On Thu, 2004-06-10 at 09:45, Steve Wampler wrote:
...
> Steve Hunter has pointed out a solution found in an old issue
> of the Icon Newsletter (Issue 18, from 1985!):
...
> procedure anagram(s)
> local c, s1
>
> s1 := "" # start with the empty string
> every c := !cset(s) do # for every character in s
> every find(c,s) do # and every time it occurs in s
> s1 ||:= c # append one
> return s1
> end
Looking at this, one advantage it has is that avoids some
type coercions that occur in some of the other solutions
(such as mine). Here's adaptations of my solutions 1 and 3
taking advantage of lessons learned from the above:
Solution 1a:
procedure ssort(s)
s ? every (ns:="") ||:= (find(c := !cset(s)),c)
return ns
end
Solution 3a (dictonary order):
procedure ssort(s)
map(s) ? every (ns:="") ||:= s[find(!cset(map(s)))]
return ns
end
> And here are its times:
(fixed the typo above...)
>
> Case 1:
> weaver% RunTests
> ------------------ icon18 -----------------------
> 1.48s user 0.23s system 100% cpu 1.705 total
>
> Case 2:
> weaver% RunTests --infile=input2.txt --valid=valid3.txt
> ------------------ icon18 -----------------------
> 3.60s user 0.07s system 99% cpu 3.687 total
And the times for solution 1a are:
Case 1:
weaver% RunTests
------------------ Steve Wampler (1a) -----------------------
1.41s user 0.30s system 100% cpu 1.708 total
Case 2:
weaver% RunTests --infile=input2.txt --valid=valid3.txt
------------------ Steve Wampler (1a) -----------------------
3.50s user 0.11s system 99% cpu 3.610 total
--
Steve Wampler -- [EMAIL PROTECTED]
The gods that smiled on your birth are now laughing out loud.
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