Hi.

 

Please, tell me where I'm wrong and how to improve my approach.

 

I'm trying to simplify algebraic expressions this way:

 

import Data.Ratio

 

data Func = Const (Ratio Int)

         | Pow (Ratio Int)

         | Add Func Func

         | Mul Func Func

         

instance Show Func where

   show (Const n) = "(" ++ show n ++ ")"

   show (Pow n) | n == 0 = "1"

                | n == 1 = "x"

                | otherwise = "(x**(" ++ show n ++ "))"

   show (Add t1 t2) ="(" ++ (show t1) ++ "+" ++ (show t2) ++ ")"

   show (Mul t1 t2) ="(" ++ (show t1) ++ "*" ++ (show t2) ++ ")"

 

 

deriv (Const _) = Const 0

deriv (Pow 1) = Const 1

deriv (Pow n) = Const n `Mul` Pow (n-1)

deriv (Add a b) = deriv a `Add` deriv b

deriv (Mul a b) = Add (deriv a `Mul` b) (a `Mul` deriv b)

 

p0 = Const 1

p1 = p0 `Add` (Mul (Pow 1) (Const 2))

p2 = p1 `Add` (Mul (Pow 2) (Const 3))

 

 

s rdc (Const x) = Const x

s rdc (Pow 0) = Const 1

s rdc (Pow x) = Pow x

s rdc (Add (Const a) (Const b)) = Const (a+b)

s rdc (Mul (Const 0) _) = Const 0

s rdc (Mul _ (Const 0)) = Const 0

s rdc (Mul (Const a) (Const b)) = Const (a*b)

s rdc (Mul (Pow n) (Pow m)) = Pow (n+m)

 

s rdc (Add x (Const 0)) =  rdc x

s rdc (Add (Const 0) x) =  rdc x

s rdc (Mul (Const m) (Mul (Const n) x)) = rdc $ Mul (Const (n*m)) (rdc x)

s rdc (Mul x (Const 1)) =  rdc x

s rdc (Mul x (Const a)) =  rdc $ Mul (Const a) (rdc x)

s rdc (Mul (Const 1) x) =  rdc x

s rdc (Mul x (Add a b)) = Mul (rdc x) (rdc a) `Add` Mul (rdc x) (rdc b)

s rdc (Mul (Add a b) x) = Mul (rdc a) (rdc x) `Add` Mul (rdc b) (rdc x)

s rdc (Mul a b) =  rdc a `Mul` rdc b

s rdc (Add a b) =  rdc a `Add` rdc b

 

fix f = f (fix f)

 

The result I got is :

 

*Main> fix s $ deriv p2

(((2 % 1)+(0 % 1))+(((6 % 1)*x)+(0 % 1)))

 

instead of the anticipated expression ((2 % 1)+((6 % 1)*x)).

And worst of all,  I must apply (fix s) repeatedly to achieve correct answer:

 

*Main> fix s $ fix s $ deriv p2

((2 % 1)+((6 % 1)*x)).

 

I'll be very much appriciated for any help and useful links.