Macalester Journal of Catalan Numbersabeverid/mjcn/mjcn2009.pdf · Macalester Journal of Catalan...

Macalester Journal ofCatalan Numbers

Volume 1 December 2009

Published by:Department of Mathematics, Statistics and Computer ScienceMacalester College1600 Grand AvenueSaint Paul, MN 55105

2009 All rights reserved

Cover art: Yenee SohEditor: Andrew Beveridge

Table of Contents

For each Catalan problem below, we enumerate the 5 elements for the n = 3 case. Moreover, eachlist appears in the same order. In other words, the natural bijection between any pair maps thefirst element of one list to the first element of the other list, and so on.

Introduction to the Catalan Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Andrew Beveridge

1 − 1 − 1− 1 − 11 −− 11 −−1− 11 − 1 −− 111 −−−

Non-Crossing Arcs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7Mo Liu

Non-Decreasing Sequences of Positive Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13Matt Kusner

123 122 113 112 111

Non-Crossing Muraski Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17Colin Keeley

(13)(24)-Free Involutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21Michael Kapernaros

(12)(34)(56) (12)(36)(45) (14)(23)(56) (16)(23)(45) (16)(25)(34)

Non-Intersecting Arcs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25James Leonard

Standard Young Tableaux of Shape (n, n) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29Jack Shirek

135246

134256

125346

124356

123456

Stacking Coins in the Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33Tom Woodward

Sequences Below the Sequence 012 . . . (n − 1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37Congcong Nie

000 001 010 011 012

Upper Triangular Ferrers Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41Katie Agnew

Pairs of Non-Crossing Lattice Paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45Drew Van Denover

Partially Connected Polygon Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53Jeanmarie Youngblood

Plane Binary Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .59David Klock

Non-Intersecting Convex Hulls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67Jorge Banuelos

Dyck Paths with No Peaks at Height Two . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71Yenee Soh

Pairs of Internally Disjoint Lattice Paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77Matthew Hurni

Dyck Paths with Odd Maximal Descents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83Emily Merrill

Macalester Journal of Catalan NumbersVolume 1 (December 2009)

An Introduction to the Catalan NumbersAndrew Beveridge∗

1 Preface

Welcome to the inaugural volume of the Macalester Journal of Catalan Numbers!This journal collects the work of the students enrolled in Math 379: Combinatoricsduring Fall 2009 at Macalester College. The 16 students were assigned problemsdrawn from the exhaustive list of Catalan exercises in Richard Stanley’s Enumer-ative Combinatorics, Volume 2. Each student produced an article proving that thecounting problem was enumerated by the Catalan numbers. Future Combina-torics classes at Macalester will tackle other Catalan problems, to be collected inlater volumes of this journal.

The articles are arranged loosely in terms of difficulty. The early articles aretypically straight forward arguments, either showing that a sequence satisfies theCatalan recurrence or presenting a simple bijection to a known Catalan problem.Later articles are a bit more involved, requiring multiple steps, more creativebijections, or the heavy machinery of generating functions. Like the class of Cata-lan problems, these articles are diverse, yet unified by common themes. We hopethat the variety of the articles reflects the beauty and flexibility of the Catalannumbers.

We thank Tom Halverson for suggesting that we collect these articles intoa journal, and for contributing a lecture on the Catalan numbers during thesemester.∗Department of Mathematics, Statistics and Computer Science, Macalester College, Saint Paul

MN 55105.

Macalester Journal of Catalan Numbers 2

2 The Catalan Numbers

The sequence of Catalan numbers is among the most well known sequences incombinatorics. The Catalan sequence Cnn≥0 begins

1, 1, 2, 5, 14, 42, 132, 429, 1340, 4862, . . .

The formula for the nth Catalan number is

Cn =1

n + 1

(2nn

).

The Catalan numbers are a combinatorial chameleon. This remarkable se-quence of numbers arises in many independent settings. In particular, it is thesolution to many counting problems. The objective of this journal is to explorethe variety of Catalan problems in enumerative combinatorics.

The key to the flexibility of the Catalan numbers lies in their recurrence rela-tion. Indeed, we have

Cn =n−1

∑k=0

CkCn−k−1 for n ≥ 1.

This recurrence relation is perfectly suited for problems that have a particulartype of natural recurrence. We will discuss this recurrence further in the nextsection.

Generating functions are a powerful tool in combinatorics, and investigationof Catalan numbers is no exception. The generating function for the Catalannumbers is

C(x) =∞

∑n=0

Cnxn =1−√

1− 4x2x

.

3 Balanced sequences of 1’s and −1’s

We consider one of the classic counting problems whose solution is given by theCatalan numbers. This particular formulation is quite useful when proving otherCatalan results. Many of the articles in this journal create a bijection from theircounting problem to this one.


Suppose that we have n copies of the symbol “1” and n copies of the symbol“−1.” How many ways are there to arrange them in a sequence s1, s2, . . . , s2n sothat the partial sums satisfy ∑

ji=1 si ≥ 0 for 1 ≤ j ≤ 2n? In other words, as we

move from left to right in such a “valid” sequence, we always encounter at least asmany 1’s as −1’s. Using “−′′ to denote “− 1′′, we enumerate all valid sequencesfor 0 ≤ n ≤ 4:

n valid sequences

0 ∅

1 1−

2 1− 1−, 11−−

3 1− 1− 1−, 1− 11−−, 11−−1−, 11− 1−−, 111−−−

4

1− 1− 1− 1−, 1− 1− 11−−, 1− 11−−1−, 1− 11− 1−−,1− 111−−−, 11−−1− 1−, 11−−11−−, 11− 1−−1−,11− 1− 1−−, 11− 11−−−, 111−−− 1−, 111−−1−−111− 1−−−, 1111−−−−

Table 3.1: An enumeration of sequences of balanced 1’s and −1’s.

Let An denote the number of valid sequences of length 2n. For 0 ≤ n ≤ 4, thevalues of An are

1, 1, 2, 5, 14,

matching the initial the Catalan numbers. Here we opt for the convention thatthere is one way to create a sequence of length 0. Continuing our enumerationfor n = 5 and n = 6 would be quite a task: there are a total 42 and 132 validsequences, respectively!


To successfully enumerate all valid sequences for n = 5, we would surelyhave to develop a systematic approach. After some trial and error, inspirationwould strike. Let us find the first time that we have seen as many −1’s as 1’s. Inmathematical terms, let 2m be the first index such that ∑2m

i=1 si = 0. (This indexmust be even since we must have equal numbers of 1’s and −1’s.) This indexnaturally partitions our sequence into two smaller valid sequences: the first withlength 2m and the second with length 2n− 2m. Note that when m = n the secondsequence has length 0. Let us reorganize our valid sequences for n = 4 accordingto this index m, where we use ∅ to denote a sequence of length 0:

m = 1 m = 2 m = 3 m = 4

1− 1− 1− 1−1− 1− 11−−1− 11−−1−1− 11− 1−−1− 111−−−

11−− 1− 1−11−− 11−−

11− 1−− 1−111−−− 1−

11− 1− 1−− ∅11− 11−−− ∅111−−1−− ∅111− 1−−− ∅1111−−−− ∅

Table 3.2: Valid sequences of balanced 1’s and −1’s, partitioned by thelength 2m of the first valid subsequence.

We make one final observation. The first 2m elements are a special type ofbalanced sequence. Namely, the index 2m is the first time that we have seen anequal number of 1’s and −1’s. (This property does not necessarily hold for thelatter sequence of length 2n − 2m.) Such an indecomposable sequence must beof the form 1, s1, s2, · · · , s2m−2,−1 where s1, s2, . . . , s2m−2 is any valid sequenceof length 2m − 2. Setting k := m − 1, we incorporate this observation into ourprevious table by highlighting the embedded sequence of length 2k:

k = 0 k = 1 k = 2 k = 3

1 ∅ − 1− 1− 1−1 ∅ − 1− 1 1−−1 ∅ − 1 1−−1−1 ∅ − 1 1− 1−−1 ∅ − 1 1 1−−−

1 1− − 1− 1−1 1− − 1 1−−

1 1− 1− − 1−1 1 1−− − 1−

1 1− 1− 1− − ∅1 1− 11−− − ∅1 1 1−−1− − ∅1 11− 1−− − ∅1 1 1 1−−− − ∅


Table 3.3: Valid sequences of balanced 1’s and −1’s, where the embedded sequenceof length 2k within the complete subsequence has been highlighted.

In summary, we can naturally partition any valid sequence into two validsubsequences. The first sequence, of length 2k+ 2, is indecomposable. The secondvalid sequence, which has length 2n− 2k− 2, has no additional restrictions. Anindecomposable sequence of length 2k + 2 corresponds to a valid sequence oflength 2k.

Every valid sequence of length 2n is built from a valid sequence of length2k and a valid sequence of length 2n − 2 − 2k. In other words, we can createa bijection from pairs pairs of valid sequences of lengths 2k and 2n − 2k − 2 tovalid sequences of length 2n. Indeed, given two valid sequences s1, s2, . . . , s2k andt1, t2, . . . t2n−2−2k, the sequence

1, s1, s2, . . . , s2k,−1, t1, t2, . . . , t2n−2−2k

is a valid sequence of length 2n.Since our choices for the first and second sequences are independent, our

counting problem satisfies the Catalan recurrence:

An =n−1

∑k=0

Ak An−k−1 for n ≥ 1.

Our initial values also match the Catalan numbers, so we can conclude that thesesequence are identical, Ann≥0 = Cnn≥0. In other words, the number of validsequences of length 2n is equal to the nth Catalan number, Cn.

The form of our systematic partition is the hallmark of Catalan problems.Every instance of size n can be decomposed into two independent substructures,of size k and n− 1− k respectively, where 0 ≤ k ≤ n− 1. Why do these numbersonly add up to n− 1? The answer is that the final “missing” index actually playsan essential role. It acts as the border between the first and second substructures.This self-partitioning property unifies all Catalan problems.

4 Dyck Paths

We describe another classic Catalan problem that is referenced by articles in thejournal. Dyck paths are lattice paths from (0, 0) to (2n, 2n) that use steps (1, 1)


and (1,−1) which never falling below the x-axis. We enumerate all Dyck pathsfor 0 ≤ n ≤ 4.

n Dyck paths of length 2n

0 ∅

1

2

3

4

Table 4.1: An enumeration of Dyck paths.

There is a simple bijection from Dyck paths to balanced sequences of 1’s and−1’s. Namely, map the step (1, 1) to the symbol 1 and map the step (1,−1) tothe symbol −1. Finding a bijection to a known Catalan problem is a particularlysatisfying form of proof.


Non-Crossing ArcsMo Liu∗

1 An Introduction

Consider n + 1 points on a horizontal line in the plane. Connect the points withn arcs such that

1. the arcs do not pass below any point

2. the graph thus formed is a tree

3. no two arcs intersect in their interiors (i.e., the arcs are noncrossing)

4. at every vertex, all the arcs exit in the same direction (left or right)

We first enumerate some small examples. Look at the graphs below:

n = 0

n = 1

∗Mo is from Beijing, China. He has been in the United States for one year. He likes jumpingup and down on his bed, crawling up and down stairs and standing upside down doing countingproblems.


n = 2

n = 3

n = 4

We show the number of such structures is governed by the Catalan Numbers.


