*Pete Chown and Dan Doel. Thank you for your solution. I actually It was not
homework . It was just a a past exam question trying to answer . *
**but your solution is very long , so I don't think he wants answer this
long in the exams. I think this answer agree100 f g = map f xs == map g xs
where xs = [1..100] from Richard O'Keefe
is do the job.
**Anyway , your solution help understand more about Haskell not only this
question . Many thanks to you guys.
On 27 May 2010 02:04, Dan Doel
On Wednesday 26 May 2010 5:38:57 pm Pete Chown wrote:
test :: (Eq a) => (Int -> a) -> (Int -> a) -> Bool test f1 f2 = unsafePerformIO $ do goodSoFar <- newIORef True forLoop 1 100 $ \i -> when (f1 i /= f2 i) $ writeIORef goodSoFar False readIORef goodSoFar
The problem with this algorithm is that it needlessly tests f1 against f2 for all i, even when a non-matching value has has already been found. Using the power of call-with-current-continuation, I have constructed an algorithm that lacks this deficiency:
import Control.Monad.Cont
test f g = flip runCont id . callCC $ \escape -> do forM_ [1..100] $ \n -> when (f n /= g n) $ escape False return True
This should perform almost 75% less work in the testFunc1 case! It certainly feels much snappier.
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