Hey all, I've found myself reaching for the following function a couple of times now, so I figured it might make a good addition. intersections :: Ord a => NonEmpty (Set a) -> Set a intersections (s :| ss) = Foldable.foldl' intersection s ss In a similar vein, we may as well add the following newtype + instance combo: newtype Intersection a = Intersection { getIntersection :: Set a } instance (Ord a) => Semigroup (Intersection a) where (Intersection a) <> (Intersection b) = Intersection $ intersection a b stimes = stimesIdempotent Would love to hear everyone's thoughts on this! Thanks Reed Mullanix _______________________________________________ Libraries mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries |
+1 from me
Seems like additional motivation for adding NonEmptyFoldable (or however people want to call it) ======= Georgi _______________________________________________ Libraries mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries |
Makes sense to me. On Sun, Dec 06, 2020 at 2:40 AM, Georgi Lyubenov <[hidden email]> wrote:
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In reply to this post by Reed Mullanix
Sounds like a good idea, could it be possible to use a typeclass instead of `NonEmpty (Set a)`? I recall needing this a few times, but not over a NonEmpty. De: Libraries <[hidden email]> em nome de Reed Mullanix <[hidden email]> Hey all, I've found myself reaching for the following function a couple of times now, so I figured it might make a good addition. intersections :: Ord a => NonEmpty (Set a) -> Set a In a similar vein, we may as well add the following newtype + instance combo: newtype Intersection a = Intersection { getIntersection :: Set a } instance (Ord a) => Semigroup (Intersection a) where Would love to hear everyone's thoughts on this! Thanks Reed Mullanix _______________________________________________ Libraries mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries |
In a perfect world we would use something like Foldable1 from semigroupoids, but sadly that is not in base. However, this is exactly why I proposed the Intersection newtype, as it would make it nice and easy to use with a Foldable1 like type class. On Sun, Dec 6, 2020 at 10:27 AM Alexandre Rodrigues Baldé <[hidden email]> wrote:
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Am So., 6. Dez. 2020 um 07:20 Uhr schrieb Reed Mullanix <[hidden email]>:
Why NonEmpty? I would expect "intersections [] = Set.empty", because the result contains all the elements which are in all sets, i.e. none. That's at least my intuition, is there some law which this would violate? _______________________________________________ Libraries mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries |
It doesn't violate any laws per say, due to the general lawlessness of Foldable, but violates aesthetics. If we restrict it to NonEmpty then we get a lovely semigroup homomorphism! If we loosen it to lists, then the identity element of the list monoid gets mapped to an absorbing element in the Set semigroup under intersection, which just feels... off. On Sun, Dec 6, 2020 at 10:50 AM Sven Panne <[hidden email]> wrote:
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First consider how `and` and `or` work for booleans. and (x ++ y) == and x && and y For this to work we need `and [] == True` or (x ++ y) == or x || or y For this to work we need `or [] == False` and and or are duals of each other. There’s an analogue here to union and intersection which are also duals of each other. This requires `union [] == []` since any list xs could be split as `[] ++ xs` We'd like to have: intersections (x ++ y) == intersect (intersections x) (intersections y) For this kind of splitting property to make sense for intersections we’d need `intersections [] == listOfAllElementsOfThisType`, but it’s not easy to construct that list of all elements in general. So instead we punt on the problem and refuse to define intersections on an empty list. -glguy
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Am So., 6. Dez. 2020 um 19:56 Uhr schrieb Reed Mullanix <[hidden email]>:
To me it's just the other way around: It violates aesthetics if it doesn't follow the mathematical definition in all cases, which is why I don't like NonEmpty here. _______________________________________________ Libraries mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries |
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I would be opposed to
