# The Universe of Discourse

Fri, 09 Nov 2018

(Previously: [1] [2])

I'm doing more work on matrix functions. A matrix represents a relation, and I am representing a matrix as a [[Integer]]. Then matrix addition is simply liftA2 (liftA2 (+)). Except no, that's not right, and this is not a complaint, it's certainly my mistake. The overloading for liftA2 for lists does not do what I want, which is to apply the operation to each pair of correponding elements. I want liftA2 (+) [1,2,3] [10,20,30] to be [11,22,33] but it is not. Instead liftA2 lifts an operation to apply to each possible pair of elements, producing [11,21,31,12,22,32,13,23,33]. And the twice-lifted version is similarly not what I want:

$$\require{enclose} \begin{pmatrix}1&2\\3&4\end{pmatrix}\enclose{circle}{\oplus} \begin{pmatrix}10&20\\30&40\end{pmatrix}= \begin{pmatrix} 11 & 21 & 12 & 22 \\ 31 & 41 & 32 & 42 \\ 13 & 23 & 14 & 24 \\ 33 & 43 & 34 & 44 \end{pmatrix}$$

No problem, this is what ZipList is for. ZipLists are just regular lists that have a label on them that advises liftA2 to lift an operation to the element-by-element version I want instead of the each-one-by-every-other-one version that is the default. For instance

    liftA2 (+) (ZipList [1,2,3]) (ZipList [10,20,30])


gives ZipList [11,22,33], as desired. The getZipList function turns a ZipList back into a regular list.

But my matrices are nested lists, so I need to apply the ZipList marker twice, once to the outer list, and once to each of the inner lists, because I want the element-by-element behavior at both levels. That's easy enough:

    matrix :: [[a]] -> ZipList (ZipList a)
matrix m = ZipList (fmap ZipList m)


(The fmap here is actually being specialized to map, but that's okay.)

Now

    (liftA2 . liftA2) (+) (matrix [[1,2],[3,4]]) (matrix [[10,20],[30, 40]])


does indeed produce the result I want, except that the type markers are still in there: instead of

    [[11,22],[33,44]]


I get

    ZipList [ ZipList [11, 22], ZipList [33, 44] ]


No problem, I'll just use getZipList to turn them back again:

    unmatrix :: ZipList (ZipList a) -> [[a]]
unmatrix m = getZipList (fmap getZipList m)


And now matrix addition is finished:

    matrixplus :: [[a]] -> [[a]] -> [[a]]
matrixplus m n = unmatrix $(liftA2 . liftA2) (+) (matrix m) (matrix n)  This works perfectly. But the matrix and unmatrix pair bugs me a little. This business of changing labels at both levels has happened twice already and I am likely to need it again. So I will turn the two functions into a single higher-order function by abstracting over ZipList. This turns this  matrix m = ZipList (fmap ZipList m)  into this:  twice zl m = zl (fmap zl m)  with the idea that I will now have matrix = twice ZipList and unmatrix = twice getZipList. The first sign that something is going wrong is that twice does not have the type I wanted. It is:  twice :: Functor f => (f a -> a) -> f (f a) -> a  where I was hoping for something more like this:  twice :: (Functor f, Functor g) => (f a -> g a) -> f (f a) -> g (g a)  which is not reasonable to expect: how can Haskell be expected to figure out I wanted two diferent functors in there when there is only one fmap? And indeed twice does not work; my desired matrix = twice ZipList does not even type-check:  <interactive>:19:7: error: • Occurs check: cannot construct the infinite type: a ~ ZipList a Expected type: [ZipList a] -> ZipList a Actual type: [a] -> ZipList a • In the first argument of ‘twice’, namely ‘ZipList’ In the expression: twice ZipList In an equation for ‘matrix’: matrix = twice ZipList • Relevant bindings include matrix :: [[ZipList a]] -> ZipList a (bound at <interactive>:20:5)  Telling GHC explicitly what type I want for twice doesn't work either, so I decide it's time to go to lunch. w I take paper with me, and while I am eating my roast pork hoagie with sharp provolone and spinach (a popular local delicacy) I work out the results of the type unification algorithm on paper for both cases to see what goes wrong. I get the same answers that Haskell got, but I can't see where the difference was coming from. So now, instead of defining matrix operations, I am looking into the type unification algorithm and trying to figure out why twice doesn't work. And that is yet another reason why I never finish my Haskell programs. (“What do you mean, λ-abstraction didn't work?”) Thu, 08 Nov 2018 I want to build an adjacency matrix for the vertices of a cube; this is a matrix that has m[a][b] = 1 exactly when vertices a and b share an edge. We can enumerate the vertices arbitrarily but a convenient way to do it is to assign them the numbers 0 through 7 and then say that vertices !!a!! and !!b!! are adjacent if, regarded as binary numerals, they differ in exactly one bit, so:  import Data.Bits a adj b = if (elem (xor a b) [1, 2, 4]) then 1 else 0  This compiles and GHC infers the type  adj :: (Bits a, Num a, Num t) => a -> a -> t  Fine. Now I want to build the adjacency matrix, which is completely straightforward:  cube = [ [a adj b | b <- [0 .. 7] ] | a <- [0 .. 7] ] where a adj b = if (elem (xor a b) [1, 2, 4]) then 1 else 0  Ha ha, no it isn't; in Haskell nothing is straightforward. This produces 106 lines of type whining, followed by a failed compilation. Apparently this is because because 0 and 7 are overloaded, and could mean some weird values in some freakish instance of Num, and then 0 .. 7 might generate an infinite list of 1-graded torsion rings or something. To fix this I have to say explicitly what I mean by 0. “Oh, yeah, by the way, that there zero is intended to denote the integer zero, and not the 1-graded torsion ring with no elements.”  cube = [ [a adj b | b <- [0 :: Integer .. 7] ] | a <- [0 .. 7] ] where a adj b = if (elem (xor a b) [1, 2, 4]) then 1 else 0  Here's another way I could accomplish this:  zero_i_really_mean_it = 0 :: Integer cube = [ [a adj b | b <- [zero_i_really_mean_it .. 7] ] | a <- [0 .. 7] ] where a adj b = if (elem (xor a b) [1, 2, 4]) then 1 else 0  Or how about this?  cube = [ [a adj b | b <- numbers_dammit [0 .. 7] ] | a <- [0 .. 7] ] where p adj q = if (elem (xor p q) [1, 2, 4]) then 1 else 0 numbers_dammit = id :: [Integer] -> [Integer]  I think there must be something really wrong with the language design here. I don't know exactly what it is, but I think someone must have made the wrong tradeoff at some point. Fri, 26 Oct 2018 In an earlier article I demanded: Maybe someone can explain to me why this is a useful behavior, and then explain why it is so useful that it should happen automatically … “This” being that instead of raising a type error, Haskell quietly accepts this nonsense:  fmap ("super"++) (++"weasel")  but it clutches its pearls and faints in horror when confronted with this expression:  fmap ("super"++) "weasel"  Nobody did explain this. But I imagined someone earnestly explaining: “Okay, but in the first case, the (++"weasel") is interpreted as a value in the environment functor, so fmap is resolved to its the environment instance, which is (.). That doesn't happen in the second example.” Yeah, yeah, I know that. Hey, you know what else is a functor? The identity functor. If fmap can be quietly demoted to its (->) e instance, why can't it also be quietly demoted to its Id instance, which is ($), so that fmap ("super"++) "weasel" can quietly produce "superweasel"?

I understand this is a terrible idea. To be clear, what I want is for it to collapse on the divan for both expressions. Pearl-clutching is Haskell's finest feature and greatest strength, and it should do it whenever possible.

Tue, 23 Oct 2018

I conplained recently about GHC not being able to infer an Applicative instance from a type that already has a Monad instance, and there is a related complaint that the Monad instance must define >>=. In some type classes, you get a choice about what to define, and then the rest of the functions are built from the ones you provided. To take a particular simple example, with Eq you have the choice of defining == or /=, and if you omit one Haskell will construct the other for you. It could do this with >>= and join, but it doesn't, for technical reasons I don't understand [1] [2] [3].

But both of these problems can be worked around. If I have a Monad instance, it seems to work just fine if I say:

    instance Applicative Tree where
pure = return
fs <*> xs = do
f <- fs
x <- xs
return (f x)


Where this code is completely canned, the same for every Monad.

And if I know join but not >>=, it seems to work just fine if I say:

    instance Monad Tree where
return = ...
x >>= f  = join (fmap f x) where
join tt = ...


I suppose these might faul foul of whatever problem is being described in the documents I linked above. But I'll either find out, or I won't, and either way is a good outcome.

[ Addendum: Vaibhav Sagar points out that my definition of <*> above is identical to that of Control.Monad.ap, so that instead of defining <*> from scratch, I could have imported ap and then written <*> = ap. ]

While I was writing up last week's long article about Traversable, I wrote this stuff about Applicative also. It's part of the story but I wasn't sure how to work it into the other narrative, so I took it out and left a remark that “maybe I'll publish a writeup of that later”. This is a disorganized collection of loosely-related paragraphs on that topic.

