how to factorize propagation of a function over a data type
Hi, Let us consider the following example: ----------------------- class FooClass a where foo1 :: a -> a foo2 :: a -> a instance FooClass Integer where foo1 v = 1 foo2 v = 2 data Bar = Bar1 Integer | Exp1 Bar Bar | Exp2 Bar Bar deriving Show instance FooClass Bar where foo1 b = case b of Bar1 i -> Bar1 (foo1 i) Exp1 b1 b2 -> Exp1 (foo1 b1) (foo1 b2) Exp2 b1 b2 -> Exp2 (foo1 b1) (foo1 b2) foo2 b = case b of Bar1 i -> Bar1 (foo2 i) Exp1 b1 b2 -> Exp1 (foo2 b1) (foo2 b2) Exp2 b1 b2 -> Exp2 (foo2 b1) (foo2 b2) main = do let a = Bar1 3 let b = Bar1 4 let c = Exp1 (Exp2 a b) b print c print $ foo1 c print $ foo2 c ----------------------- We obtain as expected: $ runghc propagate_with_duplicated_code.hs Exp1 (Exp2 (Bar1 3) (Bar1 4)) (Bar1 4) Exp1 (Exp2 (Bar1 1) (Bar1 1)) (Bar1 1) Exp1 (Exp2 (Bar1 2) (Bar1 2)) (Bar1 2) My question is related to the code inside the Fooclass instance definition for Bar: we have repeated code where only "foo1" or "foo2" changes. So the first idea is to write a higher-order function, but it does not work: ----------------------- class FooClass a where foo1 :: a -> a foo2 :: a -> a instance FooClass Integer where foo1 v = 1 foo2 v = 2 data Bar = Bar1 Integer | Exp1 Bar Bar | Exp2 Bar Bar deriving Show propagate :: FooClass a => a -> (a->a) -> a propagate v f = case v of Bar1 i -> Bar1 (f i) Exp1 b1 b2 -> Exp1 (f b1) (f b2) Exp2 b1 b2 -> Exp2 (f b1) (f b2) instance FooClass Bar where foo1 b = propagate b foo1 foo2 b = propagate b foo2 main = do let a = Bar1 3 let b = Bar1 4 let c = Exp1 (Exp2 a b) b print c print $ foo1 c print $ foo2 c ----------------------- The problem is that the type variable `a` in the definition of `propagate` cannot match both the type of i, i.e. an integer, and the type of b1 and b2, i.e. Bar. Putting the function propagate in the typeclass does not help. How to factorize my code? Thanks in advance, TP
I have no time now to answer completely, but I would say that type families
could help.
2013/12/2 TP
Hi,
Let us consider the following example:
----------------------- class FooClass a where foo1 :: a -> a foo2 :: a -> a
instance FooClass Integer where
foo1 v = 1 foo2 v = 2
data Bar = Bar1 Integer | Exp1 Bar Bar | Exp2 Bar Bar deriving Show
instance FooClass Bar where
foo1 b = case b of Bar1 i -> Bar1 (foo1 i) Exp1 b1 b2 -> Exp1 (foo1 b1) (foo1 b2) Exp2 b1 b2 -> Exp2 (foo1 b1) (foo1 b2)
foo2 b = case b of Bar1 i -> Bar1 (foo2 i) Exp1 b1 b2 -> Exp1 (foo2 b1) (foo2 b2) Exp2 b1 b2 -> Exp2 (foo2 b1) (foo2 b2)
main = do
let a = Bar1 3 let b = Bar1 4 let c = Exp1 (Exp2 a b) b
print c print $ foo1 c print $ foo2 c -----------------------
We obtain as expected:
$ runghc propagate_with_duplicated_code.hs Exp1 (Exp2 (Bar1 3) (Bar1 4)) (Bar1 4) Exp1 (Exp2 (Bar1 1) (Bar1 1)) (Bar1 1) Exp1 (Exp2 (Bar1 2) (Bar1 2)) (Bar1 2)
My question is related to the code inside the Fooclass instance definition for Bar: we have repeated code where only "foo1" or "foo2" changes. So the first idea is to write a higher-order function, but it does not work:
----------------------- class FooClass a where foo1 :: a -> a foo2 :: a -> a
