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Partitioning Sets to Decrease the Diameter

Partitioning Sets to Decrease the Diameter . N. Harvey U. Waterloo. TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: A A A A A A A. Act 1: Problem Statement & History Act 2: The Solution. Act 1: Problem Statement & History. =max.

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Partitioning Sets to Decrease the Diameter

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  1. Partitioning Setsto Decrease the Diameter N. Harvey U. Waterloo TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AAAAAAA

  2. Act 1: Problem Statement & History Act 2: The Solution

  3. Act 1: Problem Statement & History =max • Def: For any S½Rd, diam(S) =supx,y2Skx-yk2 • Question: For every S½Rd, can we partitionS into S1[S2[ [Sd+1 such thatdiam(Si) < diam(S) for all i=1,...,d+1 • Remarks: • Only makes sense if 0 < diam(S) < 1. • For simplicity, we’ll assume S is closed. Euclidean Norm:

  4. Our Question with d=1 • Question: For every S½Rd, can we partitionS into S1[S2[ [Sd+1 such thatdiam(Si) < diam(S) for all i=1,...,d+1 S d=1: Midpoint m = (min S + max S) / 2 = 0.5 0 1 2 3 -1 -2

  5. Our Question with d=1 • Question: For every S½Rd, can we partitionS into S1[S2[ [Sd+1 such thatdiam(Si) < diam(S) for all i=1,...,d+1 S Midpoint m = (min S + max S) / 2 = 0.5 0 1 2 3 -1 -2 • Let S1 = { x : x2S and x·m }Let S2 = { x : x2S and x>m } • Clearly diam(Si) ·diam(S) / 2

  6. Our Question with d=2 • Question: For every S½Rd, can we partitionS into S1[S2[ [Sd+1 such thatdiam(Si) < diam(S) for all i=1,...,d+1 Idea:Take minimum enclosing ball Partition into 90±segments Can argue diam(Si) < diam(S) S1 S2 S S3 S4

  7. Our Question with d=2 • Question: For every S½Rd, can we partitionS into S1[S2[ [Sd+1 such thatdiam(Si) < diam(S) for all i=1,...,d+1 Improved Idea:Take minimum enclosing ball Partition into 120±segments S

  8. Our Question with d=2 • Question: For every S½Rd, can we partitionS into S1[S2[ [Sd+1 such thatdiam(Si) < diam(S) for all i=1,...,d+1 • Improved Idea:Take minimum enclosing ball • Partition into 120±segments • Is diam(Si) < diam(S)? S1 S S2 S3

  9. Our Question with d=2 • Question: For every S½Rd, can we partitionS into S1[S2[ [Sd+1 such thatdiam(Si) < diam(S) for all i=1,...,d+1 • Idea:Take minimum enclosing ball • Partition into 120±segments • Is diam(Si) < diam(S)? • Yes, unless S ¼equi. triangle S S1 S2 S3

  10. Our Question with d=2 • Question: For every S½Rd, can we partitionS into S1[S2[ [Sd+1 such thatdiam(Si) < diam(S) for all i=1,...,d+1 • Idea:Take minimum enclosing ball • Partition into 120±segments • Is diam(Si) < diam(S)? • Yes, unless S ¼equi. triangle • In this case, just rotate the partitioning. S1 S S3 S2

  11. Our Question with Parameter f • Notation: f(d) = smallest integer such thatevery S½Rd can be partitioned intoS1[S2[[Sf(d) with diam(Si) < diam(S). • Our Question: Is f(d) = d+1 for all d? • What do we know so far?

  12. Lower Bound on f • Notation: f(d) = smallest integer such thatevery S½Rd can be partitioned intoS1[S2[[Sf(d) with diam(Si) < diam(S). • Trivial LB: Obviously f(d)¸ 2 for every d. Partitioning into 1 set cannot decrease diam! • What do we know so far? Optimal!

  13. Lower Bound on f • Notation: f(d) = smallest integer such thatevery S½Rd can be partitioned intoS1[S2[[Sf(d) with diam(Si) < diam(S). • Another LB: For d=2, the hard case is whenS = equilateral triangle. • The 3 corner points have pairwise distance D,where D = diam(S). • To get diam(Si) < diam(S), these 3 corner points must lie in different Si’s. • So f(2)¸ 3! S1 D D S2 S3 D

  14. Lower Bound on f • Notation: f(d) = smallest integer such thatevery S½Rd can be partitioned intoS1[S2[[Sf(d) with diam(Si) < diam(S). • What do we know so far? Optimal!

