19:31 < edwardk> bitemyapp: a transducer is recognizing that the signature of foldl splits
19:31 < edwardk> :t foldl
19:31 < lambdabot> (b -> a -> b) -> b -> [a] -> b
19:31 < bitemyapp> (b -> a -> b) -> (b -> [a] -> b) ?
19:31 < dfeuer> Splits?
19:31 < edwardk> yes, now replace a and [a] with two things and quantify over the bs
19:31 < edwardk> done
19:32 < monochrom> dfeuer: do you want -O1 or -O2?
19:32 < edwardk> type Transducer x y = forall b. (b -> y -> b) -> (b -> x -> b)
19:32 < edwardk> tada transducer
19:32 < edwardk> they compose like lenses
19:32 < edwardk> nuff said
19:32 < edwardk> now the left and right are 'of the same shape
19:32 < edwardk> :t foldl.foldl
19:32 < lambdabot> (b -> a -> b) -> b -> [[a]] -> b
19:32 < edwardk> :t foldl.foldl.foldl
19:32 < lambdabot> (b -> a -> b) -> b -> [[[a]]] -> b
19:33 < edwardk> yay transducers, next? =)
19:35 < bitemyapp> dfeuer: the last definition Hickey gave of transducers was: "a transducer is just a pre-fused
Kleisli arrows in the list monad"
19:35 < dfeuer> bitemyapp, yeeeeeeeeeeaaaaahhhhhh.
19:35 < bitemyapp> edwardk: ^^ does that definition make sense to you?
19:35 < mauke> I like git
19:35 < bitemyapp> tel added that there was ambient mutable state too.
19:35 < xplat> edwardk: what are some other transducers besides foldl and (flip.foldr.flip) or whatever it was?
19:36 < edwardk> xplat: you can define maps and concatmaps, etc.
19:36 < bitemyapp> I'm not entirely sure what the list monad has to do with it, for example.
19:36 < bitemyapp> since it seems to be about folding.
19:36 < edwardk> xplat: you basically just have to say how you explode 'y' into a bunch of 'b's and fold through it
left to right
19:36 < edwardk> er y into a bunch of x's and fold through them left to right
19:37 < edwardk> (or is it the other way around)
19:37 < bitemyapp> I guess that could be monadic.
19:37 < bitemyapp> sorta?
19:37 < bitemyapp> you're not using join though.
19:37 < xplat> x into a bunch of ys
19:37 < bitemyapp> you'd be mapping an x -> [y] and then reducing with fold.
19:37 < bitemyapp> rather than join.
19:37 < bitemyapp> based on edwardk's casual definition.
19:37 < edwardk> @let type Transducer a b = forall r. (r -> b -> r) -> r -> a -> r
19:37 < lambdabot> Defined.
19:37 < edwardk> :t foldl :: Transducer [a] a
19:37 < lambdabot> (r -> a -> r) -> r -> [a] -> r
19:37 < xplat> bitemyapp: it's 'exploding an x into a sequence of ys' i think that brings in the list monad
19:38 < bitemyapp> xplat: yeah, that's what I meant by "I guess that could be monadic"
19:38 < bitemyapp> xplat: but the thing is, it's not really.
19:38 < bitemyapp> xplat: you just need fmap for that.
19:38 < xplat> i wonder if these things have other shapes like traversals and folds do
19:38 < bitemyapp> xplat: fmap can do x -> [y], you need monad for concatMap - ie: join
19:39 < bitemyapp> xplat: since we're not using the parts that are specific to the list monad (bind/join), it's not
really about the list monadic because the reduction is whatever fold we defined.
19:39 < edwardk> :t (\f g r -> g r . f) :: (a -> b) -> Transducer a b
19:39 < lambdabot> (a -> b) -> (r -> b -> r) -> r -> a -> r
19:39 < bitemyapp> so I'm not totally convinced but I can see why the nested introduction of lists would put
somebody into thinking it was about the list monad.
19:39 < edwardk> :let tmap :: (a -> b) -> Transducer a b; tmap f g r = g r . f
19:39 < edwardk> @let tmap :: (a -> b) -> Transducer a b; tmap f g r = g r . f
19:39 < lambdabot> Defined.
19:39 < Ferdirand> hmm
19:40 < edwardk> :t foldl.tmap succ
19:40 < lambdabot> Enum b1 => (b -> b1 -> b) -> b -> [b1] -> b
19:40 < Ferdirand> so what are the other interesting Transducer types, beyond Transducer a [a] ?
19:40 < xplat> Ferdirand: well, you can use foldl or foldr for any Foldable
19:40 < edwardk> :t (\f g r -> foldl g r . f) :: (a -> [b]) -> Transducer a b
19:40 < lambdabot> (a -> [b]) -> (r -> b -> r) -> r -> a -> r
19:40 < edwardk> there you get a concatMap, etc.
19:41 < edwardk> Ferdirand: work for any Foldable, etc.
19:41 < edwardk> anyways its basically just a crippled left-biased Fold
19:41 < Ferdirand> okay, but for instance, does (Transducer [a] a) corresponds to something meaningful ?
19:42 < edwardk> by biasing left it can't handle infinite cases
19:42 < edwardk> Ferdirand: sure its a function from [a] to basically [a] ;)
19:42 < xplat> so far this all seems like stuff that would work with a Fold too; is there a difference?
19:42 < edwardk> Transducer a b is a -> [b] in disguise
19:42 < edwardk> for a finite case
19:42 < edwardk> xplat: no
19:42 < edwardk> xplat: other than the fact that this version doesn't handle infinite cases
19:42 < xplat> it seems like you could get a transducer from any fold by reading it out into Endo
19:43 < edwardk> yes
19:43 < edwardk> :t foldlOf
19:43 < lambdabot> Getting (Dual (Endo r)) s a -> (r -> a -> r) -> r -> s -> r
19:43 < edwardk> yay transducer
19:43 < edwardk> next =)
19:44 < mauke> http://jlongster.com/Transducers.js--A-JavaScript-Library-for-Transformation-of-Data are these the
same transducers?
19:44 < monochrom> dfeuer: OK, I use -O2, the experiment code is "f x y = quot x y + rem x y" and "g x y = q + r
where (q, r) = quotRem x y". whether Int, Integer, or Integral, f and g are different in GHC
core.
19:44 < edwardk> xplat: transducers themselves aren't terribly brain bending, they are just a use of the same
pattern that lens uses. finding a shape that factors through (->) when something gets quantified
and using (.) and id as the way to compose them
19:45 < edwardk> its nice that they've been rediscovered, but there isn't a lot of magic there
Martin:
Hello all,
many times I see a problem and I say to myself: "there is some state". I then play around with the state monad and often I don't get anywhere. Then at some point I realizes, that all I need is a simple fold. I don't think I ever used the state monad outside of toy examples.
Can someone give me some insights when the State Monad is beneficial and where a fold is the better choice.
John Wiegley:
martin martin.drautzburg@web.de writes:
Can someone give me some insights when the State Monad is beneficial and where a fold is the better choice.
Looking at the type of a fold:
foldr :: (a -> b -> b) -> b -> [a] -> bIf we juggle the arguments we get:
foldr :: (a -> b -> b) -> [a] -> b -> bAnd if we imagine State b () actions, we can directly rewrite this as:
foldrS :: (a -> State b ()) -> [a] -> State b ()Which generalizes to:
foldrS :: MonadState b m => (a -> m ()) -> [a] -> m ()Which is roughly the same thing as using mapM_ over our State monad:
mapM_ :: Monad m => (a -> m b) -> [a] -> m ()In other words, these two forms in our example say the same thing:
foldr f b xs
execState (mapM_ f' xs) bWith the only difference being the types of f and f':
f : a -> b -> b
f' : a -> State b ()The other question you asked is when to choose one over the other. Since they are equivalent, it's really up to you. I tend to prefer using a fold over State, to keep things on the level of functions and values, rather than pulling in monads and possibly monad transformers unnecessarily.
