊆ AM So MAM = AM
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Collapse of the AM hierarchy
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Collapse of the AM hierarchy
Intuition: Can change any MA sequence to an AM sequence
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Collapse of the AM hierarchy
Intuition: Can change any MA sequence to an AM sequence Need a notion of soundness error in each round
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Alternating Threshold TM
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Alternating Threshold TM
A generalization of ATM, with two thresholds instead of ∃ and ∀
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Alternating Threshold TM
A generalization of ATM, with two thresholds instead of ∃ and ∀
∃r’ ∃r ∃r
R
∃r’ ∃r ∃r
R R
∃r’
... ... ... ... ... ...
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Alternating Threshold TM
A generalization of ATM, with two thresholds instead of ∃ and ∀ ∃r: ! (or >) r fraction of children are 1?
∃r’ ∃r ∃r
R
∃r’ ∃r ∃r
R R
∃r’
... ... ... ... ... ...
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Alternating Threshold TM
A generalization of ATM, with two thresholds instead of ∃ and ∀ ∃r: ! (or >) r fraction of children are 1? ∃0 is ∃, and ∃1 is ∀
∃r’ ∃r ∃r
R
∃r’ ∃r ∃r
R R
∃r’
... ... ... ... ... ...
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Alternating Threshold TM
A generalization of ATM, with two thresholds instead of ∃ and ∀ ∃r: ! (or >) r fraction of children are 1? ∃0 is ∃, and ∃1 is ∀ Leaves R(x;path) = 0 or 1
∃r’ ∃r ∃r
R
∃r’ ∃r ∃r
R R
∃r’
... ... ... ... ... ...
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Alternating Threshold TM
A generalization of ATM, with two thresholds instead of ∃ and ∀ ∃r: ! (or >) r fraction of children are 1? ∃0 is ∃, and ∃1 is ∀ Leaves R(x;path) = 0 or 1 Parameters: depth (number of alternations) and size = log(#leaves) (= total length of the “messages”)
∃r’ ∃r ∃r
R
∃r’ ∃r ∃r
R R
∃r’
... ... ... ... ... ...
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Alternating Threshold TM
A generalization of ATM, with two thresholds instead of ∃ and ∀ ∃r: ! (or >) r fraction of children are 1? ∃0 is ∃, and ∃1 is ∀ Leaves R(x;path) = 0 or 1 Parameters: depth (number of alternations) and size = log(#leaves) (= total length of the “messages”) Will denote as ATTM[k,(r,r’),R] (size and individual degrees implicit)
∃r’ ∃r ∃r
R
∃r’ ∃r ∃r
R R
∃r’
... ... ... ... ... ...
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Alternating Threshold TM
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Alternating Threshold TM
We will be interested in ATTM[k,(r,r’),R] where
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Alternating Threshold TM
We will be interested in ATTM[k,(r,r’),R] where One of r, r’ is a fraction > 1/2 (called the threshold), and the
- ther is 0 or 1
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Alternating Threshold TM
We will be interested in ATTM[k,(r,r’),R] where One of r, r’ is a fraction > 1/2 (called the threshold), and the
- ther is 0 or 1
k is constant, size is polynomial and R is a polynomial time relation
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Alternating Threshold TM
We will be interested in ATTM[k,(r,r’),R] where One of r, r’ is a fraction > 1/2 (called the threshold), and the
- ther is 0 or 1
k is constant, size is polynomial and R is a polynomial time relation ATTM threshold can also be amplified using “parallel repetition”!
