1. 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. 2 Two fundamental NP-Hard problems Minimum Coloring Minimum Coloring Maximum Independent Set Maximum Independent Set

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

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

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

6. 6 Approximation algorithms Not likely to find efficient algorithms. Not likely to find efficient algorithms. Settle on efficient approximation 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. 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: App. ratio of algorithm ALG: r = ALG/OPT (min.), r = OPT/ALG (max.). r  1, the smaller the better ! r  1, the smaller the better !

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

8. 8 Approximating  (G) &  (G) Good news: Good news: Both  (G) and  (G) can be app. within ratio n(loglog n) 2 /(log n) 3 [Haldorsson, Feige]. Both  (G) and  (G) can be app. within ratio n(loglog n) 2 /(log n) 3 [Haldorsson, Feige]. Bad News: Bad News: Estimating both  (G) and  (G) up to a factor of n 1-  is “hard” (unless NP  is in random polynomial time). Estimating both  (G) and  (G) up to a factor of n 1-  is “hard” (unless NP  is in random polynomial time). [Hastad, FeigeKilian,EngebretsenHolmerin, Khot]. [Hastad, FeigeKilian,EngebretsenHolmerin, Khot]. Relatively small gap.

9. 9 What about restricted cases? Consider a graph G that is known to have small chromatic number  (G)=k. Good news: Good news: Can efficiently find coloring with: Can efficiently find coloring with: k=3  n 3/14 colors [KargerMotwaniSudan, BlumKarger]. k=3  n 3/14 colors [KargerMotwaniSudan, BlumKarger]. k=4  n 7/19 colors [HalperinNathanielZwick]. k=4  n 7/19 colors [HalperinNathanielZwick]. k  n f(k) colors (f(k)  1 as k increases) [KMS, HNZ]. k  n f(k) colors (f(k)  1 as k increases) [KMS, HNZ]. Bad news [KhannaLinialSafra, GuruswamiKhanna, Khot]: Bad news [KhannaLinialSafra, GuruswamiKhanna, Khot]: NP-hard to color 3 colorable graphs with 4 colors. NP-hard to color 3 colorable graphs with 4 colors. NP-hard to color k col. graphs with  5k/3, k  (log k) colors. NP-hard to color k col. graphs with  5k/3, k  (log k) colors. Gap is wide open.

10. 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. 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   -1/(k-1) = -1/2 = cos(120 o ) k=11   -1/(k-1) = -1/10  cos(95 o ) 120 o 95 o

12. 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(120 o ) R2R2

13. 13 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 corr. to adjacent vertices have inner product at most -1/(k-1). Vector 4-coloring: k=4  -1/(k-1) = -1/3  cos(109 o ) R3R3

14. 14 Vector coloring   (G) Every k colorable graph is also vector k-col. Every k colorable graph is also vector k-col. Identify each color class with one vertex Identify each color class with one vertex in a perfect k-1 dimensional simplex. in a perfect k-1 dimensional simplex. k = 4: k = 4: R3R3

15. 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.  for each edge (i,j)  E.  for each edge (i,j)  E. = 1 for each node i  V. = 1 for each node i  V. = -1/(k-1) = -1/(k-1)

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

17. 17 Algorithm of [KMS] cont. 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. Objective: find large IS in G. Vector 3-coloring

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

19. 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. [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. Will neglect  in remainder of talk.

20. 20 Previous work on limitation of vector coloring. As a function on n rather than . Vector 3-colorable graphs with   n 0.05 [KMS,Alon,Szegedy]. Vector 3-colorable graphs with   n 0.05 [KMS,Alon,Szegedy]. Vector k-colorable graphs with   n f(k) where f(k)  1 [KMS,Charikar,Feige]. Vector k-colorable graphs with   n f(k) where f(k)  1 [KMS,Charikar,Feige]. Our results: For k=3 we obtain   n 0.15. For k=3 we obtain   n 0.15. For other k, improve f(k). For other k, improve f(k).

