Re: [Haskell-beginners] Sankey Diagram with monads

31 May 2013

      It got a lot easier when I forgot all about the monads:

type SaState = ( P2, CircleFrac, Double )
type SaPic =   ( (Trail R2), (Trail R2) )
type SaWorld = ( SaPic, SaState )

type SaAct = SaState -> SaWorld

bigbang :: Double -> SaWorld
bigbang w = ((mempty,mempty), (origin,0,w))

(>%) :: SaWorld -> SaAct -> SaWorld
((t1,b1),s1) >% f = let ((t2,b2),s2) = f s1 in ((t1<>t2,b2<>b1),s2)

(>%=) :: SaAct -> SaAct -> SaAct
f >%= g = \s1 ->
  let ((t2,b2),s2) = f s1 in
  let ((t3,b3),s3) = g s2 in
  ((t2<>t3,b3<>b2),s3)

saVia :: Double -> SaAct
saVia l (p,a,w) =
  (
    ( hrule l # translateX (l/2) # translateY (w/2) # rotate a
    , hrule (-l) # translateX (l/2) # translateY (-w/2) # rotate a
    )
    ,
    (p .+^ (unitX # scale l # rotate a),a,w)
  )

saTo :: SaAct
saTo (p,a,w) =
  (
    ( hrule   w  # translateX (-w/2) # rotate (-0.125 :: CircleFrac) #
scale 0.7071 # translateX (w/2) # rotate a # translate (origin .-. p)
    , hrule (-w) # translateX (-w/2) # rotate ( 0.125 :: CircleFrac) #
scale 0.7071 # translateX (w/2) # rotate a # translate (origin .-. p)
    )
    ,
    (p,a,0)
  )

saTurn :: Double -> CircleFrac -> SaAct
saTurn r a' (p,a,w) = let (outr, inr, qu) = if a'>=0 then (r, -w-r,
-0.25::CircleFrac) else (-w-r, r, 0.25::CircleFrac) in
  (
    ( arc' outr (a+qu) (a+a'+qu)  # translate (unitY # rotate (a+a' )#
scale w)
    , arc' inr  (a+a'+qu) (a+qu) # translate (unitY # rotate (a+a' )# scale
w)
    )
    ,
    (p,a+a',w)
  )

saSplit :: [(Double, SaAct)] -> SaAct
saSplit fs (p,a,w) =
let (placed,_) = ( foldl ( \(l,t) -> \(i,f) -> ( l++[( ( p .+^ (unitY #
rotate a # scale (((t+i/2)-0.5)*w)), a, w*i) ,f )],t+i) ) ([],0) fs ) in
let ws = map (\(b,f)-> f b) placed in
((foldl (<>) mempty (map (\((tt,bb),_)->tt<>bb) ws),mempty),(origin,0,0))

p3 = bigbang 5 >% (saVia 10 >%= (saSplit
[ (0.80,(saVia 10 >%= saTo))
, (0.20,saTurn 1 (-0.25) >%= (saVia 2 >%= (saSplit
[ (0.10,saTurn 1 (0.25) >%= (saVia 5 >%= saTo))
, (0.80,(saVia 3 >%= saTo))
, (0.10,saTurn 1 (-0.25) >%= (saVia 5 >%= saTo))
] )))
] ))

pic3 = let p = fst $ p3 in
(strokeT $ close $ (fst p) <> (snd p) ) # fc red

But still I feel I'm missing something.

Adrian.

On 1 June 2013 00:31, Adrian May  wrote:
...
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.
...
...
= about the thing on the left. Neither do I have a reason for >>= to be
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
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  wrote:
...
On Fri, May 31, 2013 at 11:06 AM, Adrian May <
adrian.alexander.may@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@gmail.com
ballbery@sinenomine.net
unix, openafs, kerberos, infrastructure, xmonad
http://sinenomine.net
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