This document discusses using recurrence relations to model problems involving counting techniques. It provides examples of modeling problems related to bacteria population growth, rabbit population growth, the Tower of Hanoi puzzle, and valid codeword enumeration. For each problem, it defines the recurrence relation and initial conditions, derives a closed-form solution, and proves its correctness using mathematical induction. Recurrence relations provide a way to define sequences and solve problems recursively by relating terms to previous terms in the sequence.

1. MODELING WITH
RECURRENCE RELATIONS

2. CONTENTS
 COUNTING TECHNIQUES
 MOTIVATION
 RECURSIVE DEFINITIONS
 RECURRENCE RELATIONS
 MODELING WITH RECURRENCE RELATIONS
(PROBLEMS AND SOLUTIONS)

3. – Product rule
– Sum rule
– The Inclusion-Exclusion Principle
– The Pigeonhole Principle
– Permutations
– Combinations
A1A2  A1  A2  A1A2
Counting technique
Number of ways = n1n2 ··· nm
Number of ways = n1 + n2 +…+ nm
If N objects are placed into k boxes,
then there is at least N one k
box
containing at least objects.
      
!
 
, 1 2 1
!
n
P n r n n n n r
n r
     

 
!
n n
   
 
,
! !
C n r
 
r r n r
  

4. MOTIVATION
The number of bacteria in a colony doubles every hour. If the colony
begins with 5 bacteria, how many will be present in n hours?
Solution:
Let an : the number of bacteria at the end of n hours
Then;
an = 2an-1, a0=5
Since bacteria double every hour Initial condition
What we are doing actually?
By using
recursive
definition
We find a formula/model for an (the relationship between an and a0)
– this is called recurrence relations between the term of sequence.

5. RECURSIVE DEFINITIONS
 Recursion – a process to define an object explicitly
 The object can be a sequence, function or set (for BCT2078 purposes)
 The ideas – construct new elements from known elements.
 Process – specify some initial elements in a basis step and provide a rule for constructing new
elements from those we already have in the recursive step.
 Mathematical Induction can be used to prove the result.
The given
complex
problem
This
simple
version
can be
solved
This simple
version
can be
solved
Can be
solved if
Can be
solved if
Can be
solved if
an = 2an-1 an-1 = 2an-2, a1 = 2a0, a0=5

6. RECURRENCE RELATION
A recurrence relation for the sequence {an} is an equation that expresses an in terms of one or
more of the previous terms of the sequence, namely, a0, a1,…, an-1, for all integers n with n ≥ n0,
where n0 is a nonnegative integer.
A sequence is call solution of a recurrence relation if its term satisfy the recurrence relation.
an = 2an-1
Recurrence relation
a0 = 5 Initial condition
Recursive
definition
 A recursive algorithm provides the solution of a problem of size n in term of the solutions of one
or more instances of the same problem in smaller size.
 The initial condition specify the terms that precede the first term where the recurrence relation
takes effect.

7. Modeling with Recurrence Relation
 We can use recurrence relation to model a wide variety of problems
such as
 Finding a compound interest
 Counting a population of rabbits
 Determining the number of moves in the Tower of Hanoi puzzle
 Counting bit strings

8. RABBITS
A young pair of rabbits, one of each sex, is placed on an
island.
A pair of rabbits does not breed until they are 2 months old.
After they are 2 months old, each pair of rabbits produces
another pair each month.
Assume that none of the rabbits die.
How many rabbits are there after n months?

9. RABBITS
 Let 풇풏 denote the number of pairs of rabbits after n
months.
 풇ퟏ = 1 { reproducing pairs = 0, young pairs = 1 }
 풇ퟐ = 1 { reproducing pairs = 0, young pairs = 1 }
 풇ퟑ = 2 { reproducing pairs = 1, young pairs = 1 }
 풇ퟒ = 3 { reproducing pairs = 1, young pairs = 2 }
 풇ퟓ = 5 { reproducing pairs = 2, young pairs = 3 }
 풇ퟔ= 8 { reproducing pairs = 3, young pairs = 5 }

10. RABBITS
The rabbit population can be modeled by a recurrence relation.
At the end of the first month, the number of rabbits on the island is 풇ퟏ = 1.
At the end of the second month, the number of rabbits on the island is 풇ퟐ = 1.
The number of pairs of rabbits after n months 풇풏 is equal to the number of
pairs of rabbits from the previous month 풇풏−ퟏ plus the number of pairs of
newborn rabbits, which equals 풇풏−ퟐ, since each newborn pair comes from a
pair that is at least two months old, so
풇풏 = 풇풏−ퟏ + 풇풏−ퟐ for n >= 3.

