1. Application: Algorithms
Lecture 17
Section 3.8
Thu, Feb 16, 2006

2. Algorithms
An algorithm is a step-by-step procedure that is guaranteed to stop after a finite number of steps for all legitimate inputs.

3. 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?

4. 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.

5. Assignment Statements
An assignment statement is of the form
where x is a variable and expr is an expression.
Examples
found := true
count := count + 1
x := expr

6. Conditional Statements
One-way conditional statements:
Two-way conditional statements:
if condition then
sequence of statements
if condition then
sequence of statements
else

7. Iterative Statements While loops: For loops: while condition
sequence of statements
end while
for var := init expr to final expr
sequence of statements
next var

8. 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

9. 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.

10. Example: Finding the Max
Input: a, n
a is a list of numbers.
n is the size of the list, n  1.
Algorithm body:
Output: max
max := a[1]
for i := 2 to n
if a[i] > max then
max = a[i]
next i

11. 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
{7, 3, 8, 6, 4} 5 7 2 1 3 4

12. 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
{7, 3, 8, 6, 4} 5 7 2 1 3 4

13. 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
{7, 3, 8, 6, 4} 5 7 2 1 3 8 4

14. 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
{7, 3, 8, 6, 4} 5 7 2 1 3 8 4

15. 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
{7, 3, 8, 6, 4} 5 7 2 1 3 8 4 6

16. 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.

17. 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.

18. 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.

19. 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

20. The Euclidean Algorithm
Factoring is inefficient; therefore, this algorithm is inefficient.
This algorithm is O(10d), where d is the number of digits in the number.
Euclid had a much better idea.
Use the Euclidean Algorithm is O(d), where d is the number of digits in the number.

21. The Euclidean Algorithm
Input: A, B (positive integers, not both 0)
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

22. Example Apply the Euclidean Algorithm to A = 81900 and B = 54810.
Iteration A B a b r
81900
54810
27090 1 2

23. Example Apply the Euclidean Algorithm to A = 81900 and B = 54810.
Iteration A B a b r
81900
54810
27090 1
630 2

24. Example Apply the Euclidean Algorithm to A = 81900 and B = 54810.
Iteration A B a b r
81900
54810
27090 1
630 2

25. Least Common Multiples
What is the efficient way to find lcm’s?
What is the lcm of and 54810?

26. Proof of the Euclidean Algorithm
Theorem: The Euclidean Algorithm terminates for all legitimate inputs A and B.
Proof:
WOLOG, WMA B > 0.
After the first iteration of the while loop,
0  b < B
since b is the remainder of A divided by B.

27. 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.

28. 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).

29. 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.

30. 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).

31. 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.

32. 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).

33. Proof of the Euclidean Algorithm
In a later chapter, we will study mathematical induction.
At that point, we will be able to make this argument more rigorous.