But it's a very good thing to hold these isomorphisms in your mind, since you can then freely switch from one representation to another as need be. This is true of a lot of type equivalences throughout Haskell, where the more things you can see as being essentially the same, the more freedom you have to find the best abstraction for a particular context.
doubleEveryOther :: [Integer] -> [Integer]
doubleEveryOther list = reverse .doubleEveryOtherForward . reverse $ list03:28 < bitemyapp> fbernier: reverse the list, double every other number, re-reverse the list.
03:28 < bitemyapp> fbernier: the "dot" operator is just function composition.
03:28 < bitemyapp> it's nothing special, just another function.
03:28 < bitemyapp> :t (.)
03:28 < lambdabot> (b -> c) -> (a -> b) -> a -> c
03:30 < bitemyapp> fbernier: the use of $ in that function is a little idiosyncratic and unnecessary, but not problematic.
03:37 < ReinH> fbernier: there's a missing space after the . is all
03:38 < ReinH> fbernier: f x = foo $ x ==> f = foo
03:39 < ReinH> so f x = foo . bar $ x ==> f = foo . bar
03:39 < bitemyapp> fbernier: I think it's just making it point-free in this case.
03:39 < bitemyapp> @pl f x = c . b . a $ x
03:39 < lambdabot> f = c . b . a
03:39 < bitemyapp> yeah, that ^^
03:39 < bitemyapp> fbernier: identical ^^
03:40 < ReinH> fbernier: generally, when you see a $ you can wrap the things on either side with parens and get the same expression:
03:40 < ReinH> f x = foo . bar . bazz $ x ==> f x = (foo . bar . bazz) x
03:40 < ReinH> since (x) = x, ofc
03:41 < bitemyapp> @src ($)
03:41 < lambdabot> f $ x = f x
03:41 < bitemyapp> fbernier: That's the definition of $, only other thing missing is the high precedence set for it.
03:41 < ReinH> the exception is chains of $, like foo $ bar $ baz, where you have to parenthesize in the right direction
03:41 < ReinH> or the left direction, depending on how you look at it
03:42 < bitemyapp> fbernier: http://hackage.haskell.org/package/base-4.7.0.1/docs/Prelude.html ctrl-f for $ to see more
03:42 < bitemyapp> fbernier: infixr 0 is the precedence, highest there is AFAIK
03:42 < bitemyapp> fbernier: the "infixr" means it's right associative
03:42 < bitemyapp> fbernier: as opposed to infixl which would mean left associative
03:43 < ReinH> bitemyapp: or lowest, depending on how you look at it. ;)
03:43 < bitemyapp> foo $ bar $ baz ~ foo (bar (baz))
03:43 < bitemyapp> but if it was infixl
03:43 < bitemyapp> (((foo) bar) baz)
04:12 < ReinH> all infix operators can be written prefix
04:12 < ReinH> with this one weird trick. Other haskellers hate him.
04:13 < bitemyapp> > ($) id 1
04:13 < lambdabot> 1
04:13 < bitemyapp> > id $ 1
04:13 < lambdabot> 1
04:13 < bitemyapp> > id 1
04:13 < lambdabot> 1
05:00 < ReinH> pyro-: well, "reduce" already has a typeclass, depending on what you mean
05:00 < ReinH> so does "evaluation", depending on what you mean
05:02 < pyro-> ReinH: reduce is lambda calculus under strict evaluation
05:02 < ReinH> Yep, and it's also the other thing too.
05:02 < ReinH> ;)
05:03 < pyro-> :|
05:03 < pyro-> oh, like on lists?
05:04 < mm_freak_> dealing with ASTs is a real joy in haskell, because most of the code writes itself =)
05:10 < pyro-> now i am writing a cpsTransform function :D
05:10 < pyro-> it already works, but the current version introduces superflous continuations
05:10 < pyro-> so i am trying to fix :D
05:10 < ReinH> pyro-: Here's a CPS transform function: flip ($)
05:11 < pyro-> i will find out about flip
05:11 < ReinH> @src flip
05:11 < lambdabot> flip f x y = f y x
05:11 < ReinH> pyro-: the essence of CPS can be described as follows:
05:11 < ReinH> :t flip ($)
05:11 < lambdabot> b -> (b -> c) -> c
05:12 < ReinH> is the type of a function which takes a value and produces a suspended computation that takes a continuation and runs it against the value
05:12 < ReinH> for example:
05:12 < ReinH> > let c = flip ($) 3 in c show
05:12 < lambdabot> "3"
05:12 < ReinH> > let c = flip ($) 3 in c succ
05:12 < lambdabot> 4
05:13 < mm_freak_> direct style: f x = 3*x + 1
05:13 < mm_freak_> CPS: f x k = k (3*x + 1)
05:13 < mm_freak_> the rules are: take a continuation argument and be fully polymorphic on the result type
05:13 < mm_freak_> f :: Integer -> (Integer -> r) -> r
05:14 < mm_freak_> as long as your result type is fully polymorphic and doesn't unify with anything else in the type signature you can't do anything wrong other than to descend
into an infinite recursion =)
05:14 < mm_freak_> good: (Integer -> r) -> r
05:15 < mm_freak_> bad: (Integer -> String) -> String
05:15 < mm_freak_> bad: (Num r) => (Integer -> r) -> r
05:15 < mm_freak_> bad: r -> (Integer -> r) -> r
05:15 < pyro-> but flip ($) is not what i had in mind :D
05:16 < mm_freak_> that's just one CPS transform… there are many others =)
05:16 < ReinH> No, it's probably not.
05:16 < ReinH> But other things are pretty much generalizations of that
type Variable = String
data Expression = Reference Variable
| Lambda Variable Expression
| Combination Expression Expression
type Kvariable = String
data Uatom = Procedure Variable Kvariable Call
| Ureference Variable
data Katom = Continuation Variable Call
| Kreference Variable
| Absorb
data Call = Application Uatom Uatom Katom
| Invocation Katom Uatom
cpsTransform :: Expression -> Katom -> Call
cpsTransform (Reference r) k = Invocation k $ Ureference r
cpsTransform (Lambda p b) k = Invocation k $ Procedure p
"k" $
cpsTransform b $ Kreference "k"
cpsTransform (Combination a b) k = cpsTransform a $ Continuation "v" $ cpsTransform b k05:38 < ReinH> So for example, if you have an incredibly simple expression language like data Expr a = Val a | Neg a | Add a a
05:38 < ReinH> a (more) initial encoding of an expression would be Add (Val 1) (Neg (Val 1))
05:38 < ReinH> A (more) final encoding might be (1 - 1) or even 0
05:39 < ReinH> The initial encoding generally is more flexible (you can still write a double-negation elimination rule, for instance
05:39 < ReinH> the final encoding is less flexible, but also does more work up-front
05:40 < ReinH> More initial encodings tend to force you to use quantification and type-level tricks, CPS and pre-applied functions tend to appear more in final encodings
05:40 < ReinH> An even smaller example:
05:40 < ReinH> \f z -> foldr f z [1,2,3] is a final encoding of the list [1,2,3]
05:41 < ReinH> pyro-: I'm not really a lisper, but I'm always looking for good reading material
05:41 < ReinH> for bonus points, the foldr encoding is *invertible* as well :)
05:44 < ReinH> pyro-: the relevance is that you seem to be using the cps transform in a more initial encoding than I usually see it
05:44 < ReinH> not that this is at all bad
05:46 < bitemyapp> ReinH: where does the invertibility in the final encoding come from?
05:46 < ReinH> foldr (:) [] :)
05:46 < ReinH> it's not generally so
05:46 < bitemyapp> > foldr (:) [] [1, 2, 3]
05:46 < lambdabot> [1,2,3]
05:47 < bitemyapp> I may not understand the proper meaning of invertibility in this case.
