Re: [Haskell-cafe] Natural Transformations and fmap
Ryan Ingram wrote: However, the type of natural transformations comes with a free theorem, for example concat :: [[a]] - [a] has the free theorem forall f :: a - b, f strict and total, fmap f . concat = concat . fmap (fmap f) The strictness condition is needed; consider broken_concat :: [[a]] - [a] broken_concat _ = [undefined] f = const () fmap f (broken_concat []) = fmap f [undefined] = [()] broken_concat (fmap (fmap f) []) = broken_concat [] = [undefined] The 'taming selective strictness' version of the free theorem generator[1] allows removing the totality condition on f, but not the strictness condition. But in the case of concat, you can prove a stronger theorem: forall f :: a - b, fmap f . concat = concat . fmap (fmap f) My suspicion is that this stronger theorem holds for all strict and total natural transformations, but I don't know how to go about proving that suspicion. I'm a hobbyist mathematician and a professional programmer, not the other way around:) ... [1]http://www-ps.iai.uni-bonn.de/cgi-bin/polyseq.cgi There is an analogous approach to the taming selective strictness version of the free theorem generator where it is general recursion that is tamed. In that setting, you then get free theorems like that for concat without either strictness or totality side conditions. It is really very similar, indeed the taming selective strictness work takes over and develops further the much earlier taming general recursion ideas. The original source for the latter is: http://dblp.uni-trier.de/rec/bibtex/conf/esop/LaunchburyP96 Just nobody ever bothered to implement it in a tool. (Well, actually, such an implementation is essentially hidden inside the counterexample generator http://www-ps.iai.uni-bonn.de/cgi-bin/exfind.cgi) Best, Janis. -- Jun.-Prof. Dr. Janis Voigtländer http://www.iai.uni-bonn.de/~jv/ mailto:j...@iai.uni-bonn.de ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Natural Transformations and fmap
I know a bit of category theory, but I'm trying to look at it from a
fundamental perspective; I know that I intend (m :- n) to mean natural
transformation from functor m to functor n, but the type itself (forall x.
m x - n x) doesn't necessarily enforce that.
However, the type of natural transformations comes with a free theorem, for
example
concat :: [[a]] - [a]
has the free theorem
forall f :: a - b, f strict and total, fmap f . concat = concat . fmap
(fmap f)
The strictness condition is needed; consider
broken_concat :: [[a]] - [a]
broken_concat _ = [undefined]
f = const ()
fmap f (broken_concat []) = fmap f [undefined] = [()]
broken_concat (fmap (fmap f) []) = broken_concat [] = [undefined]
The 'taming selective strictness' version of the free theorem generator[1]
allows removing the totality condition on f, but not the strictness
condition.
But in the case of concat, you can prove a stronger theorem:
forall f :: a - b, fmap f . concat = concat . fmap (fmap f)
My suspicion is that this stronger theorem holds for all strict and total
natural transformations, but I don't know how to go about proving that
suspicion. I'm a hobbyist mathematician and a professional programmer, not
the other way around :)
I think it's probably easy to prove that the monoid laws imply that mult'
and one' are strict and total.
Thus, you can in principle define plenty of natural transformations which
do not have the type f :: forall X. M X - N X.
Can you suggest one? I don't see how you can get around f needing to act
at multiple types since it can occur before and after g's fmap:
fmap g . f = f . fmap g
M A --fmap_M g-- M B
| |
f_A f_B
| |
v v
N A --fmap_N g-- N B
You can have n = m, of course, but that just means f :: M :- M.
-- ryan
[1] http://www-ps.iai.uni-bonn.de/cgi-bin/polyseq.cgi
Use this term:
/\a. let flipappend =(\xs :: [a]. fix (\rec :: [a] - [a]. \ys :: [a].
case ys of { [] - xs; y:zs - y : rec zs })) in
let concat = fix (\rec :: [[a]] - [a]. \xss :: [[a]].
case xss of { [] - []_{a}; xs:yss - flipappend
(rec yss) xs}) in
concat
[2] See http://hpaste.org/56903 Summary:
-- both of these types have obvious Functor instances
newtype (f :. g) x = O (f (g x))
data Id x = Id x
class Functor m = Monad m where
one' :: Id :- m
mult' :: (m :. m) :- m
-- instances are required to satisfy monoid laws:
--one' is a left/right identity for mult':
--forall x :: m a
--mult' . O . one' . Id $ x = x = mult' . O . fmap (one' . Id)
$ x
--mult' is associative:
--forall x :: m (m (m a))).