It concerns my attempts to create various class instance definitions for the following type:

    data Tree a = Con a | Add (Tree a) (Tree a)
deriving (Eq, Show)


which notionally represents a type of very simple expression tree over values of type a.

I need some function for making Trees that isn't too simple or too complicated, and I went with:

    h n | n < 2 = Con n
h n = if even n then Add (h (n div 2)) (h (n div 2))
else Add (Con 1) (h (n - 1))


which builds trees like these:

    2 = 1 + 1
3 = 1 + (1 + 1)
4 = (1 + 1) + (1 + 1)
5 = 1 + ((1 + 1) + (1 + 1))
6 = (1 + (1 + 1)) + (1 + (1 + 1))
7 = 1 + (1 + (1 + 1)) + (1 + (1 + 1))
8 = ((1 + 1) + (1 + 1)) + ((1 + 1) + (1 + 1))


Now I wanted to traverse h [1,2,3] but I couldn't do that because I didn't have an Applicative instance for Tree. I had been putting off dealing with this, but since Traversable doesn't really make sense without Applicative I thought the day of reckoning would come. Here it was. Now is when I learn how to fix all my broken monads.

To define an Applicative instance for Tree I needed to define pure, which is obvious (it's just Con) and <*> which would apply a tree of functions to a tree of inputs to get a tree of results. What the hell does that mean?

Well, I can kinda make sense of it. If I apply one function to a tree of inputs, that's straightforward, it's just fmap, and I get a tree of results. Suppose I have a tree of functions, and I replace the function at each leaf with the tree of its function's results. Then I have a tree of trees. But a tree that has trees at its leaves is just a tree. So I could write some tree-flattening function that builds the tree of trees, then flattens out the type. In fact this is just join that I already know from Monad world. The corresponding operation for lists takes a list of lists and flattens them into a single list.) Flattening a tree is quite easy to do:

    join (Con ta) = ta


and since this is enough to define a Monad instance for Tree I suppose it is enough to get an Applicative instance also, since every Monad is an Applicative. Haskell makes this a pain. It should be able to infer the Applicative from this, and I wasn't clever enough to do it myself. And there ought to be some formulaic way to get <*> from >>= and join and fmap, the way you can get join from >>=:

    join = (>>= id)


but I couldn't find out what it was. This gets back to my original complaint: Haskell now wants every Monad instance to be an instance of Applicative, but if I give it the fmap and the join and the return it ought to be able to figure out the Applicative instance itself instead of refusing to compile my program. Okay, fine, whatever. Haskell's gonna Hask.

(I later realized that building <*> when you have a Monad instance is easy once you know the recipe; it's just:

    fs <*> xs = do
f <- fs
x <- xs
return (f x)


So again, why can't GHC infer <*> from my Monad instance, maybe with a nonfatal warning?

    Warning: No Applicative instance provided for Tree; deriving one from Monad


This is not a rhetorical question.)

(Side note: it seems like there ought to be a nice short abbreviation of the (<*>) function above, the way one can write join = (>>= id). I sought one but did not find any. One can eliminate the do notation to obtain the expression:

    fs <*> xs = fs >>= \f -> xs >>= \x -> return (f x)


but that is not any help unless we can simplify the expression with the usual tricks, such as combinatory logic and η-conversion. I was not able to do this, and the automatic pointfree converter produced (. ((. (return .)) . (>>=))) . (>>=) ARGH MY EYES.)

Anyway I did eventually figure out my <*> function for trees by breaking the left side into cases. When the tree of functions is Con f it's a single function and we can just use fmap to map it over the input tree:

    (Con f) <*> tv = fmap f tv


And when it's bigger than that we can break it up recursively:

    (Add lt rt) <*> tv = Add (lt <*> tv) (rt <*> tv)


Once this is written it seemed a little embarrassing that it took me so long to figure out what it meant but this kind of thing always seems easier from the far side of the fence. It's hard to understand until you understand it.

Actually that wasn't quite the <*> I wanted. Say we have a tree of functions and a tree of arguments.

Add (Con (* 10))
(Con (* 100))
Add (Add (Con 3) (Con 4)) (Con 5)

I can map the whole tree of functions over each single leaf on the right, like this:

Add (Add (Add (Con 30) (Con 300))


or I can map each function over the whole tree on the right, like this:

Add
(Add (Add (Con 300) (Con 400)) (Con 500))

The code I showed earlier does the second of those. You can see it from the fmap f tv expression, which takes a single function and maps it over a whole tree of values. I had actually wanted the other one, but there isn't anything quite like fmap for that. I was busy trying to understand Applicative and I was afraid if I got distracted trying to invent a reverse fmap I might lose the thread. This happens to me a lot with Haskell. I did eventually go back and figure it out. The reverse fmap is

    pamf fs v = fmap ($v) fs -- good  or  pamf = flip (fmap . flip id) -- yuck  Now there's a simple answer to this which occurs to me now that I didn't think of before, but I'm going to proceed with how I planned to do it before, with pamf. The <*> that I didn't want looked like this:  (Con f) <*> tv = fmap f tv (Add lt rt) <*> tv = Add (lt <*> tv) (rt <*> tv)  I need to do the main recursion on the values argument instead of on the functions argument:  tf <*> (Con v) = pamf tf v where pamf fs v = fmap ($ v) fs
tf <*> (Add lv rv) = Add (tf <*> lv) (tf <*> rv)


(This is an interesting example: usually the base case is trivial and the recursive clause is harder to write, but this time it's the base case that's not perfectly straightforward.)

Anyway, this worked, but there was an easier solution at hand. The difference between the first version and the second is exactly the same as the difference between

        fs <*> xs = do
f <- fs
x <- xs
return (f x)


and

        fs <*> xs = do
x <- xs
f <- fs
return (f x)


Digging deeper into why this worked this way was interesting, but it's bed time, so I'm going to cut the scroll here.

Sat, 20 Oct 2018

Haskell evolved a lot since the last time I seriously wrote any Haskell code, so much so that all my old programs broke. My Monad instances don't compile any more because I'm no longer allowed to have a monad which isn't also an instance of Applicative. Last time I used Haskell, Applicative wasn't even a thing. I had read the McBride and Paterson paper that introduced applicative functors, but that was years ago, and I didn't remember any of the details. (In fact, while writing this article, I realized that the paper I read was a preprint, and I probably read it before it was published, in 2008.) So to resuscitate my old code I had to implement a bunch of <*> functions and since I didn't really understand what it was supposed to be doing I couldn't do that. It was a very annoying experience.

Anyway I got that more or less under control (maybe I'll publish a writeup of that later) and moved on to Traversable which, I hadn't realized before, was also introduced in that same paper. (In the prepublication version, Traversable was been given the unmemorable name IFunctor.) I had casually looked into this several times in the last few years but I never found anything enlightening. A Traversable is a functor (which must also implement Foldable, but let's pass over that for now, no pun intended) that implements a traverse method with the following signature:

    traverse :: Applicative f => (a -> f b) -> t a -> f (t b)


The traversable functor itself here is t. The f thing is an appurtenance. Often one looks at the type of some function and says “Oh, that's what that does”, but I did not get any understanding from this signature.

The first thing to try here is to make it less abstract. I was thinking about Traversable this time because I thought I might want it for a certain type of tree structure I was working with. So I defined an even simpler tree structure:

    data Tree a = Con a | Add (Tree a) (Tree a)
deriving (Eq, Show)


Defining a bunch of other cases wouldn't add anything to my understanding, and it would make it take longer to try stuff, so I really want to use the simplest possible example here. And this is it: one base case, one recursive case.

Then I tried to make this type it into a Traversable instance. First we need it to be a Functor, which is totally straightforward:

    instance Functor Tree where
fmap f (Con a) = Con (f a)
fmap f (Add x y) = Add (fmap f x) (fmap f y)


Then we need it to be a Foldable, which means it needs to provide a version of foldr. The old-fashioned foldr was

    foldr :: (a -> b -> b) -> b -> [a] -> b


but these days the list functor in the third place has been generalized:

    foldr :: Foldable f => (a -> b -> b) -> b -> f a -> b


The idea is that foldr fn collapses a list of as into a single b value by feeding in the as one at a time. Each time, foldr takes the previous b and the current a and constructs a new b. The second argument is the initial value of b. Another way to think about it is that every list has the form

    e1 : e2 : .... : []


and foldr fn b applied to this list replaces the (:) calls with fn and the trailing [] with b, giving me

    e1 f e2 f .... f b


The canonical examples for lists are:

    sum = foldr (+) 0


(add up the elements, starting with zero) and

    length = foldr (\_ -> (+ 1)) 0


(ignore the elements, adding 1 to the total each time, starting with zero). Also foldr (:) [] is the identity function for lists because it replaces the (:) calls with (:) and the trailing [] with [].

Anyway for Tree it looks like this:

   instance Foldable Tree where
foldr f b (Con a) = f a b
foldr f b (Add x y) = (foldr f) (foldr f b x) y


The Con clause says to take the constant value and combine it with the default total. The Add clause says to first fold up the left-side subtree x to a single value, then use that as the initial value for folding up the right-side subtree y, so everything gets all folded up together. (We could of course do the right subtree before the left; the results would be different but just as good.)

