Supersaturation for Intersecting Families Shagnik Das University of - - PowerPoint PPT Presentation

Supersaturation for Intersecting Families

Shagnik Das

University of California, Los Angeles / ETH Z¨ urich

April 1, 2014 Joint work with Wenying Gan and Benny Sudakov

Supersaturation for Intersecting Families

Shagnik Das

University of California, Los Angeles / ETH Z¨ urich

April Fools’ Day, 2014 Joint work with Wenying Gan and Benny Sudakov

Background Counting Probabilistic Our results Conclusion

Intersecting families

Definition (Intersecting families) A family of sets F is said to be intersecting if F1 ∩ F2 = ∅ for all F1, F2 ∈ F.

Background Counting Probabilistic Our results Conclusion

Intersecting families

Definition (Intersecting families) A family of sets F is said to be intersecting if F1 ∩ F2 = ∅ for all F1, F2 ∈ F. Notation: [n] = {1, 2, . . . , n} - ground set for our set families [n]

k

• = {F ⊂ [n] : |F| = k}

[n]

≥k

• = {F ⊂ [n] : |F| ≥ k}

Background Counting Probabilistic Our results Conclusion

Non-uniform families

Observation If a family F on [n] is intersecting, |F| ≤ 2n−1.

Background Counting Probabilistic Our results Conclusion

Non-uniform families

Observation If a family F on [n] is intersecting, |F| ≤ 2n−1. Proof: F can contain at most one of F, F c for all F ⊂ [n].

Background Counting Probabilistic Our results Conclusion

Non-uniform families

Observation If a family F on [n] is intersecting, |F| ≤ 2n−1. Proof: F can contain at most one of F, F c for all F ⊂ [n]. Constructions:

Star: F = {F ⊂ [n] : 1 ∈ F}

Background Counting Probabilistic Our results Conclusion

Non-uniform families

Observation If a family F on [n] is intersecting, |F| ≤ 2n−1. Proof: F can contain at most one of F, F c for all F ⊂ [n]. Constructions:

Star: F = {F ⊂ [n] : 1 ∈ F} Large sets: F = [n] ≥ n+1

2

• ∪
• F ∈

[n]

n 2

• : 1 ∈ F

Background Counting Probabilistic Our results Conclusion

Non-uniform families

Observation If a family F on [n] is intersecting, |F| ≤ 2n−1. Proof: F can contain at most one of F, F c for all F ⊂ [n]. Constructions:

Star: F = {F ⊂ [n] : 1 ∈ F} Large sets: F = [n] ≥ n+1

2

• ∪
• F ∈

[n]

n 2

• : 1 ∈ F
• Large sets make intersections easier

Background Counting Probabilistic Our results Conclusion

Erd˝

• s-Ko-Rado

Theorem (Erd˝

• s-Ko-Rado, 1961)

Suppose n ≥ 2k, and F ⊂ [n]

k

• is intersecting. Then |F| ≤

n−1

k−1

• .

Background Counting Probabilistic Our results Conclusion

Erd˝

• s-Ko-Rado

Theorem (Erd˝

• s-Ko-Rado, 1961)

Suppose n ≥ 2k, and F ⊂ [n]

k

• is intersecting. Then |F| ≤

n−1

k−1

• .

Extremal families: stars

Background Counting Probabilistic Our results Conclusion

Erd˝

• s-Ko-Rado

Theorem (Erd˝

• s-Ko-Rado, 1961)

Suppose n ≥ 2k, and F ⊂ [n]

k

• is intersecting. Then |F| ≤

n−1

k−1

• .

Extremal families: stars 1

A star with centre 1

Background Counting Probabilistic Our results Conclusion

Beyond the threshold

Previous results answer the typical extremal problem Question How large can a structure be without containing a forbidden configuration?

Background Counting Probabilistic Our results Conclusion

Beyond the threshold

Previous results answer the typical extremal problem Question How large can a structure be without containing a forbidden configuration? Gives rise to natural extension Question How many forbidden configurations must appear in larger structures?

