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Page 1: PCP and Hardness of Approximationamanb/pdfs/complexity.pdfAman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 19 / 37. Walsh-Hadamard Code Main Idea: Bit strings

PCP and Hardness of Approximation

Aman Bansal Adwait Godbole

October 2019

Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 1 / 37

Page 2: PCP and Hardness of Approximationamanb/pdfs/complexity.pdfAman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 19 / 37. Walsh-Hadamard Code Main Idea: Bit strings

Trajectory

1 Relaxations of Hard ProblemsApproximate ProblemsGap ProblemConnection between Approximation and Gap

2 A New Proof SystemProbabilistically Checkable Proof SystemsSome PreliminariesProof of the PCP Theorem

3 Hardness of Approximation

Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 2 / 37

Page 3: PCP and Hardness of Approximationamanb/pdfs/complexity.pdfAman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 19 / 37. Walsh-Hadamard Code Main Idea: Bit strings

Trajectory

1 Relaxations of Hard ProblemsApproximate ProblemsGap ProblemConnection between Approximation and Gap

2 A New Proof SystemProbabilistically Checkable Proof SystemsSome PreliminariesProof of the PCP Theorem

3 Hardness of Approximation

Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 3 / 37

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Approximate Problems

Given an instance x ∈ X of an NP-hard optimization problem withobjective function Obj : Y → R (which is to be maximized1), a solution yis said to be an α-approximate (for α ≤ 1) solution to the instance if

α · Obj(y∗) ≤ Obj(y) ≤ Obj(y∗)

where y∗ is the true (not approximated) solution to the problem instance.

1If the problem is a minimization problem then we have a slightly different definitionAman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 4 / 37

Page 5: PCP and Hardness of Approximationamanb/pdfs/complexity.pdfAman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 19 / 37. Walsh-Hadamard Code Main Idea: Bit strings

Approximate Problems

α-approximate algorithms

An algorithm A is an α-approximate algorithm if for all x in the instancespace X , it returns an α-approximate solution y.

α-approximate problems

Problems that render (poly-time) α-approximate algorithms are calledα-approximate problems.

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Trajectory

1 Relaxations of Hard ProblemsApproximate ProblemsGap ProblemConnection between Approximation and Gap

2 A New Proof SystemProbabilistically Checkable Proof SystemsSome PreliminariesProof of the PCP Theorem

3 Hardness of Approximation

Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 6 / 37

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Promise Problem

A promise problem Π is specified by a pair of sets (YES, NO) such thatYES, NO ⊆ X and YES ∩ NO = Φ.

Any algorithm A solving Π, on input x, should output ‘yes’ if x ∈ YES,‘no’ if x ∈ NO and any output if x is a don’t care instance

Recollect the PromiseΠ problem from the midsem.

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Gap Problems

A gap problem is a promise problem parametrized by α (< 1). Let P be anNP-hard optimization problem with objective function Obj : Y → R(which is to be maximized), the corresponding gap problem gapα-P is apromise problem with (YES, NO) sets as given below:

YES = {〈x, k〉 | ∃y ∈ Y such that Obj(y) ≥ k}NO = {〈x, k〉 | ∀y ∈ Y,Obj(y) < αk}

Intuitively, the ‘gap’ refers to the interval (αk , k).

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Trajectory

1 Relaxations of Hard ProblemsApproximate ProblemsGap ProblemConnection between Approximation and Gap

2 A New Proof SystemProbabilistically Checkable Proof SystemsSome PreliminariesProof of the PCP Theorem

3 Hardness of Approximation

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Intuition

Intuitively, it seems that approximate problems and gap problems aresimilar.

α-approximation is a relaxed variant of a search problem

gapα is a relaxed variant of a decision problem

Indeed, this notion can be formalized.

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α-approximate → gapα

Connnection between α-approximation and gapαFor any problem P and 0 < α < 1, α-approximating P is at least as hardas solving gapα-P.

