Largest families of sets under conditions defined by a given poset - - PowerPoint PPT Presentation

Largest families of sets under conditions defined by a given poset

Gyula O.H. Katona R´ enyi Institute, Budapest Jiao Tong University Combinatorics Seminar October 21, 2014

An ancient combinatorial problem

Let [n] = {1, 2, . . . , n} be a finite set. Question. Find the maximum number of subsets A of [n] such that A ⊂ B holds for them.

1

Theorem (Sperner, 1928) If A is a family of distinct subsets of [n] without inclusion (A, B ∈ A implies A ⊂ B) then |A| ≤ n n

2

• .

2

Theorem (Sperner, 1928) If A is a family of distinct subsets of [n] without inclusion (A, B ∈ A implies A ⊂ B) then |A| ≤ n n

2

• .

Sharp!

3

Theorem (Sperner, 1928) If A is a family of distinct subsets of [n] without inclusion (A, B ∈ A implies A ⊂ B) then |A| ≤ n n

2

• .

Sharp! Exact maximum!

4

Theorem (Erd˝

• s, 1938)

If A is a family of distinct subsets of [n] without k + 1 distinct members satisfying A1 ⊂ A2 ⊂ . . . ⊂ Ak+1 ∈ A then |A| ≤

• k largest binomial coefficients of order n.

5

Theorem (Erd˝

• s, 1938)

If A is a family of distinct subsets of [n] without k + 1 distinct members satisfying A1 ⊂ A2 ⊂ . . . ⊂ Ak+1 ∈ A then |A| ≤

• k largest binomial coefficients of order n.

Sharp! Exact maximum!

6

Consider generalizations, where the excluded pattern is expressed only by inclusion.

7

Partially ordered set (Poset)

P = (X, <) where (i) at most one of <, =, > holds, (ii) < is transitive. (a, b ∈ X are comparable in P iff a < b, a = b or a > b.) In our case X = 2[n] and A < B in this poset iff A ⊂ B. Notation: Bn = (2[n], ⊂).

8

9

Illustration of the Sperner theorem, poset form

10

Generalizations of Sperner theorem

• Notation. La(n, P) =

the maximum number of elements of Bn such that the poset induced by these elements does not contain P as a subposet.

11

12

Example 1: P = I

Theorem (Sperner) La(n, I) = n ⌊n

2⌋

• .

13

Example 2: P = Pk+1

Theorem (Erd˝

• s)

La(n, Pk+1) = k largest binomial coefficients of order n.

14

Example 3:

Vr = {a, b1, . . . , br} where a < b1, . . . a < br

15

Construction for V2.

16

Find many sets A1, . . . , Am of size n

2

• + 1

such that |Ai ∩ Aj| < n

2

• (i = j).

This is an old open problem of coding theory. Known: n n

2

• 1

n ≤ max m ≤ n n

2

• 2

n + O 1 n2

• .

Graham and Sloane, 1980

17

Theorem (K-Tarj´ an, 1983) n n

2

• 1 + 1

n + Ω 1 n2

• ≤ La(n, V2) ≤

n n

2

• 1 + 2

n + O 1 n2

• .

Hard to find the right constant.

18

Theorem (De Bonis-K, 2007) n n

2

• 1 + r

n + Ω 1 n2

• ≤ La(n, Vr+1) ≤

n n

2

• 1 + 2r

n + O 1 n2

• .

19

Theorem (Griggs-K, 2008) n n

2

• 1 + 1

n + Ω 1 n2

• ≤ La(n, N) ≤

n n

2

• 1 + 2

n + O 1 n2

• .
• Remark. The estimates, up to the first two terms are the same as for V2.

20

But!

It is interesting to mention that the “La” function will jump if the excluded poset contains one more relation. The butterfly ⋊

⋉ contains 4 elements:

a, b, c, d with a < c, a < d, b < c, b < d. Theorem (De Bonis, K, Swanepoel, 2005) Let n ≥ 3. Then La(n, ⋊ ⋉) =

• n

⌊n/2⌋

• +
• n

⌊n/2⌋ + 1

• .

21

But!

It is interesting to mention that the “La” function will jump if the excluded poset contains one more relation. The butterfly ⋊

⋉ contains 4 elements:

a, b, c, d with a < c, a < d, b < c, b < d. Theorem (De Bonis, K, Swanepoel, 2005) Let n ≥ 3. Then La(n, ⋊ ⋉) =

• n

⌊n/2⌋

• +
• n

⌊n/2⌋ + 1

• .

