CSci 5403 Lecture 15

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CSci 5403
COMPLEXITY THEORY
LECTURE XXVII PCPs AND HARDNESS OF APPROXIMATION
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Definition. Oracle PPTM V is an (r,q)-verifier
for language L if
1. ? O,x. VO(x) uses at most r(x) random bits
and makes at most q(x) oracle queries.
2. x ? L ? ?p ? 0,1 PrrVp(xr)1 1.
(Completeness)
3. x ? L ? ?p ? 0,1, PrrVp(xr)1 ½
(Soundness)
Definition. L ? PCPr(n),q(n) iff there is a
(r,q) verifier for L.
Theorem. NP PCPO(log n), O(1).
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PCP False Proof
Traditional False Proof
Example. GNI ? PCP(O(n lg n), O(1)).
Proof. Vp(G1,G2) compute H ?(Gb), query b
p(H), and accept iff b b.
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Recall
Over the next four lectures we will prove
the following theorem.
Theorem. ?e gt 0 such that there is a polytime
reduction R from SAT to MAX-3SAT such that ? ?
SAT ? OPT(R(?)) m ? ? SAT ? OPT(R(?)) lt
(1-e)m.
Claim. MAX-3SAT theorem PCP theorem.
Proof. (?) reduce to MAX-3SAT, check random
clause. (?) each randomness value leads to a
clause.
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Definition. Let ? and ? be maximization
problems. A (c,?), (c,?) gap preserving
reduction from ? to ? is a polynomial-time
algorithm such that OPT?(x) c ? OPT?((x))
c OPT?(x) lt c/? ? OPT?((x)) lt c/?. A
gap producing reduction is a (1,1), (c,?)
gap preserving reduction for constants c,?.
Proposition. If there is a gap-producing
reduction from SAT to ? and a gap-preserving
reduction from ? to ?, then there is a
gap-producing reduction from SAT to ?.
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Definition. A gap problem is a promise
problem (?Y,?N) derived from optimization problem
(R,val) ?Y x OPT(x) c , ?N x
OPT(x) lt c/? .
Claim. If there is a reduction from 3SAT to
(?Y,?N) Then there is no c/? approximation
algorithm for (R,val) unless PNP.
Definition. MAX-INDSET (R,val) where R
((V,E), S) S ? V and ?u,v? S, (u,v) ? E
val(S) S.
Theorem. ?c gt 0 such that unless PNP there is no
nc-approximaton scheme for MAX-INDSET.
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Lemma. There exists a reduction from 3CNFs
to graphs such that OPTINDSET((?))
OPTMAX3SAT(?).
Sketch. Consider ?(x1?x2)?(?x1???x3)
Put an edge between two nodes iff they
have inconsistent assignments.
x satisfying k clauses ? ind-set S of size k
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Lemma. For every ?gt0, ? efficiently compute
such that (a(G) 2?)lg n a((G)) (a(G)
2?)lg n , where a(x) OPT(x)/n.
Note that with ? ße/8 then a(G) ß ?
a((G)) (ß(1-e/4))lg n, a(G) lt (1-e)ß ?
a((G)) lt (ß(1-3e/4))lg n . Thus the new gap
factor is O(nc).
Proof (of Lemma). We will use the
expander random walk theorem If G is an (n,d,?)
expander And x1,,xk is a k-step random walk, and
B?V(G), (B/n - 2?)k Pr?i. xi?B
(B/n2?)k.
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To make H(G), we use a (n,d,?) expander
K. Nodes(H) (u1,u2,,ul) u1?K ?
?i.(ui,ui1)? K Edges(H) (v1v2l) ? i,j.
(vi,vj) ? G
Let V(u1,,ul) u1,,ul, and V(S) ?v?S V(v)
.
Notice that T is an independent set of H iff V(T)
is an independent set of G.
So for any independent set T of H, V(T)
na(G).
Since nodes of H are just random walks in
K, T/H Pru?i.ui?V(T) (V(T)/n2?)l
(a(G)2?)l.
Let S be the maximal INDSET in G,
then Pru?i.ui?S (S/n-2?)l (a(G) 2?)l.
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