[Haskell-beginners] Sankey Diagram with monads

[Adrian May]

Thanks, but I still don't know how to fix it.

In the meantime, I'm struggling with something more basic. I plan to write
a basic monad that puts diagrams on top of each other, then I'll let State
take care of pushing the origin and angle along (turtle style). But I'm
already stuck on that basic monad.

It's >>= has to explicitly use the fact that each monad has a journey there
and a journey back. The thing on the right of >>= will be inserted in
between them.  But I always get either "something is a rigidly bound type
variable" or "Monad should have kind * -> *"

I just don't know how this is supposed to work.

I want the monad to contain two trails, in the sense of the Diagrams
module. If I bind two of them together, for the time being, I'll just stick
them on top of each other (at least I think the State monad will rescue me
from that.) I have no particular reason to tell the thing on the right of
>>= about the thing on the left. Neither do I have a reason for >>= to be
polymorphic.

Right now I'm thinking that I'll have to define a class for things that
provide a journey there and a journey back. I'd rather not, because there's
only one of them but I can't seem to restrict the game any other way, but
this way isn't helping either.

Ideally I'd be able to write something like this:

data Sankey = Sankey {there, back :: Trail R2}

instance Monad Sankey where
  return t b = Sankey t b
  l@(Sankey t b) >>= f = let (Sankey t' b') = f l in
     Sankey (t <> t') (b <> b')

although I have no reason to pass l to f. But the compiler barfs anyway. I
feel that Haskell is more complicated than what I'm trying to do. Under
duress I tried:

class Pic a where
  there :: a -> Trail R2
  back  :: a -> Trail R2

data Trails p = Trails p

instance (Pic p) => Monad (Trails p) where
  return = Trails
  (Trails l) >>= f = let (Trails r) = f l in
    ((there l <> there r),(back r <> back l))


But it doesn't like that either. What am I missing?

Adrian.














On 31 May 2013 23:42, Brandon Allbery <allbery.b at gmail.com> wrote:

> On Fri, May 31, 2013 at 11:06 AM, Adrian May <
> adrian.alexander.may at gmail.com> wrote:
>
>> Well I figured out that I should be using the State monad, but it seems
>> not to be behaving like most of the tutorials on the web. Did the syntax
>> change? ....
>>
>
> mtl-2.x changed all the base monads (State, Reader, Writer, etc.) from
> standalone to being transformers atop an Identity monad; this cleans up the
> implementation considerably (since we don't have almost exactly the same
> code for both the standalone and transformer versions, but means that all
> uses of the standalone constructors must be replaced with functions that
> build the appropriate transformer. (e.g. State becomes state, unless you
> want to spell it out as StateT Identity.)
>
> --
> brandon s allbery kf8nh                               sine nomine
> associates
> allbery.b at gmail.com
> ballbery at sinenomine.net
> unix, openafs, kerberos, infrastructure, xmonad
> http://sinenomine.net
>
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