Download Fibonacci Sequence

Lecture 4,5
Mathematical Induction
and
Fibonacci Sequences
What is Mathematical Induction?
•
•
•
•
Mathematical induction is a powerful, yet straightforward method of proving statements whose
domain is a subset of the set of integers.
Usually, a statement that is proven by induction is
based on the set of natural numbers.
This statement can often be thought of as a function
of a number n, where n = 1, 2, 3,. . .
Proof by induction involves three main steps
–
–
–
Proving the base of induction
Forming the induction hypothesis
Proving that the induction hypothesis holds true for all
numbers in the domain.
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What is Mathematical Induction?
Let P(n) be the predicate defined for any positive
integers n, and let n0 be a fixed integer. Suppose the
following two statements are true
1. P(n0) is true.
2. For any positive integers k, k  n0,
3. if P(k) is true then P(k+1)is true.
If the above statements are true then the statement:
 n  N, such that n  n0, P(n) is also true
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Steps in Proving by Induction
Claim: P(n) is true for all n  Z+, for n  n0
1. Basis
– Show formula is true when n = n0
2. Inductive hypothesis
– Assume formula is true for an arbitrary n = k
where, k  Z+ and k  n0
3. To Prove Claim
– Show that formula is then true for k+1
Note: In fact we have to prove
1) P(n0) and
2) P(k)  P(k+1)
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Proof by Induction
Example 1
• Prove that n2  n + 100
 n  11
Solution
Let P(n)  n2  n + 100
 n  11
1. P(11)  112  11 + 100  121  111, true
2. Suppose predicate is true for n = k, i.e.
P(k)  k2  k + 100, true
k  11
3. Now it can be proved that
P(k+1)  (k+1)2  (k+1) + 100,
 k2 + 2k +1  k +1 + 100  k2 + k  100 (by 1 and 2)
Hence P(k)  P(K+1)
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Validity of Proof
Example 1
• Prove that n2  n + 100
 n  11
Solution
Initially, base case
Solution set = {11}
By, P(k)  P(K+1)  P(11)  P(12), taking k = 11
Solution set = {11, 12}
Similarly, P(12)  P(13), taking k = 12
Solution set = {11, 12, 13}
And, P(13)  P(14), taking k = 13
Solution set = {11, 12, 13, 14}
And so on
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Another Easy Example
Reasoning of Proof
Example 2
Use Mathematical Induction to prove that sum of the
first n odd positive integers is n2.
Proof
n
• Let P(n) denote the proposition that  (2i  1)  n 2
i 1
2
• Basis step : P(1) is true , since 1 = 1
•
Inductive step : Let P(k) is true for a positive integer k,
i.e., 1+3+5+…+(2k-1) = k2
• Note that: 1+3+5+…+(2k-1)+(2k+1) = k2+2k+1= (k+1)2
∴ P(k+1) true, by induction, P(n) is true for all n  Z+
n
n
Another Proof
(2i  1)  2 i  n  n(n  1)  n n 2


