Catalan
Numbers
A presentation by
What is a Catalan number?
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They are a sequence of numbers that arise in various problems. The terms of the sequence
can be calculated by the formula:
Notice how the terms of the sequence generated grow rapidly:
Cn = 1, 1, 2, 5, 14, 42, 132, 429, 1430, …..
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There are other different variations of the formula but these are all equivalent, including:
Natural number proof
We shall now see that the Catalan numbers are natural numbers as this is not instantly obvious.
We shall do this by proving the previous formula:
Expressing Cn as the difference of two binomial coefficients, we have thus proved that it is in
fact a natural number.
Catalan numbers in
Pascal’s triangle
The Catalan series can be found in Pascal’s triangle:
1
1
1
1
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Looking at the numbers in the central column and the adjacent column, you will notice that the
difference of these numbers produces the Catalan sequence.
Numbers in the central
column
-
2–1= 1
6–4= 2
20 – 15 = 5
70 – 56 = 14
252 – 210 = 42
History of the Catalan
numbers
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These numbers were first discovered by Leonhard Euler in the
18th century while he was trying to see how many ways a
polygon with n+2 sides can be divided into n triangles without
any of the lines intersecting.
They were later named after Eugène Catalan in 1838 after he
defined the sequence and found a more elegant formula.
He also worked on the polygon problem but later found that
the Catalan numbers appeared when looking at the problem of
counting the number of ways a group of n letters could be
fitted into parentheses.
Eugène Charles Catalan 1814-1894
Appearances of Catalan
number
The Catalan numbers appear within combinatorical problems in mathematics. A few of the main
problems we will be looking at in closer detail include:
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The Parenthesis problem
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Rooted binary trees
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The Polygon problem
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The Grid problem
Parentheses
In 1838 Eugene Catalan solved the following problem:
How many different ways is it possible to arrange n pairs of parentheses so that they “make
sense”?
We shall say that a string of parentheses makes sense (or is valid) if it follows these rules:
i) There are an equal number of open and closed parentheses in the string.
ii) Counting from the left, the number of closed parentheses do not exceed the number of open
parentheses at any point, for example:
Parentheses Examples
By this definition, the following are valid chains of parentheses:
However these are not:
Catalan demonstrated that the number of possible ways of ordering n pairs of parentheses like
this is precisely Cn, the nth Catalan number.
Parentheses (continued)
We can check this manually for small values of n:
n=0
n=1
n=2
n=3
*
( ).
( ( ) ), ( ) ( ).
( ( ( ) ) ), ( ( ) ) ( ), ( ( ) ( ) ), ( ) ( ( ) ), ( ) ( ) ( ).
This table confirms Catalan’s results for C0 to C3.
We shall now see a proof for the general case.
1 way
1 way
2 ways
5 ways
Proof
For clarity, throughout this proof we will let O denote an open parenthesis and let C denote a
closed parenthesis.
The total number of different arrangements of n Os and n Cs is
This however includes the invalid cases (such as OCOCCO) that we are not interested with.
We must now calculate the number of invalid cases and subtract this number from
to find our answer.
Proof (continued)
Consider an invalid string of n pairs of parentheses. The first ‘fault’ is some C that is preceded
by an equal number of Os and Cs, say k of each.
Hence the ‘faulty’ C lies in the (2k +1)th position in the string of parentheses.
We can then take these first (2k+1) terms and switch each one so that every O becomes a C
and vice versa.
Following this process we now have an arrangement of (n+1) Os and (n-1) Cs.
Proof (continued)
Conversely, any arrangement of (n+1) Os and (n-1) Cs can be rewritten as an invalid sequence
of n pairs of parentheses. We do this by noting the first time Os outnumber Cs by one and
switching each term, up to and including that point.
Therefore the number of invalid sequences of n pairs of parentheses is equal to the total number of
arrangements of (n+1) Os and (n-1) Cs.
This is equal to
.