Background Counting Probabilistic Our results Conclusion

One extra set

Warm up: How many disjoint pairs must a family of 2n−1 + 1 sets contain?

Background Counting Probabilistic Our results Conclusion

One extra set

Warm up: How many disjoint pairs must a family of 2n−1 + 1 sets contain? Answer:

Background Counting Probabilistic Our results Conclusion

One extra set

Warm up: How many disjoint pairs must a family of 2n−1 + 1 sets contain? Answer: 1!

Background Counting Probabilistic Our results Conclusion

One extra set

Warm up: How many disjoint pairs must a family of 2n−1 + 1 sets contain? Answer: 1! Any maximal intersecting family can be extended with only

• ne disjoint pair:

Background Counting Probabilistic Our results Conclusion

One extra set

Warm up: How many disjoint pairs must a family of 2n−1 + 1 sets contain? Answer: 1! Any maximal intersecting family can be extended with only

• ne disjoint pair:

Let F be an intersecting family of 2n−1 sets

Background Counting Probabilistic Our results Conclusion

One extra set

Warm up: How many disjoint pairs must a family of 2n−1 + 1 sets contain? Answer: 1! Any maximal intersecting family can be extended with only

• ne disjoint pair:

Let F be an intersecting family of 2n−1 sets For minimal F0 ∈ F, add F c

0 = [n] \ F0

Background Counting Probabilistic Our results Conclusion

One extra set

Warm up: How many disjoint pairs must a family of 2n−1 + 1 sets contain? Answer: 1! Any maximal intersecting family can be extended with only

• ne disjoint pair:

Let F be an intersecting family of 2n−1 sets For minimal F0 ∈ F, add F c

0 = [n] \ F0

G ∩ F c

0 = ∅ ⇒ G ⊂ F0 ⇒ G = F0

Background Counting Probabilistic Our results Conclusion

Many extra sets

Previous argument: [n]

≥ n+1

2

• construction best to extend

Background Counting Probabilistic Our results Conclusion

Many extra sets

Previous argument: [n]

≥ n+1

2

• construction best to extend

Theorem (Frankl, 1977; Ahlswede, 1980) Suppose

• [n]

≥k+1

• ≤ m ≤
• [n]

≥k

• . Then the minimum number of

disjoint pairs in a family of m sets is attained by some F with

• [n]

≥k+1

• ⊆ F ⊆

[n]

≥k

• .

Background Counting Probabilistic Our results Conclusion

Many extra sets

Previous argument: [n]

≥ n+1

2

• construction best to extend

Theorem (Frankl, 1977; Ahlswede, 1980) Suppose

• [n]

≥k+1

• ≤ m ≤
• [n]

≥k

• . Then the minimum number of

disjoint pairs in a family of m sets is attained by some F with

• [n]

≥k+1

• ⊆ F ⊆

[n]

≥k

• .

1 . . .

n−3 2 n−1 2 n+1 2 n+3 2

. . . n − 1 n

Background Counting Probabilistic Our results Conclusion

Many extra sets

Previous argument: [n]

≥ n+1

2

• construction best to extend

Theorem (Frankl, 1977; Ahlswede, 1980) Suppose

• [n]

≥k+1

• ≤ m ≤
• [n]

≥k

• . Then the minimum number of

disjoint pairs in a family of m sets is attained by some F with

• [n]

≥k+1

• ⊆ F ⊆

[n]

≥k

• .

1 . . .

n−3 2 n−1 2 n+1 2 n+3 2

. . . n − 1 n

Background Counting Probabilistic Our results Conclusion

Many extra sets

Previous argument: [n]

≥ n+1

2

• construction best to extend

Theorem (Frankl, 1977; Ahlswede, 1980) Suppose

• [n]

≥k+1

• ≤ m ≤
• [n]

≥k

• . Then the minimum number of

disjoint pairs in a family of m sets is attained by some F with

• [n]

≥k+1

• ⊆ F ⊆

[n]

≥k

• .

