C - determine if a number is prime

Question

I am trying to come up with a method that takes an integer and returns a boolean to say if the number is prime or not and I don't know much C; would anyone care to give me some pointers?

Basically, I would do this in C# like this:

static bool IsPrime(int number)
{
    for (int i = 2; i < number; i++)
    {
        if (number % i == 0 && i != number)
            return false;
    }
    return true;
}

Answer 1

OK, so forget about C. Suppose I give you a number and ask you to determine if it's prime. How do you do it? Write down the steps clearly, then worry about translating them into code.

Once you have the algorithm determined, it will be much easier for you to figure out how to write a program, and for others to help you with it.

edit: Here's the C# code you posted:

static bool IsPrime(int number) {
    for (int i = 2; i < number; i++) {
        if (number % i == 0 && i != number) return false;
    }
    return true;
}

This is very nearly valid C as is; there's no bool type in C, and no true or false, so you need to modify it a little bit (edit: Kristopher Johnson correctly points out that C99 added the stdbool.h header). Since some people don't have access to a C99 environment (but you should use one!), let's make that very minor change:

int IsPrime(int number) {
    int i;
    for (i=2; i<number; i++) {
        if (number % i == 0 && i != number) return 0;
    }
    return 1;
}

This is a perfectly valid C program that does what you want. We can improve it a little bit without too much effort. First, note that i is always less than number, so the check that i != number always succeeds; we can get rid of it.

Also, you don't actually need to try divisors all the way up to number - 1; you can stop checking when you reach sqrt(number). Since sqrt is a floating-point operation and that brings a whole pile of subtleties, we won't actually compute sqrt(number). Instead, we can just check that i*i <= number:

int IsPrime(int number) {
    int i;
    for (i=2; i*i<=number; i++) {
        if (number % i == 0) return 0;
    }
    return 1;
}

One last thing, though; there was a small bug in your original algorithm! If number is negative, or zero, or one, this function will claim that the number is prime. You likely want to handle that properly, and you may want to make number be unsigned, since you're more likely to care about positive values only:

int IsPrime(unsigned int number) {
    if (number <= 1) return 0; // zero and one are not prime
    unsigned int i;
    for (i=2; i*i<=number; i++) {
        if (number % i == 0) return 0;
    }
    return 1;
}

This definitely isn't the fastest way to check if a number is prime, but it works, and it's pretty straightforward. We barely had to modify your code at all!

Answer 2

I'm suprised that no one mentioned this.

Use the Sieve Of Eratosthenes

Details:

1. Basically nonprime numbers are divisible by another number besides 1 and themselves
2. Therefore: a nonprime number will be a product of prime numbers.

The sieve of Eratosthenes finds a prime number and stores it. When a new number is checked for primeness all of the previous primes are checked against the know prime list.

Reasons:

1. This algorithm/problem is known as "Embarrassingly Parallel"
2. It creates a collection of prime numbers
3. Its an example of a dynamic programming problem
4. Its quick!

Answer 3

Stephen Canon answered it very well!

But

• The algorithm can be improved further by observing that all primes are of the form 6k ± 1, with the exception of 2 and 3.
• This is because all integers can be expressed as (6k + i) for some integer k and for i = −1, 0, 1, 2, 3, or 4; 2 divides (6k + 0), (6k + 2), (6k + 4); and 3 divides (6k + 3).
• So a more efficient method is to test if n is divisible by 2 or 3, then to check through all the numbers of form 6k ± 1 ≤ √n.

• This is 3 times as fast as testing all m up to √n.

  int IsPrime(unsigned int number) {
      if (number <= 3 && number > 1) 
          return 1;            // as 2 and 3 are prime
      else if (number%2==0 || number%3==0) 
          return 0;     // check if number is divisible by 2 or 3
      else {
          unsigned int i;
          for (i=5; i*i<=number; i+=6) {
              if (number % i == 0 || number%(i + 2) == 0) 
                  return 0;
          }
          return 1; 
      }
  }
  

Answer 4

1. Build a table of small primes, and check if they divide your input number.
2. If the number survived to 1, try pseudo primality tests with increasing basis. See Miller-Rabin primality test for example.
3. If your number survived to 2, you can conclude it is prime if it is below some well known bounds. Otherwise your answer will only be "probably prime". You will find some values for these bounds in the wiki page.

