Avoidance in (2+2)-Free Posets
Nihal Gowravaram
Acton Boxborough Regional High School
Mentor: Wuttisak Trongsiriwat
PRIMES Annual Conference
Massachusetts Institute of Technology
May 18-19, 2013
1 / 13
What is a Poset?
Introduction
❖ Poset
Partially Ordered Set (P, ≺)
❖ Hasse Diagrams
❖ Avoidance
Motivation
Results
Conclusion
2 / 13
What is a Poset?
Introduction
❖ Poset
❖ Hasse Diagrams
❖ Avoidance
Motivation
Results
Conclusion
Partially Ordered Set (P, ≺)
Reflexivity
i ≺ i for all i ∈ P
● Antisymmetry
If i ≺ j and j ≺ i, then i = j for i, j ∈ P
● Transitivity
If i ≺ j and j ≺ k, then i ≺ k for i, j, k ∈ P
●
2 / 13
What is a Poset?
Introduction
❖ Poset
❖ Hasse Diagrams
❖ Avoidance
Motivation
Results
Conclusion
Partially Ordered Set (P, ≺)
Reflexivity
i ≺ i for all i ∈ P
● Antisymmetry
If i ≺ j and j ≺ i, then i = j for i, j ∈ P
● Transitivity
If i ≺ j and j ≺ k, then i ≺ k for i, j, k ∈ P
●
Call i, j ∈ P comparable if i ≺ j or j ≺ i
2 / 13
Hasse Diagrams
Introduction
❖ Poset
(P ({x, y, z}) , ⊆)
❖ Hasse Diagrams
{x, y, z}
❖ Avoidance
Motivation
Results
Conclusion
{x, y}
{x}
{x, z}
{y}
{y, z}
{z}
{}
3 / 13
Hasse Diagrams
Introduction
❖ Poset
(P ({x, y, z}) , ⊆)
❖ Hasse Diagrams
{x, y, z}
❖ Avoidance
Motivation
Results
{x, y}
Conclusion
{x}
{x, z}
{y}
{y, z}
{z}
{}
{} and {x, z} are comparable
● {y} and {x, z} are incomparable
●
3 / 13
Hasse Diagrams
Introduction
❖ Poset
(P ({x, y, z}) , ⊆)
❖ Hasse Diagrams
{x, y, z}
❖ Avoidance
Motivation
Results
{x, y}
Conclusion
{x}
{x, z}
{y}
{y, z}
{z}
{}
{} and {x, z} are comparable
● {y} and {x, z} are incomparable
● {}, {x}, {x, z}, and {x, y, z} form a chain of length 4.
●
3 / 13
Avoidance in Posets
Introduction
❖ Poset
❖ Hasse Diagrams
❖ Avoidance
A poset P is said to contain a poset S if there exists some
subposet W of P that is isomorphic to S.
● P is said to avoid S if P does not contain S.
●
Motivation
Results
Conclusion
4 / 13
Avoidance in Posets
Introduction
❖ Poset
❖ Hasse Diagrams
❖ Avoidance
A poset P is said to contain a poset S if there exists some
subposet W of P that is isomorphic to S.
● P is said to avoid S if P does not contain S.
●
Motivation
Results
•
Conclusion
•
•
•
•
4 / 13
Avoidance in Posets
Introduction
❖ Poset
❖ Hasse Diagrams
❖ Avoidance
A poset P is said to contain a poset S if there exists some
subposet W of P that is isomorphic to S.
● P is said to avoid S if P does not contain S.
●
Motivation
Results
•
Conclusion
•
•
Contains (3+1)
•
•
•
Avoids (2+2)
• •
•
•
•
•
•
4 / 13
Why (2+2)-Free Posets?
Introduction
Motivation
❖ (2+2)-Free Posets
●
Interval Orders
✦
❖ Previous Results
Results
Conclusion
✦
✦
A poset is an interval order if it is isomorphic to some
set of intervals on the real line ordered by left-to-right
precedence.
Interval orders are important in mathematics,
computer science, and engineering (Ex. task
distributions in complex manufacturing processes).
(Fishburn 1970) (2+2)-Free Posets are precisely
interval orders.
5 / 13
Why (2+2)-Free Posets?
