A Fresh Look at Some Old Extremal Problems

1
A Fresh Look at SomeOld Extremal Problems

• William T. Trotter
• Georgia Institute of Technology
• November 20, 2006

2
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 trotter_at_math.gatech.edu
3
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 lt 15, (b) 5 is a maximal
element, and (c) 11 is incomparable to 19.
4
Chains

• 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.

5
Antichains

• 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.

6
Chains and Antichains

• 6,7,19,28 is a chain. It is not maximal.
• 12,13,16,30 is an antichain. It is not
  maximal.
• 8,13,34,35 is a maximal chain. It is not
  maximum.
• 12,13,30,24,16,19,14,25 is a maximal antichain.
  It is not maximum.

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Dilworths 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. Dilworths theorem is
typically grouped with other combinatorial
theorems with a common LP flavor, such as Halls
matching theorem, Tuttes 1-factor theorem, the
Konig/Egervary theorem and Mengers theorem.
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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,
Ai1 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.
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A Poset of Height 7 and aPartition into 7
Antichains
Note The red points form a maximum chain of size
7.
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The Proof of Dilworths 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 lt
a for some a in A, and U(A) x x gt 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.
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The Proof of Dilworths Theorem (2)
Then apply the inductive hypothesis to A ? U(A).
WLOG, we obtain a chain partition C1, C2, ,
Cw with ai the least element of Ci for each
i. Then Ci ? Ci 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.
12
A Poset of Width 11 and aPartition into 11
chains
Note The red points form a maximum antichain of
size 11.
13
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 gt y in L whenever x gt y in P. Example
L1 and L2 are linear extensions of the poset
P.
L1 b lt e lt a lt d lt g lt c lt f L2 a lt c lt b lt d
lt g lt e lt f
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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 gt y in Li.
L1 b lt e lt a lt d lt g lt c lt f L2 a lt c lt b lt d
lt g lt e lt f L3 a lt c lt b lt e lt f lt d lt g L4 b
lt e lt a lt c lt f lt d lt g L5 a lt b lt d lt g lt e lt
c lt f
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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 gt y in
L. Note This lemma is completely trivial for
finite posets.
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The Dimension of a Poset
L1 b lt e lt a lt d lt g lt c lt f L2 a lt c lt b lt d
lt g lt e lt f L3 a lt c lt b lt e lt f lt d lt 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
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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
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 Schndyers theorem.
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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.
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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) lt dim(P) dim(Q)
2.
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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) lt (x, y) in R if either (1) x lt
x in P, or (2) x x and y lt y in
Qx. Fact If R is the lexicographic sum of the
family F Qx x in P, then dim(R)
maxdim(P), maxdim(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.
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Basic Properties of Dimension

• Dimension is monotonic, i.e., if P is a
  subposet of Q, then dim(P) dim(Q).
• dim(Pd) dim(P), where Pd is the dual of P,
  i.e., Pd has the same ground set as P with x
  gt y in Pd if and only if x lt y in P.
• 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.

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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.
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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 gt
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 gt bi in L.
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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 lt y in P if and only
if bx lt ay in R.
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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
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Finding Interval Representations
Theorem (Greenough) Let P be an interval
order. For each x in P, let D(x) u
u lt 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.
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Canonical Interval Orders
The canonical interval order In consists of all
intervals with integer end points from 1, 2, ,
n.
I5
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Dimension of Interval Orders (1)
Theorem (Bogart and Trotter) For every t, there
exists an integer n0 so that if n gt n0, then
dim(In) gt t. Proof Let F be a realizer of
In. For each 3-element set i lt j lt k, choose
L from F with i, j gt j, k in L. If n
is sufficiently large compared to F, it
follows from Ramseys 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 gt j, k gt k, l in L which is a
contradiction since i, j lt k, l in P.
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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.
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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 gt 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 Dilworths 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.
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Update 2
Remark The width of the poset P is 7, so
dim(P) 7.
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Dimension and Width (2)
Note The inequality dim(P) width(P) is
tight, since dim(Sn) width(Sn) n.
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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.
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Dimension and Width (4)
Problem Is the question dim(P) lt 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) lt 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.
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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.
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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 lt
x whenever z lt 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).
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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 lt y whenever z lt x, and (b) w gt x
whenever w gt 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.
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Update 3
Remark Our poset has 12 points, so
dim(P) 12/2 6.
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Dimension and Cardinality (2)
Note The inequality dim(P) P/2 when
P 4 is tight, since for all n 2,
dim(Sn) n.
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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.
41
Dimension and Cardinality (4)
Theorem (Bogart and Trotter) If n 3 and
P 2n, then dim(P) lt 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.
42
Dimension and Cardinality (5)
Theorem If n 4 and P 2n 1, then
dim(P) lt 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.
43
Complements of Antichains (1)
Theorem (Kimble, Trotter) If A is an antichain
in P, then dim(P) max2, 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.
44
Complements of Antichains (2)
Note Trotter gave a complete forbidden subposet
characterization of the inequality dim(P)
max2, 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) lt n
unless P contains one of the posets from Fn.
The standard example Sn is one of these posets.
45
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) max2, P-A. to obtain a
simple proof of Hiraguchis inequality c.
dim(P) P/2 when P 4.
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Update 4
Remark The red points form an antichain A in
P and P A 5. So dim(P) 5.
47
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.
48
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 gt 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.
49
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.
50
Segment Orders Type 1
Note In a Type 1 segment order, we place c gt 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.
51
Segment Orders Type 2
Note In a Type 2 segment order, we set c gt 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.
52
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.
53
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?
54
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.
55
Realizers and Probability
L1 b lt e lt a lt d lt g lt c lt f L2 a lt c lt b lt d
lt g lt e lt f L3 a lt c lt b lt e lt f lt d lt g L4 b
lt e lt a lt c lt f lt d lt g L5 a lt b lt d lt g lt e lt
c lt f
For distinct points x and y, set Prx gt y
s/t where s is the number of linear
extensions with x gt y and t is the total
number of linear extensions. For example, here
we have Prd gt e 3/5.
56
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 Prx gt 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.
57
Fractional Dimension (2)
L1 b lt e lt a lt d lt g lt c lt f L2 a lt c lt b lt d
lt g lt e lt f L3 a lt c lt b lt e lt f lt d lt g L4 b
lt e lt a lt c lt f lt d lt g L5 a lt b lt d lt g lt e lt
c lt f
Whenever x and y are incomparable, Prx gt y
2/5, so the fractional dimension of P is at
most 5/2. In fact,
fdim(P) 5/2
58
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)
59
Weak Form of the Removable Pair Conjecturefor
Fractional Dimension
Conjecture There exists a constant e gt 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?
60
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.
61
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.
62
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.