[Haskell-cafe] lazy evaluation is not complete
A simplied version of Example 5-16 in Manna's classical book Mathematical Theory of Computation: foo x = if x == 0 then 0 else foo (x-1)*foo (x+1) If run with ghci, foo 5 does not terminate, i.e., Haskell does not look for all outermost redices in parallel. Why? For efficiency reasons? It's a pity because a parallel-outermost strategy would be complete. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] lazy evaluation is not complete
Hello Peter, Monday, February 9, 2009, 5:10:22 PM, you wrote: If run with ghci, foo 5 does not terminate, i.e., Haskell does not look for all outermost redices in parallel. Why? For efficiency reasons? of course. if you will create new thread for every cpu instruction executed, you will definitely never compute anything :D you need to use `par` -- Best regards, Bulatmailto:bulat.zigans...@gmail.com ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] lazy evaluation is not complete
On Mon, 09 Feb 2009 15:10:22 +0100 Peter Padawitz peter.padaw...@udo.edu wrote: A simplied version of Example 5-16 in Manna's classical book Mathematical Theory of Computation: foo x = if x == 0 then 0 else foo (x-1)*foo (x+1) If run with ghci, foo 5 does not terminate, i.e., Haskell does not look for all outermost redices in parallel. Why? For efficiency reasons? I believe * is implemented in the normal way and thus is always strict in both arguments. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] lazy evaluation is not complete
Peter Padawitz wrote: A simplied version of Example 5-16 in Manna's classical book Mathematical Theory of Computation: foo x = if x == 0 then 0 else foo (x-1)*foo (x+1) If run with ghci, foo 5 does not terminate, i.e., Haskell does not look for all outermost redices in parallel. Why? For efficiency reasons? It's a pity because a parallel-outermost strategy would be complete. (*) is strict in both arguments for Int. If you want to avoid this, you could do newtype X = X Int and write your own implementation of (*) that is nonstrict. -- Jochem Berndsen | joc...@functor.nl GPG: 0xE6FABFAB ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] lazy evaluation is not complete
Hi, Just for fun, here is the code that does this: newtype Int' = I Int deriving Eq instance Show Int' where show (I x) = show x instance Num Int' where I x + I y = I (x + y) I 0 * _ = I 0 I x * I y = I (x * y) I x - I y = I (x - y) abs (I x) = I (abs x) signum (I x) = I (signum x) negate (I x) = I (negate x) fromInteger n = I (fromInteger n) foo x = if x == 0 then 0 else foo (x - 1) * foo (x + 1) *Main foo 5 :: Int' 0 -Iavor On Mon, Feb 9, 2009 at 7:19 AM, Jochem Berndsen joc...@functor.nl wrote: Peter Padawitz wrote: A simplied version of Example 5-16 in Manna's classical book Mathematical Theory of Computation: foo x = if x == 0 then 0 else foo (x-1)*foo (x+1) If run with ghci, foo 5 does not terminate, i.e., Haskell does not look for all outermost redices in parallel. Why? For efficiency reasons? It's a pity because a parallel-outermost strategy would be complete. (*) is strict in both arguments for Int. If you want to avoid this, you could do newtype X = X Int and write your own implementation of (*) that is nonstrict. -- Jochem Berndsen | joc...@functor.nl GPG: 0xE6FABFAB ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] lazy evaluation is not complete
On Mon, Feb 9, 2009 at 10:50 PM, Iavor Diatchki iavor.diatc...@gmail.com wrote: I 0 * _ = I 0 I x * I y = I (x * y) Note that (*) is now non-commutative (w.r.t. _|_). Of course, that's what we need here, but it means that the obviously correct transformation of foo x = if x == 0 then 0 else foo (x - 1) * foo (x + 1) into foo' x = if x == 0 then 0 else foo' (x + 1) * foo' (x - 1) is *not* in fact correct. --Max ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] lazy evaluation is not complete
On 10 Feb 2009, at 07:57, Max Rabkin wrote: On Mon, Feb 9, 2009 at 10:50 PM, Iavor Diatchki iavor.diatc...@gmail.com wrote: I 0 * _ = I 0 I x * I y = I (x * y) Note that (*) is now non-commutative (w.r.t. _|_). Of course, that's what we need here, but it means that the obviously correct transformation of just to improve slightly: I 0 |* _ = I 0 I x |* I y = I (x * y) _ *| I 0 = I 0 I x *| I y = I (x * y) I x * | y = (I x |* I y) `unamb` (I x *| I y) Now it is commutative :) Bob ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] lazy evaluation is not complete
On Tue, 2009-02-10 at 08:03 +0100, Thomas Davie wrote: On 10 Feb 2009, at 07:57, Max Rabkin wrote: On Mon, Feb 9, 2009 at 10:50 PM, Iavor Diatchki iavor.diatc...@gmail.com wrote: I 0 * _ = I 0 I x * I y = I (x * y) Note that (*) is now non-commutative (w.r.t. _|_). Of course, that's what we need here, but it means that the obviously correct transformation of just to improve slightly: I 0 |* _ = I 0 I x |* I y = I (x * y) _ *| I 0 = I 0 I x *| I y = I (x * y) I x * | y = (I x |* I y) `unamb` (I x *| I y) Now it is commutative :) Bob See `parCommute` from the 'lub' package :) signature.asc Description: This is a digitally signed message part ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
