Hi Reichart,
> This exact same problem in other languages would be just:
> Matlab: sum(rand(100,1)>1/3)
> C: for(int i=0,j=0;i<100;i++)j+=rand(3)>1;cout<<j;
> In Perl: $i=0;$j=0;for(;$i<100;$i++){$j+=rand(3)>1;}print $j
> Rebol: w: 100 print [loop w [w: w - first random [0 0 1]]]
>
> Does not look good for Rebol.
We can use the same algorithm as you have for Perl and C, and get something
like:
s: 0 n: 100 loop n[if 1 < Random 3 [s: s + 1]]s
Which is 47 bytes -- exactly the same as your C (actually C++) example And
shorter than your Perl, though longer than Joel's improvement.
(You ask that we count the Rebol [] as an extra 8 bytes overhead. Fair
enough, but shouldn't your C++ have a standard header include?)
Unlike Perl and C, Rebol needs an explicit type cast to go from true/false
to an integer, so to follow exactly the Perl/C++ examples we'd have to write
something like:
s: 0 n: 100 loop n[s: s + to-integer (1 < Random 3)]
or
s: 0 n: 100 loop n[s: s + pick[0 1](2 < Random 3)]
but they are longer.
Rebol "loses out" on sheer shortness here because it needs spaces between
operators, and its RAND function is spelt RANDOM. Some people would see these
as sufficient compensation for coming second in such a race.
An, of course, bear in mind that the entire Rebol environment needed to run
these programs is under 252K (Rebol/Core, Windows 3.1).
If I am allowed to have a slightly larger environment (though still way
smaller than C or Perl -- I don't know about Matlab), then I'd define:
L: :Loop
R: :Random
And roll my code home in 39 bytes:
s: 0 n: 100 L n[if 1 < R 3 [s: s + 1]]s
Sunanda.
--
To unsubscribe from this list, please send an email to
[EMAIL PROTECTED] with "unsubscribe" in the
subject, without the quotes.
