Hardness of Hyper-Graph Hardness of Hyper-Graph CCoolloorriinngg

Hardness of Hyper-Graph Hardness of Hyper-Graph CCoolloorriinngg

Irit DinurIrit DinurNEC NEC

Joint work with Oded Regev and Cliff Joint work with Oded Regev and Cliff SmythSmyth

QuestionQuestion How many colors does it take to color a How many colors does it take to color a

3-colorable graph? 3-colorable graph?

4?4? 5?5? 100?100?

log n?log n? The best known algorithm uses The best known algorithm uses nn3/14 3/14

colorscolors !!!!!!

DefinitionsDefinitions

A hyper-graph, H=(V,E), E A hyper-graph, H=(V,E), E {e {e V} V} is is 3-uniform3-uniform: if each edge contains exactly 3 : if each edge contains exactly 3

vertices, |e|=3.vertices, |e|=3. 22 CCoolloorraabbllee, or has property B:, or has property B:

if there exists a red-blue coloring of the if there exists a red-blue coloring of the vertices, with no monochromatic hyper-vertices, with no monochromatic hyper-edge.edge.

Coloring - BackgroundColoring - Background Finding the chromatic number: Finding the chromatic number:

Approx within nApprox within n1-1-is hard: implies NP is hard: implies NP ZPP, ZPP, [FK][FK] Approximate Coloring… coloring graphs with Approximate Coloring… coloring graphs with

tiny chromatic numbertiny chromatic number GraphsGraphs: :

Best alg Best alg [BK][BK] - O(n - O(n3/143/14) colors ) colors NP-hard to color 3-col graph w/4 colors. NP-hard to color 3-col graph w/4 colors. [KLS, GK][KLS, GK]..

Hypergraphs:Hypergraphs: NP hard to decide 3-uniform HG is 2-col, NP hard to decide 3-uniform HG is 2-col, [Lov ’73][Lov ’73] Apx alg … Apx alg … [KNS ‘98][KNS ‘98] - O(n - O(n1/51/5) colors ) colors 4-uniform 2 vs. const is NP-hard 4-uniform 2 vs. const is NP-hard [GHS ‘98][GHS ‘98] Maximization variantMaximization variant: different for 3-unif and k>3.: different for 3-unif and k>3.

Our ResultOur Result

Theorem: Theorem:

Given a 3-uniform hypergraph, deciding whether Given a 3-uniform hypergraph, deciding whether or or ccis NP-hardis NP-hard

Corollary:Corollary:

For any constants kFor any constants k, c, c22>c>c11>1, deciding >1, deciding whether whether cc11or or cc22in a k-uniform hypergraph in a k-uniform hypergraph is NP-hard is NP-hard

Khot ’02: Khot ’02: Finding large I.S. in a 3-uniform 3-col graph is NP-hard.Finding large I.S. in a 3-uniform 3-col graph is NP-hard.

What’s aheadWhat’s ahead

1.1. The Kneser graph The Kneser graph

2.2. PCP, Layered Label-CoverPCP, Layered Label-Cover

3.3. The Hypergraph ConstructionThe Hypergraph Construction

The Kneser Graph KGThe Kneser Graph KGn,cn,c::

Vertices: ( ), Edges: disjoint subsetsVertices: ( ), Edges: disjoint subsets (KG)(KG)2c+2 : easy2c+2 : easy Kneser conj ’55: Kneser conj ’55: (KG) = 2c+2(KG) = 2c+2 (KG)(KG)2c+2: First by Lovasz ’78, using 2c+2: First by Lovasz ’78, using

Borsuk-Ulam theorem. Many other proofs, Borsuk-Ulam theorem. Many other proofs, all using topological methods.all using topological methods.

[n][n]n/2-cn/2-c

The Kneser GraphThe 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:Claim: (KG(KGn,cn,c) ) 2c+2 2c+2

ProofProof::Color #1 - all vertices v, Color #1 - all vertices v, 11vv..

Color #2 - all remaining vertices v, Color #2 - all remaining vertices v, 22vv..

……

Color #2c+1 - all remaining vertices v, Color #2c+1 - all remaining vertices v, 2c+1 2c+1 v v. .

Color #2c+2 - all remaining vertices.Color #2c+2 - all remaining vertices.