I didn't write this off the top of my head, I got it by following the types, like this:

1. In the first clause

    foldr f b (Con a) = ???


we have a function f that wants an a value and a b value, and we have both an a and a b, so put the tabs in the slots.

2. In the second clause

    foldr f b (Add x y) = ???


f needs an a value and none is available, so we can't use f by itself. We can only use it recursively via foldr. So forget f, we will only be dealing only with foldr f, which has type b -> Tree a -> b. We need to apply this to a b value and the only one we have is b, and then we need to apply that to one of the subtrees, say x, and thus we have synthesized the foldr f b x subexpression. Then pretty much the same process gets us the rest of it: we need a b and the only one we have now is foldr f b x, and then we need another tree and the only one we haven't used is y.

It turns out it is easier and more straightforward to write foldMap instead, but I didn't know that at the time. I won't go into it further because I have already digressed enough. The preliminaries are done, we can finally get on to the thing I wanted, the Traversable:

    instance Traversable Tree where
traverse = ....


and here I was stumped. What is this supposed to actually do? For our Tree functor it has this signature:

    traverse :: Applicative f => (a -> f b) -> Tree a -> f (Tree b)


Okay, a function a -> f b I understand, it turns each tree leaf value into a list or something, so at each point of the tree it gets out a list of bs, and it potentially has one of those for each item in the input tree. But how the hell do I turn a tree of lists into a single list of Tree b? (The answer is that the secret sauce is in the Applicative, but I didn't understand that yet.)

I scratched my head and read a bunch of different explanations and none of them helped. All the descriptions I found were in either prose or mathematics and I still couldn't figure out what it was for. Finally I just wrote a bunch of examples and at last the light came on. I'm going to show you the examples and maybe the light will come on for you too.

We need two Traversable functors to use as examples. We don't have a Traversable implementation for Tree yet so we can't use that. When I think of functors, the first two I always think of are List and Maybe, so we'll use those.

    > traverse (\n -> [1..n]) Nothing
[Nothing]
> traverse (\n -> [1..n]) (Just 3)
[Just 1,Just 2,Just 3]


Okay, I think I could have guessed that just from the types. And going the other way is not very interesting because the output, being a Maybe, does not have that much information in it.

    > let f x = if even x then Just (x div 2) else Nothing


If the !!x!! is even then the result is just half of !!x!!, and otherwise the division by 2 “fails” and the result is nothing. Now:

    > traverse f [ 1, 2, 3, 4 ]
Nothing
> traverse f [ 10, 4, 18 ]
Just [5,2,9]


It took me a few examples to figure out what was going on here: When all the list elements are even, the result is Just a list of half of each. But if any of the elements is odd, that spoils the whole result and we get Nothing. (traverse f [] is Just [] as one would expect.)

That pretty much exhausts what can be done with lists and maybes. Now I have two choices about where to go next: I could try making both functors List, or I could use a different functor entirely. (Making both Maybe seemed like a nonstarter.) Using List twice seemed confusing, and when I tried it I could kinda see what it was doing but I didn't understand why. So I took a third choice: I worked up a Traversable instance for Tree just by following the types even though I didn't understand what it ought to be doing. I thought I'd at least see if I could get the easy clause:

    traverse :: Applicative f => (a -> f b) -> Tree a -> f (Tree b)

instance Traversable Tree where
traverse fn (Con a) = ...


In the ... I have fn :: a -> f b and I have at hand a single a. I need to construct a Tree b. The only way to get a b is to apply fn to it, but this gets me an f b and I need f (Tree b). How do I get the Tree in there? Well, that's what Con is for, getting Tree in there, it turns a t into Tree t. But how do I do that inside of f? I tinkered around a little bit and eventually found

  traverse fn (Con a) = Con <$> (fn a)  which not only type checks but looks like it could even be correct. So now I have a motto for what <$> is about: if I have some function, but I want to use it inside of some applicative functor f, I can apply it with <$> instead of with $.

Which, now that I have said it myself, I realize it is exactly what everyone else was trying to tell me all along: normal function application takes an a -> b and applies to to an a giving a b. Applicative application takes an f (a -> b) and applies it to an f a giving an f b. That's what applicative functors are all about, doing stuff inside of f.

Okay, I can listen all day to an explanation of what an electric drill does, but until I hold it in my hand and drill some holes I don't really understand.

Encouraged, I tried the hard clause:

  traverse fn (Add x y) = ...


  traverse fn (Add x y) = Add <$> ...  The Con clause had fn a at that point to produce an f b but that won't work here because we don't have an a, we have a whole Tree a, and we don't need an f b, we need an f (Tree b). Oh, no problem, traverse fn supposedly turns a Tree a into an f (Tree b), which is just what we want. And it makes sense to have a recursive call to traverse because this is the recursive part of the recursive data structure:  traverse fn (Add x y) = Add <$> (traverse fn x) ...


Clearly traverse fn y is going to have to get in there somehow, and since the pattern for all the applicative functor stuff is

  f <$> ... <*> ... <*> ...  let's try that:  traverse fn (Add x y) = Add <$> (traverse fn x) <*> (traverse fn y)


This looks plausible. It compiles, so it must be doing something. Partial victory! But what is it doing? We can run it and see, which was the whole point of an exercise: work up a Traversable instance for Tree so that I can figure out what Traversable is about.

Here are some example trees:

 t1 = Con 3                              -- 3
t2 = Add (Con 3) (Con 4)                -- 3 + 4
t3 = Add (Add (Con 3) (Con 4)) (Con 2)  -- (3 + 4) + 2


(I also tried Add (Con 3) (Add (Con 4) (Con 2)) but it did not contribute any new insights so I will leave it out of this article.)

First we'll try Maybe. We still have that f function from before:

    f x = if even x then Just (x div 2) else Nothing


but traverse f t1, traverse f t2, and traverse f t3 only produce Nothing, presumably because of the odd numbers in the trees. One odd number spoils the whole thing, just like in a list.

So try:

    traverse f (Add (Add (Con 10) (Con 4)) (Con 18))


which yields:

          Just (Add (Add (Con 5) (Con 2)) (Con 9))


It keeps the existing structure, and applies f at each value point, just like fmap, except that if f ever returns Nothing the whole computation is spoiled and we get Nothing. This is just like what traverse f was doing on lists.

But where does that spoilage behavior come from exactly? It comes from the overloaded behavior of <*> in the Applicative instance of Maybe:

 (Just f) <*> (Just x) = Just (f x)
Nothing  <*> _        = Nothing
_  <*> Nothing  = Nothing


Once we get a Nothing in there at any point, the Nothing takes over and we can't get rid of it again.

I think that's one way to think of traverse: it transforms each value in some container, just like fmap, except that where fmap makes all its transformations independently, and reassembles the exact same structure, with traverse the reassembly is done with the special Applicative semantics. For Maybe that means “oh, and if at any point you get Nothing, just give up”.

Now let's try the next-simplest Applicative, which is List. Say,

    g n = [ 1 .. n ]


Now traverse g (Con 3) is [Con 1,Con 2,Con 3] which is not exactly a surprise but traverse g (Add (Con 3) (Con 4)) is something that required thinking about:

    [Add (Con 1) (Con 1),


This is where the light finally went on for me. Instead of thinking of lists as lists, I should be thinking of them as choices. A list like [ "soup", "salad" ] means that I can choose soup or salad, but not both. A function g :: a -> [b] says, in restaurant a, what bs are on the menu.

The g function says what is on the menu at each node. If a node has the number 4, I am allowed to choose any of [1,2,3,4], but if it has the number 3 then the choice 4 is off the menu and I can choose only from [1,2,3].

Traversing g over a Tree means, at each leaf, I am handed a menu, and I make a choice for what goes at that leaf. Then the result of traverse g is a complete menu of all the possible complete trees I could construct.

Now I finally understand how the t and the f switch places in

    traverse :: Applicative f => (a -> f b) -> t a -> f (t b)


I asked “how the hell do I turn a tree of lists into a single list of Tree b”? And that's the answer: each list is a local menu of dishes available at one leaf, and the result list is the global menu of the complete dinners available over the entire tree.

Okay! And indeed traverse g (Add (Add (Con 3) (Con 4)) (Con 2)) has 24 items, starting

      Add (Add (Con 1) (Con 1)) (Con 1)
...


and ending

      Add (Add (Con 3) (Con 4)) (Con 1)


That was traversing a list function over a Tree. What if I go the other way? I would need an Applicative instance for Tree and I didn't really understand Applicative yet so that wasn't going to happen for a while. I know I can't really understand Traversable without understanding Applicative first but I wanted to postpone the day of reckoning as long as possible.

