Maximal Chains and Antichains in Finite Partially Ordered Setsgoddard/MINI/2009/Duffus.pdf · 2009....

Maximal Chains and Antichainsin Finite Partially Ordered Sets

Dwight Duffus

Mathematics and Computer Science DepartmentEmory UniversityAtlanta GA USA

The 24th Clemson Conferenceon Discrete Mathematics and Algorithms

22 October 2009

Outline

1 Classic Problems and Results on [Maximal] Chains and Antichains

2 Maximal Antichains in the Boolean Lattice 2n

Pairwise disjoint maximal antichains in 2n

Fibres in 2n

3 Maximal Chains and Antichains in Finite Partially Ordered Sets

Conditions on chain size yield pwd maximal antichains

Conditions on antichain size yield pwd maximal chains

Outline

Fibres in 2n

Outline

Fibres in 2n

Outline

Fibres in 2n

Outline

Fibres in 2n

Outline

Fibres in 2n

Outline

Fibres in 2n

Classic problems and results

Dilworth’s Max-Min Theorem [1942]:

A partially ordered set of finite width has a partition into width-manychains.

Dedekind’s Problem [1897]:

Determine the cardinality of FD(n) the free distributive lattice on ngenerators; that is, enumerate the antichains in the Boolean lattice 2n ofall subsets of an n-set.

Sperner’s Theorem [1928]:

The maximum size of an antichain in 2n is( nbn/2c

Determine the cardinality of FD(n) the free distributive lattice on ngenerators;

that is, enumerate the antichains in the Boolean lattice 2n ofall subsets of an n-set.

Proof of Sperner’s Theorem

Call a chain X0 ≺ . . . ≺ Xh symmetric if |X0|+ |Xh| = n.

2n has a partition into symmetric chains [SCD]:

Each symmetric chain contains one set of size bn/2c so there are( nbn/2c

chains in a SCD C(n). Any antichain A intersects any chain in at most 1element. Thus:

|A| =∑

C∈C(n)

|A ∩ C| ≤(

chains in a SCD C(n). Any antichain A intersects any chain in at most 1element.

|A| =∑

C∈C(n)

|A ∩ C| ≤(

chains in a SCD C(n). Any antichain A intersects any chain in at most 1element. Thus:

|A| =∑

C∈C(n)

|A ∩ C| ≤(

Simple questions about maximal chains and antichains

How many are there?

I For chains, this is an irritating but elementary combinatorialenumeration problem.

I For maximal chains this is not so interesting: n!

I For antichains, this is Dedekind’s problem.

I For maximal antichains, this seems hard.

For particular classes of partially ordered sets,

I Are there interesting ways to generate maximal antichains?

I Are there large families of pairwise disjoint maximal antichains?

How many are there?

Antichains in 2n Pairwise disjoint maximal antichains in 2n

A family of maximal antichains

Let S ⊆ [n] have 2k − 1 elements, k < n/2 and let AS be the set of allk-subsets of S and their complements:AS is a maximal antichain in 2n.

Consequence: There is a family of more than 1.067422n of pairwisedisjoint maximal antichains in 2n .

A family of maximal antichainsLet S ⊆ [n] have 2k − 1 elements, k < n/2 and let AS be the set of allk-subsets of S and their complements:

AS is a maximal antichain in 2n.

A family of maximal antichainsLet S ⊆ [n] have 2k − 1 elements, k < n/2 and let AS be the set of allk-subsets of S and their complements:AS is a maximal antichain in 2n.

��

2k−1

��

2k−1

Maximality:

��

2k−1

Maximality:

Any set A which does not have k elementsin common with S must miss at least k

elements of S and so must be contained in

some T for a k− S subset of T

��

2k−1

Maximality:

Antichain:

Maximality:

Antichain:

Given sets U and shown, they must

intersect, so that cannot be a subset of

��

2k−1

Maximality:

Antichain:

Given sets U and shown, they must

intersect, so that cannot be a subset of

��

2k−1

Antichains in 2n Fibres in 2n

Sets that intersect all maximal antichains

Call F ⊆ 2n a fibre of 2n if for every maximal antichain M, F ∩M 6= ∅.Given A ∈ 2n, let

C(A) = {B ∈ 2n | A ⊆ B or B ⊆ A}

and note that for all A, C(A) is a fibre.