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Alternating Threshold TM
We will be interested in ATTM[k,(r,r’),R] where One of r, r’ is a fraction > 1/2 (called the threshold), and the
- ther is 0 or 1
k is constant, size is polynomial and R is a polynomial time relation ATTM threshold can also be amplified using “parallel repetition”! Takes threshold from (1/2 + c) to (1 - 1/2n)
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Alternating Threshold TM
We will be interested in ATTM[k,(r,r’),R] where One of r, r’ is a fraction > 1/2 (called the threshold), and the
- ther is 0 or 1
k is constant, size is polynomial and R is a polynomial time relation ATTM threshold can also be amplified using “parallel repetition”! Takes threshold from (1/2 + c) to (1 - 1/2n) k unchanged, size increases by a polynomial factor
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A Pair of Complementary ATTMs
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A Pair of Complementary ATTMs
Consider M+ and M- of the form ATTM[k,(r,0),R] and ATTM[k,(r,1),Rc] (where r>1/2)
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A Pair of Complementary ATTMs
Consider M+ and M- of the form ATTM[k,(r,0),R] and ATTM[k,(r,1),Rc] (where r>1/2) We’ll call it a pair of complementary (k,r) ATTMs
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A Pair of Complementary ATTMs
Consider M+ and M- of the form ATTM[k,(r,0),R] and ATTM[k,(r,1),Rc] (where r>1/2) We’ll call it a pair of complementary (k,r) ATTMs For any r>1/2, {x| M+(x)=1} and {x| M-(x)=1} are disjoint
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A Pair of Complementary ATTMs
Consider M+ and M- of the form ATTM[k,(r,0),R] and ATTM[k,(r,1),Rc] (where r>1/2) We’ll call it a pair of complementary (k,r) ATTMs For any r>1/2, {x| M+(x)=1} and {x| M-(x)=1} are disjoint M = ATTM[k,(1-r,1),Rc] is the complement of M+: {x| M+(x)=0} = {x| M(x)=1}
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A Pair of Complementary ATTMs
Consider M+ and M- of the form ATTM[k,(r,0),R] and ATTM[k,(r,1),Rc] (where r>1/2) We’ll call it a pair of complementary (k,r) ATTMs For any r>1/2, {x| M+(x)=1} and {x| M-(x)=1} are disjoint M = ATTM[k,(1-r,1),Rc] is the complement of M+: {x| M+(x)=0} = {x| M(x)=1} If r > 1-r, M- stricter than M: {x| M-(x)=1} ⊆ {x| M(x)=1}
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A Pair of Complementary ATTMs
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A Pair of Complementary ATTMs
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A Pair of Complementary ATTMs
L is said to have a pair of complementary ATTMs (M+,M-) if
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A Pair of Complementary ATTMs
L is said to have a pair of complementary ATTMs (M+,M-) if x∈L ⇔ M+(x)=1 and M-(x)=0
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A Pair of Complementary ATTMs
L is said to have a pair of complementary ATTMs (M+,M-) if x∈L ⇔ M+(x)=1 and M-(x)=0 x∉L ⇔ M-(x)=1 and M+(x)=0
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A Pair of Complementary ATTMs
L is said to have a pair of complementary ATTMs (M+,M-) if x∈L ⇔ M+(x)=1 and M-(x)=0 x∉L ⇔ M-(x)=1 and M+(x)=0 Exact threshold not critical
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A Pair of Complementary ATTMs
L is said to have a pair of complementary ATTMs (M+,M-) if x∈L ⇔ M+(x)=1 and M-(x)=0 x∉L ⇔ M-(x)=1 and M+(x)=0 Exact threshold not critical Threshold of (M+,M-) can be reduced to any r > 1/2
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A Pair of Complementary ATTMs
L is said to have a pair of complementary ATTMs (M+,M-) if x∈L ⇔ M+(x)=1 and M-(x)=0 x∉L ⇔ M-(x)=1 and M+(x)=0 Exact threshold not critical Threshold of (M+,M-) can be reduced to any r > 1/2 Reducing threshold enlarges {x| M+(x)=1} and {x| M-(x)=1}
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A Pair of Complementary ATTMs
L is said to have a pair of complementary ATTMs (M+,M-) if x∈L ⇔ M+(x)=1 and M-(x)=0 x∉L ⇔ M-(x)=1 and M+(x)=0 Exact threshold not critical Threshold of (M+,M-) can be reduced to any r > 1/2 Reducing threshold enlarges {x| M+(x)=1} and {x| M-(x)=1} And they stay disjoint
