Wed Sep 1 06:36:39 EDT 2010

```It has given me an infinite loop and out-of-memory exception. I tried
different variations but not found the valid one and don't know what to do
next.

2010/9/1 jean verdier <verdier.jean at gmail.com>

> I think that you need to reduce inner expressions before simplification.
>
> s rdc (Add a b) =  rdc a `Add` rdc b
>
> will not reduce the Add result. What you need is something like
>
> s rdc (Add a b) =  rdc (rdc a `Add` rdc b)
>
> that might not terminate but is your target.
>
>
> On Wed, 2010-09-01 at 15:19 +0600, Alexander.Vladislav.Popov wrote:
> > 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
> >
> > *Main> fix s \$ fix s \$ deriv p2
> > ((2 % 1)+((6 % 1)*x)).
> >
> > I'll be very much appriciated for any help and useful links.
> > _______________________________________________
> > Beginners mailing list