Variable overflow

Hi Kean,

the problem is that you are using (32bit) integers which are (naturally) limited in their size.
Change your code so as it uses floats and off you go...

(defun factorial (n)
   if( zerop( n ) then
      1.0
   else
      n*factorial( n-1)
   ) ; if
)

But - a factorial of 5000 is such an incredible number that even floats won't do anymore - 170 is the last one that works, after that you get 'inf'
(factorial 170)
7.257416e+306
1>
(factorial 171)
inf
1>

Just outta curiosity,- why would anybody want to compute the factorial of 5000???

Max

You can just unroll the factorials and reduce the formula to get the operations down to r (resp. n-r, whichever is smaller) multiplications and divisions (r-1 if you also skip the last division by 1). With this you can calculate the number of combinations with n=5000 and r<=161 (or >=4839) before reaching number overflow. For n<1030 it will not overflow for any r.

procedure( combinations_r_outof_n( n r )
    let( (res)
        ;C(n,r) = (n!) / (r! * (n-r)!)
        ;       = Prod((n-r+1)..n) / Prod(1..r)
        cond(
            ( n<r  || n<0   nil )
            ( n==r || n<=1  1   )
            ( t
                ; swap sides
                when( 2*r > n
                    r = n-r
                ); when

                res = 1.0
                for(x n-r+1 n
                    res = res * x / r
                    r--
                ); for
                when(res<2147483647  res=fix(res))
                res
            )
        )
    ); let
); procedure

Hi Max,

I was doing something to count the possible combinations of 2 standard cells abutment in a pool of 5000 standard cells. 
This require me to code the equation ⁿC_r.

Hi Henker,

Your solution works well for me. 

Thanks so much for the simplified algorithm.

KS.