Fast n choose k mod p for large n?

问题

What I mean by "large n" is something in the millions. p is prime.

I've tried http://apps.topcoder.com/wiki/display/tc/SRM+467 But the function seems to be incorrect (I tested it with 144 choose 6 mod 5 and it gives me 0 when it should give me 2)

I've tried http://online-judge.uva.es/board/viewtopic.php?f=22&t=42690 But I don't understand it fully

I've also made a memoized recursive function that uses the logic (combinations(n-1, k-1, p)%p + combinations(n-1, k, p)%p) but it gives me stack overflow problems because n is large

I've tried Lucas Theorem but it appears to be either slow or inaccurate.

All I'm trying to do is create a fast/accurate n choose k mod p for large n. If anyone could help show me a good implementation for this I'd be very grateful. Thanks.

As requested, the memoized version that hits stack overflows for large n:

std::map<std::pair<long long, long long>, long long> memo;

long long combinations(long long n, long long k, long long p){
   if (n  < k) return 0;
   if (0 == n) return 0;
   if (0 == k) return 1;
   if (n == k) return 1;
   if (1 == k) return n;

   map<std::pair<long long, long long>, long long>::iterator it;

   if((it = memo.find(std::make_pair(n, k))) != memo.end()) {
        return it->second;
   }
   else
   {
        long long value = (combinations(n-1, k-1,p)%p + combinations(n-1, k,p)%p)%p;
        memo.insert(std::make_pair(std::make_pair(n, k), value));
        return value;
   }  
}

回答1:

So, here is how you can solve your problem.

Of course you know the formula:

comb(n,k) = n!/(k!*(n-k)!) = (n*(n-1)*...(n-k+1))/k!

(See http://en.wikipedia.org/wiki/Binomial_coefficient#Computing_the_value_of_binomial_coefficients)

You know how to compute the numerator:

long long res = 1;
for (long long i = n; i > n- k; --i) {
  res = (res * i) % p;
}

Now, as p is prime the reciprocal of each integer that is coprime with p is well defined i.e. a^-1 can be found. And this can be done using Fermat's theorem a^p-1=1(mod p) => a*a^p-2=1(mod p) and so a^-1=a^p-2. Now all you need to do is to implement fast exponentiation(for example using the binary method):

long long degree(long long a, long long k, long long p) {
  long long res = 1;
  long long cur = a;

  while (k) {
    if (k % 2) {
      res = (res * cur) % p;
    }
    k /= 2;
    cur = (cur * cur) % p;
  }
  return res;
}

And now you can add the denominator to our result:

long long res = 1;
for (long long i = 1; i <= k; ++i) {
  res = (res * degree(i, p- 2)) % p;
}

Please note I am using long long everywhere to avoid type overflow. Of course you don't need to do k exponentiations - you can compute k!(mod p) and then divide only once:

long long denom = 1;
for (long long i = 1; i <= k; ++i) {
  denom = (denom * i) % p;
}
res = (res * degree(denom, p- 2)) % p;

EDIT: as per @dbaupp's comment if k >= p the k! will be equal to 0 modulo p and (k!)^-1 will not be defined. To avoid that first compute the degree with which p is in n*(n-1)...(n-k+1) and in k! and compare them:

int get_degree(long long n, long long p) { // returns the degree with which p is in n!
  int degree_num = 0;
  long long u = p;
  long long temp = n;

  while (u <= temp) {
    degree_num += temp / u;
    u *= p;
  }
  return degree_num;
}

long long combinations(int n, int k, long long p) {
  int num_degree = get_degree(n, p) - get_degree(n - k, p);
  int den_degree = get_degree(k, p);

  if (num_degree > den_degree) {
    return 0;
  }
  long long res = 1;
  for (long long i = n; i > n - k; --i) {
    long long ti = i;
    while(ti % p == 0) {
      ti /= p;
    }
    res = (res * ti) % p;
  }
  for (long long i = 1; i <= k; ++i) {
    long long ti = i;
    while(ti % p == 0) {
      ti /= p;
    }
    res = (res * degree(ti, p-2, p)) % p;
  }
  return res;
}