Introduction
●
Motivation
Interval Orders
✦
❖ (2+2)-Free Posets
❖ Previous Results
Results
Conclusion
✦
✦
●
A poset is an interval order if it is isomorphic to some
set of intervals on the real line ordered by left-to-right
precedence.
Interval orders are important in mathematics,
computer science, and engineering (Ex. task
distributions in complex manufacturing processes).
(Fishburn 1970) (2+2)-Free Posets are precisely
interval orders.
Ascent Sequences
✦
✦
An ascent sequence is a sequence x1 x2 · · · xn
satisfies x1 = 0 and, for all i with 1 < i ≤ n,
xi ≤ asc(x1 x2 · · · xi−1 ) + 1.
(Bousquet-Mélou et al 2009) The ascent sequences
are in bijection with the (2+2)-free posets.
5 / 13
Previous Results with (2+2)-Free posets
Introduction
Motivation
❖ (2+2)-Free Posets
❖ Previous Results
Results
Conclusion
Let Pn (x) refer to the set of posets of size n that avoid the
poset x.
● Define a function a(x) to return the ascent sequence
associated with a poset x.
● Let An (y) refer to the set of posets of size n whose ascent
sequences avoid the ascent sequence y.
●
6 / 13
Previous Results with (2+2)-Free posets
Introduction
Motivation
❖ (2+2)-Free Posets
❖ Previous Results
Results
Conclusion
Let Pn (x) refer to the set of posets of size n that avoid the
poset x.
● Define a function a(x) to return the ascent sequence
associated with a poset x.
● Let An (y) refer to the set of posets of size n whose ascent
sequences avoid the ascent sequence y.
● (Stanley 1997) |Pn (2 + 2, 3 + 1)| = Cn .
(Enum. Comb. 1) |Pn (2 + 2, N )| = Cn .
●
6 / 13
Previous Results with (2+2)-Free posets
Introduction
●
Motivation
❖ (2+2)-Free Posets
❖ Previous Results
●
Results
Conclusion
●
●
●
Let Pn (x) refer to the set of posets of size n that avoid the
poset x.
Define a function a(x) to return the ascent sequence
associated with a poset x.
Let An (y) refer to the set of posets of size n whose ascent
sequences avoid the ascent sequence y.
(Stanley 1997) |Pn (2 + 2, 3 + 1)| = Cn .
(Enum. Comb. 1) |Pn (2 + 2, N )| = Cn .
(Trongsiriwat)
Pn (2 + 2, N, p1 , · · · , pk ) = An (0101, a(p1 ), · · · , a(pk )).
6 / 13
Previous Results with (2+2)-Free posets
Introduction
●
Motivation
❖ (2+2)-Free Posets
❖ Previous Results
●
Results
Conclusion
●
●
●
●
Let Pn (x) refer to the set of posets of size n that avoid the
poset x.
Define a function a(x) to return the ascent sequence
associated with a poset x.
Let An (y) refer to the set of posets of size n whose ascent
sequences avoid the ascent sequence y.
(Stanley 1997) |Pn (2 + 2, 3 + 1)| = Cn .
(Enum. Comb. 1) |Pn (2 + 2, N )| = Cn .
(Trongsiriwat)
Pn (2 + 2, N, p1 , · · · , pk ) = An (0101, a(p1 ), · · · , a(pk )).
Question 1: Can we explicitly compute |Pn (2 + 2, p)| for
other posets p?
6 / 13
Previous Results with (2+2)-Free posets
Introduction
●
Motivation
❖ (2+2)-Free Posets
❖ Previous Results
●
Results
Conclusion
●
●
●
Let Pn (x) refer to the set of posets of size n that avoid the
poset x.
Define a function a(x) to return the ascent sequence
associated with a poset x.
Let An (y) refer to the set of posets of size n whose ascent
sequences avoid the ascent sequence y.
(Stanley 1997) |Pn (2 + 2, 3 + 1)| = Cn .
(Enum. Comb. 1) |Pn (2 + 2, N )| = Cn .
(Trongsiriwat)
Pn (2 + 2, N, p1 , · · · , pk ) = An (0101, a(p1 ), · · · , a(pk )).
Question 1: Can we explicitly compute |Pn (2 + 2, p)| for
other posets p?