--mult' . O . mult' . O $ x = mult' . O . fmap (mult' . O) $ x
On Thu, Jan 26, 2012 at 9:30 PM, wren ng thornton w...@freegeek.org wrote:
On 1/23/12 10:39 PM, Ryan Ingram wrote:
type m :- n = (forall x. m x - n x)
class Functor f where fmap :: forall a b. (a - b) - f a - f b
-- Functor identity law: fmap id = id
-- Functor composition law fmap (f . g) = fmap f . fmap g
Given Functors m and n, natural transformation f :: m :- n, and g :: a -
b, how can I prove (f . fmap_m g) = (fmap_n g . f)?
That is the defining property of natural transformations. To prove it for
polymorphic functions in Haskell you'll probably want to leverage
parametricity.
I assume you don't know category theory, based on other emails in this
thread. But the definition of a natural transformation is that it is a
family of morphisms/functions { f_X :: M X - N X | X an object/type } such
that for all g :: a - b we have that f_b . fmap_m g == fmap_n g . f_a
Thus, you can in principle define plenty of natural transformations which
do not have the type f :: forall X. M X - N X. The only requirement is
that the family of morphisms obeys that equation. It's nice however that if
a function has that type, then it is guaranteed to satisfy the equation (so
long as it doesn't break the rules by playing with strictness or other
things that make it so Hask isn't actually a category).
--
Live well,
~wren
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Re: [Haskell-cafe] Natural Transformations and fmap
On 1/27/12 7:56 PM, Ryan Ingram wrote:
Thus, you can in principle define plenty of natural transformations which
do not have the type f :: forall X. M X - N X.
Can you suggest one? I don't see how you can get around f needing to act
at multiple types since it can occur before and after g's fmap:
Right. A natural transformation is a family of functions (one for each
type).
My point was, forall is one way of defining a family of functions, but
it's not the only one. For instance, we could use a type class, or some
fancy generics library, or a non-parametric forall in languages which
allow type-case.
Or we could use some way of defining it which is outside of the language
in which the component functions exist. For example, consider the simply
typed lambda calculus. STLC doesn't have quantifiers so we can't define
(f :: forall X. M X - N X) as a natural transformation from within the
language, but we could still talk about the family of simply-typed
functions { f_X :: M X - N X | X - type }. Calling a family of
functions a natural transformation is an extralinguistic statement about
the functions; there are, in general, more natural transformations than
can be defined from within the language in question. Just as there are,
in general, more endofunctors than can be defined within the language
(let alone other functors).
The naturality behind natural transformations is just the fact that
(forall g, fmap g . f = f . fmap g). Satisfying the equation means that
the family of fs is parametric enough, regardless of how we've defined
the family or how/whether we can implement the family as polymorphism
within the language.
--
Live well,
~wren
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Re: [Haskell-cafe] Natural Transformations and fmap
I tried the free theorem generator ( http://www-ps.iai.uni-bonn.de/cgi-bin/free-theorems-webui.cgi) and it wouldn't let me use generic functors, but playing with [] and Maybe leads me to believe that the free theorem for :- is forall f :: m :- n, forall g :: a - b, g strict and total fmap g . f = f . fmap g This implies that the monad laws don't necessarily hold in situations like \m - m = const Nothing, which seems wrong to me. The counterexamples ( http://www-ps.iai.uni-bonn.de/cgi-bin/exfind.cgi), however, all rely on odd natural transformations like (\_ - Just undefined). My guess is that there is a side condition we can put on f that is implied by the monoid laws which doesn't require g to be strict or total. -- ryan On Mon, Jan 23, 2012 at 10:23 PM, Brent Yorgey byor...@seas.upenn.eduwrote: On Mon, Jan 23, 2012 at 09:06:52PM -0800, Ryan Ingram wrote: On Mon, Jan 23, 2012 at 8:05 PM, Daniel Fischer daniel.is.fisc...