OpenEuphoria

Re: Too complex!!!!!!!!!

• Posted by "Carl R. White" Jun 11, 1997
• 696 views

On Mon, 9 Jun 1997, Lucius L Hilley III wrote:

> I created this code.
> It works great.
> BUT I can't figure out how to undo it.

[Space save snip]

> HERE is the Function.

[Space save snip]

> function Flatten(sequence s)
>   sequence temp
>   sequence sh
>
>   sh = {length(s)}
>   for a = 1 to sh[1] do
>     if sequence(s[a]) then
>       temp = Flatten(s[a])
>       sh = sh & temp
>     else
>       sh = sh & {-1, s[a]}
>     end if
>   end for
>   return sh
> end function
>
> --This one of my Vain attempts
> --function DeFlatten(sequence s)
> --  atom a, p
> --  sequence temp
> --
> --  a = 2
> --  p = 1
> --  temp = repeat({}, s[1])
> --  while a <= length(s) do
> --    if s[a] = -1 then
> --      a = a + 1
> --      temp[p] = s[a]
> --      p = p + 1
> --    else
> --      temp[p] = repeat({}, s[a])
> --    end if
> --    a = a + 1
> --  end while
> --  return temp
> --end function

I don't think it *can* be undone, because both

        {{4}, 5 , {6, 7} ,8}
and
        {4, {5, 6, 7}, 8}
compact to
        {4, 5, 6, 7, 8}
.

How do you know which one was the original, when it's blatantly
obvious they both were (and several other possible combinations too)?

It's like saying sqrt(9) = 3 (which is what the function returns) but
(-3) * (-3) = 9, so how do you know what created the 9 in the first place?
Was it 3 or -3?

Sorry to burst the bubble... 

--
Carl R White   | e-mail...:                    crwhite- at -comp.brad.ac.uk
               | finger...:             crwhite- at -dcsun1.comp.brad.ac.uk
               | web......: http://www.student.comp.brad.ac.uk/~crwhite/
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