Example #1

Prime Numbers

The time has come to give you an example of a complete program. This will, hopefully, give you a better idea of how some of the features I've been talking about work in "real life". The short examples in my lessons are fine, but they don't give you the big picture.

In this example I will write a program that tests to see if a given number is a prime number. Prime numbers are integers that can't be evenly divided by any other number (except for themselves and one). Here is what I need to do.

<Ask the user to enter a number>
<Check to see if the number is prime and, if so, print a message>

I need to elaborate on this outline a bit. First, I should be sure that the user doesn't try to be cute and enter a negative number.

<Ask the user to enter a number>
IF <The number is negative> THEN
<Print an error message and end the program>
END
<Check to see if the number is prime and, if so, print a message>

Since I'm in the process of figuring out what I need to do, I'm not bothering to write things in C. Instead I'm using a pseudo language to just help me organize the steps. Later, once I'm convinced that I've got the steps right, I can translate my pseudo language into C (or, for that matter, any other reasonable programming language).

How am I going to test to see if a number is prime? How about the brute force way: I'll try to divide it by all numbers less than itself.

<Ask the user to enter a number>
IF <The number is negative> THEN
<Print an error message and end the program>
END
FOR <All numbers less than the given number> LOOP
IF <I can divide the given number by the smaller number> THEN
<The given number is not prime>
<End the program>
END
END
<If I get here the given number must be prime>

Is this logic right? Now is the time to check it. I check my programs by just going through them the way the computer does. Suppose I'm the program and I'm given a number like 117. I will loop over all the numbers less than 117, like 1, 2, 3, 4, and so forth. For each of those I will try to divide that number into 117. Oops... the value of 1 will always go into 117! I need to start my loop at 2. But then... what if the user gives us the value of 1 (or for that matter zero)? Hmmm. I should handle that case before reaching the loop.

Here is the refined program.

<Ask the user to enter a number>
IF <The number is less than 2> THEN
<Print an error message and end the program>
END
FOR <All numbers from 2 that are less than the given number> LOOP
IF <I can divide the given number by the smaller number> THEN
<The given number is not prime>
<End the program>
END
END
<If I get here the given number must be prime>

Okay, this new version looks better. If I'm given 117 it will loop from 2 to 116. That's fine. If any of those smaller numbers divide into 117 evenly then 117 can't be prime. The program will print a message and exit at once. If I get through all of those smaller numbers without triggering the print statement inside the loop, 117 must be prime. I believe the logic works.

But wait... what if the user gives us 2? The first error message won't trigger. Yet the loop won't execute either since there are no values starting at 2 that are less than 2. Thus the program will go right to the end where it will print that 2 is a prime number. Good! Two is a prime number so it works.

Now I'm ready to convert my pseudo language into C. The statements in my pseudo langauge become the comments in my C.

#include <stdio.h>

int main( void )
{
int number;   // The number I am testing.
int i;        // Loop index.

// Ask the user to enter a number.
printf( "Enter a number: " );
scanf( "%d", &number );

// Do I like it? If not, print a message and die.
if( number < 2 ) {
printf( "The value %d is not prime.\n", number );
return 1;
}

// Can I divide this number by anything less than it?
for( i = 2; i < number; i++ ) {
if( number % i == 0 ) {
printf( "I'm sorry, %d is not prime. It can be divided by %d\n", number, i );
return 0;
}
}

// If we got here, it must be prime.
printf( "The number %d is prime!\n", number );
return 0;
}

This program works, but it has at least two issues. First it is not very efficient. If you give it a large prime number it will take a while to figure it out. Most of its time will be spent in the for loop dividing your number by all the smaller ones. Consider how this will work if you give it a number in millions.

The other issue is that it contains a return in the middle of the for loop. In general, having your program end in multiple places is often not a good idea. Let me improve this program by reorganizing it so that there is only one return. To do that, I will need to introduce a "flag variable". The point of the flag is just to keep track of something that has happened. In particular, when I discover that the number is not prime, I will set the flag. The loop condition needs to look at the flag too. Here is how that looks

#include <stdio.h>

int main( void )
{
int number;    // The number I am testing.
int i;         // Loop index.
int prime = 1; // We start by assuming the number is prime.

