There are no rules written down. But the fast exponentiation algorithm used by (^) assumes that (*) is associative. I also don't think that fast exponentiation should ever multiply by 1.
-- Lennart On Sun, Jun 1, 2008 at 6:34 PM, Serge D. Mechveliani <[EMAIL PROTECTED]> wrote: > On Sun, Jun 01, 2008 at 03:34:00PM +0100, Ian Lynagh wrote: >> On Sun, Jun 01, 2008 at 05:39:49PM +0400, Serge D. Mechveliani wrote: >> > >> > This is why res and 1*res are not equivalent in Haskell-98 for >> > res :: Num a => a. >> > >> > Am I missing something? >> >> The library functions assume that class instances obey some unwritten >> laws; it's all a bit vague, but if your instances don't obey them then >> you might find that things go wrong when using library functions. For >> example, if your (*) isn't associative then (^) is going to give odd >> results, and if the type of the second argument to (^) doesn't do >> arithmetic in the normal way then very strange things could happen. >> >> Anyway, I've just tweaked the (^) definition again, so your code should >> work in 6.8.3. > > > Thank you. > This helps -- in that the public DoCon-2.11 test will run (I hope). > > But I think that you have found a bug in the DoCon test program. > Thank you. > More precisely, here are my indeas. > > > 1. Generally, for t = Num a => a, the expessions res :: t and > ((fromInteger 1) :: t) * res > are not equivalent in Haskell-98, > and the compiler has not rigth to replace the former with the latter. > Right? > > 2. But I guess, our current questions are different: > > (2.1) For t = Num a => a, has a Haskell implementation for (f^n) :: t > right to base on certain natural properties of the operation > fromInteger :: Integer -> t ? > (2.2) Has a reliable mathematical application right to base on that > for n > 0 the expression (f^n) :: Polynomial Integer > does not imply computing fromInteger 1 :: Polynomial Integer > ? > > Concerning (2.1): I do not know whether Haskell-98 allows to rely on > that f^3 is equivalent to (fromInteger 1)*(f^3). > > But concerning (2.2), I think that (independently of the question (2.1)) > a reliable mathematical application must not presume the above properties > of (^) and fromInteger. > > Concerning f^3 :: UPol Integer in my test for DoCon: > > 1) fromInteger _ :: UPol _ is defined as > error "... use fromi instead". > 2) (*) is defined via polMul ... > 3) The expression f ^ 3 relies on the Haskell-98 library for (^). > The Haskell library defines (^) via (*) -- all right. > > 4) DoCon has the function `power' to use instead of (^), > and its library avoids (^). > But for n > 0, I considered f^n :: UPol _ > as also correct. Because the Haskell library performs this via repeated > application of (*). > And I thought that if n > 0, then (fromInteger _ :: UPol _) will not > appear. > Maybe it does not appear in old GHC-s and does appear 6.8.3 ? > > I was using expressions like [(f ^ n) :: UPol Integer | n <- [2 .. 9]] > in my _test programs_ for DoCon. > But this relies on a particular property of the Haskell library definition > for (f^n) :: a > -- on that if n > 0 then (fromInteger _ :: a) does not appear in this > computation. > > Now, I think, either I need to hide the standard (^) and overload it > or to replace (^) with `power' in my examples too. > > Regards, > > ----------------- > Serge Mechveliani > [EMAIL PROTECTED] > > > > > > > _______________________________________________ > Glasgow-haskell-users mailing list > Glasgow-haskell-users@haskell.org > http://www.haskell.org/mailman/listinfo/glasgow-haskell-users > _______________________________________________ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