Subtracting this value we see that the number of ways of arranging n pairs of parentheses is equal
to
The Recurrence Relation
It is possible to determine Catalan numbers by expressing them in terms of previous values.
Here we shall prove that
=
For example we can calculate C4 using this recurrence relation.
Assume we know that C3 = 5, C2 = 2, C1 = 1, C0 = 1.
C4 = C0C3 +C1C2+ C2C1 + C3C0 = 1*5 + 1*2 + 2*1 + 5*1 = 14.
.
We shall now see that the previous problem regarding parentheses satisfies
this recurrence
relation.
Recurrence Relation Proof
It is clear that in any valid string of parentheses, the first character must be an open parenthesis ‘(‘.
Later in the string, for validity, there must be a corresponding closed parenthesis ‘)’.
Any further pairs of parentheses must lie either between the initial pair, or after them.
If we wished to arrange (n+1) pairs of parentheses so that they make sense we would place an
initial pair, and then a further n pairs of parentheses in the places marked A and B in the diagram.
A and B must be valid strings of parentheses themselves and clearly either can contain up to n
pairs, however if A contains a string of k pairs of parentheses, B must contain n-k pairs.
Proof (continued)
Hence the possibilities are:
A contains n pairs
B contains 0 pairs
CnC0 possibilities
A contains (n-1) pairs
B contains 1 pair
Cn-1C1 possibilities
A contains 0 pairs
B contains n pairs
C0Cn possibilities
Giving a total of CnC0 +Cn-1C1 + … + C0Cn possibilities.
Hence
=
Rooted Binary Trees
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A binary tree is a rooted tree where each node has two descendants, the left child and
the right child. Except for the end (shaded) nodes. Problem: How many rooted binary
trees can be made with n+1 end nodes?
n=0
n=1
n=2
n=3
The root of the problem
n=1
DUDU
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The rules of these sequences of D’s and U’s are such that no initial part of the
sequence has more U’s than D’s.
This is exactly the same as for the O’s and C’s in the Parentheses examples.
Polygons
This problem involves the number of ways an n+2-sided polygon can be divided
into n triangles by adding straight non-intersecting lines between the vertices.
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E.g. In the case of n=2:
There are clearly only two possible ways, C2=2.
Further Polygons
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n=3
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C3=5
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n=4
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n=5
C4=14
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C5=42
Connecting Polygons and
Rooted Binary Tree
This can be done by firstly planting the root of the tree to one of the edges of
the polygon:
Connecting Polygons and
Rooted Binary Tree
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Next adding nodes to each triangle and linking these together with arcs
corresponds to the branches of the binary tree with degree 3:
Connecting Polygons and
Rooted Binary Tree
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Finally by inserting the end nodes (leaves) of degree 1, it completes the
connection for the Polygon to its equivalent Rooted Binary Tree:
Grids
This problem looks at the number of ways to cross an n x n grid, in the shortest way,
starting in the bottom left corner and going to the top right, without crossing the
diagonal line.
Finish
Cannot make
moves along these
lines
Start
Must only make
moves along these
lines
A typical path:
Here is an example of a route that might be taken:
And this is equivalent to
OCOOOCCOCOCC
( ) ( ( ( ) ) ( ) ( ) )
C
O
Summary
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We have given a few examples of counting problems in which the Catalan numbers arise
but there are many other problems where they can be found.
We have seen how they are related when the problem is viewed in a different way.
If we remember the condition that ‘In a sequence of 2n items with n A’s and n B’s no initial
part of the sequence has more B’s than A’s.’ Then the number of ways of counting these is
the nth Catalan number.
We can make a new example…
Our Catalan Example
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Suppose two groups of mathematics students each perform a presentation, say one regarding
Catalan numbers (group A) and another regarding eclipses (group B).
Assume an audience of 2n people had to vote for their favourite project, n choosing group A, the
other n choosing group B. How many different ways can the votes be counted so that the eclipses
group are never ahead?
The answer is the nth Catalan number Cn.