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Hardness of Hyper-Graph C o l o r i n g

Hardness of Hyper-Graph C o l o r i n g. Irit Dinur NEC Joint work with Oded Regev and Cliff Smyth. Question. How many colors does it take to color a 3-colorable graph? 4? 5? 100? log n? The best known algorithm uses n 3/14 colors !!!. Definitions.

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Hardness of Hyper-Graph C o l o r i n g

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  1. Hardness of Hyper-Graph Coloring Irit Dinur NEC Joint work with Oded Regev and Cliff Smyth

  2. Question • How many colors does it take to color a 3-colorable graph? • 4? • 5? • 100? • log n? • The best known algorithm uses n3/14 colors!!!

  3. Definitions • A hyper-graph, H=(V,E), E  {e V} is • 3-uniform: if each edge contains exactly 3 vertices, |e|=3. • 2Colorable, or has property B: if there exists a red-blue coloring of the vertices, with no monochromatic hyper-edge.

  4. Coloring - Background • Finding the chromatic number: • Approx within n1-eis hard: implies NP  ZPP, [FK] • Approximate Coloring… coloring graphs with tiny chromatic number • Graphs: • Best alg [BK] - O(n3/14) colors • NP-hard to color 3-col graph w/4 colors. [KLS, GK]. • Hypergraphs: • NP hard to decide 3-uniform HG is 2-col, [Lov ’73] • Apx alg …[KNS ‘98] - O(n1/5) colors • 4-uniform 2 vs. const is NP-hard [GHS ‘98] • Maximization variant: different for 3-unif and k>3.

  5. Theorem: Given a 3-uniform hypergraph, deciding whether c=2 or ccis NP-hard Corollary: For any constants k3, c2>c1>1, deciding whether c=c1or cc2in a k-uniform hypergraph is NP-hard Khot ’02: Finding large I.S. in a 3-uniform 3-col graph is NP-hard. Our Result

  6. The Kneser graph PCP, Layered Label-Cover The Hypergraph Construction What’s ahead

  7. [n] n/2-c The Kneser Graph • The Kneser Graph KGn,c: • Vertices: ( ), Edges: disjoint subsets • c(KG)2c+2 : easy • Kneser conj ’55: c(KG) = 2c+2 • c(KG)2c+2: First by Lovasz ’78, using Borsuk-Ulam theorem. Many other proofs, all using topological methods.

  8. Ground set: {1,2,3,4,5} {1,2} {3,5} {3,4} {4,5} {2,5} {1,3} {2,4} {1,4} {2,3} {1,5}

  9. The Kneser Graph Claim:c(KGn,c)  2c+2 Proof: Color #1 - all vertices v, 1v. Color #2 - all remaining vertices v, 2v. … Color #2c+1 - all remaining vertices v, 2c+1  v. Color #2c+2 - all remaining vertices.

  10. {1,2} {3,5} {3,4} {4,5} {2,5} {1,3} {2,4} {1,4} {2,3} {1,5}

  11. The Kneser Graph • What if we allow only c-1 colors ? • A color class is ‘bad’ if it contains a monochromatic edge, • How small can the ‘bad’ color class be? • In the previous example, it is ~2-c,a constant. • Is this the best we can do? • No, 2 colors can already cover 1 – o(1) of vertices Combinatorial Lemma: In any c-1=2c+1 coloring of KGn,c $ ‘bad’ color class whose relative size is >1/n2c

  12. Approximation and Hardness • Optimization: Given a hypergraph H, findc(H). • Approximation: find c’ s.t.c’ < gc • Hardness is proved via a gap-problem: Given H, decide between • [Yes:] if c(H)  m • [No:] if c(H) > mg • A g-approx algorithm can distinguish between the Yes and No cases, based on whether Alg(H) > mg. • m is also a parameter… fixing m=2 makes the problem perhaps easier.

  13. Approximation and Hardness • Thm: Given H, it is NP-hard to decide between • [Yes:] if c(H)  2 • [No:] if c(H)  100 • Hardness is proven via reduction from the PCP theorem: • PCP-Thm: It is NP-hard to dist. bet. • [Yes:] … (next slide) • [No:] … • Reduction: Translate [Yes:] … to c(H)  2 (completeness) and [No:] … to c(H)  100 (soundness)

  14. 1 2 y’ 2 4 1 5 z 3 3 z’ 1 3 y 5 7 y’z’ : yz :   PCP costume: Label-Coverbi-partite Graph G=(Y,Z, E) A:( Y RY , Z RZ ) is a labeling. A covers e=(y,z) E if yz(A(y)) = A(z)). Goal: cover as many edges as possible. A:( Y RY , Z RZ ) is a label cover if every eE is covered . Z Y {1,2,…,Ry} {1,2,…,Rz}