Answer 5

this program is much efficient for checking a single number for primality check.

bool check(int n){
    if (n <= 3) {
        return n > 1;
    }

    if (n % 2 == 0 || n % 3 == 0) {
        return false;
    }
        int sq=sqrt(n); //include math.h or use i*i<n in for loop
    for (int i = 5; i<=sq; i += 6) {
        if (n % i == 0 || n % (i + 2) == 0) {
            return false;
        }
    }

    return true;
}

Answer 6

Check the modulus of each integer from 2 up to the root of the number you're checking.

If modulus equals zero then it's not prime.

pseudo code:

bool IsPrime(int target)
{
  for (i = 2; i <= root(target); i++)
  {
    if ((target mod i) == 0)
    {
      return false;
    }
  }

  return true;
}

Answer 7

After reading this question, I was intrigued by the fact that some answers offered optimization by running a loop with multiples of 2*3=6.

So I create a new function with the same idea, but with multiples of 2*3*5=30.

int check235(unsigned long n)
{
    unsigned long sq, i;

    if(n<=3||n==5)
        return n>1;

    if(n%2==0 || n%3==0 || n%5==0)
        return 0;

    if(n<=30)
        return checkprime(n); /* use another simplified function */

    sq=ceil(sqrt(n));
    for(i=7; i<=sq; i+=30)
        if (n%i==0 || n%(i+4)==0 || n%(i+6)==0 || n%(i+10)==0 || n%(i+12)==0 
           || n%(i+16)==0 || n%(i+22)==0 || n%(i+24)==0)
            return 0;

        return 1;
}

By running both functions and checking times I could state that this function is really faster. Lets see 2 tests with 2 different primes:

$ time ./testprimebool.x 18446744069414584321 0
f(2,3)
Yes, its prime.    
real    0m14.090s
user    0m14.096s
sys     0m0.000s

$ time ./testprimebool.x 18446744069414584321 1
f(2,3,5)
Yes, its prime.    
real    0m9.961s
user    0m9.964s
sys     0m0.000s

$ time ./testprimebool.x 18446744065119617029 0
f(2,3)
Yes, its prime.    
real    0m13.990s
user    0m13.996s
sys     0m0.004s

$ time ./testprimebool.x 18446744065119617029 1
f(2,3,5)
Yes, its prime.    
real    0m10.077s
user    0m10.068s
sys     0m0.004s

So I thought, would someone gain too much if generalized? I came up with a function that will do a siege first to clean a given list of primordial primes, and then use this list to calculate the bigger one.

int checkn(unsigned long n, unsigned long *p, unsigned long t)
{
    unsigned long sq, i, j, qt=1, rt=0;
    unsigned long *q, *r;

    if(n<2)
        return 0;

    for(i=0; i<t; i++)
    {
        if(n%p[i]==0)
            return 0;
        qt*=p[i];
    }
    qt--;

    if(n<=qt)
        return checkprime(n); /* use another simplified function */

    if((q=calloc(qt, sizeof(unsigned long)))==NULL)
    {
        perror("q=calloc()");
        exit(1);
    }
    for(i=0; i<t; i++)
        for(j=p[i]-2; j<qt; j+=p[i])
            q[j]=1;

    for(j=0; j<qt; j++)
        if(q[j])
            rt++;

    rt=qt-rt;
    if((r=malloc(sizeof(unsigned long)*rt))==NULL)
    {
        perror("r=malloc()");
        exit(1);
    }
    i=0;
    for(j=0; j<qt; j++)
        if(!q[j])
            r[i++]=j+1;

    free(q);

    sq=ceil(sqrt(n));
    for(i=1; i<=sq; i+=qt+1)
    {
        if(i!=1 && n%i==0)
            return 0;
        for(j=0; j<rt; j++)
            if(n%(i+r[j])==0)
                return 0;
    }
    return 1;
}

