Module II Partition and Generating Function (2).ppt

Course Title : Discrete Mathematics for IT
Course Code : CSIT 206
Module 2
Topic : Relation and function
Prepared by
Prof. (Dr.) Sanjay K Singh
Amity Institute of Information Technology
Amity University Uttar Pradesh, Lucknow Campus

2
Properties of Binary Relations, Closure of relations, Warshall’s algorithm,
Equivalence relations and partitions, Partial ordering relations. Functions,
Composition of functions, Invertible functions, Pigeonhole Principle, Discrete
numeric functions and Generating functions. Recurrence relation, Linear
Recurrence Relations with constant Coefficients.
Module II: Relations and Functions

Partition
10
A partition of a positive integer n is a multiset of positive integers that sum to n. We
denote the number of partitions of n by Pn.
Typically a partition is written as a sum, not explicitly as a multiset. Using the usual
convention that an empty sum is 0, we say that P0=1.

Generating Functions I
+ +
( ) +
( ) = ?
X1 X2
+ + X3

Problem
Ula is allowed to choose two
items from a tray containing
an apple, an orange, a pear, a
banana, and a plum. In how
many ways can she choose?
5
2

There is a
correspondence between
paths in a choice tree and
the cross terms of the
product of polynomials!

Counting with Generating Functions
x)
x)(1
x)(1
x)(1
x)(1
(1
x)
(1 5







(0 apple + 1 apple)(0 orange + 1 orange)
(0 pear + 1 pear)(0 banana + 1 banana)
(0 plum + 1 plum)
In this notation, apple2 stand for choosing
2 apples., and + stands for an exclusive or.
Take the coefficient of x2.
(1+x)5=1+5x+10x2+10x3+5x4+x5

Problem
Ula is allowed to choose two
items from a tray containing
TWO apples, an orange, a
pear, a banana, and a plum. In
how many ways can she
choose?
The two apples are identical.

x)
x)(1
x)(1
x)(1
)(1
x
x
(1 2






(0 apple + 1 apple + + 2 apple)
(0 orange + 1 orange)(0 pear + 1 pear)
(0 banana + 1 banana)(0 plum + 1 plum)
Take the coefficient of x2.

The function f(x) that has a
polynomial expansion
f(x)=a0 + a1 x + …+ an xn
is the generating function
for the sequence
a0, a1, …, an

If the polynomial (1+x+x2)4
is the generating polynomial
for ak, what is the
combinatorial meaning of ak?
The number of ways to select k
object from 4 types with at most
2 of each type.

Problem
Find the number of ways to
select R balls from a pile of
2 red, 2 green and 2 blue
balls
Coefficient by xR in
(1+x+x2)3

3
2
1
3
e
2
e
1
e e
e
e
X
X
X
X



Coefficient by xR in (1+x+x2)3
(1+x+x2)3 =
(x0+x1+x2)(x0+x1+x2)(x0+x1+x2)

Find the number of ways to select R
balls from a pile of 2 red, 2 green
and 2 blue balls
e1 + e2 + e3 = R
0 b ek b 2
3
e
2
e
1
e
X
X
X

Exercise
Find the number of integer
solutions to
x1+x2+x3+x4=21
0bxkb7
Take the coefficient by x21 in
(1+x+x2+x3+x4+x5+x6+x7)4

Exercise
Find the number of integer
solutions to
x1+x2+x3+x4 = 21
0 < xk< 7

x1+x2+x3+x4 = 21
0 < xk< 7
Observe
0 < xk < 7
-1 < xk-1 < 6
0 b xk-1 b 5
Than
(x1-1)+(x2-1)+(x3-1)+(x4-1) = 21-4

x1+x2+x3+x4=21
0<xk<7
The problem is reduced to solving
y1 + y2 + y3 + y4 = 17, 0bykb5
Solution: take the coefficient by x17 in
(1+x+x2+x3+x4+x5)4
which is 20

Problem
x1+x2+x3=9
1bx1b2
2bx2b4
1bx3b4

Solution
x1+x2+x3=9
1bx1b2 (x+x2)
2bx2b4 (x2+x3+x4)
1bx3b4 (x+x2+x3+x4)
Take the coefficient by x9 in the
product of generating functions.

Two observations
1. A generating function approach is
designed to model a selection of all
possible numbers of objects.
2. It can be used not only for counting
but for solving the linear Diophantine
equations

The 251-staff Danish party
Adam pulls a few strings…
and a large apple Danish
factory is built next to the
CMU.

The 251-staff Danish party.
Unfortunately, still only 3
cheese and 4 raspberry
pastries are available.
Raspberry pastries come in
multiples of two. What the
number of possible orders?

(1+x+x2+x3) (1+x2+x4) (1+x+x2+x3+ …)
cheese raspberry apple
There is a problem (…) with the above
expression.

(1+x+x2+x3) (1+x2+x4) (1+x+x2+x3+ …)=
cheese raspberry apple
(1+x+2x2+2x3+2x4+2x5+x6+x7) (1+x+x2+x3+ …)
=1+2x+4x2+6x3+8x4+…

The power series
a0 + a1 x + a2 x2 +…
is the generating function
for the infinite sequence
a0, a1, a2, …,


0
k
k
k x
a

A famous generating
function from calculus:
0 !
k
x
k
x
e
k


 

We can use the generating
function technique to
solve almost all the
mathematical problems you
have met in your life so far.

