A Fresh Look at Some Old Extremal Problems William T. Trotter Georgia Institute of Technology November 20, 2006.

A Fresh Look at SomeOld Extremal Problems

William T. TrotterGeorgia Institute of

TechnologyNovember 20, 2006

Note to the ReaderThese slides are an expanded version of a presentation made at the 2006 Excill Conference held November 18 – 20 at the University of Illinois at Urbana/Champaign. It is hoped that the additional material will be of value for students and postdocs – as well as those who were not present for the talk.

Happy to answer questions. Of course, corrections and coments are welcomed.

Tom Trotter

[email protected]

Partially Ordered Sets - Posets

Remark We consider finite posets like the one shown to the left, assuming that the reader is familiar with the conventions that make, for example (a) 27 < 15, (b) 5 is a maximal element, and (c) 11 is incomparable to 19.

Chains

1. A set C of points in a poset P is called a chain if any distinct pair of points from C is comparable. Any singleton set is a chain.

2. The family of all chains in a poset is partially ordered by set inclusion. The maximal elements in this poset are called maximal chains.

3. A chain C is maximum if no other chain contains more points than C. Maximal chains need not be maximum.

4. The height of P is the size of a maximum chain.

Antichains

1. A set A of points in a poset P is called an antichain if any distinct pair of points from A is incomparable. Any singleton set is an antichain.

2. The family of all antichains in a poset is partially ordered by set inclusion. The maximal elements in this poset are called maximal antichains.

3. An antichain A is maximum if no other antichain contains more points than A. Maximal antichains need not be maximum.

4. The width of P is the size of a maximum antichain.

Chains and Antichains

1. {6,7,19,28} is a chain. It is not maximal.

2. {12,13,16,30} is an antichain. It is not maximal.

3. {8,13,34,35} is a maximal chain. It is not maximum.

4. {12,13,30,24,16,19,14,25} is a maximal antichain. It is not maximum.

Dilworth’s Theorem and Its Dual

Theorem (Dilworth, 1950) A poset P of width w can be partioned into w chains.

Theorem (Folklore) A poset P of height h can be partitioned into h antichains.

Remarks As we will see, the second of these two results is trivial, while the first admits an elegant combinatorial proof. Dilworth’s theorem is typically grouped with other combinatorial theorems with a common LP flavor, such as Hall’s matching theorem, Tutte’s 1-factor theorem, the Konig/Egervary theorem and Menger’s theorem.

Proof of Dual DilworthProof For each i, let, Ai consist of those elements x from P for which the longest chain in P with x as its largest element has i elements. Evidently, each Ai is an antichain. Furthermore, the number of non-empty antichains in the resulting partition is just h, the height of P. Also, a chain C of size h can be easily found using back-tracking, starting from any element of Ah.

Algorithm A1 is just the set of minimal elements of P. Thereafter, Ai+1 is just the set of minimal elements of the poset resulting from the removal of A1, A2, …, Ai.

Example The next slide illustrates this construction for a poset of height 7.

A Poset of Height 7 and aPartition into 7 Antichains

Note The red points form a maximum chain of size 7.

The Proof of Dilworth’s Theorem (1)

Proof True when width w = 1 and thus when |P| = 1. Assume valid when |P| ≤ k. Then consider a poset P with |P| = k + 1.

For each maximal antichain A, let D(A) = {x : x < a for some a in A}, and U(A) = {x : x > a for some a in A}. Evidently, P = A D(A) U(A) is a partition into pairwise disjoint sets.

Case 1. There exists a maximum antichain A with both D(A) and U(A) non-empty.

Label the elements of A as a1, a2, …, aw. Then apply the inductive hypothesis to A D(A), which has at most k points, since U(A) is non-empty. WLOG, we obtain a chain partition C1, C2, …, Cw of A D(A) with ai the greatest element of Ci for each i = 1, 2, …, w.

The Proof of Dilworth’s Theorem (2)

Then apply the inductive hypothesis to A U(A). WLOG, we obtain a chain partition C’1, C’2, …, C’w with ai the least element of C’i for each i. Then Ci C’i is a chain for each i = 1, 2, …, w and these w chains cover P.

Case 2. For every maximum antichain A, at least one of D(A) and U(A) is empty.

Choose a maximal element y. Then choose a minimal element x with x ≤ y in P. Note that we allow x = y. Regardless, C = {x, y} is a chain – of either one or two points - and the width of P - C is w – 1. Partition P - C into w – 1 chains, and then add chain C to obtain the desired chain partition of P.