1 . . .

n−3 2 n−1 2 n+1 2 n+3 2

. . . n − 1 n

Background Counting Probabilistic Our results Conclusion

Many extra sets

Previous argument: [n]

≥ n+1

2

• construction best to extend

Theorem (Frankl, 1977; Ahlswede, 1980) Suppose

• [n]

≥k+1

• ≤ m ≤
• [n]

≥k

• . Then the minimum number of

disjoint pairs in a family of m sets is attained by some F with

• [n]

≥k+1

• ⊆ F ⊆

[n]

≥k

• .

1 . . .

n−3 2 n−1 2 n+1 2 n+3 2

. . . n − 1 n

Background Counting Probabilistic Our results Conclusion

Many extra sets

Previous argument: [n]

≥ n+1

2

• construction best to extend

Theorem (Frankl, 1977; Ahlswede, 1980) Suppose

• [n]

≥k+1

• ≤ m ≤
• [n]

≥k

• . Then the minimum number of

disjoint pairs in a family of m sets is attained by some F with

• [n]

≥k+1

• ⊆ F ⊆

[n]

≥k

• .

1 . . .

n−3 2 n−1 2 n+1 2 n+3 2

. . . n − 1 n

Background Counting Probabilistic Our results Conclusion

Many extra sets

Previous argument: [n]

≥ n+1

2

• construction best to extend

Theorem (Frankl, 1977; Ahlswede, 1980) Suppose

• [n]

≥k+1

• ≤ m ≤
• [n]

≥k

• . Then the minimum number of

disjoint pairs in a family of m sets is attained by some F with

• [n]

≥k+1

• ⊆ F ⊆

[n]

≥k

• .

1 . . .

n−3 2 n−1 2 n+1 2 n+3 2

. . . n − 1 n

Background Counting Probabilistic Our results Conclusion

Many extra sets

Previous argument: [n]

≥ n+1

2

• construction best to extend

Theorem (Frankl, 1977; Ahlswede, 1980) Suppose

• [n]

≥k+1

• ≤ m ≤
• [n]

≥k

• . Then the minimum number of

disjoint pairs in a family of m sets is attained by some F with

• [n]

≥k+1

• ⊆ F ⊆

[n]

≥k

• .

1 . . .

n−3 2 n−1 2 n+1 2 n+3 2

. . . n − 1 n

Background Counting Probabilistic Our results Conclusion

Many extra sets

Previous argument: [n]

≥ n+1

2

• construction best to extend

Theorem (Frankl, 1977; Ahlswede, 1980) Suppose

• [n]

≥k+1

• ≤ m ≤
• [n]

≥k

• . Then the minimum number of

disjoint pairs in a family of m sets is attained by some F with

• [n]

≥k+1

• ⊆ F ⊆

[n]

≥k

• .

1 . . .

n−3 2 n−1 2 n+1 2 n+3 2

. . . n − 1 n

Background Counting Probabilistic Our results Conclusion

Many extra sets

Previous argument: [n]

≥ n+1

2

• construction best to extend

Theorem (Frankl, 1977; Ahlswede, 1980) Suppose

• [n]

≥k+1

• ≤ m ≤
• [n]

≥k

• . Then the minimum number of

disjoint pairs in a family of m sets is attained by some F with

• [n]

≥k+1

• ⊆ F ⊆

[n]

≥k

• .

1 . . .

n−3 2 n−1 2 n+1 2 n+3 2

. . . n − 1 n

Background Counting Probabilistic Our results Conclusion

The k-uniform setting

Extremal families determined by pairs in [n]

k

Background Counting Probabilistic Our results Conclusion

The k-uniform setting

Extremal families determined by pairs in [n]

k

• Question (Ahlswede, 1980)

Which families in [n]

k

• minimise the number of disjoint pairs?

