[Haskell-beginners] simplifying algebraic expression
Alexander.Vladislav.Popov
alexander.vladislav.popov at gmail.com
Wed Sep 1 05:19:08 EDT 2010
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.
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