What other functors do I know? One easy one is the functor that takes type a and turns it into type (String, a). Haskell even has a built-in Applicative instance for this, so I tried it:

     > traverse (\x -> ("foo", x)) [1..3]
("foofoofoo",[1,2,3])
> traverse (\x -> ("foo", x*x)) [1,5,2,3]
("foofoofoofoo",[1,25,4,9])


Huh, I don't know what I was expecting but I think that wouldn't have been it. But I figured out what was going on: the built-in Applicative instance for the a -> (String, a) functor just concatenates the strings. In general it is defined on a -> (m, b) whenever m is a monoid, and it does fmap on the right component and uses monoid concatenation on the left component. So I can use integers instead of strings, and it will add the integers instead of concatenating the strings. Except no, it won't, because there are several ways to make integers into a monoid, but each type can only have one kind of Monoid operations, and if one was wired in it might not be the one I want. So instead they define a bunch of types that are all integers in obvious disguises, just labels stuck on them that say “I am not an integer, I am a duck”; “I am not an integer, I am a potato”. Then they define different overloadings for “ducks” and “potatoes”. Then if I want the integers to get added up I can put duck labels on my integers and if I want them to be multiplied I can stick potato labels on instead. It looks like this:

   import Data.Monoid
h n = (Sum 1, n*10)


Sum is the duck label. When it needs to combine two ducks, it will add the integers:

   > traverse h [5,29,83]
(Sum {getSum = 3},[50,290,830])


But if we wanted it to multiply instead we could use the potato label, which is called Data.Monoid.Product:

    > traverse (\n -> (Data.Monoid.Product 7, 10*n)) [5,29,83]
(Product {getProduct = 343}, [50,290,830])


There are three leaves, so we multiply three sevens and get 343.

Or we could do the same sort of thing on a Tree:

    > traverse (\n -> (Data.Monoid.Product n, 10*n)) (Add (Con 2) (Add (Con 3) (Con 4)))
(Product {getProduct = 24}, Add (Con 20) (Add (Con 30) (Con 40)))


Here instead of multiplying together a bunch of sevens we multiply together the leaf values themselves.

The McBride and Paterson paper spends a couple of pages talking about traversals over monoids, and when I saw the example above it started to make more sense to me. And their ZipList example became clearer too. Remember when we had a function that gave us a menu at every leaf of a tree, and traverse-ing that function over a tree gave us a menu of possible trees?

       > traverse (\n -> [1,n,n*n]) (Add (Con 2) (Con 3))


There's another useful way to traverse a list function. Instead of taking each choice at each leaf we make a single choice ahead of time about whether we'll take the first, second, or third menu item, and then we take that item every time:

    > traverse (\n -> Control.Applicative.ZipList [1,n,n*n]) (Add (Con 2) (Con 3))
ZipList {getZipList = [Add (Con 1) (Con 1),


There's a built-in instance for Either a b also. It's a lot like Maybe. Right is like Just and Left is like Nothing. If all the sub-results are Right y then it rebuilds the structure with all the ys and gives back Right (structure). But if any of the sub-results is Left x then the computation is spoiled and it gives back the first Left x. For example:

 > traverse (\x -> if even x then Left (x div 2) else Right (x * 10)) [3,17,23,9]
Right [30,170,230,90]
> traverse (\x -> if even x then Left (x div 2) else Right (x * 10)) [3,17,22,9]
Left 11


Okay, I think I got it.

Now I just have to drill some more holes.

Mon, 15 Oct 2018

Reading over my recent article complaining about the environment functor I realized there's yet another terminology problem that makes the discussion unnecessarily confusing. “The” environment functor isn't unique. There is a family of environment functors, one for each possible environment type e. If g is the environment functor at type e, a value of type g t is a function e → t. But e could be anything and if g and h are environment functors at two different types e and e’ they are of course different functors.

This is even obvious from the definition:

    data Environ e t = Env (e -> t)
instance Functor (Environ e) where
fmap f (Env x) = Env $\e -> f (x e)  The functor isn't Environ, it's Environ e, and the functor instance declaration, as it says on line 2. (It seems to me that the notation is missing a universal quantifier somewhere, but I'm not going to open that issue.) We should speak of Environ e as an environment functor, not the environment functor. So for example instead of: When operating in the environment functor, fmap has the type (a -> b) -> g a -> g b I should have said: When operating in an environment functor, fmap has the type (a -> b) -> g a -> g b And instead of: A function p -> q is a q parcel in the environment functor I should have said: A function p -> q is a q parcel in an environment functor or A function p -> q is a q parcel in the environment functor at p although I'm not sure I like the way the prepositions are proliferating there. The same issue affects ⸢the⸣ reader monad, ⸢the⸣ state monad, and many others. I'm beginning to find remarkable how much basic terminology Haskell is missing or gets wrong. Mathematicians have a very keen appreciation of the importance of specific and precise terminology, and you'd think this would have filtered into the Haskell world. People are forever complaining that Haskell uses unfamiliar terms like “functor”, and the community's response is (properly, I think) that these terms are pre-existing and there is no point to inventing a new term that will be just as unfamiliar, or, worse, lure people into thinking that the know what it means when they don't. You don't want to call a functor a “container”, says the argument, because many functors (environment functors for example) are nothing at all like containers. I think this is wise. But having planted their flag on that hill, the Haskell folks don't then use their own terminology correctly. I complained years ago that the term “monad” was used interchangeably for four subtly different concepts, and here we actually have a fifth. I pointed out that in the case of Environment e t, common usage refers to both Environment e and Environment e t as monads, and only the first is correct. But when people say “the environment monad” they mean that Environment itself is a monad, which it is not. Fri, 12 Oct 2018 Is there any good terminology for a value of type f a when f is an arbitrary functor? I will try calling an f t value a “t parcel” and see how that works. The more I think about “parcel” the happier I am with it. It strongly suggests container types, of course, so that a t parcel might be a boxful of ts. But it also hints at some other possible situations: • You might open the parcel and find it empty. (Maybe t) • You might open the parcel and find, instead of the t you expected, a surprising prank snake. (Either ErrorMessage t) • You might open the parcel and find that your t has been shipped with assembly required. (env -> t) • The parcel might explode when you open it. (IO t) • And, of course, a burrito is a sort of parcel of meat and beans. I coined “parcel” thinking that one would want different terminology for values of type f t depending on whether f was a functor (“parcel”) or also a monad (“mote”). Of course every mote is a parcel, but not always vice versa. Now I'm not sure that both terms are needed. Non-monadic functors are unusual, and non-applicative functors rare, so perhaps one term will do for all three. Thu, 11 Oct 2018 Here we have the well-known fmap function:  fmap :: Functor f => (a -> b) -> f a -> f b  It takes a single function and a (collection of input values / decorated input value / something something input value) and produces a (collection of output values / decorated output value / something something output value). Yow, that's not going to work. Is there any good terminology for a value of type f a when f is an arbitrary functor? A while back I discussed a similar problem and suggested the term “mote” for a value in a monadic type. I will try calling an f t value a “t parcel and see how that works. So [t], Maybe t, and IO t are all examples of t parcels, in various functors. Starting over then. Here we have the well-known fmap function:  fmap :: Functor f => (a -> b) -> f a -> f b  It takes a single function, and an a parcel, and produces a b parcel, by applying the function independently to the a values in the parcel. Here is a sort of reversed version of fmap that I call pamf:  pamf :: Functor f => f (a -> b) -> a -> f b  It takes a parcel of functions, and a single input and produces a parcel of outputs, by applying each function in the parcel independently to the single a value. It can be defined in terms of fmap:  pamf fs a = fmap ($ a) fs


So far so good. Now I ask you to predict the type of

    pamf fmap


Certainly it should start out with

    pamf fmap :: (Functor f, Functor g) => ...


because the pamf and the fmap might be operating in two different functors, right? Indeed, if I compose the functions the other way around, fmap pamf, the type does begin this way; it is:

    (Functor f, Functor g) => f (g (a -> b)) -> f (a -> g b)


The f here is the functor in which fmap operates, and the g is the functor in which pamf is operating. In general fmap takes an arbitrary function

              a       ->      b


and lifts it to a new function that operates in the f functor:

            f a       ->    f b


Here it has taken pamf, which is a function

          g (a -> b)  ->     (a -> g b)


and lifted it to a new function that operates in the f functor:

       f (g (a -> b))  ->  f (a -> g b)


This is complicated but straightforward. Okay, that was fmap pamf. What about pamf fmap though? The computed type is

        pamf fmap :: Functor f => f a -> (a -> b) -> f b


and when I saw this I said “What. Where did g go? What happened to g?”

Then I paused and for a while and said “… I bet it's that goddamn environment thing again.” Yep, that's what it was. It's the environment functor, always turning up where I don't want it and least expect it, like that one guy we all went to college with. The environment functor, by the way, is yet another one of those things that Haskell ought to have a standard name for, but doesn't. The phrase “the reader monad” is fairly common, but here I only want the functor part of the monad. And people variously say “reader monad”, “environment monad”, and “evaluation monad” to mean the same thing. In this article, it will be the environment functor.