Proof:Let A be an α-approximate algorithm for P. The following algorithm solvesgapα-P:

On input (φ, k):

1. Let k ′ = A(φ)

2. Accept iff k ′ ≥ αk

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Trajectory

1 Relaxations of Hard ProblemsApproximate ProblemsGap ProblemConnection between Approximation and Gap

2 A New Proof SystemProbabilistically Checkable Proof SystemsSome PreliminariesProof of the PCP Theorem

3 Hardness of Approximation

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Page 13: PCP and Hardness of Approximationamanb/pdfs/complexity.pdfAman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 19 / 37. Walsh-Hadamard Code Main Idea: Bit strings

Trajectory

1 Relaxations of Hard ProblemsApproximate ProblemsGap ProblemConnection between Approximation and Gap

2 A New Proof SystemProbabilistically Checkable Proof SystemsSome PreliminariesProof of the PCP Theorem

3 Hardness of Approximation

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Probabilistically Checkable Proof System

(r , q,m, t)-restricted Verifier

Let r , q,m, t : N→ N. A language L ∈ PCPc,s [r , q,m, t] if L has an(r , q,m, t) restricted verifier V such that

∀x ∈ L,∃π of size at most m(|x |),PrR [V π[x ;R] = acc] ≥ c(|x |)∀x /∈ L, ∀π of size at most m(|x |),PrR [V π[x ;R] = acc] < s(|x |)

Resource bounds

1. r(|x |) is a bound on randomness used by V

2. q(|x |) is a bound on the number of locations queried by V

3. m(|x |) is a bound the length of the proof to which V has oracle access

4. t(|x |) is a bound on the runtime of V

Completeness and Soundness constraintsc(|x |) and s(|x |) are the completeness and soundness specifications

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Remarks

The verifier could be adaptive or non-adaptive

If the verifier is non-adaptive then m(n) ≤ q(n) · 2r(n)

q(n) ≤ t(n)

PCPc,s [r , q] ⊆ NTIME (q(n) · 2r(n))NP = PCP1,0[0, poly(n)]

BPP = PCP 23, 13[poly(n), 0]

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PCP Theorem

PCP Theorem

NP = PCP1, 12[log(n), 1]

We will today present a weaker version of the PCP theorem.

PCP Theorem [Weaker]

NP = PCP1, 12[poly(n), 1]

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PCP Theorem

PCP Theorem

NP = PCP1, 12[log(n), 1]

We will today present a weaker version of the PCP theorem.

PCP Theorem [Weaker]

NP = PCP1, 12[poly(n), 1]

Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 16 / 37

Page 18: PCP and Hardness of Approximationamanb/pdfs/complexity.pdfAman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 19 / 37. Walsh-Hadamard Code Main Idea: Bit strings

Trajectory

1 Relaxations of Hard ProblemsApproximate ProblemsGap ProblemConnection between Approximation and Gap

2 A New Proof SystemProbabilistically Checkable Proof SystemsSome PreliminariesProof of the PCP Theorem

3 Hardness of Approximation

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Subset XOR

Consider the function fu(x) = x� u.Here for x, y ∈ {0, 1}n, x� y =

∑i xiyi (mod 2).

Note that fu is equivalent to choosing a subset (given by u) of ele-ments from [n] and evaluating parity over this subset.

Random Subsum Principle

For x, y ∈ {0, 1}n with y 6= 0n, Prx∈{0,1}n [x� y = 1] = 12

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Subset XOR

Consider the function fu(x) = x� u.Here for x, y ∈ {0, 1}n, x� y =

∑i xiyi (mod 2).

Note that fu is equivalent to choosing a subset (given by u) of ele-ments from [n] and evaluating parity over this subset.

Random Subsum Principle

For x, y ∈ {0, 1}n with y 6= 0n, Prx∈{0,1}n [x� y = 1] = 12

Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 18 / 37

Page 21: PCP and Hardness of Approximationamanb/pdfs/complexity.pdfAman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 19 / 37. Walsh-Hadamard Code Main Idea: Bit strings

Subset XOR

Consider the function fu(x) = x� u.Here for x, y ∈ {0, 1}n, x� y =

∑i xiyi (mod 2).