11. SOLVING RABBITS PROBLEM
The recurrence system
Initial conditions: 풇ퟏ = 1, 풇ퟐ = 1
Recurrence relation: 풇풏 = 풇풏−ퟏ + 풇풏−ퟐ for n >= 3. (Linear Homogeneous RR
with const coeff.)
For small values of n, we get
(풇ퟏ,, 풇ퟐ , 풇ퟑ,...) = (1,1,2,3,5,8,13,...)
Characteristic equation : 풓ퟐ- 풓 -1 =0
Characteristic roots: 풓ퟏ= (1+√5)/2 and 풓ퟐ = (1- √5)/2

12. SOLVING RABBITS PROBLEM
Therefore, 풇풏 = 푨ퟏ∗((1+√5)/2)
풏
+ 푨ퟐ∗((1−√5)/2)
풏
, 푨ퟏ and 푨ퟐ are
constants
푨ퟏ = √5/5 ; 푨ퟐ = −√5/5
풏
+ 푨ퟐ∗((1−√5)/2)
풇풏 = 푨ퟏ∗((1+√5)/2)
풏
=
ퟏ
√ퟓ
∗(1+√5)/2)
풏
-
ퟏ
√ퟓ
∗(1−√5)/2)
풏

13. TOWER OF HANOI
Initially, n discs are placed on the first peg. Move the n discs one at a time
from one peg to another such that no larger disc is ever placed on a
smaller disc.
Goal: Move the discs from peg 1 to peg 2.

14. TOWER OF HANOI
Let 퐻푛 denote the number of moves needed to solve the tower of Hanoi
problem with n discs.
1) Move the top n-1 discs from peg 1 to peg 3 using Hn-1 moves.

15. TOWER OF HANOI
Let 퐻푛 denote the number of moves needed to solve the tower of Hanoi
problem with n discs.
1) Move the top n-1 discs from peg 1 to peg 3 using Hn-1 moves.
2) Move the largest disc from peg 1 to peg 2.

16. TOWER OF HANOI
Let 퐻푛 denote the number of moves needed to solve the tower of Hanoi problem with n
discs.
1) Move the top n-1 discs from peg 1 to peg 3 using Hn-1 moves.
2) Move the largest disc from peg 1 to peg 2.
3) Move the n-1 discs from peg 3 to peg 2 using Hn-1 moves.

17. TOWER OF HANOI
Let 퐻푛 denote the number of moves needed to solve the tower of Hanoi problem with n
discs.
1) Move the top n-1 discs from peg 1 to peg 3 using 퐻푛−1 moves.
2) Move the largest disc from peg 1 to peg 2.
3) Move the n-1 discs from peg 3 to peg 2 using 퐻푛−1 moves.

18. TOWER OF HANOI
Let 퐻푛 denote the number of moves needed to solve the tower of Hanoi
problem with n discs. Thus the no of moves is given by
Where:
1 1 2 1 ; 1 n n H H H    
Hn-1 moves
Moves the largest
disk to peg 2
transfer the top of
n-1 disks to peg 2
transfer the top of
n-1 disks to peg 3
+ +
Hn-1 moves 1 move

19. SOLVING TOWER OF HANOI PROBLEM
Initial condition: 퐻1 = 1
Recurrence relation: 퐻푛 = 2퐻푛−1+1 for n >= 2.
For small values of n, we get
(퐻1퐻2퐻3퐻4,...) = (1,3,7,15,...)
Therefore, we can guess that 퐻푛 = ퟐ풏 - 1

20. SOLVING TOWER OF HANOI PROBLEM
How can we prove that 푯풏 = ퟐ풏-1 holds for all n>=1?
By induction.
Base Step: 푯ퟏ = 1 = ퟐퟏ − ퟏ, so our claim holds for n=1.
Inductive Step: Suppose that 퐻푛 = ퟐ풏 - 1 holds for some n>=1.
It follows that
푯풏+ퟏ = 2퐻푛 + 1 = 2(ퟐ풏- 1 ) + 1 = ퟐ풏+ퟏ-2+1= ퟐ풏+ퟏ-1
Therefore, the claim follows by induction on n.