A BOOK proof by P´ eter Burcsi and D´ aniel T. Nagy, 2013.

22

The height h(P) of the poset P is the length of the longest chain. Theorem (Burcsi, D.T. Nagy, 2013) La(n, P) ≤ |P| + h(P) 2 − 1 n ⌊n/2⌋

• .

23

The height h(P) of the poset P is the length of the longest chain. Theorem (Burcsi, D.T. Nagy, 2013) La(n, P) ≤ |P| + h(P) 2 − 1 n ⌊n/2⌋

• .

They also found infinitely many P with equality.

24

The height h(P) of the poset P is the length of the longest chain. Theorem (Burcsi, D.T. Nagy, 2013) La(n, P) ≤ |P| + h(P) 2 − 1 n ⌊n/2⌋

• .

They also found infinitely many P with equality. Improvements of the bound by Hong-Bin Chen and Wei-Tian Li.

25

asymmetric butterfly

26

asymmetric butterfly

Theorem (Abishek Methuku and Casey Tompkins) Let n ≥ 3. Then La(n, ) =

• n

⌊n/2⌋

• +
• n

⌊n/2⌋ + 1

• .

27

Results for trees

A poset is a tree if it Hasse diagram is a tree.

28

Results for trees

A poset is a tree if it Hasse diagram is a tree. Theorem (Griggs-Linyuan Lincoln Lu, 2009) Let T be a tree and suppose that it has two levels, then n ⌊n

2⌋

1 + Ω 1 n

• ≤ La(n, T) ≤

n ⌊n

2⌋

1 + O 1 n

• .

Theorem (Bukh, 2010) Let T be a tree. Then (h(T)−1) n ⌊n

2⌋

1 + Ω 1 n

• ≤ La(n, T) ≤ (h(T)−1)

n ⌊n

2⌋

1 + O 1 n

• .

29

The diamond problem

Let P be the following poset: P = D is called the diamond.

• n

⌊n/2⌋

• +
• n

⌊n/2⌋ + 1

• ≤ La(n, D).

(Two middle levels.)

30

Theorem (Axenovich-Manske-Martin, 2012) La(n, D) ≤ 2.283 n ⌊n

2⌋

1 + O 1 n

• .

31

Theorem (Axenovich-Manske-Martin, 2012) La(n, D) ≤ 2.283 n ⌊n

2⌋

1 + O 1 n

• .

Theorem (Griggs-Lu Linyuan-Li Wei-Tian, 2012) La(n, D) ≤ 2.273 n ⌊n

2⌋

1 + O 1 n

• .

32

Theorem (Axenovich-Manske-Martin, 2012) La(n, D) ≤ 2.283 n ⌊n

2⌋

1 + O 1 n

• .

Theorem (Griggs-Lu Linyuan-Li Wei-Tian, 2012) La(n, D) ≤ 2.273 n ⌊n

2⌋

1 + O 1 n

• .

Theorem (Kramer-Martin-Young, 2012) La(n, D) ≤ 2.25 n ⌊n

2⌋

1 + O 1 n

• .

33

An easier problem: restrict ourselves to the 3 middle levels.

34

An easier problem: restrict ourselves to the 3 middle levels. Theorem (Manske-Shen, 2013) If F is in the 3 middle levels and contains no diamond then |F| ≤ (2.15471 + o(1)) n ⌊n

2⌋

• .

35

An easier problem: restrict ourselves to the 3 middle levels. Theorem (Manske-Shen, 2013) If F is in the 3 middle levels and contains no diamond then |F| ≤ (2.15471 + o(1)) n ⌊n

2⌋

• .

Theorem (Balogh-Hu-Lidick´ y-Liu, 2014) If F is in the 3 middle levels and contains no diamond then |F| ≤ (2.15121 + o(1)) n ⌊n

2⌋

• .

36

Czabarka, Dutle, Johnston and Sz´ ekely recently gave a class of very nice algebraic constructions with asymptotic equality.

37

Conjecture (everybody) La(n, P) n

⌊n

2⌋

• tends to an integer for every poset P.

38

Conjecture (everybody) La(n, P) n

⌊n

2⌋

• tends to an integer for every poset P.

The first application of these type of theorems (namely Bukh’s theorem) is due to Patk´

• s.

39

A new type of generalization of the Sperner theorem. Given a ”small” poset P, find the maximum number of copies of P in Bn in such a way that no two elements in different copies are comparable.

40

A new type of generalization of the Sperner theorem. Given a ”small” poset P, find the maximum number of copies of P in Bn in such a way that no two elements in different copies are comparable.

