2014-03-21 - [ haskell, lens ]

# Intro

In Semantic editor combinators, Conal has shown us a simple but powerful trick of walking into a portion of data and then modifying it.

When I saw this simple idea, I persuaded myself to take it for grainted and tried it out. Even though it works well in practice, I'd like to explain to myself what's happening, which is the motivation of this article.

I'll implement first, second, result and other combinators step by step, and come up with the way to walking into user defined data, which is easy but not presented in Conal's post.

# Editing a pair

First we need to think about what the type would be for first and second:

• When we reach for the portion of data, we need to modify it, so we need a function f :: a -> b

• We also need a pair of data (x,y) :: (c,d) for this editor's input.

• For first, we want type c = a, and the return value should be of type (b,d)

• For second, we want type d = a, and the return value should be of type (c,b)

Put them together, we end up with type signature for first and second:

first  :: (a -> b) -> (a,d) -> (b,d)
second :: (a -> b) -> (c,a) -> (c,b)

Introduce f and (x,y) for two arguments, and the implementation is straightforward:

first  :: (a -> b) -> (a,d) -> (b,d)
first  f (x,y) = (f x, y)

second :: (a -> b) -> (c,a) -> (c,b)
second f (x,y) = (x, f y)

Let's bring up GHCi for some tests:

λ> first (const 1) ("foo","bar") -- (1, "bar")
(1,"bar")
λ> second (* 20) (Just 1, 1) -- (Just 1, 20)
(Just 1,20)

# Editing a function

There's two way of editing a function: we can either modify the return value (the result combinator), or modify the argument (the argument combinator).

## the result combinator

Let's think about the type of result:

• How to modify a function is given by f :: a -> b
• The function we want to edit is g :: c -> d
• We want to modify on the return value of g, therefore d = a.
• Then the resulting function should be of type c -> b
result :: (a -> b) -> (c -> a) -> (c -> b)

We know that -> is right-associative, (e.g. p -> q -> r and p -> (q -> r) are equivalent) so we can also think result as a function that accepts 3 arguments, and its type signature becomes:

result :: (a -> b) -> (c -> a) -> c -> b

Introduce f g x for these 3 arguments, and we want a value of type b.

result :: (a -> b) -> (c -> a) -> c -> b
result f g x = undefined
• To produce a value of b, we might need f :: a -> b.
• To produce a value of a, we might need g :: c -> a.
• x is of type c

Chaining these facts together, we yield the implementation of result:

result :: (a -> b) -> (c -> a) -> c -> b
result f g x = f (g x)

This is exactly the definition of (.), function composition.

So we end up with:

result :: (a -> b) -> (c -> a) -> c -> b
result = (.)

Test it:

λ> result (+2) (*5) 1 -- 1 * 5 + 2 = 7
7

## the argument combinator

Use the same approach, let’s implement argument combinator:

• f :: a -> b to modify the argument
• g :: c -> d the function to be modified
• b = c so that the type of f and g matches.
• the resulting function should be of type a -> d.

Write down the type signature, introduce f, g, x, the goal is to produce a value of type d

argument :: (a -> b) -> (b -> d) -> a -> d
argument f g x = undefined
• To produce a value of d, we might need g :: b -> d.
• To produce a value of b, we might need f :: a -> b.
• x is of type a

Therefore the definition of argument is:

argument :: (a -> b) -> (b -> d) -> a -> d
argument f g x = g (f x)

To simplify this, we do some transformations:

g (f x)
> (g . f) x           -- definition of (.)
> (.) g f x           -- infix operator
> ((.) g f) x         -- property of curried function
> ((flip (.)) f g) x  -- flip
> flip (.) f g x      -- rule of function application

Finally:

argument :: (a -> b) -> (b -> d) -> a -> d
-- point-free style of: argument f g x = flip (.) f g x
argument = flip (.)

Test it:

λ> argument show (\x -> x ++ "!") 2048 -- 2048!
"2048!"

# Walking into data

For now we have first and second to work with pairs, and argument and result with functions, but sometimes we want to go deeper inside data, say:

• Goal 1: we want 2 to be 5 in (7,(2,1))
• Goal 2: we want a+b to be (a+b)*2 in \a -> (a+1 ,\b -> (a+b, 1))

Here I try to give some explanation of how to do this.

