Classi cation problems of good distance sets and classi cation...

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Classification problems of good distance setsand

classification problems of good families

Masashi Shinohara（Shiga University)

Osaka Combinatorics SeminarApril 16, 2016

Masashi Shinohara(Shiga University) Classification problems of distance sets and families

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.. Contents

Connections of two topics

Distance sets

DefinitionsConvex planar distance sets (survey)Planar distance sets (survey, S.)Three-distance sets in R3 (S.)Two-distance sets in Rd (Nozaki-S.)Distance sets on circles (Momihara-S.)

Families of sets

Erdos-Ko-Rado’s theorem and Katona’s proof (survey)Some generalizations of Erdos-Ko-Rado’s theorem (survey)Union families (Frankl-Tokushige)r -wise union families (Frankl-S.-Tokushige)


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.. s-code

s-code in Sd−1 s-code in Fn2

X ⊂ Sd−1 is s-code if C ⊂ Fn2 is s-code if

p · q ≤ s for ∀p,q ∈ X (p = q) dH(x, y) ≥ s for ∀x, y ∈ C (x = y)

p · q: usual inner product. dH(x, y) := |{i | xi = yi}|

max{|X | | X ⊂ Sd−1 is an s-code}? max{|C | | C ⊂ Fn2 is an s-code}?


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.. s-code in Fn2 and Intersecting families

[n] = {1, 2, . . . , n}2[n] := {F | F ⊂ [n]}([n]k

):= {F ∈ 2[n] | |F | = k}

For x ∈ Fn2, we define x ∈ 2n by

x := {i | xi = 1}. (support of x)

1 2 3 4 5

x 1 1 0 0 1

y 1 0 1 0 0

x ∨ y = (1, 1, 1, 0, 1), where(x ∨ y)i := max{xi , yi}

12

4

35

x y[5]

dH(x, y) = |x|+ |y| − 2|x ∩ y|= 2k − 2|x ∩ y| (if |x| = |y| = k)


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.. Contents


Distance sets


Families of sets



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.. Definition of distance sets

Rd : d-dimensional Euclidean space.

Sd−1(r) (or Sd−1): a sphere in Rd with radius r .

For P = (p1, p2, . . . , pd), Q = (q1, q2, . . . , qd) ∈ Rd ,

PQ =

√√√√ d∑i=1

(pi − qi )2

.Definition..

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For X ⊂ Rd (|X | <∞), A(X ) = {PQ : P ,Q ∈ X ,P = Q}and k(X ) = |A(X )|.X is called an s-distance set if k(X ) = s.

Two subsets are called isomorphic if there exists a similartransformation form one to the others.


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.. Examples of distance sets

X1

X2 X3 X4

A(X1) = {1} A(X2) = {1,√2} A(X3) = {1, τ} A(X4) = {1,

√2,√3}

1-distance set 2-distance set 2-distance set 3-distance set(τ = 1+

√5

2 )


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X1

X2 X3 X4

A(X1) = {1}

A(X2) = {1,√2} A(X3) = {1, τ} A(X4) = {1,

√2,√3}


√5

2 )


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X1

X2 X3 X4

A(X1) = {1}

A(X2) = {1,√2} A(X3) = {1, τ} A(X4) = {1,

√2,√3}

1-distance set

2-distance set 2-distance set 3-distance set(τ = 1+

√5

2 )


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X1 X2

X3 X4

A(X1) = {1}

A(X2) = {1,√2} A(X3) = {1, τ} A(X4) = {1,

√2,√3}

1-distance set


√5

2 )


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X1 X2

X3 X4

A(X1) = {1} A(X2) = {1,√2}

A(X3) = {1, τ} A(X4) = {1,√2,√3}

1-distance set


√5

2 )


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X1 X2

X3 X4

A(X1) = {1} A(X2) = {1,√2}

A(X3) = {1, τ} A(X4) = {1,√2,√3}

1-distance set 2-distance set

2-distance set 3-distance set(τ = 1+

√5

2 )


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X1 X2 X3

X4

A(X1) = {1} A(X2) = {1,√2}

A(X3) = {1, τ} A(X4) = {1,√2,√3}


2-distance set 3-distance set(τ = 1+

√5

2 )


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X1 X2 X3

X4

A(X1) = {1} A(X2) = {1,√2} A(X3) = {1, τ}

A(X4) = {1,√2,√3}



(τ = 1+√5

2 )


. . . . . .


X1 X2 X3

X4

A(X1) = {1} A(X2) = {1,√2} A(X3) = {1, τ}

A(X4) = {1,√2,√3}

1-distance set 2-distance set 2-distance set

3-distance set

(τ = 1+√5

2 )


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X1 X2 X3 X4

A(X1) = {1} A(X2) = {1,√2} A(X3) = {1, τ}

A(X4) = {1,√2,√3}


3-distance set

(τ = 1+√5

2 )


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X1 X2 X3 X4

A(X1) = {1} A(X2) = {1,√2} A(X3) = {1, τ} A(X4) = {1,

√2,√3}


3-distance set

(τ = 1+√5

2 )


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X1 X2 X3 X4

A(X1) = {1} A(X2) = {1,√2} A(X3) = {1, τ} A(X4) = {1,

√2,√3}


√5

2 )


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.. Goodness for distance sets

.Problem (数学セミナー 11 月号, Nozaki).... ..

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Classify 2-distance sets in R2 which have maximum cardinality.

.Definition..

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gd(k) := max{|X | : X is a k-distance set in Rd}.A k-distance set in Rd is said to be optimal if |X | = gd(k).

.Problem..

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k , d : given.

Determine gd(k).Classify k-distance sets in Rd with gd(k) points or close togd(k) points.When are there finitely many k-distance sets in Rd .

d , n: given . . .

n, k: given . . .


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.Definition..

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gd(k) := max{|X | : X is a k-distance set in Rd}.

A k-distance set in Rd is said to be optimal if |X | = gd(k).

.Problem..

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k , d : given.


d , n: given . . .

n, k: given . . .


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.Definition..

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.Problem..

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k , d : given.


d , n: given . . .

n, k: given . . .


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.Definition..

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.Problem..

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k , d : given.

Determine gd(k).

Classify k-distance sets in Rd with gd(k) points or close togd(k) points.When are there finitely many k-distance sets in Rd .

d , n: given . . .

n, k: given . . .


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.Definition..

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.Problem..

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k , d : given.

Determine gd(k).Classify k-distance sets in Rd with gd(k) points or close togd(k) points.

When are there finitely many k-distance sets in Rd .

d , n: given . . .

n, k: given . . .


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.Definition..

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.Problem..

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k , d : given.


d , n: given . . .

n, k: given . . .


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.. Contents


Distance sets


Families of sets



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.. Erdos’s problem (On sets of distances of n points, AMM, 1946)

.Conjecture (Erdos)..