2 The Catalan Numbers and Recurrence Relations

Define An is the number of ways of drawing n + 1 such points connected by arcs.The size of a graph is the number of points. We have A0 = 1, A1 = 1, A2 = 2,A3 = 5, A4 = 14. We know that the Catalan numbers start with 1, 1, 2, 5, 14, 42....We show that An is the nth Catalan number.

The formula for the nth Catalan number is Cn =1

n + 1

(2nn

). There is also a

recurrence ralation for the Catalan numbers:

Cn = C0Cn−1 + C1Cn−2 + ... + Cn−2C1 + Cn−1C0

We prove that the sequence of An also satisfies this recurrence:

An = A0An−1 + A1An−2 + ... + An−2A1 + An−1A0

3 Proof

Take n = 4 as an example. A0 = 1 is simply true from the first graph drawn.When n ≥ 1, we can see that the leftmost point is always connected to the right-most point. Indeed, the arcs connected to the leftmost point can only exit to theright and the arcs connected to the rightmost point can only exit to the left. Ifthese two points are not connected, there must be another point connecting both

of them, which would be like . In this case, the middle point is con-nected to two different directions of arcs, which is not allowed, so the leftmostand the rightmost points are always connected. As the outmost arc exists for ev-ery graph, we can just remove it and focus on inner arcs.

We partition the graphs in the following four catagories:

1. P0 and P1 are disconnected:

P0 P1 P2 P3 P4 P0 P1 P2 P3 P4


P0 P1 P2 P3 P4 P0 P1 P2 P3 P4

P0 P1 P2 P3 P4

The graph is split into two components: the first component with a size of 1and the second component with a size of 4, so the number of ways to creategraphs of this type is A1−1A4−1 = A0A3.


P0 P1 P2 P3 P4 P0 P1 P2 P3 P4



P0 P1 P2 P3 P4 P0 P1 P2 P3 P4



P0 P1 P2 P3 P4 P0 P1 P2 P3 P4


P0 P1 P2 P3 P4 P0 P1 P2 P3 P4

P0 P1 P2 P3 P4


So the number of ways of drawing 5 points connected by arcs is A0A3 +A1A2 + A2A1 + A3A0 = 14.

Now consider the general case. Suppose we have a valid graph of size n + 1.Removing the arc connecting P0, Pn+1 disconnects the graph into 2 intervals:(P0, P1, P2, ..., Pk), (Pk+1, Pk+2, ..., Pn+1). The number of different graphs is Ak An−k−1.We sum over for all possible k, and we get the recurrence relation

An =n−1

∑k=0

(Ak An−1−k) = A0An−1 + A1An−2 + ... + An−2A1 + An−1A0

Since the An satisfies the Catalan recurrence and A0 = C0 = 1, the numbers areequal for all n.


Non-Decreasing Sequences of Positive IntegersMatthew Kusner∗

1 Introduction

Consider sequences of integers, a1a2 · · · an, such that,

1 ≤ a1 ≤ · · · ≤ an

andai ≤ i

We define Sn as the set containing all of the sequences of length n that satisfythese two constraints. We will prove that |Sn| = Cn, the nth Catalan number.

2 Enumeration for n ≥ 4

We list the sets Sn for 0 ≤ n ≤ 4.

n Sn Cn

0 ∅ 11 1 12 11 12 23 111 112 113 122 123 54 1111 1112 1113 1114 14

1124 1134 1224 12341123 1223 1133 1122

1222 1233

∗Matthew is a junior majoring in mathematics and computer science. Born in Cleveland, OHhe currently calls home the city of Iowa City, IA. If Matthew had the opportunity to be endowedwith the skills of any basketball player he would choose Carmelo Anthony for his smooth movesand steady jump shot.


3 Proof

We will construct a bijection from our problem to a known Catalan problem.Consider the set Tn of all northeast lattice paths which do not go above the maindiagonal for a n x n grid. For example, here are the elements of T3:

T3

It is vital to note that there must be three horizontal sections of each path. In otherwords, a path must move along the bottom edge of three different squares withinthe lattice path. These horizontal movements may occur at any height below themain diagonal of the lattice. A path in Tn must begin with a horizontal move tothe right. We will define movements along the bottom of the lattice as having aheight of one. Thus, it is the case that h0 = 1. In general, the height of the ithhorizontal movement of a path in Tn is equal to:

hi = v + 1

where v is the total number of vertical movements that occurred before the hori-zontal movement.

We now define our function f : Tn → Sn and show that it is a bijection. Fort ∈ Tn, define f (t) to be:

f (t) = h0h1h2 · · · hn

For example, the corresponding heights for T3 are 111, 112, 113, 122, and 123.First, we show that f (t) ∈ Sn. Consider a lattice path t ∈ Tn with a sequence

of n horizontal movement heights h0h1h2 · · · hn. It must be the case that:

1 ≤ h1 ≤ h2 ≤ · · · ≤ hn (1)

because no downward movements are allowed. Thus, the horizontal movementheights may only increase. As well, we have:

hi ≤ i (2)


Otherwise the point (i, hi) in our path is above the main diagonal. Therefore,f (t) ∈ Sn.

We claim that f is onto. Indeed, the horizontal movement heights of anyt ∈ Tn are always nondecreasing and positive. Therefore, given a sequence s ∈ Snthere exists at least one t ∈ Tn whose horizontal movement heights correspond tothe numbers in s.

The function f must also be one-to-one. Given two lattice paths t1, t2 ∈ T,such that t1 6= t2, there is a specific point at which the paths diverge. Say thisoccurs at position j. It is then the case that hj of t1 is not equal to hj of t2. As such,f (t1) 6= f (t2).

Thus, f is indeed a bijection. We know that the Catalan numbers count thenumber of northeast lattice paths that do not go above the main diagonal. There-fore they also count the number of nondecreasing sequences of positive integers.


Non-Crossing Muraski DiagramsColin Keeley∗

1 Introduction

Noncrossing Muraski diagrams consist of n equally spaced vertical lines. Thelines are then connected to each other from the top of one to the top of another orwe can choose not to connect them at all. The horizontal lines are not allowed tocross each other. This can be achieved by varying the heights of the vertical lines.We show that the number of noncrossing Muraski diagrams of size n is equal tothe n Catalan number, Cn.

2 Examples for 0 ≤ n ≤ 4

2.1 n = 0, C0 = 1

∅

2.2 n = 1, C1 = 1

C1 = C0C0

∗Colin, class of 2011, is an economics and mathematics major. He plays on the basketball teamand is from Kildeer, IL.


2.3 n = 2, C2 = 2

C2 = C0C1 + C1C0

2.4 n = 3, C3 = 5

C3 = C0C2 + C1C1 + C2C0

2.5 n = 4, C4 = 14

C3C0: Do not connect 1st line to any other line.


C0C3: Connect 1st line to 2nd line

C1C2: Connect 1st line to 3rd line.

C2C1: Connect 1st line to 4th line.

C4 = C3C0 + C1C2 + C2C1 + C0C3

3 General Solution

For n lines we connect the 1st line to the kth line. This divides the diagramup into 2 separate problems because no line from 1 to k can be connected to anyline that is after k without crossing another line. The number of diagrams on the


left side up to k is Ck because it is equal to only having the lines from 1 to k. Thenumber of diagrams on the right side of k is Cn−1−k because it is equivalent toonly being able to use the lines from k+1 to n so the number of available lines isn − 1 − k.

Consider k = 1. We begin by not connecting the 1st line to any other line andthen we find all the possible diagrams of the other n − 1 lines. By the productprinciple, the total number of diagrams is C0Cn−1. When k = 2, we connect the1st line to the 2nd line and find all the possible diagrams of the other n − 2 lines.By the product principle, the total number of diagrams is C1Cn−2. We continuethis method from k = 0 to when k = n − 1 and add all the values together to getCn.Therefore we have:

C0 = 1

Cn =n−1

∑k=0

CkC(n−1)−k for n ≥ 0.

So the number of noncrossing Muraski diagrams satisfies the Catalan recur-rence.


(13)(24)-Free InvolutionsMichael Kapernaros∗

1 Problem Statement

We seek to show that the number of fixed-point-free involutions of [2n] such thatif i < j < k < l and w(i) = k, then w(j) 6= l is equivalent to the series known asthe Catalan Numbers, generated recursively as follows:

Cn = Σnk=0(Ck ∗ Cn−k−1)

C0 = 1

The Catalan Numbers are defined for n ≥ 0. Recall that an involution of [m]is an m-permutation w such that w2 is the identity permutation.

2 Valid involutions through n=4

The following subsections enumerate the involutions that satisfy the conditionsof the problem statement for n = 1, 2, 3, 4.

n = 1 (12)n = 2 (12)(34), (14)(23)n = 3 (12)(34)(56), (12)(36)(45), (14)(23)(56),

(16)(23)(45), (16)(25)(34)n = 4 (12)(34)(56)(78), (12)(34)(58)(67), (12)(36)(45)(78)

(12)(38)(45)(67), (12)(38)(47)(56), (14)(23)(56)(78)(14)(23)(58)(67), (16)(78)(23)(45), (16)(78)(25)(43)(18)(23)(45)(67), (18)(23)(47)(56), (18)(25)(34)(67)

(18)(27)(34)(56), (18)(27)(36)(45)

∗Michael, class of 2010, is a Mathematics major and an avid Scrabble player. He hails fromEssex, CT.


3 Proof

Let Vn be the set of fixed-point free involutions of [2n] such that if i < j < k < land w(i) = k, then w(j) 6= l. Let |V|n = Vn. We show that the following recursionholds:

Vn = Σnk=0(Vk ∗Vn−k−1)

Since all these are involutions and fixed-point-free, they cannot contain any-thing but 2-cycles.

We will partition Vn into two sets, as follows:

An = w ∈ Vn|w(1) = 2m + 1, m ≥ 1

Bn = w ∈ Vn|w(1) = 2m, m ≥ 1

An

Note that any remaining number less than 2m+ 1 cannot map to a number greaterthan 2m + 1, or we break the conditions of the problem. Thus, all numbers2, 3, ..., 2m must map to another number within their own set. However, thereare 2m− 1 such numbers. Thus, there are no possible combinations of 2-cycleswithin these. Thus, An = ∅.

Bn

Any remaining number less than 2m cannot map to a number greater than 2m, orwe break the conditions of the problem. Thus,w must map the set 2, 3, ..., 2m− 1must map to itself. However, this is simply a reduced form of our problem: thereare Vm−1 such legal mappings. (It is obvious that shifting the numbers to theright will not affect their position relative to the others).

Similarly, w must map the set 2m + 1, 2m + 2, ..., 2n to itself. Again, this is areduced form of the problem and there are Vn−m legal mappings.