## Goal 1

To modify d in (a,(c,d)), that means we need to first focus on (c,d), this can be achieved by second, let’s pass a const function const "Focus" to second, so we know we are focusing on which portion of the data:

λ> let d1 = (7,(2,1))
λ> let toFocus = const "Focus"
λ> second toFocus d1
(7,"Focus")

We now know that missing (2,1) is passed to toFocus, we can use a lambda expression to capture it:

λ> second (\x -> x) d1
(7,(2,1))

Let's do something more interesting:

λ> second (\x -> let (a,b) = x in (a+1,b,b)) d1
(7,(3,1,1))

The observation is: inside the body of lambda expression \x -> ..., we can do something to (2,1). That means we can apply semantic editor combinators on it!

Now assume we are dealing with (2,1), to reach the goal, we apply const 5 to its fst part:

λ> first (const 5) (2,1)
(5,1)

Now replace the body of \x -> ... with what we want:

λ> second (\x -> first (const 5) x) d1
(7,(5,1))

Done!

## Some intuition

Let's not hurry to Goal 2, I’ll show you some transformations:

second (\x -> first (const 5) x) d1
> second (first (const 5)) d1         -- eta reduction / point-free
> (second (first (const 5))) d1       -- rule of function application
> ((second . first) (const 5)) d1     -- definition of (.)

I don’t simplify it to (second . first) (const 5) d1 on purpose, and I’m going to show you why.

It might not be straightforward at first glance, since function composition is usually “right to left”, that is, (g . f) means “first apply f, on its result, apply g. But here, second . first can be read as”walk to the snd part, and then fst part of it".

My explanation is: the function does compose from right to left, but what’s “carrying” with the composition is not the data.

The observation is: no matter how many functions are composed together, the resulting function just takes one argument:

λ> let u = undefined
λ> :t u . u
u . u :: a -> c
λ> :t u . u . u . u
u . u . u . u :: a -> c
λ> :t u . u . u . u . u
u . u . u . u . u :: a -> c

So what is the argument taken by this resulting function? In our Goal 1, it is const 5, the “how we modify it” part. So despite that (second . first) (const 5) d1 is simple, to understand it, what you really need would be: ((second . first) (const 5)) d1

The observation is: functions are composed from right to left and the mutator is buried more deeply as the composition goes.

To see this, let's just pick up combinators randomly to form a chain of function composition, and check its type:

λ> :t first (const 5)
first (const 5) :: Num b => (a, d) -> (b, d)
λ> :t (second . first) (const 5)
(second . first) (const 5) :: Num b => (c, (a, d)) -> (c, (b, d))
λ> :t (result . second . first) (const 5)
(result . second . first) (const 5)
:: Num b => (c -> (c1, (a, d))) -> c -> (c1, (b, d))
λ> :t (second . result . second . first) (const 5)
(second . result . second . first) (const 5)
:: Num b => (c, c1 -> (c2, (a, d))) -> (c, c1 -> (c2, (b, d)))
λ> :t (first . second . result . second . first) (const 5)
(first . second . result . second . first) (const 5)
:: Num b =>
((c, c1 -> (c2, (a, d1))), d) -> ((c, c1 -> (c2, (b, d1))), d)

Reading the composition from right to left, we are constructing / definiting the “shape” of the data this combinator would like to work on. In contrast, reading from left to right, we are deconstructing the data until we reach for the position that we want to modify. These two views does not conflict with each other.

Therefore, IMHO, when we are thinking about how to walking into the interesting portion, it's helpful thinking about build up combinator from left to right. But to answer the question about why, we should instead read from right to left.

In addition, here is another interesting observation:

(<comb1> . <comb2>) (<comb3> . <mutator>) d1
<=> (<comb1> . <comb2> . <comb3>) <mutator> d1

The shape of semantic editor combinators are always: {comb} {mut} {dat}

So if we see a combinator appear as the head of the (.) chain in {mut}, we can move it to the tail of the (.) chain in {comb}, and the other way around also holds true.