A Poset of Width 11 and aPartition into 11 chains

Note The red points form a maximum antichain of size 11.

Linear Extensions

L1 = b < e < a < d < g < c < f

L2 = a < c < b < d < g < e < f

Let L be a linear order on the ground set of a poset P. We call L a linear extension of P if x > y in L whenever x > y in P.

Example L1 and L2 are linear extensions of the poset P.

Realizers of Posets

L1 = b < e < a < d < g < c < f

L2 = a < c < b < d < g < e < f

L3 = a < c < b < e < f < d < g

L4 = b < e < a < c < f < d < g

L5 = a < b < d < g < e < c < f

A family F = {L1, L2, …, Lt} of linear extensions of P is a realizer of P if P = F, i.e., whenever x is incomparable to y in P, there is some Li in F with x > y in Li.

Every Poset Has a Realizer

Lemma (Szpilrajn) If F is the family of all linear extensions of P, then F is a realizer of P, i.e., whenever x is incomparable to y in P, there is some L in F with x > y in L.

Note This lemma is completely trivial for finite posets.

The Dimension of a Poset

L1 = b < e < a < d < g < c < f

L2 = a < c < b < d < g < e < f

L3 = a < c < b < e < f < d < gThe dimension of a poset is the minimum

size of a realizer. This realizer shows dim(P) ≤ 3. In fact, it is an easy exercise to show that

dim(P) = 3

Complexity Issues

Theorem (Yannakakis) For fixed t ≥ 3, the question dim(P) ≤ t? is NP-complete.

Fact There is a simple poly-time algorithm for testing whether dim(P) ≤ 2.

Remark The question dim(P) ≤ t? is NP-complete even for height 2 posets when t ≥ 4. However, it is not known whether testing dim(P) ≤ 3 is NP-complete for height 2 posets. This anomaly may result from the connection with Schndyer’s theorem.

A Continuing Example

Remark The poset shown below has a very large number of linear extensions and finding its dimension may be a daunting challenge. As the talk continues, we will get – step by step – improved bounds on its dimension. But if you want to give it a go, try to determine its dimension just from the definition.

Cartesian ProductsRemark If P and Q are posets, the cartesian product P × Q of P and Q is the poset whose ground set is the cartesian product of the ground sets of P and Q with (x, y) ≤ (x’, y’) in P × Q if and only if x ≤ x’ in P and y ≤ y’ in Q.

Fact dim(P × Q) ≤ dim(P) + dim(Q).

Theorem (Baker) dim(P × Q) = dim(P) + dim(Q) if both P and Q have distinct greatest and least elements.

Note Trotter showed that dim(Sn × Sn) = 2n – 2 for all n ≥ 3, but it is not known whether there exist posets P and Q for which dim(P × Q) < dim(P) + dim(Q) – 2.

Lexicographic SumsRemark If P is a poset and F = {Qx : x in P} is a family of posets indexed by the ground set of P, we define the lexicographic sum of F over P as the poset R whose ground set consists of all ordered pairs of the form (x, y) with y in Qx. The ordering on R is defined by (x, y) < (x’, y’) in R if either (1) x < x’ in P, or (2) x = x’ and y < y’ in Qx.

Fact If R is the lexicographic sum of the family F = {Qx : x in P}, then dim(R) = max{dim(P), max{dim(Qx) : x in P}}.

Remark A poset P is irreducible if the removal of any point from P lowers the dimension. Clearly an irreducible poset cannot be written as a lexicographic sum – except in a trivial way, i.e., when either (1) |P| = 1, or (b) |Qx| = 1 for all x in P.

Basic Properties of Dimension

1. Dimension is monotonic, i.e., if P is a subposet of Q, then dim(P) ≤ dim(Q).

2. dim(Pd) = dim(P), where Pd is the dual of P, i.e., Pd has the same ground set as P with x > y in Pd if and only if x < y in P.

3. Dimension is “continuous”, i.e., dim(P) ≤ 1 + dim(P – x) for all x in P.

Remark The first and second properties are trivial. The third requires a little bit of work.

Update 1

Remark The fact that the removal of a point decreases the dimension by at most one implies that dim(P) ≤ |P| for every poset P. In this case, we have 12 points, so dim(P) ≤ 12.

