1/17 Optimal Long Test with One Free Bit Nikhil Bansal (IBM) Subhash Khot (NYU)

2/17 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. 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

3/17 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 (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

4/17 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.

5/17 1|prec|  j w j C j Problem Given n jobs, arbitrary weights and sizes (job j: wt. w j, size p j ) 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

6/17 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

7/17 Outline Introduction Free Bit Complexity of PCPs Background (UGC, Influences, …) Dictatorship Test & Proof

8/17 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 PCP c,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.

9/17 Free bit Complexity f = log 2 (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) = log 2 4 =2 Thm [FGLSS, BGS]: PCP c,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.

10/17 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)

11/17 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 = (x 1,…,x n ) 2 {0,1} n Labeling: f ! {0,1} n ! {0,1} Dictator (co-ordinate) labeling: f(x) = x i x1x1 x3x3 x2x 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

12/17 Influences Infl i (f) = Pr[ f(x)  f(x © i) ] Infl i (f) =  S: i 2 S (S) 2 Actually: Deg-k influence Infl k i (f) =  S:i 2 S, |S| · k (S) 2 Eg: 1. Dictator f(x) = x i has influence 1 for co-ordinate i, 0 for others 2. Majority function f(x) has small influences (  (1/n 1/2 )) Soundness: For any f s.t. Infl i (f) ·  for all co-ordinates i. Test must fail (pass with prob. · s) (for list-decoding purposes) i-th coordinate

13/17 Dictatorship Test Pick a random sub-cube on  n co-ordinates f(x 1, *, x 3, *, …., x n ) Accept if mono-chromatic (all 0’s or all 1’s). Huge number of queries (2  n ), but 1 free bit! Completeness: For any dictator function f(x 1,…,x n ) = x i Random sub-cube on  n co-ordinates is mono-chromatic with probability 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)

14/17 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,  s.t. if Infl k 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)

15/17 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 f S (x) = max (y : y i = x i for all i 2 [n]\S } Subcube (x, S) contains a 1 iff f S (x) = 1 Claim: For random S, with prob. 1 -  /10 E[f S ] ¸ 1-  /10

16/17 Proof (continued) To show: For 1-  /10 fraction of S, E[f S ] ¸ 1-  /10 Proof: Define f = f 0, f 1, …., f k = f S k = |S| =  n For i =i 1,…,i k 2 S, f j (x) = max ( f j-1 (x), f j-1 (x © i j )) E[f j ] = E[f j-1 ] + (f j-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 f j-1.)

17/17 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

18/17 Thanks !

19/17 Influences Infl i (f) = Pr[ f(x)  f(x © i) ] Infl i (f) =  S: i 2 S (S) 2 Deg-k influence Infl k i (f) =  S: i 2 S, |S| · k (S) 2 Crucial point:  i Infl k i (f) · k Far from dictator: For all i, Infl k i (f) ·  If dictatorship test accepts with prob ¸ s +  some i s.t. Infl k i (f) ¸  Can list decode