I assume I did not optimize the code, but it's fair. Now, the tests. Because so many dynamic memory, I expected the list 2 3 5 to be a little slower than the 2 3 5 hard-coded. But it was ok as you can see bellow. After that, time got smaller and smaller, culminating the best list to be:

2 3 5 7 11 13 17 19

With 8.6 seconds. So if someone would create a hardcoded program that makes use of such technique I would suggest use the list 2 3 and 5, because the gain is not that big. But also, if willing to code, this list is ok. Problem is you cannot state all cases without a loop, or your code would be very big (There would be 1658879 ORs, that is || in the respective internal if). The next list:

2 3 5 7 11 13 17 19 23

time started to get bigger, with 13 seconds. Here the whole test:

$ time ./testprimebool.x 18446744065119617029 2 3 5
f(2,3,5)
Yes, its prime.
real    0m12.668s
user    0m12.680s
sys     0m0.000s

$ time ./testprimebool.x 18446744065119617029 2 3 5 7
f(2,3,5,7)
Yes, its prime.
real    0m10.889s
user    0m10.900s
sys     0m0.000s

$ time ./testprimebool.x 18446744065119617029 2 3 5 7 11
f(2,3,5,7,11)
Yes, its prime.
real    0m10.021s
user    0m10.028s
sys     0m0.000s

$ time ./testprimebool.x 18446744065119617029 2 3 5 7 11 13
f(2,3,5,7,11,13)
Yes, its prime.
real    0m9.351s
user    0m9.356s
sys     0m0.004s

$ time ./testprimebool.x 18446744065119617029 2 3 5 7 11 13 17
f(2,3,5,7,11,13,17)
Yes, its prime.
real    0m8.802s
user    0m8.800s
sys     0m0.008s

$ time ./testprimebool.x 18446744065119617029 2 3 5 7 11 13 17 19
f(2,3,5,7,11,13,17,19)
Yes, its prime.
real    0m8.614s
user    0m8.564s
sys     0m0.052s

$ time ./testprimebool.x 18446744065119617029 2 3 5 7 11 13 17 19 23
f(2,3,5,7,11,13,17,19,23)
Yes, its prime.
real    0m13.013s
user    0m12.520s
sys     0m0.504s

$ time ./testprimebool.x 18446744065119617029 2 3 5 7 11 13 17 19 23 29
f(2,3,5,7,11,13,17,19,23,29)                                                                                                                         
q=calloc(): Cannot allocate memory

---

PS. I did not free(r) intentionally, giving this task to the OS, as the memory would be freed as soon as the program exited, to gain some time. But it would be wise to free it if you intend to keep running your code after the calculation.

---

BONUS

int check2357(unsigned long n)
{
    unsigned long sq, i;

    if(n<=3||n==5||n==7)
        return n>1;

    if(n%2==0 || n%3==0 || n%5==0 || n%7==0)
        return 0;

    if(n<=210)
        return checkprime(n); /* use another simplified function */

    sq=ceil(sqrt(n));
    for(i=11; i<=sq; i+=210)
    {    
        if(n%i==0 || n%(i+2)==0 || n%(i+6)==0 || n%(i+8)==0 || n%(i+12)==0 || 
   n%(i+18)==0 || n%(i+20)==0 || n%(i+26)==0 || n%(i+30)==0 || n%(i+32)==0 || 
   n%(i+36)==0 || n%(i+42)==0 || n%(i+48)==0 || n%(i+50)==0 || n%(i+56)==0 || 
   n%(i+60)==0 || n%(i+62)==0 || n%(i+68)==0 || n%(i+72)==0 || n%(i+78)==0 || 
   n%(i+86)==0 || n%(i+90)==0 || n%(i+92)==0 || n%(i+96)==0 || n%(i+98)==0 || 
   n%(i+102)==0 || n%(i+110)==0 || n%(i+116)==0 || n%(i+120)==0 || n%(i+126)==0 || 
   n%(i+128)==0 || n%(i+132)==0 || n%(i+138)==0 || n%(i+140)==0 || n%(i+146)==0 || 
   n%(i+152)==0 || n%(i+156)==0 || n%(i+158)==0 || n%(i+162)==0 || n%(i+168)==0 || 
   n%(i+170)==0 || n%(i+176)==0 || n%(i+180)==0 || n%(i+182)==0 || n%(i+186)==0 || 
   n%(i+188)==0 || n%(i+198)==0)
            return 0;
    }
    return 1;
}