● Question 2: For what posets p is it true that
Pn (2 + 2, p) = An (a(p))?
●
6 / 13
(2+2)-Free Posets and Ascent Sequences
Introduction
Motivation
Results
❖ Posets and Ascent
Sequences
❖ (2+2, ∨)-Free and
(2+2, ∧)-Free
❖ (2+2, 3)-Free
❖ Bijection
❖ (2+2, 4)-Free and
(2+2, Y)-Free
Ascent Sequence a(p)
|Pn (2 + 2, p)|
012
2n−1
•
010
2n−1
•
•
001
2n−1
011
2n−1
(2+2)-Free Poset p
•
•
•
•
•
•
Conclusion
•
•
•
7 / 13
(2+2, ∨)-Free and (2+2, ∧)-Free Posets
Introduction
●
|Pn (2 + 2, ∨)| = |Pn (2 + 2, ∧)| (Inverting Procedure)
Motivation
Results
❖ Posets and Ascent
Sequences
❖ (2+2, ∨)-Free and
(2+2, ∧)-Free
❖ (2+2, 3)-Free
❖ Bijection
❖ (2+2, 4)-Free and
(2+2, Y)-Free
Conclusion
8 / 13
(2+2, ∨)-Free and (2+2, ∧)-Free Posets
Introduction
Motivation
|Pn (2 + 2, ∨)| = |Pn (2 + 2, ∧)| (Inverting Procedure)
● Pn (2 + 2, ∨)
●
Results
❖ Posets and Ascent
Sequences
❖ (2+2, ∨)-Free and
(2+2, ∧)-Free
❖ (2+2, 3)-Free
❖ Bijection
❖ (2+2, 4)-Free and
(2+2, Y)-Free
Conclusion
8 / 13
(2+2, ∨)-Free and (2+2, ∧)-Free Posets
Introduction
Motivation
Results
❖ Posets and Ascent
Sequences
❖ (2+2, ∨)-Free and
(2+2, ∧)-Free
|Pn (2 + 2, ∨)| = |Pn (2 + 2, ∧)| (Inverting Procedure)
● Pn (2 + 2, ∨)
●
✦
✦
Add a free node.
Add a maximal node.
❖ (2+2, 3)-Free
•
❖ Bijection
❖ (2+2, 4)-Free and
(2+2, Y)-Free
•
•
•
Conclusion
•
•
•
•
•
•
•
or
•
•
•
8 / 13
(2+2, ∨)-Free and (2+2, ∧)-Free Posets
Introduction
Motivation
|Pn (2 + 2, ∨)| = |Pn (2 + 2, ∧)| (Inverting Procedure)
● Pn (2 + 2, ∨)
●
Results
❖ Posets and Ascent
Sequences
✦
✦
❖ (2+2, ∨)-Free and
(2+2, ∧)-Free
Add a free node.
Add a maximal node.
❖ (2+2, 3)-Free
•
❖ Bijection
❖ (2+2, 4)-Free and
(2+2, Y)-Free
•
•
•
Conclusion
•
•
•
●
•
•
•
•
or
•
•
•
|Pn (2 + 2, ∨)| = |Pn (2 + 2, ∧)| = 2n−1
8 / 13
(2+2, 3)-Free Posets
Introduction
●
Pn (2 + 2, 3) are posets with level at most 2.
Motivation
Results
❖ Posets and Ascent
Sequences
❖ (2+2, ∨)-Free and
(2+2, ∧)-Free
❖ (2+2, 3)-Free
❖ Bijection
❖ (2+2, 4)-Free and
(2+2, Y)-Free
Conclusion
9 / 13
(2+2, 3)-Free Posets
Introduction
●
Motivation
✦
✦
Results
❖ Posets and Ascent
Sequences
❖ (2+2, ∨)-Free and
(2+2, ∧)-Free
❖ (2+2, 3)-Free
❖ Bijection
❖ (2+2, 4)-Free and
(2+2, Y)-Free
Pn (2 + 2, 3) are posets with level at most 2.
Level 2: a nodes: x1 , x2 , · · · , xa .
Level 1: b nodes: y1 , y2 , · · · , yb .
Define Si = {yj |yj ≺ xi }.
