在C中找到最接近的素数到无符号长整数（32位宽）的方法？

Question

I'm looking for a way to find the closest prime number. Greater or less than, it doesn't matter, simply the closest ( without overflowing, preferably. ) As for speed, if it can compute it in approximately 50 milliseconds on a 1GHz machine ( in software, running inside Linux ), I'd be ecstatic.

Answer 1

The largest prime gap in the range up to (2^32 - 1) is (335). There are (6542) primes less than (2^16) that can be tabulated and used to sieve successive odd values after a one-time setup. Obviously, only primes <= floor(sqrt(candidate)) need be tested for a particular candidate value.

Alternatively: The deterministic variant of the Miller-Rabin test , with SPRP bases: {2, 7, 61} is sufficient to prove primality for a 32-bit value. Due to the test's complexity (requires exponentiation, etc), I doubt it would be as fast for such small candidates.

Edit: Actually, if multiply/reduce can be kept to 32-bits in exponentiation (might need 64-bit support), the MR test might be better. The prime gaps will typically be much smaller, making the sieve setup costs excessive. Without large lookup tables, etc., you might also get a boost from better cache locality.

Furthermore: The product of primes {2, 3, 5, 7, 11, 13, 17, 19, 23} = (223092870). Explicitly test any candidate in [2, 23]. Calculate greatest common divisor: g = gcd(u, 223092870UL) . If (g != 1) , the candidate is composite. If (g == 1 && u < (29 * 29)) , the candidate ( u > 23 ) is definitely prime. Otherwise, move on to the more expensive tests. A single gcd test using 32-bit arithmetic is very cheap, and according to Mertens' (?) theorem, this will detect ~ 68.4% of all odd composite numbers.

Answer 2

UPDATE 2 : Fixed (in a heavy-handed way) some bugs that caused wrong answers for small n. Thanks to Brett Hale for noticing! Also added some asserts to document some assumptions.

UPDATE : I coded this up and it seems plenty fast enough for your requirements (solved 1000 random instances from [2^29, 2^32-1] in <100ms, on a 2.2GHz machine -- not a rigorous test but convincing nonetheless).

It is written in C++ since that's what my sieve code (which I adapted from) was in, but the conversion to C should be straightforward. The memory usage is also (relatively) small which you can see by inspection.

You can see that because of the way the function is called, the number returned is the nearest prime that fits in 32 bits, but in fact this is the same thing since the primes around 2^32 are 4294967291 and 4294967311.

I tried to make sure there wouldn't be any bugs due to integer overflow (since we're dealing with numbers right up to UINT_MAX); hopefully I didn't make a mistake there. The code could be simplified if you wanted to use 64-bit types (or you knew your numbers would be smaller than 2^32-256) since you wouldn't have to worry about wrapping around in the loop conditions. Also this idea scales for bigger numbers as long as you're willing to compute/store the small primes up to the needed limit.

I should note also that the small-prime-sieve runs quite quickly for these numbers (4-5 ms from a rough measurement) so if you are especially memory-starved, running it every time instead of storing the small primes is doable (you'd probably want to make the mark[] arrays more space efficient in this case)

#include <iostream>
#include <cmath>
#include <climits>
#include <cassert>

using namespace std;

typedef unsigned int UI;

const UI MAX_SM_PRIME = 1 << 16;
const UI MAX_N_SM_PRIMES = 7000;
const UI WINDOW = 256;

void getSMPrimes(UI primes[]) {
  UI pos = 0;
  primes[pos++] = 2;

  bool mark[MAX_SM_PRIME / 2] = {false};
  UI V_SM_LIM = UI(sqrt(MAX_SM_PRIME / 2));
  for (UI i = 0, p = 3; i < MAX_SM_PRIME / 2; ++i, p += 2)
    if (!mark[i]) {
      primes[pos++] = p;
      if (i < V_SM_LIM)
        for (UI j = p*i + p + i; j < MAX_SM_PRIME/2; j += p)
          mark[j] = true;
      }
  }

UI primeNear(UI n, UI min, UI max, const UI primes[]) {
  bool mark[2*WINDOW + 1] = {false};

  if (min == 0) mark[0] = true;
  if (min <= 1) mark[1-min] = true;

  assert(min <= n);
  assert(n <= max);
  assert(max-min <= 2*WINDOW);

  UI maxP = UI(sqrt(max));
  for (int i = 0; primes[i] <= maxP; ++i) {
    UI p = primes[i], k = min / p;
    if (k < p) k = p;
    UI mult = p*k;
    if (min <= mult)
      mark[mult-min] = true;
    while (mult <= max-p) {
      mult += p;
      mark[mult-min] = true;
      }
    }

  for (UI s = 0; (s <= n-min) || (s <= max-n); ++s)
    if ((s <= n-min) && !mark[n-s-min])
      return n-s;
    else if ((s <= max-n) && !mark[n+s-min])
      return n+s;

  return 0;
  }

int main() {
  UI primes[MAX_N_SM_PRIMES];
  getSMPrimes(primes);

  UI n;
  while (cin >> n) {
    UI win_min = (n >= WINDOW) ? (n-WINDOW) : 0;
    UI win_max = (n <= UINT_MAX-WINDOW) ? (n+WINDOW) : UINT_MAX;

    if (!win_min)
      win_max = 2*WINDOW;
    else if (win_max == UINT_MAX)
      win_min = win_max-2*WINDOW;

    UI p = primeNear(n, win_min, win_max, primes);
    cout << "found nearby prime " << p << " from window " << win_min << ' ' << win_max << '\n';
    }
  }

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You can sieve intervals in that range if you know primes up to 2^16 (there are only 6542 <= 2^16; you should go a bit higher if the prime itself could be greater than 2^32 - 1). Not necessarily the fastest way but very simple, and fancier prime testing techniques are really suited to much larger ranges.

Basically, do a regular Sieve of Eratosthenes to get the "small" primes (say the first 7000). Obviously you only need to do this once at the start of the program, but it should be very fast.

Then, supposing your "target" number is 'a', consider the interval [an/2, a+n/2) for some value of n. Probably n = 128 is a reasonable place to start; you may need to try adjacent intervals if the numbers in the first one are all composite.

For every "small" prime p, cross out its multiples in the range, using division to find where to start. One optimization is that you only need to start crossing off multiples starting at p*p (which means that you can stop considering primes once p*p is above the interval).

Most of the primes except the first few will have either one or zero multiples inside the interval; to take advantage of this you can pre-ignore multiples of the first few primes. The simplest thing is to ignore all even numbers, but it's not uncommon to ignore multiples of 2, 3, and 5; this leaves integers congruent to 1, 7, 11, 13, 17, 19, 23, and 29 mod 30 (there are eight, which map nicely to the bits of a byte when sieving a large range).

...Sort of went off on a tangent there; anyway once you've processed all the small primes (up till p*p > a+n/2) you just look in the interval for numbers you didn't cross out; since you want the closest to a start looking there and search outward in both directions.