Time:

$ time ./testprimebool.x 18446744065119617029 7
h(2,3,5,7)
Yes, its prime.
real    0m9.123s
user    0m9.132s
sys     0m0.000s

Answer 8

I would just add that no even number (bar 2) can be a prime number. This results in another condition prior to for loop. So the end code should look like this:

int IsPrime(unsigned int number) {
    if (number <= 1) return 0; // zero and one are not prime
    if ((number > 2) && ((number % 2) == 0)) return 0; //no even number is prime number (bar 2)
    unsigned int i;
    for (i=2; i*i<=number; i++) {
        if (number % i == 0) return 0;
    }
    return 1;
}

Answer 9

Avoid overflow bug

unsigned i, number;
...
for (i=2; i*i<=number; i++) {  // Buggy
for (i=2; i*i<=number; i += 2) {  // Buggy
// or
for (i=5; i*i<=number; i+=6) { // Buggy

These forms are incorrect when number is a prime and i*i is near the max value of the type.

Problem exists with all integer types, signed, unsigned and wider.

Example:

Let UINT_MAX_SQRT as the floor of the square root of the maximum integer value. E.g. 65535 when unsigned is 32-bit.

With for (i=2; i*i<=number; i++), this 10-year old failure occurs because when UINT_MAX_SQRT*UINT_MAX_SQRT <= number and number is a prime, the next iteration results in a multiplication overflow. Had the type been a signed type, the overflow is UB. With unsigned types, this itself is not UB, yet logic has broken down. Interations continue until a truncated product exceeds number. An incorrect result may occur. With 32-bit unsigned, try 4,294,967,291‬ which is a prime.

If some_integer_type_MAX been a Mersenne Prime, i*i<=number is never true.

---

To avoid this bug, consider that number%i, number/i is efficient on many compilers as the calculations of the quotient and remainder are done together, thus incurring no extra cost to do both vs. just 1.

A simple full-range solution:

bool IsPrime(unsigned number) {
    for(unsigned i = 2; i <= number/i; i++){
        if(number % i == 0){
            return false;
        }
    }
    return number >= 2;
}

Answer 10

int is_prime(int val)
{
   int div,square;

   if (val==2) return TRUE;    /* 2 is prime */
   if ((val&1)==0) return FALSE;    /* any other even number is not */

   div=3;
   square=9;    /* 3*3 */
   while (square<val)
   {
     if (val % div == 0) return FALSE;    /* evenly divisible */
     div+=2;
     square=div*div;
   }
   if (square==val) return FALSE;
   return TRUE;
}

Handling of 2 and even numbers are kept out of the main loop which only handles odd numbers divided by odd numbers. This is because an odd number modulo an even number will always give a non-zero answer which makes those tests redundant. Or, to put it another way, an odd number may be evenly divisible by another odd number but never by an even number (E*E=>E, E*O=>E, O*E=>E and O*O=>O).

A division/modulus is really costly on the x86 architecture although how costly varies (see http://gmplib.org/~tege/x86-timing.pdf). Multiplications on the other hand are quite cheap.

Answer 11

To check if a number is prime i'm using a Miller/Rabin Algorithm.