  15. Generalize to d>2? • For d=2, the hard case is an equilateral triangle • Corner points have distance D,where D = diam(S) • What is a d-dimensional analog? D D D D D D D Simplex(Generalized Tetrahedron) D

  16. Generalize to d>2? • For d=2, the hard case is an equilateral triangle • Corner points have distance D,where D = diam(S) • What is a d-dimensional analog? • A Simplex: The convex hull of d+1 points with pairwise distances all equal to D • To get diam(Si) < diam(S), these d+1 pointsmust lie in different Si’s. • So f(d)¸d+1! D D D D D

  17. Lower Bound on f • Notation: f(d) = smallest integer such thatevery S½Rd can be partitioned intoS1[S2[[Sf(d) with diam(Si) < diam(S). • What do we know so far? Optimal!

  18. Generalize Upper Bound to d>2? Idea:Take minimum enclosing ball Partition using orthants Can argue diam(Si) < diam(S) # orthants is 2d ) f(d) ·2d S S1 S5 S3 S6 S2 S4

  19. History • Conjecture [Borsuk ‘33]:f(d)=d+1. Karol Borsuk, 1905-1982Field: Topology

  20. History • Conjecture [Borsuk ‘33]:f(d)=d+1. • Special Cases: • f(d)=d+1 when S = d-dimensional sphere [Borsuk ‘33] • f(d)=d+1 for all smooth sets S [Borsuk ‘33] Open! Huge Gap!

  21. History • Conjecture (Borsuk 1933):f(d)=d+1. • Theorem (Kahn-Kalai 1993): • Corollary: Borsuk’s conjecture is false! [Kahn-Kalai ‘93]

  22. History • Conjecture [Borsuk ‘33]:f(d)=d+1. • Theorem [Kahn-Kalai ‘93]: • Corollary: Borsuk’s conjecture is false (for d¸1325). • Improvements: False for d¸946 [Nilli ’94], ...,d¸298 [Hinrichs & Richter ‘02]. [Kahn-Kalai ‘93]

  23. Intermission A Little Story

  24. Act 2: Kahn & Kalai’s Theorem • Definition: f(d) = smallest integer such thatevery S½Rd (with 0 < diam(S) < 1)can be partitioned into S1[S2[[Sf(d)with diam(Si) < diam(S)8i. • Theorem: , for large enough d. • Corollary:Borsuk’s conjecture is false.

  25. How could Kahn-Kalai prove this? • Every smooth set satisfies f(d)=d+1, sof(d) can only be large for non-smooth sets • What’s a very non-smooth set? A finite set! • Can we reduce Borsuk’s problem for finite setsto a nice combinatorics problem? Smooth SetBorsuk’s Conjecture is True Finite SetBorsuk’s Conjecture is False?

  26. H1 011 111 H0 010 110 • Natural idea: look at {0,1}d instead of Rd • Trick of the trade: Look at points with same # of 1’s • Notation:Hk = { x : x 2 {0,1}d, x has k 1’s } • Each point x2Hk corresponds to a set X½{1,2,...,d}with |X|=k, i.e., X = { i : xi=1 } H3 001 101 000 100 H2 [d] Notation:

  27. Combinatorial Setup • Hk = { x : x 2 {0,1}d, x has k 1’s } • Each point x2Hk corresponds to a set X½{1,2,...,d}with |X|=k, i.e., X = { i : xi=1 } • Let x,y2Hk. Let X,Yµ[d] be corresponding sets • Note: We don’t care about actual distances, just whether diameter shrinks or not. It is more convenient to use “modified distances”:(this is L1-norm / 2) [d]

  28. Combinatorial Setup • Points x,y2 Hk correspond to sets X,Yµ[d], |X|=|Y|=k • A set of pointsS = {x1, x2, ...} µHk corresponds toa family of sets {X1, X2, ...}, where each Xiµ[d] and |Xi|=k • Main Goal: find S µHk½ {0,1}d and p ¸ d+1 such that,letting , For all partitions S = S1[[ Spsome Si contains X, Y with k-|X \ Y|=D Get rid of k bysetting M = k-D