Exercise
Count the number of N-letter
combinations of MATH in
which M and A appear with
repetitions and T and H only
once.
Coefficient by xN in
(1+x+x2+…)2(1+x)2

What is the coefficient of Xk
in the expansion of:
( 1 + X + X2 + X3 + X4 + . . . . )n ?
A solution to:
e1 + e2 + . . . + en = k
ek ≥ 0

What is the coefficient of Xk
in the expansion of:
( 1 + X + X2 + X3 + X4 + . . . . )n







 

k
1
k
n











 





0
k
k
n
2
x
k
1
k
n
...)
x
x
(1
GF for the pirates and gold problem
x
1
1
...
x
x
1 2





But what is the LHS?

...
x
x
1
y
x
-
1
1
1
x)
-
y(1
1
y
x
-
y
...
x
x
y
x
...
x
x
1
y
2
2
2

























 


0
k
k
x
k
1
k
n
n
x)
-
(1
1
GF for the pirates and gold problem

1
-
x
x
x
...
x
x
1
n
1
-
n
2 1






x
-
1
x
x
...
x
x
1 n
1
-
n
2 1
... 






(when x ≠ 1)
(when |x| < 1)

We will be dealing with formal
power series (aka
generating functions)


0
k
k
kx
a

The coefficients ak can
be any complex numbers,
though we will be mainly
working over the
integers.


0
k
k
kx
a

Power series have no analytic
interpretation.


0
k
k
kx
a

FORMAL MANIPULATION:
We deal with power series as we
deal we numbers!

1/P(X) is defined to be the
polynomial Q(X) such that
Q(X) P(X) = 1.
Th: P(X) will have a reciprocal
if and only if a0  0. In this
case, the reciprocal is unique.

Power series make a
commutative ring!

The derivative of
is defined by:

Power Series = Generating Function
Given a sequence of integers a0, a1, …, an
We will associate with it a function




0
k
k
kx
a
f(x)
The On-Line Encyclopedia of Integer Sequences

Generating Functions
Sequence: 1, 1, 1, 1, …
Generating function:
x
1
1
x
x
a
f(x)
0
k
k
0
k
k
k



 






Find a generating function f(x)
1, 2, 3, 4, 5, …
...
4x
3x
2x
1
f(x) 3
2





2
x)
(1
1
f(x)







0
k
k
kx
F
f(x)
F0=0, F1=1,
Fn=Fn-1+Fn-2 for n≥2

= x




0
k
k
kx
F
f(x)
(x+x2+2 x3+3 x4+5 x5+8 x6+13 x7+…)(1-x-x2)
= x + x2 + 2 x3 + 3 x4 + 5 x5 + 8 x6 + 13 x7+…
- x2 - x3 - 2 x4 - 3 x5 - 5 x6 - 8 x7-…
- x3 - x4 - 2 x5 - 3 x6 - 5 x7-…
2
0
k
k
k
x
x
1
x
x
F
f(x)



 



If a sequence satisfies a
linear recurrence relation,
we can always find its
generating function.

Use generating
functions to show that
Proving Identities with GF


















 n
n
k
n
n
k
2
0
2

Prove by induction
Prove algebraically


















 n
n
k
n
n
k
2
0
2
Prove combinatorially
Prove by Manhattan walk (see previous lecture)
15-355
Computer
Algebra

•
Consider the right hand side:
n
n
x
x
n
n 2
)
1
](
[
2




























 n
n
k
n
n
k
2
0
2
Coefficient by xn

Compute a coefficient of xk
Use the binomial theorem to obtain


















 n
n
k
n
n
k
2
0
2
2
2
)
...
1
0
(
)
1
( n
n
x
n
n
x
n
n
x 





























Find a closed form
??












n
0
k k
n
k
Computing Combinatorial Sums

Don't try
to evaluate the sum that you're
looking at. Instead, find the
generating
function for it, then read off
the coefficients.
Computing Combinatorial Sums

??












n
0
k k
n
k
Multiply it by xn and sum it over n
 

 









0
n
n
0
k
n
k
n
k
x
Interchange the sums

Take r = n - k as the new dummy variable of
inner summation
We recognize the inner sum as xk (1 + x)k

 





 


















 0
k k
n
n
0
n
n
0
k
n
k
n
k
x
k
n
k
x



 






















 0
k 0
r
k
r
0
k k
n
n
r
k
x
k
n
k
x
k

This is a geometric series




0
k
k
k
x)
(1
x
2
0
k
k
k
x
x
1
1
x)
x(1
1
1
x)
(1
x










The RHS is our old friend …

What did we find?
2
0
n
n
0
k
n
x
x
1
1
k
n
k
x












 

 
2
0
k
k
k
x
x
1
x
x
F






2
0
n
n
1
n
x
x
1
1
x
F








What did we find?

 




 









 0
n
n
1
n
0
n
n
0
k
n
x
F
k
n
k
x
1
n
n
0
k
F
k
n
k














It’s one of the
most important
mathematical
ideas of all time!

Study Bee
• Counting with generating
functions
• Solving the Diophantine
equations
• Proving combinatorial
identities

Module II Partition and Generating Function (2).ppt

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