The problem is to find the n-th Catalan number mod m, where m is NOT prime, m = (10^14 + 7). Here are the list of methods that I have tried: (max N = 10,000)
- Dynamic programming for table look-up, too slow
- Use Catalan formula
ncr(2*n, n)/(n + 1), again it wasn’t fast enough due to thencrfunction, can’t speed up using exponentiation squaring becausemis not prime. - Hardcode a table of pre-generated
Catalans, but it failed due to the file size limit. - Recurrence relation
C(i,k) = C(i-1,k-1) + C(i-1,k), this is way too slow
So I wonder is there any other faster algorithm to find the n-th Catalan number that I’m not aware of?
Using Dynamic Programming
void generate_catalan_numbers() {
catalan[1] = 1;
for (int i = 2; i <= MAX_NUMBERS; i++) {
for (int j = 1; j <= i - 1; j++) {
catalan[i] = (catalan[i] + ((catalan[j]) * catalan[i - j]) % MODULO) % MODULO;
}
catalan[i] = catalan[i] % MODULO;
}
}
Using original formula
ull n_choose_r(ull n, ull r) {
if (n < r)
return 0;
if (r > n/2) {
r = n - r;
}
ull result = 1;
ull common_divisor;
for (int i = 1; i <= r; ++i) {
common_divisor = gcd(result, i);
result /= common_divisor;
result *= (n - i + 1) / (i / common_divisor);
}
return result;
}
Using recurrence relation
ull n_choose_r_relation(ull n, ull r) {
for (int i = 0; i <= n + 1; ++i) {
for (int k = 0; k <= r && k <= i; ++k) {
if (k == 0 || k == i) {
ncr[i][k] = 1;
}
else {
ncr[i][k] = (ncr[i - 1][k - 1] + ncr[i - 1][k]) % MODULO;
}
}
}
return ncr[n][r];
}
Easy, peasy. Compute the prime factors of the binomial coefficient. A simple task using a sieve. I won’t get into the rest of it, but a powermod computation is trivial, and you don’t even need a divide.
For N = 10000, I get 42224403014400 in short order.
But, if you want the full number itself, the 10000’th Catalan number itself is …