Interesting * I think it is a Bad Idea for an application to assume that the implementation of (f^n) will not multiply by 1. Implementations of numeric algorithms probably make all sorts of ill-documented assumptions about the algebraic properties of numeric operations * On the other hand, it's a bit stupid for the implementation of (f^n) to multiply by 1, and Ian has, I think, already fixed this. Simon | -----Original Message----- | From: glasgow-haskell-users-bounces@haskell.org [mailto:glasgow-haskell-users-bounces@haskell.org] On | Behalf Of Serge D. Mechveliani | Sent: 01 June 2008 18:34 | To: Ian Lynagh | Cc: glasgow-haskell-users@haskell.org | Subject: Re: ANNOUNCE: GHC 6.8.3 Release Candidate | | 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 | mechvel@botik.ru | | | | | | | _______________________________________________ | Glasgow-haskell-users mailing list | Glasgow-haskell-users@haskell.org | http://www.haskell.org/mailman/listinfo/glasgow-haskell-users