Here's what happened. Here are fmap and pamf again:

    fmap :: Functor f => (p -> q) -> f p -> f q

pamf :: Functor g => g (a -> b) -> a -> g b


The first argument to pamf should be a parcel in the g functor. But fmap is not a parcel, so pamf fmap will be a type error, right? Wrong! If you are committed enough, there is a way to construe any function as a parcel. A function p -> q is a q parcel in the environment functor. Say that g denotes an environment functor. In this functor, a parcel of type g t is a function which consults an “environment” of type e and yields a result of type t. That is, $$g\ t \equiv e \to t.$$

When operating in the environment functor, fmap has the type (a -> b) -> g a -> g b, which is shorthand for (a -> b) -> (e -> a) -> (e -> b). This instance of fmap is defined this way:

    fmap f x = \e -> f (x e)


or shorter and more mysteriously

    fmap = (.)


which follows by η-reduction, something Haskell enthusiasts never seem to get enough of.

In fmap f x, the x isn't the actual value to give to f; instead it's a parcel, as it always is with fmap. In the context of the environment functor, x is a function that consults the environment e and returns an a. The result of fmap f x is a new parcel: it uses x to consult the supplied environment for a value of type a, which it then feeds to f to get the required value of type b.

In the application pamf fmap, the left side pamf wants fmap to be a parcel. But it's not a parcel, it's a function. So, type error, right? No! Any function is a parcel if you want it to be, it's a parcel in the environment functor! And fmap is a function:

    fmap :: Functor f => (p -> q) -> f p -> f q


so it can be understood as a parcel in the environment functor, where the environment e has type p -> q. Then pamf is operating in this environment functor, so $$g\ t = (p \to q) \to t.$$ A g t parcel is a function that consults an “environment” of type p -> q and somehow produces a t value. (Haskell folks, who are obsessed with currying all the things, will write this as the nearly unreadable g = ((->) (p -> q)).)

We wanted pamf to have this type:

    pamf :: Functor g =>            g (a -> b)  -> a ->            g b


and since Haskell has decided that g must be the environment functor with !!g\ x \equiv (p \to q) \to x!!, this is an abbreviation for:

    pamf ::              ((p -> q) -> (a -> b)) -> a -> ((p -> q) -> b)


To apply this to fmap, we have to unify the type of pamf's argument, which is (p -> q) -> (a -> b), and the type of fmap, which is (p -> q) -> (f p -> f q). Then !!a\equiv f\ p!! and !!b \equiv f\ q!!, so the result of pamf fmap is

    pamf fmap :: Functor f => f p -> ((p -> q) -> f q)


Where did g go? It was specialized to mean the environment functor ((->) (p -> q)), so it's gone.

The funny thing about the type of pamf fmap is that it is exactly the type of flip fmap, which is fmap with the order of its two arguments reversed:

   (flip fmap) x f ≡ fmap f x


and indeed, by some theorem or other, because the types are identical, the functions themselves must be identical also! (There are some side conditions, all of which hold here.) The two functions pamf fmap and flip fmap are identical. Analogous to the way fmap, restricted to the environment functor, is identical to (.), pamf, when similarly restricted, is exactly flip. You can even see this from its type:

    pamf :: ((p -> q) -> (a -> b)) -> a -> ((p -> q) -> b)


Or, cleaning up some superfluous parentheses and inserting some new ones:

    pamf :: ((p -> q) ->  a -> b) -> (a ->  (p -> q) -> b)


And putting !!c = p\to q!!:

    pamf :: (c        -> a -> b) -> (a -> c        -> b)
flip :: (                 the same                 )


Honestly, I would have preferred a type error: “Hey, dummy, fmap has the wrong type to be an argument to pamf, which wants a functorial value.” Instead I got “Okay, if you want functions to be a kind of functor I can do that, also wouldn't it be simpler if the universe was two-dimensional and there were only three kinds of quarks? Here you go, no need to thank me!” Maybe someone can explain to me why this is a useful behavior, and then explain why it is so useful that it should happen automatically and implicitly instead of being triggered by some lexical marker like:

    newtype Environment e a = Environment (e -> a)
instance Functor (Environment e) where
fmap f (Environment x) = Environment $\e -> f (x e)  I mean, seriously, suppose you wrote a + b where b was accidentally a function instead of a number. What if when you did that, instead of a type error, Haskell would silently shift into some restricted domain in which it could implicitly interpret b as a number in some weird way and give you something totally bizarre? Isn't the whole point of Haskell supposed to be that it doesn't implicitly convert things that way? [ Addendum 20181111: Apparently, everyone else hates it too. ] Sat, 08 Sep 2018 Here's something else that often goes wrong when I am writing a Haskell program. It's related to the problem in the previous article but not the same. Let's say I'm building a module for managing polynomials. Say Polynomial a is the type of (univariate) polynomials over some number-like set of coefficients a. Now clearly this is going to be a functor, so I define the Functor instance, which is totally straightforward:  instance Functor Polynomial where fmap f (Poly a) = Poly$ map f a


Then I ask myself if it is also going to be an Applicative. Certainly the pure function makes sense; it just lifts a number to be a constant polynomial:

       pure a = Poly [a]


But what about <*>? This would have the type:

    (Polynomial (a -> b)) -> Polynomial a -> Polynomial b


The first argument there is a polynomial whose coefficients are functions. This is not something we normally deal with. That ought to be the end of the matter.

But instead I pursue it just a little farther. Suppose we did have such an object. What would it mean to apply a functional polynomial and an ordinary polynomial? Do we apply the functions on the left to the coefficients on the right and then collect like terms? Say for example

\begin{align} \left((\sqrt\bullet) \cdot x + \left(\frac1\bullet\right) \cdot 1 \right) ⊛ (9x+4) & = \sqrt9 x^2 + \sqrt4 x + \frac19 x + \frac14 \\ & = 3x^2 + \frac{19}{9} x + \frac 14 \end{align}

Well, this is kinda interesting. And it would mean that the pure definition wouldn't be what I said; instead it would lift a number to a constant function:

    pure a = Poly [λ_ -> a]


Then the ⊛ can be understood to be just like polynomial multiplication, except that coefficients are combined with function composition instead of with multiplication. The operation is associative, as one would hope and expect, and even though the ⊛ operation is not commutative, it has a two-sided identity element, which is Poly [id]. Then I start to wonder if it's useful for anything, and how ⊛ interacts with ordinary multiplication, and so forth.

This is different from the failure mode of the previous article because in that example I was going down a Haskell rabbit hole of more and more unnecessary programming. This time the programming is all trivial. Instead, I've discovered a new kind of mathematical operation and I abandon the programming entirely and go off chasing a mathematical wild goose.

[ Addendum 20181109: Another one of these. ]

Mon, 03 Sep 2018

Whenever I try to program in Haskell, the same thing always goes wrong. Here is an example.

I am writing a module to operate on polynomials. The polynomial !!x^3 - 3x + 1!! is represented as

    Poly [1, -3, 0, 1]


[ Addendum 20180904: This is not an error. The !!x^3!! term is last, not first. Much easier that way. Fun fact: two separate people on Reddit both commented that I was a dummy for not doing it the easy way, which is the way I did do it. Fuckin' Reddit, man. ]

I want to add two polynomials. To do this I just add the corresponding coefficients, so it's just

    (Poly a) + (Poly b) = Poly $zipWith (+) a b  Except no, that's wrong, because it stops too soon. When the lists are different lengths, zipWith discards the extra, so for example it says that !!(x^2 + x + 1) + (2x + 2) = 3x + 3!!, because it has discarded the extra !!x^2!! term. But I want it to keep the extra, as if the short list was extended with enough zeroes. This would be a correct implementation:  (Poly a) + (Poly b) = Poly$ addup a b   where


and I can write this off the top of my head.

But do I? No, this is where things go off the rails. “I ought to be able to generalize this,” I say. “I can define a function like zipWith that is defined over any Monoid, it will combine the elements pairwise with mplus, and when one of the lists runs out, it will pretend that that one has some memptys stuck on the end.” Here I am thinking of something like ffff :: Monoid a => [a] -> [a] -> [a], and then the (+) above would just be

    (Poly a) + (Poly b) = Poly (ffff a b)


as long as there is a suitable Monoid instance for the as and bs.

I could write ffff in two minutes, but instead I spend fifteen minutes looking around in Hoogle to see if there is already an ffff, and I find mzip, and waste time being confused by mzip, until I notice that I was only confused because mzip is for Monad, not for Monoid, and is not what I wanted at all.

So do I write ffff and get on with my life? No, I'm still not done. It gets worse. “I ought to be able to generalize this,” I say. “It makes sense not just for lists, but for any Traversable… Hmm, or does it?” Then I start thinking about trees and how it should decide when to recurse and when to give up and use mempty, and then I start thinking about the Maybe version of it.

Then I open a new file and start writing

    mzip :: (Traversable f, Monoid a) => f a -> f a -> f a
mzip as bs = …


And I go father and farther down the rabbit hole and I never come back to what I was actually working on. Maybe the next step in this descent into madness is that I start thinking about how to perform unification of arbitrary algebraic data structures, I abandon mzip and open a new file for defining class Unifiable

Actually when I try to program in Haskell there a lot of things that go wrong and this is only one of them, but it seems like this one might be more amenable to a quick fix than some of the other things.