Note that fu is equivalent to choosing a subset (given by u) of ele-ments from [n] and evaluating parity over this subset.

Random Subsum Principle

For x, y ∈ {0, 1}n with y 6= 0n, Prx∈{0,1}n [x� y = 1] = 12

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Walsh-Hadamard Code

Main Idea: Bit strings u ∈ {0, 1}n are encoded as the truth table ofa linear function over F2

WH encoding

For an n-bit string u ∈ {0, 1}n, WH(u) is the 2n bit string representing thetruth table of the function f (x) = x� u for x ∈ {0, 1}n.

Walsh-Hadamard codeword

f ∈ {0, 1}2n such that f = WH(u) for some u ∈ {0, 1}n.

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Walsh-Hadamard Code

Main Idea: Bit strings u ∈ {0, 1}n are encoded as the truth table ofa linear function over F2

WH encoding

For an n-bit string u ∈ {0, 1}n, WH(u) is the 2n bit string representing thetruth table of the function f (x) = x� u for x ∈ {0, 1}n.

Walsh-Hadamard codeword

f ∈ {0, 1}2n such that f = WH(u) for some u ∈ {0, 1}n.

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Page 24: PCP and Hardness of Approximationamanb/pdfs/complexity.pdfAman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 19 / 37. Walsh-Hadamard Code Main Idea: Bit strings

Walsh-Hadamard Code

Main Idea: Bit strings u ∈ {0, 1}n are encoded as the truth table ofa linear function over F2

WH encoding

For an n-bit string u ∈ {0, 1}n, WH(u) is the 2n bit string representing thetruth table of the function f (x) = x� u for x ∈ {0, 1}n.

Walsh-Hadamard codeword

f ∈ {0, 1}2n such that f = WH(u) for some u ∈ {0, 1}n.

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Properties of WH

EC Error correcting with minimum distance 12 .

This means that for x, y ∈R {0, 1}n with x 6= y, WH(x) and WH(y)differ in 1/2 the bits.

LIN Linearity of WH(u)f = WH(u) when viewed as a function from {0, 1}n to {0, 1} is infact linear (over F2)

LT Locally TestableGiven access to a function f : {0, 1}n → {0, 1}, we can check whetherit is a Walsh-Hadamard code-word by querying a constant number ofplaces.

LD Locally DecodableGiven f and an x ∈ {0, 1}n, we can find f̃ (x) in constant queries to f ,where f̃ is the true codeword.

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Properties of WH

EC Error correcting with minimum distance 12 .

This means that for x, y ∈R {0, 1}n with x 6= y, WH(x) and WH(y)differ in 1/2 the bits.

LIN Linearity of WH(u)f = WH(u) when viewed as a function from {0, 1}n to {0, 1} is infact linear (over F2)

LT Locally TestableGiven access to a function f : {0, 1}n → {0, 1}, we can check whetherit is a Walsh-Hadamard code-word by querying a constant number ofplaces.

LD Locally DecodableGiven f and an x ∈ {0, 1}n, we can find f̃ (x) in constant queries to f ,where f̃ is the true codeword.

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Properties of WH

EC Error correcting with minimum distance 12 .

This means that for x, y ∈R {0, 1}n with x 6= y, WH(x) and WH(y)differ in 1/2 the bits.

LIN Linearity of WH(u)f = WH(u) when viewed as a function from {0, 1}n to {0, 1} is infact linear (over F2)

LT Locally TestableGiven access to a function f : {0, 1}n → {0, 1}, we can check whetherit is a Walsh-Hadamard code-word by querying a constant number ofplaces.

LD Locally DecodableGiven f and an x ∈ {0, 1}n, we can find f̃ (x) in constant queries to f ,where f̃ is the true codeword.

Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 20 / 37

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Properties of WH

EC Error correcting with minimum distance 12 .

This means that for x, y ∈R {0, 1}n with x 6= y, WH(x) and WH(y)differ in 1/2 the bits.