Background Counting Probabilistic Our results Conclusion

The k-uniform setting

Extremal families determined by pairs in [n]

k

• Question (Ahlswede, 1980)

Which families in [n]

k

• minimise the number of disjoint pairs?

Natural construction: union of stars 1

Background Counting Probabilistic Our results Conclusion

The k-uniform setting

Extremal families determined by pairs in [n]

k

• Question (Ahlswede, 1980)

Which families in [n]

k

• minimise the number of disjoint pairs?

Natural construction: union of stars 1

Background Counting Probabilistic Our results Conclusion

The k-uniform setting

Extremal families determined by pairs in [n]

k

• Question (Ahlswede, 1980)

Which families in [n]

k

• minimise the number of disjoint pairs?

Natural construction: union of stars 1

Background Counting Probabilistic Our results Conclusion

The k-uniform setting

Extremal families determined by pairs in [n]

k

• Question (Ahlswede, 1980)

Which families in [n]

k

• minimise the number of disjoint pairs?

Natural construction: union of stars 1 2

Background Counting Probabilistic Our results Conclusion

The k-uniform setting

Extremal families determined by pairs in [n]

k

• Question (Ahlswede, 1980)

Which families in [n]

k

• minimise the number of disjoint pairs?

Natural construction: union of stars 1 2

Background Counting Probabilistic Our results Conclusion

The k-uniform setting

Extremal families determined by pairs in [n]

k

• Question (Ahlswede, 1980)

Which families in [n]

k

• minimise the number of disjoint pairs?

Natural construction: union of stars 1 2

Background Counting Probabilistic Our results Conclusion

The k-uniform setting

Extremal families determined by pairs in [n]

k

• Question (Ahlswede, 1980)

Which families in [n]

k

• minimise the number of disjoint pairs?

Natural construction: union of stars 1 2 . . . s

Background Counting Probabilistic Our results Conclusion

The k-uniform setting

Extremal families determined by pairs in [n]

k

• Question (Ahlswede, 1980)

Which families in [n]

k

• minimise the number of disjoint pairs?

Natural construction: union of stars 1 2 . . . s

Background Counting Probabilistic Our results Conclusion

The k-uniform setting

Extremal families determined by pairs in [n]

k

• Question (Ahlswede, 1980)

Which families in [n]

k

• minimise the number of disjoint pairs?

Natural construction: union of stars 1 2 . . . s

Background Counting Probabilistic Our results Conclusion

A conjecture

Conjecture (Bollob´ as-Leader, 2000) For small families, a union of stars minimises the number of disjoint pairs.

Background Counting Probabilistic Our results Conclusion

A conjecture

Conjecture (Bollob´ as-Leader, 2000) For small families, a union of stars minimises the number of disjoint pairs. A family is optimal iff its complement is

Conjecture ⇒ for large families, a clique is optimal

Background Counting Probabilistic Our results Conclusion

A conjecture

Conjecture (Bollob´ as-Leader, 2000) For small families, a union of stars minimises the number of disjoint pairs. A family is optimal iff its complement is

Conjecture ⇒ for large families, a clique is optimal

Ahlswede-Katona resolved the case k = 2

Background Counting Probabilistic Our results Conclusion

A conjecture

Conjecture (Bollob´ as-Leader, 2000) For small families, a union of stars minimises the number of disjoint pairs. A family is optimal iff its complement is

Conjecture ⇒ for large families, a clique is optimal

Ahlswede-Katona resolved the case k = 2 Theorem (Ahlswede-Katona, 1978) The n-vertex graph with m edges and the minimum number of disjoint pairs of edges is: a union of stars if m < 1

2

n

2

• − n

2, and

a clique if m > 1

2

n

2

• + n

2.

Background Counting Probabilistic Our results Conclusion

Probabilistic supersaturation

Definition (Random subfamilies) Given 0 ≤ p ≤ 1, and a set family F, let Fp denote the random subfamily where every set is retained independently with probability p.

