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How to find the Nth prime number

I need to find the Nth prime number. I.E. if the input is 3 it should return 5 because the first three prime numbers are 2,3, and 5. 4 should return 7 and so on. I just need an algorithm or some pseudocode here right now because i'm writing this in assembler. So far this is what i have.

for(divisor = 2; divisor < num; divis++)
{
     if((num mod divisor) == 0)
          num is not prime
          break out of the loop
}
if((num - divisor) == 1)
     num is a prime
//if num - divisor == 1 that means that it made it the whole way through the for loop without being broken out of the loop and there was no number that would divide the number therefore it is prime.


     
0
toymachiner62
Asked:
toymachiner62
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1 Solution
 
HardiCommented:
There is a site for calculating the n-th prime, and it explains a bit of the algorithm you may be interested
http://primes.utm.edu/nthprime/algorithm.php
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peprCommented:
I would recommend to have a look at wikipedia "Prime number" (http://en.wikipedia.org/wiki/Prime_number).  You will find there some theory and also a lot of references.

See the section "5.1 Finding prime numbers" (http://en.wikipedia.org/wiki/Prime_number#Finding_prime_numbers). The very first references in the section point to the "easy, brute force" and "fast" algorithms:

Sieve of Eratosthenes (http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes)
Sieve of Atkin (http://en.wikipedia.org/wiki/Sieve_of_Atkin -- contains reference to implementation in Python and in C)
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TalmashCommented:
beyond the info on web:

find = 4 // find first 4 primes
found=1 // still only "2" found
print "2\n"
num = 3 // check if num is a prime

while (found < find) {
  for(divisor = 2; divisor < num; divis++) {
     if((num mod divisor) == 0) {
          num is not prime
          break out of the "for" loop
     }
   
    if(divisor*divisor > num) {
    //if((num - divisor) == 1)
     num is a prime
     print num
     found++
     break out of the "for" loop
    }
  } // close "for"
} // close while

this code is closer to what you need.

tal
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