LIN Linearity of WH(u)f = WH(u) when viewed as a function from {0, 1}n to {0, 1} is infact linear (over F2)

LT Locally TestableGiven access to a function f : {0, 1}n → {0, 1}, we can check whetherit is a Walsh-Hadamard code-word by querying a constant number ofplaces.

LD Locally DecodableGiven f and an x ∈ {0, 1}n, we can find f̃ (x) in constant queries to f ,where f̃ is the true codeword.

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LT: Local Testability

LIN: WH(u) for u ∈ {0, 1}n captures all n-bit linear functions on F2.

ρ-closeness of functions

Functions f , g are ρ-close if Prx∈R{0,1}n [f (x) = g(x)] ≥ ρ.

A function f is ρ-close to a linear function if there exists a linear functiong such that f and g are ρ-close

Let f be such that

Prx,y∈R{0,1}n [f (x + y) = f (x) + f (y)] ≥ ρ

for some ρ > 12 . Then f is ρ-close to a linear function.

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LT: Local Testability

LIN: WH(u) for u ∈ {0, 1}n captures all n-bit linear functions on F2.

ρ-closeness of functions

Functions f , g are ρ-close if Prx∈R{0,1}n [f (x) = g(x)] ≥ ρ.

A function f is ρ-close to a linear function if there exists a linear functiong such that f and g are ρ-close

Let f be such that

Prx,y∈R{0,1}n [f (x + y) = f (x) + f (y)] ≥ ρ

for some ρ > 12 . Then f is ρ-close to a linear function.

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LT: Local Testability

LIN: WH(u) for u ∈ {0, 1}n captures all n-bit linear functions on F2.

ρ-closeness of functions

Functions f , g are ρ-close if Prx∈R{0,1}n [f (x) = g(x)] ≥ ρ.

A function f is ρ-close to a linear function if there exists a linear functiong such that f and g are ρ-close

Let f be such that

Prx,y∈R{0,1}n [f (x + y) = f (x) + f (y)] ≥ ρ

for some ρ > 12 . Then f is ρ-close to a linear function.

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LD: Local Decodability

EC: For x 6= y, WH(x) and WH(y) differ in 1/2 the bits

Let f be (1− δ)-close to a linear function f̃ for some δ < 1/4. Then byEC, f uniquely determines f̃ .

So, given a (possibly illegal) f , having a corresponding f̃ , we want to findf̃ (x). Here we have oracle access only to f . The idea is to once again userandomness.

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LD: Local Decodability

EC: For x 6= y, WH(x) and WH(y) differ in 1/2 the bits

Let f be (1− δ)-close to a linear function f̃ for some δ < 1/4. Then byEC, f uniquely determines f̃ .

So, given a (possibly illegal) f , having a corresponding f̃ , we want to findf̃ (x). Here we have oracle access only to f . The idea is to once again userandomness.

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LD: Local Decodability

EC: For x 6= y, WH(x) and WH(y) differ in 1/2 the bits

Let f be (1− δ)-close to a linear function f̃ for some δ < 1/4. Then byEC, f uniquely determines f̃ .

So, given a (possibly illegal) f , having a corresponding f̃ , we want to findf̃ (x). Here we have oracle access only to f . The idea is to once again userandomness.

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LD: Local Decodability

Objective: With oracle access only to f , given an x ∈ {0, 1}n, find f̃ (x).The idea is to use randomness and linearity.

Choose x′ ∈R {0, 1}n

Set x′′ ← x + x′

Let y′ = f (x′) and y′′ = f (x′′)

Output y′ + y′′

With probability atleast 1− 2δ we have y′ = f (x′) and y′′ = f (x′′) andhence f̃ (x) = y′ + y′′.

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LD: Local Decodability

Objective: With oracle access only to f , given an x ∈ {0, 1}n, find f̃ (x).The idea is to use randomness and linearity.