Background Counting Probabilistic Our results Conclusion

Probabilistic supersaturation

Definition (Random subfamilies) Given 0 ≤ p ≤ 1, and a set family F, let Fp denote the random subfamily where every set is retained independently with probability p. Question (Katona-Katona-Katona, 2012) Which families F of m subsets of [n] maximise P(Fp is intersecting)?

Background Counting Probabilistic Our results Conclusion

Previous results

Non-uniform setting (Katona et al, 2012) Solved for m ≤ 2n−1 +

• n−1

(n−3)/2

Background Counting Probabilistic Our results Conclusion

Previous results

Non-uniform setting (Katona et al, 2012) Solved for m ≤ 2n−1 +

• n−1

(n−3)/2

• (Katona et al, 2012) Conjectured the existence of nested

extremal families

Background Counting Probabilistic Our results Conclusion

Previous results

Non-uniform setting (Katona et al, 2012) Solved for m ≤ 2n−1 +

• n−1

(n−3)/2

• (Katona et al, 2012) Conjectured the existence of nested

extremal families (Russell, 2012) Extremal families are nested between the families [n]

≥k

Background Counting Probabilistic Our results Conclusion

Previous results

Non-uniform setting (Katona et al, 2012) Solved for m ≤ 2n−1 +

• n−1

(n−3)/2

• (Katona et al, 2012) Conjectured the existence of nested

extremal families (Russell, 2012) Extremal families are nested between the families [n]

≥k

• (Russell-Walters, 2013) No nested sequence from

[n]

≥3

• to

[n]

≥2

Background Counting Probabilistic Our results Conclusion

Previous results

Non-uniform setting (Katona et al, 2012) Solved for m ≤ 2n−1 +

• n−1

(n−3)/2

• (Katona et al, 2012) Conjectured the existence of nested

extremal families (Russell, 2012) Extremal families are nested between the families [n]

≥k

• (Russell-Walters, 2013) No nested sequence from

[n]

≥3

• to

[n]

≥2

• Uniform setting

Background Counting Probabilistic Our results Conclusion

Previous results

Non-uniform setting (Katona et al, 2012) Solved for m ≤ 2n−1 +

• n−1

(n−3)/2

• (Katona et al, 2012) Conjectured the existence of nested

extremal families (Russell, 2012) Extremal families are nested between the families [n]

≥k

• (Russell-Walters, 2013) No nested sequence from

[n]

≥3

• to

[n]

≥2

• Uniform setting

(Katona et al, 2012) Determined answer for m = n−1

k−1

• + 1

Background Counting Probabilistic Our results Conclusion

Link to counting

The two measures are closely related

Background Counting Probabilistic Our results Conclusion

Link to counting

The two measures are closely related Definition (Number of intersecting subfamilies) Given a set family F, let int(F, t) = |{G ⊂ F : |G| = t, G intersecting}|.

Background Counting Probabilistic Our results Conclusion

Link to counting

The two measures are closely related Definition (Number of intersecting subfamilies) Given a set family F, let int(F, t) = |{G ⊂ F : |G| = t, G intersecting}|. P(Fp is intersecting) =

• G⊆F intersecting

P(Fp = G) =

• G⊆F intersecting

p|G|(1 − p)|F\G| =

• t

int(F, t)pt(1 − p)m−t

Background Counting Probabilistic Our results Conclusion

Counting ⇒ probability

If int(F∗, t) ≥ int(F, t) for all other families F, then

Background Counting Probabilistic Our results Conclusion

Counting ⇒ probability

If int(F∗, t) ≥ int(F, t) for all other families F, then P(F∗

p is intersecting) =

• t

int(F∗, t)pt(1 − p)m−t ≥

• t

int(F, t)pt(1 − p)m−t = P(Fp is intersecting)

Background Counting Probabilistic Our results Conclusion

Counting ⇒ probability

If int(F∗, t) ≥ int(F, t) for all other families F, then P(F∗

p is intersecting) =

• t

int(F∗, t)pt(1 − p)m−t ≥

• t

int(F, t)pt(1 − p)m−t = P(Fp is intersecting) Hence F∗ is most probably intersecting for all p!