@googlemail.com wrote: On Tuesday 24 January 2012, 04:39:03, Ryan Ingram wrote: At the end of that paste, I prove the three Haskell monad laws from the functor laws and monoid-ish versions of the monad laws, but my proofs all rely on a property of natural transformations that I'm not sure how to prove; given type m :- n = (forall x. m x - n x) class Functor f where fmap :: forall a b. (a - b) - f a - f b -- Functor identity law: fmap id = id -- Functor composition law fmap (f . g) = fmap f . fmap g Given Functors m and n, natural transformation f :: m :- n, and g :: a - b, how can I prove (f . fmap_m g) = (fmap_n g . f)? Unless I'm utterly confused, that's (part of) the definition of a natural transformation (for non-category-theorists). Alright, let's pretend I know nothing about natural transformations and just have the type declaration type m :- n = (forall x. m x - n x) And I have f :: M :- N g :: A - B instance Functor M -- with proofs of functor laws instance Functor N -- with proofs of functor laws How can I prove fmap g. f :: M A - N B = f . fmap g :: M A - N B I assume I need to make some sort of appeal to the parametricity of M :- N. This is in fact precisely the free theorem you get from the parametricity of f. Parametricity means that f must act uniformly for all x -- which is an intuitive way of saying that f really is a natural transformation. -Brent ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Natural Transformations and fmap
On 1/23/12 10:39 PM, Ryan Ingram wrote:
type m :- n = (forall x. m x - n x)
class Functor f where fmap :: forall a b. (a - b) - f a - f b
-- Functor identity law: fmap id = id
-- Functor composition law fmap (f . g) = fmap f . fmap g
Given Functors m and n, natural transformation f :: m :- n, and g :: a -
b, how can I prove (f . fmap_m g) = (fmap_n g . f)?
That is the defining property of natural transformations. To prove it
for polymorphic functions in Haskell you'll probably want to leverage
parametricity.
I assume you don't know category theory, based on other emails in this
thread. But the definition of a natural transformation is that it is a
family of morphisms/functions { f_X :: M X - N X | X an object/type }
such that for all g :: a - b we have that f_b . fmap_m g == fmap_n g . f_a
Thus, you can in principle define plenty of natural transformations
which do not have the type f :: forall X. M X - N X. The only
requirement is that the family of morphisms obeys that equation. It's
nice however that if a function has that type, then it is guaranteed to
satisfy the equation (so long as it doesn't break the rules by playing
with strictness or other things that make it so Hask isn't actually a
category).
--
Live well,
~wren
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[Haskell-cafe] Natural Transformations and fmap
I've been playing around with the relationship between monoids and monads (see http://www.jonmsterling.com/posts/2012-01-12-unifying-monoids-and-monads-with-polymorphic-kinds.htmland http://blog.sigfpe.com/2008/11/from-monoids-to-monads.html), and I put together my own implementation which I'm quite happy with, that you can see at http://hpaste.org/56903 ; relying only on the extensions RankNTypes, TypeOperators, NoImplicitPrelude, ScopedTypeVariables; At the end of that paste, I prove the three Haskell monad laws from the functor laws and monoid-ish versions of the monad laws, but my proofs all rely on a property of natural transformations that I'm not sure how to prove; given type m :- n = (forall x. m x - n x) class Functor f where fmap :: forall a b. (a - b) - f a - f b -- Functor identity law: fmap id = id -- Functor composition law fmap (f . g) = fmap f . fmap g Given Functors m and n, natural transformation f :: m :- n, and g :: a - b, how can I prove (f . fmap_m g) = (fmap_n g . f)? Is there some more fundamental law of natural transformations that I'm not aware of that I need to use? Is it possible to write a natural transformation in Haskell that violates this law? -- ryan ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Natural Transformations and fmap
On Tuesday 24 January 2012, 04:39:03, Ryan Ingram wrote: At the end of that paste, I prove the three Haskell monad laws from the functor laws and monoid-ish versions of the monad laws, but my proofs all rely on a property of natural transformations that I'm not sure how to prove; given type m :- n = (forall x. m x - n x) class Functor f where fmap :: forall a b. (a - b) - f a - f b -- Functor identity law: fmap id = id -- Functor composition law fmap (f . g) = fmap f . fmap g Given Functors m and n, natural transformation f :: m :- n, and g :: a - b, how can I prove (f . fmap_m g) = (fmap_n g . f)? Unless I'm utterly confused, that's (part of) the definition of a natural transformation (for non-category-theorists). Is there some more fundamental law of natural transformations that I'm not aware of that I need to use? Is it possible to write a natural transformation in Haskell that violates this law? -- ryan ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Natural Transformations and fmap