instance FooClass Integer where
foo1 v = 1 foo2 v = 2
data Bar = Bar1 Integer | Exp1 Bar Bar | Exp2 Bar Bar deriving Show
propagate :: FooClass a => a -> (a->a) -> a propagate v f = case v of Bar1 i -> Bar1 (f i) Exp1 b1 b2 -> Exp1 (f b1) (f b2) Exp2 b1 b2 -> Exp2 (f b1) (f b2)
instance FooClass Bar where
foo1 b = propagate b foo1 foo2 b = propagate b foo2
main = do
let a = Bar1 3 let b = Bar1 4 let c = Exp1 (Exp2 a b) b
print c print $ foo1 c print $ foo2 c -----------------------
The problem is that the type variable `a` in the definition of `propagate` cannot match both the type of i, i.e. an integer, and the type of b1 and b2, i.e. Bar. Putting the function propagate in the typeclass does not help. How to factorize my code?
Thanks in advance,
TP
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You can replace your `propagate` function by this one:
propagate :: Bar -> (Integer -> Integer) -> Bar
propagate v f = case v of
Bar1 i -> Bar1 (f i)
Exp1 b1 b2 -> Exp1 (propagate b1 f) (propagate b2 f)
Exp2 b1 b2 -> Exp2 (propagate b1 f) (propagate b2 f)
In your code, you were applying the same (w.r.t. to its type) `f` to
Bar and Integer.
Also, your Bar data type contains, at its leaf, an Intger, not a `a`.
You might want to look at functors, and `fmap` too.
2013/12/2 TP
Hi,
Let us consider the following example:
----------------------- class FooClass a where foo1 :: a -> a foo2 :: a -> a
instance FooClass Integer where
foo1 v = 1 foo2 v = 2
data Bar = Bar1 Integer | Exp1 Bar Bar | Exp2 Bar Bar deriving Show
instance FooClass Bar where
foo1 b = case b of Bar1 i -> Bar1 (foo1 i) Exp1 b1 b2 -> Exp1 (foo1 b1) (foo1 b2) Exp2 b1 b2 -> Exp2 (foo1 b1) (foo1 b2)
foo2 b = case b of Bar1 i -> Bar1 (foo2 i) Exp1 b1 b2 -> Exp1 (foo2 b1) (foo2 b2) Exp2 b1 b2 -> Exp2 (foo2 b1) (foo2 b2)
main = do
let a = Bar1 3 let b = Bar1 4 let c = Exp1 (Exp2 a b) b
print c print $ foo1 c print $ foo2 c -----------------------
We obtain as expected:
$ runghc propagate_with_duplicated_code.hs Exp1 (Exp2 (Bar1 3) (Bar1 4)) (Bar1 4) Exp1 (Exp2 (Bar1 1) (Bar1 1)) (Bar1 1) Exp1 (Exp2 (Bar1 2) (Bar1 2)) (Bar1 2)
My question is related to the code inside the Fooclass instance definition for Bar: we have repeated code where only "foo1" or "foo2" changes. So the first idea is to write a higher-order function, but it does not work:
----------------------- class FooClass a where foo1 :: a -> a foo2 :: a -> a
instance FooClass Integer where
foo1 v = 1 foo2 v = 2
data Bar = Bar1 Integer | Exp1 Bar Bar | Exp2 Bar Bar deriving Show
propagate :: FooClass a => a -> (a->a) -> a propagate v f = case v of Bar1 i -> Bar1 (f i) Exp1 b1 b2 -> Exp1 (f b1) (f b2) Exp2 b1 b2 -> Exp2 (f b1) (f b2)
instance FooClass Bar where
foo1 b = propagate b foo1 foo2 b = propagate b foo2
main = do
let a = Bar1 3 let b = Bar1 4 let c = Exp1 (Exp2 a b) b
print c print $ foo1 c print $ foo2 c -----------------------
The problem is that the type variable `a` in the definition of `propagate` cannot match both the type of i, i.e. an integer, and the type of b1 and b2, i.e. Bar. Putting the function propagate in the typeclass does not help. How to factorize my code?