41

A new type of generalization of the Sperner theorem. Given a ”small” poset P, find the maximum number of copies of P in Bn in such a way that no two elements in different copies are comparable.

42

This maximum is denoted by LA(n, P). The longest path in P is h(P). Theorem LA(n, P) ≤ n−h(P )+1

⌊n−h(P )+1

2

⌋

• .

This is sharp when P is a subposet of Bh(P )−1:

43

44

Observation Suppose P ∗ is an embedding into Bn, a < b < c ∈ Bn, a, c ∈ P ∗. If d is incomparable with the elements of P ∗ then b and d are also incomparable.

45

Observation Suppose P ∗ is an embedding into Bn, a < b < c ∈ Bn, a, c ∈ P ∗. If d is incomparable with the elements of P ∗ then b and d are also incomparable. Proof Indirect way:

46

Observation Suppose P ∗ is an embedding into Bn, a < b < c ∈ Bn, a, c ∈ P ∗. If d is incomparable with the elements of P ∗ then b and d are also incomparable. Proof Indirect way:

47

Convex closure of P ∗: add all such bs cc(P ∗) = P ∗ ∪ {b : a < b < c for some a, c ∈ P ∗}.

48

Convex closure of P ∗: add all such bs cc(P ∗) = P ∗ ∪ {b : a < b < c for some a, c ∈ P ∗}. Consequence If P ∗

1 and P ∗ 2 are two incomparable embeddings of P

then cc(P ∗

1 ) and cc(P ∗ 2 ) are also incomparable.

The embedding P ∗ is called convex if cc(P ∗) = P ∗.

49

50

51

52

One can prove that

m

• i=1

|Q∗

i | ≤

n n

2

• asymptotically holds.

53

One can prove that

m

• i=1

|Q∗

i | ≤

n n

2

• asymptotically holds.

What is known about Q∗

i ?

54

One can prove that

m

• i=1

|Q∗

i | ≤

n n

2

• asymptotically holds.

What is known about Q∗

i ?

Q∗

i ⊃ P convex embedding.

55

One can prove that

m

• i=1

|Q∗

i | ≤

n n

2

• asymptotically holds.

What is known about Q∗

i ?

Q∗

i ⊃ P convex embedding.

In other words: Q ⊃ P having a convex embedding Q∗

i .

56

One can prove that

m

• i=1

|Q∗

i | ≤

n n

2

• asymptotically holds.

What is known about Q∗

i ?

Q∗

i ⊃ P convex embedding.

In other words: Q ⊃ P having a convex embedding Q∗

i .

Needed: min |Q∗

i | under these conditions.

57

Example

58

Example

59

Example

60

Example

61

The convex embedding number of the poset P is cen(P) = min{|Q| : Q ⊃ P, Q has a convex embedding }.

62

The convex embedding number of the poset P is cen(P) = min{|Q| : Q ⊃ P, Q has a convex embedding }.

m

• i=1

|Q∗

i | ≤

n n

2

• implies

mcen(P) ≤ n n

2

• ,

asymptotically.

63

Theorem (Dove-Griggs, Katona-D.T. Nagy, independently) LA(n, P) = n n

2

• 1

cen(P) + o(n)

• .

64

Theorem (Dove-Griggs, Katona-D.T. Nagy, independently) LA(n, P) = n n

2

• 1

cen(P) + o(n)

• .

Particular case: P can be embedded into Bh(P )−1.

65

Theorem (Dove-Griggs, Katona-D.T. Nagy, independently) LA(n, P) = n n

2

• 1

cen(P) + o(n)

• .

Particular case: P can be embedded into Bh(P )−1. Then the smallest Q is Bh(P )−1.

66

Theorem (Dove-Griggs, Katona-D.T. Nagy, independently) LA(n, P) = n n

2

• 1

cen(P) + o(n)

• .

Particular case: P can be embedded into Bh(P )−1. Then the smallest Q is Bh(P )−1. n − h(P) + 1 ⌊n−h(P )+1

2

⌋

• ∼

1 2h(P )−1 n n

2

• 67

A construction for P = V

68

A construction for P = V

69

A construction for P = V

70

A construction for P = V

71

A construction for P = V

72

A construction for P = V

73

A construction for P = V

74

A construction for P = V

n − 2i − 2

n 2 − 2i − 1

• 75

Conjecture If n is even then LA(n, V ) =

• i=0

n − 2i − 2

n 2 − 2i − 1

• .

76

77

78