Transcript A Fresh Look at Some Old Extremal Problems

A Fresh Look at Some
Old Extremal Problems
William T. Trotter
Georgia Institute of Technology
November 20, 2006
Note to the Reader
These 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.
2.
3.
4.
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.
The family of all chains in a poset is partially ordered by
set inclusion. The maximal elements in this poset are
called maximal chains.
A chain C is maximum if no other chain contains more
points than C. Maximal chains need not be maximum.
The height of P is the size of a maximum chain.
Antichains
1.
2.
3.
4.
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.
The family of all antichains in a poset is partially ordered
by set inclusion. The maximal elements in this poset are
called maximal antichains.
An antichain A is maximum if no other antichain
contains more points than A. Maximal antichains need
not be maximum.
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 Dilworth
Proof 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 a
Partition 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 a
Partition into 11 chains
Note The red points form a maximum antichain of size 11.
Linear Extensions
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.
L1 = b < e < a < d < g < c < f
L2 = a < c < b < d < g < e < f
Realizers of Posets
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.
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
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 < g
The 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
Fact There is a simple poly-time algorithm for testing
whether dim(P) ≤ 2.
Theorem (Yannakakis) For fixed t ≥ 3, the question
dim(P) ≤ t? is NP-complete.
Remark The question dim(P) ≤ t? is NPcomplete 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 Products
Remark 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 Sums
Remark 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
Sn
Fact For n ≥ 2, the standard example Sn is a poset of
dimension n.
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 Orders
A 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 Orders
The 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)
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.
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.
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)? NPcomplete?
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.
Sketch of the proof. It is relatively easy to show that for every
poset P with |P| ≥ 5, either
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).
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 4dimensional 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 Conjecture
for 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 Dimension
Conjecture
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 4
Theorem (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 Comment
Remark 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.