Sorry, I realized that it does not cover the
shorter [1..5] (shorter [2..] [3..])
case...
Eugene Crosser wrote:
> On Tue, 10 Oct 2006 [EMAIL PROTECTED] wrote:
>> Hi all,
>> I'm trying to implement a function that returns the shorter one of two given
>> lists,
>> something like
>> shorter :: [a]
On Tue, 10 Oct 2006 [EMAIL PROTECTED] wrote:
> Hi all,
> I'm trying to implement a function that returns the shorter one of two given
> lists,
> something like
> shorter :: [a] -> [a] -> [a]
> such that shorter [1..10] [1..5] returns [1..5],
> and it's okay for shorter [1..5] [2..6] to return eithe
Nicolas,
I think you gave a fine explanation. Just a few minor remarks...
l1 = Cons 3 Nil
l2 = Cons 3 _|_
l2 > l1 because l1 is more defined.
Surely you mean l2 < l1, then.
Moreover, are you sure you need to define your order in such a way
that l2 < l1. I'd say, for these purposes, it's e
On 11/10/06, Nicolas Frisby <[EMAIL PROTECTED]> wrote:
Intuitively
q = lfp f = f(f(f(f(f(f (f(f(f _|_)))...)(*)
r = lfg g = g(g(g(g(g(g (g(g(g _|_)))...) (**)
This way of writing it would imply (at least to me) that the number of
f's and g's involved is finit
For my own edification (experts: please remark if I stray from any
conventional notions--be as picky as possible) and perhaps for anyone
who didn't quite grok Ross Paterson's punchline, I'm going to try to
answer the question "How does seq affect the definedness CPO?"
At the heart of these two re
On Oct 11, 2006, at 11:14 AM, Ross Paterson wrote:
On Wed, Oct 11, 2006 at 11:04:49AM -0400, Robert Dockins wrote:
let q = seq q (FinCons 3 q) in q(beta)
We have (from section 6.2):
seq _|_ y = _|_
seq x y = yiff x /= _|_
Now, here we have an interesting dilemma
On Wed, Oct 11, 2006 at 11:04:49AM -0400, Robert Dockins wrote:
> let q = seq q (FinCons 3 q) in q(beta)
>
> We have (from section 6.2):
>seq _|_ y = _|_
>seq x y = yiff x /= _|_
>
> Now, here we have an interesting dilemma.
The meaning of a recursive definition i
t 2006 15:14:23 +0100
> Subject: Re: [Haskell-cafe] beginner's problem about lists
>
> Matthias Fischmann <[EMAIL PROTECTED]> wrote:
>
> > > No, your Fin type can also hold infinite values.
> >
> > let q = FinCons 3 q in case q of FinCons i _ -> i ==>
On Oct 11, 2006, at 10:14 AM, Malcolm Wallace wrote:
Matthias Fischmann <[EMAIL PROTECTED]> wrote:
No, your Fin type can also hold infinite values.
let q = FinCons 3 q in case q of FinCons i _ -> i ==> _|_
does that contradict, or did i just not understand what you are
saying?
That may
Malcolm Wallace wrote:
Matthias Fischmann <[EMAIL PROTECTED]> wrote:
No, your Fin type can also hold infinite values.
let q = FinCons 3 q in case q of FinCons i _ -> i ==> _|_
does that contradict, or did i just not understand what you are
saying?
That may be the result in
Matthias Fischmann <[EMAIL PROTECTED]> wrote:
> > No, your Fin type can also hold infinite values.
>
> let q = FinCons 3 q in case q of FinCons i _ -> i ==> _|_
>
> does that contradict, or did i just not understand what you are
> saying?
That may be the result in ghc, but nhc98 gives the ans
On Wed, Oct 11, 2006 at 11:40:59AM +0100, Malcolm Wallace wrote:
> To: haskell-cafe@haskell.org
> From: Malcolm Wallace <[EMAIL PROTECTED]>
> Date: Wed, 11 Oct 2006 11:40:59 +0100
> Subject: Re: [Haskell-cafe] beginner's problem about lists
>
> ihope <[EMAI
On Wed, 2006-10-11 at 11:40 +0100, Malcolm Wallace wrote:
> ihope <[EMAIL PROTECTED]> wrote:
>
> > It's possible to make both "infinite list" and "finite list"
> > datatypes:
> >
> > data Inf a = InfCons a (Inf a)
> > data Fin a = FinCons a !(Fin a) | FinNil
> >
> > At least, I think the Fin typ
ihope <[EMAIL PROTECTED]> wrote:
> It's possible to make both "infinite list" and "finite list"
> datatypes:
>
> data Inf a = InfCons a (Inf a)
> data Fin a = FinCons a !(Fin a) | FinNil
>
> At least, I think the Fin type there has to be finite...
