Download Lecture 10: Combinatorics 1 Binomial coefficient and Pascals triangle

DD2458, Problem Solving and Programming Under Pressure
Lecture 10: Combinatorics
Date: 2008-11-17
Scribe(s): John Eriksson, Tony Karlsson
Lecturer: Douglas Wikström
This lecture will go through different areas in combinatorics such as binomial coefficients, multinomial coefficients, permutations, Stirling numbers, Bell numbers
and Catalan numbers.
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n
k
Binomial coefficient and Pascals triangle
- is called the binomial coefficient or sometimes the binomial numbers and is
defined as the number of ways you can choose a subset of k elements from a set of
n elements. The binomial coefficient nk can be computed as:
n!
n
=
k
k!(n − k)!
(1)
An alternative way to compute the binomial coefficient is the following:
(
1
n
=
n−1
k
k−1 +
n−1
k
if n = k or k = 0
otherwise
(2)
Since using the first way to calculate a binomial coefficient would produce very
large numbers it is better to simply use the second way to recursively calculate the
binomial coefficient the same way it’s done in Pascals triangle.
We can see that the construction of Pascals triangle is related to the binomial
numbers by the recursive definition above. The definition is exactly how Pascals
triangle is built from top to bottom, where an element is the sum of the two closest
elements above it.
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1.1
DD2458 – Popup HT 2008
Binomial coefficient algorithm
The following algorithm will calculate the binomial coefficient with the help of
dynamic programming and memoization. Elements are saved in a matrix T [i][j]
where they are initially set to ⊥ . The table
is then reused between calls
the
where
n−1
and
results of previous calls to calculate nk , and its recursive calls n−1
k ,
k−1
have been saved in the matrix T [i][j]. The technique of reusing results calculated
by previous calls is called memoization.
Input: Integers
n and k such that 0 ≤ k ≤ n
n
Output: k
Bin(n,k)
(1)
if T [n, k] = ⊥
(2)
if n = k ∨ k = 0
(3)
T [n, k] = 1
(4)
else
(5)
T [n, k] ← Bin(n − 1, k − 1) + Bin(n − 1, k)
(6)
return T [n, k]
1.2
Binomial coefficient algorithm
The following algorithm will use only dynamic programming to calculate the binomial coefficient using recursion.
Input: Integers
n and k such that 0 ≤ k ≤ n
Output: nk
Bin(n, k)
(1)
for i = 0 to n
(2)
T [i, 0] ← T [i, i] ← 1
(3)
for i = 2 to n
(4)
for j = 1 to min(i − 1, k)
(5)
T [i, j] ← T [i − 1, j − 1] + T [i − 1, j]
(6)
return T [n, k]
2
Multinomial coefficient
Suppose we have t boxes of sizes k1 , . . . , kt . Then the multinomial coefficient
n
k1 ,...,kt can be interpreted as the number of ways to place the elements 1, . . . , n
Pt
in these boxes where n = i=1 ki . This can be computed as:
n
k1 , . . . , kt
=
n!
k1 ! · · · kt !
(3)
The multinomial coefficient gets its name from the following equality, where it
defines a coefficient.
Combinatorics
3
X
t
n
xi
X
=
Pt
i=1
i=1 ki =n
Y
t
n
xki
k1 , . . . , kt i=1 i
(4)
Since it is too expensive to use equation 3 to calculate the multinomial coefficient due to the large numbers that will be computed by the factorial expressions
the following expression can be used, where we just have to use one of the recursive
algorithms described in the previous section to calculate the binomial coefficients
and multiply them.
n
k1 , . . . , kt
Pt
=
i=1
k1
ki
Pt
i=2
k2
ki
kt
...
kt
(5)
Equation 5 can be derived from the following argument. Fill the first box by
choosing k1 elements from the set of n elements, this can be done in kn1 ways.
Next fill the second box by choosing
k2 elements from the remaining set of n − k1
1
elements, this can be done in n−k
ways and so on until all boxes have been filled.
k2
The total number of ways to do this can be seen as the product of the individual
binomial expressions.
3
Permutations
A permutation of a string x = x1 x2 . . . xn is a rearrangement of that string into a
new string. To find all the permutations of the string x one have to find all the
possible rearrangements such that no two rearrangements are equal.
If x = x1 x2 . . . xn is a string of length n then we can calculate the number of
permutations of the string x in the following way. Let fa = number of times the
letter a is present in the string x.
n
(6)
fa , fb , . . .
An example is the string x = cbaac , then we get fa = 2,
fb =51,fc = 2. Which
will give us the following number of permutations fa ,fnb ,fc = 2,1,2
= 30.
In the special case where all the letters, xi of string x are distinct we get the
number of permutations as n!.
3.1
Lexicographical order generation
A string x1 x2 . . . xk is given, then the following algorithm will generate all permutations of the string x1 x2 . . . xk in lexicographical order. If the string is sorted
then the complete set of permutations in lexicographical order will be generated,
otherwise the permutation following the one in lexicographical order of the string
will be generated.
The algorithm Next first try to find a suffix in reverse lexicographical order
then swapping the element xi just before the suffix with the rightmost element
larger than xi and finally reversing the suffix.
If we for example run the algorithm Next on the string bcfecb we get i = 2
and j = 4. Running Swap we get the string befccb and then running reverse will
get the next permutation as bebccf.
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DD2458 – Popup HT 2008
If we in another example run the algorithm Next on the string dcba we cannot
find any such i since this is the last permutation in lexicographical order.
