From: Victor Reyes <victor.reyes@...>
Date: 2008-07-14T21:53:53+09:00
Subject: Re: Factorials

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On Sun, Jul 13, 2008 at 8:04 PM, Serabe <serabe@gmail.com> wrote:

> The easy way:
>
> def factorial(n)
>  if n==1
>    1
>  else
>    n*factorial(n-1)
>  end
> end
>
> puts factorial(5)
>
> It prints 120.
>
> A more ruby way method:
>
> def factorial(n)
>  (1..n).inject(1) {|product, n| product * n }
> end
>
> puts factorial(5)
>
> It does also print 120.
>
> Regards,
>
> Serabe
>
> --


> http://www.serabe.com
>
>

The "Rubish" version of your factorial solutions appear to be much faster!
Although I find it more confusing than the "traditional" recursive solution.
I guess my problem is understanding the role of the *inject* method.
I use 100 instead of 5 to make it run a bit longer.

def factorial(n)
 if n==1
   1
 else
   n*factorial(n-1)
 end
end

puts factorial(100)

timex ruby factorial01
93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000

real 0.05
user 0.00
sys  0.00
====================================================
def factorial(n)
 (1..n).inject(1) {|product, n| product * n }
end

puts factorial(100)

timex ruby factorial02
93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000

real 0.02
user 0.00
sys  0.00


Regards,

Victor

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