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 either.
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] - [a] -
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
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. The
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 type there has to be
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 [EMAIL PROTECTED] wrote:
It's possible to make
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 answer 3.
It
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
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 be
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 == _|_
does that contradict, or did i just not understand what you are
saying
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 is the
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
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
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
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
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 one of
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
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 either.
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..]
Thanks for your reply. I tried afew ways butnoneworked.One islike: 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..]
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?
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 -
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 type there
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