Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Check if number is prime number

I would just like to ask if this is a correct way of checking if number is prime or not? because I read that 0 and 1 are NOT a prime number.

int num1;

Console.WriteLine("Accept number:");
num1 = Convert.ToInt32(Console.ReadLine());
if (num1 == 0 || num1 == 1)
{
    Console.WriteLine(num1 + " is not prime number");
    Console.ReadLine();
}
else
{
    for (int a = 2; a <= num1 / 2; a++)
    {
        if (num1 % a == 0)
        {
            Console.WriteLine(num1 + " is not prime number");
            return;
        }

    }
    Console.WriteLine(num1 + " is a prime number");
    Console.ReadLine();
}
like image 943
user1954418 Avatar asked Apr 01 '13 12:04

user1954418


People also ask

How do you check if a number is a prime?

To prove whether a number is a prime number, first try dividing it by 2, and see if you get a whole number. If you do, it can't be a prime number. If you don't get a whole number, next try dividing it by prime numbers: 3, 5, 7, 11 (9 is divisible by 3) and so on, always dividing by a prime number (see table below).


3 Answers

var number;

Console.WriteLine("Accept number:");
number = Convert.ToInt32(Console.ReadLine());

if (IsPrime(number))
{
    Console.WriteLine("It is prime");
}
else
{
    Console.WriteLine("It is not prime");
}       

public static bool IsPrime(int number)
{
    if (number <= 1) return false;
    if (number == 2) return true;
    if (number % 2 == 0) return false;

    var boundary = (int)Math.Floor(Math.Sqrt(number));
          
    for (int i = 3; i <= boundary; i += 2)
        if (number % i == 0)
            return false;
    
    return true;        
}

I changed number / 2 to Math.Sqrt(number) because from in wikipedia, they said:

This routine consists of dividing n by each integer m that is greater than 1 and less than or equal to the square root of n. If the result of any of these divisions is an integer, then n is not a prime, otherwise it is a prime. Indeed, if n = a*b is composite (with a and b ≠

  1. then one of the factors a or b is necessarily at most square root of n
like image 53
Soner Gönül Avatar answered Oct 24 '22 00:10

Soner Gönül


Using Soner's routine, but with a slight variation: we will run until i equals Math.Ceiling(Math.Sqrt(number)) that is the trick for the naive solution:

boolean isPrime(int number)
{
    if (number == 1) return false;
    if (number == 2) return true;

    var limit = Math.Ceiling(Math.Sqrt(number)); //hoisting the loop limit

    for (int i = 2; i <= limit; ++i)  
       if (number % i == 0)  
           return false;
    return true;

}
like image 44
0x90 Avatar answered Oct 24 '22 00:10

0x90


Here's a nice way of doing that.

    static bool IsPrime(int n)
    {
        if (n > 1)
        {
            return Enumerable.Range(1, n).Where(x => n%x == 0)
                             .SequenceEqual(new[] {1, n});
        }

        return false;
    }

And a quick way of writing your program will be:

        for (;;)
        {
            Console.Write("Accept number: ");
            int n = int.Parse(Console.ReadLine());
            if (IsPrime(n))
            {
                Console.WriteLine("{0} is a prime number",n);
            }
            else
            {
                Console.WriteLine("{0} is not a prime number",n);
            }
        }
like image 12
Raz Megrelidze Avatar answered Oct 24 '22 01:10

Raz Megrelidze