[Haskell-beginners] Tying the knot

[Heinrich Apfelmus]

aditya siram wrote:
> My brain turns into strange braid when I see this kind of thing. I
> don't quite understand it and I've never used it in real world code
> but I'll try and explain anyway. Caveat emptor.
> 
> [..]
> 
> Now look at 'n' and imagine it was a memory location. Mentally
> substitute some hex address (like 0x0000) if it makes it easier.

Another way to look at it is to observe that the result  n  does not 
depend on the input  n , even though the notation might suggest otherwise.

To see that, let  _|_  denote an expression that is  undefined , i.e. 
raises an error when you try to evaluate at it. Using the definition

     data Tree a = Branch (Tree a) (Tree a) | Leaf a

     label n (Branch a b) = (na+nb, Branch a' b')
         where
         (na,a') = label n a
         (nb,b') = label n b
     label n (Leaf _)     = (1, Leaf n)

we have

     label _|_ (Branch (Leaf 'c') (Leaf 'd'))
   = (na+nb, Branch a' b')
     where (na,a') = label _|_ (Leaf 'c')
           (nb,b') = label _|_ (Leaf 'd')
   = (1+1, Branch (Leaf _|_) (Leaf _|_))

So, even though the argument  n  is undefined, the function  label 
still produces a partially defined result.



Regards,
Heinrich Apfelmus

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