Question

I attempted Project Euler’s problem 10 using the very easy algorithm and the running time looks like hours. So I googled for an efficient algorithm and found this by Shlomif Fish.
The code is reproduced below:

int main(int argc, char * argv[])
{
    int p, i;
    int mark_limit;
    long long sum = 0;

    memset(bitmask, '\0', sizeof(bitmask));
    mark_limit = (int)sqrt(limit);

    for (p=2 ; p <= mark_limit ; p++)
    {
        if (! ( bitmask[p>>3]&(1 << (p&(8-1))) ) )
        {
            /* It is a prime. */
            sum += p;
            for (i=p*p;i<=limit;i+=p)
            {
                bitmask[i>>3] |= (1 << (i&(8-1)));
            }
        }
    }
    for (; p <= limit; p++)
    {
        if (! ( bitmask[p>>3]&(1 << (p&(8-1))) ) )
        {
            sum += p;
        }
    }

I have problems understanding the code. Specifically, how does this bit shifting code able to determine whether a number is prime or not.

   if (! ( bitmask[p>>3]&(1 << (p&(8-1))) ) )
        {
            /* It is a prime. */
            sum += p;
            for (i=p*p;i<=limit;i+=p)
            {
                bitmask[i>>3] |= (1 << (i&(8-1)));
            }
        }

Can someone please explain this code block to me, especially this part ( bitmask[p>>3]&(1 << (p&(8-1)? Thank you very much.

Answer 1

The code is a modified Sieve of Eratosthenes. He is packing one number into one bit: 0 = prime, 1 = composite. The bit shifting is to get to the correct bit in the byte array.

 bitmask[p>>3]

is equivalent to

 bitmask[p / 8]

which selects the correct byte in the bitmask[] array.

(p&(8-1))

equals p & 7, which selects the lower 3 bits of p. This is equivalent to p % 8

Overall we are selecting bit (p % 8) of byte bitmask[p / 8]. That is we are selecting the bit in the bitmask[] array which represents the number p.

The 1 << (p % 8) sets up a 1 bit correctly located in a byte. This is then AND’ed with the bitmask[p / 8] byte to see if that particular bit is set or not, thus checking whether p is a prime number.

The overall statement equates to if (isPrime(p)), using the already completed part of the sieve to help extend the sieve.