Algorithms Lecture 3: Analysis of Algorithms II

Analysis and Design of Algorithms
Analysis of Algorithms II

Maximum Pairwise Product
Fibonacci
Greatest Common Divisors

Maximum Pairwise Product

Given a sequence of non-negative integers a0,…,an−1, find
the maximum pairwise product, that is, the largest integer
that can be obtained by multiplying two different elements
from the sequence (or, more formally, max aiaj where
0≤i≠j≤n−1). Different elements here mean ai and aj with
i≠j (it can be the case that ai=aj).

Constraints 2≤n≤2 * 105 & 0≤a0,…,an−1≤105.

 Sample 1
 Input: 1 2 3
 Output:6
 Sample 2
 Input: 7 5 14 2 8 8 10 1 2 3
 Output:140

 Sample 3
 Input: 4 6 2 6 1
 Output:36

 Assume the following array
2 4 3 5 1

2 4 3 5 1
i j
Result=0

2 4 3 5 1
i j
If a[i]*a[j] > result
result=a[i]*a[j]=8

2 4 3 5 1
i j
result=8

2 4 3 5 1
i j
result= a[i]*a[j] =10

 Naive algorithm

 Python Code:

 Time complexity O(n2)

we need a faster algorithm. This is because our program
performs about n2 steps on a sequence of length n. For the
maximal possible value n=200,000 = 2*105, the number
of steps is 40,000,000,000 = 4*1010. This is too much.
Recall that modern machines can perform roughly 109
basic operations per second

 Find maximum number1
2 4 3 5 1
max1

 Find maximum number2 but not maximum number1
2 4 3 5 1
max2 max1

 Find maximum number2 but not maximum number1
2 4 3 5 1
max2 max1
return max1*max2

 Efficient
algorithm

 Efficient algorithm

 Time complexity O(n)

Fibonacci

0,1,1,2,3,5,8,13,21,34, …

Definition:
𝐹𝑛 =
0 𝑛 = 0
1 𝑛 = 1
𝐹𝑛−2 + 𝐹𝑛−1 𝑛 > 1

Examples:
𝐹8 = 21
𝐹20 = 6765
𝐹50 = 12586269025
𝐹100 = 354224848179261915075

 Examples:
𝐹500 =
1394232245616978801397243828
7040728395007025658769730726
4108962948325571622863290691
557658876222521294125

 Naive algorithm
Very Long Time why????

F6
F5
F4
F3
F2
F1
F0
F0
F1
F0
F2
F1
F0
F0
F3
F2
F1
F0
F0
F1
F0
F4
F3
F2
F1
F0
F0
F1
F0
F2
F1
F0
F0

Fib algorithm is very slow because of
recursion
Time complexity = O(2n)

0 1 1 2 3 5
Create array then insert fibonacci

short Time why????

Fib_Fast algorithm is fast because of
loop + array
Time complexity = O(n2)

 Try
Very long Time why????

Advanced algorithm
No array
Need two variable + Loop

 Advanced algorithm
 Compute F6
 a=0, b=1
0 1
a b

 Compute F6
 a=b, b=a+b
1 1
a b

 Compute F6
 a=b, b=a+b
1 2
a b

 Compute F6
 a=b, b=a+b
2 3
a b

 Compute F6
 a=b, b=a+b
3 5
a b

 Compute F6
 a=b, b=a+b
5 8
a b

 Compute F6=8
5 8
a b

Very short Time why????

Fib_Faster algorithm is faster because
of loop + two variables
Time complexity = O(n)

 Try
Short Time why????

Greatest Common Divisors

In mathematics, the greatest common divisor
(gcd) of two or more integers, which are not all
zero, is the largest positive integer that divides
each of the integers.

Input Integers a,b >=0
Output gcd(a,b)

 What is the greatest common divisor of 54 and 24?
 The divisors of 54 are: 1,2,3,6,9,18,27,54
 Similarly, the divisors of 24 are: 1,2,3,4,6,8,12,24
 The numbers that these two lists share in common are the common
divisors of 54 and 24: 1,2,3,6
 The greatest of these is 6. That is, the greatest common divisor of 54
and 24. gcd(54,24)=6

gcd algorithm is slow because of loop
Time complexity = O(n)
n depend on a,b

gcd_fast((3918848, 1653264))
gcd_fast((1653264, 612320))
gcd_fast((612320, 428624))
gcd_fast((428624, 183696))
gcd_fast((183696, 61232))
return 61232

Efficient algorithm
Take 5 steps to solve gcd_fast( ( 3918848, 1653264 ) )
Time complexity = O(log(n))
 n depend on a,b

Naive algorithm is too slow.
The Efficient algorithm is much better.
Finding the correct algorithm requires knowing
something interesting about the problem.

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Algorithms Lecture 3: Analysis of Algorithms II

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Algorithms Lecture 3: Analysis of Algorithms II