05:47 < bitemyapp> Do you mean invertibility from final to initial encoding?
05:47 < ReinH> Just that, yes
05:47 < bitemyapp> how would it get you back to final from initial?
05:47 < ReinH> I'm not sure if that's the correct term
05:47 < bitemyapp> I don't think it is, but the intent is understood and appreciated.
05:48 < bitemyapp> invertibility implies isomorphism, implies ability to go final -> initial -> final
05:48 < ReinH> well, there is an isomorphism
05:48 < bitemyapp> well, we've established final -> initial, where's initial -> final for this example?
05:49 < bitemyapp> I figured it was a morphism of some sort, but with only a final -> initial and not a way to get back, I wasn't sure which.
05:49 < ReinH> toInitial k = k (:) []; toFinal xs = \f z -> foldr f z xs
05:49 < bitemyapp> thank you :)
05:51 < ReinH> bitemyapp: usually one loses information going from initial to final though
05:51 < ReinH> there's probably an adjunction here
05:51 < ReinH> there's always an adjunction
05:52 < ReinH> lol of course there's an adjunction
Asker:
I am teaching myself haskell. The first impression is very good. But phrase "haskell is polynomially reducible" is making me sad :(. Anyway I am trying to backport my algorithm written in C. The key to performance is to have ability to remove element from the end of a list in O(1). But the original haskell functions last and init are O(n). My questions are:
- Is last function is something like "black box" written in C++ which perform O(1)? So I shouldn't even try to imagine some haskell O(1) equivalent.
- Or will optimizer (llvm?) reduce init&last complexity to 1?
- Some people suggest to use sequences package, but still how do they implement O(1) init&last sequences equivalent in haskell?
Tom Ellis:
I'm rather confused about your question. If you want a Haskell data structure that supports O(1) head, tail, init and last why not indeed use Data.Sequence as has been suggested? As for how it's implemented, it uses the (very cool) fingertree datastructure. See here for more details:
Asker:
Tom said that finger tree gives us O(1) on removing last element, but in haskell all data is persistent. So function should return list as is minus last element. How it could be O(1)? This is just blows my mind...
My hypothesis is that somehow compiler reduces creating of a new list to just adding or removing one element. If it is not so. Then even ':' which is just adding to list head would be an O(n) operation just because it should return brand new list with one elem added. Or maybe functional approach uses pretty much different complexity metric, there copying of some structure "list" for example is just O(1)? If so then Q about compiler is still exists.
Tom Ellis:
Sounds like magic doesn't it :)
But no, there's no compiler magic, just an amazing datastructure. The caveat is that the complexity is amortised, not guaranteed for every operation. Have a look at the paper if you learn about how it works. It's linked from the Hackage docs.
http://hackage.haskell.org/package/containers-0.2.0.1/docs/Data-Sequence.html
Asker:
Jake It would be great if you give some examples when find your notebook :) And link to the book about pure functional data structures which you are talking about. Also If some "haskell.org" maintainers are here I'd like to recommend them to pay more attention to optimality/performance questions. Because almost first question which is apeared in head of standart C/C++ programmer is "Do I get same perfomance?" (even if he do not need it). Maybe some simple and cool PDF tutorial which describes why haskell could be as fast as others will be great to have.
Richard A. O'Keefe:
I am teaching myself haskell. The first impression is very good... Anyway I am trying to backport my algorithm written in C. The key to performance is to have ability to remove element from the end of a list in O(1).
You can't. Not in any programming language. That's because lists are one of many possible implementations of the "sequence" concept, and they are optimised to support some operations at the expense of others. At the beginning level, you should think of all Haskell data structures as immutable; fixed; frozen; forever unchanged. You can't even remove an element from the front of a Haskell list, at all. All you can do is to forget about the original list and concentrate on its tail.
But the original haskell functions last and init are O(n).
Haskell lists are singly linked lists. Even by going to assembly code, you could not make these operations O(1) without using a different data structure.
My questions are:
- Is last function is something like "black box" written in C++ which perform O(1)?
No.
- Or will optimizer (llvm?) reduce init&last complexity to 1?
No.
- Some people suggest to use sequences package, but still how do they implement O(1) init&last sequences equivalent in haskell?
Well, you could try reading Chris Okasaki's functional data structures book.
There is a classic queue representation devised for Lisp last century which represents <a,b,c,d,e> by ([a,b],[e,d,c]) so that you can push and pop at either end. When the end you are working on runs out, you reverse the other end, e.g., ([],[e,d,c]) -> ([c,d,e],[]).
That can give you a queue with amortised constant time. (There is a technical issue which I'll avoid for now.)
But let's start at the beginning. You have an interesting problem, P. You have an algorithm for it, A, written in C. You want an algorithm for it, H, written in Haskell. Your idea is to make small local syntactic changes to A to turn in into H. That's probably going to fail, because C just loves to smash things, and Haskell hates to. Maybe you should be using quite a different approach, one that would be literally unthinkable in C. After all, being able to do things that are unthinkable in C is one of the reasons for learning Haskell.
Why not tell us what problem P is?
Tony Morris:
data SnocList a = SnocList ([a] -> [a])
Inserts to the front and end in O(1).
Edward Kmett:
Note: all of the options for playing with lists and queues and fingertrees come with trade-offs.
Finger trees give you O(log n) appends and random access, O(1) cons/uncons/snoc/unsnoc etc. but cost you infinite lists.
Realtime queues give you the O(1) uncons/snoc. There are catenable output restricted deques that can preserve those and can upgrade you to O(1) append, but we've lost unsnoc and random access along the way.
Skew binary random access lists give you O(log n) drop and random access and O(1) cons/uncons, but lose the infinite lists, etc.
Tarjan and Mihaescu's deque may get you back worst-case bounds on more of the, but we still lose O(log n) random access and infinite lists.
Difference lists give you an O(1) append, but alternating between inspection and construction can hit your asymptotics.
Lists are used by default because they cleanly extend to the infinite cases, anything more clever necessarily loses some of that power.
20:26 < ifesdjee_> hey guys, could anyone point me to the place where I could read up on how `listen` of writer monad works?
20:26 < ifesdjee_> can't understand it from type signature, don't really know wether it does what i want..
20:30 < ReinH> :t listen
20:30 < lambdabot> MonadWriter w m => m a -> m (a, w)
20:31 < mm_freak_> ifesdjee_: try this: runWriterT (listen (tell "abc" >> tell "def") >>= liftIO . putStrLn . snd)
20:33 < mm_freak_> in any case 'listen' really just embeds a writer action and gives you access to what it produced
20:33 < ifesdjee_> most likely i misunderstood what happens in `listen`...
20:34 < ifesdjee_> i thought i could access current "state" of writer
20:34 < mm_freak_> remember that the embedded writer's log still becomes part of the overall log
20:34 < mm_freak_> execWriter (listen (tell "abc") >> tell "def") = "abcdef"
20:35 < mm_freak_> all you get is access to that "abc" from within the writer action
20:35 < ifesdjee_> yup, I see
20:35 < ifesdjee_> thank you a lot!
20:35 < mm_freak_> my pleasure
20:37 < mm_freak_> i wonder why there is no evalWriter*
20:37 < ifesdjee_> not sure, really
21:32 < sclv> does anyone have a citation for the introduction of free monads?
21:33 < sclv> they’re so universally used in the literature nobody cites where they came from anymore
21:33 < sclv> in a computational context goes back to ’91 at least
21:40 < sclv> found it
21:40 < sclv> coequalizers and free triples, barr, 1970
http://link.springer.com/article/10.1007%2FBF01111838#page-1
Note: Seeing a paper on free monoids dating to 1972 by Eduardo J. Dubuc.
03:13 < shachaf> dolio: Do you know what people mean when they say rank-2 types are inferrable?