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Every convex n-gon (n ≥ 6) has at least ⌊n/2⌋ difference distancesbetween vertices where ⌊n⌋ is a greatest integer at most n.

Rn: the vertex set of a regular n-gon

R8 R9

Rn is a ⌊n/2⌋-distance set.


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.. Convex distance sets (n-point k-distance set)

Altman (1963)n ≤ 2k + 1.If n = 2k + 1, then X = R2k+1.

Fishburn (1995)If n = 2k (k ≥ 4), then X = R2k or X ⊂ R2k+1.If (n, k) = (7, 4), then X ⊂ R2k or X ⊂ R2k+1.

.Conjecture.... ..

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If n = 2k − 1 (k ≥ 4), then X ⊂ R2k or X ⊂ R2k+1.

Erdos-Fishburn (1996)If (n, k) = (9, 5), then X ⊂ R2k or X ⊂ R2k+1.If (n, k) = (11, 6), then X ⊂ R2k or X ⊂ R2k+1 (comment).

Let s := (2k + 1)− n.

s \ k 1 2 3 4 5 6 · · ·0 ⃝1 × × ⃝2 × ⃝ ⃝ △ ?


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Fishburn (1995)If n = 2k (k ≥ 4), then X = R2k or X ⊂ R2k+1.

If (n, k) = (7, 4), then X ⊂ R2k or X ⊂ R2k+1..Conjecture.... ..

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Let s := (2k + 1)− n.

s \ k 1 2 3 4 5 6 · · ·0 ⃝1 × × ⃝2 × ⃝ ⃝ △ ?


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Fishburn (1995)If n = 2k (k ≥ 4), then X = R2k or X ⊂ R2k+1.If (n, k) = (7, 4), then X ⊂ R2k or X ⊂ R2k+1.

.Conjecture.... ..

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Let s := (2k + 1)− n.

s \ k 1 2 3 4 5 6 · · ·0 ⃝1 × × ⃝2 × ⃝ ⃝ △ ?


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.. Infinitely many classes

6-point 4-distance set 8-point 6-distance set

.Example..

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Let n ≥ 4 and

Mn =

{3t, if n = 4t or 4t − 1,

3t − 2, if n = 4t − 2 or 4t − 3

Then there exist (infinitely many) n-point Mn-distance sets in S1

which are not subsets of any regular polygon.


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.. Infinitely many classes

6-point 4-distance set 8-point 6-distance set.Example..

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Let n ≥ 4 and

Mn =

{3t, if n = 4t or 4t − 1,

3t − 2, if n = 4t − 2 or 4t − 3

Then there exist (infinitely many) n-point Mn-distance sets in S1

which are not subsets of any regular polygon.


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.. Contents


Distance sets


Families of sets



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.. Planar distance sets

Rk : the vertex set of the regular k-gon in S1.

R2k+1 and R2k is a k-distance set.

Optimal planar k-distance sets

k = 1R3

k = 2 (Kelly, 1947)R5

k = 3 (Erdos-Fishburn, 1996)R7 and R6∪ {O}

Is every optimal planar k-distance set R2k+1 for large k?


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k = 4 (Erdos-Fishburn, 1996)R9 and three sets in Fig. 1.

Fig. 1

k = 5 (Erdos-Fishburn, S. 2008)

Fig. 2


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k = 6, 7 (Candidates of optimal planar 6-, 7- distance sets,Wei(2012))

Fig. 3

.Conjecture (Erdos-Fishburn)..

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Every optimal k-distance sets for k ≥ 7 are subsets of L∆ whereL∆ = {a(1, 0) + b(1/2,

√3/2) : a, b ∈ Z} (the triangular lattice).


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.. Distance sets on a plane

k \ n 3 4 5 6 7 8 9 10 11 12 13 14

2 ∞ 6 1 ×3 ∞ 34 9 2 ×4 ∞ 42 15 4 ×5 ∞ ? 4 1 ×6 ∞ ? ≥ 2 ×

The number of planar k-distance sets with n points.

Kelly(1947), Einhorn-Schoenberg(1966), Erdos-Fishburn(1996), Harborth

-Piepmeyer(1996), S.(2004, 2008), Lan-Wei (2013), Wei (2011, 2012)

Examples of n-point k-distance sets(n, k) = (5, 3) (7, 4) (12, 5) (13, 6)


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.. Contents


Distance sets


Families of sets



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.. Finiteness of two-distance sets

There are infinitely many (d + 1)-point two-distance sets in Rd .

.Theorem (Einhorn-Schoenberg, 1966)..

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There are finitely many (d + 2)-point two-distance sets in Rd .

For any simple graph G (= Kn,Nn) of order n,∃!X ⊂ Rd such that G ↔ X and d ≤ n − 2. (function view)

X : n-point two-distance set in Rd .

d \ n 3 4 5 6 7

2 ∞ 6 1 ×3 − ∞ 26 6 ×


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.. Finiteness of two-distance sets

There are infinitely many (d + 1)-point two-distance sets in Rd ..Theorem (Einhorn-Schoenberg, 1966)..

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There are finitely many (d + 2)-point two-distance sets in Rd .

For any simple graph G (= Kn,Nn) of order n,∃!X ⊂ Rd such that G ↔ X and d ≤ n − 2. (function view)

X : n-point two-distance set in Rd .

d \ n 3 4 5 6 7

2 ∞ 6 1 ×3 − ∞ 26 6 ×


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.. Three-distance sets in R2, R3

.Theorem (S, 2004)..

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There are exactly 34 three-distance sets with 5 points.

g3(2) = 7 and optimal three-distance sets in R2 is R7 and R+6 .

.Conjecture (Einhorn-Schoenberg, 1966)..

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Every 12-point three-distance set is isomorphic to the vertex set ofa regular icosahedron. In particular, g3(3) = 12.

.Theorem (S.)..

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Every 12-point three-distance set is isomorphic to the vertex set ofa regular icosahedron. In particular, g3(3) = 12.


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.. A relation between their results and conjecture

.Proposition..

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Every 12-point three-distance set containing a 5-point two-distanceset is isomorphic to the vertex set of a regular icosahedron.

.Proposition..

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Every 14-point three-distance set in R3 contains a 5-pointtwo-distance set in R3.

.Proposition..

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Every 12-point three-distance set in R3 contains a 5-pointtwo-distance set in R3.


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.. Definition of diameter graphs

.Definition..

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D(X ) := maxA(X ) :the diameter of X

G := DG (X ) :the diameter graph of X

X

{V (G ) = X ,

For P ,Q ∈ X , P ∼ Q if PQ=D(X).

.Example.... ..

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DG (R2m+1) = C2m+1, DG (R2m) = m · P2


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.. Independent set of diameter graph

Let X be a k-distance set.