Thanks in advance,
TP
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Vo Minh Thu wrote:
You can replace your `propagate` function by this one:
propagate :: Bar -> (Integer -> Integer) -> Bar propagate v f = case v of Bar1 i -> Bar1 (f i) Exp1 b1 b2 -> Exp1 (propagate b1 f) (propagate b2 f) Exp2 b1 b2 -> Exp2 (propagate b1 f) (propagate b2 f)
In your code, you were applying the same (w.r.t. to its type) `f` to Bar and Integer. Also, your Bar data type contains, at its leaf, an Intger, not a `a`.
You are right, I made a stupid error in my code. The following version indeed works: ---------------- class FooClass a where foo1 :: a -> a foo2 :: a -> a instance FooClass Integer where foo1 v = 1 foo2 v = 2 data Bar = Bar1 Integer | Exp1 Bar Bar | Exp2 Bar Bar deriving Show -- The following line works because there are only integers in the leaves. propagate :: Bar -> (Integer -> Integer) -> Bar propagate v f = case v of Bar1 i -> Bar1 (f i) Exp1 b1 b2 -> Exp1 (propagate b1 f) (propagate b2 f) Exp2 b1 b2 -> Exp2 (propagate b1 f) (propagate b2 f) instance FooClass Bar where foo1 b = propagate b foo1 foo2 b = propagate b foo2 main = do let a = Bar1 3 let b = Bar1 4 let c = Exp1 (Exp2 a b) b print c print $ foo1 c print $ foo2 c ---------------- However, if we add another type in the leaves, we cannot use the solution above. ---------------- class FooClass a where foo1 :: a -> a foo2 :: a -> a instance FooClass Integer where foo1 v = 1 foo2 v = 2 instance FooClass Float where foo1 v = 0.25 foo2 v = 0.5 data Bar = Bar1 Integer | Bar2 Float | Exp1 Bar Bar | Exp2 Bar Bar deriving Show -- This time the following line does not work. propagate :: Bar -> (Integer -> Integer) -> Bar -- The following line does not work either. -- propagate :: FooClass a => Bar -> (a->a) -> Bar propagate v f = case v of Bar1 i -> Bar1 (f i) Bar2 i -> Bar2 (f i) Exp1 b1 b2 -> Exp1 (propagate b1 f) (propagate b2 f) Exp2 b1 b2 -> Exp2 (propagate b1 f) (propagate b2 f) instance FooClass Bar where foo1 b = propagate b foo1 foo2 b = propagate b foo2 main = do let a = Bar1 3 let b = Bar1 4 let c = Exp1 (Exp2 a b) b print c print $ foo1 c print $ foo2 c ----------------
There are several ways to approach this problem. What you are basically
trying to do is map a function over the leaves of your datastructure. So
naturally a function that comes to mind is:
mapBar :: (Integer -> Integer) -> (Float -> Float) -> Bar -> Bar
mapBar f _ (Bar1 i) = Bar1 (f i)
mapBar _ g (Bar2 r) = Bar2 (g r)
mapBar f g (Exp1 e1 e2) = Exp1 (mapBar f g e1) (mapBar f g e2)
mapBar f g (Exp2 e1 e2) = Exp2 (mapBar f g e1) (mapBar f g e2)
And the Bar instance becomes
instance FooClass Bar where
foo1 = mapBar foo1 foo1
foo2 = mapBar foo2 foo2
As far as I understand this is not what you're looking for, as you want the
mapBar function to be agnostic wrt what type the leaves contain. The
minimal assumption that this requires is that the leaf types are a member
of FooClass, and indeed you can write such a map:
mapBar :: (forall a. FooClass a => a -> a) -> Bar -> Bar
mapBar f (Bar1 i) = Bar1 (f i)
mapBar f (Bar2 r) = Bar2 (f r)
mapBar f (Exp1 e1 e2) = Exp1 (mapBar f e1) (mapBar f e2)
mapBar f (Exp2 e1 e2) = Exp2 (mapBar f e1) (mapBar f e2)
instance FooClass Bar where
foo1 = mapBar foo1
foo2 = mapBar foo2
I think this is closer to what you were looking for. The above map requires
-XRankNTypes, because mapBar requires a function that is fully polymorphic
('a' will instantiate to Integer and Float respectively). If you haven't
used higher ranked types before I think it is instructive to think about
why the above signature works and the one you wrote doesn't. In particular
think about at which point the polymorphic type 'a' is instantiated in both
cases, or rather what the "scope" of 'a' is.