// Ask the user to enter a number.
printf( "Enter a number: " );
scanf( "%d", &number );

// Do I like it? If not, print a message and die.
if( number < 2 ) {
printf( "The value %d is not prime.\n", number );
return 1;
}

// Can I divide this number by anything less than it?
for( i = 2; (i < number) && prime; i++ ) {
if( number % i == 0 ) {
prime = 0;
}
}

// What is the result of my test?
if( prime ) {
printf( "The number %d is prime!\n", number );
}
else {
printf( "I'm sorry, %d is not prime. It can be divided by %d\n", number, i - 1 );
}
return 0;
}

Although the flag variable, prime, is an integer, it is being used to hold a true/false value. Initially I assume that the given number will be prime and I set prime to 1 (true). The for loop executes "as long as" both i < number AND prime is true. If I discover inside the for loop that the given number is not prime I set prime to 0 (false). This will cause the loop to end because the loop condition will be false as well.

There are two reasons why I could be at the code below the for loop. Either the loop ran all the way (meaning that the given number was prime) or the loop aborted because I discovered that the number was not prime. I can distinguish between those two cases easily by just looking at the flag variable. If it's still set to "true" than the loop must have run all the way. Notice that in this case when I print out the value of i (when the number is not prime) I actually print i - 1. This is because the end expression of the for loop did execute and thus the statement executed inside the loop.

This new version has only one return statement. If I wanted to later do something special just before this program ends, it would be easy to see where that could be done. Furthermore, this version cleanly divides the analysis from the output. The for loop figures out if the number is prime or not. Then, later, I print out the result. I think this is much better than the earlier program.

Okay... what about speed? This program is still horribly slow. Well, if the given number is even it can't be prime. Only odd numbers can be prime (except for 2) and odd numbers can't be divided evenly by even numbers. (I leave the proof as an exercise). Thus I could speed up the program by a factor of two if I just skip all even numbers in the for loop.

Actually I can do far better than that. Consider 117 again as an example. If 117 can't be divided by 2, it certainly can't be divided by anything bigger than 117/2 (such as, say, 96). It sure seems silly when you think about it to try doing 117/110, 117/111, 117/112, etc! Obviously those aren't going to work. Furthermore, if 117 can't be divided by 3, it can't be divided by anything bigger than 117/3. Thus each time we test a number, we can lower the upper bound of the loop. If 117 isn't evenly divisable by four, we know nothing over 117/4 will work either. Thus we get this version of the program

#include <stdio.h>

int main( void )
{
int number;    // The number I am testing.
int i;         // Loop index.
int prime = 1; // We start by assuming the number is prime.
int upper;     // Upper bound of the for loop.

// Ask the user to enter a number.
printf( "Enter a number: " );
scanf( "%d", &number );

// Do I like it? If not, print a message and die.
if( number < 2 ) {
printf( "The value %d is not prime.\n", number );
return 1;
}

// Can I divide this number by anything less than it?
upper = number;
for( i = 2; (i < upper) && prime; i++ ) {

// If it divides evenly, it's not prime. Otherwise I can adjust upper.
if( number % i == 0 ) {
prime = 0;
}
else {
upper = number / i;
}
}

// What is the result of my test?
if( prime ) {
printf( "The number %d is prime!\n", number );
}
else {
printf( "I'm sorry, %d is not prime. It can be divided by %d\n", number, i - 1 );
}
return 0;
}

This version is not just a little bit faster than the earlier one, it is a LOT faster. In fact, the amount of time it takes to test a number is proportional to the square root of the number. The earlier version required a time proportional to the number itself. Thus to test a number near 1000000, the first first would require 1000000 passes of the for loop. The latest version only requires 1000. It would be about 1000 times faster!

We could, of course, speed up the program by another factor of two by skipping all even numbers while we were at it. However, to do that we would have to add some special case handling for the value 2. I leave that as an exercise.

If you decide you want to play around with this program, it might be handy to know some large prime numbers. Here are a few for your reference.

141,359
1,234,577
33,876,329
357,666,983
2,147,483,629

There are actually quite a few prime numbers. You can probably find a few large ones of your own by just trying some numbers out.

My final program is prime.c.

© Copyright 2016 by Peter C. Chapin.
Last Revised: February 1, 2016