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Graphs with tiny vector chromatic numbers and huge chromatic numbers

Graphs with tiny vector chromatic numbers and huge chromatic numbers. Michael Langberg. Weizmann Institute of Science. Joint work with U. Feige and G. Schechtman. Two fundamental NP-Hard problems. Minimum Coloring Maximum Independent Set . Minimum coloring.

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Graphs with tiny vector chromatic numbers and huge chromatic numbers

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  1. Graphs with tiny vector chromatic numbers and huge chromatic numbers Michael Langberg Weizmann Institute of Science Joint work with U. Feige and G. Schechtman

  2. Two fundamental NP-Hard problems • Minimum Coloring • Maximum Independent Set

  3. Minimum coloring • Vertex-coloring: Assignment of colors to V s.t. endpoints of each edge have diff. colors. G=(V,E) (G)=3 • Chromatic number(G): Minimum number of colors needed.

  4. Maximum independent set • IS: Set of vertices that do not share any edges. G=(V,E) (G)=3 • (G): Size of maximum IS.

  5. Coloring vs. IS Every color class in a coloring of G is an IS. • Coloring  finding a cover of G with disjoint IS. • (G)(G) n. • Algorithms for IS  algorithms for coloring.

  6. Approximation algorithms • Not likely to find efficient algorithms. • Settle on efficient approximation algorithms. • Provide solutions whose value is guaranteed to be within a ratio no worse than r from the value of the optimal solution. • App. ratio of algorithm ALG: • r= ALG/OPT (min.), r= OPT/ALG (max.). • r 1, the smaller the better !

  7. This talk • [KargerMotwaniSudan] introduce the notion of vector coloring. • Plays major role in approximation algorithms for IS and Coloring. • Our work: present tight results on the limitation of vector coloring. • Structure: • Background on IS and Coloring. • Vector coloring. • Our results.

  8. Approximating (G) & (G) • Good news: • Both (G) and (G) can be app. within ratio n(loglog n)2/(log n)3[Haldorsson, Feige]. • Bad News: • Estimating both (G) and (G) up to a factor of n1- is “hard” (unless NPis in random polynomial time). [Hastad, FeigeKilian ,EngebretsenHolmerin, Khot]. Relatively small gap.

  9. What about restricted cases? Consider a graph G that is known to have small chromatic number (G)=k. • Good news: • Can efficiently find coloring with: • k=3  n3/14 colors [KargerMotwaniSudan, BlumKarger]. • k=4  n7/19 colors [HalperinNathanielZwick]. • k  nf(k) colors (f(k)1 as k increases)[KMS, HNZ]. • Bad news [KhannaLinialSafra, GuruswamiKhanna, Khot]: • NP-hard to color 3 colorable graphs with 4 colors. • NP-hard to color k col. graphs with  5k/3, k(log k) colors. Gap is wide open.

  10. Vector coloring [KMS] Plays a major role in approximation algorithms for coloring and IS (restricted cases). Definition:G=(V,E) is vector k-colorable if one can assign unit vectors to its vertices, s.t. every two adjacent vertices are embedded far apart.

  11. Vector coloring cont. Definition:G=(V,E) is vector k-colorable if one can assign unit vectors to its vertices, s.t. every two vectors corresponding to adjacent vertices have inner product at most -1/(k-1). k=3 <vi,vj>  -1/(k-1) = -1/2 = cos(120o) 120o 95o k=11 <vi,vj> -1/(k-1) = -1/10  cos(95o)

  12. Vector coloring – example Definition: G=(V,E) is vector k-colorable if one can assign unit vectors to its vertices, s.t. every two vectors corresponding to adjacent vertices have inner product at most -1/(k-1). Vector 3-coloring: • k=3 -1/(k-1) = -1/2 = cos(120o) R2

  13. Vector coloring – example Definition: G=(V,E) is vectork-colorable if one can assign unit vectors to its vertices, s.t. every two vectors corr. to adjacent vertices have inner product at most -1/(k-1). Vector 4-coloring: • k=4 -1/(k-1) = -1/3  cos(109o) R3

  14. Vector coloring (G) • Every k colorable graph is also vector k-col. • Identify each color class with one vertex in a perfect k-1 dimensional simplex. • k = 4: R3

  15. Vector coloring in P If G=(V,E) is vector k-colorable, such a vector coloring can be computed in polynomial time (semidefinite programming). Min  s.t. • <vi,vj>   for each edge (i,j)  E. • <vi,vi> = 1 for each node i V. •  = -1/(k-1)

  16. Algorithm of [KMS] Use vector coloring to color graphs with small chromatic number. • Input: Graph G which satisfies (G)=3. • Output: Coloring of G with few colors. • ALG: • (G)=3  G is vector 3-colorable. • Find vector 3-coloring of G (SDP). • Use geometrical structure to find good coloring of G.