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A Pair of Complementary ATTMs
L is said to have a pair of complementary ATTMs (M+,M-) if x∈L ⇔ M+(x)=1 and M-(x)=0 x∉L ⇔ M-(x)=1 and M+(x)=0 Exact threshold not critical Threshold of (M+,M-) can be reduced to any r > 1/2 Reducing threshold enlarges {x| M+(x)=1} and {x| M-(x)=1} And they stay disjoint So they do not change (as they were already a partitioning)
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A Pair of Complementary ATTMs
L is said to have a pair of complementary ATTMs (M+,M-) if x∈L ⇔ M+(x)=1 and M-(x)=0 x∉L ⇔ M-(x)=1 and M+(x)=0 Exact threshold not critical Threshold of (M+,M-) can be reduced to any r > 1/2 Reducing threshold enlarges {x| M+(x)=1} and {x| M-(x)=1} And they stay disjoint So they do not change (as they were already a partitioning) By parallel repetition, can increase threshold to exponentially close to 1, starting from 1/2 + c
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AM and ATTM-pairs
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AM and ATTM-pairs
A language L has an AM[k,r] protocol iff L has a pair of complementary (k,r) ATTMs for r>1/2+c
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AM and ATTM-pairs
A language L has an AM[k,r] protocol iff L has a pair of complementary (k,r) ATTMs for r>1/2+c Guarantees on probability of acceptance translated to threshold guarantees, and vice versa
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AM and ATTM-pairs
A language L has an AM[k,r] protocol iff L has a pair of complementary (k,r) ATTMs for r>1/2+c Guarantees on probability of acceptance translated to threshold guarantees, and vice versa AM[k,r] protocol → (k,r’) ATTM pair: natural conversion works if r > 1-2-2k and r’ = 3/ 4 [Exercise]
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AM and ATTM-pairs
A language L has an AM[k,r] protocol iff L has a pair of complementary (k,r) ATTMs for r>1/2+c Guarantees on probability of acceptance translated to threshold guarantees, and vice versa AM[k,r] protocol → (k,r’) ATTM pair: natural conversion works if r > 1-2-2k and r’ = 3/ 4 [Exercise] (k,r’) ATTM pair → AM[k,r] protocol: natural conversion works if r’ > 1-1/ 4k and r = 3/ 4 [Exercise]
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SLIDE 96
AM and ATTM-pairs
A language L has an AM[k,r] protocol iff L has a pair of complementary (k,r) ATTMs for r>1/2+c Guarantees on probability of acceptance translated to threshold guarantees, and vice versa AM[k,r] protocol → (k,r’) ATTM pair: natural conversion works if r > 1-2-2k and r’ = 3/ 4 [Exercise] (k,r’) ATTM pair → AM[k,r] protocol: natural conversion works if r’ > 1-1/ 4k and r = 3/ 4 [Exercise] Enough, because we can reduce error (increase thresholds) for both AM protocols and ATTMs
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AM[k] = AM
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AM[k] = AM
In terms of ATTM-pairs
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SLIDE 99
AM[k] = AM
In terms of ATTM-pairs Flipping MA to AM: reduces depth, does not change size, but requires threshold to be reduced from 1 - 1/2m+2 to 3/ 4
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SLIDE 100
AM[k] = AM
In terms of ATTM-pairs Flipping MA to AM: reduces depth, does not change size, but requires threshold to be reduced from 1 - 1/2m+2 to 3/ 4 Amplifying again: Threshold increased to 1 - 1/2m+2, but size increased by a polynomial factor
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AM[k] = AM
In terms of ATTM-pairs Flipping MA to AM: reduces depth, does not change size, but requires threshold to be reduced from 1 - 1/2m+2 to 3/ 4 Amplifying again: Threshold increased to 1 - 1/2m+2, but size increased by a polynomial factor Repeat ~k/2 times to reduce to AM[2]
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One-Sided Error
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Recall BPP ⊆ Σ2P
One-Sided Error
