22.02.2012, 11:20, wren ng thornton w...@freegeek.org:
On 2/22/12 1:45 AM, Miguel Mitrofanov wrote:
However, there is no free ordering on:
{ (a0,b) | b- B } \cup { (a,b0) | a- A }
What? By definition, since, a0= a and b0= b, we have (a0, b0)= (a0, b)
and (a0, b0)= (a0, b0), so,
On 2/22/12 2:20 AM, wren ng thornton wrote:
On 2/22/12 1:45 AM, Miguel Mitrofanov wrote:
However, there is no free ordering on:
{ (a0,b) | b - B } \cup { (a,b0) | a - A }
What? By definition, since, a0 = a and b0 = b, we have (a0, b0)
= (a0,b) and (a0, b0) = (a0, b0), so, (a0, b0) is clearly
On 2/22/12 2:37 AM, Dan Doel wrote:
unless I'm still sketchy on what you mean by domain. I don't think it
matters that we're only considering strict homomorphisms.
I think part of the problem is that there are many different ideas of
what exact properties a domain has. The one I'm most
On 2/21/12 10:44 AM, wren ng thornton wrote:
but domain products do not form domains! In order to
get a product which does form a domain, we'd need to use the smash
product[2] instead. Unfortunately we can't have our cake and eat it too
Bah, I don't know why my wires were crossed yesterday.
On 2/21/12 2:17 AM, Roman Cheplyaka wrote:
* Sebastian Fischerfisc...@nii.ac.jp [2012-02-21 00:28:13+0100]
On Mon, Feb 20, 2012 at 7:42 PM, Roman Cheplyakar...@ro-che.info wrote:
Is there any other interpretation in which the Reader monad obeys the
laws?
If selective strictness (the seq
Ehm... why exactly don't domain products form domains?
On 21 Feb 2012, at 19:44, wren ng thornton wrote:
On 2/21/12 2:17 AM, Roman Cheplyaka wrote:
* Sebastian Fischerfisc...@nii.ac.jp [2012-02-21 00:28:13+0100]
On Mon, Feb 20, 2012 at 7:42 PM, Roman Cheplyakar...@ro-che.info wrote:
Is
On Tue, Feb 21, 2012 at 10:44 AM, wren ng thornton w...@freegeek.org wrote:
That's a similar sort of issue, just about whether undefined ==
(undefined,undefined) or not. If the equality holds then tuples would be
domain products[1], but domain products do not form domains!
...
[1] Also a
On 2/21/12 11:27 AM, MigMit wrote:
Ehm... why exactly don't domain products form domains?
One important property of domains[1] is that they have a unique bottom
element. Given domains A and B, let us denote the domain product as:
(A,B) def= { (a,b) | a - A, b - B }
Which will inherit
On 2/21/12 11:54 AM, Dan Doel wrote:
On Tue, Feb 21, 2012 at 10:44 AM, wren ng thorntonw...@freegeek.org wrote:
That's a similar sort of issue, just about whether undefined ==
(undefined,undefined) or not. If the equality holds then tuples would be
domain products[1], but domain products do
22.02.2012, 09:30, wren ng thornton w...@freegeek.org:
On 2/21/12 11:27 AM, MigMit wrote:
Ehm... why exactly don't domain products form domains?
One important property of domains[1] is that they have a unique bottom
element. Given domains A and B, let us denote the domain product as:
On 2/21/12 11:54 AM, Dan Doel wrote:
You don't have to get rid of bottom entirely (I think). If you make
matches against products irrefutable, then you're again in the
situation of seq being the only thing able to distinguish between _|_
and (_|_, _|_), so we could keep the current
On 2/22/12 1:45 AM, Miguel Mitrofanov wrote:
However, there is no free ordering on:
{ (a0,b) | b- B } \cup { (a,b0) | a- A }
What? By definition, since, a0= a and b0= b, we have (a0, b0)= (a0, b) and (a0,
b0)= (a0, b0), so, (a0, b0) is clearly the bottom of A\times B.
Sorry, the
On Wed, Feb 22, 2012 at 1:40 AM, wren ng thornton w...@freegeek.org wrote:
It's a category-theoretic product, but not for the category of domains. Let
Set be the category of sets and set-theoretic functions. And let pDCPO be
the category of (pointed) domains and their homomorphisms.
The
I just realised that many common monads do not obey the monad laws when
it comes to bottoms.
E.g. for the Reader monad:
undefined = return /= undefined
return () = undefined /= undefined
return () = const undefined /= undefined
return undefined = \x - case x of () - return () /=
On Mon, Feb 20, 2012 at 7:42 PM, Roman Cheplyaka r...@ro-che.info wrote:
Is there any other interpretation in which the Reader monad obeys the
laws?
If selective strictness (the seq combinator) would exclude function
types, the difference between undefined and \_ - undefined could not
* Sebastian Fischer fisc...@nii.ac.jp [2012-02-21 00:28:13+0100]
On Mon, Feb 20, 2012 at 7:42 PM, Roman Cheplyaka r...@ro-che.info wrote:
Is there any other interpretation in which the Reader monad obeys the
laws?