intersections [] = [] as it is true that *all* things have the property of being a member of every element of []. In set theory without classes I've usually seen this worked around by never using intersections on an empty set (ie using NonEmpty) ======= Georgi _______________________________________________ Libraries mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries |
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intersections [] = empty is unacceptable to me. The options that make sense to me are the nonempty (ideally Foldable1) one, one producing a Maybe, and I guess even one that throws an error on an empty list (ouch). On Sun, Dec 6, 2020, 1:50 PM Sven Panne <[hidden email]> wrote:
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Intersection is a lattice operation, whose identity element is the "whole
universe" not the empty set. Imagine if we had an intersections :: [Set a] -> Set a we would expect it have the law intersections (a ++ b) = (intersections a) `intersect` (intersections b) now let b be [] and aa == intersections a, and we get aa == aa `intersect` (intersections []) and this must hold for all aa, which is impossible unless we have some kind of universe set. David -- David Casperson, PhD, R.P., | [hidden email] Associate Professor and Chair, | (250) 960-6672 Fax 960-5544 Computer Science | 3333 University Way University of Northern British Columbia | Prince George, BC V2N 4Z9 | CANADA Sven Panne, on 2020-12-06, you wrote: > From: Sven Panne <[hidden email]> > To: Reed Mullanix <[hidden email]> > Date: Sun, 6 Dec 2020 10:50:13 > Cc: Haskell Libraries <[hidden email]> > Subject: Re: containers: intersections for Set, along with Semigroup newtype > Message-ID: > <CANBN=muqE8tE11UCGgYYuUEKP=SN19TAQ7_3wE68v46583H-=[hidden email]> > > > CAUTION: This email is not from UNBC. Avoid links and attachments. Don't buy gift cards. > > Am So., 6. Dez. 2020 um 07:20 Uhr schrieb Reed Mullanix <[hidden email]>: > [...] > intersections :: Ord a => NonEmpty (Set a) -> Set a > intersections (s :| ss) = Foldable.foldl' intersection s ss > [...] > > > Why NonEmpty? I would expect "intersections [] = Set.empty", because the result contains all the elements which are > in all sets, i.e. none. That's at least my intuition, is there some law which this would violate? > > Libraries mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries _______________________________________________ Libraries mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries |
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On Sun, 6 Dec 2020, Reed Mullanix wrote: > In a perfect world we would use something like Foldable1 from > semigroupoids, but sadly that is not in base. Is it necessary to have it in base for your use? For me, 'NonEmpty.foldl1 Set.intersection' is perfect. _______________________________________________ Libraries mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries |
That implementation doesn't short-circuit if the accumulator goes empty. It also relies on the optimizer to accumulate strictly. On Mon, Dec 7, 2020, 3:44 AM Henning Thielemann <[hidden email]> wrote:
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Am 06.12.20 um 19:58 schrieb Sven Panne:
> To me it's just the other way around: It violates aesthetics if it doesn't > follow the mathematical definition in all cases, which is why I don't like > NonEmpty here. I think you've got that wrong. x `elem` intersections [] = {definition} forall xs in []. x `elem` xs = {vacuous forall} true Any proposition P(x) is true for all x in []. So the mathematical definition of intersections::[Set a]-> Set a would not be the empty set but the set of all x:a, which in general we have no way to construct. Cheers Ben _______________________________________________ Libraries mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries |
On Sun, Dec 20, 2020 at 11:05:39PM +0100, Ben Franksen wrote:
> Am 06.12.20 um 19:58 schrieb Sven Panne: > > To me it's just the other way around: It violates aesthetics if it doesn't > > follow the mathematical definition in all cases, which is why I don't like > > NonEmpty here. > > I think you've got that wrong. > > x `elem` intersections [] > = {definition} > forall xs in []. x `elem` xs > = {vacuous forall} > true > > Any proposition P(x) is true for all x in []. So the mathematical > definition of intersections::[Set a]-> Set a would not be the empty set > but the set of all x:a, which in general we have no way to construct. Yes, and to bring this back to Sven's original claim | Why NonEmpty? I would expect "intersections [] = Set.empty", because | the result contains all the elements which are in all sets, | i.e. none. That's at least my intuition, is there some law which | this would violate? the correct definition of "intersections []" should be "all elements which are in all of no sets" i.e. _every_ value of the given type! Tom _______________________________________________ Libraries mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries |
One *could* write
import Data.Universe.Class (Finite (..)) class (Ord a, Finite a) => OrderedFinite a where orderedFiniteUniverse :: Set a orderedFiniteUniverse = S.fromList universeF data SMaybe a = SJust !a | SNothing wackyIntersections :: OrderedFinite a => [Set a] -> Set a wackyIntersections = \ss -> foldr go stop ss SNothing where stop SNothing = orderedFiniteUniverse stop (SJust s) = s go s r SNothing = r (SJust s) go s r (SJust acc) | null acc = empty | otherwise = r (SJust acc `intersection` s) No such things will be going in `containers`, but perhaps they would fit in `universe`. On Mon, Dec 21, 2020 at 5:12 AM Tom Ellis <[hidden email]> wrote: > > On Sun, Dec 20, 2020 at 11:05:39PM +0100, Ben Franksen wrote: > > Am 06.12.20 um 19:58 schrieb Sven Panne: > > > To me it's just the other way around: It violates aesthetics if it doesn't > > > follow the mathematical definition in all cases, which is why I don't like > > > NonEmpty here. > > > > I think you've got that wrong. > > > > x `elem` intersections [] > > = {definition} > > forall xs in []. x `elem` xs > > = {vacuous forall} > > true > > > > Any proposition P(x) is true for all x in []. So the mathematical > > definition of intersections::[Set a]-> Set a would not be the empty set > > but the set of all x:a, which in general we have no way to construct. > > Yes, and to bring this back to Sven's original claim > > | Why NonEmpty? I would expect "intersections [] = Set.empty", because > | the result contains all the elements which are in all sets, > | i.e. none. That's at least my intuition, is there some law which > | this would violate? > > the correct definition of "intersections []" should be "all elements > which are in all of no sets" i.e. _every_ value of the given type! > > Tom > _______________________________________________ > Libraries mailing list > [hidden email] > http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries Libraries mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries |
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why are we equating the lattice operators for True and false with the lattice operators for set? (for both structures, we have the dual partial order is also a lattice, so unless we have ) (i'm going to get the names of these equations wrong, but ) the "identity" law is going to be max `intersect` y == y , min `union` y === y the "absorbing" law is going to be min `intersect` y == min , max `union` y == max these rules work the same for (min = emptyset, max == full set, union == set union, intersect == set intersecct) OR for its dual lattice (min == full set, max == emtpy set, union = set intersection, intersect == set union) at some level arguing about the empty list case turns into artifacts of "simple" definitions that said, i suppose a case could be made that for intersect :: [a] -> a , that as the list argument gets larger the result should be getting *smaller*, so list intersect of lattice elements should be "anti-monotone", and list union should be monotone (the result gets bigger). I dont usually see tht either way, I do strongly feel that either way, arguing by how we choose to relate the boolean lattice and seet lattices is perhaps the wrong choice... because both lattices are equivalent to theeir dual lattice On Mon, Dec 21, 2020 at 5:12 AM Tom Ellis <[hidden email]> wrote: On Sun, Dec 20, 2020 at 11:05:39PM +0100, Ben Franksen wrote: _______________________________________________ Libraries mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries |
edit: neglected to mention that choosing which lattice (and or dual) to use only really matters when considering products/sums of lattices to form new lattices On Mon, Dec 21, 2020 at 11:12 AM Carter Schonwald <[hidden email]> wrote:
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All this talk about lattices reminds me of one of my pet gripes I have
