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

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

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

Coloring - Background Finding the chromatic number: Finding the chromatic number: Approx within n 1-   is hard: implies NP  ZPP, [FK] Approx within n 1-   is hard: implies NP  ZPP, [FK] Approximate Coloring… coloring graphs with tiny chromatic number Approximate Coloring… coloring graphs with tiny chromatic number Graphs: Graphs: Best alg [BK] - O(n 3/14 ) colors Best alg [BK] - O(n 3/14 ) colors NP-hard to color 3-col graph w/4 colors. [KLS, GK]. NP-hard to color 3-col graph w/4 colors. [KLS, GK]. Hypergraphs: Hypergraphs: NP hard to decide 3-uniform HG is 2-col, [Lov ’ 73] NP hard to decide 3-uniform HG is 2-col, [Lov ’ 73] Apx alg … [KNS ‘ 98] - O(n 1/5 ) colors Apx alg … [KNS ‘ 98] - O(n 1/5 ) colors 4-uniform 2 vs. const is NP-hard [GHS ‘ 98] 4-uniform 2 vs. const is NP-hard [GHS ‘ 98] Maximization variant: different for 3-unif and k>3. Maximization variant: different for 3-unif and k>3.

Our Result Theorem: Given a 3-uniform hypergraph, deciding whether  or  c  is NP-hard Corollary: For any constants k , c 2 >c 1 >1, deciding whether  c 1  or  c 2  in a k-uniform hypergraph is NP-hard Khot ’ 02: Finding large I.S. in a 3-uniform 3-col graph is NP-hard.

What’s ahead 1. The Kneser graph 2. PCP, Layered Label-Cover 3. The Hypergraph Construction

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

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

Claim:  (KG n,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. The Kneser Graph

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

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

Approximation and Hardness Optimization: Given a hypergraph H, find  (H). Optimization: Given a hypergraph H, find  (H). Approximation: find  ’ s.t.  ’ < g   Approximation: find  ’ s.t.  ’ < g   Hardness is proved via a gap-problem: Hardness is proved via a gap-problem: Given H, decide between [Yes:] if  (H)  m [Yes:] if  (H)  m [No:] if  (H) > m  g [No:] if  (H) > m  g 1. A g-approx algorithm can distinguish between the Yes and No cases, based on whether Alg(H) > m  g. 2. m is also a parameter… fixing m=2 makes the problem perhaps easier.

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

PCP costume: Label-Cover bi-partite Graph G=(Y,Z, E) Y Z A:( Y  R Y, Z  R Z ) is a labeling. A covers e=(y,z)  E if. A covers e=(y,z)  E if  y  z (A(y)) = A(z)). Goal: cover as many edges as possible. A:( Y  R Y, Z  R Z ) is a label cover if every e  E is covered.  y  z :  y z  y’  z’ :  y’ z’ {1,2,…,R y }{1,2,…,R z }

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

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

New Layered Label-Cover multi-partite Graph G=(X 0,X 1,..,X L,E) X0X0 X1X1 X2X2 XLXL x Y  x  y : R X 0  R X2 x’ Y’  x’  y’ : R X 0  R X1 {1,…,R L }{1,…,R 2 }{1,…,R 1 }{1,…,R 0 }

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: NP-hard to distinguish between the following: 1. [Yes:]  label-cover for the graph 2. [No:] For every i,j any label-cover of X i and X j covers <  of the edges between them Moreover, for any k>0 layers i 1 0 layers i 1 <…<i k, and subsets S j  X i j of relative size 2/k,  S j,S j’, with 1/k 2 of all the edges between X i j and X i j’ X0X0 X1X1 X2X2 X L-1 XLXL S2S2 S1S1 S3S3

p

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

The Hypergraph Construction X0X0 XiXi X This is really the “Long-Code” V =  (X i  ( )) R i / RiRiRiRi

3-uniform Hyper-Edges X0X0 XiXi X 10000

3-uniform Hyper-Edges

Y  y  z :  {1,2,…,R y }{1,2,…,R z } z RYRY RzRz {v 1,v 2,u}  E iff: v 1  v 2 =  and  x  y ( R\(v 1  v 2 ))  u Note that v 1,v 2, are connected to a constant fraction of the u

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

[Yes] maps to [Yes] X0X0 XiXi X Red(x) = {v | a x  v} Blue(x) = {v | a x  v} If v 1, v 2 blue, then a x  R\(v 1  v 2 ) thus, a y =  x  y (a x )  u, so u is red If v 1, v 2 blue, then a x  R\(v 1  v 2 ) thus, a y =  x  y (a x )  u, so u is red Two disjoint vertices cannot both be red

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 X i and X j and a labeling that covers >  of edges between them

Combinatorial Lemma: In every  coloring of KG R,49  `special’ color class whose relative size is > 1/R 98 Given a 99-coloring of the HG, we find for each special block, a ‘special’ color-class that is: 1. Large > 1/R Contains two vertices v 1  v 2 =  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,  X i,X j with many edges between S i and S j X0X0 X1X1 X2X2 X L-1 XLXL SiSi SjSj Concentrate on S i, S j S2S2 S1S1 S3S3

Define  x  S i : A(x)  R R\(v 1  v 2 ) Define  y  S j : A(y) so as to maximize cover size SiSi SjSj x y We prove: A(y) is “popular” among its neighbors Left blocks x: Each contain blue v 1,v 2 disjoint – preventing all right hand u, {v 1,v 2, u}  E from being blue. These are u containing the “hole” R i \(v 1  v 2 ) and are a constant fraction of the Kneser block. Key point: the holes must be `aligned’ Define a labeling for S i, S j as follows: (last slide of proof )

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

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