Optimal Long Test with
One Free Bit
Nikhil Bansal (IBM)
Subhash Khot (NYU)
1/17
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.
Vertex Cover
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|.
2/17
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
Vertex Cover
Equivalently, hard to find independent set of size  n even if
graph has two disjoint independent sets of size (½-) n
3/17
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.
4/17
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)
A valid schedule
Precedence Graph
Goal: Schedule jobs to minimize weighted completion time.
Various 2-approximations:
Potts linear ordering formulation
Completion time Formation
Time indexed Formulation
Sidney’s decomposition
Since the 70’s and
continuing until recently
5/17
Scheduling Problem
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)
6/17
Outline
•
•
•
•
Introduction
Free Bit Complexity of PCPs
Background (UGC, Influences, …)
Dictatorship Test & Proof
7/17
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 proof
s: Soundness, prob. of accepting wrong proof
q: Number of queries
Another parameter: number of free bits.
8/17
Free bit Complexity
f = log2 (number of accepting configurations for query)
Eg: Hastad’s test: Accept if x © y © z = 0
3 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.
9/17
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 s
will translate to PCP with same properties)
10/17
Dictatorship tests
(1,1,0)
(1,0,0)
x1
(1,1,1)
(1,0,1)
(0,1,0)
(0,1,1)
x2
x3 (0,0,0)
Vertices are x = (x1,…,xn) 2 {0,1}n
Labeling: f ! {0,1}n ! {0,1}
(0,0,1)
Dictator (co-ordinate) labeling:
f(x) = xi
0
1
0
0
1
0
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
11/17
1
Influences
1
0
Infli(f) = Pr[ f(x)  f(x© i) ]
Infli(f) = S: i 2 S (S)2
0
0
0
1
1
1
i-th coordinate
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)
Actually: Deg-k influence Inflki(f) = S:i2S, |S|·k (S)2
(for list-decoding purposes)
12/17
Dictatorship Test
(1,1,0)
Pick a random sub-cube on n co-ordinates
f(x1, *, x3, *, …., xn)
Accept if mono-chromatic (all 0’s or all 1’s).
(1,0,0)
(1,1,1)
(1,0,1)
(0,1,0)
(0,0,0)
(0,1,1)
(0,0,1)
Huge number of queries (2n) , 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-
1
0
0
0
1
0
1
1
13/17
Soundness
Soundness: 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,  = O(1) s.t. if Inflki(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)
14/17
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
15/17
Proof (continued)
To show: For 1-/10 fraction of S,
Proof: Define f = f0 , f1, …., fk = fS
For i =i1,…,ik 2 S,
fj(x) = max ( fj-1(x) , fj-1(x © ij))
E[fj] = E[fj-1] +
(fj-1)/2
E[fS] ¸ 1-/10
k = |S| =  n
1
1
0
0
10
1 0
1
1
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.)
16/17
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
17/17
Thanks !
18/17
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. Inflki(f) ¸ 
Can list decode
19/17