03:14 < dolio> Not really. I've never taken the time to understand it.
03:16 < dolio> One reading makes no sense, I think. Because rank-2 is sufficient to lack principal types, isn't it?
03:17 < dolio> Or perhaps it isn't....
03:17 < shachaf> Well, you can encode existentials.
03:17 < dolio> Can you?
03:17 < dolio> forall r. (forall a. a -> r) -> r
03:17 < dolio> I guess that's rank-2.
03:18 < shachaf> You can give rank-2 types to expressions like (\x -> x x)
03:18 < shachaf> What type do you pick for x?
03:19 < dolio> forall a. a -> β
03:19 < dolio> Presumably.
03:20 < shachaf> Does β mean something special here?
03:20 < dolio> It's still open.
03:20 < dolio> Greek for unification variables.
03:21 < shachaf> OK, but what type do you infer for the whole thing?03:21 < dolio> forall r. (forall a. a -> r) -> r
03:23 < dolio> (\f -> f 6) : forall r. (Int -> r) -> r
03:23 < dolio> Is that a principal type?
03:23 < shachaf> Do you allow type classes?
03:24 < dolio> People who say rank-2 is decidable certainly shouldn't be thinking about type classes.
03:24 < shachaf> I guess with impredicativity the type you gave works... Well, does it?
03:25 < dolio> Maybe rank-2 is sufficient to eliminate all ambiguities.
03:25 < dolio> Like, one common example is: [id]
03:25 < dolio> Is that forall a. [a -> a] or [forall a. a -> a]
03:25 < dolio> But, we're not talking about Haskell, we're talking about something like system f.
03:26 < dolio> So you'd have to encode.
03:26 < dolio> And: (forall r. ((forall a. a -> a) -> r -> r) -> r -> r) is rank-3.
03:27 < shachaf> I guess...
03:27 < dolio> If I had to guess, that's what the answer is.
-
Practical type inference for arbitrary-rank types - Peyton Jones, Vytinotis, Weirich, Shields
-
https://stackoverflow.com/questions/9259921/haskell-existential-quantification-in-detail
02:17 < bartleby> so I understand sum and product types, but why does a -> b have b^a cardinality?
02:23 < Iceland_jack> How many functions are there of type
02:23 < Iceland_jack> () -> b
02:23 < Iceland_jack> if b has 5 inhabitants?
02:23 < bartleby> 5
02:24 < Iceland_jack> which is 5^1 right?
02:24 < Iceland_jack> You'll want to look at Chris's blog: http://chris-taylor.github.io/blog/2013/02/10/the-algebra-of-algebraic-data-types/
02:24 < bartleby> yes
02:24 < bartleby> purple link, hm... I've been there, might've missed that.
02:25 < Iceland_jack> Now what about
02:25 < Iceland_jack> Bool -> b
02:25 < Iceland_jack> if b has 3 inhabitants
02:25 < Iceland_jack> You can gain your intuition by working these things out for increasingly more involved types
02:26 < bartleby> I was trying this, but it looked like a product type... I'm doing something wrong
02:26 < bartleby> let me see this case
02:26 < Iceland_jack> sure
02:27 < bartleby> wait, if I have one pattern for True and another for False, does it count as a single function? or two?
02:28 < Iceland_jack> If they're two patterns in the same function then it's the same function
02:28 < Iceland_jack> I.e. in the function definition
02:28 < Iceland_jack> f True = ...
02:28 < Iceland_jack> f False = ...
02:28 < Iceland_jack> 'f' is a single function
02:29 < Iceland_jack> and for the first ellipsis '...' you have one of three choices (b = {b1, b2, b3}) and same for the second one
02:29 < pyro-> does b^a include non total functions?
02:29 < Iceland_jack> no
02:29 < pyro-> why is that?
02:30 < Iceland_jack> Because it breaks all sorts of reasoning and makes it more complicated
02:30 < pyro-> :D
02:30 < bartleby> no? I thought that was what I was missing...
02:30 < Iceland_jack> bartleby: How many functions of type
02:30 < Iceland_jack> Bool -> ()
02:31 < bartleby> yes, that's where I'm confused. I'd guess one?
02:31 < Iceland_jack> Right, because the only choice is
02:31 < Iceland_jack> fn True = ()
02:31 < Iceland_jack> fn False = ()
02:31 < bartleby> matching True and False, but only returning ()
02:32 < Iceland_jack> so the number of function |Bool -> ()| is |()| ^ |Bool|
02:32 < Iceland_jack> |()| ^ |Bool|
02:32 < Iceland_jack> = 1 ^ 2
02:32 < Iceland_jack> = 1
02:32 < bartleby> ah, I think I get it
02:33 < Iceland_jack> And there are 2 functions from
02:33 < Iceland_jack> Bool -> ()
02:33 < Iceland_jack> conversely
02:33 < Iceland_jack> oops, () -> Bool I meant
02:33 < pyro-> Just by sitting in this channel I a learning things :D bartleby, how is it that cardinality of a type has interested you? I haven't even heard the term before
02:33 < Iceland_jack> 'const False' and 'const True' respectively
02:33 < bartleby> Iceland_jack: because 2^1
02:33 < Iceland_jack> Precisely
02:34 < Iceland_jack> pyro-: You should definitely read up on the 'Algebra of Algebraic Data Types'
http://chris-taylor.github.io/blog/2013/02/10/the-algebra-of-algebraic-data-types/
02:34 < pyro-> thanks
02:34 < Iceland_jack> Lated parts discuss some more advanced uses
02:34 < Iceland_jack> *Later
02:34 < bartleby> pyro-: Algebraic Data Types, means you have an algebra for dealing with them.
02:35 < Iceland_jack> Just like you knew that
02:35 < Iceland_jack> 1 + 2 = 2 + 1
02:35 < Iceland_jack> in grade school so you can know that
02:35 < Iceland_jack> Either () Bool ≅ Either Bool ()
02:35 < bartleby> blowed my mind when I read about zippers, but I hadn't seen it with functions yet
02:36 < Iceland_jack> viewing (+) = Either, 1 = () and 2 = Bool
02:36 < Iceland_jack> It also means that you can define Bool as
02:36 < Iceland_jack> type Bool = Either () ()
02:36 < Iceland_jack> rather than
02:36 < Iceland_jack> data Bool = False | True
02:36 < Iceland_jack> since
02:36 < Iceland_jack> 1 + 1 ≅ 2
02:37 < Iceland_jack> Given the recent pattern synonyms extensions (PatternSynonyms) you can even use the same constructors and pattern match
02:37 < pyro-> Thats interesting
02:37 < Iceland_jack> type (+) = Either
02:37 < Iceland_jack> type BOOL = () + ()
02:37 < Iceland_jack> pattern TRUE = Right () :: BOOL
02:37 < Iceland_jack> pattern FALSE = Left () :: BOOL
02:38 < Iceland_jack> and then
02:38 < Iceland_jack> not :: BOOL -> BOOL
02:38 < Iceland_jack> not TRUE = FALSE
02:38 < Iceland_jack> not FALSE = TRUE
02:38 < pyro-> what abut values instead of types? 1 + 2 = 2 + 1 works for Int. what about algebra for values of other type?
02:38 < Iceland_jack> pyro-: You're not actually using numbers
02:38 < Iceland_jack> 1 is just a nice and confusing way to refer to the type ()
02:38 < pyro-> i understand
02:38 < bartleby> whoa, easy there boy! I'm overheating with 2^2 here
02:38 < Iceland_jack> not the value 1
02:38 < bartleby> :-D
02:38 < pyro-> thanks
02:39 < Iceland_jack> bartleby: Slowing down :)
02:39 < pyro-> actually that i'm not using numbers is kind of the point right?
