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The Euclidean Algorithm

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I remember being in school and learning about fractions. In particular, I remember the problems we had when trying to add and subtract fractions. This problem also presented itself when we tried to multiply fractions, although we still received partial credit if we couldn’t reduce fractions to their lowest terms.

What I’m referring to is the problem of finding the Greatest Common Divisor (GCD) of two numbers. The GCD of two numbers is the largest number that divides into both numbers with a remainder of zero each time. This problem has many applications, but for most of us we can relate to it because of our frustrations with fractions.

The algorithm that I’m writing about and have written a script for is called the Euclidean Algorithm, which solves precisely this problem. To be more precise, the Euclidean Algorithm finds the greatest common divisor between two integer numbers.

There are different versions of the algorithm, but the one I have implemented finds the GCD by subtracting the smaller number from the larger number, and if the result is greater than 0, the procedure repeats itself with the two lower numbers. Otherwise, the result (the GCD) is the final number that was greater than 0.

Lets see an example:
Consider the number 9 and 21.
21 – 9 = 12. 12 is greater than 0, so we repeat the procedure with 9 and 12.
12 – 9 = 3. 3 is greater than 0, so we repeat the procedure with 3 and 9.
9 – 3 = 6. 6 is greater than 0, so we repeat the procedure with 3 and 6.
6 – 3 = 3. 3 is greater than 0, so we repeat the procedure with 3 and 3.
3 – 3 = 0. 0 is not greater than 0, so can exit the loop portion of the algorithm.
Since 3 was the last positive number that we arrived at in this procedure, we see that the GCD of 9 and 12 is 3.

To learn more and see more examples, check out My Script on The Euclidean Algorithm.

Examples

Sieve of Eratosthenes

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Prime numbers are an important concept in Number Theory and Cryptography which often uses the difficulty of finding prime numbers as a basis for building encryption systems that are difficult to break without going through all (or a very large number of) possible choices.

Remember that a prime number is a number greater than 1 whose only divisors are 1 and that number itself. One of the most famous algorithms for searching for prime numbers is the Sieve of Eratosthenes. I added a script which implements the Sieve of Eratosthenes to my Examples page.

This algorithm prints out all prime numbers less than a given number by first canceling out all multiples of 2 (the smallest prime), then all multiples of 3 (the second smallest prime), then all multiples of 5 (the third smallest prime – multiples of 4 do not need to be considered because they are also multiples of 2), etc until we have reached a number which cannot be a divisor of this maximum number.

So if we are given a number, n, the first step of the algorithm is write out a table that lists all the numbers that are less than n. For example lets run this Sieve on 50. So all numbers less than 50 are

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50.

So since 1 is not a prime number (by the definition of prime numbers), we cancel that number out.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50.

Next, we look at the list and the first number that is not crossed out is a prime. That number is 2. We will put a mark by 2 and cancel out all of 2’s multiples.

1, 2*, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50.

Again, we look at the list and the first number that is not marked or crossed out is 3, so that number is prime. We will put a mark by 3 and cancel out all of 3’s multiples.

1, 2*, 3*, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50.

Once again, we look at the list and the first number that is not marked or crossed out is 5, so that number is prime. We will put a mark by 5 and cancel out all of 5’s multiples.

1, 2*, 3*, 4, 5*, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50.

We look at the list and the first number that is not marked or crossed out is 7, so that number is prime. We will put a mark by 7 and cancel out all of 7’s multiples.

1, 2*, 3*, 4, 5*, 6, 7*, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50.

Now, we look and the first number that is not crossed out is 11. However, since 11 is greater than sqrt(50) we know that each of 11’s multiples that are less than 50 will have been cancelled out by a previous prime number. So we have finished the algorithm.

Check out my script which implements the Sieve of Eratosthenes for more examples.