On 2 December 2013 19:37, TP
Vo Minh Thu wrote:
You can replace your `propagate` function by this one:
propagate :: Bar -> (Integer -> Integer) -> Bar propagate v f = case v of Bar1 i -> Bar1 (f i) Exp1 b1 b2 -> Exp1 (propagate b1 f) (propagate b2 f) Exp2 b1 b2 -> Exp2 (propagate b1 f) (propagate b2 f)
In your code, you were applying the same (w.r.t. to its type) `f` to Bar and Integer. Also, your Bar data type contains, at its leaf, an Intger, not a `a`.
You are right, I made a stupid error in my code. The following version indeed works:
---------------- class FooClass a where foo1 :: a -> a foo2 :: a -> a
instance FooClass Integer where
foo1 v = 1 foo2 v = 2
data Bar = Bar1 Integer | Exp1 Bar Bar | Exp2 Bar Bar deriving Show
-- The following line works because there are only integers in the leaves. propagate :: Bar -> (Integer -> Integer) -> Bar propagate v f = case v of Bar1 i -> Bar1 (f i) Exp1 b1 b2 -> Exp1 (propagate b1 f) (propagate b2 f) Exp2 b1 b2 -> Exp2 (propagate b1 f) (propagate b2 f)
instance FooClass Bar where
foo1 b = propagate b foo1 foo2 b = propagate b foo2
main = do
let a = Bar1 3 let b = Bar1 4 let c = Exp1 (Exp2 a b) b
print c print $ foo1 c print $ foo2 c ----------------
However, if we add another type in the leaves, we cannot use the solution above.
---------------- class FooClass a where foo1 :: a -> a foo2 :: a -> a
instance FooClass Integer where
foo1 v = 1 foo2 v = 2
instance FooClass Float where
foo1 v = 0.25 foo2 v = 0.5
data Bar = Bar1 Integer | Bar2 Float | Exp1 Bar Bar | Exp2 Bar Bar deriving Show
-- This time the following line does not work. propagate :: Bar -> (Integer -> Integer) -> Bar -- The following line does not work either. -- propagate :: FooClass a => Bar -> (a->a) -> Bar propagate v f = case v of Bar1 i -> Bar1 (f i) Bar2 i -> Bar2 (f i) Exp1 b1 b2 -> Exp1 (propagate b1 f) (propagate b2 f) Exp2 b1 b2 -> Exp2 (propagate b1 f) (propagate b2 f)
instance FooClass Bar where
foo1 b = propagate b foo1 foo2 b = propagate b foo2
main = do
let a = Bar1 3 let b = Bar1 4 let c = Exp1 (Exp2 a b) b
print c print $ foo1 c print $ foo2 c ----------------
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Andras Slemmer wrote:
As far as I understand this is not what you're looking for, as you want the mapBar function to be agnostic wrt what type the leaves contain. The minimal assumption that this requires is that the leaf types are a member of FooClass, and indeed you can write such a map:
mapBar :: (forall a. FooClass a => a -> a) -> Bar -> Bar mapBar f (Bar1 i) = Bar1 (f i) mapBar f (Bar2 r) = Bar2 (f r) mapBar f (Exp1 e1 e2) = Exp1 (mapBar f e1) (mapBar f e2) mapBar f (Exp2 e1 e2) = Exp2 (mapBar f e1) (mapBar f e2)
instance FooClass Bar where foo1 = mapBar foo1 foo2 = mapBar foo2
I think this is closer to what you were looking for. The above map requires -XRankNTypes, because mapBar requires a function that is fully polymorphic ('a' will instantiate to Integer and Float respectively). If you haven't used higher ranked types before I think it is instructive to think about why the above signature works and the one you wrote doesn't. In particular think about at which point the polymorphic type 'a' is instantiated in both cases, or rather what the "scope" of 'a' is.