02:39 < Iceland_jack> well it makes the analogy with elementary arithmetic clearer
02:39 < bartleby> pyro-: you are counting possible values of that type
02:40 < Iceland_jack> So you can write '2' for Bool because Bool has two things
02:40 < bartleby> so Either () Bool has three because: Left (), or Right True, or Right False
02:40 < Iceland_jack> Maybe Bool would be 3
02:40 < Iceland_jack> Yes exactly
02:40 < Iceland_jack> and thus
02:40 < Iceland_jack> Either () Bool ≅ Maybe Bool
02:41 < Iceland_jack> and also
02:41 < Iceland_jack> Maybe a ≅ Either () a
02:41 < Iceland_jack> If you define
02:41 < Iceland_jack> Maybe b = 1 + b
02:41 < Iceland_jack> Either a b = a + b
02:41 < Iceland_jack> then it becomes fairly clear
02:44 < bartleby> ah, I think it clicked here. I managed to list Bool -> Bool, four different functions
02:46 < Iceland_jack> and then for Bool -> Three where |Three| = 3 you have 3 independent choices for True and False so you have 3 * 3 = 3^2
02:46 < Iceland_jack> and so forth
02:46 < Iceland_jack> hope this clears things up a bit
02:46 < bartleby> I was unsure about partial fuctions, but now it makes sense. It's just a permutations of b I think (not sure if permutation is the right word)
02:47 < bartleby> how many arrangements with `a` elements of type `b` can I make?
02:51 < bartleby> Iceland_jack: thank you. I see that I have that page bookmarked, but I think I didn't get that Functions sections at the time
02:52 < bartleby> in fact, it's still confusing...
02:52 < bartleby> "Then each of First, Second and Third can map to two possible values, and in total there are 2⋅2⋅2 = 2^3 = 8 functions of type Trio -> Bool"
02:53 < bartleby> counting like this I was only seeing First->True, First->False, Second->True, Second->False... 6, like a product
02:54 < Iceland_jack> You have to map all the values
02:54 < Iceland_jack> so the first function might be
02:54 < Iceland_jack> f1 First = False
02:54 < Iceland_jack> f1 Second = False
02:54 < Iceland_jack> f1 Third = False
02:54 < Iceland_jack> And the second function might be
02:54 < Iceland_jack> f2 First = True
02:54 < Iceland_jack> f2 Second = False
02:54 < Iceland_jack> f2 Third = False
02:54 < bartleby> yeah, I missed that. Thinking about combinations is easier IMO. True True True, True True False, ...
02:55 < bartleby> reminds me of truth tables :)
02:55 < Iceland_jack> writing False as 0 and True as 1 you get
02:55 < Iceland_jack> Trio -> Bool = { 000, 001, 010, 011, 100, 101, 110, 111 }
02:55 < Iceland_jack> with
02:55 < Iceland_jack> |Trio -> Bool|
02:56 < Iceland_jack> = |Bool| ^ |Trio|
02:56 < dibblego> a function of the type X -> Y has Y^X possibilites
02:56 < Iceland_jack> = 2 ^ 3 = 8
02:56 < Iceland_jack> right :)
02:57 < Iceland_jack> so a function from
02:57 < Iceland_jack> Trio -> Bool
02:57 < Iceland_jack> has the following implementations
02:57 < Iceland_jack> > replicateM 3 [0, 1]
02:57 < lambdabot> [[0,0,0],[0,0,1],[0,1,0],[0,1,1],[1,0,0],[1,0,1],[1,1,0],[1,1,1]]
02:58 < Iceland_jack> and
02:58 < Iceland_jack> Quad -> Bool
02:58 < Iceland_jack> > replicateM 4 [0, 1] -- etc.
02:58 < lambdabot> [[0,0,0,0],[0,0,0,1],[0,0,1,0],[0,0,1,1],[0,1,0,0],[0,1,0,1],[0,1,1,0],[0,1,...
02:58 < Iceland_jack> > [ length (replicateM domainSize [0,1]) | domainSize <- [0..6] ]
02:58 < lambdabot> [1,2,4,8,16,32,64]
02:59 < Iceland_jack> > [ 2^domainSize | domainSize <- [0..6] ]
02:59 < lambdabot> [1,2,4,8,16,32,64]
03:01 < bartleby> > replicateM 2 [0,1,2]
03:01 < lambdabot> [[0,0],[0,1],[0,2],[1,0],[1,1],[1,2],[2,0],[2,1],[2,2]]
03:01 < bartleby> so that's Bool -> Trio. nice
03:01 < Iceland_jack> Which has 3^2 = 9 elements not to put too fine a point on it
03:02 * bartleby is counting subarrays
03:02 < bartleby> yup, nine
03:02 < bartleby> now it makes sense, thanks
03:04 < spion> so basically, you want the number of the possible tables, rather than the number of items in a table?
03:04 < spion> :)
03:04 < dibblego> this is why you find there are 4 implementations of (Bool -> Bool)
03:05 < Iceland_jack> yes since you can interpret each table as a function definition
03:05 < Iceland_jack> True | False
03:05 < Iceland_jack> -----+------
03:05 < Iceland_jack> a | b
03:05 < spion> right
03:05 < Iceland_jack> and
03:05 < Iceland_jack> replicateM (length xs) xs
03:05 < Iceland_jack> should always have n^n elements given n = length xs
03:06 < Iceland_jack> can also be rewritten as
03:06 < Iceland_jack> (length >>= replicateM) xs
03:07 < Iceland_jack> > map (length . (length>>=replicateM) . flip replicate ()) [0..7]
03:07 < lambdabot> [1,1,4,27,256,3125,46656,823543]
03:07 < Iceland_jack> > [ n^n | n <- [0..7] ]
03:07 < lambdabot> [1,1,4,27,256,3125,46656,823543]
02:42 < dibblego> > liftA2 (+) [1,2,3] [30,40,50]
02:42 < lambdabot> [31,41,51,32,42,52,33,43,53]
02:42 < blueclaude> Thanks dibblego
02:42 < dibblego> ! [1+30,1+40,1+50,2+30,2+40,2+50,3+30,3+40,3+50]
02:43 < benzrf> blueclaude: (<*>) on the list applicative is cartesian
product, but applying the first item to the second
02:43 < benzrf> > [(++"foo"), (++"bar")] <*> ["test", "othertest", "more"]
02:43 < lambdabot>
["testfoo","othertestfoo","morefoo","testbar","othertestbar","morebar"]
02:44 < dibblego> > join (Just (Just 4))
02:44 < lambdabot> Just 4
02:44 < dibblego> > join (Just Nothing)
02:44 < lambdabot> Nothing
02:44 < benzrf> > join []
02:45 < lambdabot> []
02:45 < damncabbage> > [(+ 1), (+ 2)] <*> [1,2,3]
02:45 < lambdabot> [2,3,4,3,4,5]
02:45 < dibblego> Maybe is cosemimonad, but not a comonad
02:47 < dibblego> bitemyapp: [] is also cosemimonad but not comonad
liftA2 :: Applicative f => (a -> b -> c) -> f a -> f b -> f c
(<*>) :: Applicative f => f (a -> b) -> f a -> f b
-- by itself, cosemimonad
class Functor w =>
extend :: (w a -> b) -> w a -> w bmyFunc :: (forall a. a -> (a -> a) -> a) -> Int
myFunc f = f 0 (+1)
-- won't work
-- otherFunc :: (a -> (a -> a) -> a) -> Int
-- otherFunc f = f 0 (+1)
-- use:
-- myFunc (flip ($))22:42 < mm_freak_> because 'f' is polymorphic, myFunc gets to apply it
to its own choice of types
22:42 < mm_freak_> in particular it can make different choices in
different places
22:43 < mm_freak_> the "forall" really just means that the function
implicitly takes a type argument
22:44 < bitemyapp> mm_freak_: I think part of the problem is the difference between
22:44 < bitemyapp> (forall a. a -> (a -> a) -> a) -> Int
22:44 < bitemyapp> vs.