Thus, since any legal mapping from the set 2, 3, ..., 2m− 1 can pair with anylegal mapping from the set 2m + 1, 2m + 2, ..., 2n,

Vn = Σnm=1(Vm−1 ∗Vn−m)


Reindexing, k = m− 1,

Vn = Σnk=0(Vk ∗Vn−k−1),

V0 = 1.

Vn and Cn satisfy the same recurrence, and V0=C0. Thus, Vn = Cn for all n.


Non-Intersecting ArcsJames Leonard∗

1 Introduction

How many ways are there to connect 2n points in the plane lying on a horizontalline by n nonintersecting arcs, each arc connecting two of the points and lyingabove the points? We show that the answer is the n’th Catalan number.

2 Examples

Here are the ways to solve the problem up to n = 4.

n = 1

n = 2

n = 3

∗James, class of 2010, is a Mathematics major.


n = 4

3 Proof

To prove that these are counted by the Catalan numbers, we make a bijectionfrom the problem of counting the number of ways to make n pairs of balancedparentheses. A set of parentheses are balanced if one can take pairs of left andright parentheses such that each left parenthesis is to the left of each right paren-


thesis. To make our bijection, we start with a set of points and arcs. Then wewe turn each left endpoint into a left parenthesis and each right endpoint into aright parenthesis. The parentheses are balanced because we can take the pairs tobe the pairs of endpoints that are connected by arcs.

Now we must show the other direction. To do this I will first show that thereis at least one way to do this, and then I will show that there is at most one wayto do this. Since our pairs of parentheses are balanced, we can find correspond-ing pairs such that each left parenthesis is to the left of each right parenthesis.Now, when we turn the parentheses into dots and connect the correspondingpairs, there may or may not be intersections. Since the number of arcs is finite,the number of intersections is finite. Suppose we have an intersection that lookslike this.

Then we can reconnect the dots such that the two arcs don’t intersect.

Since the resulting arcs have been shortened, no new intersections are made, butat least one intersection is removed. Therefore we can repeat this process untilthere are no intersections.

Now to show that there is at most one way to connect the dots, we go fromleft to right such that each time we reach a right parenthesis we connect it to therightmost of the unused left parentheses that is to the left of the right parenthesis.If at any point we don’t follow this method, we are bound to fail. Suppose wehave a choice of left parentheses and we don’t choose the rightmost one. Thenwe have something like this.

Then all the unpaired points inside must be left parentheses, because we havedealt with all the right parentheses. In order to connect any of these left parenthe-ses, we must cross over an existing arc which is not allowed. Therefore we mustfollow this method, and since we have shown there is at least one way to do this,this must be the only way. Therefore our mapping is a bijection.


Standard Young Tableaux of Shape (n, n).Jack Shirek∗

1 Standard Young Tableaux

A Standard Young Tableaux, Tn of the form (n, n) consists of two rows of lengthn, such that the numbers are increasing in each row and column. We enumeratethe cases up through n = 4:

n Tn

0 ∗1

12

212 1334 24

3123 124 125 134 135456 356 346 256 246

4

1234 1235 1236 1237 1245 1246 12475678 4678 4578 4568 3678 3578 3568

1256 1257 1345 1346 1347 1356 13573478 3468 2678 2578 2568 2478 2468

We see that at n = 0, there is one possible standard young tableaux diagram;at n = 1 there is one; at n = 2 there are 2; at n = 3, 5 possibilities; and at n = 4,we see 14 variations. The number of variations of standard young tableaux of size(n, n) appear to be equal to the Catalan numbers at term Cn. We will prove thisin the subsequent section.∗Jack is a Junior majoring in mathematics at Macalester college. Hailing from Minnetonka,

Minnesota he loves cheese, Scrabble, and all manner of challenging puzzles. Jack, a pharmacytechnician at HealthPartners, also enjoys playing with his dogs, video games, and the companyof his amazing friends.


2 A Bijection to Balanced Parentheses

In order to prove that our counting problem is simply another guise of the Cata-lan numbers, we will create a bijection between our problem and the balancedparentheses problem. We know the balanced parentheses problem is counted bythe Catalan numbers.

Instances of the balanced parentheses problem consist of sequences of n (’sand n )’s such that each open parentheses has a matching closed parentheses,which comes after it. Additionally, as you move from left to right, you never seeore )’s than (’s. We’ll call the set of all 2n balanced parentheses, Pn. We list thefirst few terms:

n Pn

0 *1 ()2 ()(), (())3 ((())), (()()), (())(), ()(()), ()()()4 (((()))), ((()())), ((())()), ((()))(), (())(()), (())()(), (()(()))

(()()()), (()())(), ()((())), ()(())(), ()(()()), ()()(()), ()()()()

Proof:Let Tn be the set of possible Standard Young Tableaux of size (n,n) and let

t ∈ Tn. Define f : Tn → Pn as f (t) = p = p1p2p3 . . . p2n, where pi = ( for all i inthe top row of the Standard Young Tableaux diagram and pi =) for all i in the

bottom row. For example: for t =1234

we have f (t) = (()).

Let t ∈ Tn and f (t) be the sequence of n open and n closed parentheses. Weclaim that f (t) ∈ Pn. The rows of t are increasing to the right and the columnsincreasing downward, therefore t must always begin its first row with one andend its second with 2n. This ensures that f (t) always has an open parentheses asthe first term of its sequence and a closed parentheses as its last. In other words,p1 = ( and p2n = ).

Similarly, because the rows are always increasing, as are the columns, we seethat the ith open parentheses must always come before the ith closed parentheses.

Let t =t11 . . . t1nt21 . . . t2n

. The ith open parentheses occurs at t1i while the ith closed

parentheses occurs at t2i and t1i < t2i for 1 ≤ i ≤ n.


We prove that f is a bijection by shoing that f is invertible. Define g : Pn → Tn

as g(p) = t =ta1 . . . tantb1 . . . tbn

, where ta1 = 1 and tbn = 2n and a1 ≤ a2 ≤ . . . ≤ an are

the indices of open parentheses and b1 ≤ b2 ≤ . . . ≤ bn are the indices of closedparentheses. The mapping g simply reverses the mapping of f , so g is the inversefunction for f . For example, given the sequence p = (()()) we can reconstruct theStandard Young Tableaux diagram, by simply reversing the steps taken to create

the sequence p. Thus, g(p) =124356

Since f is a bijection, we have |Tn| = |Pn| for all n.


Stacking Coins in the PlaneTom Woodward∗

1 Problem Description

Let COINn be the number of ways to stack coins in the plane with the bottomrow consisting of n consecutive coins. In order for a stacking to be valid, if row iconsists of j coins then row i + 1 can only have k coins where 0 ≤ k ≤ j − 1. Wewill show that that COINn is equal to the nth Catalan number.

2 Some Examples

Below is a table showing the all possible stackings of coins for 0 ≤ n ≤ 4. Notethat for n = 0 there is one way to stack the coins, namely, it is the empty stackingin which the bottom row consists of 0 coins.

n = 0: * 1 wayn = 1: 1 way

n = 2:, 2 ways

n = 3:, , , , 5 ways

n = 4:, , , , , , , ,

, , , , ,

14 ways

Table 1: A table of possible stackings for n ≤ 4.

∗Tom, class of 2010, is a mathematics major. He hails from Lincoln, NE.


3 Proof that the Catalan numbers are the solution

Let’s find a recurrence for COINn. Our recurrence will be based on the locationof the first gap in the second row of coins. Suppose this gap appears between thekth and (k + 1)st coins. If there is no gap, then take k = n. We treat the kth coin asspecial and therefore we color it black. The black coin is crucial in that it dividesour problem into two pieces. This will become more clear as we proceed throughthe proof. Below are the stackings for n = 4 arranged in increasing order of k.

, , , , , , ,

, , , , , ,

Without loss of generality, assume we want to build our stacks one coin at atime, starting at the bottom left corner. We begin by placing a single coin (k = 1).This essentially splits the problem into two parts. To the left of (and above) thiscoin, we place all possible stackings for k − 1 = 1 − 1 = 0 coins. To the right ofthe first coin we place all valid stackings for n − k = n − 1 coins. So we have atotal of COIN0 · COINn−1 possibilites here.

we place 0 coins here

we place every stacking for n − 1 coins here

It is clear that this will give a valid stacking of n because the bottom row containsn − 1 + 1 = n coins.

Now let’s consider starting with 2 coins. To the left of and above the second(kth) coin, we can place any valid stacking for k − 1 = 2 − 1 = 1 coins. To theright of the second coin we place all of the possible stackings for n − k = n − 2coins. In this case we have COIN1 · COINn−2 total stackings.

we can place any stacking for n − 2 coins here


Consider k = 3 wherein we place a third coin. We can choose any validstacking for n = 2, which again is k − 1, to the left and above the newly placedcoin and any valid stacking for n − k = n − 3 coins to the right. Here we haveCOIN2 · COINn−3 stackings.

we place every stacking for n − 3 coins here every stacking for n − 3

If k = 4 and we place a fourth coin, we can choose any valid stacking forn = 3, which, yet again, is k − 1, to the left and above the fourth (kth) coin andany valid stacking for n − k = n − 4 to the right of this coin. Here we haveCOIN3 · COINn−4 stackings.

we place every stacking for n − 4 coins to the right of each of these 5 stackings

Now it might be obvious what happens after we have placed k coins on thebottom row. We place all valid stackings for k − 1 above and to the left of the kth

coin and all valid stackings for n − k to the right.

we place every stacking for k − 1 here

we place every stacking for n − k here

Finally, when k = n we place all possible stackings for n − 1 above and to theleft of the nth coin and nothing to the right, giving a total of COINn−1 · COIN0stackings.


If we add up all possible stackings as outlined above, we obtain the followingrecurrence relation

COIN0 = 1

COINn = COIN0 · COINn−1 + COIN1 · COINn−2 + · · ·+ COINn−1 · COIN0

=n

∑k=1

COINk−1COINn−k n ≥ 1

Also, recall that the Catalan numbers satisfy the recurrence relation below

C0 = 1

Cn = C0Cn−1 + C1Cn−2 + C2Cn−3 + . . . + Cn−1C0

=n

∑k=1

Ck−1Cn−k n ≥ 1

Notice that these two recurrences are the same. We saw in Table 1 that for the firstfive values of n, COINn = Cn. But all we really need here is that COIN0 = C0.Because this is true, and both sequences follow the same recurrence relation, itfollows by induction that they are the same.


Sequences Below the Sequence 012 . . . (n − 1)Congcong Nie∗

1 The Interesting Sequence w

Consider sequences w= w1w2w3 . . . wn, such that w1 = 0 and each digit of w, wihas the following property,

0 ≤ wi+1 ≤ wi + 1.

We show that the number of such sequences is equal to the nth Catalan numberCn.

