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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. - PowerPoint PPT Presentation
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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:
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• Let S1 = { x : x2S and x·m }
Let S2 = { x : x2S and x>m }
• Clearly diam(Si) · diam(S) / 2
Our Question with d=2• Question: For every S½Rd, can we partition
S into S1 [ S2 [ [ Sd+1 such thatdiam(Si) < diam(S) for all i=1,...,d+1
S
Idea:Take minimum enclosing ballPartition into 90± segmentsCan argue diam(Si) < diam(S)
S1 S2
S3 S4
Our Question with d=2• Question: For every S½Rd, can we partition
S into S1 [ S2 [ [ Sd+1 such thatdiam(Si) < diam(S) for all i=1,...,d+1
S
Improved Idea:Take minimum enclosing ballPartition into 120± segments
Our Question with d=2• Question: For every S½Rd, can we partition
S into S1 [ S2 [ [ Sd+1 such thatdiam(Si) < diam(S) for all i=1,...,d+1
S
Improved Idea:Take minimum enclosing ballPartition into 120± segmentsIs diam(Si) < diam(S)?S1 S2
S3
Our Question with d=2• Question: For every S½Rd, can we partition
S into S1 [ S2 [ [ Sd+1 such thatdiam(Si) < diam(S) for all i=1,...,d+1
S
Idea:Take minimum enclosing ballPartition into 120± segmentsIs diam(Si) < diam(S)?Yes, unless S ¼ equi. triangle
S1 S2
S3
Our Question with d=2• Question: For every S½Rd, can we partition
S into S1 [ S2 [ [ Sd+1 such thatdiam(Si) < diam(S) for all i=1,...,d+1
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
every 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?d Lower Bound on f(d) Upper Bound on f(d)1 2 22 2 3
Optimal!
Lower Bound on f• Notation: f(d) = smallest integer such that
every S½Rd can be partitioned intoS1 [ S2 [ [ Sf(d) with diam(Si) < diam(S).
• 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
Lower Bound on f• Notation: f(d) = smallest integer such that
every S½Rd can be partitioned intoS1 [ S2 [ [ Sf(d) with diam(Si) < diam(S).
• 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
Lower Bound on f• Notation: f(d) = smallest integer such that
every S½Rd can be partitioned intoS1 [ S2 [ [ Sf(d) with diam(Si) < diam(S).
• 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
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]
• Theorem (Kahn-Kalai 1993):• Corollary: Borsuk’s conjecture is false!
• Conjecture (Borsuk 1933): f(d)=d+1.
1:2p d[Kahn-Kalai ‘93]
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]
d d+1 [Borsuk ‘33] 2d [Knast ‘74]1.23d [Schramm ‘88]
• 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
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 ,
For all partitions S = S1 [ [ Sp
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 ,
For all partitions S = S1 [ [ Sp
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 ,
For all partitions S = S1 [ [ Sp
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