Next(x1 x2 . . . xn )
(1)
Find i such that xi < xi+1 ≥ . . . ≥ xn
(2)
Find maximum j such that xj > xi
(3)
Swap(xj , xj )
(4)
Reverse(xi+1 . . . xn )
(5)
return x1 x2 . . . xn
3.2
Generating k-subsets
If we have a set {1, . . . n} and we want to generate all possible subsets of size k from
that set then we can represent the k-subset S as binary string x1 x2 . . . xn where
xi = 1 if i ∈ S and xi = 0 if i ∈ S.
For example if we have the set {1, 2, 3} the 2-subsets {1, 2}, {1, 3}, {2, 3} can be
coded as the binary strings 110, 101, 011.
There exists a couple of ways to generate these k-subsets:
1. Generate all subsets, that is all elements of the power set and filter out the
ones that are not of size k.
2. Use the binary representation in a string and generate with Next.
3. Generate with a binary version of Next called BinaryNext which uses
binary operations on the bits.
BinaryNext(x1 x2 . . . xn )
(1)
t ← x ∧ (−x)
(2)
r ←x+t
(3)
return ((r ⊕ x)/4t) ∨ r
4
Partitions
Definition 4.1 A set B = { B1 , B2 , . . . , Bk } is a partition of the set [n] = {1, 2,
. . . ,n} if and only if:
1. ∪ki=1 Bi = [n]
2. Bi ∩ Bj = ∅
3. Bi 6= ∅
That is, we divide the set into one or more blocks where each block contains at
least one element and each element belongs to exactly one block.
An example partition of [10]: {{1, 5, 8}, {2, 3, 10}, {4}, {6, 7, 9}}.
Combinatorics
4.1
5
Stirling numbers
How many partitions of [n] in k blocks exists? These numbers are known as Stirling
numbers of the second kind.
Definition 4.2 The Stirling numbers of the second kind, S(n, k), describe the number of possible ways to partition a set of n elements into k blocks.
S(n, 0) = 0 if n > 0
S(n, 1) = 1
S(n, n) = 1
S(n, k) = S(n − 1, k − 1) + kS(n − 1, k)
The base cases are natural: You cannot put n elements into 0 blocks. You can put
n elements in a single block in exactly one way, and you can put n elements in n
blocks in exactly one way (one element in each block).
The last line can be explained as follows. When putting n elements in k blocks, we
can either put n − 1 of the first elements in k − 1 blocks, letting the last element
form its own block. Alternatively, we can put the n − 1 elements in k blocks and
put the last element in one of the first blocks.
4.2
Bell numbers
What is the total number of ways to partition a set? These numbers are known as
Bell numbers.
The Bell numbers can be described using the Stirling numbers as follows:
B(n) =
n
X
S(n, i)
(7)
i=0
Alternatively, they can be specified using this recursion formula:
B(n + 1) =
n X
n
B(i)
n−i
i=0
B(0) = 1
(8)
(9)
The formula can be explained as follows. Consider the block containing the
number n + 1. Suppose that block contains i elements
other than n + 1. Then we
n
can choose the elements in the other blocks in n−i
ways. We can also partition
the rest of the elements in that block in B(i) ways. We can do this for every choice
of i from 0 to n.
5
Catalan numbers
The Catalan numbers can be used for different purposes. Among them is describing
the number of balanced parenthesis expressions for a given number of parenthesis
pairs. We will concentrate on the definition where they are explained as follows.
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Definition 5.1 The Catalan numbers describes the number of ordered binary trees
with n nodes.
A binary tree is a tree where each node has at most two children.
The Catalan numbers can be computed recursively as:
(
1
if n = 0
C(n) = Pn−1
C(k)C(n
−
1
−
k)
otherwise
k=0
(10)
The formula can be explained as follows. Let Ck (n) denote the number of ordered binary trees that has a total of n nodes, with k nodes in its left subtree. If
the left subtree has k nodes, then the right subtree will have n − k − 1 nodes (the
total number of nodes minus the ones in the left subtree and the root node). The
subtrees can be chosen independently, so we will have Ck (n) = C(k)C(n − k − 1)
Pn−1
possible trees. We also know that C(n) = k=0 Ck (n), because the left subtree
can be chosen to have 0 to n − 1 nodes.
However, it is usually easier to use the fact that the Catalan numbers can be
described as:
C(n) =
2n
n
(11)
n+1
The ten first (C(0) − C(9)) Catalan numbers are: 1, 1, 2, 5, 14, 42, 132, 429,
1430, 4862.
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Inclusion-Exclusion
In order to compute the cardinality of a union A1 ∪ A2 , we can first add the sizes of
the two sets and then subtract the number of elements that are contained in both
sets: |A1 | + |A2 | − |A1 ∩ A2 |.
When computing the cardinality of a union A1 ∪ A2 ∪ A3 , we add the sizes of the
individual sets, subtract their pairwise intersection sizes and finally add their total
intersection size: |A1 |+|A2 |+|A3 |−|A1 ∩A2 |−|A1 ∩A3 |−|A2 ∩A3 |+|A1 ∩A2 ∩A3 |.
The general formula for counting the union cardinality of n sets A1 ∪ A2 ... ∪ An is:
X
\
(−1)|I|+1 |
Ai |
(12)
|A1 ∪ A2 ... ∪ An | =
I⊆[n]
i∈I
This formula is only
T useful in programming contexts for small values of n (up
to n ' 20), or when | i∈I Ai | can be computed efficiently.