Блог пользователя Eramoni

Fastest way to get factorial modulo a prime
Автор Eramoni, 6 лет назад, По-английски

I have learnt from this site that n! modulo p where p is a prime, can be calculated using Wilson's Theorem and FFT with complexity sqrt(p).(log p)^3

There is also a problem using the idea in SPOJ. link

But I am not being able to implement the idea. If anyone have any implementation of the problem please feel free to share.

Thanks for your time. Happy Coding.

  • Проголосовать: нравится
  • +43
  • Проголосовать: не нравится

»
6 лет назад, # |
  Проголосовать: нравится +36 Проголосовать: не нравится

It has many parts. Try implementing / debugging one part at a time maybe?
First part is calculating (x + 1)(x + 2)...(x + m). Try this here: 960G - Bandit Blues

The later one is the harder part: Multipoint Evaluation. You can try submitting here: POLYEVAL -Evaluate the polynomial. It has weak judge data that allows sub optimal solutions.

Again, for that multipoint evaluation you need polynomial inverse. You can test that here: 438E - The Child and Binary Tree (which also needs square root of a polynomial, also has good editorial on how to do that).

If you don't know how multipoint evaluation is done, read this: On Fast Fourier Transform

→ Ответить
  • »
    »
    6 лет назад, # ^ |
      Проголосовать: нравится +16 Проголосовать: не нравится

    Hmmm, but multipoint evaluation is , isn't it?

    → Ответить
  • »
    »
    6 лет назад, # ^ |
    ← Rev. 3   Проголосовать: нравится +33 Проголосовать: не нравится

    Your approach could be improved to , where .

    That is, if you want to calculate fn(d), fn(2d), ..., fn(nd) where , you may calculate fn / 2(d), fn / 2(2d), ..., fn / 2(nd / 2) first, use FFT and binomial coefficient to calculate fn / 2((n / 2 + 1)d), fn / 2((n / 2 + 2)d), ..., fn / 2(nd), fn / 2(d + n / 2), fn / 2(2d + n / 2), ..., fn / 2(nd + n / 2) then, and finally obtain fn(x) by fn / 2(x)·fn / 2(x + n / 2).

    If you need more details, please refer to 階乗 mod 素数 — Min_25's memo.

    → Ответить
»
6 лет назад, # |
← Rev. 3   Проголосовать: нравится +38 Проголосовать: не нравится

Relatedly, this is also how you get the fastest (in terms of asymptotics) deterministic integer factorization algorithm: if n is a composite it has a prime factor that is at most n1 / 2, so compute ⌈ n1 / 2⌉! modulo n in O(n1 / 4polylog(n)) time and take the GCD of the result with n.

→ Ответить
»
6 лет назад, # |
  Проголосовать: нравится +8 Проголосовать: не нравится

Mainly for future reference: There's also a solution in that uses generating functions to efficiently evaluate a polynomial of degree d at {0, 2k, 2·2k, 3·2k, ..., (m - 1)·2k} in .

Let f be a polynomial of degree d, then the sequence (an)n defined by an = f(n) satisfies a linear recurrence with characteristic polynomial (X - 1)d + 1. From this, we get a generating function

where P0 is a polynomial of degree at most d + 1 that can be computed by

We extend the left side by (X + 1)d + 1 and get

If we decompose the numerator into even and odd terms and let P1(X2) denote the even terms of the numerator, then we get by considering only even terms

i.e.

so we found a generating function for f(0), f(2), f(4), .... By repeditively defining

we get generating functions

so we can compute the first m terms on the left by power series inversion.

If we want to compute N!, then we pick , let , take

F(X) = X·(X - 1)·...·(X - 2k + 1)

with d = 2k and initial values

F(0) = F(1) = ... = F(2k - 1) = 0, F(2k) = (2k)!, F(2k + 1) = (2k + 1)!

and use the above algorithm to compute F(2k), F(2·2k), ..., F(m·2k). Then multiply the results together with the numbers from m2k + 1 to N. The total runtime is .

→ Ответить
  • »
    »
    5 лет назад, # ^ |
    ← Rev. 2   Проголосовать: нравится +3 Проголосовать: не нравится

    Can you share your implementation of this problem from SPOJ as I am not able to implement my own. I saw that you got AC for this Factmodp problem on Spoj.

    → Ответить