X

⊃ Y

↕ ↕V (DG (X )) ⊃ H : an independent set

Then Y is an at most (k − 1)-distance set..Proposition..

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Let G = DG (X ) be the diameter graph of X ⊂ R3 with |X | = 12.Then α(G ) ≥ 5.


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X

⊃ Y

↕

↕

V (DG (X ))

⊃ H : an independent set


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X

⊃ Y

↕

↕

V (DG (X )) ⊃ H : an independent set


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X

⊃

Y



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X ⊃ Y



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X ⊃ Y


Then Y is an at most (k − 1)-distance set.

.Proposition..

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X ⊃ Y



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.. Diameter graphs for X ⊂ R2

.Proposition..

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Let G = DG (X ) for X ⊂ R2. Then(i) G contains no C2k for any k ≥ 2;(ii) if G contains C2k+1, then any two vertices in V (G ) \V (C2k+1)are not adjacent.In particular, G contains at most one cycle.

.Proof...

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Two segments whose lengths are the diameter of X mustintersect.

isolated vertices


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.. Applications for planar distance sets

.Proposition..

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Let G = DG (X ) be the diameter graph of X ⊂ R2 with |X | = n.If G = Cn, then we have α(G ) ≥

⌈n2

⌉where

⌈n2

⌉is the smallest

integer at least n2 .

.Application..

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Let X be a 9-point 4-distance set in R2 and G = DG (X ).

If G = C9, then X = R9 (∵ X is a convex 4-distance set)

If G = C9, then α(G ) ≥⌈92

⌉= 5. Therefore X contains a

5-point (at most) 3-distance set in R2.


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.. Diameter graphs for X ⊂ R3

.Theorem (Dol’nikov(2000))..

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Let G = DG (X ) be the diameter graph of X ⊂ R3. If G containstwo cycles of odd length C and C ′, then V (C ) ∩ V (C ′) = ∅.

.Corollary..

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Let G = DG (X ) be the diameter graph of X ⊂ R3 with |X | = n.Let G contain a triangle C. Then G − C is a bipartite graph. Inparticular, α(G ) ≥

⌈n−32

⌉.

.Corollary..

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Let G = DG (X ) be the diameter graph of X ⊂ R3. Then G doesnot contain two disjoint 5-cycles.


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.. α(DG (X )) ≥ 5 for X ⊂ R3 with |X | = 12

.Proposition..

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Let G = DG (X ) be the diameter graph of X ⊂ R3 with |X | = 12.Then α(G ) ≥ 5, where α(G ) is a independence number of G.

Let G = DG (X ) for X ⊂ R3 with n (≥ 12) points.By Corollary, if G contains a triangle,

α(G ) ≥⌈12− 3

2

⌉= 5.

We assume a simple graph G satisfy the following conditions.|V (G )| = 12

α(G ) < 5

triangle− free

Then we can prove that G contains disjoint 5-cycles C and C ′.


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.. Contents


Distance sets


Families of sets



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.. Example of a two-distance set in Sd−1 ⊂ Rd

g2(d) := max{|X | : X is a 2-distance set in Rd}g∗2 (d) := max{|X | : X is a 2-distance set in Sd−1}

Vd : the set of all mid-points of edges of a regular simplex in Rd .Then

|Vd | =(d + 1

2

)=

d(d + 1)

2

and Vd is a two-distance set in Rd .

.Proof...

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Let P1 ↔ E1, P2 ↔ E2 for P1,P2 ∈ Vd . (Ei : edge)(a) E1 and E2 have a common vertex ⇒ P1P2 = 1/2(b) E1 and E2 don’t have a common vertex ⇒ P1P2 =

√2/2

d(d + 1)

2≤ g∗

2 (d) ≤ g2(d)


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.. Example of a two-distance set in Sd−1 ⊂ Rd

g2(d) := max{|X | : X is a 2-distance set in Rd}g∗2 (d) := max{|X | : X is a 2-distance set in Sd−1}

Vd : the set of all mid-points of edges of a regular simplex in Rd .Then

|Vd | =(d + 1

2

)=

d(d + 1)

2

and Vd is a two-distance set in Rd ..Proof...

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Let P1 ↔ E1, P2 ↔ E2 for P1,P2 ∈ Vd . (Ei : edge)(a) E1 and E2 have a common vertex ⇒ P1P2 = 1/2(b) E1 and E2 don’t have a common vertex ⇒ P1P2 =

√2/2

d(d + 1)

2≤ g∗

2 (d) ≤ g2(d)


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.. Tight two-distance sets

Upper boundsDelsarte-Goethals-Seidel 1977

d(d + 1)

2≤ g∗

2 (d) ≤(d + 2

2

)− 1 =: T (d)

Blokhuis 1983, Bannai-Bannai-Stanton 1983

d(d + 1)

2≤ g2(d) ≤

(d + 2

2

)−1 =: T (d)

A two-distance set X in Sd−1 (resp. Rd) is said to be tight if

|X | = T (d)(resp. |X | =

(d+22

)= T (d) + 1

).

.Remark..

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Tight two-distance sets in Sd−1 are known only ford = 2, 6, 22.

It is known some conditions for d to exist tight two-distanceset in Sd−1. (Bannai-Damerell, Bannai-Munemasa-Venkov)


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.. Optimal two-distance sets and other results

d 1 2 3 4 5 6 7 8

g∗2 (d) 2 5 6 10 16 27 28 (36)

g2(d) 3 5 6 10 16 27 29 45

.Theorem (Musin(2008), JCTA)..

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g∗2 (d) =

d(d+1)2 for 8 ≤ d ≤ 39 (d = 22, 23) and

g∗2 (22) = 275 = T (22), g∗

2 (23) = 276 or 277.

.Theorem (Nozaki-S. (2010), JCTA)..

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∃ tight two-distance set in Sd−2(⊂ Rd−1) ⇐⇒∃ proper locally two-distance set in Rd with more than d(d+1)

2points ..Theorem (Nozaki-S. (2012), LAA)..

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G: strongly regular graph of order n ⇐⇒d(G ) + d(G ) = n − 1 and both X (G ) and X (G ) are spherical.


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.. Contents


Distance sets


Families of sets



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.. Distance sets on circles.Theorem (Momihara-S., to appear in AMM)..

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Let X ⊂ S1 be a k-distance set with n points.If k < Mn, then X ⊂ R2k or X ⊂ R2k+1.

.Proposition 1..

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Let X ⊂ S1 be a k-distance set with n points.If k < Mn, then X ⊂ Rm for some m..Proposition 2..

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Let X ⊂ Rm be a k-distance set with n points. Assume that⟨X ⟩ = Rm. If k < Mn, then m ∈ {2k, 2k + 1}.

