Lecture 19: PCP contd.: Equivalence of the
proof and the gap problem formulations
Amleshwar Kumar1 and Chandan Saha2
1
2
BS, Undergraduate Department, Indian Institute of Science
[email protected]
Instructor, Computer Science and Automation, Indian Institute of Science
Abstract. In this lecture, we’ll state the two ways in which we can
view the P CP theorem. The first version of the P CP theorem is called
the proof version, which shows that the P CP theorem constructs locally testable proof systems, i.e. we can transform every mathematical
proof into a form which allows us to verify the proof by looking at very
few positions, chosen randomly. The second version of the P CP theorem is called the hardness of approximation version which shows that
approximation of optimization version of many N P -complete problems
is also hard. We’ll then introduce the notion of constraint satisfaction
problems(CSP) and using N P - hardness of a certain gap version of a
CSP problem, we’ll show the equivalence of the two versions of the P CP
theorem.
Keywords: q-ary Constraint Satisfaction Problems
1
Two Versions of the P CP theorem
Theorem 1 (P CP Theorem(Proof version/ Verifier version)). N P =
P CP (log n, 1)
Theorem 2 (P CP Theorem(Hardness of approximation version)). ∃ρ <
1 such that for every language L in N P , ∃ a polynomial time reduction from L
to 3SAT(x 7→ ϕx ) such that:
– if x ∈ L then val(ϕx ) = 1
– if x ∈
/ L then val(ϕx ) < ρ
Definition 1 (ρ Gap 3SAT Problem).
Input: ϕ ∈ 3CNF such that either val(ϕx ) = 1 or val(ϕx ) < ρ.
Output: Given a ϕ as above find out which of the 2 cases does it belong to.
Note that Theorem 2 implies ∃ρ < 1 such that ρ Gap 3SAT is Hard.
2
Equivalence of the two versions
Definition 2 (Constraint Satisfaction Problem). A q-ary CSP instance in
n variables consists of m boolean functions { ϕ1 , ..., ϕm }, where ϕi : {0, 1}q 7→
{0, 1}, such that every ϕi is a function of at most q variables xi1 , ..., xiq . Let
ϕ = {ϕ1 , ..., ϕm },
– We say that ϕ is satisfiable if ∃u ∈ {0, 1}n such that every ϕi evaluates to 1
under the assignment u.
– val(ϕ) is the maximum fraction of the ϕi s that can be satisfied by any assignment u ∈ {0, 1}.
Definition 3 (-Gap q-CSP Problem).
Input: A q-CSP instance ϕ with the promise that:
– either val(ϕ) = 1
– or val(ϕ) < Output: Given such a q-CSP instance, find out whether val(ϕ) = 1 or val(ϕ) <
.
Theorem 3. ∃ constants q and < 1 such that -Gap q-CSP problem is NP
Hard.
Claim. Theorem 1 and Theorem 2 are equivalent.
Proof. We’ll prove the above claim by proving the following:
1. Theorem 1 ≡ Theorem 3
2. Theorem 2 ≡ Theorem 3
Proof (1. Theorem 1 ≡ Theorem 3). (⇒) Let L ∈ N P , then by Theorem 1, L
has an (O(log n), O(1)) verifier M π for a proof string π. For any given r, the
query locations of π are fixed and thus M π becomes a fixed boolean function in
the bits of these q locations, i.e. M π x,r = ϕ(yi1 , ..., yiq ) = ϕx,r . If x ∈ L, then ∃π
such that P r(M π (x) = 1) = 1, i.e. ∧r∈{0,1}c log n {M π x,r } = ∧ϕx,r is satisfiable.
Thus for every x ∈ {0, 1}n , the collection ϕx = {ϕx,r (π)} is a polynomial sized
q-CSP instance. Observe that if x ∈ L then the q-CSP instance ϕx in 2c log n q
variables and 2c log n boolean functions is satisfiable and if x ∈
/ L then ∀π at
most half of the 2c log n boolean functions in ϕx is satisfied.
(⇐) Suppose that ρ-Gap q-CSP is N P -hard for some constants q, ρ < 1, then
this easily translates into a P CP system with q queries, ρ soundness and logarithmic randomness for every language L ∈ N P : Given an input x and a
proof π, the verifier M π first computes the reduction f (x) to a q-CSP instance
ϕx = {ϕ1 , ..., ϕm }. It then tosses coins for log m time to choose a ϕi randomly.
Suppose ϕi = ϕi (yi1 , ..., yiq ). It then reads locations i1 , ..., iq from the proof π
and checks the value of ϕi (πi1 , ..., πiq ). If the value of ϕi (πi1 , ..., πiq ) is 1, then
M π ouputs 1, else it outputs 0. Clearly, if x ∈ L then the verifier will accept
with probability 1, while if x ∈
/ L, it will accept with probability ρ which can
then be boosted to 1/2 at the expense of a constant factor in the randomness
and number of queries.
Proof (Theorem 2 ≡ Theorem 3). (⇒)Since 3CNF formulas are a special case
of 3CSP instances, Theorem 2 implies Theorem 3.
(⇐) Let > 0 and q be such that ()-Gap q-CSP is N P -hard. Let L ∈ N P . For
any input x ∈ {0, 1}∗ we perform the reduction x 7→ ϕx = {ϕ1 , ..., ϕn } such that
ϕx is a q-CSP and if x ∈ L then val(ϕx ) = 1 else valϕx < . We can then convert
ϕi , which is a boolean function in q variables, to a q-CNF with a blowup to 2q
clauses and then to a 3-CNF by adding more variables. After these conversions
we get ϕi = ψi1 ∧ ... ∧ ψiki where ki ≈ q2q . Therefore ϕx = ∧m
i=1 (ψi1 ∧ ... ∧ ψiki )
has mq2q clauses. If ϕx is satisfiable then so is ψx . But if (1 − ) fraction of ϕi s
are not satisfied then, (1 − )m/mq2q fraction of ψj ’s are not satisfied.
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References
1. Arora, S., Barak, B.: Computational Complexity: A Modern Approach Cambridge
University Press New York, NY, USA