22:44 < bitemyapp> forall a. (a -> (a -> a) -> a) -> Int
22:44 < bitemyapp> yes?
22:44 < bitemyapp> the latter being implicitly the case in Haskell.
22:44 < mm_freak_> yes, but think about it… think really really simple in this case
22:45 < mm_freak_> in the former case myFunc receives a polymorphic function, so myFunc
gets to choose the type
22:45 < mm_freak_> in the latter case myFunc itself is polymorphic, so the applier of
myFunc gets to choose it
22:45 < mm_freak_> notice that in the former case myFunc is
monomorphic!
22:46 < mm_freak_> yeah… its type isn't quantified over any type
variables
22:46 < bitemyapp> mm_freak_: but the lambda passed to it is?
22:46 < mm_freak_> yeah
22:46 < bitemyapp> okay, yes.
22:46 < bitemyapp> so we're assigning/shifting around polymorphism
22:46 < bitemyapp> between the top level function the func arg
22:46 < bitemyapp> based on the ranks/nesting
22:46 < bitemyapp> / scope'ish
bitemyapp edited 4 days ago | link | delete | reply
I realize this is partly because the examples are in Scala, but none of this gets at what a Functor really is.
Functor is an algebra.
Functor is an algebra with one operation, usually called map.
That one operation has a type something like:
(a -> b) -> f a -> f bThat one operation should respect identity:
map id = id
And that one operation should be distributive with respect to function composition:
map (p . q) = (map p) . (map q)
That’s it people. That’s it. Functor is a very weak structure. Many things can be functor. Many of those things will not look anything like a “list”, “collection”, or even a “data structure”.
Understanding free objects, free versions of these algebraic structures, can lend a more faithful intuition for what these things are.
Glancing at Coyoneda (the free functor) should give one some idea of why you’re not dealing with something that has anything to do with lists.
Want to know more?
You know the drill: https://github.com/bitemyapp/learnhaskell
Edit:
Since I take great satisfaction in excising misunderstandings, I’m going to include a Functor instance that should help drop the “collections” oriented view of what they are.
-- (->) or -> is the type constructor for functions
-- a -> a, the identity function's type is a type of
-- -> taking two parameters of the same type (a and a)
-- (->) a a analogous to Either a b
instance Functor ((->) r) where
map = (.)
-- (.) or . is function composition
-- (.) :: (b -> c) -> (a -> b) -> a -> c
-- more on this Functor instance:
-- https://stackoverflow.com/questions/10294272/confused-about-function-as-instance-of-functor-in-haskellBonus round for upvoting me:
http://www.haskellforall.com/2012/09/the-functor-design-pattern.html http://hackage.haskell.org/package/kan-extensions-3.7/docs/Data-Functor-Coyoneda.html http://oleksandrmanzyuk.wordpress.com/2013/01/18/co-yoneda-lemma/ https://www.reddit.com/r/haskell/comments/17a33g/free_functors_the_reason_free_and_operational_are/c83p8k2 https://gist.github.com/thoughtpolice/5843762
tel 4 days ago | link | reply
Understanding free objects, free versions of these algebraic
structures, can lend a more faithful intuition for what these things
are.
This is a super great point—it also, meaningfully, applies to other structures like Monads, Applicatives, or Monoids, Categories, Arrows. Really quickly, here’s Yoneda and Coyoneda (the “two” free functors)
newtype Yoneda f a = Yoneda { runYoneda :: forall b . (a -> b) -> f b }
data Coyoneda f b where Coyoneda :: f a -> (a -> b) -> Coyoneda f bIn each case we see that functor tends to mean having a parametric structure (the f) and a method of transforming the parameter to something else (the functions a -> b). When we “collapse” this free view of a functor we get to decide if, how, when, and why we combine that structure and its mapping function. For lists we, well, map it. For something like
data Liar a = Liar -- note that `a` does not appear on the right sidewe just throw the mapping function away.
(Another key point that’s a bit harder to see is that if you map the Yoneda/Coyoneda formulation repeatedly it does not store each and every mapping function but instead composes them all together and retains only that composition. This ensures that functors cannot “see” how many times fmap has been called. That would let you violate the functor laws!)
gclaramunt 3 days ago | link | reply
Do you have any reference of functor being an algebra? I’m intrigued
Since we’re clarifying what a functor is, I guess is worth noting that you’re talking about endofunctors in the (idealized) Hask category. In category theory, a functor is defined by two mappings: one for objects in the category and one for arrows, that must preserve identity and composition (the laws you mention). Since the mapping of objects is already given by the type constructor, here one needs to provide only the mapping of functions but it kind of irks me when ppl. say a functor is only defined by “map” :)
tel 2 days ago | link | reply
Functor is definitely an algebra. Its rules mean that it has tight relation to certain functors in CT.
gclaramunt edited 2 days ago | link | reply
Interesting… any refereces I can read? Or you’re talking about F-algebras?
tel 2 days ago | link | reply
I mean “algebra” as “set of operations and equalities”.
gclaramunt 2 days ago | link | reply
Ok. To be honest, I need to familiarize myself with the definition of algebra, is just that I had never heard this before :)
tel 1 day ago | link | reply
It’s an incredibly overloaded term, tbh. In the context of abstract algebra you’d probably want to think of a (G, L)-algebra as a set inductively defined by generators G and laws L. For instance, here’s a “free” monoid algebra (note that this isn’t a free monoid, but a “free monoid algebra” or a “free algebra of the monoid type” or a “(monoid, {})-algebra” maybe)
data FMonoid where
Fmempty :: FMonoid
Fmappend :: FMonoid -> FMonoid -> FMonoid
class Monoid FMonoid where -- this is wrong! doesn't follow laws!
mempty = Fmempty
mappend = Fmappendnote that it has all the “generators” of the typeclass Monoid but follows none of the rules (mempty <> mempty != mempty). Typically we also want to add a set of constants to form the smallest free algebra over a set
data FMonoid a where
Embed :: a -> FMonoid a
Fmempty :: FMonoid a
Fmappend :: FMonoid a -> FMonoid a -> FMonoid agclaramunt 1 day ago | link | reply
Really interesting, thanks a lot! Now I’m trying to see how this ties to the Functor typeclass: G are the instance constructors and the functor laws make L ? I think I’m missing an important piece of the puzzle here :)
tel 1 day ago | link | reply
You’re not, that’s basically it.
data FFunctor f a where
EmbedFunctor :: f a -> FFunctor f a
Ffmap :: (a -> b) -> FFunctor f a -> FFunctor f bThis lets you build the free (Functor, {})-algebra over some initial type f. If we translate it naively then it doesn’t follow the laws
class Functor (FFunctor f) where -- wrong!
fmap = Ffmapbut we can implement it properly if we’re a little more clever
class Functor (FFunctor f) where
fmap f x = case x of
EmbedFunctor fa -> Ffmap f x
Ffmap g fa -> Ffmap (f . g) faWe need one more function, though, since we can’t use EmbedFunctor directly without exposing information about whether or not we’ve ever fmaped this functor (which shouldn’t be possible to access, that’s what fmap id = id says)
embed :: f a -> FFunctor f a
embed fa = Ffmap id (EmbedFunctor fa)And now, if we think about it, we can see that every value of FFunctor constructed using embed and fmap is of the form
Ffmap fun (EmbedFunctor fa)And so that EmbedFunctor constructor is totally superfluous. Let’s remove it
data FFunctor f a where
Ffmap :: (a -> b) -> f a -> FFunctor f b
embed :: f a -> FFunctor f a
embed fa = Ffmap id faAnd—well—this is just CoYoneda again!
lower :: Functor f => FFunctor f a -> f a
lower (Ffmap f fa) = fmap f fagclaramunt about 9 hours ago | link | reply
Nice Haven’t digested it properly but I see the trick is to capture the functor with a datatype (is the same thing with free monads, right?) Now is easier to see from where CoYoneda comes, thanks! (you did show me an important piece of the puzzle :P )
edwardkmett 7 points an hour ago
Much of Guy Steele's work here pertained to a desire to be able to parallelize calculation. This is a laudable goal. The main issue with a naïve magma approach Steele proposed for Fortress is that you have zero guarantees about efficient splittability. All the mass of your magma could be on one side or the other.