21. 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]. There are graphs for which ratio between  (G) and vector chromatic number  n/2 [Feige]. G is 2 vec. col.  (G)  n/2 large gap n/polylog(n). Our results improve gap to n/polylog(n). (k=log(n)/loglog(n)). (k=log(n)/loglog(n)). Vector coloring does not app.  within factor better than n/polylog(n). O(log 1/2 (n))

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

23. 23 Our work Use different graphs. Use different graphs. We use graphs presented in that addresses a SDP relaxation of the Max-Cut problem [GoemansWilliamson]. 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). 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. 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. 120 o

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

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

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

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

28. 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. Vector 3-colorable. Large chromatic number. Large chromatic number. Problem: Discrete graph has large degree (n 1-  ), can not show    1/3. Solution: Take random sample. Max. degree decreases. Max. degree decreases. Will show that  remains large. Will show that  remains large.

29. Expansion properties of continuous graph

30. 30 The continuous graph G c Vertex set: all points in unit sphere S d-1. Vertex set: all points in unit sphere S d-1. Edge set: (v i,v j )  E iff  -1/2 = cos(120 o ) (corresponds to vector 3-coloring). Edge set: (v i,v j )  E iff  -1/2 = cos(120 o ) (corresponds to vector 3-coloring). Use natural measure for subsets of V, E. Use natural measure for subsets of V, E. 120 o measure = 1/2

31. 31 Main theorem Let d = dimension of sphere,   (1-  ) d. Let A and B be two subsets of G c 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. Two random subsets of measure  are expected to share  2 |E| edges. A B

32. 32 Proof outline Theorem: Let A and B be two subsets of G c 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 1: Subsets A, B which share the least measure of edges are caps (shifting). Step 2: Analyze measure of edges between caps. Step 2: Analyze measure of edges between caps. A B

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

34. 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. 35 Step 2: edges between caps First show that caps A and B that share minimal measure of edges satisfy A = B. First show that caps A and B that share minimal measure of edges satisfy A = B. Then compute measure of edges in cap. Then compute measure of edges in cap. Good estimates are known. Good estimates are known. A B A=B

36. 36 Theorem restated Let A and B be two subsets of G c 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. 38 Discrete graph Partition G c into many small cells each of small diameter and equal measure. Partition G c into many small cells each of small diameter and equal measure. Discrete graph G d Discrete graph G d V = cells. V = cells. E = pairs of cells which E = pairs of cells which share edges in G c. share edges in G c. Vector 3-colorable. Vector 3-colorable. Inherits expansion properties of G c. Inherits expansion properties of G c. A B

39. 39 Theorem – discrete graph Let G d = (V,E) be the discrete graph. Let G d = (V,E) be the discrete graph. Let A and B be two subsets of G d of size Let A and B be two subsets of G d of size  |V| =  n (  n 1-  ).  |V| =  n (  n 1-  ). Theorem: The number of edges between A and B is at least  4 |E|. A B

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

41. Random Sample

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

43. 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 small random sample has large . Use property testing on discrete graph to prove that random sample does not have large IS. Use property testing on discrete graph to prove that random sample does not have large IS. Consider property P: “having large IS”. Consider property P: “having large IS”. “having small  ”. “having small  ”.

44. 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  n 2 edges. W.h.p. a random subset R of size s  /  4 will not have IS of size  s   (R) > 1/ . Theorem [GoldreichGoldwasserRon] : Let G be a graph in which each subset of size  n induces at least  n 2 edges. W.h.p. a random subset R of size s  /  4 will not have IS of size  s   (R) > 1/ . Theorem [AlonKrivelevich]: Let G be a graph in which at least  n 2 edges 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/ . Theorem [AlonKrivelevich]: Let G be a graph in which at least  n 2 edges 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. 45 Naïve approach Theorem [GoldreichGoldwasserRon] : Let G be a graph in which each subset of size  n induces at least  n 2 edges. W.h.p. a random subset R of size s  /  4 will not have IS of size  s   (R) > 1/ . Theorem [GoldreichGoldwasserRon] : Let G be a graph in which each subset of size  n induces at least  n 2 edges. W.h.p. a random subset R of size s  /  4 will not have IS of size  s   (R) > 1/ . Our case G d satisfies  n 2 =  4 |E|. Our case G d satisfies  n 2 =  4 |E|. s  /  4 does not suffice (for our proof). s  /  4 does not suffice (for our proof). Will yield graphs which are vector 3-colorable and have chromatic number  0.038 (<<  1/3 ). Will yield graphs which are vector 3-colorable and have chromatic number  0.038 (<<  1/3 ).