[ Addendum 20180904: A lobste.rs user points out that I don't need Monoid, but only Semigroup, since I don't need mempty. True that! I didn't know there was a Semigroup class. ]

[ Addendum 20181109: More articles in this series: [2] [3] ]

Wed, 08 Aug 2018

[ Previously: [1] [2] ]

In my original article, I said:

I was fairly confident I had seen something like this somewhere before, and that it was not original to me.

Jeremy Yallop brought up an example that I had definitely seen before.

In 2008 Conor McBride and Ross Paterson wrote an influential paper, “Idioms: applicative programming with effects” that introduced the idea of an applicative functor, a sort of intermediate point between functors and monads. It has since made its way into standard Haskell and was deemed sufficiently important to be worth breaking backward compatibility.

McBride and Paterson used several notations for operations in an applicative functor. Their primary notation was !!\iota!! for what is now known as pure and !!\circledast!! for what has since come to be written as <*>. But the construction

$$\iota f \circledast is_1 \circledast \ldots \circledast is_n$$

came up so often they wanted a less cluttered notation for it:

We therefore find it convenient, at least within this paper, to write this form using a special notation

$$[\![ f is_1 \ldots is_n ]\!]$$

The brackets indicate a shift into an idiom where a pure function is applied to a sequence of computations. Our intention is to provide a sufficient indication that effects are present without compromising the readability of the code.

On page 5, they suggested an exercise:

… show how to replace !![\![!! and !!]\!]!! by identifiers iI and Ii whose computational behaviour delivers the above expansion.

They give a hint, intended to lead the reader to the solution, which involves a function named iI that does some legerdemain on the front end and then a singleton type data Ii = Ii that terminates the legerdemain on the back end. The upshot is that one can write

iI f x y Ii


and have it mean

(pure f) <*> x <*> y


The haskell wiki has details, written by Don Stewart when the McBride-Paterson paper was still in preprint. The wiki goes somewhat further, also defining

 data J = J


so that

iI f x y J z Ii


now does a join on the result of f x y before applying the result to z.

I have certainly read this paper more than once, and I was groping for this example while I was writing the original article, but I couldn't quite put my finger on it. Thank you, M. Yallop!

[ By the way, I am a little bit disappointed that the haskell wiki is not called “Hicki”. ]

In the previous article I described a rather odd abuse of the Haskell type system to use a singleton type as a sort of pseudo-keyword, and asked if anyone had seen this done elsewhere.

Joachim Breitner reported having seen this before. Most recently in LiquidHaskell, which defines a QED singleton type:

 data QED = QED
infixl 2 ***

(***) :: a -> QED -> Proof
_ *** _ = ()


so that they can end every proof with *** QED:

singletonP x
=   reverse [x]
==. reverse [] ++ [x]
==. [] ++ [x]
==. [x]
*** QED


This example is from Vazou et al., Functional Pearl: Theorem Proving for All, p. 3. The authors explain: “The QED argument serves a purely aesthetic purpose, allowing us to conclude proofs with *** QED.”.

Or see the examples from the bottom of the LH splash page, proving the associative law for ++.

I looked in the rest of the LiquidHaskell distribution but did not find any other uses of the singleton-type trick. I would still be interested to see more examples.

A friend asked me the other day about techniques in Haskell to pretend to make up keywords. For example, suppose we want something like a (monadic) while loop, say like this:

      while cond act =
cond >>= \b -> if b then act >> while cond act
else return ()


This uses a condition cond (which might be stateful or exception-throwing or whatever, but which must yield a boolean value) and an action act (likewise, but its value is ignored) and it repeates the action over and over until the condition is false.

Now suppose for whatever reason we don't like writing it as while condition action and we want instead to write while condition do action or something of that sort. (This is a maximally simple example, but the point should be clear even though it is silly.) My first suggestion was somewhat gross:

      while c _ a = ...


Now we can write

      while condition "do" action


and the "do" will be ignored. Unfortunately we can also write while condition "wombat" action and you know how programmers are when you give them enough rope.

But then I had a surprising idea. We can define it this way:

      data Do = Do
while c Do a = ...


Now we write

      while condition
Do action


and if we omit or misspell the Do we get a compile-time type error that is not even too obscure.

For a less trivial (but perhaps sillier) example, consider:

    data Exception a = OK a | Exception String

data Catch = Catch
data OnSuccess = OnSuccess
data AndThen = AndThen

try computation Catch handler OnSuccess success AndThen continuation =
case computation of OK a        -> success >> (OK a) >>= continuation
Exception e ->            (handler e) >>= continuation


The idea here is that we want to try a computation, and do one thing if it succeeds and another if it throws an exception. The point is not the usefulness of this particular and somewhat contrived exception handling construct, it's the syntactic sugar of the Catch, OnSuccess, and AndThen:

    try (evaluate some_expression)
Catch (\error -> case error of "Divison by zero" -> ...
... )
OnSuccess ...
AndThen ...


I was fairly confident I had seen something like this somewhere before, and that it was not original to me. But I've asked several Haskell experts and nobody has said it was familar. I thought perhaps I had seen it somewhere in Brent Yorgey's code, but he vehemently denied it.

So my question is, did I make up this technique of using a one-element type as a pretend keyword?

[ Addendum: At least one example of this trick appears in LiquidHaskell. I would be interested to hear about other places it has been used. ]

[ Addendum: Jeremy Yallop points out that a similar trick was hinted at in McBride and Paterson “Idioms: applicative programming with effects” (2008), with which I am familiar, although their trick is both more useful and more complex. So this might have been what I was thinking of. ]

Fri, 24 Apr 2015

Suppose you would like to perform an exhaustive search. Let's say for concreteness that we would like to solve this cryptarithm puzzle:

    S E N D
+   M O R E
-----------
M O N E Y


This means that we want to map the letters S, E, N, D, M, O, R, Y to distinct digits 0 through 9 to produce a five-digit and two four-digit numerals which, when added in the indicated way, produce the indicated sum.

(This is not an especially difficult example; my 10-year-old daughter Katara was able to solve it, with some assistance, in about 30 minutes.)

If I were doing this in Perl, I would write up either a recursive descent search or a solution based on a stack or queue of partial solutions which the program would progressively try to expand to a full solution, as per the techniques of chapter 5 of Higher-Order Perl. In Haskell, we can use the list monad to hide all the searching machinery under the surface. First a few utility functions:

    import Control.Monad (guard)

digits = [0..9]

to_number = foldl (\a -> \b -> a*10 + b) 0
remove rs ls = foldl remove' ls rs
where remove' ls x = filter (/= x) ls


to_number takes a list of digits like [1,4,3] and produces the number they represent, 143. remove takes two lists and returns all the things in the second list that are not in the first list. There is probably a standard library function for this but I don't remember what it is. This version is !!O(n^2)!!, but who cares.

Now the solution to the problem is:

    --     S E N D
--   + M O R E
--   ---------
--   M O N E Y

solutions = do
s <- remove [0] digits
e <- remove [s] digits
n <- remove [s,e] digits
d <- remove [s,e,n] digits
let send = to_number [s,e,n,d]
m <- remove [0,s,e,n,d] digits
o <- remove [s,e,n,d,m] digits
r <- remove [s,e,n,d,m,o] digits
let more = to_number [m,o,r,e]
y <- remove [s,e,n,d,m,o,r] digits
let money = to_number [m,o,n,e,y]
guard $send + more == money return (send, more, money)  Let's look at just the first line of this:  solutions = do s <- remove [0] digits …  The do notation is syntactic sugar for  (remove [0] digits) >>= \s -> …  where “…” is the rest of the block. To expand this further, we need to look at the overloading for >>= which is implemented differently for every type. The mote on the left of >>= is a list value, and the definition of >>= for lists is:  concat$ map (\s -> …) (remove [0] digits)


where “…” is the rest of the block.

So the variable s is bound to each of 1,2,3,4,5,6,7,8,9 in turn, the rest of the block is evaluated for each of these nine possible bindings of s, and the nine returned lists of solutions are combined (by concat) into a single list.