Differences in A ⊂ Zm.A = {0, 1, 4} ⊂ Z9.A− A = {0,±1,±3,±4}.|A− A| = 7↔ 3-distance set


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Let X ⊂ S1 be a k-distance set with n points.If k < Mn, then X ⊂ R2k or X ⊂ R2k+1..Proposition 1..

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Let X ⊂ S1 be a k-distance set with n points.If k < Mn, then X ⊂ Rm for some m.

.Proposition 2..

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Let X ⊂ S1 be a k-distance set with n points.If k < Mn, then X ⊂ R2k or X ⊂ R2k+1..Proposition 1..

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Let X ⊂ S1 be a k-distance set with n points.If k < Mn, then X ⊂ Rm for some m..Proposition 2..

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.. Sum set (cf. B. Nathanson, Additive number theory, Inverse problmems...)

.Definition..

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G: a finite abelian group A,B: subsets of G .A+ B = {a+ b | a ∈ A, b ∈ B}.

.Example..

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We take subsets A,B,C of Z11 where ,A = {0, 1, 2},B = {0, 2, 4},C = {3, 4, 5}.

Then

A+ B = {0, 1, 2, 3, 4, 5, 6},A+ C = {3, 4, 5, 6, 7}.

.Remark.... ..

.

.

We may assume 0 ∈ A ∩ B.

.Theorem (Cauchy(1893)-Davenport(1935))..

.

. ..

.

.

Let G = Zp (p : prime).|A+ B| ≥ min{p, |A|+ |B| − 1}.


. . . . . .

.. Vosper’s theorem

.Theorem (Cauchy(1893)-Davenport(1935))..

.

. ..

.

.

Let G = Zp (p : prime).|A+ B| ≥ min{p, |A|+ |B| − 1}.

.Theorem (Vosper(1956))..

.

. ..

.

.

Let G = Zp (p : prime). If |A+ B| = |A|+ |B| − 1, then one ofthe followings hold:

|A| = 1 or |B| = 1.

|A+ B| = p − 1.

A and B are arithmetic progressions with same commondifference.

A = {a+ id | i = 0, 1, . . . , k−1},B = {b+ id | i = 0, 1, . . . , ℓ−1}.

ThenA+ B = {a+ b + id | i = 0, 1, . . . , k + ℓ− 2}.Masashi Shinohara(Shiga University) Classification problems of distance sets and families

. . . . . .

.. Kneser’s theorem.Definition..

.

. ..

.

.

G: finite abelian group,A,B ⊂ G (subset)

A+ B = {a+ b | a ∈ A, b ∈ B}..Theorem (Cauchy-Davenport)..

.

. ..

.

.

Let G = Zp (p : prime).

|A+ B| ≥ min{p, |A|+ |B| − 1}..Theorem (Kneser)..

.

. ..

.

.

G: finite abelian group∃H < G such that

|A+ B| ≥ min{|G |, |A|+ |B| − |H|}.Corollary (Kneser)..

.

. ..

.

.

∃H < G such that

|A+ B| ≥ |A+ H|+ |B + H| − |H|Masashi Shinohara(Shiga University) Classification problems of distance sets and families

. . . . . .

.. Bounds for sizes of Rm.Corollary (Kneser)..

.

. ..

.

.

∃H < G such that

|A+ B| ≥ |A+ H|+ |B + H| − |H|.Proposition..

.

. ..

.

.

Let X be an n-subset of Zm such that ⟨X ⟩ = Zm.Then |X − X | ≥ min{m, sn}

where

sn =

{3n/2, if n ≡ 0 (mod 2),

3(n + 1)/2, if n ≡ 1 (mod 2).

|X − X | ≥ |X + H|+ |− X + H| − |H| = 2|X + H| − |H|We may assume (G : H) ≥ 2. Then |X + H| ≥ 2|H| by ⟨X ⟩ = Zm.

|X − X | ≥ 3

2|X + H| ≥ 3

2n


. . . . . .

.. Bounds for sizes of Rm.Proposition (even case)..

.

. ..

.

.

Let X be an 2n-subset of Zm such that ⟨X ⟩ = Zm.Then |X − X | ≥ min{m, s2n(= 3n)}..Proposition 2 (even case)..

.

. ..

.

.

Let X be a k-distance set with 2n points on Rm with n ≥ 2.Assume ⟨X ⟩ = Rm.If k < M2n(= ⌈(3n − 1)/2⌉), then m ∈ {2k, 2k + 1}.

.Proof...

.

. ..

.

.

If X ⊂ Rm with |X | = 2n > ⌊m/2⌋, then k = ⌊m/2⌋.|X − X | ∈ {2k , 2k + 1}, since X is a k-distance set.

2k + 1 ≥ |X − X | ≥ min{m, 3n}⇐⇒ k ≥ min{⌈(m − 1)/2⌉, ⌈(3n − 1)/2⌉}⌈(m − 1)/2⌉ ≤ k < ⌈(3n − 1)/2⌉ since k < ⌈(3n − 1)/2⌉.m < 3n⇐⇒ |X | = 2n > 2

3m.


. . . . . .


.

. ..

.

.


.

. ..

.

.

Let X be a k-distance set with 2n points on Rm with n ≥ 2.Assume ⟨X ⟩ = Rm.If k < M2n(= ⌈(3n − 1)/2⌉), then m ∈ {2k, 2k + 1}..Proof...

.

. ..

.

.

If X ⊂ Rm with |X | = 2n > ⌊m/2⌋, then k = ⌊m/2⌋.

|X − X | ∈ {2k , 2k + 1}, since X is a k-distance set.


3m.


. . . . . .


.

. ..

.

.


.

. ..

.

.


.

. ..

.

.



3m.


. . . . . .


.

. ..

.

.


.

. ..

.

.


.

. ..

.

.


2k + 1 ≥ |X − X | ≥ min{m, 3n}⇐⇒ k ≥ min{⌈(m − 1)/2⌉, ⌈(3n − 1)/2⌉}

⌈(m − 1)/2⌉ ≤ k < ⌈(3n − 1)/2⌉ since k < ⌈(3n − 1)/2⌉.m < 3n⇐⇒ |X | = 2n > 2

3m.


. . . . . .


.

. ..

.

.


.

. ..

.

.


.

. ..

.

.


2k + 1 ≥ |X − X | ≥ min{m, 3n}⇐⇒ k ≥ min{⌈(m − 1)/2⌉, ⌈(3n − 1)/2⌉}⌈(m − 1)/2⌉ ≤ k < ⌈(3n − 1)/2⌉ since k < ⌈(3n − 1)/2⌉.

m < 3n⇐⇒ |X | = 2n > 23m.


. . . . . .


.

. ..

.

.


.

. ..

.

.


.

. ..

.

.



3m.


. . . . . .