Standard Examples

Fact For n ≥ 2, the standard example Sn is a poset of dimension n.

Sn

Note If L is a linear extension of Sn, there can only be one value of i for which ai > bi in L. Furthermore, if F is a family of linear extensions of Sn, then F is a realizer if and only if for each i = 1, 2, …, n, there is some L in F with ai > bi in L.

Interval OrdersA poset P is an interval order if there exists a function I assigning to each x in P a closed interval I(x) = [ax, bx] of the real line R so that x < y in P if and only if bx < ay in R.

Characterizing Interval Orders

Theorem (Fishburn) A poset is an interval order if and only if it does not contain the standard example S2.

S2 = 2 + 2

Finding Interval Representations

Theorem (Greenough) Let P be an interval order. For each x in P, let D(x) = {u : u < x in P}. Note that any two of these down sets are ordered by inclusion. Label the distinct down sets D(x) so that

D1 D2 … Dm

Also, label the up sets so that

Un Un-1 … U1

Then m = n. Furthermore, if x in P is associated with the interval I(x) = [i, j] where D(x) = Di and U(x) = Uj, then we have an interval representation of P using the minimum number of distinct end points.

Canonical Interval OrdersThe canonical interval order In consists of all intervals with integer end points from {1, 2, …, n}.

I5

Dimension of Interval Orders (1)

Theorem (Bogart and Trotter) For every t, there exists an integer n0 so that if n > n0, then dim(In) > t.

Proof Let F be a realizer of In. For each 3-element set {i < j < k}, choose L from F with [i, j] > [j, k] in L. If n is sufficiently large compared to |F|, it follows from Ramsey’s theorem that there is some 4-element subset H = {i, j, k, l} and some L in F so that all 3-elements of H are associated with L. This requires

[i, j] > [j, k] > [k, l] in L

which is a contradiction since [i, j] < [k, l] in P.

Dimension of Interval Orders (2)

Theorem (Füredi, Hajnal, Rödl and Trotter)

dim(In) = lg lg n + (1/2 + o(1)) lg lg lg n

Note All logarithms here are base 2.

Note The most important aspect of the preceding theorem is that there exist interval orders of large dimension. In some sense, this is analogous to the statement that there exist triangle-free graphs with large chromatic number. For a more accurate estimate on the growth rate of dim(In), we have the following asymptotic formula.

Dimension and Width (1)Lemma (Hiraguchi) If C is chain in a poset P,

then there exists a linear extension L of P with x > y in L whenever x is in C and x is incomparable to y in P.

Proof Start with C and then insert the points of P – C one at a time as low as possible in the linear order, consistent with the requirement that L be a linear extension of P.

Corollary (Hiraguchi) dim(P) ≤ width(P), for every poset P.

Proof If w = width(P), use Dilworth’s theorem to find a partition of P into w chains. Then apply the lemma to each of these chains to obtain a realizer of size w.

Update 2

Remark The width of the poset P is 7, so dim(P) ≤ 7.

Dimension and Width (2)

Note The inequality dim(P) ≤ width(P) is tight, since dim(Sn) = width(Sn) = n.

Dimension and Width (3)

Fact For n ≥ 2, the dimension and the width of this poset is n + 1. When n ≥ 3, it is irreducible, i.e., remove any point and the dimension drops to n.

Dimension and Width (4)

Problem Is the question dim(P) < width(P)? NP-complete?

Problem For fixed w, are there only finitely many irreducible posets satisifying dim(P) = width(P) = w.

Note If the answer to the second problem is yes, then the question dim(P) < width(P) admits a poly-time solution. On the other hand, if the answer to the second question is no, then the answer to the first could still go either way – unless of course P = NP.

Dimension and Cardinality (1)

Theorem (Hiraguchi) If |P| ≥ 4, then dim(P) ≤ |P|/2.

a. There exist x, y in P such that dim(P) ≤ 1 + dim(P – x – y).

b. There exist x, y, z, w in P such that dim(P) ≤ 2 + dim(P – x – y – z – w).

Sketch of the proof. It is relatively easy to show that for every poset P with |P| ≥ 5, either

As a result, it is straightforward to complete the proof by induction on |P|, once the result is known to hold for small values, say |P| ≤ 5.

Removable Pair Conjecture (1)

Lemma (Hiraguchi) If x is a minimal element in P, y is a maximal element in P and x is incomparable to y, then

dim(P) ≤ 1 + dim(P – x – y).

