[Haskell-beginners] oddsFrom3 function

[Rein Henrichs]

> 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 at 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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