Choose x′ ∈R {0, 1}n

Set x′′ ← x + x′

Let y′ = f (x′) and y′′ = f (x′′)

Output y′ + y′′

With probability atleast 1− 2δ we have y′ = f (x′) and y′′ = f (x′′) andhence f̃ (x) = y′ + y′′.

Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 23 / 37

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LD: Local Decodability

Objective: With oracle access only to f , given an x ∈ {0, 1}n, find f̃ (x).The idea is to use randomness and linearity.

Choose x′ ∈R {0, 1}n

Set x′′ ← x + x′

Let y′ = f (x′) and y′′ = f (x′′)

Output y′ + y′′

With probability atleast 1− 2δ we have y′ = f (x′) and y′′ = f (x′′) andhence f̃ (x) = y′ + y′′.

Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 23 / 37

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Trajectory

1 Relaxations of Hard ProblemsApproximate ProblemsGap ProblemConnection between Approximation and Gap

2 A New Proof SystemProbabilistically Checkable Proof SystemsSome PreliminariesProof of the PCP Theorem

3 Hardness of Approximation

Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 24 / 37

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Proof of the PCP Theorem [Weaker]

PCP Theorem [Weaker]

NP = PCP1, 12[poly(n), 1]

We show that QUADEQ - the satisfiability problem for quadraticequations over F2 - has a PCP[poly(n), 1] proof system

QUADEQ over variables u1, u2, · · · , un is is of the form AU = b, where Ais an m × n2 matrix and b ∈ {0, 1}m. U = u⊗ u is the tensor product (orthe Hadamard product).

Claim

QUADEQ, the language of all satisfiable instances is NP-complete

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Proof of the PCP Theorem [Weaker]

PCP Theorem [Weaker]

NP = PCP1, 12[poly(n), 1]

We show that QUADEQ - the satisfiability problem for quadraticequations over F2 - has a PCP[poly(n), 1] proof system

QUADEQ over variables u1, u2, · · · , un is is of the form AU = b, where Ais an m × n2 matrix and b ∈ {0, 1}m. U = u⊗ u is the tensor product (orthe Hadamard product).

Claim

QUADEQ, the language of all satisfiable instances is NP-complete

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Proof of the PCP Theorem [Weaker]

PCP Theorem [Weaker]

NP = PCP1, 12[poly(n), 1]

We show that QUADEQ - the satisfiability problem for quadraticequations over F2 - has a PCP[poly(n), 1] proof system

QUADEQ over variables u1, u2, · · · , un is is of the form AU = b, where Ais an m × n2 matrix and b ∈ {0, 1}m. U = u⊗ u is the tensor product (orthe Hadamard product).

Claim

QUADEQ, the language of all satisfiable instances is NP-complete

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Proof of the PCP Theorem [Weaker]

PCP Theorem [Weaker]

NP = PCP1, 12[poly(n), 1]

We show that QUADEQ - the satisfiability problem for quadraticequations over F2 - has a PCP[poly(n), 1] proof system

QUADEQ over variables u1, u2, · · · , un is is of the form AU = b, where Ais an m × n2 matrix and b ∈ {0, 1}m. U = u⊗ u is the tensor product (orthe Hadamard product).

Claim

QUADEQ, the language of all satisfiable instances is NP-complete

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π and V

What is the proof π and what does the verifier V do?

π The proof is 〈WH(u),WH(u⊗ u)〉V Denote the proof by f = WH(u) and g = WH(u⊗ u).

The verifier does the following

1) Check linearity of f and g

2) Verify that g encodes u⊗ u

3) Verify that f encodes a satisfying assignment

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π and V

What is the proof π and what does the verifier V do?

π The proof is 〈WH(u),WH(u⊗ u)〉

V Denote the proof by f = WH(u) and g = WH(u⊗ u).The verifier does the following

1) Check linearity of f and g

2) Verify that g encodes u⊗ u

3) Verify that f encodes a satisfying assignment

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π and V

What is the proof π and what does the verifier V do?

π The proof is 〈WH(u),WH(u⊗ u)〉V Denote the proof by f = WH(u) and g = WH(u⊗ u).