The next line is the same:

      e <- remove [s] digits


for each of the nine possible values for s, we loop over nine value for e (this time including 0 but not including whatever we chose for s) and evaluate the rest of the block. The nine resulting lists of solutions are concatenated into a single list and returned to the previous map call.

      n <- remove [s,e] digits
d <- remove [s,e,n] digits


This is two more nested loops.

      let send = to_number [s,e,n,d]


At this point the value of send is determined, so we compute and save it so that we don't have to repeatedly compute it each time through the following 300 loop executions.

      m <- remove [0,s,e,n,d] digits
o <- remove [s,e,n,d,m] digits
r <- remove [s,e,n,d,m,o] digits
let more = to_number [m,o,r,e]


Three more nested loops and another computation.

      y <- remove [s,e,n,d,m,o,r] digits
let money = to_number [m,o,n,e,y]


Yet another nested loop and a final computation.

      guard $send + more == money return (send, more, money)  This is the business end. I find guard a little tricky so let's look at it slowly. There is no binding (<-) in the first line, so these two lines are composed with >> instead of >>=:  (guard$ send + more == money) >> (return (send, more, money))


which is equivalent to:

      (guard $send + more == money) >>= (\_ -> return (send, more, money))  which means that the values in the list returned by guard will be discarded before the return is evaluated. If send + more == money is true, the guard expression yields [()], a list of one useless item, and then the following >>= loops over this one useless item, discards it, and returns yields a list containing the tuple (send, more, money) instead. But if send + more == money is false, the guard expression yields [], a list of zero useless items, and then the following >>= loops over these zero useless items, never runs return at all, and yields an empty list. The result is that if we have found a solution at this point, a list containing it is returned, to be concatenated into the list of all solutions that is being constructed by the nested concats. But if the sum adds up wrong, an empty list is returned and concated instead. After a few seconds, Haskell generates and tests 1.36 million choices for the eight bindings, and produces the unique solution:  [(9567,1085,10652)]  That is:  S E N D 9 5 6 7 + M O R E + 1 0 8 5 ----------- ----------- M O N E Y 1 0 6 5 2  It would be an interesting and pleasant exercise to try to implement the same underlying machinery in another language. I tried this in Perl once, and I found that although it worked perfectly well, between the lack of the do-notation's syntactic sugar and Perl's clumsy notation for lambda functions (sub { my ($s) = @_; … } instead of \s -> …) the result was completely unreadable and therefore unusable. However, I suspect it would be even worse in Python because of semantic limitations of that language. I would be interested to hear about this if anyone tries it.

[ Addendum: Thanks to Tony Finch for pointing out the η-reduction I missed while writing this at 3 AM. ]

[ Addendum: Several people so far have misunderstood the question about Python in the last paragraph. The question was not to implement an exhaustive search in Python; I had no doubt that it could be done in a simple and clean way, as it can in Perl. The question was to implement the same underlying machinery, including the list monad and its bind operator, and to find the solution using the list monad.

[ Peter De Wachter has written in with a Python solution that clearly demonstrates that the problems I was worried about will not arise, at least for this task. I hope to post his solution in the next few days. ]

[ Addendum 20150803: De Wachter's solution and one in Perl ]

Thu, 26 Aug 2010

I think one problem (of many) that beginners might have with Haskell monads is the confusing terminology. The word "monad" can refer to four related but different things:

2. When a type constructor T of kind ∗ → ∗ is an instance of Monad we say that T "is a monad". For example, "Tree is a monad"; "((→) a) is a monad". This is the only usage that is strictly corrrect.

3. Types resulting from the application of monadic type constructors (#2) are sometimes referred to as monads. For example, "[Integer] is a monad".

Usage #1 is not a real problem; it does not occur that often, and is readily distinguished by context, capitalization, type font, and other markers. #2 is actually correct, so there is no problem there. #3 seems to be an uncommon colloquialism.

The most serious problem here is #4, that people refer to individual values of monadic types as "monads". Even when they don't do this, they are hampered by the lack of a good term for it. As I know no good alternative has been proposed. People often say "monadic value" (I think), which is accurate, but something of a mouthful.

One thing I have discovered in my writing life is that the clarity of a confusing document can sometimes be improved merely by replacing a polysyllabic noun phrase with a monosyllable. For example, chapter 3 of Higher-Order Perl discussed the technique of memoizing a function by generating an anonymous replacement for it that maintains a cache and calls the real function on a cache miss. Early drafts were hard to understand, and improved greatly when I replaced the phrase "anonymous replacement function" with "stub". The Perl documentation was significantly improved merely by replacing "associative array" everywhere with "hash" and "funny punctuation character" with "sigil".

I think a monosyllabic replacement for "monadic value" would be a similar boon to discussion of monads, not just for beginners but for everyone else too. The drawback, of introducing yet another jargon term, would in this case be outweighed by the benefits. Jargon can obscure, but sometimes it can clarify.

The replacement word should be euphonious, clear but not overly specific, and not easily confused with similar jargon words. It would probably be good for it to begin with the letter "m". I suggest:

mote

So return takes a value and returns a mote. The >>= function similarly lifts a function on pure values to a function on motes; when the mote is a container one may think of >>= as applying the function to the values in the container. [] is a monad, so lists are motes. The expression on the right-hand side of a var ← expr in a do-block must have mote type; it binds the mote on the right to the name on the left, using the >>= operator.

I have been using this term privately for several months, and it has been a small but noticeable success. Writing and debugging monadic programs is easier because I have a simple name for the motes that the program manipulates, which I can use when I mumble to myself: "What is the type error here? Oh, commit should be returning a mote." And then I insert return in the right place.

I'm don't want to oversell the importance of this invention. But there is clearly a gap in the current terminology, and I think it is well-filled by "mote".

(While this article was in progress I discovered that What a Monad is not uses the nonceword "mobit". I still prefer "mote".)

Sun, 03 Jan 2010

A short bibliography of probability monads
Several people helpfully wrote to me to provide references to earlier work on probability distribution monads. Here is a summary:

My thanks to Stephen Tetley, Gaal Yahas, and Luke Palmer for these.

I did not imagine that my idea was a new one. I arrived at it by thinking about List as a representation of non-deterministic computation. But if you think of it that way, the natural interpretation is that every list element represents an equally likely outcome, and so annotating the list elements with probabilities is the obvious next step. So the existence of the Erwig library was not a big surprise.

A little more surprising though, were the references in the Erwig paper. Specifically, the idea dates back to at least 1981; Erwig cites a paper that describes the probability monad in a pure-mathematics context.

Nobody responded to my taunting complaint about Haskell's failure to provide support a good monad of sets. It may be that this is because they all agree with me. (For example, the documentation of the Erwig package says "Unfortunately we cannot use a more efficient data structure because the key type must be of class Ord, but the Monad class does not allow constraints for result types.") But a number of years ago I said that the C++ macro processor blows goat dick. I would not have put it so strongly had I not naïvely believed that this was a universally-held opinion. But no, plenty of hapless C++ programmers wrote me indignant messages defending their macro system. So my being right is no guarantee that language partisans will not dispute with me, and the Haskell community's failure to do so in this case reflects well on them, I think.

Thu, 31 Dec 2009

A monad for probability and provenance
I don't quite remember how I arrived at this, but it occurred to me last week that probability distributions form a monad. This is the first time I've invented a new monad that I hadn't seen before; then I implemented it and it behaved pretty much the way I thought it would. So I feel like I've finally arrived, monadwise.

Suppose a monad value represents all the possible outcomes of an event, each with a probability of occurrence. For concreteness, let's suppose all our probability distributions are discrete. Then we might have:

	data ProbDist p a = ProbDist [(a,p)] deriving (Eq, Show)
unpd (ProbDist ps) = ps

Each a is an outcome, and each p is the probability of that outcome occurring. For example, biased and unbiased coins:

    unbiasedCoin = ProbDist [ ("heads", 0.5),
("tails", 0.5) ];

biasedCoin   = ProbDist [ ("heads", 0.6),
("tails", 0.4) ];


Or a couple of simple functions for making dice:

    import Data.Ratio

d sides = ProbDist [(i, 1 % sides) | i <- [1 .. sides]]
die = d 6


d n is an n-sided die.

The Functor instance is straightforward:

    instance Functor (ProbDist p) where
fmap f (ProbDist pas) = ProbDist $map (\(a,p) -> (f a, p)) pas  The Monad instance requires return and >>=. The return function merely takes an event and turns it into a distribution where that event occurs with probability 1. I find join easier to think about than >>=. The join function takes a nested distribution, where each outcome of the outer distribution specifies an inner distribution for the actual events, and collapses it into a regular, overall distribution. For example, suppose you put a biased coin and an unbiased coin in a bag, then pull one out and flip it:  bag :: ProbDist Double (ProbDist Double String) bag = ProbDist [ (biasedCoin, 0.5), (unbiasedCoin, 0.5) ]  The join operator collapses this into a single ProbDist Double String:  ProbDist [("heads",0.3), ("tails",0.2), ("heads",0.25), ("tails",0.25)]  It would be nice if join could combine the duplicate heads into a single ("heads", 0.55) entry. But that would force an Eq a constraint on the event type, which isn't allowed, because (>>=) must work for all data types, not just for instances of Eq. This is a problem with Haskell, not with the monad itself. It's the same problem that prevents one from making a good set monad in Haskell, even though categorially sets are a perfectly good monad. (The return function constructs singletons, and the join function is simply set union.) Maybe in the next language. Perhaps someone else will find the >>= operator easier to understand than join? I don't know. Anyway, it's simple enough to derive once you understand join; here's the code:  instance (Num p) => Monad (ProbDist p) where return a = ProbDist [(a, 1)] (ProbDist pas) >>= f = ProbDist$ do
(a, p) <- pas
let (ProbDist pbs) = f a
(b, q) <- pbs
return (b, p*q)

So now we can do some straightforward experiments:

	liftM2 (+) (d 6) (d 6)