2 Examples for n ≤ 4

wn Sequences Countw0 * 1w1 0 1w2 00 01 2w3 000 001 010 011 012 5w4 0000 0001 0010 0011 0012 14

0100 01010110 0111

0112 0120 0121 0122 0123

∗Congcong, class of 2011, was born and raised in Beijing, and she likes spacing out, bubbletea, collecting cards, making colorful posters, reading in a comfy chair in a cloudy afternoon, andrandom things. She <3s Combinatorics as much as she <3s Macalester. :-)


3 Proof: Catalan numbers are the answers

We prove the number of sequences is Cn by providing a bijection to a well knownCatalan problem: balanced sequences of n 1’s and n −1’s. Note that we’ll neversee more −1’s than 1’s at any digit. To prove this, we define b=b1b2 . . . bn, where

bi = wi − wi+1 + 1

for 1 ≤ i ≤ n

where we set wn+1 = 0 for convenience.We create our sequence S(b) of 1’s and −1’s as follows. For each i that 1 ≤

i ≤ n, write a single 1 followed by bi −1’s. The sequence S(b) will be a balancedsequence.

3.1 Numerical Examples

Consider the 4-digits sequence w1w2w3w4 = 0122. We calculate,

b1 = w1 − w2 + 1 = 0 − 1 + 1 = 0

b2 = w2 − w3 + 1 = 1 − 2 + 2 = 0

b3 = w3 − w4 + 1 = 2 − 2 + 1 = 1

b4 = w4 − w5 + 1 = 2 − 0 + 1 = 3

The resulting sequence is 111(−1)1(−1)(−1)(−1). This is a balanced sequenceof four 1’s and four −1’s.

This algorithm works in the reverse direction too. For example, consider thesequence 1(−1)11(−1)1(−1)(−1), we know that each 1 marks the start of a wi.Recall that w1 = 0. Since the first 1 is followed by a −1, we have b1 = 1 so that

b1 = w1 − w2 + 1 = 1

Thus w2 = w1 = 0. Since nothing follows w2, b2 = 0. Therefore,

b2 = w2 − w3 + 1 = 0

which means that w3 = w2 + 1 = 1. Similarly, we can get

b3 = w3 − w4 + 1 = 1,


so that w4 = w3 = 1.The corresponding 4-digit sequence is 0011. In the next section, we show that

this is a bijection.

3.2 The Proof

Given a sequence w = w1w2 . . . wn, let b = b1b2 . . . bm. The total number of 1’s inS(b) is n. Again, for convenience we define wn+1 = 0

The total number of −1’s is,

n

∑i=1

bi =n

∑i=1

(wi − wi+1 + 1)

= w1 − wn+1 + n= 0 − 0 + n = n.

Therefore S(b) has length of 2n with equal numbers of 1’s and −1’s. Meanwhile,

i

∑k=1

bk =i

∑k=1

(wi − wi+1 + 1)

= w1 − wi+1 + i ≤ i

For 1 ≤ i ≤ n, the number of −1’s does not exceed the number of previouslyseen 1’s. This algorithm works in the reverse direction too. Given the values ofb1b2 . . . bn, we can get below n + 2 equations, in n + 1 unknowns w1w2w3 . . . wn+1,

w1 = 0

bi = wi − wi+1 + 1

wn+1 = 0

1 ≤ i ≤ n

However, the equation for bn is redundant.

bn = n −n−1

∑k=1

bk = wn + 1


n−1

∑k=1

bk =n−1

∑k=1

(wk − wk+1 + 1)

n−1

∑k=1

bk = −wn + (n − 1)

Therefore,

bn = n − (−wn + (n − 1)) = 1 − wn

In other words, we have a unique solution w1, w2, . . . , wn+1 for each sequenceb.

We now must show the solution w1, w2, . . . , wn is a valid sequence. First, wehave,

wi+1 = wi − bi + 1 ≤ wi + 1

We know that

k

∑i=1

bi ≤ k

k

∑i=1

(wi − wi+1 + 1) ≤ k

−wk+1 + k ≤ k

Thus we have wk+1 ≥ 0. For 1 ≤ k ≤ nIn summary, the solution both unique and valid, so the mapping is invertible.

Therefore the number of valid sequences of length n is the nth Catalan numberCn.


Upper Triangular Ferrers DiagramsKatie Agnew∗

1 Introduction

A Ferrers Diagram is a collection of left-justifed boxes where each row has nomore boxes than the row above. Let Fn be the set of Ferrers Diagrams that fit inthe shape (n− 1, n− 2, ..., 1). We show that |Fn| = Cn, the nth Catalan number.

2 Examples

The sets Fn for 1 ≤ n ≤ 4 are as follows. In the table below, we give the shape(n − 1, n − 2, ..., 1), and then list the elements of Fn. Note that we include the“empty diagram” consisting of no boxes as a member of each Fn.

∗Katie is Mathematics major and Environmental Studies minor at Macalester College. She wasborn and raised in Minnesota, plays rugby, and enjoys a good blizzard.


n (n− 1, n− 2, . . . , 1) Fn

1 ∅ ∅

2

3

4

3 Bijection

We will create a bijection from Fn to another set that is counted by the Catalannumbers. Let Sn be the sequences 1 ≤ a1 ≤ ... ≤ an of integers with ai ≤ i. Forexample, S3 = 111, 112, 113, 122, 123. In his paper in this journal, Matt Kusnerproves that the size of the set Sn is the nth Catalan number.

We now define f : Fn → Sn. Given F ∈ Fn, we always consider F to haven − 1 rows by allowing a row to have 0 boxes. We create an (n − 1) tuple asfollows: starting from the bottom row, we create an (n− 1)-tuple whose entriesare the number of elements in each row. So, for F3


we would get:00 01 02 11 12.

Next, we prepend a 0 to this number to get a string of length n. In our example,we get

000 001 002 011 012.

Finally, we add 1 to each digit of the number. The resulting sequences for ourexample are the elements of S3:

111 112 113 122 123.

This mapping is invertible. We simply follow these three steps in reverse. Forexample, if we begin with S3, and then subtract 1 from each digit of the numbers(111 in this case), we get: 000 001 002 011 012. Removing the leading 0gives 00 01 02 11 12. Which corresponds to F3.

Now, we will show that this holds for any value of n. Given F ∈ Fn, we canrepresent F with the (n − 1)-tuple ( f1, f2, ..., fn−1) where fi is the length of the(n− i)th row. Since F is a Ferrers Diagram, we have fi ≤ i and fi ≤ fi+1.

Our mapping φ : Fn → Sn is given by φ(F) = s where s = (s1, ..., sn) =(1, f1 + 1, ..., fn−1 + 1). We show that this mapping is well defined by showingthat s ∈ S. We have

s1 = 1,

si = 1 + fi−1 ≤ 1 + (i− 1) = i for 2 ≤ i ≤ n,

si = 1 + fi−1 ≤ 1 + fi = si+1 for 2 ≤ i ≤ n− 1,

sn = 1 + fn−1 ≤ 1 + (n− 1) = n.

Therefore s ∈ Sn. Because φ is a bijection from Fn to Sn, the size of Fn is the nthCatalan Number.


Pairs of Non-Crossing Lattice PathsDrew Van Denover∗

1 Introduction

Our task is to count the set of unordered pairs of lattice paths with n− 1 stepseach. Both paths must start at the origin and end at the same point, using onlysteps of (1,0) or (0,1) such that one path never crosses the other. For example, thepossible pairs when n = 2 are:

The leftmost paths begin at (0,0) and end at (0,1). The paths in the rightmostpair begin at (0,0) and end at (1,0). In both cases, the paths are identical, so theytechnically overlap one another. They are drawn slightly apart so they can beeasily distinguished.

We prove that the number of pairs of lattice paths of this type is given by thenth Catalan number, Cn. We demonstrate a bijection between the set of thesepaths and the set of balanced groupings of 2n parentheses, where a "balanced"grouping contains equal numbers of open and closed parentheses arranged suchthat if you begin at the left and move to the right, you have never passed moreclosed parentheses than open ones. The existence of such a bijection proves thatboth sets have the same cardinality. Since the number of balanced groupings ofparentheses is known to be the Catalan numbers, the sequence Cn must also de-scribe the number of possible pairs of lattice paths that meet our conditions.

∗Drew, class of 2011, is a Mathematics and Philosophy major from Denver, Colorado.


2 Solutions for n ≤ 4

n = 0:

∅

n = 1:

. .

n = 2:

n = 3:

n = 4:


3 Defining the Mapping

We define a function f : A → B, where A is the set of our lattice pairs oflength n− 1 and B is the set of balanced groupings of 2n parentheses. First, weseparate the two lattice paths and describe each one as strings of U’s and R’s,which represent steps up and right, respectively. For each step of (0,1) we markU, and for each step of (1,0) we mark R. If the two paths are identical, then wecan arbitrarily label one the "top" path and the other the "bottom." If they are notidentical, there is a first step where they disagree. In that case, we call the pathwith behavior (0,1) at that step the top path and the path with (1,0) the bottom.

We then write the string for the top path above the string for the bottom pathto create a 2 x (n− 1) matrix, where each row describes the behavior of one ofthe paths, and the ith column describes the directions of both paths at the ith stepfor 1 ≤ i ≤ n− 1. For example:

→ →→

R U RR R U

Next, we replace each column of that matrix with a particular string of twoparentheses based on the column’s contents. The columns map to pairs of paren-theses as follows:

UU

⇐⇒ ( )RR

⇐⇒ ) (

UR

⇐⇒ ( (RU

⇐⇒ ) )(1)

Using this key, we can convert our example into a string of parentheses like this:

→ →→

R U RR R U

→ ) ( ( ( ) )

Finally, we add an open parenthesis to the beginning of the string, and a closedparenthesis to the end. The end result is a balanced grouping of 2n parentheses:

→ →→

R U RR R U

→ ) ( ( ( ) )→ ( ) ( ( ( ) ) )


4 The Mapping is Well-defined

The result of the function f outlined above must be a balanced group of paren-theses because of two properties of the 2 x n− 1 matrix:

Claim 1: There must be equal numbers ofUR

steps andRU

steps.

Let x be the vertical distance between the two paths after a certain number ofsteps. How does each of the four possible path behaviors affect this value? Let ytrepresent the y-coordinate of the top path, and yb represent the y-coordinate ofthe bottom path. Then, each at step ∆x = ∆yt − ∆yb. There are four cases:

–UU

: Both paths move up one unit. Thus, ∆yt = ∆yb = 1, and ∆x = 0.

Movements of this type do not change the vertical distance between thepaths.

–RR

: Neither path moves in the vertical direction. ∆yt = ∆yb = 0, and

∆x = 0. This behavior does not change the vertical distance either.

–UR

: The top path moves one unit upwards, but the bottom path does not.

So ∆yt = 1 and ∆yb = 0, and ∆x = 1. These steps increase the verticaldistance between the paths by one unit.

–RU

: The opposite is true. ∆yt = 0 and ∆yt = 1, so ∆x = −1. This behavior

reduces the vertical distance between the paths by one unit.

So, for a path with a steps ofUR

and b steps ofRU

, x = a− b. However, in

order to belong to the set of our pairs of paths, both paths must end the samepoint. This means that the total value of x after all steps have been completedmust be 0. So x = a− b = 0, and a = b. In summary, all of our pairs of paths

must have equal numbers ofUR

steps andRU

steps.


