Re: [Haskell-beginners] Simplify (normalize) symbolic polynom-like expression

Not for equality testing, only for prettyprinting. So I want some simple way to apply all the rules with smallest number of expression tree traverses As for ipoly, it looks like CAS. I will use Maxima as CAS and I don't want to do extra Maxima call for something, that can be easily (expecting) done in Haskell. 2012/6/17 Máté Kovács :

Hi Daniel,

It depends on what you want to use the normalized / canonical form for.

If it's to reduce semantic equivalence testing to simple syntactic equality, e.g. (A == B) = (canonize(A) == canonize(B)), then you could just use the fully expanded form, which isn't really simplification. :)

I'm doing something similar here (for polynomial expressions over an inner product space): https://github.com/mkovacs/ipoly/blob/master/Poly.hs

Cheers, Máté

On Sat, Jun 16, 2012 at 2:17 PM, Daniel Hlynskyi wrote:

...

Hello.

I am new to typefull programming, so I've got a question. I have a simple mathematical expression (addition, product and exponentiation only):

...

data Expr = I Int -- integer constant           | V -- symbolic variable           | Sum [Expr]           | Prod [Expr]           | Pow Expr Expr

What I want is normalize\simplify this expression. Eg `Prod [Pow V (I 0), Pow V (I 1)] ` must be simplified to just `V`. What techniques should I use to write `normalize` function? Simplification rules are quite simple:

...

normalize (Sum [a]) = normalize a normalize (Sum xs) | (I 0) `elem` xs = map nomalize . Sum $ filter (/= I 0) xs                          | otherwise = map normalize xs normalize (Prod xs) | (I 0) `elem` xs = I 0 normalize (Prod xs) | (I 1) `elem` xs = map nomalize . Prod $ filter (/= I 1) xs                          | otherwise = map normalize xs normalize (Pow a (I 0)) = I 1 normalize (Pow a (I 1)) = normalize a

and so on. But rules like theese cannot simplify some expressions, for example, `Prod [Pow V (I 0), Pow V (I 1)] `.

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