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Optimal Long Test with One Free Bit

Nikhil Bansal (IBM)

Subhash Khot (NYU)

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Min Vertex Cover

Vertex Cover: Given G=(V,E), a set S of vertices s.t. each edge has at least one end point in S.

Hardness: 1.36 assuming P NP [Dinur-Safra 02]

2 - assuming UGC [Khot-Regev 03]

S is vertex cover iff V\S is independent set.

[Khot-Regev 03]: Even if graph has independent set of

size (1/2 - ) |V| cannot find one of size |V|.

Vertex Cover

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Our results

Thm: Assuming UGC, it is NP-Hard to get 2- approx. for vertex cover, even for essentially bipartite graphs.

(even though min-VC is easy for bipartite graphs).

Independent Independent(1/2-) n (1/2-) n

Equivalently, hard to find independent set of size n even if graph has two disjoint independent sets of size (½-) n

Vertex Cover

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Additional Features

1. Previous results on VC use biased long code.(most naturally viewed as combinatorial constructions)

2. Our result is most naturally viewed as a PCP. (in fact, assuming UGC we show a PCP with 1 free bit, and

completeness = 1-, soundness = )

3. Very natural dictatorship test with simple analysis.

4. Unlike [Khot-Regev], do not need a (equivalent) version of UGC with special properties.

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1|prec|j wjCj Problem

Given n jobs, arbitrary weights and sizes (job j: wt. wj, size pj)Precedence constraints: DAG (edge (i,j) ) j cannot start before i finishes)

Goal: Schedule jobs to minimize weighted completion time.

Precedence Graph

A valid schedule

Various 2-approximations: Potts linear ordering formulation Completion time Formation Time indexed Formulation Sidney’s decomposition

Since the 70’s and continuing until recently

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No (1+) approximation [Ambuhl-Mastrolilli-Svensson 07]

(assuming SAT cannot be solved in sub-exponential time)

Special case of Vertex Cover [Chudak-Hochbaum 99, Correa-Schulz 05, …]

Belief that < 2 might be possible.

Thm: Hardness of 2- assuming a variant of UGC.

Our dictatorship test inspired by certain hard instances proposed by Woeginger [W 03] (do not discuss this connection in talk)

Scheduling Problem

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Outline

• Introduction

• Free Bit Complexity of PCPs

• Background (UGC, Influences, …)

• Dictatorship Test & Proof

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PCP Theorem [AS, ALMSS]

X 2 SAT can be verified by writing proof of length poly(n)And querying only O(1) positions in the proof.

PCP Thm: NP ½ PCPc,s(O(log n), q=O(1)) c: Completeness, prob. of accepting correct proofs: Soundness, prob. of accepting wrong proofq: Number of queries

Another parameter: number of free bits.

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Free bit Complexity

f = log2 (number of accepting configurations for query)

Eg: Hastad’s test: Accept if x © y © z = 03 queries, but accepts on 4 answers: (0,0,0), (0,1,1), (1,0,1), (1,1,0)

Free bits (f) = log2 4 =2

Thm [FGLSS, BGS]: PCPc,s with f=0 is equivalent to Reduction: 3-SAT to graph on V vertices, s.t. NP-Hard to tell

if independent Set has size at least c |V| or at most s |V|.

(f=0 means verifier expects unique answer for each query.)

Corollary: Implies (1-s)/(1-c) hardness for vertex cover.

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Our Result

Thm: Assuming UGC, there is a PCP with 1 free bit, completeness = 1-, soundness = ,

Cor: PCP with 0 free bits, completeness = ½ - , soundness = Pf: Take PCP with 1 free bit (has 2 good answers per query),

verifier can choose one of these answers randomly.

(1-s)/(1-c) = 2- hardness for V.C. (+ almost-bipartite property)

Using usual UGC techniques, suffices to give a related dictatorship test on the boolean hypercube.

