Reverse factorial

Question

Well, we all know that if N is given it\'s easy to calculate N!. But what about the inverse?

N! is given and you are about to find N - Is that possible ? I\'m curio

Answer 1

1. Set X=1.
2. Generate F=X!
3. Is F = the input? If yes, then X is N.
4. If not, then set X=X+1, then start again at #2.

You can optimize by using the previous result of F to compute the new F (new F = new X * old F).

It's just as fast as going the opposite direction, if not faster, given that division generally takes longer than multiplication. A given factorial A! is guaranteed to have all integers less than A as factors in addition to A, so you'd spend just as much time factoring those out as you would just computing a running factorial.

Answer 2

Based on:

Full inverted factorial valid for x>1

Use the suggested calculation. If factorial is expressible in full binary form the algorithm is:

Suppose input is factorial x, x=n!

1. Return 1 for 1
2. Find the number of trailing 0's in binary expansion of the factorial x, let us mark it with t
3. Calculate x/fact(t), x divided by the factorial of t, mathematically x/(t!)
4. Find how many times x/fact(t) divides t+1, rounded down to the nearest integer, let us mark it with m
5. Return m+t

__uint128_t  factorial(int  n);

int invert_factorial(__uint128_t fact)
{
    if (fact == 1) return 1;

    int t = __builtin_ffs(fact)-1;
    int res = fact/factorial(t);

    return t + (int)log(res)/log(t+1);
}

128-bit is giving in on 34!

Answer 3

int p = 1,i;
//assume variable fact_n has the value n!
for(i = 2; p <= fact_n; i++) p = p*i;
//i is the number you are looking for if p == fact_n else fact_n is not a factorial

I know it isn't a pseudocode, but it's pretty easy to understand

Answer 4

If you do not know whether a number M is N! or not, a decent test is to test if it's divisible by all the small primes until the Sterling approximation of that prime is larger than M. Alternatively, if you have a table of factorials but it doesn't go high enough, you can pick the largest factorial in your table and make sure M is divisible by that.

Answer 5

If you have Q=N! in binary, count the trailing zeros. Call this number J.

If N is 2K or 2K+1, then J is equal to 2K minus the number of 1's in the binary representation of 2K, so add 1 over and over until the number of 1's you have added is equal to the number of 1's in the result.

Now you know 2K, and N is either 2K or 2K+1. To tell which one it is, count the factors of the biggest prime (or any prime, really) in 2K+1, and use that to test Q=(2K+1)!.

For example, suppose Q (in binary) is

1111001110111010100100110000101011001111100000110110000000000000000000

(Sorry it's so small, but I don't have tools handy to manipulate larger numbers.)

There are 19 trailing zeros, which is

10011

Now increment:

1: 10100
2: 10101
3: 10110 bingo!

So N is 22 or 23. I need a prime factor of 23, and, well, I have to pick 23 (it happens that 2K+1 is prime, but I didn't plan that and it isn't needed). So 23^1 should divide 23!, it doesn't divide Q, so

N=22