Subject: fixed point - msg#00200

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Re: Regex parser for Dfa

G'day. Quoting Graham Klyne <GK@xxxxxxxxxxxxxx>: > I tackled this as an exercise in writing a lightweight Parsec parser. A > copy is attached. Thanks, I've checked it in. > I do have a sourceforge count (GrahamK), but I'm not familiar with the > procedure for submitting a file. I'll mail you offlist. Cheers, Andrew Bromage

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Re: fixed point

On Sun, 26 Oct 2003 19:24:26 -0500 "Harris, Andrew" <Andrew.Harris@xxxxxxxxxx> wrote: > Hi - > > I am trying to do Exercise 9.9 in HSOE; and I've come up with an > answer that works but I'm not sure if it answers the question > properly. The problem is: > > The Question: > ------------- > > Suppose we define a function "fix" as: > > fix f = f (fix f) > > Suppose further we have a recursive function: > > remainder :: Integer -> Integer -> Integer > remainder a b = if a < b then a > else remainder (a - b) b > > Rewrite this function using "fix" so that it's not recursive. > > My Answer: > ---------- > > wierdFunc x y z = if y - z > z then ((\a b c -> a b c) x (y-z) z) > else ((\a b c -> a b c) (\d e -> d) (y-z) z) > > myRemainder = fix wierdFunc > > My Question: > ------------ > > Does the ((\a b c -> a b c) x (y-z) z) mean that I've not removed the > recursion? I was assuming that I was returning a function that is to > be evaluated and not actually doing any recursion. That's why I > thought I answered the question. However, I have a headache now and > would like another opinion. Notice that, (\x -> x) a reduces to a, so (\a b c -> a b c) x (y-z) z reduces to x (y-z) z. You can therefore simplify your function quite a bit. wierdFunc x y z = if y-z > z then x (y-z) z else (\d e -> d) (y-z) z and you can still apply that lambda abstraction (beta-reduce) wierdFunc x y z = if y-z > z then x (y-z) z else y-z None of these (except, of course, fix and remainder) are recursive. A recursive function is just one that calls itself. For wierdFunc to be recursive, the identifier wierdFunc would have to occur in the right-hand side of it's definition. This all said, you are making the problem much more difficult than it needs to be. Try matching up your x,y,z's to things in remainder and I think the expected answer will become obvious. Also, you may want to look at the type of fix and wierdFunc (you can use :type in Hugs or GHCi).

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random numbers

I'm very much a beginning programmer, studying Haskell. I have two questions - Firstly, I am trying to generate random numbers in Haskell, but although I have found a 'random' library on www.zvon.org, I don't really know how to include library functions, and the documentation given doesn't really tell me the effective difference between all the functions provided in the library. In another interesting development, I couldn't find any mention of this library in haskell.org's list of libraries. Secondly, but perhaps more importantly, is there a more 'beginnerish' list that I should be addressing this to? I've been following discussions on this one and they don't seem to be at quite this level. Thanks Jac

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Re: fixed point

On Sun, 26 Oct 2003 19:24:26 -0500 "Harris, Andrew" <Andrew.Harris@xxxxxxxxxx> wrote: > Hi - > > I am trying to do Exercise 9.9 in HSOE; and I've come up with an > answer that works but I'm not sure if it answers the question > properly. The problem is: > > The Question: > ------------- > > Suppose we define a function "fix" as: > > fix f = f (fix f) > > Suppose further we have a recursive function: > > remainder :: Integer -> Integer -> Integer > remainder a b = if a < b then a > else remainder (a - b) b > > Rewrite this function using "fix" so that it's not recursive. > > My Answer: > ---------- > > wierdFunc x y z = if y - z > z then ((\a b c -> a b c) x (y-z) z) > else ((\a b c -> a b c) (\d e -> d) (y-z) z) > > myRemainder = fix wierdFunc > > My Question: > ------------ > > Does the ((\a b c -> a b c) x (y-z) z) mean that I've not removed the > recursion? I was assuming that I was returning a function that is to > be evaluated and not actually doing any recursion. That's why I > thought I answered the question. However, I have a headache now and > would like another opinion. Notice that, (\x -> x) a reduces to a, so (\a b c -> a b c) x (y-z) z reduces to x (y-z) z. You can therefore simplify your function quite a bit. wierdFunc x y z = if y-z > z then x (y-z) z else (\d e -> d) (y-z) z and you can still apply that lambda abstraction (beta-reduce) wierdFunc x y z = if y-z > z then x (y-z) z else y-z None of these (except, of course, fix and remainder) are recursive. A recursive function is just one that calls itself. For wierdFunc to be recursive, the identifier wierdFunc would have to occur in the right-hand side of it's definition. This all said, you are making the problem much more difficult than it needs to be. Try matching up your x,y,z's to things in remainder and I think the expected answer will become obvious. Also, you may want to look at the type of fix and wierdFunc (you can use :type in Hugs or GHCi).