i 1
i 1
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Reasoning of Proof
Example 3
• Use mathematical Induction to prove that
n < 2n for all n  Z+
Proof
• Let P(n) be the proposition that n < 2n
• Basis step : P(1) is true since 1 < 21 .
• Inductive step :
Assume that P(n) is true for a positive integer n = k,
i.e., k < 2k.
• Now consider for P(k+1) :
Since, k + 1 < 2k + 1  2k + 2k = 2.2k = 2k + 1
∴ P(k+1) is true.
It proves that P(n) is true for all n  Z+.
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Example 4: Harmonic Numbers
The harmonic numbers Hk, k = 1, 2, 3, …, are
defined by H  1  1  1  ...  1
k
2 3
k
n
H 2n  1 
Use mathematical induction to show that
2
whenever n is a nonnegative integer.
Proof
Let P(n) be the proposition that
H 2n  1  n / 2
Basis step :
P(0) is true, since, H 0  H1  1  1  0 / 2  1
2
Inductive step
Assume that P(k) is true for some k,
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H 2k  1  k / 2
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Example 4: Harmonic Numbers
Now consider
H 2k 1
1 1
1
1
1
1
 1   k  k
 k
   k 1
2 3
2
2 1 2  2
2  2k  2k
1
1
1
 H 2k  k
 k
   k 1
2 1 2  2
2
k
1
1
1
 (1  )  k
 k
   k 1
2 2 1 2  2
2
k
1
1
1
 (1  )  k
 k
 k
k
k
2 2 2
2 2
2  2k
k
2k
k 1
k 1
 (1  )  k
  1
k  1
2 2 2
2 2
2
∴P(k+1) is true.
Hence the statement is true for all n  Z+.
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Fibonacci Sequences
Dr Nazir A. Zafar
Advanced Algorithms Analysis and Design
Today Covered
In this lecture we will cover the following:
•
Fibonacci Problem and its Sequence
•
Construction of Mathematical Model
•
Recursive Algorithms
•
Generalizations of Rabbits Problem and
Constructing its Mathematical Models
•
Applications of Fibonacci Sequences
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Fibonacci Sequence
• By studying Fibonacci numbers and constructing
Fibonacci sequence we can imagine how
mathematics is connected to apparently unrelated
things in this universe.
• Even though these numbers were introduced in
1202 in Fibonacci’s book Liber abaci.
• Fibonacci, who was born Leonardo da Pisa gave a
problem in his book whose solution was the
Fibonacci sequence as we will discuss it today.
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Fibonacci’s Problem
Statement:
• Start with a pair of rabbits, one male and one female,
born on January 1.
• Assume that all months are of equal length and that
rabbits begin to produce two months after their own birth.
• After reaching age of two months, each pair produces
another mixed pair, one male and one female, and then
another mixed pair each month, and no rabbit dies.
How many pairs of rabbits will there be after one year?
Answer: The Fibonacci Sequence!
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, . . .
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Construction of Mathematical Model
F0 = 0
F1 = 1
end of month 1
end of month 2
F2 = 1
end of month 3
F3 = 2
F4 = 3
end of month 4
end of month 5
F5 = 5
end of month 6
F6 = 8
end of month 7
...
...
F7 = 13
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...
end of month 12
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Construction of Mathematical Model
• Total pairs at level k = Total pairs at level k-1 + Total
pairs born at level k
(1)
• Since
Total pairs born at level k = Total pairs at level k-2
(2)
• Hence by equation (1) and (2)
Total pairs at level k = Total pairs at level k-1 + Total
pairs at level k-2
• Now let us denote
Fk = Total pairs at level k
• Now our recursive mathematical model will become
Fk = Fk-1 + Fk-2
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Computing Values using Mathematical Model
Since Fk = Fk-1 + Fk-2
• F2 = F1 + F0= 1 + 0 = 1
• F 3 = F 2 + F 1= 1 + 1 = 2
• F4 = F3 + F2= 2 + 1 = 3
• F5 = F4 + F3= 3 + 2 = 5
• F6 = F5 + F4= 5 + 3 = 8
• F7 = F6 + F5= 8 + 5 = 13
• F8 = F7 + F6= 13 + 8 = 21
• F9 = F8 + F7= 21 + 13 = 34
• F10 = F9 + F8= 34 + 21 = 55
• F11 = F10 + F9= 55 + 34 = 89
• F12 = F11 + F10= 89 + 55 = 144 . . .
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F0 = 0, F1= 1
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Explicit Formula Computing Fibonacci Numbers
Theorem:
The fibonacci sequence F0,F1, F2,…. Satisfies
the recurrence relation
Fk  Fk 1  Fk 2
k  2
with initial condition F0  F1  1
Find the explicit formula for this sequence.
Solution:
The given fibonacci sequence
Fk  Fk 1  Fk  2
Let tk is solution to this, then characteristic equation
t 2  t 1  0
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Fibonacci Sequence
t1
1
1 4
2
1
5
1
5

, t2 
2
2
t 
For some real C and D fibonacci sequence satisfies the relation
n
1 5 
1 5 
  D

Fn  C 



 2 
 2 
n0
0
1 5 
1 5 




F0  C 

D

 2 
 2 


 F0  C  D
n
n0
0
 C  D  0 1  F0  0
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Fibonacci Sequence
Now
n 1
1 5 
1 5 
  D

F1  C 



2
2





1 5
1 5
C 
D 1
2
2
2
 F1  1
Solving 1 and 2  simultaneo usly we get
1
1
C
,D  
5
5
Hence
n
 1  1  5   1  1  5 
Fn  




 5  2   5  2 5 
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Dr Nazir A. Zafar
n
20
Advanced Algorithms Analysis and Design
Fibonacci Sequence
After simplifying we get
n
1 1 5 
1 1 5 

 


Fn 
5  2 
5  2 
n
which is called the explicit formula for the
Fibonacci sequence recurrence relation.
1 5 
1 5 


 and   
 then
Let   

 2 
2




1 n
1 n
Fn 
 

5
5
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Verification of the Explicit Formula
Example: Compute F3
1 5 
1 5 
1 n 1 n