  17. Algorithm of [KMS] cont. • Objective: find large IS in G. Pick random cap. • Consider vertices corr. to vectors in cap. • Small cap small IS. • Large cap  large set with many edges. [KMS] optimize size of cap. Vector 3-coloring

  18. [KMS] results : maximum degree in G. • Graphs which are vector 3-colorable can be colored eff. in 1/3 colors (+1 trivial). • As function of n: obtain n1/4 [Wigderson]. • Graphs which are vector k-colorable can be colored efficiently in min(1-2/k,n1-3/(k+1))colors. Improving these results will yield improved results in[BlumKarger, AlonKahale, HalperinNathanielZwick].

  19. Our results Negative in nature. Prove that the results of [KMS] are tight. [KMS]: Graphs which are vector k-colorable can be colored efficiently in 1-2/k colors. Present vector k-colorable graphs with chromatic number at least 1-2/k-. • Will neglect  in remainder of talk.

  20. Previous work on limitation of vector coloring. As a function on n rather than. • Vector 3-colorable graphs with   n0.05[KMS,Alon,Szegedy]. • Vector k-colorable graphs with   nf(k) where f(k)  1 [KMS,Charikar,Feige]. Our results: • For k=3 we obtain  n0.15. • For other k, improve f(k).

  21. How large is the gap? How good of an app. is vector coloring to (G)? • There are graphs for which ratio between (G) and vector chromatic number n/2 [Feige]. O(log1/2(n)) large gap O(log1/2(n)) O(log1/2(n)) (G)  n/2 G is 2 vec. col. Our results improve gap to n/polylog(n). • (k=log(n)/loglog(n)). Vector coloringdoes not app. within factor better than n/polylog(n).

  22. Previous work – graphs used All previous work use similar graphs G=(V,E): • V:{0,1}n (“hypercube”). • E: vertices u and v are connected iff Hamming distance is large. • Natural embedding in unit sphere (ensures small vector chromatic number). • Known bounds on maximum IS (ensures large chromatic num). 01 11 00 10

  23. Our work • Use different graphs. • We use graphs presented in [FeigeSchechtman] that addresses a SDP relaxation of the Max-Cut problem [GoemansWilliamson]. • Goal:G is vector 3-colorable, (G) is large (k=3). • Our graph G: place n random points on the unit sphere, connect each two points that are far apart. I.e. inner product at most -1/2. 120o

  24. Main theorem • G is vector 3-colorable (by definition). • G has chromatic number 1/3.

  25. Analyzing (G) Do not know how to analyze (G) directly. Follow ideas of [FS]: construct G in three steps. • Start with a continuous graph. • V = all points on unit sphere. • E = pairs of points far appart. • Analyze expansion properties. • Switch to discrete version. • Take random sample.

  26. Proof outline – (G) is large. • Step 1: continuous graph. • Continuous graph is vector 3-col. • Continuous graph has nice expansion properties. • Step 2: discrete graph. • Discrete graph is vector 3-col. • Inherits expansion properties. • Step 3: random sample. • Random sample is vector 3-col. • Expansion properties of discrete graph imply random sample has large . A B

  27. Remainder of this talk • Step 1: continuous graph. • Continuous graph is vector 3-col. • Continuous graph has nice expansion properties (isoperimetric inequalities on the sphere). • Step 2: discrete graph inherits expansion properties of continuous graph. • Step 3: random sample. • Random sample is vector 3-col. • Expansion prop. of discrete graph imply random sample has large  (property testing). A B

  28. Wait a minute ! Continuous  Discrete  Random Why do we need the random graph? Doesn’t the discrete version suffice? Properties of discrete graph (easy to prove): • Vector 3-colorable. • Large chromatic number. Problem:Discrete graph has large degree (n1-), can not show   1/3. Solution:Take random sample. • Max. degree decreases. • Will show that  remains large.

  29. Expansion properties of continuous graph

  30. The continuous graph Gc • Vertex set: all points in unit sphere Sd-1. • Edge set: (vi,vj) E iff <vi,vj> -1/2 = cos(120o) (corresponds to vector 3-coloring). • Use natural measure for subsets of V, E. measure = 1/2 120o

  31. Main theorem Let d = dimension of sphere,   (1-)d. Let A and B be two subsets of Gc of measure . Theorem: The measure of edges between A and B is at least 4|E|. • Two random subsets of measure  are expected to share 2|E| edges. A B

  32. Proof outline Theorem: Let A and B be two subsets of Gc of measure .The measure of edges between A and B is at least 4|E|. • Step 1: Subsets A, B which share the least measure of edges are caps (shifting). • Step 2: Analyze measure of edges between caps. A B

  33. Step 1: caps share few edges • Step 1: Subsets A, B which share the least measure of edges are caps (of same measure). Would like a shifting procedurethat converts any two sets A and B to caps while preserving measure and decreasing the amount of edges between A and B. • Use [BaernsteinTaylor] two point symmetrization procedure: • Choose arbitrary hyperplane. • Consider each point and its mirror image. • “Shift up” if possible. • Procedure converges into cap. A A*