Background Counting Probabilistic Our results Conclusion

Probability ⇒ counting

P(Fp is intersecting) =

• t

int(F, t)pt(1 − p)t

Background Counting Probabilistic Our results Conclusion

Probability ⇒ counting

P(Fp is intersecting) =

• t

int(F, t)pt(1 − p)t Let p ≪ 2−n, so |Fp| ≤ 2 wvhp

Background Counting Probabilistic Our results Conclusion

Probability ⇒ counting

P(Fp is intersecting) =

• t

int(F, t)pt(1 − p)t Let p ≪ 2−n, so |Fp| ≤ 2 wvhp int(F, 0) = 1, int(F, 1) = m

Background Counting Probabilistic Our results Conclusion

Probability ⇒ counting

P(Fp is intersecting) =

• t

int(F, t)pt(1 − p)t Let p ≪ 2−n, so |Fp| ≤ 2 wvhp int(F, 0) = 1, int(F, 1) = m int(F, 2) = m

2

• − ( number of disjoint pairs )

Background Counting Probabilistic Our results Conclusion

Probability ⇒ counting

P(Fp is intersecting) =

• t

int(F, t)pt(1 − p)t Let p ≪ 2−n, so |Fp| ≤ 2 wvhp int(F, 0) = 1, int(F, 1) = m int(F, 2) = m

2

• − ( number of disjoint pairs )

Thus F is most probably intersecting iff it minimises the number of disjoint pairs

Background Counting Probabilistic Our results Conclusion

Our results

Counting supersaturation (with W. Gan, B. Sudakov)

Background Counting Probabilistic Our results Conclusion

Our results

Counting supersaturation (with W. Gan, B. Sudakov) Verify the Bollob´ as-Leader conjecture for small families

Background Counting Probabilistic Our results Conclusion

Our results

Counting supersaturation (with W. Gan, B. Sudakov) Verify the Bollob´ as-Leader conjecture for small families Obtain extensions to other settings

Background Counting Probabilistic Our results Conclusion

Our results

Counting supersaturation (with W. Gan, B. Sudakov) Verify the Bollob´ as-Leader conjecture for small families Obtain extensions to other settings Probabilistic supersaturation (with B. Sudakov)

Background Counting Probabilistic Our results Conclusion

Our results

Counting supersaturation (with W. Gan, B. Sudakov) Verify the Bollob´ as-Leader conjecture for small families Obtain extensions to other settings Probabilistic supersaturation (with B. Sudakov) (k = 2) Find most probably intersecting graphs that are not too dense

Background Counting Probabilistic Our results Conclusion

Our results

Counting supersaturation (with W. Gan, B. Sudakov) Verify the Bollob´ as-Leader conjecture for small families Obtain extensions to other settings Probabilistic supersaturation (with B. Sudakov) (k = 2) Find most probably intersecting graphs that are not too dense (k ≥ 3) For small families, show that most probably intersecting families are approximately unions of stars

Background Counting Probabilistic Our results Conclusion

Our results

Counting supersaturation (with W. Gan, B. Sudakov) Verify the Bollob´ as-Leader conjecture for small families Obtain extensions to other settings Probabilistic supersaturation (with B. Sudakov) (k = 2) Find most probably intersecting graphs that are not too dense (k ≥ 3) For small families, show that most probably intersecting families are approximately unions of stars (k ≥ 3) Determine extremal families exactly for specific sizes

Background Counting Probabilistic Our results Conclusion

Main theorem

We count disjoint pairs in small families of sets

Background Counting Probabilistic Our results Conclusion

Main theorem

We count disjoint pairs in small families of sets Theorem (D.-Gan-Sudakov, 2013+) Given n > 108k2s(k + s), and n

k

• −

n−s+1

k

• ≤ m ≤

n

k

• −

n−s

k

• ,

then the minimum number of disjoint pairs for a family of m sets in [n]

k

• is attained by taking s − 1 full stars and a partial star.