● Assign x1 , x2 , · · · , xa to the a nodes such that
|S1 | ≥ |S2 | ≥ · · · ≥ |Sa |.
●
Conclusion
9 / 13
(2+2, 3)-Free Posets
Introduction
●
Motivation
✦
✦
Results
❖ Posets and Ascent
Sequences
❖ (2+2, ∨)-Free and
(2+2, ∧)-Free
❖ (2+2, 3)-Free
❖ Bijection
❖ (2+2, 4)-Free and
(2+2, Y)-Free
Pn (2 + 2, 3) are posets with level at most 2.
Level 2: a nodes: x1 , x2 , · · · , xa .
Level 1: b nodes: y1 , y2 , · · · , yb .
Define Si = {yj |yj ≺ xi }.
● Assign x1 , x2 , · · · , xa to the a nodes such that
|S1 | ≥ |S2 | ≥ · · · ≥ |Sa |.
● S1 ⊇ S2 ⊇ · · · ⊇ Sa .
●
Conclusion
•
•
•
•
•
b
a+b−1
● {|S1 |, |S2 |, · · · , |Sa |} →
=
.
a
a
9 / 13
(2+2, 3)-Free Posets
Introduction
●
Motivation
✦
✦
Results
❖ Posets and Ascent
Sequences
❖ (2+2, ∨)-Free and
(2+2, ∧)-Free
❖ (2+2, 3)-Free
❖ Bijection
❖ (2+2, 4)-Free and
(2+2, Y)-Free
Pn (2 + 2, 3) are posets with level at most 2.
Level 2: a nodes: x1 , x2 , · · · , xa .
Level 1: b nodes: y1 , y2 , · · · , yb .
Define Si = {yj |yj ≺ xi }.
● Assign x1 , x2 , · · · , xa to the a nodes such that
|S1 | ≥ |S2 | ≥ · · · ≥ |Sa |.
● S1 ⊇ S2 ⊇ · · · ⊇ Sa .
●
Conclusion
•
•
•
•
•
b
a+b−1
● {|S1 |, |S2 |, · · · , |Sa |} →
=
.
a a
n−1
P
P n−1
a+b−1
● |Pn (2 + 2, 3)| =
=
= 2n−1
a
a
a=0
a+b=n
9 / 13
Bijection between (2+2, 3) and (2+2, ∧)-Free
Posets
Introduction
Motivation
Results
❖ Posets and Ascent
Sequences
●
Poset P → (A, B).
✦
✦
A = {Maximal Nodes in P }.
B = P \ A.
❖ (2+2, ∨)-Free and
(2+2, ∧)-Free
❖ (2+2, 3)-Free
❖ Bijection
❖ (2+2, 4)-Free and
(2+2, Y)-Free
Conclusion
10 / 13
Bijection between (2+2, 3) and (2+2, ∧)-Free
Posets
Introduction
●
Motivation
✦
✦
Results
❖ Posets and Ascent
Sequences
❖ (2+2, ∨)-Free and
(2+2, ∧)-Free
Poset P → (A, B).
●
A = {Maximal Nodes in P }.
B = P \ A.
In (2+2, ∧)-Free, B forms a chain.
❖ (2+2, 3)-Free
•
❖ Bijection
❖ (2+2, 4)-Free and
(2+2, Y)-Free
Conclusion
•
•
•
•
•
10 / 13
Bijection between (2+2, 3) and (2+2, ∧)-Free
Posets
Introduction
●
Motivation
✦
✦
Results
❖ Posets and Ascent
Sequences
❖ (2+2, ∨)-Free and
(2+2, ∧)-Free
Poset P → (A, B).
●
A = {Maximal Nodes in P }.
B = P \ A.
In (2+2, ∧)-Free, B forms a chain.
❖ (2+2, 3)-Free
•
❖ Bijection
•
•
❖ (2+2, 4)-Free and
(2+2, Y)-Free
•
•
•
Conclusion
●
In (2+2, 3)-Free, B forms the lower level.
• • •
• • •
10 / 13
Bijection between (2+2, 3) and (2+2, ∧)-Free
Posets
Introduction
●
Motivation
✦
✦
Results
❖ Posets and Ascent
Sequences
❖ (2+2, ∨)-Free and
(2+2, ∧)-Free
Poset P → (A, B).
