PCP and Hardness of Approximation
Aman Bansal Adwait Godbole
October 2019
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 1 / 37
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
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
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
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 5 / 37
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
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 7 / 37
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).
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 8 / 37
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 9 / 37
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 10 / 37
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
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 11 / 37
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 12 / 37
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 13 / 37
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
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 14 / 37
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]
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 15 / 37
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
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
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 17 / 37
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
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
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
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 19 / 37
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 19 / 37
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 19 / 37
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
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
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
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
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 21 / 37
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 21 / 37
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 21 / 37
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 22 / 37
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 22 / 37
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 22 / 37
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
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
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
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
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
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 25 / 37
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
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 25 / 37
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
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 25 / 37
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
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 25 / 37
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
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 26 / 37
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
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 26 / 37
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
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 26 / 37
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 27 / 37
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 27 / 37
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 27 / 37
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 28 / 37
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 28 / 37
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 28 / 37
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 29 / 37
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 29 / 37
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 29 / 37
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 29 / 37
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).
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 30 / 37
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 31 / 37
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
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 32 / 37
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 33 / 37
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
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 34 / 37
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.
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 35 / 37
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/
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 36 / 37
Shukria
Aman Bansal Adwait Godbole PCP and Hardness of Approximation October 2019 37 / 37
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