Background Counting Probabilistic Our results Conclusion

Heuristic: small covers

Almost all intersections take place within the cover

Background Counting Probabilistic Our results Conclusion

Heuristic: small covers

Almost all intersections take place within the cover X

Background Counting Probabilistic Our results Conclusion

Heuristic: small covers

Almost all intersections take place within the cover X F v

Background Counting Probabilistic Our results Conclusion

Heuristic: small covers

Almost all intersections take place within the cover X F v

# good intersections = deg(v) = Ω n−1

k−1

Background Counting Probabilistic Our results Conclusion

Heuristic: small covers

Almost all intersections take place within the cover X F v

# good intersections = deg(v) = Ω n−1

k−1

• # bad intersections ≤ k|X|

n−2

k−2

• = o

n−1

k−1

Background Counting Probabilistic Our results Conclusion

Outline of proof

• I. Induction

If there is a full star, we can remove it This enables us to shift sets

Background Counting Probabilistic Our results Conclusion

Outline of proof

• I. Induction

If there is a full star, we can remove it This enables us to shift sets

• II. High degrees

Double-counting ⇒ there are high-degree vertices

Background Counting Probabilistic Our results Conclusion

Outline of proof

• I. Induction

If there is a full star, we can remove it This enables us to shift sets

• II. High degrees

Double-counting ⇒ there are high-degree vertices

• III. Small covers

Shifting ⇒ every set has a high-degree vertex There cannot be many high-degree vertices

Background Counting Probabilistic Our results Conclusion

Outline of proof

• I. Induction

If there is a full star, we can remove it This enables us to shift sets

• II. High degrees

Double-counting ⇒ there are high-degree vertices

• III. Small covers

Shifting ⇒ every set has a high-degree vertex There cannot be many high-degree vertices

• IV. Exact structure

Inclusion-exclusion gives precise structure

Background Counting Probabilistic Our results Conclusion

Further results

Characterisation of all extremal families

Background Counting Probabilistic Our results Conclusion

Further results

Characterisation of all extremal families t-disjoint pairs:

We say F1, F2 are t-disjoint if |F1 ∩ F2| < t

Background Counting Probabilistic Our results Conclusion

Further results

Characterisation of all extremal families t-disjoint pairs:

We say F1, F2 are t-disjoint if |F1 ∩ F2| < t Using similar methods, we determine which small families minimise the number of t-disjoint pairs

Background Counting Probabilistic Our results Conclusion

Further results

Characterisation of all extremal families t-disjoint pairs:

We say F1, F2 are t-disjoint if |F1 ∩ F2| < t Using similar methods, we determine which small families minimise the number of t-disjoint pairs When t = k − 1, this is known as the Kleitman-West problem, and arises in connection to information theory

Background Counting Probabilistic Our results Conclusion

Further results

Characterisation of all extremal families t-disjoint pairs:

We say F1, F2 are t-disjoint if |F1 ∩ F2| < t Using similar methods, we determine which small families minimise the number of t-disjoint pairs When t = k − 1, this is known as the Kleitman-West problem, and arises in connection to information theory

Can also minimise the number of q-matchings

Background Counting Probabilistic Our results Conclusion

Open problems

Counting: Which medium-sized families are extremal?

Spectral arguments ⇒ stars and cliques are near-extremal Bollob´ as-Leader have a more general conjecture

Background Counting Probabilistic Our results Conclusion

Open problems

Counting: Which medium-sized families are extremal?

Spectral arguments ⇒ stars and cliques are near-extremal Bollob´ as-Leader have a more general conjecture

Probabilistic: Can we develop more probabilistic techniques?

What about when the extremal families depend on p?

Background Counting Probabilistic Our results Conclusion