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# Efficient prime test

, 12 Nov 2013 CPOL
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The IsPrime algorithm will always find a number that divides n if it is composed.

## Introduction

Prime numbers are essential in many fields of mathematics and computer science, especially cryptography. An interesting problem arises when we need to decide whether an integer is a prime number.

The straightforward, naïve algorithm for deciding if a number n is prime follows a procedure in which a loop from 2 to n-1 checks every time whether the number representing the step of the loop divides n and in that case returns false.

The previous algorithm runs in O (n). An improvement to the running time of the Naïve prime test can be achieved if we ask ourselves the question: Is it really necessary to loop from 2 to n-1? The answer to this question is no, is not necessary, it would be enough to loop only from 2 to squart(n).

Correctness of the IsPrime algorithm: Let’s assume that all numbers that divide n are greater than squart(n). If this is the case then the smallest number that can divide n is squart(n)+1, but, if n is composed then the smallest numbers dividing n must be greater or equal than [squart(n)+1] [squart(n)+1], but this product is greater than n, which is a contradiction, though there must be at least a number dividing n less than or equal to squart(n) proving that the IsPrime algorithm will always find a number that divides n if it is composed.

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 First Prev Next
 This is not "efficient" prime test fulloflove28-Feb-14 17:15 fulloflove 28-Feb-14 17:15
 Not the prime example of optimization CaldasGSM12-Nov-13 7:37 CaldasGSM 12-Nov-13 7:37
 i you really want to optimize, besides only checking up to sqrt, there are also other optimizations you can do to the naive prime validation.. like not calculations the sqrt at every iteration here is the implementation that I have in my lib.. I think the comments are self explanatory ```public static bool IsPrime(uint nNumber) { //exceptional case that the function doesn't validate if (nNumber == 2) return true;   //even numbers are never prime (except 2), and neither 1 (by convention) if (nNumber % 2 == 0 || nNumber == 1) return false;   //the last prime factor possibility for some number N would be Prime(m) where Prime(m + 1) squared exceeds N uint nSqrt = (uint)Sqrt(nNumber);   //skip even numbers because if its not divisible by 2 its not divisible by any even number for (uint i = 3; i <= nSqrt; i += 2) if ((nNumber % i) == 0)//if divisible, not prime return false;   return true; } ``` You don't want to know...
 Re: Not the prime example of optimization arnaldo.skywalker12-Nov-13 8:36 arnaldo.skywalker 12-Nov-13 8:36
 Message Removed bharat_h0312-Nov-13 22:49 bharat_h03 12-Nov-13 22:49
 Re: Not the prime example of optimization arnaldo.skywalker14-Nov-13 4:07 arnaldo.skywalker 14-Nov-13 4:07
 Re: Not the prime example of optimization bharat_h0321-Nov-13 0:01 bharat_h03 21-Nov-13 0:01
 There are several prime sieves. Pete O'Hanlon12-Nov-13 5:59 Pete O'Hanlon 12-Nov-13 5:59
 Re: There are several prime sieves. arnaldo.skywalker12-Nov-13 6:53 arnaldo.skywalker 12-Nov-13 6:53
 Re: There are several prime sieves. PIEBALDconsult12-Nov-13 11:06 PIEBALDconsult 12-Nov-13 11:06
 Re: There are several prime sieves. arnaldo.skywalker14-Nov-13 4:11 arnaldo.skywalker 14-Nov-13 4:11
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