Let f (n) be the minimum size of a fibre in 2n

For any A of size n/2, C(A) shows that f (n) ≤ 2n/2.

(Duffus, Sands, Winkler (1990)) Every fibre in 2n contains at leastn!/2n−1 distinct maximal chains.

( Luczak (1996)) 2n/3 ≤ f (n)

Improve these bounds on f (n):

2n/3 ≤ f (n) ≤ 2n/2

Sets that intersect all maximal antichainsCall F ⊆ 2n a fibre of 2n if for every maximal antichain M, F ∩M 6= ∅.Given A ∈ 2n, let

C(A) = {B ∈ 2n | A ⊆ B or B ⊆ A}

( Luczak (1996)) 2n/3 ≤ f (n)

2n/3 ≤ f (n) ≤ 2n/2

C(A) = {B ∈ 2n | A ⊆ B or B ⊆ A}

( Luczak (1996)) 2n/3 ≤ f (n)

2n/3 ≤ f (n) ≤ 2n/2

C(A) = {B ∈ 2n | A ⊆ B or B ⊆ A}

( Luczak (1996)) 2n/3 ≤ f (n)

2n/3 ≤ f (n) ≤ 2n/2

C(A) = {B ∈ 2n | A ⊆ B or B ⊆ A}

( Luczak (1996)) 2n/3 ≤ f (n)

2n/3 ≤ f (n) ≤ 2n/2

C(A) = {B ∈ 2n | A ⊆ B or B ⊆ A}

( Luczak (1996)) 2n/3 ≤ f (n)

2n/3 ≤ f (n) ≤ 2n/2

C(A) = {B ∈ 2n | A ⊆ B or B ⊆ A}

( Luczak (1996)) 2n/3 ≤ f (n)

2n/3 ≤ f (n) ≤ 2n/2

C(A) = {B ∈ 2n | A ⊆ B or B ⊆ A}

( Luczak (1996)) 2n/3 ≤ f (n)

2n/3 ≤ f (n) ≤ 2n/2

Chains and antichains in general Conditions on chain size yield pwd maximal antichains

Chain length and pairwise disjoint maximal antichains

For any finite partially orderedset P, if all maximal chains inP have the same size, say n,then P has n pairwise disjointmaximal antichains, namely,its levels.

length(P) = l(P) = 5

If we want 2 disjoint maximal antichains, we need only avoid 1-elementchains - AKA isolated elements. But to ensure 3 pwd maximal antichains,we must restrict the sizes of maximal chains more carefully.

!!!! !

P (n, 3)

P(n, 3) has maximal chains of sizes n and2n − 2, and no 3 pwd maximal antichains.

However, this is as bad as things can get.

For all n ≥ 3, if every maximal chain C ofa partially ordered set P satisfies

n ≤ |C | ≤ 2n − 3

then P has 3 pwd maximal antichains.

In fact, there is a condition that yields k pwd maximal antichains, thoughit took some time to find it.

!!!! !

P (n, 3)

n ≤ |C | ≤ 2n − 3

!!!! !

P (n, 3)

n ≤ |C | ≤ 2n − 3

!!!! !

P (n, 3)

n ≤ |C | ≤ 2n − 3

!!!! !

P (n, 3)

n ≤ |C | ≤ 2n − 3

Theorem [Duffus and Sands - 2009]

Let n and k be integers with 3 ≤ k ≤ n and let P be a finite partiallyordered set. If every maximal chain C of P satisfies

n ≤ |C | ≤ n +n − k

k − 2

then P contains k pairwise disjoint maximal antichains.

Alternative formulation:

Let the shortest maximal chains of P have m elements and the longesthave M elements. Then P contains at least

⌊m − 2

M −m + 1

pairwise disjoint maximal antichains.

Theorem [Duffus and Sands - 2009]

Let n and k be integers with 3 ≤ k ≤ n and let P be a finite partiallyordered set. If every maximal chain C of P satisfies

n ≤ |C | ≤ n +n − k

k − 2

then P contains k pairwise disjoint maximal antichains.

Alternative formulation:

Let the shortest maximal chains of P have m elements and the longesthave M elements. Then P contains at least

⌊m − 2

M −m + 1

pairwise disjoint maximal antichains.

The theorem is sharp: for 4 ≤ k ≤ n, P(n, k) has all maximal chains withsize between n and n + n−k

k−2 + 1, and no k pwd maximal antichains.