21. CODEWORD ENUMERATION
A computer system considers a string of decimal digits a valid
codeword if it contains an even number of 0 digits.
For instance, 1230407869 is valid, whereas 120987045608 is not valid.
Let 풂풏 be the number of valid n-digit codewords.

22. CODEWORD ENUMERATION
Note that 풂ퟏ = 9
A recurrence relation can be derived for this sequence by
considering how a valid n-digit string can be obtained from
strings of n-1 digits.
Two ways to form a valid string with n digits from a string with
one fewer digit.

23. CODEWORD ENUMERATION
First, a valid string of n digits can be obtained by appending a valid
string of n − 1 digits with a digit other than 0. This appending can be
done in nine ways. Hence, a valid string
with n digits can be formed in this manner in 9푎푛−1 ways

24. CODEWORD ENUMERATION
Second, a valid string of n digits can be obtained by appending a 0 to a string of
length n − 1 that is not valid. The number of ways that this can be done equals
the number of invalid (n − 1)-digit strings. Because there are ퟏퟎ풏−ퟏ strings of length
n − 1, and 풂풏−ퟏ are valid, there are ퟏퟎ풏−ퟏ − 풂풏−ퟏ valid n-digit strings obtained by
appending an invalid string of length n − 1 with a 0.
Since all valid strings of length n are produced in one of these two ways, it follows
that there are
풂풏 = 9풂풏−ퟏ + (ퟏퟎ풏−ퟏ − 풂풏−ퟏ)
= 8풂풏−ퟏ + ퟏퟎ풏−ퟏ
valid strings of length n.

25. SOLVING CODEWORD ENUMERATION
Initial condition: 푎1 = 1
Recurrence relation: 푎푛 = 8푎푛−1 + 10푛−1 for n >= 2. (Non-homogeneous RR
with const coeff.)
=> 푎- 8푎푛−1
푛 푛−1 = 10We need to solve 풂풂(풉) 풏 = 풏
+ 풂(풑)
풏
Characteristic equation: r - 8 = 0
Characteristic root: 푟1 = 8
β = 10

26. SOLVING CODEWORD ENUMERATION
(풉) = 푨ퟏ∗ ퟖ풏 , where 푨ퟏ is constant
Therefore, 풂풏
(풑) = 풓ퟎ ∗(푷ퟏ ∗ ퟏퟎ풏−ퟏ) , where 푷ퟏ is a constant
And, 풂풏
=> 푷(ퟏퟎ풏−ퟏ − ퟖ ∗ ퟏퟎ풏−ퟐ풏−ퟏ ퟏ) = ퟏퟎ, since 푎푛 - 8푎= 10푛−1
푛−1 => 푷ퟏ = 5
=> 풂(풑) 풏
= 5∗ ퟏퟎ풏−ퟏ
(풉) + 풂풏
Therefore, 풂풏 = 풂풏
(풑) = (푨ퟏ∗ ퟖ풏) + (5 ∗ ퟏퟎ풏−ퟏ)
 => 푨ퟏ = ½ (풂ퟏ = 9)

27. SOLVING CODEWORD ENUMERATION
Therefore,
(풉) + 풂풏
풂풏 = 풂풏
(풑)
= (푨ퟏ∗ ퟖ풏) + (5 ∗ ퟏퟎ풏−ퟏ)
= (1/2 ∗ ퟖ풏) + (5 ∗ ퟏퟎ풏−ퟏ)
=
ퟏ
ퟐ
(ퟖ풏 + ퟏퟎ풏)

28. THAT’S ALL : THANK YOU