[Haskell-cafe] Calculating with list comprehension

[Dan Weston]

Keep in mind this is a *lexical* rewrite. In the generator rule x and e 
are not independent: x is a pattern (which introduces a bind variable) 
and e is an expression (with free variables, one of which may be bound by x)

After one application of the generator rule, we get (using a lambda 
expression instead of introducing a fresh function name f):

concatMap (\a -> [(a,b) | b <- [1..2]]) [1..3]

After another:

concatMap (\a -> concatMap (\b -> [(a,b)]) [1..2]) [1..3]

Note that the "a <-" and "b <-" map into \a -> and \b -> and bind the 
free variables a and b in the expression (a,b).

Dan


R J wrote:
> I can calculate non-nested list comprehensions without a problem, but am 
> unable to calculate nested comprehensions involving, for example, the 
> generation of a list of pairs where the first and separate elements are 
> drawn from two separate lists, as in:
> 
>    [(a, b) | a <- [1..3], b <- [1..2]]
> 
> How does one calculate the expansion of this list?  The two rules for 
> expanding list comprehensions are:
> 
> 1.  Generator rule:  [e | x <- xs, Q]  =  concat (map f xs)
>                                           where
>                                               f x = [e | Q]
> 
> 2.  Guard rule:      [e | p, Q]        =  if p then [e | Q] else []
> 
> 
> There is a third rule that I've seen on the Internet, not in an 
> authoritative text:
> 
>    [e | Q1 , Q2]     =  concat [ [e | Q 2] | Q1 ]
> 
> I don't understand what this third rule means, or whether it's relevant.
> 
> Concat and map are defined as:
> 
> concat           :: [[a]] -> [a]
> concat []        =  []
> concat (xs:xss)  =  xs ++ concat xss
> 
> map              :: (a -> b) -> [a] -> [b]
> map f []         =  []
> map f (x:xs)     =  f x : (map f xs)
> 
> Any help is appreciated.
> 
> 
> 
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