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Application:
Algorithms
Lecture 19
Section 3.8
Tue, Feb 20, 2007
Algorithms

An algorithm is a step-by-step procedure
that is guaranteed to stop after a finite
number of steps for all legitimate inputs.
Example: Max Algorithm
Describe an algorithm for finding the
maximum value in a list of numbers.
 Special cases:

What if the list is empty?
 Others?

An Algorithmic Language
An algorithmic language is much like a
computer language.
 Its primary purpose is to express
unambiguously the steps of an algorithm,
including all decisions and special cases
that may arise.

Assignment Statements

An assignment statement is of the form
x := expr
where x is a variable and expr is an
expression.
 Examples
found := true
 count := count + 1

Conditional Statements

One-way conditional statements:
if condition then
sequence of statements

Two-way conditional statements:
if condition then
sequence of statements
else
sequence of statements
Iterative Statements

While loops:
while condition
sequence of statements
end while

For loops:
for var := init expr to final expr
sequence of statements
next var
For Loops

A for loop can always be rewritten as a
while loop.
var := init expr
while var  final expr
sequence of statements
var := var + 1
end while
A Notation for Algorithms

An algorithm will be organized into three
parts:
Input – List all input variables and any
special assumptions about them.
 Algorithm body – Describe the step-by-step
procedure.
 Output – List the output variables.

Example: Finding the Max

Input: a, n
a is a list of numbers.
 n is the size of the list, n  1.


Algorithm body:
max := a[1]
for i := 2 to n
if a[i] > max then
max = a[i]

Output: max
Tracing an Algorithm

Write a trace table for the Max algorithm
and the input a = {7, 3, 8, 6, 4}, n = 5.
Iteration
a
n
max
i
0
{7, 3, 8, 6, 4}
5
7
2
1
2
3
4
Tracing an Algorithm

Write a trace table for the Max algorithm
and the input a = {7, 3, 8, 6, 4}, n = 5.
Iteration
a
n
max
i
0
{7, 3, 8, 6, 4}
5
7
2
7
3
1
2
3
4
Tracing an Algorithm

Write a trace table for the Max algorithm
and the input a = {7, 3, 8, 6, 4}, n = 5.
Iteration
a
n
max
i
0
{7, 3, 8, 6, 4}
5
7
2
1
7
3
2
8
4
3
4
Tracing an Algorithm

Write a trace table for the Max algorithm
and the input a = {7, 3, 8, 6, 4}, n = 5.
Iteration
a
n
max
i
0
{7, 3, 8, 6, 4}
5
7
2
1
7
3
2
8
4
3
8
5
4
Tracing an Algorithm

Write a trace table for the Max algorithm
and the input a = {7, 3, 8, 6, 4}, n = 5.
Iteration
a
n
max
i
0
{7, 3, 8, 6, 4}
5
7
2
1
7
3
2
8
4
3
8
5
4
8
6
Greatest Common Divisors
Let a and b be integers that are not both 0.
 The greatest common divisor of a and b,
denoted gcd(a, b), is the unique positive
integer d with the following properties:

d | a and d | b.
 For every integer c, if c | a and c | b, then
c | d.

Least Common Multiples
Let a and b be nonzero integers.
 The least common multiple of a and b,
denoted lcm(a, b), is the unique positive
integer m with the following properties:

a | m and b | m.
 For every integer c, if a | c and b | c, then
m | c.

The High-school gcd
Algorithm
The high-school method is very inefficient.
 Factor each number into standard form.
 For each prime that appears in both
factorizations, use it as a factor of the gcd
along with the smaller of the two
exponents.

Example
Find the gcd of 81900 and 54810.
 We factor them as

81900 = 22 32  52  71  131  290
 54810 = 21  33  51  71  130  291


Therefore, the gcd is

2  32  5  7 = 630
The Euclidean Algorithm

Factoring is inefficient; therefore, this
algorithm is inefficient.


The run time of this algorithm is O(10d),
where d is the number of digits in the
number.
Euclid had a much better idea.

The run time of the Euclidean Algorithm is
O(d), where d is the number of digits in the
number.
The Euclidean Algorithm

Input: A, B (positive integers, not both 0)
The Euclidean Algorithm

Algorithm body:

Output: b
a := A
b := B
if b = 0 then
swap a and b
r := a mod b
while r > 0
a := b
b := r
r := a mod b
end while
Example

Apply the Euclidean Algorithm to A = 81900
and B = 54810.
Iteration
A
B
a
b
r
0
81900
54810
81900
54810
27090
1
2
Example

Apply the Euclidean Algorithm to A = 81900
and B = 54810.
Iteration
A
B
a
b
r
0
81900
54810
81900
54810
27090
54810
27090
630
1
2
Example

Apply the Euclidean Algorithm to A = 81900
and B = 54810.
Iteration
A
B
a
b
r
0
81900
54810
81900
54810
27090
1
54810
27090
630
2
27090
630
0
Least Common Multiples
What is the efficient way to find lcm’s?
 What is the lcm of 81900 and 54810?

Proof of the Euclidean
Algorithm
Theorem: The Euclidean Algorithm
terminates for all legitimate inputs A and B.
 Proof:

We may assume that B > 0.
 After the first iteration of the while loop,
0b<B
since b is the remainder of A divided by B.

Proof of the Euclidean
Algorithm
Each iteration produces a nonnegative
remainder that is smaller than the previous
remainder.
 This cannot happen more than B times
before the remainder is 0.

Proof of the Euclidean
Algorithm
Lemma 1: If b > 0, then gcd(b, 0) = b.
 Proof:

b | b and b | 0.
 For all integers c, if c | 0 and c | b, then
c | b.
 Therefore, b = gcd(b, 0).

Proof of the Euclidean
Algorithm
Lemma 2: If a and b are integers, with b 
0, and q and r are integers such that
a = qb + r
then gcd(a, b) = gcd(b, r).
 Proof:

Let d = gcd(b, r).
 Then d | b and d | r and any integer that
divides b and r must also divide d.

Proof of the Euclidean
Algorithm
We must show that d | a and d | b and any
integer that divides a and b must also
divide d.
 We already know that d | b.
 Since a = qb + r, it follows that d | a.
 Let c be an integer such that c | a and c | b.
 Since r = a – qb, it follows that c | r and so
c | d.
 Therefore, d = gcd(a, b).

Proof of the Euclidean
Algorithm
Theorem: The Euclidean Algorithm
produces the gcd of A and B.
 Proof:

After the final iteration of the while loop,
r = 0.
 By Lemma 1, the output b is the gcd of b
and r.
 By Lemma 2, that is equal to the gcd of “a”
and “b” in the final iteration.

Proof of the Euclidean
Algorithm
But “a” and “b” on the last iteration were “b”
and “r” on the previous iteration.
 Therefore, gcd(a, b) on the last iteration
equals gcd(b, r) on the previous iteration,
which equals gcd(a, b) on the previous
iteration, and so on.
 Following this argument all the back to the
first iteration, we see that the output is
gcd(A, B).

Proof of the Euclidean
Algorithm
In the next chapter, we will study
mathematical induction.
 At that point, we will be able to make this
argument more rigorous.