Lemma (Bogart and Trotter) If x and y are maximal elements in P and z < x whenever z < y, then

dim(P) ≤ 1 + dim(P – x – y).

Conjecture (Trotter, 1971)

If |P| ≥ 3, then there exist x, y in P such that

dim(P) ≤ 1 + dim(P – x – y).

Removable Pair Conjecture (2)

Remark The Removable Pair Conjecture may admit a simple proof. On the other hand, a counter-example will require a quite clever construction.

Remark An incomparable pair (x, y) in a poset P is called a critical pair when (a) z < y whenever z < x, and (b) w > x whenever w > y. Bogart and Trotter conjectured that removing a critical pair always decreases the dimension by at most 1. This was disproved by Reuter, who found a 4-dimensional poset containing a critical pair whose removal left a 2-dimensional subposet. Subsequently, Kierstead and Trotter found such posets for every dimension larger than 4.

Update 3

Remark Our poset has 12 points, so

dim(P) ≤ 12/2 = 6.

Dimension and Cardinality (2)

Note The inequality dim(P) ≤ |P|/2 when |P| ≥ 4 is tight, since for all n ≥ 2, dim(Sn) = n.

Dimension and Cardinality (3)

Fact These posets are 3-irreducible, i.e., they have dimension 3 and the removal of any point lowers the dimension to 2. The full list of all 3-irreducible posets is known. It consists (up to duality) of 7 infinite families and 10 other examples.

Dimension and Cardinality (4)

Theorem (Bogart and Trotter) If n ≥ 3 and |P| = 2n, then dim(P) < n unless P is Sn except when n = 3 and P (or its dual) is the chevron.

Remark Most of the difficulty in proving this theorem comes just with establishing that S8 is the only 4-irreducible poset on 8 vertices. Once this is one, the full result follows from application of removal theorems.

Dimension and Cardinality (5)

Theorem If n ≥ 4 and |P| ≤ 2n + 1, then dim(P) < n unless P contains Sn.

Note The proof of this theorem is very lengthy, and no entirely complete version has ever been written down. Part of the difficulty stems from the fact that it is difficult to show that there are no 4-irreducible posets on 9 points, but even if this is assumed – say on the basis of computer search - the general argument is still complicated.

Complements of Antichains (1)

Theorem (Kimble, Trotter) If A is an antichain in P, then dim(P) ≤ max{2, |P – A|}.

Sketch of the Proof The argument is by induction starting with the case |P - A| = 2, where the inequality follows from the observation that, up to duality, there are only two posets which (a) are indecomposable with respect to lexicographic sums, and (b) consist of an antichain and two additional points. These are shown below. Both have dimension 2.

Complements of Antichains (2)

Note Trotter gave a complete “forbidden subposet characterization” of the inequality

dim(P) ≤ max{2, |P – A|}

where A is an antichain in P.

When |P – A| = n ≥ 4, there is a family Fn consisting of 2n – 1 irreducible posets so that if P is a poset consisting of an antichain A and n other points, then dim(P) < n unless P contains one of the posets from Fn. The standard example Sn is one of these posets.

Complements of Antichains (3)

Remark We can combine the preceding two inequalities:

a. dim(P) ≤ width(P).

b. If A is an antichain in P, then

dim(P) ≤ max{2, |P-A|}.

to obtain a simple proof of Hiraguchi’s inequality:

c. dim(P) ≤ |P|/2 when |P| ≥ 4.

Update 4

Remark The red points form an antichain A in P and |P – A| = 5. So dim(P) ≤ 5.

Removable Pairs in Some Posets

Lemma The removable pair conjecture holds for a poset P provided either

a. dim(P) ≤ 3, or

b. P is an interval order.

Proof If dim(P) = 2, there is nothing to prove. If dim(P) = 3, then P must contain a 3-element antichain A and at least 3 other points. Choose two of the points not in A. Their removal leaves a poset of dimension at least 2.

Now let P be an interval order. Then let x and y be distinct elements whose down sets are as large as possible. Then removing x and y decreases the dimension by at most 1.

Posets of Dimension at Most 2

Remark Consider a family of line segments with end points on the axes as shown above. Then they determine a poset P in a natural way. For example, c > e because the segment for c is always above the segment for e. It is easy to see that a poset P has such a representation if and only if it has dimension at most 2.