.. Distance sets on a half circle

d(P ,Q)←→ al(P,Q)(arc length)

Mn − 1 =

{3t − 1, if n = 4t or 4t − 1,

3t − 3, if n = 4t − 2 or 4t − 3.

X : (4m − 1)-point (3m − 1)-distance set in S1

X ′: 2m-point k-distance set in R1 for k ≤ 3m − 1


. . . . . .

.. Distance sets in R1

a1 a2 a3 a4 a5 a6 a7

b1 b2 b3 b4 b5 b6

.Definition..

.

. ..

.

.

X = (b1, b2, · · · , bn−1) is rational if bi/b1 ∈ Q for 1 ≤ ∀i ≤ n − 1.X is irrational if X is not rational.

.Example..

.

. ..

.

.

(n = 8) X = (1, 1, 1, c , 1, 1, 1) (c /∈ Q)Then k = 10 since A(X ) = {1, 2, 3} ∪ {i + c | 0 ≤ i ≤ 6}.

.Lemma..

.

. ..

.

.

Let X be an irrational k-distance set with n points.Then

k ≥ ⌊3n − 3

2⌋.


. . . . . .

.Lemma.... ..

.

.

X = {a1, . . . , an}: irrational. Then |A(X ) \ A(X \ {a1, an})| ≥ 3.

We may assume d(a1, a2) = d(an−1, an) = 1.Let bs (resp. bt) be irrational interval with smallest (resp. largest)index.

bs

a1 a2 at+1 anas as+1

bt

at

d1∗

d2∗

ai aj

d1∗

Without loss of generality, we may assume d1∗ ≥ d2

∗.If there exists 2 ≤ i ≤ s and t + 1 ≤ j ≤ n − 1 such thatd(ai , aj) = d1

∗ then bt ∈ Q.Therefore @ ai , aj ∈ X ( 2 ≤ i ≤ s and t + 1 ≤ j ≤ n − 1) suchthat d(ai , aj) = d1

∗.Therefore {d(a1, an), d(a1, an−1), d

∗1} ⊂ A(X ) \ A(X \ {a1, an}).


. . . . . .

.. Irrational distance sets

n-point k-distance set in R1

k⌊3n−52 ⌋n − 1

rational setnonexistance

We classified irrational n-point k-distance set for(n, k) = (2m, 3m − 2), (2m, 3m − 1) and (2m + 1 , 3m).

.Lemma..

.

. ..

.

.

Let X be a 2m-point (3m − 1)-distance set for m ≥ 7. Then X isequivalent to one of the followings.

(1, 1, . . . , 1︸︷︷︸m

, c , 1, 1, . . . , 1︸︷︷︸m−2

).

(1 + c , 1, c , 1, c , . . . , 1, c︸︷︷︸2m−2

).

(1 + c , . . . , 1 + c︸︷︷︸m1+1

, 1, c , 1, c , . . . , c , 1︸︷︷︸2m2−1

, 1 + c , . . . , 1 + c︸︷︷︸m1−1

)


. . . . . .

.. Classification of irrational distance sets

.Lemma..

.

. ..

.

.

Let X be an irrational k-distance set with n point.Then the followings hold:

(1) If (n, k) = (2m, 3m − 2) for m ≥ 2, then

X = [m] ∪ τc([m]);

(2) If (n, k) = (2m + 1, 3m) for m ≥ 5, then

X = [m + 1] ∪ τc([m]);

(3) If (n, k) = (2m, 3m − 1) for m ≥ 7, then

X = [m + 1] ∪ τc([m − 1]),

where c is any irrational number, [m] = {1, 2, . . . ,m} andτc(x) = x + c for x ∈ R.


. . . . . .

ℓX := (ℓ, b1, b2, . . . , bn−1).Lemma..

.

. ..

.

.

Let X = (b1, b2, . . . , bn−1) and ℓ > 0. If |D(ℓX ) \D(X )| ≤ 2, thenℓ ∈ D(X )..Lemma..

.

. ..

.

.

Let X = (b1, b2, . . . , bn−2) be a rational k-distance set with n − 1points. If ℓX is an irrational k ′-distance set, then k ′ ≥ 2n − 4..Lemma..

.

. ..

.

.

Let X be an irrational k-distance set with n points. Then thefollowings hold:

(1) If (n, k) = (3, 3), then X = (1, c);

(2) If (n, k) = (4, 4), then X = (1, c , 1);

(3) If (n, k) = (4, 5), then X = (c , 1, 1 + c), (1, 1, c);

(4) If (n, k) = (5, 6), thenX = (c , 1, c , 1), (1, 1, c , 1), (1, c , c, 1), (1 + c , 1, c , 1).


. . . . . .

.. Distance sets obtained inductively from the starter (1, c , 1)

m = 3 m = 4 m = 5 m = 6(1, 1, c , 1, 1) → (1, . . . , 1) → A–I

→ (c , . . . , c) → (1, . . . , 1) → (1, . . . , 1)→ (1 + c, . . . , 1 + c)

→ (1 + c , . . . , 1 + c) → (1, . . . , 1) → (2 + c , . . . , 2 + c)→ (2 + c , . . . , 2 + c)

Table: Distance sets obtained inductively from the starter (1, 1, c , 1, 1)

(c, 1, c, 1, c) → (1, . . . , 1) → A–II, A–III, B–III→ (1 + c, . . . , 1) → B–III→ (1 + c, . . . , 1 + c) → A–III→ (c, . . . , c)

(1 + c, 1, c, 1, 1 + c) → (1 + c, . . . , 1 + c) → A–III→ (1, . . . , 1) → (1 + c, . . . , 1 + c) → (2 + c, . . . , 2 + c)→ (2 + c, . . . , 2 + c)

(1 + c, 1, c, 1, c) → (1 + c, . . . , 1) → B–III(−1 + c, 1, c, 1,−1 + c) → (1, . . . , 1) → (c, . . . , c) → (c, . . . , c)

(2 + c, 1, c, 1, 2 + c) → ×


. . . . . .

.. Half circle with large points

.Lemma..

.

. ..

.

.

Let X be a set of n points in S1.

If n is odd, then ∃ai ∈ X such that |R(ai )|=|L(ai )|.If n is even, then ∃ai , ai+1 ∈ X such that|R(ai )| > n/2 and |L(ai+1)| > n/2.


. . . . . .

.. Rational distance sets → Circular distance set.Lemma..

.

. ..

.

.

Let X = {a0, a1, . . . , an−1} be a k-distance set on S1.Assume that al(ai , ai + 1) ∈ Q for i = 1, 2, . . . , n − 1.If k < n − 1, then X ⊂ Rm for some m.

Since k < n − 1,∃ai , aj ∈ X such that al(a0, ai ) = al(a0, aj)

Then al(a0, a1) = al(a0, ai )− al(a1, ai )

= al(a0, aj)− al(a1, ai ) ∈ Q>0


. . . . . .

