Partitioning Sets to Decrease the Diameter

Partitioning Setsto Decrease the Diameter

N. HarveyU. Waterloo

Act 1: Problem Statement & History

Act 2: The Solution

Act 1: Problem Statement & History• Def: For any S½Rd, diam(S) = supx,y2S kx-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:

=max

Our Question with d=1• Question: For every S½Rd, can we partition

S into S1 [ S2 [ [ Sd+1 such thatdiam(Si) < diam(S) for all i=1,...,d+1

0 1 2 3-1-2

S

Midpoint m = (min S + max S) / 2 = 0.5

d=1:



0 1 2 3-1-2

S

Midpoint m = (min S + max S) / 2 = 0.5• Let S1 = { x : x2S and x·m }

Let S2 = { x : x2S and x>m }

• Clearly diam(Si) · diam(S) / 2



S

Idea:Take minimum enclosing ballPartition into 90± segmentsCan argue diam(Si) < diam(S)

S1 S2

S3 S4



S

Improved Idea:Take minimum enclosing ballPartition into 120± segments



S

Improved Idea:Take minimum enclosing ballPartition into 120± segmentsIs diam(Si) < diam(S)?S1 S2

S3



S

Idea:Take minimum enclosing ballPartition into 120± segmentsIs diam(Si) < diam(S)?Yes, unless S ¼ equi. triangle

S1 S2

S3



S

Idea:Take minimum enclosing ballPartition into 120± segmentsIs diam(Si) < diam(S)?Yes, unless S ¼ equi. triangleIn this case, just rotate the partitioning.

S1

S2S3

Our Question with Parameter f• Notation: f(d) = smallest integer such that

every 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?d Lower Bound on f(d) Upper Bound on f(d)1 ? 22 ? 3

Lower Bound on f• Notation: f(d) = smallest integer such that


• Trivial LB: Obviously f(d) ¸ 2 for every d. Partitioning into 1 set cannot decrease diam!

• What do we know so far?d Lower Bound on f(d) Upper Bound on f(d)1 2 22 2 3

Optimal!



• Another LB: For d=2, the hard case is when S = 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

S2S3

D

DD



• What do we know so far?d Lower Bound on f(d) Upper Bound on f(d)1 2 22 3 3

Optimal!

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

DD

Simplex(Generalized Tetrahedron)

D

DD

DD

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 points

must lie in different Si’s.• So f(d) ¸ d+1!

D

DD

DD



• What do we know so far?d Lower Bound on f(d) Upper Bound on f(d)1 2 22 3 3d d+1 ?

Optimal!

Generalize Upper Bound to d>2?

SIdea:Take minimum enclosing ballPartition using orthantsCan argue diam(Si) < diam(S)

# orthants is 2d

) f(d) · 2d

S1

S2

S3

S4

S5

S6

d Lower Bound on f(d) Upper Bound on f(d)1 2 22 3 3d d+1 2d

History• Conjecture [Borsuk ‘33]: f(d)=d+1.

Karol Borsuk, 1905-1982Field: Topology

History

d Lower Bound on f(d) Upper Bound on f(d)1 2 22 3 3 [Borsuk ‘33]

3 4 4 [Perkal-Eggleston ‘47-’55]

4 5 9 [Lassak ‘82]

d d+1 [Borsuk ‘33] 2d [Knast ‘74]1.23d [Schramm ‘88]

• Special Cases:– f(d)=d+1 when S = d-dimensional sphere [Borsuk ‘33]

– f(d)=d+1 for all smooth sets S [Borsuk ‘33]

• Conjecture [Borsuk ‘33]: f(d)=d+1.

Open!

Huge Gap!

100 101 978388059150 151 30603191218909200 201 957243195505543289

History



4 5 9 [Lassak ‘82]


• Theorem (Kahn-Kalai 1993):• Corollary: Borsuk’s conjecture is false!

• Conjecture (Borsuk 1933): f(d)=d+1.

1:2p d[Kahn-Kalai ‘93]

History




• 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].

• Conjecture [Borsuk ‘33]: f(d)=d+1.

1:2p d[Kahn-Kalai ‘93]

Intermission

A Little Story

http://gilkalai.wordpress.com/2009/06/22/a-little-story-regarding-borsuks-conjecture/

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) 8 i.

• Theorem: , for large enough d.• Corollary: Borsuk’s conjecture is false.

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?

• 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 }

000

[d]Notation:

H0

H1

H2

H3

100

010

001 101

011

110

111

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]

Combinatorial Setup• Points x,y 2 Hk correspond to sets X,Yµ[d], |X|=|Y|=k

• A set of points S = {x1, x2, ...} µ Hk corresponds toa family of sets {X1, X2, ...}, where each Xiµ[d] and |Xi|=k

•

• Main Goal: find S µ Hk ½ {0,1}d and p ¸ d+1 such that,letting ,

For all partitions S = S1 [ [ Sp

some Si contains X, Y with k-|X \ Y|=D

Get rid of k bysetting M = k-D

Combinatorial Setup• Points x,y 2 Hk correspond to sets X,Yµ[d], |X|=|Y|=k

• A set of points S = {x1, x2, ...} µ Hk corresponds toa family of sets {X1, X2, ...}, where each Xiµ[d] and |Xi|=k

•

• Main Goal: find S µ Hk ½ {0,1}d and p ¸ d+1 such that,letting ,


some Si contains X, Y with |X \ Y|=M

• Main Goal: find S µ Hk ½ {0,1}d and p ¸ d+1 such that letting ,


some 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”

Aiming at the Goal

• (Extremal Set) Theory“How many subsets of a finite set are there with a Property X?”

• Extreme (Set Theory)

The Reals areUncountable!

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:

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

– |Bi Å Bj| n/4 8ijThen |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, Bj 2 B with |Bi Å Bj| = n/4.

• Goal: find S µ Hk ½ {0,1}d and p ¸ d+1 such that letting ,


some 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,Bj2B s.t. |BiÅBj|=n/4.

• How to connect Goal & Hammer?– Not obvious...– Kahn & Kalai need a clever idea... (fortunately they are very clever)

Goal & Hammer

• 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 vU } (“Cut”)• Let S = { ±(B) : B2B } (“Bisection Cuts”)• For any ±(B), ±(B’)2S, we have |±(B) Å ±(B’)| ¸ n2/8

Equality holds iff |B Å B’|=n/4

a

db

c B = { {a,b}, {a,c}, {a,d}, {b,c}, {b,d}, {c,d} }n = 4

U = {a,b}

S = { ±({a,b}), ±({a,c}), ±({a,d}), ... } = { {ac,ad,bc,bd}, {ab,ad,bc,cd}, ... }

±({a,b}) = {ac,ad,bc,bd} ±({a,c}) = {ab,ad,bc,cd}

{a,c}

±({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}, ... }

• 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 vU }

(“Cut”)• Let S = { ±(B) : B2B } (“Bisection Cuts”)• For any ±(B), ±(B’)2S, we have |±(B) Å ±(B’)| ¸ n2/8

Equality holds iff |B Å B’|=n/4• Consider a partition S = S1 [ ... [ Sp

and corresponding partition B = B1 [ ... [ Bp

• If p·1.005n then 9i s.t.• Hammer: 9B, B’ 2 Bi such that |BÅB’|=n/4 ) ±(B), ±(B’)2Si and |±(B) Å ±(B’)| = n2/8• Goal Solved:

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 [ [ Sp 9i with diam(Si) = diam(S).

Borsuk Kahn Kalai

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Partitioning Sets to Decrease the Diameter