
Code for some of the algorithms discussed exists in
various programming languages
such as
Algol, C, Haskell, Java, Pascal and SML.
The Bibliography
also contains references on algorithms and data structures
in journals and books.


Algorithm: a process or set of rules
used for calculation or problemsolving, esp. with a computer.
Program: a series of coded instructions
to control the operation of a computer or other machine.
[concise OED '91]
Example
Problem:
Find the greatest common divisor (GCD) of two integers, m and n.
Euclid's Algorithm:
while m is greater than zero:
If n is greater than m,
swap m & n.
Subtract n from m.
n is the GCD
Program (in C):
int gcd(int m, int n)
/* precondition: m>0 & n>0.
Let g=gcd(m,n). */
{ while( m > 0 )
{ /* invariant: gcd(m,n)=g */
if( n > m ) /* swap */
{ int t=m; m=n; n=t; }
/* m >= n > 0 */
m = n;
}
return n;
}
Correctness
Why do we believe that this algorithm
devised thousands of years ago, is correct?
Given m>0 and n>0, let g = gcd(m,n).
(i) If m=n then m=n=gcd(m,n) and the algorithm
sets m to zero and returns n, obviously correct.
(ii) Otherwise, suppose m>n.
Then m=p×g and n=q×g where p and q are coprime,
from the definition of greatest common divisor.
We claim that gcd(mn,n)=g.
Now mn=p×gq×g=(pq)g.
so we must show that (pq) and q are coprime.
If not then pq=a×c and q=b×c for some a,b,c>1.
But then
p=q+a×c
=b×c+a×c
=(a+b)×c and
because q=b×c,
p and q would not have been coprime ... contradiction.
Hence (pq) and q are coprime,
and gcd(mn,n)=gcd(m,n).
(iii) If m<n, the algorithm swaps them so that m>n and
that case has been covered.
So the algorithm is correct, provided that it terminates.
Termination
At the start of each iteration of the loop,
either n>m or m≥n.
(i) If m≥n,
then m is replaced by mn which is smaller than the previous
value of m, and still nonnegative.
(ii) If n>m, m and n are exchanged,
and at the next iteration case (i) will apply.
So at each iteration,
max(m,n) either remains unchanged (for just one iteration)
or it decreases.
This cannot go on for ever because m and n are integers
(this fact is important), and eventually
a lower limit is reached,
when m=0 and n=g.
So the algorithm does terminate.
Testing
Having proved the algorithm to be correct,
one might argue that there is no need to test it.
But there might be an error in the proof
or maybe the program has been coded wrongly.
Good test values would include:
 special cases where m or n equals 1, or
 m, or n, or both equal small primes 2, 3, 5, ..., or
 products of two small primes such as p1×p2 and p3×p2,
 some larger values, but ones where you know the answers,
 swapped values, (x,y) and (y,x), because gcd(m,n)=gcd(n,m).
The objective in testing is to "exercise"
all paths through the code, in different combinations.
Debugging code be inserted to print the values of m and n
at the end of each iteration to confirm
that they behave as expected.
Complexity
We are interested in how much time and space (computer memory)
a computer algorithm uses; i.e. how efficient it is.
This is called time and space complexity.
Typically the complexity is a function of the values of the inputs
and we would like to know what function.
We can also consider the best, average, and worstcases.
Time
The time to execute one iteration of the loop
depends on whether m>n or not,
but both cases take constant time:
one test, a subtraction and 4 assignments v.
one test, a subtraction and one assignment.
So the time taken for one iteration of the loop is bounded by a constant.
The real question then is, how many iterations take place?
The answer depends on m and n.
If m=n, there is just one iteration; this is the bestcase.
If n=1, there are m iterations; this is the worstcase
(equivalently, if m=1 there are n iterations).
The averagecase timecomplexity of this algorithm is difficult to analyse.
Space
The spacecomplexity of Euclid's algorithm is a constant,
just space for three integers: m, n and t.
We shall see later that this is 'O(1)'.
Exercises
 Devise a quicker version of Euclid's algorithm
that does not sit in the loop subtracting
individual copies of n from m when m>>n.