.. Rational distance sets → Circular distance set

odd even

.Proposition..

.

. ..

.

.

Let X be a k-distance set with n points on S1 with k < Mn.

If n is even, then X ⊂ Rm for some m.

If n is odd and both L(ai ) and R(ai ) are rational, thenX ⊂ Rm.


. . . . . .

.. Irrational distance set → Circular distance set

.Proposition..

.

. ..

.

.

Let X be a k-distance set with n points on S1 with k < Mn.Then both R(ai ) and L(ai ) are rational.


. . . . . .

.. Main results for distance sets on circles

Mn =

{3t, if n = 4t or 4t − 1,

3t − 2, if n = 4t − 2 or 4t − 3

.Theorem (Momihara-S. to appear in AMM)..

.

. ..

.

.

Let X ⊂ S1 be a k-distance set with n points.If k < Mn, then X ⊂ R2k or X ⊂ R2k+1.

.Problem..

.

. ..

.

.

k , d : given.

Determine gd(k).Classify k-distance sets in Rd with gd(k) points or close togd(k) points.When are there finitely many k-distance sets in Rd

d , n: given . . .

n, k: given . . .


. . . . . .

.. Contents


Distance sets


Families of sets



. . . . . .

.. Intersecting families.Definition..

.

. ..

.

.

A family F is called intersecting (resp. t-intersecting) ifF ∩ F ′ = ∅ (resp. |F ∩ F ′| ≥ t)

holds for all F ,F ′ ∈ F ..Example (Fano plane)..

.

. ..

.

.

F = {{1, 2, 6}, {1, 3, 7}, {1, 4, 5},{2, 3, 5}, {2, 4, 7}, {3, 4, 6}, {5, 6, 7}}

.Example (trivial intersecting family)..

.

. ..

.

.

F0 =

{[t] ∪ F | F ∈

([n] \ [t]k − t

)}.

Then F0 is a t-intersecting family with |F| =(n−tk−t

).


. . . . . .

.. Erdos-Ko-Rado’s theorem.Theorem (Erdos-Ko-Rado(1961))..

.

. ..

.

.

F ⊂([n]k

): intersecting

If n ≥ 2k, then

|F| ≤(n − 1

k − 1

).

.Definition..

.

. ..

.

.

For a circular permutation C = (a1, a2, . . . , an) on [n],

C (k) := {{ai , ai+1, · · · , ai+k−1} | 1 ≤ i ≤ n}.

Let C = (1, 5, 7, 2, 4, 6, 3).

C (3) = {{1, 5, 7}, {2, 5, 7}, {2, 4, 7},{2, 4, 6}, {3, 4, 6}, {1, 3, 6}, {1, 3, 5}}.


. . . . . .

.. Katona’s proof

.Lemma..

.

. ..

.

.

Let n ≥ 2k. Let C be a cyclic permutation on [n]. If F ⊂ C (k) isintersecting, then |F| ≤ k.


. . . . . .

.. Katona’s proof

By a double counting of

p(F) := |{(C ,F ) | F ∈ F ,C is a circ . perm. on [n],F ∈ C (k)}| .

|F| · (n − k)! · k! = p(F) ≤ (n − 1)! · k.

|F| ≤ (n − 1)! · k(n − k)! · k!

=(n − 1)!

(k − 1)!(n − k)!=

(n − 1

k − 1

).


. . . . . .

.. Contents


Distance sets


Families of sets



. . . . . .

.. Erdos-Ko-Rado’s theorem (general case)

.Theorem (Erdos-Ko-Rado(1961))..

.

. ..

.

.

F ⊂([n]k

): t-intersecting family

If n ≥ n0(k , t), then|F| ≤

(n − t

k − t

).

.Theorem (Frankl (1978))..

.

. ..

.

.

The best possible bound for n0(k, t) is (k − t + 1)(t + 1) fort ≥ 15.

.Theorem (Wilson (1984)).... ..

.

.

The best possible bound for n0(k, t) is (k − t + 1)(t + 1) for all t.


. . . . . .

.. Intersecting families when n < (k − t + 1)(t + 1)

.Example (non-trivial t-intersecting family)..

.

. ..

.

.

Fi = {S ∈([n]

k

)| |S ∩ [t + 2i ]| ≥ t + i} (0 ≤ i ≤ ⌊(n − t)/2⌋)

is t intersecting.

1 · · · i i + 1 · · · 2i 1 + 2i · · · t + 2i · · · n

× · · · × ⃝ · · · ⃝× · · · × ⃝ · · · ⃝

.Theorem (Ahlswede-Khachatrian (1996))..

.

. ..

.

.

Let F ⊂([n]k

)be a t-intersecting. Then

|F| ≤ max0≤i≤ n−t

2

|Fi |

holds. Moreover, equality holds only for F = Fi for some i.


. . . . . .

.. Other problems

t-intersecting family F ⊂([n]k

).

Non-trivial intersecting family for n > n0(k, t).(Hilton-Milner(1967), Ahlswede-Khachatrian(1996))r -wise t-intersecting family (Brace-Daykin(1971),Frankl-Tokushige(2002,2005), Tokushige(2005,2007,2010))

Other situations

Intersecting family for F ⊂ 2[n] (Katona(1964))Multisets (Furedi-Gerbner-Vizer(2014))Cross-intersecting familyJohnson graph, Hamming graph (other (Q-polynomial)distance regular graphs)


. . . . . .

.. Contents


Distance sets


Families of sets



. . . . . .

.. Intersecting families and Union families

.Theorem (Erdos-Ko-Rado(1961), Frankl (1978), Wilson (1984))..

.

. ..

.

.

Given n ≥ k ≥ t > 0 and a t-intersecting family F ⊂([n]k

). If

n ≥ (k − t + 1)(t + 1), then|F| ≤

(n − t

k − t

).

.Definition..

.

. ..

.

.

A family F is called s-union if|F ∪ F ′| ≤ s

holds for all F ,F ′ ∈ F ..Remark..

.

. ..

.

.

Let F ⊂([n]k

). Since |F ∪ F ′| = 2k − |F ∩ F ′| for F ,F ′ ∈

([n]k

),

F : t-intersecting ⇐⇒ F : (2k − t)-union.


. . . . . .

.. Union for q-ary codes

1 2 3 4 5

x 1 1 0 0 1

y 1 0 1 0 0

x ∨ y = (1, 1, 1, 0, 1), where(x ∨ y)i := max{xi , yi}

12

4

35

x y[5]

.Definition (join)..

.

. ..

.

.

For x, y ∈ {0, 1, . . . , q − 1}n, x ∨ y is defined by(x ∨ y)i := max{xi , yi}

(2, 3, 1) (1, 2, 2) (2, 3, 2)


. . . . . .