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Recall BPP ⊆ Σ2P
One-Sided Error
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 }
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SLIDE 105
Recall BPP ⊆ Σ2P Using “shifts” of random tapes
One-Sided Error
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 }
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SLIDE 106
Recall BPP ⊆ Σ2P Using “shifts” of random tapes x∈L ⇒ ∃P P(Yesx) = {0,1}m
One-Sided Error
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 }
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SLIDE 107
Recall BPP ⊆ Σ2P Using “shifts” of random tapes x∈L ⇒ ∃P P(Yesx) = {0,1}m x∉L ⇒ ∀P |P(Yesx)| < 2m/ 4
One-Sided Error
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 }
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SLIDE 108
Recall BPP ⊆ Σ2P Using “shifts” of random tapes x∈L ⇒ ∃P P(Yesx) = {0,1}m x∉L ⇒ ∀P |P(Yesx)| < 2m/ 4 As an MAM protocol
One-Sided Error
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 }
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SLIDE 109
Recall BPP ⊆ Σ2P Using “shifts” of random tapes x∈L ⇒ ∃P P(Yesx) = {0,1}m x∉L ⇒ ∀P |P(Yesx)| < 2m/ 4 As an MAM protocol Merlin sends P
One-Sided Error
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 }
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SLIDE 110
Recall BPP ⊆ Σ2P Using “shifts” of random tapes x∈L ⇒ ∃P P(Yesx) = {0,1}m x∉L ⇒ ∀P |P(Yesx)| < 2m/ 4 As an MAM protocol Merlin sends P Arthur picks r←{0,1}m
One-Sided Error
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 }
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SLIDE 111
Recall BPP ⊆ Σ2P Using “shifts” of random tapes x∈L ⇒ ∃P P(Yesx) = {0,1}m x∉L ⇒ ∀P |P(Yesx)| < 2m/ 4 As an MAM protocol Merlin sends P Arthur picks r←{0,1}m Merlin sends s ∈ Yesx s.t. r ∈ P(s)
One-Sided Error
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 }
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SLIDE 112
Recall BPP ⊆ Σ2P Using “shifts” of random tapes x∈L ⇒ ∃P P(Yesx) = {0,1}m x∉L ⇒ ∀P |P(Yesx)| < 2m/ 4 As an MAM protocol Merlin sends P Arthur picks r←{0,1}m Merlin sends s ∈ Yesx s.t. r ∈ P(s) One-sided error
One-Sided Error
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 }
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SLIDE 113
Perfect Completeness
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SLIDE 114
Converting MA protocol to perfectly complete MA
Perfect Completeness
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SLIDE 115
Converting MA protocol to perfectly complete MA Consider Yesx,a where a is the message from Merlin
Perfect Completeness
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SLIDE 116
Converting MA protocol to perfectly complete MA Consider Yesx,a where a is the message from Merlin x∈L ⇒ ∃a,P P(Yesx,a) = {0,1}m
Perfect Completeness
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SLIDE 117
Converting MA protocol to perfectly complete MA Consider Yesx,a where a is the message from Merlin x∈L ⇒ ∃a,P P(Yesx,a) = {0,1}m x∉L ⇒ ∀a,P |P(Yesx,a)| < 2m/ 4
Perfect Completeness
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SLIDE 118
Converting MA protocol to perfectly complete MA Consider Yesx,a where a is the message from Merlin x∈L ⇒ ∃a,P P(Yesx,a) = {0,1}m x∉L ⇒ ∀a,P |P(Yesx,a)| < 2m/ 4 Perfectly complete MA protocol
Perfect Completeness
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SLIDE 119
Converting MA protocol to perfectly complete MA Consider Yesx,a where a is the message from Merlin x∈L ⇒ ∃a,P P(Yesx,a) = {0,1}m x∉L ⇒ ∀a,P |P(Yesx,a)| < 2m/ 4 Perfectly complete MA protocol Merlin sends a, P
Perfect Completeness
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SLIDE 120
Converting MA protocol to perfectly complete MA Consider Yesx,a where a is the message from Merlin x∈L ⇒ ∃a,P P(Yesx,a) = {0,1}m x∉L ⇒ ∀a,P |P(Yesx,a)| < 2m/ 4 Perfectly complete MA protocol Merlin sends a, P Arthur picks r←{0,1}m
Perfect Completeness
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SLIDE 121