EDIT: There is one more optimization that can be added to the solution above - instead of computing the inverse number of each multiple in k!, we can compute k!(mod p) and then compute the inverse of that number. Thus we have to pay the logarithm for the exponentiation only once. Of course again we have to discard the p divisors of each multiple. We only have to change the last loop with this:

long long denom = 1;
for (long long i = 1; i <= k; ++i) {
  long long ti = i;
  while(ti % p == 0) {
    ti /= p;
  }
  denom = (denom * ti) % p;
}
res = (res * degree(denom, p-2, p)) % p;

回答2:

For large k, we can reduce the work significantly by exploiting two fundamental facts:

If p is a prime, the exponent of p in the prime factorisation of n! is given by (n - s_p(n)) / (p-1), where s_p(n) is the sum of the digits of n in the base p representation (so for p = 2, it's popcount). Thus the exponent of p in the prime factorisation of choose(n,k) is (s_p(k) + s_p(n-k) - s_p(n)) / (p-1), in particular, it is zero if and only if the addition k + (n-k) has no carry when performed in base p (the exponent is the number of carries).
Wilson's theorem: p is a prime, if and only if (p-1)! ≡ (-1) (mod p).

The exponent of p in the factorisation of n! is usually calculated by

long long factorial_exponent(long long n, long long p)
{
    long long ex = 0;
    do
    {
        n /= p;
        ex += n;
    }while(n > 0);
    return ex;
}

The check for divisibility of choose(n,k) by p is not strictly necessary, but it's reasonable to have that first, since it will often be the case, and then it's less work:

long long choose_mod(long long n, long long k, long long p)
{
    // We deal with the trivial cases first
    if (k < 0 || n < k) return 0;
    if (k == 0 || k == n) return 1;
    // Now check whether choose(n,k) is divisible by p
    if (factorial_exponent(n) > factorial_exponent(k) + factorial_exponent(n-k)) return 0;
    // If it's not divisible, do the generic work
    return choose_mod_one(n,k,p);
}

Now let us take a closer look at n!. We separate the numbers ≤ n into the multiples of p and the numbers coprime to p. With

n = q*p + r, 0 ≤ r < p

The multiples of p contribute p^q * q!. The numbers coprime to p contribute the product of (j*p + k), 1 ≤ k < p for 0 ≤ j < q, and the product of (q*p + k), 1 ≤ k ≤ r.

For the numbers coprime to p we will only be interested in the contribution modulo p. Each of the full runs j*p + k, 1 ≤ k < p is congruent to (p-1)! modulo p, so altogether they produce a contribution of (-1)^q modulo p. The last (possibly) incomplete run produces r! modulo p.

So if we write

n   = a*p + A
k   = b*p + B
n-k = c*p + C

we get

choose(n,k) = p^a * a!/ (p^b * b! * p^c * c!) * cop(a,A) / (cop(b,B) * cop(c,C))

where cop(m,r) is the product of all numbers coprime to p which are ≤ m*p + r.

There are two possibilities, a = b + c and A = B + C, or a = b + c + 1 and A = B + C - p.

In our calculation, we have eliminated the second possibility beforehand, but that is not essential.

In the first case, the explicit powers of p cancel, and we are left with

choose(n,k) = a! / (b! * c!) * cop(a,A) / (cop(b,B) * cop(c,C))
            = choose(a,b) * cop(a,A) / (cop(b,B) * cop(c,C))

Any powers of p dividing choose(n,k) come from choose(a,b) - in our case, there will be none, since we've eliminated these cases before - and, although cop(a,A) / (cop(b,B) * cop(c,C)) need not be an integer (consider e.g. choose(19,9) (mod 5)), when considering the expression modulo p, cop(m,r) reduces to (-1)^m * r!, so, since a = b + c, the (-1) cancel and we are left with

choose(n,k) ≡ choose(a,b) * choose(A,B) (mod p)

In the second case, we find

choose(n,k) = choose(a,b) * p * cop(a,A)/ (cop(b,B) * cop(c,C))

since a = b + c + 1. The carry in the last digit means that A < B, so modulo p

p * cop(a,A) / (cop(b,B) * cop(c,C)) ≡ 0 = choose(A,B)

(where we can either replace the division with a multiplication by the modular inverse, or view it as a congruence of rational numbers, meaning the numerator is divisible by p). Anyway, we again find

choose(n,k) ≡ choose(a,b) * choose(A,B) (mod p)

Now we can recur for the choose(a,b) part.