The benefit is that without those guarantees infinite magmas make sense in a lazy language. You can have infinitely large trees just fine, that go off to infinity at any point not just at the right.
This has a certain pleasing structure to it. Why? Well, lists aren't really the free monoid if you allow for infinitely recursive use of your monoid! You have unit and associativity laws and by induction you can apply them a finite number of times, but reassociating an infinite tree from the left to the right requires an infinite number of steps, taking us out of the constructive world we can program. So ultimately a free Monoid (allowing for infinite monoids) is something like Sjoerd Visscher's
newtype Free p = Free { runFree :: forall r. p r => (a -> r) -> r }
type List = Free MonoidHere we borrow the assumption of unit and association from the target r and generate something using it. It is an almost vacuous but now correct construction, whereas the association to the right to make a list required us to be able to right associate infinite trees. You can view this as a sort of quotient on a magma, where you guarantee to only consume it with monoidal reductions.
Binding/substituting on a (unital) magma can now take longer than O(n), why? Because now I have to walk past all the structure. You can replace this with Oleg and Atze's "Reflection without Remorse", but walking down a unital Magma structure doesn't decrease n necesssarily.
In the absence of infinite trees, you usually want some form of balance depending on what you want to do with the structure. e.g. turning it into a catenable deque gives you efficient access to both ends and lets you still glue in O(1) or O(log n).
Switching to a finger tree gives you guaranteed O(log n) splits, but now merges go from O(1) to O(log n) In a general magma the split is potentially completely lopsided. You can 'steal work' but as often as not you likely steal a single unit, or in a unital magma, possibly nothing.
The cost of these richer structures is you lose the continuous extension to the infinite case, but when trading O(n) or worse for O(log n) it is often worth making that trade-off.
23:20 < slack1256> It is folklore that lazy evaluation compose better, usually showing how it supports separated
generators/consumers as in "Why functional programming matters", but does this notion of
composition goes further?
23:20 < Cale> slack1256: In what sense?
23:21 < slack1256> as in, if some function is not based on generator/consumer based can still benefit from
laziness?.
23:21 < slack1256> (this is a problem of me having lack of imagination)
23:22 < Cale> slack1256: Most functions are consuming or generating something. Those which produce or consume
larger structures with many parts that could be evaluated separately tend to benefit from laziness.
23:22 < tabemann> from what I gather, though, lazy lists aren't as useful as that paper purported
23:23 < Cale> We use them all the time
23:23 < haasn> Laziness can play well into parallelism
23:23 < haasn> Or, rather, nonstrictness
23:23 < haasn> Only evaluate the strict parts immediately, evaluate parts that don't have to be forced yet in parallel
23:23 < tabemann> what I mean is that there are better lazy sequence data structures than the list
23:23 < Cale> Oh?
23:23 < Cale> Lists are absolutely perfect for what they are
23:24 < slack1256> Mmm you're right, most functions are consuming or generating something, specially in a pure
language.
23:24 < Cale> If you plan on iterating through a list of things in order, then lists present you with about as
concise a representation as possible.
23:24 < Cale> slack1256: Lists are essentially our loops
23:25 < Cale> and it helps sometimes that they can be infinite, or combinatorially large
23:25 < Cale> for the same reason that you might want to have an infinite loop, or one which potentially iterates
more times than you'll practically ever want to actually have the loop body occur.
23:26 < slack1256> In "more points for lazy evaluation" augustss shows that laziness enable efficient higher-order
functions, and bob concedes that point that in strict languages that really hurts because you
have to use manual recursion.
23:26 < Cale> yep
23:26 < slack1256> Maybe I should really learn SML to appreciate more the benefits of laziness
23:27 < josephle> then you'll lament the lack of backpack in Haskell ;)
23:28 < Cale> It really needs to be the default for that reason: if the default is to be strict, when you find the
functions that you want to compose in your library, the chances are good that whoever wrote it won't
have thought about your use case, and you'll need to rewrite it to be explicitly lazy, which defeats
a lot of the point of having things be compositional.
23:31 < Cale> Whereas strictness is just slightly more rarely required, and tends to be the kind of thing that you
can't ignore when you really need it, because your program's performance will suffer dramatically. So
it just becomes a matter of learning to spot the places where it'll be important. The rule of thumb I
use is this: if you're collapsing many individual bits of data down into a single thing which depends
on all of them
23:31 < Cale> and can't be partially evaluated, that's where you want some sort of strict higher-order function
like foldl' or some explicit strictness annotations.
23:32 < Cale> Basically, things you'd think of as accumulations of some sort. You want to avoid building up large
expressions in memory composed of strict functions (functions that must pattern match on their input
to produce any part of their result).
23:34 < Cale> So for instance, when you're repeatedly updating the contents of an IORef, or recursively updating an
accumulating parameter without matching on it, you want to be careful there.
< OscarZ> instance Functor ((->) t) where
< ajcoppa> OscarZ: you know about (->) from the types of functions, right?
< OscarZ> ajcoppa: yes
< ajcoppa> ((->) t) is a partial definition that describes a function that takes in a value of type t
< OscarZ> ajcoppa: im a bit confused.. so does "instance Functor ((->) t) where" mean that were defining that any function that takes a parameter is a Functor instance?
< ajcoppa> OscarZ: yep!
< OscarZ> ajcoppa: is that different than saying instance Functor (a -> b) where ?
< ajcoppa> OscarZ: it is different, because functors need to have a kind of * -> *
< ajcoppa> as an example, when defining a functor for Optional, you don't say instance Functor (Optional Int) where -- you say instance Functor Optional where
< pjdelport> OscarZ: "Functor ((->) r)" is a lot like "Functor (Either a)" and "Functor ((,) a)"
< pjdelport> OscarZ: You're declaring an instance for a partially-applied type, which itself will get applied to a remaining type (which is the "slot" that the functor operates on)
< pjdelport> OscarZ: So in the same sense that "instance Functor Maybe where ..." means: fmap :: (a -> b) -> Maybe a -> Maybe b
< pjdelport> "instance Functor ((,) x) where ..." means: fmap (a -> b) -> (x,a) -> (x,b)
< pjdelport> Or in equivalent syntax: fmap :: (a -> b) -> ((,) x) a -> ((,) x) b
< pjdelport> where ((,) x) is what "f" gets replaced with
< pjdelport> So for "instance Functor (Either x) where ...", you have fmap :: (a -> b) -> Either x a -> Either x b
< pjdelport> and for "instance Functor ((->) r) where ...", you have fmap :: (a -> b) -> (->) r a -> (->) r b
< pjdelport> or equivalently: fmap :: (a -> b) -> (r -> a) -> (r -> b)
< OscarZ> oh.. i think i finally get it
< pjdelport> OscarZ: To read that more intuitively, just remember what the "slot" that fmap operates on is for each Functor instance: for lists, the "slots" are all of a list's elements. For Maybe, the "slots" are Just values (if any). For Either, the "slots" are Right values.
< pjdelport> OscarZ: For functions, (->), the "slot" that you're operating on is the *result* of the function.
< pjdelport> So applying "fmap f" to a function value like (r -> a) uses f to "replace" the result value 'a' with a 'b', yielding a new function (r -> b).
< pjdelport> (Just like how applying "fmap f" to a list value [a] uses f to "replace" each element 'a' with a 'b', yielding a new list [b].)