Claim 2: There can never be moreRU

steps thanUR

steps as we move from

left to right.If the two paths are at the same point after n steps, the n + 1th step cannot be

RU

, or else the paths will cross (which our lattice paths cannot). Our paths can

move in aRU

step without crossing only if the paths are in different positions after

the previous step. The paths will only be in different positions if theRU

step is

preceded by an unmatchedUR

step, since that is the only other type of behavior

capable of separating the paths.

These two properties taken in conjunction demonstrate that the function fwill always generate a balanced set of parentheses. That is, f is well-defined. Totest whether a group of parentheses is balanced, we keep track of a sum as wemove from left to right, adding 1 each time we pass an open parentheses, andsubtracting 1 each time we pass a closed one. The grouping is balanced as longas that sum is never negative and the total sum is zero.

The first parenthesis in the output from f will always be open, so our sumafter one character is 1. Some combination of "( )", ") (", "( (", and ") )" follow.The groupings "( )" and ") (" each contain one open parenthesis and one closed,so they have no net effect on our sum when we pass them. We could add anynumber of either of these pairings without risking an invalid grouping.

"( (" adds 2 to the sum, and ") )" subtracts 2. These could potentially cause thesum to become negative if we pass a ") )" pair before a "( (" pair. However, weknow this is impossible. We demonstrated above that there are always at least

as manyUR

steps asRU

steps at any point, so there will always be at least as

many "( (" pairings as ") )" pairings as we move from left to right. Thus, we willonly subtract 2 from the sum when we have already added 2, and the sum willnever become negative.

We also know that there are equal numbers of "( (" pairings and ") )" pairings,so the +2s and −2s they generate will ultimately cancel each other out. After


adding any number of "( )" pairings, any number of ") (" pairings, and an equalnumber of "( (" and ") )" pairings, the net sum is 1 (thanks to the first unmatchedopen parentheses). Then, the last parentheses will always be closed, taking thesum back down to 0, indicating that the grouping is balanced. Clearly, any com-bination of steps that corresponds to a valid pair of lattice paths will necessarilygenerate a balanced grouping of parentheses.

5 Defining the Inverse Mapping

We define a function g : B → A which reverses the function f . Given abalanced grouping of 2n parentheses, we can follow the same steps in reverseto produce a unique lattice pair. To demonstrate, we will undo the string ofparentheses we just created.

First, we strip the first and last parentheses from the string, leaving 2(n− 1)parentheses remaining:

( ) ( ( ( ) ) )→ ) ( ( ( ) )

Now, we take the first two parentheses in the string and replace them with a 2x 1 matrix containing U’s and R’s, using the mappings defined in (1). We repeatthis process for the next two parentheses and so on, until the entire string hasbeen replaced. Note that there must be an even number of parentheses becausethe string contained 2(n− 1) parentheses when we began replacing them. Afterreplacing them all, we combine the 2 x 1 columns into an 2 x n− 1 matrix. In ourexample:

( ) ( ( ( ) ) )→ ) ( ( ( ) )→ R U RR R U

Finally, we can use the rows of the resulting matrix to reconstruct each pathindividually and combine them:

( ) ( ( ( ) ) )→ ) ( ( ( ) )→ R U RR R U

→→ →

We have arrived at our original pair of lattices.


6 Conclusion

We have defined a function f : A → B to transform our lattice pairs into abalanced grouping of parentheses, as well the function g : B → A. We have alsoshown that g f : A→ A, which is the identity mapping idA. That is, the functiong is the inverse function for f . Therefore, f is a bijection and |A| = |B|. SinceCn counts balanced groupings of 2n parentheses, it must also count our pairs oflattices with length n− 1.


Partially Connected Polygon GraphsJeanmarie Youngblood∗

1 Problem

How many ways are there to connect some pairs of vertices of an (n− 1)−gonwith non-crossing edges and then circle a subset of the remaining vertices?

This problem is in effect asking how many graphs of an (n − 1)−gon havevertices of maximum degree 1 (and non-crossing edges); each vertex may connectto exactly one other vertex, connect to itself (in other words be circled), or remainisolated. For consistency, we will orient each polygon such that the (n− 1)th ver-tex is in the bottom right corner, and label v1 through vn−2 clockwise, beginningto the immediate left of vn−1. We will show via a recurrence relation that thesolution to this problem for any (n− 1)−gon is the nth Catalan number Cn.

2 Examples

Here are the solutions for 1 ≤ n ≤ 4:

n = 1: 0−gon

∅ =1 way = C1

n = 2: 1−gon

= 2 ways = C2

∗Jeanmarie, class of 2012, has been trying to find a closed formula for primes since her fourthgrade teacher told her she would be on the front page of the New York Times if she did. She caneat more ice cream than you. She hails from Brooklyn, NY.


n = 3: 2−gon

= 5 ways = C3

n = 4: 3−gon

= 14 ways = C4

= 14 ways = C4

For each of these values of 1 ≤ n ≤ 4, the number of possible desired graphsis the corresponding Catalan number Cn. We will now show that these graphssatisfy the recurrence relation for the Catalan numbers.

3 Recurrence Relation

The recurrence relation for the Catalan numbers is

C0 = 1

Cn+1 =n

∑i=0

CiCn−i for n ≥ 0.

To show that the number of possible graphs for each (n− 1)−gon in fact corre-lates to Cn, we must partition each set of graphs such that the parts correspondto C0Cn−1, C1Cn−2,. . . , Cn−1C0, according to the recurrence relation. Before con-sidering the general case, we explain how to partition these graphs when n = 4.


3.1 For the case n = 4

In this case, we must partition the 14 triangles as follows:

– C0C3 = 1 ∗ 5: 5 triangles




The characteristics of vertex v3 provide the basis for creating the partition.Observe that v3 must either share an edge with v1, v2, or itself (meaning circled),or remain isolated.

Consider the graphs in which v3 is circled or isolated as special cases. In eachcase, the number of possible graphs of the remaining 2−gon can be drawn C3ways (as shown in the example where n = 2). Because C0 = 1, these two sets ofgraphs can be counted by C0C3 and C3C0 respectively.

In all other cases, (when an edge is drawn from v3 to another vertex), the edgedivides the remaining vertices into two distinct groups A and B. Because thereare only three vertices in this case, one group will contain the remaining vertexand the other will contain no vertices.

Number of of possible graphs when

– v3 remains isolated: 1∗[number of possible graphs with 2 vertices]= 1 ∗ 5 =C0C3.

v1

v2

v3 v1

v2

v3 v1

v2

v3 v1

v2

v3 v1

v2

v3

– v3 connects to v1: [Side A: number possible graphs with 0 vertices]∗[Side B:number possible graphs with 1 vertex] = 1 ∗ 2 = C1C2.

ABv1

v2

v3ABv1

v2

v3


– v3 connects to v2: [Side A: number possible graphs with 1 vertex]∗[Side B:number possible graphs with 0 vertices]= 1 ∗ 2 = C2C1.

BAv1

v2

v3

BAv1

v2

v3

– v3 is circled: [number of possible graphs with 2 vertices]∗1 = 5 ∗ 1 = C3C0.

v1

v2

v3 v1

v2

v3 v1

v2

v3 v1

v2

v3 v1

v2

v3

This shows that by classifying parts according to which vertex connects to v3,we partition the set of graphs of a 3−gon such that each part corresponds to aterm in the recurrence relation summation.

3.2 Extention to any (n− 1)−gon:

The partition for an (n− 1)−gon depends on the state of vn−1. There are threecases:

– When vn−1 connects to vi, 1 ≤ i ≤ n− 2:The vertices on either side of the edge between vn−1 and vi (1 ≤ i ≤ n− 2)form two separate sets of points A and B, where |A| = i − 1 and |B| =(n − 1) − (i − 1) − 2 = n − i − 2 (The 2 being subtracted corresponds tothe two vertices vn−1 and vi). More specifically, A = 1, 2, . . . , i − 1, andB = i + 1, i + 2, . . . , vn−2.By the recurrence relation, the ways to draw acceptable graphs is Ci forthe vertices in A and Cn−i−1 for those in B. Because these processes areindependent of one another, multiplying them gives the number of ways todo both at once. So, the number of desired graphs of an (n − 1)−gon isCiCn−i−1 for all 0 < i < n− 1.


A B

v1

v2

v3

v4 vn−4

vn−3

vn−2vn−1

vi

– When vn−1 is circled:n− 2 vertices remain. We can therefore draw the rest of the graph in Cn−1 =1 ∗ Cn−1 = C0Cn−1 ways.

– vn−1 is isolated:As in the previous case, we can draw the rest of the graph in Cn−1 = Cn−1 ∗1 = Cn−1C0 ways.

Summing up the number of graphs for each i gives C0Cn−1 + C1Cn− 2 + ... +Cn−2C1 + Cn−1C0, which we know is equal to Cn. In conclusion,

C0 = 1

Cn+1 =n

∑i=0

CiCn−i for n ≥ 0.

so the number of our graphs satisfies the Catalan recurrence.Therefore, the Catalan numbers count the ways to connect some vertices of an

(n− 1)−gon and circle a subset of the remaining vertices.


Plane Binary TreesDavid Klock∗

1 Problem: Plane Binary Trees with n Internal Nodes

We wish to determine how many different structures are possible for a planebinary tree with 2n+1 nodes, or, equivalently, n internal (non-leaf) nodes. Weprove that this number is equivalent to Cn, the nth Catalan number.

2 Solution: Catalan Numbers

2.1 Proof Idea

We already know that the number of valid groupings of 2n parentheses is Cn. Wewill show that the parentheses problem is equivalent to the binary tree problemby defining a bijection from the set of all binary trees B to the set of all groupingsof parentheses P, represented as strings of parentheses. Our bijection will bebased on the following three rules:

1. A node is represented by a pair of parentheses.

2. If a node has a left child, that child is represented as the first pair of paren-theses inside the pair representing the parent node-that is, (( )) represents aparent and a left child.

3. If a node has a right child, that child is represented as the first pair of paren-theses after the pair representing the parent node-that is, ( )( ) represents aparent and a right child.

∗David, class of 2010, is a double major in Political Science and Computer Science, with aminor in Mathematics.


Note that these rules are recursive. For example, if we have the string ofparentheses (( )( )), this represents a parent with a left child which, in turn, has aright child. (It does not represent a parent with two left children–this is why therules emphasize the word “first”.)

For example, if we have the binary tree

we would represent it as “((( ))( ))( )”.Intuitively, these rules will lead to a unique, and valid, representation of a

binary tree as a string of parentheses. Each “(” has a matching “)”. Invalidstrings like “)()(” do not have a chance to occur. Moreover, any given binary treehas one unambiguous parenthetical representation. Below, we prove rigorouslythat these rules indeed lead to a bijection.

2.2 Enumeration of Trees for n ≤ 4

n = 1

n = 2

n = 3


n = 4

2.3 The Bijection

We now explain exactly how our bijection maps B onto P. The problem concernsstrings and nodes. Because these are concepts treated in computer science, it willbe useful for us to think of the bijection as a computer procedure. Let us call itWriteTree(tree).