If selective strictness (the seq combinator) would exclude function
I've got this expression
expression = do
w - hello
y - to you
return w
I wanna know how can I reduce it using monad laws
--
View this message in context:
http://haskell.1045720.n5.nabble.com/Monad-Laws-and-Do-Notation-tp4828598p4828598.html
Sent from the Haskell -
On 22 September 2011 11:46, diazepan spanishbizar...@yahoo.com wrote:
I've got this expression
expression = do
w - hello
y - to you
return w
I wanna know how can I reduce it using monad laws
I don't think you can: the best you can do is minimise it with other
monadic
Hi,
thanks for all the comments.
In ghc 6.12.1 the behavior is strange:
If f is defined in the interpreter via let, then the seq-expression converges
GHCi, version 6.12.1: http://www.haskell.org/ghc/ :? for help
let f = \x - (undefined::IO Bool)
seq (f True) True
True
but if you build a
Hi Luke,
I wrote:
For this reason, I consider it a bug in GHC that return :: IO a
is lazy.
Luke Palmer wrote:
Wait a minute...
return undefined = const (return 42)
= const (return 42) undefined
= return 42
But if return undefined = undefined, then that equals;
undefined = const
Hi,
when checking the first monad law (left unit) for the IO-monad (and also
for the ST monad):
return a = f ≡ f a
I figured out that there is the distinguishing context (seq [] True)
which falsifies the law
for a and f defined below
let a = True
let f = \x - (undefined::IO Bool)
seq
On Tue, Mar 2, 2010 at 1:17 PM, David Sabel
sa...@ki.informatik.uni-frankfurt.de wrote:
Hi,
when checking the first monad law (left unit) for the IO-monad (and also for
the ST monad):
return a = f ≡ f a
I figured out that there is the distinguishing context (seq [] True) which
falsifies
On 2 March 2010 15:44, Luke Palmer lrpal...@gmail.com wrote:
On Tue, Mar 2, 2010 at 1:17 PM, David Sabel
sa...@ki.informatik.uni-frankfurt.de wrote:
Hi,
when checking the first monad law (left unit) for the IO-monad (and also
for
the ST monad):
return a = f ≡ f a
I figured out
David Sabel wrote:
when checking the first monad law (left unit) for the IO-monad (and also for
the ST monad):
I figured out that there is the distinguishing context (seq [] True) which
falsifies the law...
It's worse than that - Haskell types and functions do not
even form a category under
On Tue, Mar 2, 2010 at 4:37 PM, Yitzchak Gale g...@sefer.org wrote:
For this reason, I consider it a bug in GHC that return :: IO a
is lazy.
Wait a minute...
return undefined = const (return 42)
= const (return 42) undefined
= return 42
But if return undefined = undefined, then that
On Fri, 2006-09-08 at 01:43 -0400, Albert Lai wrote:
Deokhwan Kim [EMAIL PROTECTED] writes:
What is the practical meaning of monad laws?
1. (return x) = f == f x
2. m = return == m
3. (m = f) = g == m (\x - f x = g)
I offer to re-write the laws in do-notation. (Please view
Deokhwan Kim wrote:
What is the practical meaning of monad laws?
(M, return, =) is not qualified as a category-theoretical monad, if
the following laws are not satisfied:
1. (return x) = f == f x
2. m = return == m
3. (m = f) = g == m (\x - f x = g)
But what practical problems can
Brian,
Are you really sure Haskell compilers do that optimization?
I would regard a compiler that does optimizations that are justified
by laws that the compiler cannot check as broken.
-- Lennart
On Sep 7, 2006, at 08:50 , Brian Hulley wrote:
Deokhwan Kim wrote:
What is the
On Sep 7, 2006, at 9:04 AM, Lennart Augustsson wrote:
Brian,
Are you really sure Haskell compilers do that optimization?
I would regard a compiler that does optimizations that are
justified by laws that the compiler cannot check as broken.
What, like list fusion?
;-)
Although, more
Lennart Augustsson wrote:
On Sep 7, 2006, at 08:50 , Brian Hulley wrote:
Deokhwan Kim wrote:
What is the practical meaning of monad laws?
Afaiu the monad laws are needed so the compiler can do various
optimizations, especially in regard to the do notation. Consider:
g c = do
G'day all.
Quoting Deokhwan Kim [EMAIL PROTECTED]:
What is the practical meaning of monad laws?
Interesting philosophical question. There will be an article on
this topic in the next The Monad.Reader, so watch this space.
But what practical problems can unsatisfying them cause?
Pretty much
What is the practical meaning of monad laws?
(M, return, =) is not qualified as a category-theoretical monad, if
the following laws are not satisfied:
1. (return x) = f == f x
2. m = return == m
3. (m = f) = g == m (\x - f x = g)
But what practical problems can unsatisfying them cause?
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