46. 46 Naïve approach #2 Theorem [AlonKrivelevich]: Let G be a graph in which at least  n 2 edges 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/ . Theorem [AlonKrivelevich]: Let G be a graph in which at least  n 2 edges 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: G d satisfies  n 2 =  3 |E|. Can prove: G d satisfies  n 2 =  3 |E|. s  1/  2  does not suffice (for our proof). s  1/  2  does not suffice (for our proof). Will yield graphs which are vector 3-colorable and have chromatic number  0.087. Will yield graphs which are vector 3-colorable and have chromatic number  0.087. Need s to be much smaller (s   /  ). Need s to be much smaller (s   /  ).

47. 47 Main theorem Let G be a graph in which each two subsets A and B of size  n share at least  n 2 edges (as G d ). 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. Properties of our graphs G are stronger.

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

49. 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  n 2 edges. W.h.p. a random subset of size s  /  4 will not have IS of size  s. Theorem: Let G be a graph in which each subset of size  n induces at least  n 2 edges. W.h.p. a random subset of size s  4 /  3 will not have IS of size  s.

50. 50 Putting things together G d : Every two subsets A and B of size  n share  n 2 edges. G d : Every two subsets A and B of size  n share  n 2 edges.    4 |E|/n 2   4 (  /n).    4 |E|/n 2   4 (  /n). R  G d is of size s   /   (n/  )1/  3 satisfies: R  G d is of size s   /   (n/  )1/  3 satisfies:  (R)   s   (R)  1/ .  (R)   s   (R)  1/ .  R  (  /n)s  1/  3.  R  (  /n)s  1/  3.  (R)  (  R ) 1/3

51. 51 Concluding remarks Present tight bounds on the chromatic number of vector k-colorable graphs. Present tight bounds on the chromatic number of vector k-colorable graphs. Open problems: Open problems: Consider stronger relaxations (strict vector coloring  Lovasz  function). Consider stronger relaxations (strict vector coloring  Lovasz  function). Do they improve [KMS] ? Do they improve [KMS] ? Do our negative results extend ? Do our negative results extend ? Prove similar expansion on “hypercube”. Prove similar expansion on “hypercube”. Further improve sample size in theorem of [GGR] (property testing framework) to  1/  2. Further improve sample size in theorem of [GGR] (property testing framework) to  1/  2. Thank you.

52. Expansion properties of continuous graph

53. 53 Main theorem Let A and B be two subsets of G c of measure  (  < 1). 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. Two random subsets of measure  are expected to share  2 |E| edges. A B

54. 54 Symmetrization procedure [BT] Choose arbitrary hyperplane. Choose arbitrary hyperplane. Consider each point x and its Consider each point x and its mirror image. mirror image. Shift up if possible. Shift up if possible. Procedure converges into cap. Procedure converges into cap.

55. 55 Proof sketch Definition: E(A,B) = measure of edges between A and B. Definition:  2 = all pairs of closed subsets in S d-1 (compact w.r.t. Hausdorff metric).  Theorem: If A and B are of measure  then E(A,B) ≥ E(C,C) where C is a cap of measure . Proof: Let 2 be all pairs ( ,  )  2 that satisfy  (A) =  (  ),  (B) =  (  ).  (A) =  (  ),  (B) =  (  ).  (A  ) ≥  (   ),  (B  ) ≥  (   ).  (A  ) ≥  (   ),  (B  ) ≥  (   ). E(A,B) ≥ E( ,  ). E(A,B) ≥ E( ,  ). Will show that (C,C)  2.

56. 56 Proof outline Let 2 be all pairs ( ,  )  2 that satisfy  (A) =  (  ),  (B) =  (  ).  (A) =  (  ),  (B) =  (  ).  (A  ) ≥  (   ),  (B  ) ≥  (   ).  (A  ) ≥  (   ),  (B  ) ≥  (   ). E(A,B) ≥ E( ,  ). E(A,B) ≥ E( ,  ). Claim: (C,C)  2 : 2 is closed under symmetrization. 2 is closed under symmetrization. 2 is a closed subset of  2. 2 is a closed subset of  2. (C,C)  2 (two steps: first show (C,*)  2 ). (C,C)  2 (two steps: first show (C,*)  2 ).