Thanks a lot. This solution has already been proposed to me in the afternoon by JC Mincke in a private communication. Indeed I did not know RankNTypes. I think I understand your explanation in terms of "scope" of 'a': In the type signature propagate :: (FooClass a)=> Bar -> (a->a) -> Bar which is in fact implicitly propagate :: forall a. (FooClass a)=> Bar -> (a->a) -> Bar it is supposed that the type signature of propagate is valid for a given value of the type variable a, i.e. a given type. Thus we obtain an error if we apply recursively propagate to different types in the code of propagate. Whereas in the type signature propagate :: Bar -> (forall a. (FooClass a) => a->a) -> Bar the type signature of propagate is such that it allows several values for the type variable `a` in its second argument `a->a`. PS: a working code corresponding to my last example: ------------- {-# LANGUAGE RankNTypes #-} class FooClass a where foo1 :: a -> a foo2 :: a -> a instance FooClass Integer where foo1 v = 1 foo2 v = 2 instance FooClass Float where foo1 v = 0.25 foo2 v = 0.5 data Bar = Bar1 Integer | Bar2 Float | Exp1 Bar Bar | Exp2 Bar Bar deriving Show propagate :: Bar -> (forall a. (FooClass a) => a->a) -> Bar propagate v f = case v of Bar1 i -> Bar1 (f i) Bar2 fl -> Bar2 (f fl) Exp1 b1 b2 -> Exp1 (propagate b1 f) (propagate b2 f) Exp2 b1 b2 -> Exp2 (propagate b1 f) (propagate b2 f) -- The two previous lines may be replaced by: -- Exp1 b1 b2 -> Exp1 (f b1) (f b2) -- Exp2 b1 b2 -> Exp2 (f b1) (f b2) instance FooClass Bar where foo1 b = propagate b foo1 foo2 b = propagate b foo2 main = do let a = Bar1 3 let b = Bar1 4 let c = Bar2 0.4 let d = Exp1 (Exp2 a c) b print d print $ foo1 d print $ foo2 d ---------------
TP wrote:
propagate :: Bar -> (forall a. (FooClass a) => a->a) -> Bar
In fact, I do not understand why we have to add the typeclass constraint (FooClass a). Indeed, there is no mention to foo1 and foo2 functions (members of the FooClass typeclass) in the code of propagate: ------- propagate v f = case v of Bar1 i -> Bar1 (f i) Exp1 b1 b2 -> Exp1 (propagate b1 f) (propagate b2 f) Exp2 b1 b2 -> Exp2 (propagate b1 f) (propagate b2 f) ------- propagate deals with any function a priori, not only foo1 and foo2. So how to understand the need for this typeclass constraint? TP
The reason you need the typeclass constraint is not in the use of 'f' but
rather in the use of propagate and what you pass in to it (foo1/foo2 here).
If you leave away the typeclass constraint what you're left with is:
propagate' Bar -> (forall a. a->a) -> Bar
The implementation of this function is the same as before, however your use
of propagate' is restricted:
forall a. a->a is a very "strict" type, in fact the only inhabitant of this
type is 'id' (and bottom, but disregard that here), which means the only
way to call propagate is to pass in 'id'. Try it yourself!
