[Haskell-beginners] Re: [Haskell-cafe] running and understanding a lifting program

[Patrick LeBoutillier]

Patrick,

I found this program interesting and decided to spend a bit of time on
it, even though I'm still a newbie.
I did a few things to simplify the code, here are some comments:

1) I chose to rename the arithmetic functions in the Number class
instead of trying to overload the "real" ones, I'm not that good at
Haskell yet...

2) The program had some errors, namely I think the definition of the
Point type should be:

      data Point a = Point a a

     to allow for different types of Points.

3) The Points class seemed useless in the end, I simply defined the
dist function at the top level.

4) If you import Control.Monad, it makes functions (and therefore
"Changing v") into a Monad (maybe my terminology is off here...) and
allows you to use the general liftM and liftM2 lifting functions
instead of defining your own.


Here's the complete program:

{-# LANGUAGE TypeSynonymInstances, FlexibleInstances #-}

data Point a = Point { x ::a, y :: a }
type Time = Float

-- Functor Changing, which adds time parameter t to its input value.
-- For example, Changing Float indicates a changing floating number
-- (i.e. a function of time).
type Changing v = Time -> v

-- Lifting functions
lift1 op a = \t -> op (a t)
lift2 op a b = \t -> op (a t) (b t)

class Number a where
 add, sub, mul :: a -> a -> a
 square, squareRoot :: a -> a
 square a = a `mul` a

instance Number Float where
  add = (+)
  sub = (-)
  mul = (*)
  squareRoot = sqrt

instance Number (Changing Float) where
  add = lift2 add
  sub = lift2 sub
  mul = lift2 mul
  squareRoot = lift1 squareRoot

-- The distance operation is defined as follow
dist :: Number a => Point a -> Point a -> a
dist a b = squareRoot $ square((x a) `sub` (x b)) `add` square ((y a)
`sub` (y b))

-- Running the code
-- If p1 and p2 are two 2D static points,
-- their distance d is calculated as follows:
p1, p2 :: Point Float
p1 = Point 3.4 5.5
p2 = Point 4.5 4.5

-- distance between p1 and p2 --> 1.55
d = dist p1 p2

-- For 2D moving points mp1 and mp2, their distance md,
-- which is a function of time, is calculated as follows:
mp1, mp2 :: Point (Changing Float)
mp1 = Point (\t -> 4.0 + 0.5 * t) (\t -> 4.0 - 0.5 * t)
mp2 = Point (\t -> 0.0 + 1.0 * t) (\t -> 0.0 - 1.0 * t)
--  distance between mp1 and mp2
md = dist mp1 mp2
-- distance md for time 2 ----> 5.83


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Patrick

-- 
=====================
Patrick LeBoutillier
Rosemère, Québec, Canada