On Mon, Jan 23, 2012 at 8:05 PM, Daniel Fischer daniel.is.fisc...@googlemail.com wrote: On Tuesday 24 January 2012, 04:39:03, Ryan Ingram wrote: At the end of that paste, I prove the three Haskell monad laws from the functor laws and monoid-ish versions of the monad laws, but my proofs all rely on a property of natural transformations that I'm not sure how to prove; given type m :- n = (forall x. m x - n x) class Functor f where fmap :: forall a b. (a - b) - f a - f b -- Functor identity law: fmap id = id -- Functor composition law fmap (f . g) = fmap f . fmap g Given Functors m and n, natural transformation f :: m :- n, and g :: a - b, how can I prove (f . fmap_m g) = (fmap_n g . f)? Unless I'm utterly confused, that's (part of) the definition of a natural transformation (for non-category-theorists). Alright, let's pretend I know nothing about natural transformations and just have the type declaration type m :- n = (forall x. m x - n x) And I have f :: M :- N g :: A - B instance Functor M -- with proofs of functor laws instance Functor N -- with proofs of functor laws How can I prove fmap g. f :: M A - N B = f . fmap g :: M A - N B I assume I need to make some sort of appeal to the parametricity of M :- N. Is there some more fundamental law of natural transformations that I'm not aware of that I need to use? Is it possible to write a natural transformation in Haskell that violates this law? -- ryan ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Natural Transformations and fmap
Have you tried generating a free theorem for :- ? (I haven't as I'm writing from my phone) 24.01.2012, в 9:06, Ryan Ingram ryani.s...@gmail.com написал(а): On Mon, Jan 23, 2012 at 8:05 PM, Daniel Fischer daniel.is.fisc...@googlemail.com wrote: On Tuesday 24 January 2012, 04:39:03, Ryan Ingram wrote: At the end of that paste, I prove the three Haskell monad laws from the functor laws and monoid-ish versions of the monad laws, but my proofs all rely on a property of natural transformations that I'm not sure how to prove; given type m :- n = (forall x. m x - n x) class Functor f where fmap :: forall a b. (a - b) - f a - f b -- Functor identity law: fmap id = id -- Functor composition law fmap (f . g) = fmap f . fmap g Given Functors m and n, natural transformation f :: m :- n, and g :: a - b, how can I prove (f . fmap_m g) = (fmap_n g . f)? Unless I'm utterly confused, that's (part of) the definition of a natural transformation (for non-category-theorists). Alright, let's pretend I know nothing about natural transformations and just have the type declaration type m :- n = (forall x. m x - n x) And I have f :: M :- N g :: A - B instance Functor M -- with proofs of functor laws instance Functor N -- with proofs of functor laws How can I prove fmap g. f :: M A - N B = f . fmap g :: M A - N B I assume I need to make some sort of appeal to the parametricity of M :- N. Is there some more fundamental law of natural transformations that I'm not aware of that I need to use? Is it possible to write a natural transformation in Haskell that violates this law? -- ryan ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Natural Transformations and fmap
On Mon, Jan 23, 2012 at 09:06:52PM -0800, Ryan Ingram wrote: On Mon, Jan 23, 2012 at 8:05 PM, Daniel Fischer daniel.is.fisc...@googlemail.com wrote: On Tuesday 24 January 2012, 04:39:03, Ryan Ingram wrote: At the end of that paste, I prove the three Haskell monad laws from the functor laws and monoid-ish versions of the monad laws, but my proofs all rely on a property of natural transformations that I'm not sure how to prove; given type m :- n = (forall x. m x - n x) class Functor f where fmap :: forall a b. (a - b) - f a - f b -- Functor identity law: fmap id = id -- Functor composition law fmap (f . g) = fmap f . fmap g Given Functors m and n, natural transformation f :: m :- n, and g :: a - b, how can I prove (f . fmap_m g) = (fmap_n g . f)? Unless I'm utterly confused, that's (part of) the definition of a natural transformation (for non-category-theorists). Alright, let's pretend I know nothing about natural transformations and just have the type declaration type m :- n = (forall x. m x - n x) And I have f :: M :- N g :: A - B instance Functor M -- with proofs of functor laws instance Functor N -- with proofs of functor laws How can I prove fmap g. f :: M A - N B = f . fmap g :: M A - N B I assume I need to make some sort of appeal to the parametricity of M :- N. This is in fact precisely the free theorem you get from the parametricity of f. Parametricity means that f must act uniformly for all x -- which is an intuitive way of saying that f really is a natural transformation. -Brent ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