57. Property testing

58. 58 Main theorem Let G be a graph in which each subsets A and B of size  n share at least  n 2 edges. Theorem: W.h.p. a random subset R of size s  /  will satisfy  (R) < 2  s. Will show: Let G be a graph in which each subset A size  n induces at least  n 2 edges. Theorem: W.h.p. a random subset R of size s  1/  will satisfy  (R) < 2  s. Proof used ideas from [AlonKrivelevich].

59. 59 Proof outline:  (R) < 2  s Let R be random sample of G. Let R be random sample of G. Goal: Every subset of size 2  s Goal: Every subset of size 2  s has at least one edge. has at least one edge. Consider one such subset. Consider one such subset. Choose vertices one by one. Choose vertices one by one. Each vertex defines set of neighbors Each vertex defines set of neighbors (forbidden vertices). (forbidden vertices). Once set of neighboors is very large, few additional vertices suffice. Once set of neighboors is very large, few additional vertices suffice. We show that properties of G imply set of neighbors grows fast  s is small. We show that properties of G imply set of neighbors grows fast  s is small. R x x x xx x x x x x

60. 60 Notation Every subset in G has few vertices of low degree. Definition: For each subset I of G, let N(I) be the vertices adjacent to a vertex in I and F(I) be the remaining vertices. Definition: A vertex v is GOOD w.r.t. a subset I if v  N(I). v  N(I). v  F(I), and F(I) >  n, and v has at least (  /  )n neighbors in F(I). v  F(I), and F(I) >  n, and v has at least (  /  )n neighbors in F(I). Lemma: For every small I, the probability that a random vertex in V-I is GOOD (w.r.t. I) is at least 1- .

61. 61 Tree of independent sets Consider the tree defined by choosing R={r 1,…,r s } one by one. Node is indexed by a couple (X,Y), where X and Y are subsets of R. If X is an IS then node is OPEN, otherwise CLOSED. Root = ( ,  ) (OPEN). Root = ( ,  ) (OPEN). Inductively define: Inductively define: (X,Y) is OPEN and r i  R is GOOD w.r.t. X: (X,Y) is OPEN and r i  R is GOOD w.r.t. X: If |Y|<s-2  s define two sons: (X  {r i },Y) and (X, Y  {r i }). If |Y|<s-2  s define two sons: (X  {r i },Y) and (X, Y  {r i }). Otherwise define one son (X  {r i },Y). Otherwise define one son (X  {r i },Y). (X,Y) is CLOSED, do nothing. (X,Y) is CLOSED, do nothing. Properties: for each node X and Y are disjoint, |Y|  s-2  s.

62. 62 Properties of tree Depth: bounded by Depth: bounded by s - 2  s +  / . s - 2  s +  / . All leaves CLOSED  All leaves CLOSED  no IS of size 2  s. no IS of size 2  s. For s  1/ , w.h.p. all leaves are CLOSED. For s  1/ , w.h.p. all leaves are CLOSED. … …… … (,)(,)(,)(,) ({r 1 },  ) ( ,{r 1 }) ( ,{r 1,r 2 }) ({r 1,r 3 },  ) (X,Y) ({r 2 },{r 1 })

63. 63 Proof sketch Claim: For s  1/ , w.h.p. all leaves are CLOSED. Consider potential leaf (left right choices). Consider potential leaf (left right choices). Compute the probability that the path Compute the probability that the path remains OPEN after s random vertices. remains OPEN after s random vertices. The path is surely CLOSED if we hit s-2  s+  /  GOOD vertices out of the possible s. The path is surely CLOSED if we hit s-2  s+  /  GOOD vertices out of the possible s. Pr[path OPEN]  Pr[did not hit s-2  s+  /  ]. Pr[path OPEN]  Pr[did not hit s-2  s+  /  ]. This probability is << 1/#of leaves if s is  1/ . This probability is << 1/#of leaves if s is  1/ . … …… … (,)(,)(,)(,) (X,Y)