The verifier does the following

1) Check linearity of f and g

2) Verify that g encodes u⊗ u

3) Verify that f encodes a satisfying assignment

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Check linearity of f and g

Note that f = WH(u) and g = WH(u⊗ u)

V performs a 0.99-close (high-probability) linearity test on both f and g .This is done by the LT property described earlier.

Note crucially that a high but nevertheless constant closeness suffices.This is since we eventually plan to query π at only a small constantnumber of points.

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Check linearity of f and g

Note that f = WH(u) and g = WH(u⊗ u)

V performs a 0.99-close (high-probability) linearity test on both f and g .This is done by the LT property described earlier.

Note crucially that a high but nevertheless constant closeness suffices.This is since we eventually plan to query π at only a small constantnumber of points.

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Check linearity of f and g

Note that f = WH(u) and g = WH(u⊗ u)

V performs a 0.99-close (high-probability) linearity test on both f and g .This is done by the LT property described earlier.

Note crucially that a high but nevertheless constant closeness suffices.This is since we eventually plan to query π at only a small constantnumber of points.

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Verify that g encodes u⊗ u

V chooses r, r′ independently at random from {0, 1}n and assert thatf (r)f (r′) = g(r⊗ r′).

Let W be an n × n matrix representing the entries of w and U be such amatrix for u⊗ u. Then

1. g(r⊕ r′) = w� (r⊗ r′) = rW r′

2. f (r)f (r′) = (u� r)(u� r′) = rUr′

By the random subsum principle, we claim this test rejects atleast 1/4 ofthe time on instances where w 6= u⊗ u. Repeating this 3 times, we getprobability of rejection as 37/64.

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Verify that g encodes u⊗ u

V chooses r, r′ independently at random from {0, 1}n and assert thatf (r)f (r′) = g(r⊗ r′).

Let W be an n × n matrix representing the entries of w and U be such amatrix for u⊗ u. Then

1. g(r⊕ r′) = w� (r⊗ r′) = rW r′

2. f (r)f (r′) = (u� r)(u� r′) = rUr′

By the random subsum principle, we claim this test rejects atleast 1/4 ofthe time on instances where w 6= u⊗ u. Repeating this 3 times, we getprobability of rejection as 37/64.

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Verify that g encodes u⊗ u

V chooses r, r′ independently at random from {0, 1}n and assert thatf (r)f (r′) = g(r⊗ r′).

Let W be an n × n matrix representing the entries of w and U be such amatrix for u⊗ u. Then

1. g(r⊕ r′) = w� (r⊗ r′) = rW r′

2. f (r)f (r′) = (u� r)(u� r′) = rUr′

By the random subsum principle, we claim this test rejects atleast 1/4 ofthe time on instances where w 6= u⊗ u. Repeating this 3 times, we getprobability of rejection as 37/64.

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Verify that f encodes a satisfying assignment

Now we are assured that the form of π is 〈WH(u),WH(u⊗ u)〉 forsome u ∈ {0, 1}n.

1 All that remains is to check that u is a satisfying assignment

2 Wonderfully, we also have O(1) access to Ai · (u⊗ u), the value ofthe i th equation in the QUADEQ instance and can match it with bi .

But but but ... how do we check all the m equations of the QUADEQinstance in constant number of queries? /

Use the random subsum principle AGAIN! Choose a subset of equationsrandomly from [k] and add them together to create a new quadraticequation. If u did not satisfy even one equation of the original system, itwill not satisfy the new equation with probability at least 1/2.

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Verify that f encodes a satisfying assignment

Now we are assured that the form of π is 〈WH(u),WH(u⊗ u)〉 forsome u ∈ {0, 1}n.

1 All that remains is to check that u is a satisfying assignment

2 Wonderfully, we also have O(1) access to Ai · (u⊗ u), the value ofthe i th equation in the QUADEQ instance and can match it with bi .