ProbDist [(2,1 % 36),(3,1 % 36),(4,1 % 36),(5,1 % 36),(6,1 %
36),(7,1 % 36),(3,1 % 36),(4,1 % 36),(5,1 % 36),(6,1 %
36),(7,1 % 36),(8,1 % 36),(4,1 % 36),(5,1 % 36),(6,1 %
36),(7,1 % 36),(8,1 % 36),(9,1 % 36),(5,1 % 36),(6,1 %
36),(7,1 % 36),(8,1 % 36),(9,1 % 36),(10,1 % 36),(6,1 %
36),(7,1 % 36),(8,1 % 36),(9,1 % 36),(10,1 % 36),(11,1 %
36),(7,1 % 36),(8,1 % 36),(9,1 % 36),(10,1 % 36),(11,1 %
36),(12,1 % 36)]

This is nasty-looking; we really need to merge the multiple listings of the same event. Here is a function to do that:

        agglomerate :: (Num p, Eq b) => (a -> b) -> ProbDist p a -> ProbDist p b
agglomerate f pd = ProbDist $foldr insert [] (unpd (fmap f pd)) where insert (k, p) [] = [(k, p)] insert (k, p) ((k', p'):kps) | k == k' = (k, p+p'):kps | otherwise = (k', p'):(insert (k,p) kps) agg :: (Num p, Eq a) => ProbDist p a -> ProbDist p a agg = agglomerate id  Then agg$ liftM2 (+) (d 6) (d 6) produces:

        ProbDist [(12,1 % 36),(11,1 % 18),(10,1 % 12),(9,1 % 9),
(8,5 % 36),(7,1 % 6),(6,5 % 36),(5,1 % 9),
(4,1 % 12),(3,1 % 18),(2,1 % 36)]

Hey, that's correct.

There must be a shorter way to write insert. It really bothers me, because it looks look it should be possible to do it as a fold. But I couldn't make it look any better.

You are not limited to calculating probabilities. The monad actually will count things. For example, let us throw three dice and count how many ways there are to throw various numbers of sixes:

        eq6 n = if n == 6 then 1 else 0
agg $liftM3 (\a b c -> eq6 a + eq6 b + eq6 c) die die die ProbDist [(3,1),(2,15),(1,75),(0,125)]  There is one way to throw three sixes, 15 ways to throw two sixes, 75 ways to throw one six, and 125 ways to throw no sixes. So ProbDist is a misnomer. It's easy to convert counts to probabilities:  probMap :: (p -> q) -> ProbDist p a -> ProbDist q a probMap f (ProbDist pds) = ProbDist$ (map (\(a,p) -> (a, f p))) pds

normalize :: (Fractional p) => ProbDist p a -> ProbDist p a
normalize pd@(ProbDist pas) = probMap (/ total) pd where
total = sum . (map snd) $pas normalize$ agg $probMap toRational$
liftM3 (\a b c -> eq6 a + eq6 b + eq6 c) die die die

ProbDist [(3,1 % 216),(2,5 % 72),(1,25 % 72),(0,125 % 216)]

I think this is the first time I've gotten to write die die die in a computer program.

The do notation is very nice. Here we calculate the distribution where we roll four dice and discard the smallest:

        stat = do
a <- d 6
b <- d 6
c <- d 6
d <- d 6
return (a+b+c+d - minimum [a,b,c,d])

probMap fromRational \$ agg stat

ProbDist [(18,1.6203703703703703e-2),
(17,4.1666666666666664e-2), (16,7.253086419753087e-2),
(15,0.10108024691358025),   (14,0.12345679012345678),
(13,0.13271604938271606),   (12,0.12885802469135801),
(11,0.11419753086419752),   (10,9.41358024691358e-2),
(9,7.021604938271606e-2),   (8,4.7839506172839504e-2),
(7,2.9320987654320986e-2),  (6,1.6203703703703703e-2),
(5,7.716049382716049e-3),   (4,3.0864197530864196e-3),
(3,7.716049382716049e-4)]


One thing I was hoping to get didn't work out. I had this idea that I'd be able to calculate the outcome of a game of craps like this:

	dice = liftM2 (+) (d 6) (d 6)

point n = do
roll <- dice
case roll of 7 -> return "lose"
_ | roll == n  = "win"
_ | otherwise  = point n

craps = do
roll <- dice
case roll of 2 -> return "lose"
3 -> return "lose"
4 -> point 4
5 -> point 5
6 -> point 6
7 -> return "win"
8 -> point 8
9 -> point 9
10 -> point 10
11 -> return "win"
12 -> return "lose"

This doesn't work at all; point is an infinite loop because the first value of dice, namely 2, causes a recursive call. I might be able to do something about this, but I'll have to think about it more.

It also occurred to me that the use of * in the definition of >>= / join could be generalized. A couple of years back I mentioned a paper of Green, Karvounarakis, and Tannen that discusses "provenance semirings". The idea is that each item in a database is annotated with some "provenance" information about why it is there, and you want to calculate the provenance for items in tables that are computed from table joins. My earlier explanation is here.

One special case of provenance information is that the provenances are probabilities that the database information is correct, and then the probabilities are calculated correctly for the joins, by multiplication and addition of probabilities. But in the general case the provenances are opaque symbols, and the multiplication and addition construct regular expressions over these symbols. One could generalize ProbDist similarly, and the ProbDist monad (even more of a misnomer this time) would calculate the provenance automatically. It occurs to me now that there's probably a natural way to view a database table join as a sort of Kleisli composition, but this article has gone on too long already.

Happy new year, everyone.

[ Addendum 20100103: unsurprisingly, this is not a new idea. Several readers wrote in with references to previous discussion of this monad, and related monads. It turns out that the idea goes back at least to 1981. ]

My thanks to Graham Hunter for his donation.

Tue, 16 Jun 2009

The Haskell folks have chosen a new logo.

Ouch.

Thu, 03 Jan 2008

Note on point-free programming style
This old comp.lang.functional article by Albert Y. C. Lai, makes the point that Unix shell pipeline programming is done in an essentially "point-free" style, using the shell example:

    grep '^X-Spam-Level' | sort | uniq | wc -l


    length . nub . sort . filter (isPrefixOf "X-Spam-Level")

Neither one explicitly mentions its argument, which is why this is "point-free". In "point-free" programming, instead of defining a function in terms of its effect on its arguments, one defines it by composing the component functions themselves, directly, with higher-order operators. For example, instead of:

  foo x y = 2 * x + y

one has, in point-free style:

  foo = (+) . (2 *)

where (2 *) is the function that doubles its argument, and (+) is the (curried) addition function. The two definitions of foo are entirely equivalent.

As the two examples should make clear, point-free style is sometimes natural, and sometimes not, and the example chosen by M. Lai was carefully selected to bias the argument in favor of point-free style.

Often, after writing a function in pointful style, I get the computer to convert it automatically to point-free style, just to see what it looks like. This is usually educational, and sometimes I use the computed point-free definition instead. As I get better at understanding point-free programming style in Haskell, I am more and more likely to write certain functions point-free in the first place. For example, I recently wrote:

        soln = int 1 (srt (add one (neg (sqr soln))))

and then scratched my head, erased it, and replaced it with the equivalent:

        soln = int 1 ((srt . (add one) . neg . sqr) soln)

I could have factored out the int 1 too:
        soln = (int 1 . srt . add one . neg . sqr) soln

I could even have removed soln from the right-hand side:

        soln = fix (int 1 . srt . add one . neg . sqr)

but I am not yet a perfect sage.

Sometimes I opt for an intermediate form, one in which some of the arguments are explicit and some are implicit. For example, as an exercise I wrote a function numOccurrences which takes a value and a list and counts the number of times the value occurs in the list. A straightforward and conventional implementation is:

        numOccurrences x []     = 0
numOccurrences x (y:ys) =
if (x == y) then 1 + rest
else                 rest
where rest = numOccurrences x ys

but the partially point-free version I wrote was much better:

        numOccurrences x = length . filter (== x)

Once you see this, it's easy to go back to a fully pointful version:

        numOccurrences x y = length (filter (== x) y)

Or you can go the other way, to a point-free version:

        numOccurrences = (length .) . filter . (==)

which I find confusing.

Anyway, the point of this note is not to argue that the point-free style is better or worse than the pointful style. Sometimes I use the one, and sometimes the other. I just want to point out that the argument made by M. Lai is deceptive, because of the choice of examples. As an equally biased counterexample, consider:

        bar x = x*x + 2*x + 1

which the automatic converter informs me can be written in point-free style as:

        bar = (1 +) . ap ((+) . join (*)) (2 *)

Perusal of this example will reveal much to the attentive reader, including the definitions of join and ap. But I don't think many people would argue that it is an improvement on the original. (Maybe I'm wrong, and people would argue that it was an improvement. I won't know for sure until I have more experience.)

For some sort of balance, here is another example where I think the point-free version is at least as good as the pointful version: a recent comment on Reddit suggested a >>> operator that composes functions just like the . operator, but in the other order, so that:

        f >>> g = g . f

or, if you prefer:

        >>> f g x = g(f(x))

The point-free definition of >>> is:

        (>>>) = flip (.)

where the flip operator takes a function of two arguments and makes a new function that does the same thing, but with the arguments in the opposite order. Whatever your feelings about point-free style, it is undeniable that the point-free definition makes perfectly clear that >>> is nothing but . with its arguments in reverse order.