Oh, it is a valid value (I think I implied this by saying they'll compile and you can even evaluate).  Just not useful in any given scenario (an inductive structure where you don't have terminals).
 

On Mon, Aug 17, 2015 at 10:57 PM, akash g <akaberto@gmail.com> wrote:

@Rein:

Perhaps I should have been a bit more clear.  There is no way to get a terminal value from said function. 


oddsFrom3 :: [Integer]
oddsFrom3 = map (+2) oddsFrom3

Try a head for it perhaps.

oddsFrom3 = map (+2) oddsFrom3

<=> ((head  oddsFrom3) + 2) : map (+2) ((tail  oddsFrom3) + 2)

<=> ((head (map (+2) oddsFrom3) + 2) : map (+2) ((tail  oddsFrom3) + 2)

Sure, it doesn't hang until you try to evaluate this (in lazy language evaluators).  However, for any inductive structure, there needs to be a (well any finite number of terminals) terminal (base case) which can be reached  from the starting state in a finite amount of computational (condition for termination).  Type sigs don't/can't guarantee termination.  If they don't have a terminal value, you'll never get to the bottom (bad pun intended) of it.

Take an infinite list as an example.

x a = a :  x a

Here, one branch of the tree (representing the list as a highly unbalanced tree where every left branch is of depth one at any given point).  If such a structure is not present, you can never compute it to a value and you'll have to infinitely recurse.

Try x a = x a ++ x a

And think of the getting the head from this.  You're stuck in an infinite loop.

You may also think of the above as a small BNF and try to see if termination is possible from the start state.  A vaguely intuitive way of looking at it for me, but meh, I might be missing something.

On Mon, Aug 17, 2015 at 10:23 PM, Rein Henrichs <rein.henrichs@gmail.com> wrote:

> The initial version which the OP posted doesn't have a terminal value. 

The point is that it doesn't need a terminal value. Infinite lists like oddsFrom3 and (repeat "foo") and (let xs = 1 : xs) are all perfectly valid Haskell values.

On Mon, Aug 17, 2015 at 6:17 AM Doug McIlroy <doug@cs.dartmouth.edu> wrote:

> > oddsFrom3 :: [Integer]
> > oddsFrom3 = 3 : map (+2) oddsFrom3
> >
> >
> > Thanks for your help.
>
> Try to expand a few steps of the recursion by hand e.g.:
>
>    3 : (map (+2) (3 : map (+2) (3 : map (+2) ...)))
>
>
> As you can see, the deeper you go more 'map (+2)' are applied to '3'.

Some more ways to describe the program, which may be useful:

As with any recursive function, assume you know the whole series and
then confirm that by verifying the inductive step. In this case
        oddsFrom3          = [3,5,7,9,11,...]
        map (+2) oddsFrom3 = [5,7,9,11,13,...]
voila
        oddsFrom3 = 3 : map (+2) oddsFrom3

Assuming we have the whole series, we see its tail is
computed from the whole by adding 2 to each element.
Notice that we don't actually have to know the values in the
tail in order to write the formula for the tail.

Yet another way to describe the program: the "output"  is taken
as "input". This works because the first element of the output,
namely 3, is provided in advance. Each output element can then
be computed before it is needed as input.

In an imperative language this would be done so:
        integer oddsFrom3[0:HUGE]
        oddsFrom3[0] := 3
        for i:=1 to HUGE do
                oddsFrom3[i] = oddsFrom3[i-1] + 2
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