#include <stdlib.h>

typedef size_t positive_number; // also try __uint128_t

static inline positive_number multiplication_modulo(positive_number lhs, positive_number rhs, positive_number mod) {
    positive_number res = 0; // we avoid overflow in modular multiplication
    for (lhs %= mod, rhs %= mod; rhs; (rhs & 1) ? (res = (res + lhs) % mod) : 0, lhs = (lhs << 1) % mod, rhs >>= 1);
    return res; // <= (lhs * rhs) % mod
}

static int is_prime(positive_number n, int k) {
    positive_number a = 0, b, c, d, e, f, g; int h, i;
    if ((n == 1) == (n & 1)) return n == 2;
    if (n < 51529) // fast constexpr check for small primes (removable)
        return (n & 1) & ((n < 6) * 42 + 0x208A2882) >> n % 30 && (n < 49 || (n % 7 && n % 11 && n % 13 && n % 17 && n % 19 && n % 23 && n % 29 && n % 31 && n % 37 && (n < 1369 || (n % 41 && n % 43 && n % 47 && n % 53 && n % 59 && n % 61 && n % 67 && n % 71 && n % 73 && ( n < 6241 || (n % 79 && n % 83 && n % 89 && n % 97 && n % 101 && n % 103 && n % 107 && n % 109 && n % 113 && ( n < 16129 || (n % 127 && n % 131 && n % 137 && n % 139 && n % 149 && n % 151 && n % 157 && n % 163 && n % 167 && ( n < 29929 || (n % 173 && n % 179 && n % 181 && n % 191 && n % 193 && n % 197 && n % 199 && n % 211 && n % 223))))))))));
    for (b = c = n - 1, h = 0; !(b & 1); b >>= 1, ++h);
    for (; k--;) {
        for (g = 0; g < sizeof(positive_number); ((char*)&a)[g++] = rand()); // random number
        do for (d = e = 1 + a % c, f = n; (d %= f) && (f %= d););
        while (d > 1 && f > 1);
        for (d = f = 1; f <= b; f <<= 1);
        for (; f >>= 1; d = multiplication_modulo(d, d, n), f & b && (d = multiplication_modulo(e, d, n)));
        if (d == 1) continue;
        for (i = h; i-- && d != c; d = multiplication_modulo(d, d, n));
        if (d != c) return 0;
    }
    return 1;
}

The test is to print the first primes :

#include <stdio.h>
int main() {
    // C Fast Iterative Algorithm
    // The First 10,000 Primes
    for (int i = 0 ; i < 104730 ; ++i)
        if (is_prime(i, 20))
            printf("%d %c", i, (i+1) % 10 ? ' ' : '\n');

    if (is_prime(9223372036854775783UL, 12))
        if (is_prime(9223372036854775643UL, 12))
            if (!is_prime(3037000493ULL * 3037000453ULL, 12))
                printf("Done.\n");
}

You can put it into a primes.c file then compile + execute :

gcc -O3 -std=c99 -Wall -pedantic primes.c ; ./a.out ;

This variant of the Fermat test has a polynomial runtime in log(n).
The __uint128_t type may be available with 128-bit integers GCC extension.

Answer 12

Using Sieve of Eratosthenes, computation is quite faster compare to "known-wide" prime numbers algorithm.

By using pseudocode from it's wiki (https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes), I be able to have the solution on C#.

public bool IsPrimeNumber(int val) {
    // Using Sieve of Eratosthenes.
    if (val < 2)
    {
        return false;
    }

    // Reserve place for val + 1 and set with true.
    var mark = new bool[val + 1];
    for(var i = 2; i <= val; i++)
    {
        mark[i] = true;
    }

    // Iterate from 2 ... sqrt(val).
    for (var i = 2; i <= Math.Sqrt(val); i++)
    {
        if (mark[i])
        {
            // Cross out every i-th number in the places after i (all the multiples of i).
            for (var j = (i * i); j <= val; j += i)
            {
                mark[j] = false;
            }
        }
    }

    return mark[val];
}

IsPrimeNumber(1000000000) takes 21s 758ms.

NOTE: Value might vary depend on hardware specifications.