We begin by defining a subprocedure, WriteNode(node, string), which Write-Tree will use to do the actual representation. WriteNode converts a node (includ-ing its children, recursively) into a parenthesized representation. node is the nodeto be converted, and string is the string to write the parenthesized representationout to.

WriteNode(node, string):

1. Write an opening parenthesis, “(”, to string.

2. If node has a left child left, invoke WriteNode(left, string).

3. Write a closing parenthesis, “)”, to string.

4. If node has a right child right, invoke WriteNode(right, string).

Given this procedure, WriteTree(tree) works as follows:WriteTree(tree):

1. Let string equal an empty string, “ ”.

2. Let node equal the root node of tree.

3. Invoke WriteNode(node, string).

4. Return string as the result.

In other words, WriteTree(tree) takes the root of tree, invokes WriteNode on it,and returns the resulting string of parentheses.

2.4 The Proof

We claim that this function will produce a bijection from B to P such that theimage of a tree with n nodes is a string with n pairs of parentheses. We do so byshowing that WriteTree is a valid, invertible function.

2.4.1 Validity

For WriteTree to be a valid function, the string of parentheses given by Write-Tree(tree) must be within P for every tree in B. A string of parentheses is valid(within P) if there are an equal number of open and closed parentheses and if,beginning at the left and moving right, one at no point encounters more closedparentheses than open parentheses. We observe that:


1. The procedure WriteNode(node, string) writes only two characters–one “(”and one “)”. No other part of the function writes a character. Thus, thenumber of opening parenthesis must equal the number of closing parenthe-sis.

2. WriteNode(node, string) always writes a “(” first, followed by a matching“)” at some point later in the string. Thus, at any point in the string, theremust be at least as many opening parentheses as closing parentheses in theportion of the string coming before that point.

Thus, WriteTree produces strings which are valid members of P. It is, therefore,a valid function from B into P.

2.4.2 Invertibility

We must also show that WriteTree is invertible. To do so, we define its inverse,ReadString. We begin by creating a procedure, ReadParens(string, currentNode,index), which reverses the operation of WriteNode(node, string). The currentNodein ReadParens(string, currentNode, index) requires some explanation; in essence, ifone expects a pair of parentheses is about to come up in string (by looking ahead),one creates the node represented by those parentheses in advance, then invokesReadParens on that node to process its parentheses and, potentially, add childrento it. The index tracks our position in the string.

ReadParens(string, currentNode, index):

1. Begin reading from index.

2. Read the opening parenthesis representing currentNode, “(”, from string.

3. Increment index.

4. If the character now at index in string is “(”:

(a) Create a new node, left. Make this currentNode’s left child.(b) Invoke ReadParens(string, left, index).

5. Read the closing parenthesis representing currentNode, “)”, from string.

6. Increment index.

7. If the character now at index in string is “(”:


(a) Create a new node, right. Make this currentNode’s right child.(b) Invoke ReadParens(string, right, index).

Given this procedure, ReadString(string) for string a valid string of parentheses(as defined above) works as follows. (We ignore the edge case where P is empty,representing a tree with only one node.)

ReadString(string):

1. Create a new node root, which will serve as the root node of our tree.

2. Invoke ReadParens(string, root, 0).

3. Return a binary tree with root as root.

We will prove that ReadString(string) does, in fact, invert WriteTree by induc-tion on n, the number of internal nodes in our tree.

We will make use of the fact that a node with two leaves will be representedby WriteNode as “(( ))( )”, regardless of where it is in the tree. More generally,WriteNode will depict any internal node in the form “(L)R”, where L and R arethe representations of the subtrees of the left and right child of the given node.(When L and R are both leaves, this reduces to “(( ))( )”.)

Our base case in the induction will be that n = 1. In this case, we have abinary tree tree consisting of the root root and two leaves. WriteTree(tree) willtake root and invoke WriteNode on it to get a string of parentheses. By the noteabove, this string will simply be “(( ))( )”. Call this string string. If we invokeReadString(string), we will get back tree.Thus, ReadString inverts WriteTree for this base case.

We now consider the induction case. Suppose that ReadString((WriteTree(tree)))= tree for all trees tree with n - 1 internal nodes. We show that ReadString(WriteTree(tree))= tree for tree tree with n internal nodes.

Suppose we have treen, a tree with n internal nodes. Moreover, suppose node isan internal node with two leaves. At least one such node must exist. If we replacenode with a leaf, we get a tree treen−1 with n - 1 internal nodes. Let f(treen−1) =stringn−1. Then, by our assumption, ReadString(stringn−1) = treen−1.

Now, let WriteTree(treen) = stringn. We show that ReadString(stringn) = treen.Our proof depends on whether node is a left or a right child of nodeparent. Thetwo cases are similar; we consider the case where node is a left child, and omit the


other case.

nodeparent

leaf right child R (potentially has children)

The subtree of nodeparent for treen−1.WriteNode represents this as “(( ))R”.

nodeparent

node right child R (potentially has children)

leaf leaf

The subtree of nodeparent for treen.WriteNode represents this as “((( ))( ))R”.

Consider how nodeparent is represented in stringn−1. To obtain the tree bn−1, wereplaced node with a leaf. Thus, in treen−1, nodeparent has a leaf for its left child. Asobserved above, invoking WriteNode on nodeparent represents it as “(L)R”; sincethe left child is a leaf in treen−1, this reduces to “(( ))R” in stringn−1.

Now, consider how nodeparent is represented in stringn. To obtain thetree treen,nodeparent has node as its left child. Now, node itself has two leaves as children;as noted before, this situation is represented as “(( ))( )”. Thus, WriteNode willrepresent nodeparent in stringn as “((( ))( ))R”.

We can now consider how ReadString acts on stringn−1 and stringn. We showthat ReadString acts the same way on each string except when processing themodified part of the string, and, moreover, uses the modified part of the stringto create a leaf in treen−1 and the node node in treen. This proves that ReadStringworks as expected.

ReadString(stringn−1) and ReadString(stringn) both begin in the same way;they create a root node root, and invoke ReadParens(root, stringn−1, index) andReadParens(root, stringn, index), respectively. Since ReadParens operates in a lin-ear fashion and only ever looks ahead one character, both functions will createthe same structure until they begin reading the modified part of the string repre-senting nodeparent: “(( ))R” in stringn−1, and “((( ))( ))R” in stringn.

At this point, the two functions will differ. ReadString(stringn−1) will interpret


"(( ))R" to create a nodeparent with a leaf on the left and the subtree R on the right,as desired. ReadString(stringn), on the other hand, will interpret "((( ))( ))R" tocreate a nodeparent with the internal node node (with two leaves) as a left child andthe subtree R as a right child. This is also as desired.

From here on, the two functions are parsing the same string. As a result, therest of the tree structure they create is identical. Thus, the only difference betweenthe two trees they generate comes at nodeparent. At this point, when ReadStringparses stringn−1, it gives nodeparent a leaf for a left child. On the other hand, whenReadString parses stringn, it gives nodeparent an internal node for a left child. Thisis exactly what we want to occur. Thus, the induction step is valid in the casewhere node is a left child of nodeparent. The proof for a right child is similar.

Thus, by induction, ReadString is the inverse of WriteTree when WriteTreeacts on binary trees for which n ≥ 1. (We may define a special case to handle thetrivial situation where n = 0; we ignore it here.) Since WriteTree has an inverse,it is a bijection between B and P. Since we can define a bijection between the setof all binary trees with n internal nodes and the set of all groupings of n pairs ofparentheses, the sizes of these sets must be the same. Finally, since the numberof groupings of n pairs of parentheses is Cn, the nth Catalan number, the numberof binary trees with n internal nodes must also be Cn, and our claim is proven.


Non-Intersecting Convex HullsJorge Banuelos∗

1 Counting Problem

Consider the set of partitions B(n) = B1, . . . , Bk of [n] such that if the numbers1, 2, . . . , n are arranged in order around a circle, then the convex hulls of theblocks B1 . . . , Bk are pairwise disjoint. For example, when n = 3 we have 5such partitions:

3 2

1

3 2

1

3 2

1

3 2

1

3 2

1

We show the number of such partitions is equal to the nth Catalan number.

2 Bijection with the Balanced Parentheses Problem

We will prove a bijection exists between the set C(n) of partitions of BalancedParentheses, which we know has a bijection with the Catalan numbers, and ourgiven set of non-intersecting convex hulls B(n) by demonstrating that a bijectionexists for both sets with the set A(n)=(i1, . . . , in) : ∑n

k=1 ik = n and ∑kj=1 ij ≤

k. We begin by proving a bijection between B(n) and A(n). We then prove abijection between C(n) and A(n).

2.1 Non-Intersecting Convex Hulls

We enumerate the set B(n) for n ∈ 1, 2, 3, 4∗Jorge, class of 2011, is a mathematics major from Zacatecas, Mexico.


n=1 1

1 partition

n=2 1 2 1 2

2 partitions

n=3 3 2

1

3 2

1

3 2

1

3 2

1

5 partitions

3 2

1

3 2

1

n=4

4 21

34 2

1

34 2

1

34 2

1

3

14 partitions

4 21

34 2

1

34 2

1

34 2

1

3

4 21

34 2

1

34 2

1

34 2

1

3

4 21

34 2

1

3

Partitions in B(n) : n ∈ 1, 2, 3, 4

Now, we define a bijection g : B(n)→ A(n). If B ∈ B(n) then g(B) = (i1, i2, . . . , in)where the ik are defined as follows. If the element k ∈ [n] is in a z-gon by itselfthen g(k) = 1. If k is the largest element in the z-gon then g(k) = z, otherwise


g(k) = 0.

Example 1: g : B 7→ A

4 21

37→ (0, 1 , 2 , 1)

The inverse function g−1 : A(n) → B(n) simply reverses the process outlinedabove. If ik = 1 then g−1 places the element k in a 1-gon by itself, otherwise it isin a z-gon with other elements. Considering the elements in the domain of g−1

where ik 6= 0, 1 in ascending order, g−1 represents the last element in a ik-gon withthe directly preceding ik elements for which ik = 0, which have not been used inthe creation of another z-gon. g−1 evidently maps each element in its domain toa unique element in B(n). Since our function g is invertible, it is a bijection fromB(n) to A(n).

2.2 Balanced Parentheses

Now we define an invertible function f : C(n) → A(n). We have f (C) =(i1, i2, . . . , in) where the ik are defined as follows. For each partition C ∈ C(n)number the left parentheses in ascending order with the element k ∈ [n]. Fork < n, let ik be the number of right parentheses between the kth and k + 1th leftparentheses. For k = n, let ik be the number of right parentheses after the nth leftparenthesis.

