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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 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 • 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.
Maximum independent set • IS: Set of vertices that do not share any edges. G=(V,E) (G)=3 • (G): Size of maximum IS.
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.
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 !
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.
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 NPis in random polynomial time). [Hastad, FeigeKilian ,EngebretsenHolmerin, Khot]. Relatively small gap.
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.
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.
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)
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
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
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
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)
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.
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
[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].
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.
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).
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).
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
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
Main theorem • G is vector 3-colorable (by definition). • G has chromatic number 1/3.
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.
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
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
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.
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
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
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
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*
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
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
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|.
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
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|.
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.
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. ).
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 ”.
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 s1/2will satisfy (R) > 1/.
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).
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 s1/2will satisfy (R) > 1/. • Can prove: Gd satisfies n2=3|E|. • s1/2does 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 /).
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.
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
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 s4/3will not have IS of size s.
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