 and   
 then
Since Fn 
 
 where   

 2 
5
5
 2 


3
3




1 1 5
1 1 5




F3 




5 2 
5  2 
1  1  3.12. 5  3.1.5  5 5  1  1  3.12. 5  3.1.5  5 5 




Now, F3 


8
8
5 
5 





1
1
F3 
1  3.1. 5  3.1.5  5 5 
1  3.1. 5  3.1.5  5 5
8. 5
8. 5



1
F3 
1  3.1. 5  3.1.5  5 5  1  3.1. 5  3.1.5  5 5  2
8. 5
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Recursive Algorithm Computing Fibonacci Numbers
Fibo-R(n)
if n = 0
then 0
Terminating conditions
if n = 1
then 1
else Fibo-R(n-1) + Fibo-R(n-2) Recursive calls
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Running Time of Recursive Fibonacci Algorithm
• Least Cost: To find an asymptotic bound of computational
cost of this algorithm.
if n  2
1
T ( n)  
T (n  1)  T (n  2) n  2
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Drawback in Recursive Algorithms
Recursion Tree
F(n)
F(n-1)
F(n-2)
F(0)
F(n-2)
F(n-3)
F(1)
F(n-3)
F(1)
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F(n-4)
F(0)
25
Generalization of Rabbits Problem
Statement:
• Start with a pair of rabbits, one male and one female,
born on January 1.
• Assume that all months are of equal length and that
rabbits begin to produce two months after their own birth.
• After reaching age of two months, each pair produces
two other mixed pairs, two male and two female, and
then two other mixed pair each month, and no rabbit dies.
How many pairs of rabbits will there be after one year?
Answer: Generalization of Fibonacci Sequence!
0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, . . .
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Construction of Mathematical Model
F0 = 0
F1 = 1
F2 = 1
F3 = 3
F4 = 5
F5 = 11
F6 = 21
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Construction of Mathematical Model
• Total pairs at level k =
Total pairs at level k-1 + Total pairs born at level k
(1)
• Since
Total pairs born at level k =
2 x Total pairs at level k-2
(2)
• By (1) and (2), Total pairs at level k =
Total pairs at level k-1 + 2 x Total pairs at level k-2
• Now let us denote
Fk = Total pairs at level k
• Our recursive mathematical model:
Fk = Fk-1 + 2.Fk-2
• General Model (m pairs production): Fk = Fk-1 + m.Fk-2
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Generalization
• Recursive mathematical model
(one pair production)
Fk = Fk-1 + Fk-2
• Recursive mathematical model
(two pairs production)
Fk = Fk-1 + 2.Fk-2
• Recursive mathematical model
(m pairs production)
Fk = Fk-1 + m.Fk-2
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Computing Values using Mathematical Model
Since Fk = Fk-1 + 2.Fk-2
F0 = 0, F1 = 1
• F2 = F1 + 2.F0= 1 + 0 = 1
• F3 = F2 + 2.F1= 1 + 2 = 3
• F4 = F3 + 2.F2= 3 + 2 = 5
• F5 = F4 + 2.F3= 5 + 6 = 11
• F6 = F5 + F4= 11 + 10 = 21
• F7 = F6 + F5= 21 + 22 = 43
• F8 = F7 + F6= 43 + 42 = 85
• F9 = F8 + F7= 85 + 86 = 171
• F10 = F9 + F8= 171 + 170 = 341
• F11 = F10 + F9= 341 + 342 = 683
• F12 = F11 + F10= 683 + 682 = 1365 . . .
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Another Generalization of Rabbits Problem
Statement:
• Start with a different kind of pair of rabbits, one male and
one female, born on January 1.
• Assume all months are of equal length and that rabbits
begin to produce three months after their own birth.
• After reaching age of three months, each pair produces
another mixed pairs, one male and other female, and then
another mixed pair each month, and no rabbit dies.
How many pairs of rabbits will there be after one year?
Answer: Generalization of Fibonacci Sequence!
0, 1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, . . .
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Construction of Mathematical Model
F0 = 0
F1 = 1
F2 = 1
F3 = 1
F4 = 2
F5 = 3
F6 = 4
F7 = 6
F8 = 9
F9 = 13
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F10 = 19
32
Construction of Mathematical Model
• Total pairs at level k =
Total pairs at level k-1 + Total pairs born at level k
• Since
Total pairs born at level k = Total pairs at level k-3
• By (1) and (2)
Total pairs at level k =
Total pairs at level k-1 + Total pairs at level k-3
• Now let us denote
Fk = Total pairs at level k
• This time mathematical model:
Fk = Fk-1 + Fk-3
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(1)
(2)
33
Computing Values using Mathematical Model
Since Fk = Fk-1 + Fk-3
• F3 = F2 + F0= 1 + 0 = 1
• F 4 = F 3 + F 1= 1 + 1 = 2
• F5 = F4 + F2= 2 + 1 = 3
• F6 = F5 + F3= 3 + 1 = 4
• F7 = F6 + F4= 4 + 2 = 6
• F8 = F7 + F5= 6 + 3 = 9
• F9 = F8 + F6= 9 + 4 = 13
• F10 = F9 + F7= 13 + 6 = 19
• F11 = F10 + F8= 19 + 9 = 28
• F12 = F11 + F9= 28 + 13 = 41 . . .
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F0 = 0, F1= F2= 1
34
More Generalization
• Recursive mathematical model
(one pair, production after three months)
Fk = Fk-1 + Fk-3
• Recursive mathematical model
(two pairs, production after three months)
Fk = Fk-1 + 2.Fk-3
• Recursive mathematical model
(m pairs, production after three months)
Fk = Fk-1 + m.Fk-3
• Recursive mathematical model
(m pairs, production after n months)
Fk = Fk-1 + m.Fk-n
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Applications of Fibonacci Sequences
Fibonacci sequences
• Are used in trend analysis
• By some pseudorandom number generators
• Many plants show the Fibonacci numbers in the
arrangements of the leaves around the stems.
• Seen in arrangement of seeds on flower heads
• Consecutive Fibonacci numbers give worst case
behavior when used as inputs in Euclid’s algorithm.
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