with the standard libraries (base), namely that the boolean operators aren't overloaded. I don't want to open endless discussions about the perfect generalization, because there are lots of valid generalizations and the lattice package is fine for those. But most generalizations don't have a 'not' (complement) operator. So, a simple type class Boolean with instances for Bool and functions returning Booleans should cover the majority of use cases; more instances could be added of course. Something like {-# LANGUAGE NoImplicitPrelude #-} module Data.Boolean where import Prelude hiding ((&&),(||),not) import qualified Prelude infixr 3 && infixr 2 || class Boolean a where -- laws: that of a boolean algebra, i.e. -- complemented distributive lattice, see -- https://en.wikipedia.org/wiki/Boolean_algebra#Laws (&&) :: a -> a -> a (||) :: a -> a -> a not :: a -> a top :: a bottom :: a instance Boolean Bool where (&&) = (Prelude.&&) (||) = (Prelude.||) not = Prelude.not top = True bottom = False instance Boolean b => Boolean (a->b) where (f && g) x = f x && g x (f || g) x = f x || g x (not f) x = not (f x) top = const top bottom = const bottom IMHO this would be absolutely benign, no problems with type inference, fully Haskell98, no breakage of existing code I can think of. (I didn't check that last point but I would be very surprised if there were.) Cheers Ben Am 21.12.20 um 17:14 schrieb Carter Schonwald: > edit: neglected to mention that choosing which lattice (and or dual) to use > only really matters when considering products/sums of lattices to form new > lattices > > On Mon, Dec 21, 2020 at 11:12 AM Carter Schonwald < > [hidden email]> wrote: > >> why are we equating the lattice operators for True and false with the >> lattice operators for set? (for both structures, we have the dual partial >> order is also a lattice, so unless we have ) >> (i'm going to get the names of these equations wrong, but ) >> >> the "identity" law is going to be max `intersect` y == y , min `union` y >> === y >> >> the "absorbing" law is going to be min `intersect` y == min , max >> `union` y == max >> >> these rules work the same for (min = emptyset, max == full set, union == >> set union, intersect == set intersecct) >> OR for its dual lattice (min == full set, max == emtpy set, union = set >> intersection, intersect == set union) >> >> at some level arguing about the empty list case turns into artifacts of >> "simple" definitions >> >> that said, i suppose a case could be made that for intersect :: [a] -> a , >> that as the list argument gets larger the result should be getting >> *smaller*, so list intersect of lattice elements should be "anti-monotone", >> and list union should be monotone (the result gets bigger). I dont usually >> see tht >> >> either way, I do strongly feel that either way, arguing by how we choose >> to relate the boolean lattice and seet lattices is perhaps the wrong >> choice... because both lattices are equivalent to theeir dual lattice >> >> On Mon, Dec 21, 2020 at 5:12 AM Tom Ellis < >> [hidden email]> wrote: >> >>> On Sun, Dec 20, 2020 at 11:05:39PM +0100, Ben Franksen wrote: >>>> Am 06.12.20 um 19:58 schrieb Sven Panne: >>>>> To me it's just the other way around: It violates aesthetics if it >>> doesn't >>>>> follow the mathematical definition in all cases, which is why I don't >>> like >>>>> NonEmpty here. >>>> >>>> I think you've got that wrong. >>>> >>>> x `elem` intersections [] >>>> = {definition} >>>> forall xs in []. x `elem` xs >>>> = {vacuous forall} >>>> true >>>> >>>> Any proposition P(x) is true for all x in []. So the mathematical >>>> definition of intersections::[Set a]-> Set a would not be the empty set >>>> but the set of all x:a, which in general we have no way to construct. >>> >>> Yes, and to bring this back to Sven's original claim >>> >>> | Why NonEmpty? I would expect "intersections [] = Set.empty", because >>> | the result contains all the elements which are in all sets, >>> | i.e. none. That's at least my intuition, is there some law which >>> | this would violate? >>> >>> the correct definition of "intersections []" should be "all elements >>> which are in all of no sets" i.e. _every_ value of the given type! >>> >>> Tom >>> _______________________________________________ >>> Libraries mailing list >>> [hidden email] >>> http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries >>> >> > > > _______________________________________________ > Libraries mailing list > [hidden email] > http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries > _______________________________________________ Libraries mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries |
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