●
A = {Maximal Nodes in P }.
B = P \ A.
In (2+2, ∧)-Free, B forms a chain.
❖ (2+2, 3)-Free
•
❖ Bijection
•
•
❖ (2+2, 4)-Free and
(2+2, Y)-Free
•
•
•
Conclusion
●
In (2+2, 3)-Free, B forms the lower level.
• • •
• • •
●
Maintains all order relations between A and B.
10 / 13
(2+2, 4) and (2+2, Y)-Free Posets
Introduction
●
Pn (2 + 2, Y ) ↔ Pn (2 + 2, 4).
Motivation
Results
❖ Posets and Ascent
Sequences
❖ (2+2, ∨)-Free and
(2+2, ∧)-Free
Theorem.
|Pn(2 +
2, Y )| = |Pn (2+ 2, 4)| = 1 +
X
r+m<n
n + mr + 1
n−m−r
−
n + m(r − 1) + 1
n−m−r
−
n + r(m − 1)
n−m−r
+
n + (r − 1)(m − 1)
n−m−r
where r ≥ 0 and m > 0.
❖ (2+2, 3)-Free
❖ Bijection
❖ (2+2, 4)-Free and
(2+2, Y)-Free
Conclusion
11 / 13
,
Future Directions of Research
Introduction
Motivation
Pn (2 + 2, ∨) ↔ Pn (2 + 2, 3).
● Pn (2 + 2, Y ) ↔ Pn (2 + 2, 4).
●
Results
Conclusion
❖ Future Directions
❖ Acknowledgements
12 / 13
Future Directions of Research
Introduction
Motivation
Pn (2 + 2, ∨) ↔ Pn (2 + 2, 3).
● Pn (2 + 2, Y ) ↔ Pn (2 + 2, 4).
●
Results
Conclusion
❖ Future Directions
❖ Acknowledgements
Conjecture. Define a function Y (n), n ≥ 3 as follows.
✦
✦
Y (3) = ∨.
Y (n) is the result of adding a minimal node to
Y (n − 1).
Then, Pn (2 + 2, Y (k)) ↔ Pn (2 + 2, k).
12 / 13
Future Directions of Research
Introduction
Motivation
Pn (2 + 2, ∨) ↔ Pn (2 + 2, 3).
● Pn (2 + 2, Y ) ↔ Pn (2 + 2, 4).
●
Results
Conjecture. Define a function Y (n), n ≥ 3 as follows.
Conclusion
❖ Future Directions
✦
✦
❖ Acknowledgements
Y (3) = ∨.
Y (n) is the result of adding a minimal node to
Y (n − 1).
Then, Pn (2 + 2, Y (k)) ↔ Pn (2 + 2, k).
|Pn (2 + 2, 3 + 1)| = |Pn (2 + 2, N )|.
● |Pn (2 + 2, Y )| = |Pn (2 + 2, 4)|
●
12 / 13
Future Directions of Research
Introduction
Motivation
Pn (2 + 2, ∨) ↔ Pn (2 + 2, 3).
● Pn (2 + 2, Y ) ↔ Pn (2 + 2, 4).
●
Results
Conjecture. Define a function Y (n), n ≥ 3 as follows.
Conclusion
❖ Future Directions
✦
✦
❖ Acknowledgements
Y (3) = ∨.
Y (n) is the result of adding a minimal node to
Y (n − 1).
Then, Pn (2 + 2, Y (k)) ↔ Pn (2 + 2, k).
|Pn (2 + 2, 3 + 1)| = |Pn (2 + 2, N )|.
● |Pn (2 + 2, Y )| = |Pn (2 + 2, 4)|
●
Query. Do there exist other nontrivial Wilf-Equivalences
in (2+2)-Free Posets? What other posets p, q exist such
that |Pn (2 + 2, p)| = |Pn (2 + 2, q)| for all n ∈ N?
12 / 13
Acknowledgements
Introduction
Motivation
Results
Conclusion
❖ Future Directions
❖ Acknowledgements
Thanks to
My mentor Wuttisak Trongsiriwat for his valuable insight
and guidance.
● The PRIMES program for making this experience
possible.
● My parents for their support.
●
Thanks to all of you for listening.
13 / 13