Segment Orders

Note In the material to follow, we allow extension of segments into the first quadrant. Farhad Shahroki proposed two different ways to define – in a natural way – a partial order associated with such a family of segments.

Segment Orders – Type 1

Note In a Type 1 segment order, we place c > e because the projection of c contains the projection of e, and where the projections overlap, the segment for c is always on top. For this family of segments, there are no other comparabilities.

Segment Orders – Type 2

Note In a Type 2 segment order, we set c > d because the segment for c starts and ends before the segment for d. Also, they are disjoint and where their projections overlap, the segment for c is always on top. For this family of segments, there are no other comparabilities.

Properties of Segment Orders

Theorem For i = 1 and 2, a poset P is a Type i segment order if any of the following three conditions hold:

a. Dim(P) ≤ 3.

b. P is an interval order.

c. P = Sn for some n ≥ 2.

Theorem Almost all posets P with dimension at least 4 are not segment orders of either type.

Remark There is a lot of content to this slide. The three (actually six) parts of the first theorem all involve clever constructions, while the second theorem requires the Alon/Scheinerman degrees of freedom theory.

Segment Orders and Removable Pairs

In view of the properties listed on the last slide, the family of segment orders of either type consists of the union of two families, each satisfying the removable pair conjecture – plus a handful of other posets, some of which have large dimension. So it is natural to ask:

Question Does the removable pair conjecture hold for segment orders of either type?

Are The Two Types are Distinct

Question 1 Are the two types of segment orders distinct, i.e., does there exist a Type 1 segment order P that is not a Type 2 segment order? Does there exist a Type 2 segment order Q that is not a Type 1 segment order.

Question 2 For i = 1, 2, if P is a Type i segment order, is the dual of P also a Type i segment order?

Note After the conference, Biro and Trotter showed that there are Type 1 segment orders that are not Type 2, and there are Type 2 segment orders that are not Type 1.

Realizers and Probability

L1 = b < e < a < d < g < c < f

L2 = a < c < b < d < g < e < f

L3 = a < c < b < e < f < d < g

L4 = b < e < a < c < f < d < g

L5 = a < b < d < g < e < c < f

For distinct points x and y, set Pr[x > y] = s/t where s is the number of linear extensions with x > y and t is the total number of linear extensions. For example, here we have Pr[d > e] = 3/5.

Fractional Dimension (1)Definition When P is a poset and F = {L1, L2, …, Lt} is a multi-set of linear extensions of P, we call F a multi-realizer of P if P = F. Then let q(F) = min Pr[x > y] taken over all incomparable pairs (x, y) from P.

In turn, let q(P) = max q(F) taken over all multi-realizers F of P.

Finally, let the fractional dimension of P, denoted fdim(P), be the reciprocal 1/q(P).

Evidently, fdim(P) ≤ dim(P) for all P.

Fractional Dimension (2)

L1 = b < e < a < d < g < c < f

L2 = a < c < b < d < g < e < f

L3 = a < c < b < e < f < d < g

L4 = b < e < a < c < f < d < g

L5 = a < b < d < g < e < c < f

Whenever x and y are incomparable, Pr[x > y] ≥ 2/5, so the fractional dimension of P is at most 5/2. In fact,

fdim(P) = 5/2

Removable Pair Conjecturefor Fractional Dimension

Conjecture

If |P| ≥ 3, then there exist x, y in P such that

fdim(P) ≤ 1 + fdim(P – x – y).

Theorem (Brightwell and Scheinerman) The conjecture holds for any poset on n points provided

fdim(P) ≤ (n2 – 5n + 6)/(4n – 6)

Weak Form of the Removable Pair Conjecture

for Fractional DimensionConjecture

There exists a constant e > 0 so that if |P| ≥ 3, then there exist x, y in P such that

fdim(P) ≤ 2 - e + fdim(P – x – y).

Question Does this weak form of the removable pair conjecture hold for either of the two types of segment orders?

Update 4Theorem (Trotter and Moore) If the diagram of P is a tree, then dim(P) ≤ 3.

Note The diagram of our poset is a tree, so dim(P) ≤ 3.

Update 5

Theorem If the diagram of P is planar, and P has both a greatest and a least element, then dim(P) ≤ 2.

Note The diagram shows that dim(P) ≤ 2, so dim(P) = 2.

A Closing CommentRemark To end the presentation on a different note, I could have used the following example – and used all the same updates except the very last one.

Now, it is easy to see that dim(P) = 3.