1 1 1 1 1 1 1 1 1

n n n n n n n n n

C2 C3 · · · Ci+1 Ci+2 · · · Ck!2 Ck!1

"!(i+1)n

!(k!3)n

! ! ! ! ! ! ! ! ! !

! ! ! ! ! ! ! ! !! ! ! ! ! ! ! ! !! ! ! ! ! ! ! ! !

! ! ! ! ! ! ! ! !

P (n, k)

The theorem is sharp: for 4 ≤ k ≤ n, P(n, k) has all maximal chains withsize between n and n + n−k

k−2 + 1, and no k pwd maximal antichains.

1 1 1 1 1 1 1 1 1

n n n n n n n n n

C2 C3 · · · Ci+1 Ci+2 · · · Ck!2 Ck!1

"!(i+1)n

!(k!3)n

! ! ! ! ! ! ! ! ! !

! ! ! ! ! ! ! ! !! ! ! ! ! ! ! ! !! ! ! ! ! ! ! ! !

! ! ! ! ! ! ! ! !

P (n, k)

Chains and antichains in general Conditions on antichain size yield pwd maximal chains

Idle thought . . .

What about the dual question? Can we guarantee the existence of k pwdmaximal chains by bounding the sizes of maximal antichains?

It’s true that

if every maximal antichain contains at least 2 elements then thepartially ordered set has at least 2 pwd maximal chains,

if all maximal antichains have size n then the Dilworth decompositionyields n pwd maximal chains,

there are several famous examples of valid statements with validduals, and

the partially ordered sets P(n, k) are all 2-dimensional.

Meaning?

Idle thought . . .

It’s true that

Meaning?

Idle thought . . .

It’s true that

Meaning?

Idle thought . . .

It’s true that

Meaning?

Idle thought . . .

It’s true that

Meaning?

Idle thought . . .

It’s true that

Meaning?

Idle thought . . .

It’s true that

Meaning?

Given a partially ordered set P of dimension 2, that is, whose order is theintersection of 2 linear orders ≤1 and ≤2, there is a complementaryordered set Q – its ordering is ≤1 ∩ ≤d

2 , where ≤d2 is the dual of ≤2:

!!!! !

P (n, 3)

P’s max chains = Q’s max antichains

######

$$$$$$

######

1 2 · · · n! 1

! ! ! !! ! ! !

Q(n, 3)

P’s max antichains = Q’s max chains

!!!! !

P (n, 3)

######

$$$$$$

######

1 2 · · · n! 1

! ! ! !! ! ! !

Q(n, 3)

!!!! !

P (n, 3)

######

$$$$$$

######

1 2 · · · n! 1

! ! ! !! ! ! !

Q(n, 3)

Observation

Given 3 ≤ k ≤ n and a 2-dimensional partially ordered set Q all of whosemaximal antichains A satisfy

n ≤ |A| ≤ n +n − k

k − 2,

Q has k pwd maximal chains. Morever, the ordered sets Q(n, k) show thatthe result is sharp.

Question

For 3 ≤ k , what are the maximal intervals Ik so that whenever the size ofevery maximal antichain of an ordered set P lies in Ik , P contains k pwdmaximal chains?

Surely, we thought, the intervals are smaller in the general case.

Observation

n ≤ |A| ≤ n +n − k

k − 2,

Question

Observation

n ≤ |A| ≤ n +n − k

k − 2,

Question

To our surprise . . .

Theorem [Howard and Trotter - 2009]

Let n and k be integers with 3 ≤ k ≤ n and let P be a finite partiallyordered set. If every maximal antichain A of P satisfies

n ≤ |A| ≤ n +n − k

k − 2

then P contains k pairwise disjoint maximal chains.

Moreover, the result remains sharp, due to Q(n, k), the complementary(2-dimensional) partially ordered set to P(n, k).

To our surprise . . .

Theorem [Howard and Trotter - 2009]

Let n and k be integers with 3 ≤ k ≤ n and let P be a finite partiallyordered set. If every maximal antichain A of P satisfies

n ≤ |A| ≤ n +n − k

k − 2

then P contains k pairwise disjoint maximal chains.

Moreover, the result remains sharp, due to Q(n, k), the complementary(2-dimensional) partially ordered set to P(n, k).

Maximal Chains and Antichains in Finite Partially Ordered Setsgoddard/MINI/2009/Duffus.pdf · 2009....

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