  29. Combinatorial Setup • Points x,y2 Hk correspond to sets X,Yµ[d], |X|=|Y|=k • A set of pointsS = {x1, x2, ...} µHk corresponds toa family of sets {X1, X2, ...}, where each Xiµ[d] and |Xi|=k • Main Goal: find S µHk½ {0,1}d and p ¸ d+1 such that,letting , For all partitions S = S1[[ Spsome Si contains X, Y with |X \ Y|=M

  30. Aiming at the Goal • Main Goal: find S µHk½ {0,1}d and p ¸ d+1 such that letting , For all partitions S = S1[[ Spsome Si contains X, Y with |X \ Y|=M • How to argue about all partitions of S? No structure... • By Pigeonhole Principle, one Si is “large”: |Si| ¸ |S|/p • Need a theorem saying “Every large family must contain two sets with a given intersection size” • Such questions are the topic of “Extremal Set Theory”

  31. (Extremal Set) Theory“How many subsets of a finite set are there with a Property X?” • Extreme (Set Theory) The Reals areUncountable!

  32. Looking for a Hammer • Need a Theorem saying “Every large family must contain two sets with a given intersection size” • My favorite book on Extremal Set Theory:

  33. Looking for a Hammer • Need a Theorem saying “Every large family must contain two sets with a given intersection size” • Theorem [Frankl-Rodl]: Let n be a multiple of 4.Let B={B1, B2, ...} be a family of subsets of [n] with • |Bi| = n/2 8i • |BiÅBj|  n/4 8ij Then |B| ·1.99n. • Contrapositive: Let B={B1, B2, ...} be a family of subsets of [n] with each |Bi|=n/2. If |B|>1.99n then 9Bi, Bj2B with |BiÅBj| = n/4.

  34. Goal & Hammer • Goal: find S µHk½ {0,1}d and p ¸ d+1 such that letting , For all partitions S = S1[[ Spsome Si contains X, Y with |X Å Y|=M • Hammer: Let B={B1, B2, ...} be a family of subsets of [n]with |Bi|=n/2. If |B|>1.99n then 9Bi,Bj2Bs.t. |BiÅBj|=n/4. • How to connect Goal & Hammer? • Not obvious... • Kahn & Kalai need a clever idea... (fortunately they are very clever)

  35. Kn = (V,E) is complete graph; |V|=n is a multiple of 4 • Let B = { B : B½V and |B|=n/2 } (“Bisections”) • For any UµV, let ±(U) = { uv : u2U and vU } (“Cut”) • Let S = { ±(B) : B2B } (“Bisection Cuts”) • For any ±(B), ±(B’)2S, we have |±(B) ű(B’)| ¸n2/8Equality holds iff |B Å B’|=n/4 {a,c} n = 4 a c B = { {a,b}, {a,c}, {a,d}, {b,c}, {b,d}, {c,d} } ±({a,b}) ={ac,ad,bc,bd} ±({a,c}) ={ab,ad,bc,cd} ±({a,b}) ={ac,ad,bc,bd} ±({a,c}) ={ab,ad,bc,cd} S = { ±({a,b}), ±({a,c}), ±({a,d}), ... } = { {ac,ad,bc,bd}, {ab,ad,bc,cd}, ... } S = { ±({a,b}), ±({a,c}), ±({a,d}), ... } = { {ac,ad,bc,bd}, {ab,ad,bc,cd}, ... } b d U = {a,b}

  36. Kn = (V,E) is complete graph; |V|=n is a multiple of 4 • Let B = { B : B½V and |B|=n/2 } (“Bisections”) • For any UµV, let ±(U) = { uv : u2U and vU } (“Cut”) • Let S = { ±(B) : B2B } (“Bisection Cuts”) • For any ±(B), ±(B’)2S, we have |±(B) ű(B’)| ¸n2/8Equality holds iff |B Å B’|=n/4 • Consider a partition S = S1[ ... [Spand corresponding partitionB = B1[ ... [Bp • If p·1.005n then 9i s.t. • Hammer:9B, B’2Bisuch that |BÅB’|=n/4 )±(B), ±(B’)2Si and |±(B) ű(B’)| = n2/8 • Goal Solved:

  37. Conclusion • Borsuk’s Conjecture: For every S½Rd, we can partition S into S1[S2[ [Sd+1 such thatdiam(Si) < diam(S) for all i=1,...,d+1 • Kahn-Kalai: No! There exists S½Rd and s.t. for every partition S into S1[S2[ [Sp9i with diam(Si) = diam(S). Borsuk Kahn Kalai

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