Example 2: f : C 7→ A

( ) ( ( ( ( ) ) ( ) ) ) 7→ (1,0,0,0,2,3)

The function f is one-to-one: it maps each element of C(n) to a different elementof A(n).We define f−1 : A(n) → C(n) by reversing the above process. First for eachelement k ∈ [n] draw a corresponding left parenthesis in ascending order. Thenafter each parenthesis draw ik right parentheses. Since ∑k

j=1 ij ≤ k : 1 ≤ k ≤ n


we are assured that a partition of balanced parentheses is created. Furthermore,each element A evidently maps to a different element C, since the number ofright parenthesis after a certain left parenthesis will not be the same for any 2distinct elements.Since the function f : C(n) → A(n) is invertible it follows that a bijection existsbetween the sets C(n) and A(n).

3 Final Result

It follows that since a bijection exists between the sets A(n) and C(n) and A(n)and B(n) that a bijection must exists between the sets C(n) and B(n). Since thenumbers of elements in the set C(n), of partitions of Balanced Parentheses, isknown to have a bijection with the Catalan numbers, the set B(n) does as well.


Dyck Paths With No Peaks At Height TwoYenee Soh∗

1 Dyck Paths: An Introduction To The Problem

A mountain range (also called a Dyck path) of length 2n is a path from (0, 0)to (2n, 0) which uses only steps (1, 1) and (1,−1) and does not go beneath thex-axis. The resulting number of paths is the nth Catalan number. In this paper,we consider Dyck paths from (0, 0) to (2n + 2, 0) with no peaks at height two. Apeak occurs when a (1, 1) step is followed by a (−1,−1) step. We prove that thenumber of ways to create paths from (0, 0) to (2n + 2, 0) with this restriction isequal to the nth Catalan number Cn .

2 Enumeration of Examples for n ≤ 4

We enumerate all such Dyck paths for n ≤ 4. From the examples below we cansee that the number of paths equal the Catalan numbers for C0, C1 C2, C3 and C4.

For n = 0, we have 1 possible path:

For n = 1, we have 1 possible path:

For n = 2, we have 2 possible paths:

∗Yenee grew up in Seattle, WA and currently lives in Seoul, South Korea. She enjoys travel-ing and has been to Micronesia, Guam, Japan, the Philippines, Cambodia, Vanuatu, Australia,Vietnam and Canada. She loves experiencing new cultures, learning languages and meeting newpeople.


3 Proof: Catalan Numbers Are The Solution

We give a bijection between regular Dyck paths from 0 to 2n and our problem,Dyck paths of length 2n + 2 with no peaks at height 2. It is known that thenumber of regular Dyck paths of length 2n is Cn. By proving a bijection, weconclude that our problem also results in the Catalan numbers.

Let S be the set of regular Dyck paths created with n upstrokes and n down-strokes that all stay above the x-axis. Let T be the set of Dyck paths from (0, 0) to(2n + 2, 0) with no peaks at height two.

We now define our function f : S→ T. First, starting with a Dyck path s fromS, we obtain a Dyck path t from T using the following steps.


(1) We attach a Dyck path of length 2 to the left of s to create s∗, which gives usthe increased length of 2 necessary for T.

(2) A maximal sub-Dyck path refers to a sub-Dyck path with no peaks at height 1that is followed directly by a sub-Dyck path of height 1. Let p∗ be a maximalsub-Dyck path of s∗. To each such p∗ add a (1, 1) step at the beginning and a(−1,−1) step at the end to create sub-Dyck path p∗∗ with no peaks at height2. This step produces a Dyck path s∗∗ which may be larger than 2n + 2.

(3) From s∗∗ eliminate each Dyck path of length 2 that is to the immediate leftof each p∗∗. Removing these peaks will result in a Dyck path of length2n + 2. By lengthening the height of the sub-Dyck paths to ensure no peaksat height 2, we have overall increased the length. If there is one maximalsub-Dyck path, we just get rid of the Dyck path of length 2 in step (1),since both increased the Dyck path by length 2. If we have more than onemaximal sub-Dyck path, we can always get rid of the Dyck path of length2, with height 1 right before it. Indeed a peak of height 1 exists betweenthe maximal sub-Dcyk paths. The result is in T since we have changed theregular Dyck path of length n to a Dyck path of length 2n + 2 that has nopeaks at height 2 and is a unique element t ∈ T.

We now define the reverse function g : T → S. To obtain s from t, we do theabove steps in reverse as follows:

(1) Let t∗ be a sub-Dyck path of t between two consecutive points on the x-axiswith t∗ having no peaks at height 1. To each t∗ add a Dyck path of length 2immediately to the left. This step produces a Dyck path s∗∗.

(2) Let p∗∗ be a maximal sub-Dyck path of s∗∗. From each such p∗∗ remove theleft-most (1, 1) step and the right-most (−1,−1) step to produce a sub-Dyckpath p∗. This step produces a Dyck path s∗ of length 2n + 2.

(3) From s∗, remove the left-most Dyck path of length 2 to produce s.

Through these steps, we regain the Dyck paths of length 2 and of height 1that we had gotten rid of in step (3) of the original function, and decrease theheight of the maximal sub-Dyck paths to their original height by getting rid ofthe left-most (1, 1) step and the right-most (−1,−1) step, and remove the step we


had added in step(1) of the original function. This results in a regular Dyck pathwith length 2n.

In conclusion, f is a bijection since the reverse function g gives us back withthe Dyck path we started with.

For example, with n = 3 we have a regular Dyck path with length 6 and add aDyck path of length 2 to the left. We will define this as s∗. This can be illustratedas the following:

s

becomes

s∗

Next, we ensure that we have no peaks at height two. We would look at thepeaks in s∗. First we do not have to consider those that have peaks of height one,since this does not affect our requirements for T. So we would focus on peakswhere the height is greater than one. To eliminate any possibilities of having anypeaks at height 2 we would increase the regular Dyck path by adding a step onboth ends of these paths, creating s∗∗ as illustrated below:

s∗

becomes

s∗∗

By doing the two operations above, we have provided the length 2n require-ment and the requirement that there are no peaks of height 2. However, by doingthe second operation we have increased the length to 2n+ 4, so we eliminate eachDyck path of length 2 to the immediate left of each s∗∗. This gives us a uniquet ∈ T, as illustrated below:

t


It seems that the adding a Dyck path and then removing it is unnecessary, yetit is an important step when there are multiple maximal sub-Dyck paths wherewe increase the height. In this case, the number of Dyck paths of height 1 equalsthe number of mountain ranges we increase (so the two cancel out each other).

With the same example, if we go through the second set of steps that definethe reverse function, we start with a t ∈ T that gives us back the s ∈ S that westarted with.

First on the t that we acquired as a result of the first operation, we add a Dyckpath of length 2 immediately to the left to produce Dyck path s∗∗.

t

becomes

s∗∗

We have regained the Dyck paths of length 2 that we had gotten rid of. Next,we back step the increased height of the maximal sub-Dyck paths, to produce s∗,as illustrated below.

s∗∗

becomes

s∗

Then we have to get rid of the Dyck path of length 2 added to the left-most ofthe Dyck path to end with our original Dyck path.

s∗

becomes

s

As we can see from the example above, the result of the reverse function givesus back what we started with in the beginning.


Pairs of Internally Disjoint Lattice PathsMatthew Hurni∗

1 Counting Problem

Let Ln be the set of unordered pairs of lattice paths with n + 1 steps each, startingat (0, 0), using steps (1, 0) or (0, 1), and ending at the same point. We show |Ln|is the nth Catalan number.

2 A Bijection with Sequences of 1’s and (-1)’s

Let Sn be the set of balanced sequences of n 1’s and n (-1)’s. This is a well knownCatalan Problem. We will describe a bijection from Ln to Sn. Begin by breakingthe lattice paths down into a grid of squares with sides of length 1. Supposethis grid has k columns C1, . . . , Ck. Let ai be the number of squares in column Ci,for 1 ≤ i ≤ k, and let bi be the number of rows in common to Ci and Ci+1, for1 ≤ i ≤ k− 1. Define a sequence Ω of 1’s and (-1)’s as follows (where exponenti-ation denotes repitition):

Ω = 1a1(−1)a1−b1+11a2−b1+1(−1)a2−b2+11a3−b2+1...1ak−bk−1+1(−1)ak .

The following table lists the lattice paths along with its corresponding se-quence and the values of (a1, . . . , ak) and (b1, . . . , bk−1).

∗Matt, class of 2011, is from Little Falls, MN. He is a mathematics major and a physics minor,and is also a member of the Macalester football team.


n grid of squares sequence of 1’s and (-1)’s ai bi

1 1− (1) ∅

2 1− 1− (1, 1) (1)

11−− (2) ∅

1− 1− 1− (1, 1, 1) (1, 1)

1− 11−− (1, 2) (1)

3 11− 1−− (2, 1) (1)

11−−1− (2, 2) (2)

111−−− (3) ∅

1− 1− 1− 1− (1, 1, 1, 1) (1, 1, 1)

1− 1− 11−− (1, 1, 2) (1, 1)

1− 11− 1−− (1, 2, 1) (1, 1)

4 1− 11−−1− (1, 2, 2) (1, 2)

1− 111−−− (1, 3) (1)

11−−1− 1− (2, 1, 1) (1, 1)

11−−11−− (2, 2) (1)


n grid of squares sequence of 1’s and (-1)’s ai bi

11− 1−−1− (2, 2, 1) (2, 1)

11− 1− 1−− (2, 2, 2) (2, 2)

11− 11−−− (2, 3) (2)

4 111− 1−−− (3, 1) (1)

111−−1−− (3, 2) (2)

111−−− 1− (3, 3) (3)

1111−−−− (4) ∅

This mapping is well defined due to the following:

– The total number of 1’s and (-1)’s are the same. All that must be done is toadd up the exponents of each. Indeed, the sum of the exponents of the 1’sis

a1 + (a2 − b1 + 1) + (a3 − b2 + 1) + · · ·+ ak − bk−1 + 1 =k

∑i=1

ai −k−1

∑j=1

bj + k

and the sum of the exponents of the (-1)’s is

(a1− b1 + 1)+ (a2− b2 + 1)+ (ak−1− bk−1 + 1)+ · · ·+ ak =k

∑i=1

ai−k−1

∑j=1

bj + k.

– At any given point in the sequence of 1’s and (-1)’s there are at least as many1’s as (-1)’s. The most likely point to find more (-1)’s than 1’s occurs at the


end of a sequence of (-1)’s. For 1 ≤ r ≤ k− 1, by the rth pair of 1’s and (-1)’s,the total number of 1’s we have encountered is:

a1 +r

∑j=2

(aj − bj−1 + 1) =r

∑j=1

aj −r−1

∑j=1

bj + (r− 1)

and the total number of (-1)’s we have encountered is:

r

∑j=1

(aj − bj−1 + 1) =r

∑j=1

aj −r

∑j=1

bj + r ≤r

∑j=1

aj −r−1

∑j=1

bj + (r− 1)

since br ≥ 1.

3 Bijection: From Squares to 1’s and (-1)’s

We show the mapping is one-one and onto.