(Dictatorship test with 1 free bit, completeness c and soundness swill translate to PCP with same properties)

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Dictatorship tests

(0,0,0) (0,0,1)

(1,0,1)(1,0,0)

(1,1,1)(1,1,0)

(0,1,1)(0,1,0)

Vertices are x = (x1,…,xn) 2 {0,1}n

Labeling: f ! {0,1}n ! {0,1}

Dictator (co-ordinate) labeling:f(x) = xi

x1

x3

x20

0

0

0

1

11

1

Dictatorship Test: 1. Completeness: Accept any dictator labeling with prob ¸ c 2. Soundness: Accept any function “far” from dictator with prob. · s

Far from dictator = All variables have small degree-k influences

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Influences

Infli(f) = Pr[ f(x) f(x© i) ]Infli(f) = S: i 2 S (S)2

Actually: Deg-k influence Inflki(f) = S:i2S, |S|·k (S)2

Eg: 1. Dictator f(x) = xi has influence 1 for co-ordinate i, 0 for others 2. Majority function f(x) has small influences ((1/n1/2))

Soundness: For any f s.t. Infli(f) · for all co-ordinates i. Test must fail (pass with prob. · s)

(for list-decoding purposes)

0

0

0

0

1

11

1

i-th coordinate

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Dictatorship Test

Pick a random sub-cube on n co-ordinates

f(x1, *, x3, *, …., xn)

Accept if mono-chromatic (all 0’s or all 1’s).

Huge number of queries (2n) , but 1 free bit!

Completeness: For any dictator function f(x1,…,xn) = xi

Random sub-cube on n co-ordinates

is mono-chromatic with probability 1-

0

0

0

0

1

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1

(0,0,0) (0,0,1)

(1,0,1)(1,0,0)

(1,1,1)(1,1,0)

(0,1,1)(0,1,0)

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SoundnessSoundness: Low influences ) · fraction subcubes mono-chromatic (not true say if f=0 everywhere or f=1 everywhere)Folding Trick: Consider subcube C at x, and subcube at Accept if both monochromatic and have different colors.

Soundness: If 1/3 · E[f] · 2/3, for any > 0, 9 k, s.t. if Inflk

i(f) · for each i, then · fraction of subcubes monochromatic.

Proof follows from [Mossel-O’donnell-Oleszkiewicz’05 ]

Invariance Principle: Low deg-k influence ) f is “random-like”

[MOO’05] Ain’t Over Till It’s Over (proposed by Freidgut-Kalai)

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Alternate Proof

Soundness: Small influences, then · fraction of subcubes monochromatic.

Will Show: Random subcube contains a 1 with prob. ¸ 1- /2(symmetric argument implies it has 0 also with prob. ¸ 1-/2)Hence non-monochromatic with prob. ¸ 1-

Pf: Random subcube (x,S)= Pick random x, n random coordinates S

Define fS(x) = max (y : yi = xi for all i 2 [n]\S }Subcube (x, S) contains a 1 iff fS(x) = 1

Claim: For random S, with prob. 1 - /10 E[fS] ¸ 1- /10

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Proof (continued)To show: For 1-/10 fraction of S, E[fS] ¸ 1-/10

Proof: Define f = f0 , f1, …., fk = fS k = |S| = n

For i =i1,…,ik 2 S,

fj(x) = max ( fj-1(x) , fj-1(x © ij))

E[fj] = E[fj-1] + (fj-1)/2

If total influence >> 1/, if choose fraction of co-ordinates, expect to add up 1 unit of influences.

Freidgut, KKL: If f is balanced, and has all influences small (· ), then total influence is high ¸ (1/log())

(Technical issue: Only f had high sum of influences, not fj-1.)

1

0

0

0

1

01

11

1

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Concluding Remarks

Assuming UGC, min-VC is 2- hard even in almost bipartite graphs

New: k- hardness of vertex cover in k-uniform hypergraphs that are almost k-partite.

Implies optimum hardness for some scheduling problems.

Could be useful in other contexts (such as coloring ?)

Min-VC: 1.36 hardness still best assuming P NP

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Thanks !

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Influences

Infli(f) = Pr[ f(x) f(x© i) ]Infli(f) = S: i 2 S (S)2

Deg-k influence Inflki(f) = S: i 2 S, |S| · k (S)2

Crucial point: i Inflki(f) · k

Far from dictator: For all i, Inflki(f) ·

If dictatorship test accepts with prob ¸ s + some i s.t. Inflk

i(f) ¸ Can list decode