.. Union families in Nn ({0, 1, . . . , q − 1}nfor large q])

.Definition..

.

. ..

.

.

N = {0, 1, 2, . . .}.For a = (a1, a2, . . . , an) ∈ Nn, |a| := a1 + a2 + · · ·+ an.

For a,b ∈ Nn, we define the join a ∨ b by

(a ∨ b)i := max{ai , bi}.A ⊂ Nn is s-union if

|a ∨ b| ≤ s for all a,b ∈ A.

wn(s) := max{|A| | A ⊂ Nn is s-union}.For a,b ∈ Nn, we let a ≺ b iff ai ≤ bi for all 1 ≤ i ≤ n.

a ∈ A is maximal if @ b ∈ A such that a ≺ b.

For x ∈ Nn, D(x) := {y ∈ Nn | y ≺ x} (down set)..Remark..

.

. ..

.

.

If A ⊂ Nn is s-union, then (D(A) :=)∪a∈AD(a) is also s-union.


. . . . . .

.. D(a) and balanced partition

D(x) := {y ∈ Nn | y ≺ x}

.Example (10-union in N3)..

.

. ..

.

.

Let a = (4, 3, 3). Then D(a) is 10-union with|D(a)| = (4 + 1)× (3 + 1)2 = 80. Therefore w3(10) ≥ 80.

.Definition (balanced partition)..

.

. ..

.

.

b = (b1, b2, . . . , bn) ∈ Nn is called a balanced partition iff|bi − bj | ≤ 1 for 1 ≤ i < j ≤ n.

.Lemma..

.

. ..

.

.

If b is a balanced partition and a is not a balanced partition with|b| = |a|, then

|D(a)| < |D(b)|.


. . . . . .

.. Balanced partition

.Proposition..

.

. ..

.

.

wn(s) := max{|A| | A ⊂ Nn is s-union}.

w1(s) = |D(a1)| = s + 1,

w2(2s) = |D(a2)| = (s + 1)2,

w2(2s + 1) = |D(a3)| = (s + 2)(s + 1).

where ai is a balanced partition with |ai | = s, 2s, 2s + 1,respectively. i. e. a1 = (s), a2 = (s, s), a3 = (s + 1, s).


. . . . . .

.. Examples and upper set S(a, d).Example (Another example of 10-union)..

.

. ..

.

.

A := {(4, 2, 2), (2, 4, 2), (2, 2, 4), (3, 3, 3)}.Then A is 10-union. (We will check soon.) Moreover,

|D(A)| = |{(i , j , k) : i , j , k ∈ {0, 1, 2, 3}}|+ |{(4, j , k) : j , k ∈ {0, 1, 2}}| × 3

= 43 + 33 = 91 > 80 = |D((4, 3, 3))|.

A = {a+ 2e1, a+ 2e2, a+ 2e3} ∪ {a+ 1},where ei is the i-the standard base of Rn, 1 :=

∑ei , a = (2, 2, 2).

.Definition (upper set at distance d from a ∈ Nn)..

.

. ..

.

.

S(a, d) = {a+ ϵ : ϵ ∈ Nn, |ϵ| = d}

D(A) = D (S(a, 2)) ∪ D (S(a+ 1, 0)) where a = (2, 2, 2).


. . . . . .

.. D(A) = D (S(a, 2)) ∪ D (S(a+ 1, 0)) where a = (2, 2, 2)

If p,q ∈ D (S(a+ 1, 0)), then clearly |p ∨ q| ≤ 9.

If p,q ∈ D (S(a, 2)), thenp ∨ q ∈ D (S(a, 2 + 2)) .

Therefore |p ∨ q| ≤ 10.

If p ∈ D (S(a, 2)) and q ∈ D (S(a+ 1, 0)),

p ∨ q ∈ D (S(a+ 1, 2− (3− 2))) .

Therefore |p ∨ q| ≤ 10..Definition..

.

. ..

.

.

Kn(a, d) :=

⌊ dn−1

⌋∪i=0

D (S (a+ i1, d − (n − 1)i)) .

K3(a, 2) =1∪

i=0

D (S (a+ i1, 2− 2i)) where a = (2, 2, 2).

.Proposition.... ..

.

.

Kn(a, d) is (|a|+ 2d)-union.


. . . . . .

.. Good construction of s-union

Kn(a, d) :=

⌊ dn−1

⌋∪i=0

D (S (a+ i1, d − (n − 1)i))

.Proposition.... ..

.

.

Kn(a, d) is (|a|+ 2d)-union..Proof...

.

. ..

.

.

Let 0 ≤ i ≤ j ≤ ⌊ dn−1⌋, and

b ∈ D (S (a+ i1, d − (n − 1)i)) ,c ∈ D (S (a+ j1, d − (n − 1)j)) .

|b \ c| : =∑

1≤l≤n

max{bl − cl , 0}

≤ d − (n − 1)i − (j − i),

|b ∨ c| = |c|+ |b \ a|≤ (|a|+ d + j) + d − (n − 2)i − j

≤ |a|+ 2d − (n − 2)i ≤ |a|+ 2d .


. . . . . .

.. |Kn(a, d)| and balanced partition b

.Lemma..

.

. ..

.

.

|Kn(a, d)| =n∑

j=0

(d + j

j

)σn−j(a)

+

⌊ dn−1

⌋∑i=1

((d − (n − 1)i + n

n

)−

(d − (n − 1)i + n − 1

n

)).

whereσk(a) =

∑K∈([n]k )

∏i∈K

ai .

.Lemma..

.

. ..

.

.

Let |a| = |b|. If b is a balanced partition, then|Kn(a, d)| ≤ |Kn(b, d)|.


. . . . . .

.. Known Fact

.Conjecture (Frankl-Tokushige, to appear in JCTA)..

.

. ..

.

.

Let n,s be given. Then it follows that

wn(s) = max0≤d≤⌊s/2⌋

|Kn(a, d)|

where a ∈ Nn is a balanced partition with |a| = s − 2d .

.Remark..

.

. ..

.

.

Frankl-Tokushige[1] verified the conjecture for the following cases:

(i) s = 3, (It is somewhat surprising that the case n = 3 is not so easy,

and the formula for w3(s) is rather involved.)

(ii) n > n0(s),

(iii) Under two suppositions;

a is well-defined{P1,P2, . . . ,Pn} ⊂ A

where a and Pi ’s are defined from an s-union family A ⊂ Nn.


. . . . . .

.. Polytope Pn(a, d) with |a| = s − nd

Let n, s be given, and a = (a1, a2, . . . , an) ∈ Nn with |a| = s − 2dfor some d ∈ N. We define convex polytope P = Pn(a, d) ⊂ Rn bythe following equations:

xi ≥ 0 (1 ≤ i ≤ n),

xi ≤ ai + d (1 ≤ i ≤ n),

xi + xj ≤ ai + aj + d (1 ≤ i < j ≤ n).