Related note: there is a proof that in fact the only inhabitant of (forall
a. a -> a) is 'id' and it is the consequence of the "parametricity"
property. It is a very neat result I suggest you look it up!
On 2 December 2013 22:28, TP
TP wrote:
propagate :: Bar -> (forall a. (FooClass a) => a->a) -> Bar
In fact, I do not understand why we have to add the typeclass constraint (FooClass a). Indeed, there is no mention to foo1 and foo2 functions (members of the FooClass typeclass) in the code of propagate:
------- propagate v f = case v of Bar1 i -> Bar1 (f i) Exp1 b1 b2 -> Exp1 (propagate b1 f) (propagate b2 f) Exp2 b1 b2 -> Exp2 (propagate b1 f) (propagate b2 f) -------
propagate deals with any function a priori, not only foo1 and foo2. So how to understand the need for this typeclass constraint?
TP
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Andras Slemmer wrote:
The reason you need the typeclass constraint is not in the use of 'f' but rather in the use of propagate and what you pass in to it (foo1/foo2 here). If you leave away the typeclass constraint what you're left with is:
propagate' Bar -> (forall a. a->a) -> Bar
The implementation of this function is the same as before, however your use of propagate' is restricted: forall a. a->a is a very "strict" type, in fact the only inhabitant of this type is 'id' (and bottom, but disregard that here), which means the only way to call propagate is to pass in 'id'. Try it yourself!
Indeed I have tried, it works as you say.
Related note: there is a proof that in fact the only inhabitant of (forall a. a -> a) is 'id' and it is the consequence of the "parametricity" property. It is a very neat result I suggest you look it up!
Interesting. I have tried to google on the topic, but I find mainly research articles. For example: https://www.google.fr/search?client=ubuntu&channel=fs&q=haskell+%22parametricity+property%22&ie=utf-8&oe=utf-8&gws_rd=cr&ei=P_acUvboDse_0QX9mYDADQ#channel=fs&q=%22parametricity+property%22+haskell Are there textbooks where a proof of this fact could be found? I'm an autodidact (who also benefits from help of guys like you), I don't know what lectures on type theory at university level could look like. Thanks TP
On Mon, Dec 02, 2013 at 10:30:11PM -0400, TP wrote:
Andras Slemmer wrote:
Related note: there is a proof that in fact the only inhabitant of (forall a. a -> a) is 'id' and it is the consequence of the "parametricity" property. It is a very neat result I suggest you look it up!
Are there textbooks where a proof of this fact could be found? I'm an autodidact (who also benefits from help of guys like you), I don't know what lectures on type theory at university level could look like.
I guess the most accessible reference might be Wadler 1989 "Theorems for Free". http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.38.9875 Tom
class FooClass a where foo1 :: a -> a foo2 :: a -> a
instance FooClass Integer where
foo1 v = 1 foo2 v = 2
data Bar = Bar1 Integer | Exp1 Bar Bar | Exp2 Bar Bar deriving Show
instance FooClass Bar where
foo1 b = case b of Bar1 i -> Bar1 (foo1 i) Exp1 b1 b2 -> Exp1 (foo1 b1) (foo1 b2) Exp2 b1 b2 -> Exp2 (foo1 b1) (foo1 b2)
foo2 b = case b of Bar1 i -> Bar1 (foo2 i) Exp1 b1 b2 -> Exp1 (foo2 b1) (foo2 b2) Exp2 b1 b2 -> Exp2 (foo2 b1) (foo2 b2)
I think you're really asking for a generics library like 'uniplate': http://community.haskell.org/~ndm/darcs/uniplate/uniplate.htm Greetings, Daniel
participants (6)
-
Alejandro Serrano Mena -
Andras Slemmer -
Daniel Trstenjak -
Tom Ellis -
TP -
Vo Minh Thu