Converting MA protocol to perfectly complete MA Consider Yesx,a where a is the message from Merlin x∈L ⇒ ∃a,P P(Yesx,a) = {0,1}m x∉L ⇒ ∀a,P |P(Yesx,a)| < 2m/ 4 Perfectly complete MA protocol Merlin sends a, P Arthur picks r←{0,1}m Checks if there exists s ∈ P-1(r) s.t. s ∈ Yesx,a
Perfect Completeness
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SLIDE 122
Converting MA protocol to perfectly complete MA Consider Yesx,a where a is the message from Merlin x∈L ⇒ ∃a,P P(Yesx,a) = {0,1}m x∉L ⇒ ∀a,P |P(Yesx,a)| < 2m/ 4 Perfectly complete MA protocol Merlin sends a, P Arthur picks r←{0,1}m Checks if there exists s ∈ P-1(r) s.t. s ∈ Yesx,a Converting AM protocols
Perfect Completeness
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SLIDE 123
Converting MA protocol to perfectly complete MA Consider Yesx,a where a is the message from Merlin x∈L ⇒ ∃a,P P(Yesx,a) = {0,1}m x∉L ⇒ ∀a,P |P(Yesx,a)| < 2m/ 4 Perfectly complete MA protocol Merlin sends a, P Arthur picks r←{0,1}m Checks if there exists s ∈ P-1(r) s.t. s ∈ Yesx,a Converting AM protocols Yesx = {r| ∃a s.t. Arthur accepts x on transcript (r,a) }
Perfect Completeness
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SLIDE 124
Converting MA protocol to perfectly complete MA Consider Yesx,a where a is the message from Merlin x∈L ⇒ ∃a,P P(Yesx,a) = {0,1}m x∉L ⇒ ∀a,P |P(Yesx,a)| < 2m/ 4 Perfectly complete MA protocol Merlin sends a, P Arthur picks r←{0,1}m Checks if there exists s ∈ P-1(r) s.t. s ∈ Yesx,a Converting AM protocols Yesx = {r| ∃a s.t. Arthur accepts x on transcript (r,a) } A one-sided error MAM protocol: (P, r, a)
Perfect Completeness
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SLIDE 125
Converting MA protocol to perfectly complete MA Consider Yesx,a where a is the message from Merlin x∈L ⇒ ∃a,P P(Yesx,a) = {0,1}m x∉L ⇒ ∀a,P |P(Yesx,a)| < 2m/ 4 Perfectly complete MA protocol Merlin sends a, P Arthur picks r←{0,1}m Checks if there exists s ∈ P-1(r) s.t. s ∈ Yesx,a Converting AM protocols Yesx = {r| ∃a s.t. Arthur accepts x on transcript (r,a) } A one-sided error MAM protocol: (P, r, a) But MAM = AM (and preserves completeness)
Perfect Completeness
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SLIDE 126
Perfect Completeness
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SLIDE 127
Perfect Completeness
Therefore requiring perfect completeness does not change the classes MA or AM
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SLIDE 128
Perfect Completeness
Therefore requiring perfect completeness does not change the classes MA or AM Contrast with RP vs. BPP
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SLIDE 129
Today
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SLIDE 130
Today
MA ⊆ AM. MAM = AM.
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SLIDE 131
Today
MA ⊆ AM. MAM = AM. AM[k] = AM for k ! 2
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SLIDE 132
Today
MA ⊆ AM. MAM = AM. AM[k] = AM for k ! 2 Using alternate characterization in terms of pairs of complementary ATTMs
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SLIDE 133
Today
MA ⊆ AM. MAM = AM. AM[k] = AM for k ! 2 Using alternate characterization in terms of pairs of complementary ATTMs
- ne-sided-error-AM = AM
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SLIDE 134
Today
MA ⊆ AM. MAM = AM. AM[k] = AM for k ! 2 Using alternate characterization in terms of pairs of complementary ATTMs
- ne-sided-error-AM = AM
Coming up:
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SLIDE 135
Today
MA ⊆ AM. MAM = AM. AM[k] = AM for k ! 2 Using alternate characterization in terms of pairs of complementary ATTMs
- ne-sided-error-AM = AM
Coming up: A little more of AM (and where it fits into the zoo)
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SLIDE 136
Today
MA ⊆ AM. MAM = AM. AM[k] = AM for k ! 2 Using alternate characterization in terms of pairs of complementary ATTMs
- ne-sided-error-AM = AM
Coming up: A little more of AM (and where it fits into the zoo) Some other concepts in interactive proofs
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