The Kneser GraphThe 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 What if we allow only colors ? colors ? A color class is ‘bad’ if it contains a A color class is ‘bad’ if it contains a

monochromatic edge, monochromatic edge, How small can the ‘bad’ color class be?How small can the ‘bad’ color class be? In the previous example, it is ~2In the previous example, it is ~2--,,a constant.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 verticesNo, 2 colors can already cover 1 – o(1) of vertices

Combinatorial LemmaCombinatorial Lemma: : In any In any =2c+1 coloring of =2c+1 coloring of KGKGn,cn,c

‘ ‘bad’ color class whose relative size is >bad’ color class whose relative size is > 1/n1/n2c2c

The Kneser GraphThe Kneser Graph

Approximation and Approximation and HardnessHardness

Optimization: Given a hypergraph Optimization: Given a hypergraph HH, find, find (H)(H).. Approximation: find Approximation: find ’ ’ s.t.s.t. ’ < g’ < g Hardness is proved via a gap-problem: Hardness is proved via a gap-problem:

Given Given HH, decide between, decide between [Yes:][Yes:] if if (H) (H) m m [No:][No:] if if (H) > m(H) > mgg

1.1. A A g-g-approx algorithm can distinguish between the approx algorithm can distinguish between the YesYes and and NoNo cases, based on whether cases, based on whether Alg(H) > Alg(H) > mmgg..

2.2. m m is also a parameter… fixing is also a parameter… fixing m=2m=2 makes the makes the problem perhaps easier.problem perhaps easier.

Approximation and Approximation and HardnessHardness

ThmThm: Given : Given HH, it is NP-hard to decide between, it is NP-hard to decide between [Yes:][Yes:] if if (H) (H) 2 2 [No:][No:] if if (H) (H) 100 100

Hardness is proven via reduction from the PCP Hardness is proven via reduction from the PCP theorem: theorem:

PCP-ThmPCP-Thm: It is NP-hard to dist. bet.: It is NP-hard to dist. bet. [Yes:][Yes:] … … (next slide)(next slide)

[No:][No:] … … Reduction:Reduction:

Translate Translate [Yes:][Yes:] … to … to (H) (H) 2 2 (completeness)(completeness)

and and [No:][No:] … to … to (H) (H) 100 100 (soundness)(soundness)

PCP costume: Label-CoverPCP costume: Label-Coverbi-partite Graph G=(Y,Z, E)bi-partite Graph G=(Y,Z, E)

Y Z

A:( YA:( Y R RY Y , Z, Z R RZ Z )) is a is a labeling labeling ..A covers e=(y,z)A covers e=(y,z) E if E if yz(A(y)) = A(z)) ..Goal: cover as many edges as possible.Goal: cover as many edges as possible.

A:( YA:( Y R RY Y , Z, Z R RZ Z )) is a is a label coverlabel cover if every e if every eE is covered E is covered ..

yz :

yz

y’z’ :

y’

z’

11

22

55

33

11

77

22

44

11

33

33

55

{1,2,…,Ry} {1,2,…,Rz}

Label-CoverLabel-Coverbi-partitebi-partite Graph G=(Y,Z, E)Graph G=(Y,Z, E)

yz :

Theorem[ALMSS,AS,Raz]: It is NP-hard to distinguishIt is NP-hard to distinguish1.1. [Yes:][Yes:] label-cover for the graph.label-cover for the graph.

2.2. [No:] [No:] Any labeling covers < Any labeling covers < of the edges. of the edges.

Theorem[ALMSS,AS,Raz]: It is NP-hard to distinguishIt is NP-hard to distinguish1.1. [Yes:][Yes:] label-cover for the graph.label-cover for the graph.

2.2. [No:] [No:] Any labeling covers < Any labeling covers < of the edges. of the edges.

Y Z{1,2,…,Ry} {1,2,…,Rz}

Do we really need layers??Do we really need layers??

The hypergraph is built in the following The hypergraph is built in the following way: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 – Hyperedges are based on the edges –

always between Y and Zalways between Y and Z

Therefore, without layers, the hypergraph Therefore, without layers, the hypergraph is always 2-colorable !is always 2-colorable !

New Layered Label-Cover New Layered Label-Cover multi-partitemulti-partite Graph Graph G=(XG=(X00,X,X11,..,X,..,XLL,E),E)

X0 X1 X2 XL

x

Y

xy : RX0 RX2

x’

Y’

x’y’ : RX0 RX1

{1,…,RL}{1,…,R

2}{1,…,R

1}{1,…,R

0}

Layered Label-CoverLayered Label-Cover

TheoremTheorem: [D., Guruswami, Khot, Regev, ’01] : [D., Guruswami, Khot, Regev, ’01]

L>0,L>0,>0>0 in an (L+1)-partite graph it is in an (L+1)-partite graph it is

NP-hard to distinguish between the following:NP-hard to distinguish between the following:

1.1. [Yes:][Yes:] label-cover for the graphlabel-cover for the graph

2.2. [No:][No:] For every i,j any For every i,j any label-coverlabel-cover of X of Xi i and Xand Xj j covers < covers < of the edges between themof the edges between them

Moreover, for any k>0 layers iMoreover, for any k>0 layers i11<…<i<…<ikk, and subsets S, and subsets Sjj XXiijj

of relative size 2/k, of relative size 2/k, SSjj,,SSj’j’, with 1/k, with 1/k22 of all the edges of all the edges between between XXiijj

and X and Xiij’j’

X0 X1 X2 XL-1 XL

S2

S1

S3

p

Reducing Label-Cover to Reducing Label-Cover to Hyper Graph ColoringHyper Graph Coloring

Reduction:Reduction: Translate multi-partite G into a hyper- Translate multi-partite G into a hyper-graph H s.t.graph H s.t.