Example:

choose(144,6) (mod 5)
144 = 28 * 5 + 4
  6 =  1 * 5 + 1
choose(144,6) ≡ choose(28,1) * choose(4,1) (mod 5)
              ≡ choose(3,1) * choose(4,1) (mod 5)
              ≡ 3 * 4 = 12 ≡ 2 (mod 5)

choose(12349,789) ≡ choose(2469,157) * choose(4,4)
                  ≡ choose(493,31) * choose(4,2) * choose(4,4
                  ≡ choose(98,6) * choose(3,1) * choose(4,2) * choose(4,4)
                  ≡ choose(19,1) * choose(3,1) * choose(3,1) * choose(4,2) * choose(4,4)
                  ≡ 4 * 3 * 3 * 1 * 1 = 36 ≡ 1 (mod 5)

Now the implementation:

// Preconditions: 0 <= k <= n; p > 1 prime
long long choose_mod_one(long long n, long long k, long long p)
{
    // For small k, no recursion is necessary
    if (k < p) return choose_mod_two(n,k,p);
    long long q_n, r_n, q_k, r_k, choose;
    q_n = n / p;
    r_n = n % p;
    q_k = k / p;
    r_k = k % p;
    choose = choose_mod_two(r_n, r_k, p);
    // If the exponent of p in choose(n,k) isn't determined to be 0
    // before the calculation gets serious, short-cut here:
    /* if (choose == 0) return 0; */
    choose *= choose_mod_one(q_n, q_k, p);
    return choose % p;
}

// Preconditions: 0 <= k <= min(n,p-1); p > 1 prime
long long choose_mod_two(long long n, long long k, long long p)
{
    // reduce n modulo p
    n %= p;
    // Trivial checks
    if (n < k) return 0;
    if (k == 0 || k == n) return 1;
    // Now 0 < k < n, save a bit of work if k > n/2
    if (k > n/2) k = n-k;
    // calculate numerator and denominator modulo p
    long long num = n, den = 1;
    for(n = n-1; k > 1; --n, --k)
    {
        num = (num * n) % p;
        den = (den * k) % p;
    }
    // Invert denominator modulo p
    den = invert_mod(den,p);
    return (num * den) % p;
}

To calculate the modular inverse, you can use Fermat's (so-called little) theorem

If p is prime and a not divisible by p, then a^(p-1) ≡ 1 (mod p).

and calculate the inverse as a^(p-2) (mod p), or use a method applicable to a wider range of arguments, the extended Euclidean algorithm or continued fraction expansion, which give you the modular inverse for any pair of coprime (positive) integers:

long long invert_mod(long long k, long long m)
{
    if (m == 0) return (k == 1 || k == -1) ? k : 0;
    if (m < 0) m = -m;
    k %= m;
    if (k < 0) k += m;
    int neg = 1;
    long long p1 = 1, p2 = 0, k1 = k, m1 = m, q, r, temp;
    while(k1 > 0) {
        q = m1 / k1;
        r = m1 % k1;
        temp = q*p1 + p2;
        p2 = p1;
        p1 = temp;
        m1 = k1;
        k1 = r;
        neg = !neg;
    }
    return neg ? m - p2 : p2;
}

Like calculating a^(p-2) (mod p), this is an O(log p) algorithm, for some inputs it's significantly faster (it's actually O(min(log k, log p)), so for small k and large p, it's considerably faster), for others it's slower.

Overall, this way we need to calculate at most O(log_p k) binomial coefficients modulo p, where each binomial coefficient needs at most O(p) operations, yielding a total complexity of O(p*log_p k) operations. When k is significantly larger than p, that is much better than the O(k) solution. For k <= p, it reduces to the O(k) solution with some overhead.

来源：https://stackoverflow.com/questions/10118137/fast-n-choose-k-mod-p-for-large-n

标签

c++

algorithm

modular

binomial-coefficients