No, your Fin type can also hold infinite values
[EMAIL PROTECTED] wrote:
Hi all,
I'm trying to implement a function that returns the shorter one of two
given lists,
something like
shorter :: [a] -> [a] -> [a]
such that shorter [1..10] [1..5] returns [1..5],
and it's okay for shorter [1..5] [2..6] to return either.
Simple, right?
However,
On 10/10/06, Nicolas Frisby <[EMAIL PROTECTED]> wrote:
> data Fin
> data Inf
> data List l a = Cons a (List l a) | Nil
It's possible to make both "infinite list" and "finite list" datatypes:
data Inf a = InfCons a (Inf a)
data Fin a = FinCons a !(Fin a) | FinNil
At least, I think the Fin typ
On Tue, Oct 10, 2006 at 08:58:05AM -0400, Lennart Augustsson wrote:
> >a function that takes two lists and decides whether one of them is
> >finite or not , without being given further information on the lists,
> >does not exist.
>
> A function that takes two lists and decides if one is finite do
I suppose using indicative types (dependent style) is out of the
question? I presume i) that would over-simplify the problem and ii)
we're tied to the [-] type.
It deserves mention no less.
data Fin
data Inf
data List l a = Cons a (List l a) | Nil
shorter :: List Inf a -> List Inf a -> Li
Hi
However this will result in a non-terminating loop for shorter [1..] [2..],
since the first two patterns of f shall never match.
The specification of your problem makes this a guarantee. How do you
know that a list is finite? You find the [] at the end. How do you
know a list is infinite? Y
Thanks for your reply. I tried a few ways but none worked.One is like: shorter as bs = f id id as bs wheref ca cb [] _ = ca [] f ca cb _ [] = cb []f ca cb (a:as) (b:bs) = f (ca.(a:)) (cb.(b:)) as bs
However this will result in a non-terminating loop for shorter [1..] [2..],since the f
On Tue, 10 Oct 2006, Henning Thielemann wrote:
>
> On Tue, 10 Oct 2006 [EMAIL PROTECTED] wrote:
>
> > Hi all,
> >
> > I'm trying to implement a function that returns the shorter one of two given
> > lists,
> > something like
> > shorter :: [a] -> [a] -> [a]
> > such that shorter [1..10] [1..5]
On 10/10/06, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
I'm trying to implement a function that returns the shorter one of two given
lists,
something like
shorter :: [a] -> [a] -> [a]
However, it becomes difficult when dealing with infinite lists, for example,
shorter [1..5] (shorter [2..]
On Tue, 10 Oct 2006 [EMAIL PROTECTED] wrote:
> Hi all,
>
> I'm trying to implement a function that returns the shorter one of two given
> lists,
> something like
> shorter :: [a] -> [a] -> [a]
> such that shorter [1..10] [1..5] returns [1..5],
> and it's okay for shorter [1..5] [2..6] to return
On Oct 10, 2006, at 08:49 , Matthias Fischmann wrote:
On Tue, Oct 10, 2006 at 08:10:44PM +0800, [EMAIL PROTECTED] wrote:
To: haskell-cafe@haskell.org
From: [EMAIL PROTECTED]
Date: Tue, 10 Oct 2006 20:10:44 +0800
Subject: [Haskell-cafe] beginner's problem about lists
Hi all,
I'm trying to imp
On Tue, Oct 10, 2006 at 08:10:44PM +0800, [EMAIL PROTECTED] wrote:
> To: haskell-cafe@haskell.org
> From: [EMAIL PROTECTED]
> Date: Tue, 10 Oct 2006 20:10:44 +0800
> Subject: [Haskell-cafe] beginner's problem about lists
>
> Hi all,
>
> I'm trying to implement a function that returns the shorter
Hi,
The trick is not call "length", since length demands the whole of a
list, and won't terminate on an infinite list. You will want to
recurse down the lists.
Is this a homework problem? It's best to declare if it is, and show
what you've managed to do so far.
Thanks
Neil
On 10/10/06, [EMAIL
26 matches
Mail list logo