  34. Measure of edges decreases At each step measure of edges between A and B does not increase. • Consider two vetrices and their mirror image. • Vertices may be in A or B in both or not in any. • Check number of edges before and after shifting. • Case analysis. • Step 1: OK A B

  35. Step 2: edges between caps • First show that caps A and B that share minimal measure of edges satisfy A = B. • Then compute measure of edges in cap. • Good estimates are known. A=B A B

  36. Theorem restated Let A and B be two subsets of Gc of measure  (  (1-)d). Theorem: The measure of edges between A and B is at least 4|E|.

  37. Continuous graph to discrete graph

  38. Discrete graph • Partition Gc into many small cells each of small diameter and equal measure. • Discrete graph Gd • V = cells. • E = pairs of cells which share edges in Gc. • Vector 3-colorable. • Inherits expansion properties of Gc. B A

  39. B A Theorem – discrete graph • Let Gd = (V,E) be the discrete graph. • Let A and B be two subsets of Gd of size |V| = n ( n1-). Theorem: The number of edges between A and B is at least 4|E|.

  40. Recall Goal: vector 3-colorable graphs with chromatic number at least 1/3 (k=3). Shown: Discrete graph Gd: Every two subsets A, B of size n n1- share many edges. •  IS of Gdis less than n  (Gd)  1/ n. Problem: d(max. degree in Gd) is very large, will not imply desired bounds. Solution: Take random sample R of Gd.

  41. Random Sample

  42. Expansion  bounds on  • Discrete graph Gd: • Nice expansion on sets of size |Gd|. • (Gd)  1/. • Random sample R: • (R)= (1/). • The smaller the sample the better (better relation  vs. ).

  43. Property testing [GGR] G is far from having property P small random sample of G will not have property P. • Use property testing on discrete graph to prove that small random sample has large . • Use property testing on discrete graph to prove that random sample does not have large IS. • Consider property P: “having large IS”. “having small ”.

  44. Property testing [GGR] G is far from having property P small random sample of G will not have property P. • Theorem [GoldreichGoldwasserRon]:Let G be a graph in which each subset of size n induces at least n2edges. W.h.p. a random subset R of size s/4will not have IS of size s  (R) > 1/. • Theorem[AlonKrivelevich]:Let G be a graph in which at least n2edges need to be removed in order to color G with 1/ colors. W.h.p. a random subset R of size s1/2will satisfy (R) > 1/.

  45. Naïve approach • Theorem [GoldreichGoldwasserRon]:Let G be a graph in which each subset of size n induces at least n2edges. W.h.p. a random subset R of size s/4will not have IS of size s  (R) > 1/. • Our case Gd satisfies n2=4|E|. • s/4does not suffice (for our proof). • Will yield graphs which are vector 3-colorable and have chromatic number 0.038 (<<1/3).

  46. Naïve approach #2 • Theorem[AlonKrivelevich]:Let G be a graph in which at least n2edges need to be removed in order to color G with 1/ colors. W.h.p. a random subset R of size s1/2will satisfy (R) > 1/. • Can prove: Gd satisfies n2=3|E|. • s1/2does not suffice (for our proof). • Will yield graphs which are vector 3-colorable and have chromatic number 0.087. • Need s to be much smaller (s  /).

  47. Main theorem Let G be a graph in which each two subsets A and B of size n share at least n2edges (as Gd). Theorem: W.h.p. a random subset R of size s/ will satisfy (R) = (1/). Theorem: W.h.p. a random subset R of size s/ will satisfy (R) = O(s). • Properties of our graphs G are stronger.

  48. Proof outline: (R) = O(|R|) • Use ideas appearing in [GGR,AK]. • Let R be random sample of G. • Goal: Every subset of size  |R| has at least one edge. • Consider one such subset. • Choose vertices one by one. • Each vertex defines set of neighbors (forbidden vertices). • Once set of neighboors is very large, few additional vertices suffice. • We show that properties of G imply set of neighbors grows fast  |R| is small. R x x x x x x x x x x

  49. Additional result Applying our proof tech. (extension of [AK]) on graphs G with properties as defined in [GGR] yields improved results. Theorem [GGR]:Let G be a graph in which each subset of size n induces at least n2edges. W.h.p. a random subset of size s/4will not have IS of size s. Theorem:Let G be a graph in which each subset of size n induces at least n2edges. W.h.p. a random subset of size s4/3will not have IS of size s.

  50. Putting things together • Gd: Every two subsets A and B of size n share n2 edges. •   4|E|/n2  4(/n). • R  Gdis of size s  /  (n/)1/3satisfies: • (R)  s  (R)  1/. • R  (/n)s  1/3. (R)  (R)1/3

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