3.1 One-One

For any valid shape a unique series of 1’s and (-1)’s exists. For instance, supposetwo shapes are described by a1, b1, a2, b2, . . . , an and c1, d1, c2, d2, . . . , cm. If n 6= mthere are a different number of columns. So the resulting sequences are distinct.Now consider the two sequences a1, b1, a2, b2, . . . , an and c1, d1, c2, d2, . . . , cn. If anyai 6= ci or any bi 6= di then there is obviosly a distinct series of 1’s and (-1)’scorresponding to that term as a result.

3.2 Onto

Consider a balanced sequence 1r1(−1)s11r2(−1)s2 · · · 1rk(−1)sk where ri ≥ 1 andsi ≥ 1. The exponents satisfy ∑k

i=1 ri = ∑ki=1 sk = n and ∑

ji=1 si ≤ ∑

ji=1 ri for

1 ≤ j ≤ k.

We claim that a balanced sequence of 1’s and (-1)’s, has a unique solution fora1, · · · , ak, b1, · · · bk−1. Indeed, the exponents result in 2k equations. However, wehave 2k − 1 unknown variables (the a’s and the b’s). This would be a problem


except for the fact that the last equation is redundant.We show that the constraint sk = ak is redundant. If the other 2k − 1 equationsare satisfied, then

sk =k

∑i=1

ri −k−1

∑i=1

si = a1 +k

∑i=2

(ai − bi−1 + i)−k−1

∑i=1

ai − bi + 1 = ak.

So we have 2k− 1 constraints and 2k− 1 unknown variables. So given a set ofr’s and s’s, we are able to obtain the a’s and the b’s.Now we must show that the a’s and the b’s we’ve obtained lead to a valid shape.A pair of sequences (a1, . . . , ak),(b1, . . . , bk−1) create a valid shape if and only ifthe following conditions hold.

1. aj ≥ 1, 1 ≤ j ≤ k

2. bj ≥ 1, 1 ≤ j ≤ k− 1

3. bj ≤ aj, 1 ≤ j ≤ k− 1

4. bj−1 ≤ aj, 2 ≤ j ≤ k

We now show that the a’s and b’s cooresponding to the balanced sequencesatisfy all 4 conditions.

– For 1 ≤ i ≤ k − 1, we have 1 ≤ si ≤ ai − bi + 1. Therefore bi ≤ ai andcondition 3 holds.

– For 2 ≤ i ≤ k, we have 1 ≤ ri ≤ ai − bi−1 + 1. Therefore bi−1 ≤ ai andcondition 4 holds.

We now show that condition 2 holds. We have

j

∑i=1

si ≤j

∑i=1

ri


j

∑i=1

(ai − bi + 1) ≤ a1 +j

∑i=2

(ai − bi−1 + 1)

j

∑i=1

ai −j

∑i=1

bi + j ≤j

∑i=1

ai −j−1

∑i=1

bj + (j− 1)

−bj ≤ −1

Therefore bj ≥ 1. Finally condition 1 also holds since a1 = r1 ≥ 1 and 1 ≤ bj−1 ≤aj for 2 ≤ j ≤ r.

In conclusion, our mapping is a bijection, so |Ln| is the nth Catalan number.


Dyck Paths With Odd Maximal DescentsEmily Merrill∗

1 Problem

We will consider a specific group of Dyck paths and prove that they can be enu-merated by the Catalan numbers. Dyck paths are lattice paths with steps (1, 1)and (1,−1) which never fall below the x-axis. We consider Dyck paths from (0, 0)to (2n + 2, 0) such that any maximal sequence of consecutive steps (1,−1) end-ing on the x-axis has odd length. Let Ω be the set of all such Dyck paths. Letωn be the subset of paths of length 2n + 2, and let zn = |ωn|. As an example,both figures below are Dyck paths from (0, 0) to (4, 0). However, only the one onthe left is a member of Ω, since the one on the right has two consecutive (1,−1)steps leading to the x-axis. This proof will demonstrate that zn = cn for all n ≥ 0,where cn is the nth Catalan number.

2 Enumeration of Solutions for n ≤ 4

We enumerate all members of ω for 0 ≤ n ≤ 4.

z0 = 1

∗Emily, class of 2010, is a math and environmental studies major with a geography minor. Shewas raised in Illinois, Indiana, and Texas.


z1 = 1

z2 = 2

z3 = 5


z4 = 14


3 Proof

3.1 Establishing a Generating Function

Let αn be the set of all Dyck paths from (0, 0) to (2n, 0) which never touch thex-axis between 0 and 2n and have an odd number of consecutive (1,−1) steps atthe end. Let A =

⋃n≥0 αn. We can see that A is a subset of Ω, and αn is a subset

of ωn for any given n. More specifically, A is the set of all paths in Ω which nevertouch the x-axis in between their endpoints.

Now, we will try to construct Ω entirely from A, which will be easier to workwith in the long-term. If we cut up a path in Ω at each point where it touchesthe x-axis, each smaller component will be a member of A. Indeed, each of thesmaller paths does not touch the x-axis between its endpoints. Furthermore, thedefinition of Ω requires that each sequence of (1,−1) which ends on the x-axisbe of odd length.

In order to write out Ω in terms of A, we must define a notation for ap-pended paths. The path which is composed of some member of αm followed bysome member of αn (connected at their endpoints) will be represented by αmαn.We must also introduce a way to refer to members of ωn by the number of inter-sections which they have with the x-axis. We will in the future refer to ω

pn for

paths in ωn which touch the x-axis p times between their endpoints:For any given n, ωn consists of the union of:

– ω0n, paths which never touch the x-axis between the two endpoints

– ω1n, paths which touch the x-axis once between the two endpoints

– ω2n, paths which touch the x-axis twice between the two endpoints

– et cetera.

In other words,

ωn =n⋃

k=0

ωkn.


Furthermore, we have

ω0n = αn+1,

ω1n =

n⋃

k=1

αkαn−k+1,

ω2n =

n⋃

j=1

j⋃

k=1

αkαj−k+1

αn−j,

and so on.We now develop formulae for an = |αn| and zn = |ωn|. The above equations

lead to analogous equations for the sizes of these sets.

zn =n

∑k=0

zkn,

z0n = an+1,

z1n =

n

∑k=1

akan−k+1,

z2n =

n

∑j=1

(j

∑k=1

akaj−k+1

)an−j.

We now introduce the generating functions for these sequences:

A(x) =∞

∑i=0

aixi,

Z(x) =∞

∑i=0

zixi.

It will be useful to employ the product rule for generating functions. Given twoordinary generating functions F(x) and G(x), whose coefficients of xn are fn andgn, respectively, we can find the coefficients of H(x) = F(x)G(x) in terms of fnand gn using the so-called product formula:

hn =n

∑i=0

fign−i.


This sum looks similar to our equation for z1n, with the main difference coming

from the indexing. By this, z1n should be a coefficient of the generating function

A(x)A(x) = A(x)2.At this point, though, it is important to recall that paths in ωn extend from 0 to

2(n + 1), while paths in αn travel from 0 to 2n. This means that z1n is actually the

(n + 1)th coefficient of A(x)2. We can express this as: z1n = [xn+1]A(x)2 From our

knowledge of the product rule and the formula for z2n above, we can extrapolate

that z2n = [xn+1]A(x)3. More generally, zk

n = [xn+1]A(x)k+1. This leads us to

Z(x)k =A(x)k+1

x.

This is the generating function for Ωk. We know that the set Ω is the union ofΩks, k ≥ 0 (all paths which touch zero times, one time, two times, etc.) Therefore:

Z(x) =∞

∑k=0

A(x)k+1

x

xZ(x) =∞

∑k=1

A(x)k

xZ(x) + A(x)0 =∞

∑k=0

A(x)k

xZ(x) + 1 =∞

∑k=0

A(x)k

xZ(x) + 1 =1

1− A(x). (1)

3.2 Relating Generating Functions for Dyck Paths

We now have Z(x) in terms of the geometric power series of A(x). We want torelate A(x) to the Catalan numbers, cn, and their generating function, C(x). TheCatalan numbers enumerate the general Dyck paths, with cn counting paths from(0, 0) to (2n, 0).

First, we’ll need to define βn, the set of all Dyck paths which never hit thex-axis between 0 and 2n and end in an even number of (1,−1) steps down. Next,we’ll define B(x) as the generating function for bn = |βn|. The sum (an + bn)


counts all Dyck paths from (1, 1) to (2n− 1, 1) with one leg added to each side.In other words, an + bn = cn−1, so A(x) + B(x) = xC(x).

We can also prove a few more equalities about A(x) and B(x):

A(x) = x(1 + C(x)B(x)), (2)

B(x) = xC(x)A(x). (3)

From the product rule, C(x)B(x) is a series consisting of all paths from theset counted by B(x) appended to all paths from the set counted by C(x). Inpractice, this encompasses all paths which end in an even number of steps down.If we add 1x0 to that (for paths strictly in B with an empty Catalan path onthe left) and multiply the result by x, we have added a leg to both sides, andwhat formerly ended in an even number of steps now ends in an odd number ofsteps. These new paths do not touch the x-axis between their endpoints, makingthem members of A(x). In the same way, placing C(x) and A(x) together andmultiplying by x gives us all of the paths counted by B(x).

Combining (2) and (3) gives:

A(x) =x

1− x2C(x)2 . (4)

3.3 The Catalan Generating Function

Let C(x) be the generating function for the Catalan numbers. Starting from therecursive formula of the Catalan numbers and using the product formula, wefind that:

∞

∑n=1

cnxn =∞

∑n=1

(n−1

∑k=0

ckcn−1−k

)xn

C(x)− 1 = x∞

∑n=1

(n−1

∑k=0

ckcn−1−k

)xn−1

C(x)− 1 = x∞

∑m=0

(m

∑k=0

ckcm−k

)xm

C(x) = 1 + xC(x)2.

Now, by moving terms around, we see:


1 = C(x)− xC(x)2

1 = C(x)(

1− xC(x)2)

C(x) =1

1− xC(x)

xC(x) =x

1− xC(x)

0 = xC(x)− x1− xC(x)

1 = 1 + xC(x)− x1− xC(x)

1 = (1 + xC(x))1− xC(x)1− xC(x)

− x1− xC(x)

1 =(1 + xC(x)) (1− xC(x))

1− xC(x)− x

1− xC(x)

1 =1− x2C(x)2

1− xC(x)− x

1− xC(x)

1 =1− x2C(x)2 − x

1− xC(x)

1 =1− xC(x)

1− x2C(x)2 − x. (5)

3.4 Proving that C(x) = Z(x)

Now, we’ll use the relationship among A(x), C(x), and Z(x) to prove C(x) =Z(x). By (1) and (4), we have:


1 + xZ(x) =1

1− A(x)=

11− x

1−x2C(x)2

=1− x2C(x)2

1− x2C(x)2 − x

= (1 + xC(x))1− xC(x)

1− x2C(x)2 − x= 1 + xC(x),

where the last equality follows from (5). Therefore, Z(x) = C(x) and |ωn| = zn =cn for n ≥ 0. We have now proven that our variation of the Dyck paths are in factenumerated by the Catalan numbers.

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