But but but ... how do we check all the m equations of the QUADEQinstance in constant number of queries? /

Use the random subsum principle AGAIN! Choose a subset of equationsrandomly from [k] and add them together to create a new quadraticequation. If u did not satisfy even one equation of the original system, itwill not satisfy the new equation with probability at least 1/2.

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Verify that f encodes a satisfying assignment

Now we are assured that the form of π is 〈WH(u),WH(u⊗ u)〉 forsome u ∈ {0, 1}n.

1 All that remains is to check that u is a satisfying assignment

2 Wonderfully, we also have O(1) access to Ai · (u⊗ u), the value ofthe i th equation in the QUADEQ instance and can match it with bi .

But but but ... how do we check all the m equations of the QUADEQinstance in constant number of queries? /

Use the random subsum principle AGAIN! Choose a subset of equationsrandomly from [k] and add them together to create a new quadraticequation. If u did not satisfy even one equation of the original system, itwill not satisfy the new equation with probability at least 1/2.

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Verify that f encodes a satisfying assignment

Now we are assured that the form of π is 〈WH(u),WH(u⊗ u)〉 forsome u ∈ {0, 1}n.

1 All that remains is to check that u is a satisfying assignment

2 Wonderfully, we also have O(1) access to Ai · (u⊗ u), the value ofthe i th equation in the QUADEQ instance and can match it with bi .

But but but ... how do we check all the m equations of the QUADEQinstance in constant number of queries? /

Use the random subsum principle AGAIN! Choose a subset of equationsrandomly from [k] and add them together to create a new quadraticequation. If u did not satisfy even one equation of the original system, itwill not satisfy the new equation with probability at least 1/2.

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QED

With this, we have proved that NP ⊆ PCP[poly(n), 1]. The otherdirection is trivial.

The stronger theorem makes further observations regarding the form ofthe proof π given here. Then it uses further results such as gapamplification and alphabet reduction to prove the general statementNP = PCP(log(n), 1).

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Trajectory

1 Relaxations of Hard ProblemsApproximate ProblemsGap ProblemConnection between Approximation and Gap

2 A New Proof SystemProbabilistically Checkable Proof SystemsSome PreliminariesProof of the PCP Theorem

3 Hardness of Approximation

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Hardness of Approximation

A Sad Result

gapα-MAX3SAT is NP-hard

Just as SAT captured the essence of hardness of exact solution, gapα-MAX3SAT, or it’s more general formulation, qCSP, captures the es-sence of hardness of approximation

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Hardness of Approximation

Proof Method

PCP Theorem =⇒ gapα-MAX3SAT is NP-hard

Proof: Consider any L ∈ PCP1, 12[c · log(n),Q]. The idea is to encode the

Verifier’s possible actions by a Boolean formula Ψ.

Ψ =∧

coins R

hR

But hr is an arbitrary predicate over Q variables.

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Hardness of Approximation

Fact

∀q, ∃l(q), k(q) such that any q-ary Boolean function h can be encoded bya 3-CNF formula ψh with k(q) clauses over q + l(q) variablesx1, . . . , xq, z1, . . . , zl(q) such that

h(x) = 1 =⇒ ∃z , ψh(x , z) = 1

h(x) = 0 =⇒ ∀z , ψh(x , z) = 0

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Hardness of Approximation

Ψ =∧

coins R

ψhR

1. If x ∈ L then ∃ proof π such that ∀R, hR(π) = 1

2. If x /∈ L then at least 12 choices of R accept make hR(π) = 0. If the

total number of clauses are M ( = 2Rk(q)) then maximum fraction ofclauses that can be satisfied is (1− 1

k ).

This proves that PCP-theorem leads to the hardness ofgapα-MAX3SAT which in turn as presented earlier leads toNP-hardness of α-approximating MAX3SAT.

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References

1 Arora, Sanjeev, and Boaz Barak. Computational complexity: amodern approach. Cambridge University Press, 2009.

2 Limits of approximation algorithms : PCPs and Unique Games -Spring Semester (2009-10) at TIFR, IMSchttp://www.tcs.tifr.res.in/ prahladh/teaching/2009-10/limits/

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Shukria

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