< OscarZ> to implement Functor functions with signature a -> a -> a, should it be instance Function ((->) r t) where ?
< pjdelport> OscarZ: a -> a -> a, or (a -> (a -> a)), is just a special case of (a -> b) where b = (a -> a)
< OscarZ> pjdelport: so like Functor (Either a) matches (Either a) part in Either a b, Functor ((->) r) matches just the first a in a -> (a -> a) ? in this case the slot would be of type (a -> a) ?
< pjdelport> OscarZ: Yep.
< OscarZ> ok thanks.. i think i get.. apparently im not used to functions as first-class citizens :) somehow got confused when it was a function instead of something like Either a b
< ajcoppa> it is very normal to need some time to wrap your head around this instance in particular
< pjdelport> Yeah, it can be mind-bending. :)
< pjdelport> but it's good exercise
< OscarZ> you guys are good at explaining :)
< OscarZ> you could teach FP to a rotten cabbage
-- n.b. * means we're showing kind signatures, not type signatures.
-- application of type arguments
data (->) a b
(->) :: * -> * -> *
((->) e) :: * -> *
instance Functor ((->) r) where
fmap = (.)
instance Functor ((->) r) where
fmap f g = (\x -> f (g x))
data (,) a b = (,) a b
(,) :: * -> * -> *
((,) a) :: * -> *
instance Functor ((,) a) where
fmap f (x,y) = (x, f y)
newtype Reader r a = Reader { runReader :: r -> a }
Reader :: * -> * -> *
(Reader r) :: * -> *
instance Functor (Reader r) where
fmap f m = Reader $ \r -> f (runReader m r)
instance Monad (Reader r) where
return a = R $ \_ -> a
m >>= k = R $ \r -> runReader (k (runReader m r)) r
class Functor (f :: * -> *) where
fmap :: Functor f => (a -> b) -> f a -> f b
class Monad (m :: * -> *) where
(>>=) :: m a -> (a -> m b) -> m b
return :: a -> m a18:51 < OscarZ> about join... im wondering about join (+) 7 = 14
19:14 < pjdelport> OscarZ: join for Reader becomes even more Intuitive when you
pronounce Reader e a as "function from an environment e to a"
19:15 < pjdelport> Then "Reader e (Reader e a)" is just "function from an environment e
to function from an environment e to a"
19:16 < pjdelport> And joining means turning that into just one "function from an
environment e to a"
19:16 < pjdelport> And saying it like that should (hopefully) make the implementation and what
it does more obvious :)
19:23 < pjdelport> I still think just pondering "type Reader e a = e -> a" is a great way to
to get that "a ha" moment
λ> join (+) 7
14
λ> join (*) 7
49SpacemanInBikini:
I made a function in Haskell that takes in a number and returns a list of integers from 1 to that number.
gen x = [1..x]
gen 4
-> [1, 2, 3, 4]As I was learning about the monads I noticed that I could bind a list into this function as it has the type of (a -> m a).
[2, 3, 6, 1, 4] >>= gen
-> [1, 2, 1, 2, 3, 1, 2, 3, 4, 5, 6, 1, 1, 2, 3, 4]Then I wanted to write a function that would reverse the (>>= gen) and i came up with this function.
revgen (x:xs) =
case (null xs) of
False -> case ((head xs) - x) of
1 -> 1 + revgen xs
otherwise -> 0 : revgen xs
True -> 1 : []But it doesn't work, I think because of the revgen returning both numbers and lists. But I can't come up with a way to do this any other way with one function. Also the nested cases look ugly as hell but I quess they work as intented because at least it compiles.
meditans:
Given the definitions:
gen x = [1..x]
f xs = xs >>= genIf you want a function g such that g . f = id, I think you could use:
g xs = map last $ groupBy (<) xsλ> groupBy (<) [1, 2, 1, 2, 3, 1, 2, 3, 4, 5, 6, 1, 1, 2, 3, 4]
[[1,2],[1,2,3],[1,2,3,4,5,6],[1],[1,2,3,4]]
λ> map last $ groupBy (<) [1, 2, 1, 2, 3, 1, 2, 3, 4, 5, 6, 1, 1, 2, 3, 4]
[2,3,6,1,4]tel:
If the range is finite and the domain has decidable equality
class Finite a where
-- x :: a iff elem x universe = True
universe :: [a]
inv :: (Eq b, Finite a) => (a -> b) -> (b -> a)
inv f b = let [a] = [ a | a <- universe, f a == b ] in aThe pattern matching above will fail if f is not bijective.
16:24 < Lindrian> sparks in haskell seem so... unintuitive
16:24 < Lindrian> It doesnt really follow the traditional sense of how you use threads,
which might be why I find it odd
16:25 < qu1j0t3> this might also be a very good thing :-)
16:26 < Lindrian> why?
16:28 < Lindrian> I need to remember the functionality of seq, par and pseq.
16:28 < Lindrian> seq forces evaluation of the first argument, before returning the second?
par creates a "spark" that allows the first argument to be evaluated at
the same time the second is being returned?
16:28 < Lindrian> I thinkkkkk
16:30 < Lindrian> damn cant remember pseq, gotta look it up
16:31 < Lindrian> oh right, pseq evaluates the left before returning the right, guaranteed.
16:32 < Lindrian> So the difference between seq and pseq is that pseq guarantees it while seq doesnt?
16:51 < qu1j0t3> well, thread API has proven to be a poor fit for many purposes.
16:53 < zwer_z> > seq (error "one") (error "two")
16:53 < lambdabot> *Exception: one
16:53 < Lindrian> > pseq (error "first") (error "second")
16:53 < lambdabot> Not in scope: ‘pseq’
16:53 < lambdabot> Perhaps you meant ‘seq’ (imported from Prelude)
16:54 < zwer_z> IIRC in ghc seq will always evaluate to error "one", while
haskell (the language) allows either
16:55 < Lindrian> i read seq as "try to evaluate the first argument, but compiler might think otherwise"
16:55 < ski> Lindrian : yes, `pseq a b' guarantees `a' is forced before `b'. `seq a b' doesn't
16:56 < ski> (i think in `seq a b' you're not even guaranteed that `a' is forced before `b' is
returned .. as long as it will happen eventually)
16:56 < zwer> ski I think there's a lack of guarantee only if both a and b are bottom.
16:57 < zwer> > seq (error "foo") 10
16:57 < lambdabot> *Exception: foo
16:57 < zwer> whereas that should always evaluate to bottom
16:58 < benzrf> iirc
16:58 < ski> zwer : consider `seq (seq (error "foo") ()) (error "bar")' -- i think this can legally
raise `error "bar"'
16:58 < benzrf> the haskell spec only says
16:58 < benzrf> seq bottom anything = bottom
16:58 < ski> yes
16:58 < benzrf> seq anything bottom = bottom
16:58 < zwer> ski I think so too. if b is bottom a does not have to be evaluated at all
17:00 < ski> zwer : in this case "`b'" for the inner call is not bottom
17:02 < Lindrian> so seq will try to evaluate to WHNF?
17:02 < Lindrian> pseq guarantees WHNF
17:03 < zwer> ski a is entire inner seq, which may not be evaluated
17:05 < ski> yes
17:05 < ski> (i think i used a different argument, though. i can't recall it atm)
17:06 < ski> Lindrian : both will evaluate to WHNF. `pseq' guarantees sequential *ordering*.
`seq' just guarantees that both will be forced
17:07 < Lindrian> pseq guarantees the first argument is evaluated to WHNF before the second,
while seq doesnt. Ok.
17:07 < ski> `seq' is specified using "denotational semantics", which only talks about the
final value/denotation of an evaluation, not how you can get to it
17:08 < ski> `pseq' would need to be specified using "operational semantics", which talks
in terms of rewriting steps (so there's an inherent ordering that we talk
about here)
17:08 * ski nods to Lindrian