  15. Z Y Theorem[ALMSS,AS,Raz]: It is NP-hard to distinguish [Yes:]label-cover for the graph. [No:] Any labeling covers <  of the edges. yz :  Label-Coverbi-partiteGraph G=(Y,Z, E) {1,2,…,Ry} {1,2,…,Rz}

  16. Do we really need layers?? The hypergraph is built in the following way: • New vertices are created from Y • New vertices are created from Z • Hyperedges are based on the edges – always between Y and Z • Therefore, without layers, the hypergraph is always 2-colorable !

  17. xy : RX0 RX2 x’y’ : RX0 RX1 x’ Y’ x Y New Layered Label-Cover multi-partite Graph G=(X0,X1,..,XL,E) X0 X1 X2 XL {1,…,R0} {1,…,R1} {1,…,R2} {1,…,RL}

  18. S1 S2 S3 X0 X1 X2 XL-1 XL Layered Label-Cover Theorem: [D., Guruswami, Khot, Regev, ’01] L>0,>0 in an (L+1)-partite graph it is NP-hard to distinguish between the following: [Yes:]label-cover for the graph [No:]For every i,j any label-cover of Xi and Xj covers <  of the edges between them Moreover, for any k>0 layers i1<…<ik, and subsets SjÍ Xij of relative size 2/k, $Sj,Sj’, with 1/k2 of all the edges between Xij and Xij’

  19. p p

  20. Reducing Label-Cover to Hyper Graph Coloring Reduction: Translate multi-partite G into a hyper-graph H s.t. [Yes:]label-cover for G c(H) = 2 [No:]Every labelingcovers <  of the edges c(H)  100

  21. V =  (Xi ( )) Ri Ri/2 - 49 The Hypergraph Construction This is really the “Long-Code” X10000 X0 Xi

  22. 3-uniform Hyper-Edges X10000 X0 Xi

  23. 3-uniform Hyper-Edges

  24. yz :  3-uniform Hyper-Edges {1,2,…,Ry} {1,2,…,Rz} Y z RY Rz {v1 ,v2,u} ÎE iff: v1Ç v2 = f and xy( R\(v1È v2))Íu Note that v1 ,v2, are connected to a constant fraction of the u

  25. Proof Part I – [Yes] maps to [Yes] A label-cover of G translates to a legal 2-coloring of H.

  26. Red(x) = {v | ax v} Blue(x) = {v | ax v} 2 3 1 4 1 2 2 1 1 [Yes] maps to [Yes] Two disjoint vertices cannot both be red X10000 X0 Xi If v1, v2 blue, then axÎ R\(v1È v2) thus, ay=xy(ax)Îu, so u is red If v1, v2 blue, then axÎ R\(v1È v2) thus, ay=xy(ax)Îu, so u is red

  27. Part II – [No] maps to [No] If there’s no –cover for G, then (H)=100. Given a 99-coloring of H, We find in G, 2 layers Xi and Xj and a labeling that covers >  of edges between them

  28. Part II – [No] maps to [No] Combinatorial Lemma: In every 99 coloring of KGR,49 $ `special’ color class whose relative size is >1/R98 Given a 99-coloring of the HG, we find for each block, a ‘special’ color-class that is: Large > 1/R98 Contains two vertices v1Ç v2 = f

  29. S1 S2 S3 X0 X1 X2 XL-1 XL Concentrate on Si , Sj Sj Si Part II – [No] maps to [No] Given a 99-coloring, find “special” colors Assume blue is the prevalent special color. By layered label-cover theorem, $ Xi,Xj with many edges between Si and Sj

  30. (last slide of proof) Left blocks x: Each contain blue v1,v2disjoint – preventing all right hand u, {v1,v2, u}  E from being blue. These are u containing the “hole” Ri\(v1v2) and are a constant fraction of the Kneser block. Key point: the holes must be `aligned’ Define a labeling for Si , Sj as follows: y x Sj Si Define  x  Si : A(x) R R\(v1v2) Define  y  Sj : A(y) so as to maximize cover size We prove: A(y) is “popular” among its neighbors

  31. Summary • Kneser graph • PCP and Layered Label-Cover • Hypergraph Construction • Proof of Reduction

  32. Open Questions • Coloring: • We still can’t color a 2-col 3-uniform H.G. with less than ne colors, or prove a matching hardness • Worse situation for graphs: 3 vs. anything bigger than 5 • Maximization Versions of Coloring (e.g. max-cut)

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