[Yes:][Yes:] label-cover for Glabel-cover for G (H) = 2(H) = 2

[No:][No:] Every labelingEvery labeling covers < covers < of the edges of the edges (H) (H) 100100

The Hypergraph The Hypergraph ConstructionConstruction

X0 XiX10000

This is really the “Long-Code”

V = V = (X (Xii ( )) ( ))RRii/2 - 49/2 - 49RRii

3-uniform Hyper-Edges3-uniform Hyper-Edges

X0 XiX10000

3-uniform Hyper-Edges3-uniform Hyper-Edges

3-uniform Hyper-Edges3-uniform Hyper-Edges

Y

yz : {1,2,…,Ry} {1,2,…,Rz}

z

RY

Rz

{v1 ,v2,u} E iff: v1 v2 = and xy( R\(v1 v2)) u

Note that v1 ,v2, are connected to a constant fraction of the u

Proof Proof Part I – Part I – [Yes] [Yes] maps to maps to [Yes][Yes]

A label-cover of G translates to a A label-cover of G translates to a legal 2-coloring of H.legal 2-coloring of H.

[Yes] [Yes] maps to maps to [Yes][Yes]

X0 XiX10000

1 2

2 4

3

1

1 2 1

Red(x) = {v | aRed(x) = {v | axx v} v}

Blue(x) = {v | aBlue(x) = {v | axx v} v}

If v1, v2 blue, then ax R\(v1 v2)

thus, ay=xy(ax)u, so u is red

If v1, v2 blue, then ax R\(v1 v2)

thus, ay=xy(ax)u, so u is red

Two disjoint vertices cannot both be red

Part II – Part II – [No] [No] maps tomaps to [No] [No]

If there’s no If there’s no –cover for G, then –cover for G, then (H)=100.(H)=100.

Given a 99-coloring of H, Given a 99-coloring of H,

We find in G, 2 layers XWe find in G, 2 layers Xii and X and Xjj and a labeling and a labeling

that covers > that covers > of edges between them of edges between them

Combinatorial LemmaCombinatorial Lemma: In every : In every coloring of KG coloring of KGR,49R,49

`special’ color class whose relative size is >`special’ color class whose relative size is > 1/R1/R9898

Given a 99-coloring of the HG, we find for each

block, a ‘specialspecial’ color-class that is:

1. Large > 1/R98

2. Contains two vertices v1 v2 =

Part II – Part II – [No][No] maps to maps to [No][No]

Part II – Part II – [No] [No] maps tomaps to [No] [No]

Given a 99-coloring, find “special” colorsAssume blue is the prevalentspecial color.

By layered label-cover theorem, Xi,Xj with many edges

between Si and Sj

X0 X1 X2 XL-1 XL

Si

SjConcentrate on Si , Sj

S2

S1

S3

Define x Si : A(x) R R\(v1v2)

Define y Sj : A(y) so as to maximize cover size

Si Sj

xy

We prove: A(y) is “popular” among its neighbors

Left blocks x: Each contain blue v1,v2 disjoint –preventing all right hand u, {v1,v2, u} E from being blue.

These are u containing the “hole” Ri\(v1v2) and are a constant fraction of the Kneser block.

Key point: the holes must be `aligned’

Define a labeling for Si , Sj as follows:

(last slide (last slide of proofof proof))

SummarySummary

Kneser graphKneser graph PCP and Layered Label-CoverPCP and Layered Label-Cover Hypergraph ConstructionHypergraph Construction Proof of ReductionProof of Reduction

Open QuestionsOpen Questions

Coloring:Coloring: We still can’t color a 2-col 3-uniform We still can’t color a 2-col 3-uniform

H.G. with less than nH.G. with less than ncolors, or prove a colors, or prove a matching hardness matching hardness

Worse situation for graphs: 3 vs. Worse situation for graphs: 3 vs. anything bigger than 5anything bigger than 5

Maximization Versions of Coloring Maximization Versions of Coloring (e.g. max-cut)(e.g. max-cut)