L = Ln(a, d) := {x ∈ Nn : x ∈ P}.Lemma.... ..

.

.Two set K and L are the same, and s-union.

Kn(a, d) =

⌊ dn−1

⌋∪i=0

D (S (a+ i1, d − (n − 1)i))

In [1], they prove K = L by (i) K ⊂ L and (ii) |K | = |L|.


. . . . . .

.. Definition of Pi ’s for s-union A(⊂ Nn)

m = m(A) := (m1,m2, . . . ,mn) where mi := max{xi : x ∈ A}

d = d(A) :=|m| − s

n − 2. (d ∈ N?)

a = a(A) := (a1, a2, . . . , an) where ai = mi − d . (a ∈ Nn?)

Then s = |a|+ 2d since d =|m| − s

n − 2and |a| = |m|+ nd .

Pi := a+ dei .

.Remark.... ..

.

.If ai < 0, then Pi is not defined. For n = 3, we have ai ≥ 0.

WLOG, we may assume that m1 ≥ m2 ≥ m3. Then a1 ≥ a2 ≥ a3.Since A is s-union, m1 +m2 ≤ s. Then |m| − s ≤ m3.d = |m| − s ≤ m3. Therefore a3 ≥ m3 − d ≥ 0.


. . . . . .

.. A ⊂ L3(a, d)

L = L3(a, d) := {x ∈ N3 : x satisfies (1), (2), (3)}xi ≥ 0 (1 ≤ i ≤ 3), (1)

xi ≤ ai + d (1 ≤ i ≤ 3), (2)

xi + xj ≤ ai + aj + d (1 ≤ i < j ≤ 3). (3)

mi := max{xi : x ∈ A}, d = |m|−s, ai := mi−d , s = |a|+2d ..Lemma..

.

. ..

.

.

Let A ⊂ N3 be s-union, and a := a(A) and d := d(A). Then

A ⊂ L3(a, d)..Proof...

.

. ..

.

.

(1) Trivial since A ⊂ N3.(2) For any x ∈ (x1, x2, x3) ∈ A, xi ≤ mi = ai + d .(3) For any x ∈ (x1, x2, x3) ∈ A, take y = (∗, ∗,m3) ∈ A.Then |a|+ 2d = s ≥ x1 + x2 +m3 = x1 + x2 + a3 + d .This implies a1 + a2 + d ≥ x1 + x2.


. . . . . .

.. Summary for s-union family in Nn

.Lemma..

.

. ..

.

.

Ln(a, d) = Kn(a, d) and they are (|a|+ 2d)-union.

Let |a| = |b|. If b is a balanced partition, then|Kn(a, d)| ≤ |Kn(b, d)|.

1. If A ⊂ Ln(a, d), then

2. A ⊂ Kn(a, d) since Ln(a, d) = Kn(a, d).

3. |A| ≤ |Kn(b, d)| for a balanced partition b with |b| = |a|.

|A| ≤ |Ln(a, d)| = |Kn(a, d)| ≤ |Kn(b, d)|

|m|−sn−2 ∈ N? mi − d ≥ 0? {Pi} ⊂ A? A ⊂ L!=⇒

Yes=⇒Yes

=⇒Yes

⇓ ⇓ ⇓? |A|:small? |A|:small?No No No


. . . . . .

.. Contents


Distance sets


Families of sets



. . . . . .

.. r -wise union families in Nn

.Definition..

.

. ..

.

.

a,b, . . . , z ∈ Nn, we define the join a ∨ b ∨ · · · ∨ z by

(a ∨ b ∨ · · · ∨ z)i := max{ai , bi , . . . , zi}.A ⊂ Nn is r -wise s-union if

|a1 ∨ · · · ∨ ar | ≤ s for all a1, a2, . . . ar ∈ A.

.Definition..

.

. ..

.

.

Kn(r , a, d) :=

⌊d/u⌋∪i=0

D (S (a+ i1, d − ui))

where u = n − r + 1.

.Conjecture..

.

. ..

.

.

Let r ≥ 2 and A be a r -wise s-union in Nn. Then

|A| ≤ max0≤d≤⌊s/r⌋

|Kn(r , a, d)|

where a ∈ Nn is a balanced partition with |a| = s − 2d .


. . . . . .

.. r -wise s-union families in Nk for k ≤ r

.Proposition (s-union family for n ≤ 2)..

.

. ..

.

.

wn(s) := max{|A| | A ⊂ Nn is s-union}.w1(s) = |D(a1)| = s + 1,

w2(2s) = |D(a2)| = (s + 1)2,

w2(2s + 1) = |D(a3)| = (s + 2)(s + 1).

where ai is a balanced partition with |ai | = s, 2s, 2s + 1,respectively. i. e. a1 = (s), a2 = (s, s), a3 = (s + 1, s).

.Proposition..

.

. ..

.

.

Let A be r-wise s-union in Nk for k ≤ r . Then

|A| ≤ |D(a)|

where a ∈ Nn is a balanced partition with |a| = s.


. . . . . .

.. Main Theorem.Theorem (Framkl-Tokushige)..

.

. ..

.

.

Conjecture for s-union is true for the following cases:

(i) s = 3,

(ii) n > n0(s),



where a and Pi ’s are defined from A..Theorem (Frankl-S-Tokushige)..

.

. ..

.

.

Conjecture for r -wise s-union is true for the following cases:(i) s = r + 1,

(ii) n > n0(r , s),



where a and Pi ’s are defined from A.Masashi Shinohara(Shiga University) Classification problems of distance sets and families

. . . . . .

.. Correspondence table

u := n − r + 1

(2-wise) s-union 2-wise s-union|a1 ∨ a2| ≤ s (∀a1, a2 ∈ A) |a1 ∨ · · · ∨ ar | ≤ s (∀a1, . . . , ar ∈ A)

K

⌊ dn−1 ⌋∪i=0

D (S (a+ i1, d − (n − 1)i))

⌊ du ⌋∪

i=0

D (S (a+ i1, d − ui))

Lxi ≥ 0 (1 ≤ i ≤ n),xi ≤ ai + d (1 ≤ i ≤ n),xi+xj ≤ ai+aj+d (1 ≤ i < j ≤ n)

xi ≥ 0(1 ≤ i ≤ n),∑i∈I

xi ≤∑i∈I

ai + d

(1 ≤ I ≤ n − r + 1, I ⊂ [n]),

d |m|−sn−2

|m|−sn−r

|a| s + 2d s + rd

|Kn(r , a, d)| =n∑

j=0

(d + j

j

)σn−j(a)

+

⌊d/u⌋∑i=1

n∑j=u+1

((d − ui + j

j

)−(d − ui + u

j

))σn−j(a+ i1)


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