## Goal 2

Recall Goal 2: we want a+b to be (a+b)*2 in \a -> (a+1, \b -> (a+b, 1))

Based on the previous understand, we need to:

• walk into (a+1, \b -> (a+b, 1)), using result
• walk into \b -> (a+b, 1), using second
• walk into (a+b, 1), using result
• walk into (a+b), using first
• apply (* 2)

Try it out:

λ> let d2 = \a -> (a+1, \b -> (a+b, 1))
λ> let m2 = (result . second . result . first) (*2) d2
λ> (snd (d2 10)) 20
(30,1)
λ> (snd (m2 10)) 20
(60,1)
λ> fst ((snd (d2 10)) 20)
30
λ> fst ((snd (m2 10)) 20)
60
λ> (fst . (\$ 20) . snd . (\$ 10)) d2
30
λ> (fst . (\$ 20) . snd . (\$ 10)) m2
60

Last two programs are just rewriting the two programs right before it, Do you notice the symmetry between fst . (\$ 20) . snd . (\$ 10) and result . second . result . first?

I believe this is not an accident, but to explain it might beyond my reach, this is what I have for now:

With the help of Lambdabot:

(fst . (\$ 20) . snd . (\$ 10)) (result . second . result . first) (*2) d2
>  fst (snd (result (second (result (first 10)))) 20) (2 *) d2

I guess when fst and first meet together, or when (\$ 10) and result meet together, etc. they will be somehow “cancelled”. If someone knows how to finish this explanation, please let me know.

# Play with lists

Conal has shown us element, which walks into every element of a list.

It’s easy to work out its type signature:

• how to modify: f :: a -> b
• data to be modified: [a]
• resulting data: [b]

and this is exactly map.

element :: (a -> b) -> [a] -> [b]
element = map

Here I show you 3 extra combinators:

• how to modify: f :: a -> a (list can only contain elements of same types)
• data to be modified: [a]
• resulting data: [a]
inHead :: (a -> a) -> [a] -> [a]
inHead f (x:xs) = f x : xs

inTail modifies the tail of a list:

• how to modify: f :: [a] -> [a]
• data to be modified: [a]
• resulting data: [a]
inTail :: ([a] -> [a]) -> [a] -> [a]
inTail f [] = []
inTail f (x:xs) = x : f xs

inPos modifies the element at a given position:

• we first need the position: Int
• how to modify: f :: a -> a
• data to be modified: [a]
• resulting data: [a]
inPos :: Int -> (a -> a) -> [a] -> [a]
inPos n f xs
| null xs    = xs
| n < 0      = xs
| n == 0     = let (y:ys) = xs
in f y : ys
| otherwise  = let (y:ys) = xs
in y : inPos (n - 1) f ys

inSome is like inPos, but modifies multiple positions:

• we first need the indices: [Int]
• how to modify: f :: a -> a
• data to be modified: [a]
• resulting data: [a]
inSome :: [Int] -> (a -> a) -> [a] -> [a]
inSome ixs f xs = foldl (\acc i -> inPos i f acc) xs ixs

Some examples:

λ> let d1 = ("foo",[1..10])
λ> (second . element) (+ 1) d1
("foo",[2,3,4,5,6,7,8,9,10,11])
λ> (first . element) ord d1
([102,111,111],[1,2,3,4,5,6,7,8,9,10])
λ> (second . inTail) (take 2) d1
("foo",[1,2,3])
λ> (second . inHead) (* 200) d1
("foo",[200,2,3,4,5,6,7,8,9,10])
λ> (second . inPos 3) (* 200) d1
("foo",[1,2,3,800,5,6,7,8,9,10])
λ> let d2 = replicate 3 (replicate 4 0)
λ> (inPos 1 . inPos 2) (const 255) d2
[[0,0,0,0],[0,0,255,0],[0,0,0,0]]

# Walking into user defined data

Suppose user has defined a binary tree:

data BinTree a = Node a
| Tree (BinTree a) (BinTree a) a
deriving (Show, Eq, Functor)

To write a combinator is easy, just follow the pattern <combinator> <mutator> <data> = <modified-data>:

treeV :: (a -> a) -> BinTree a -> BinTree a
treeV f (Node x) = Node (f x)
treeV f (Tree l r x) = Tree l r (f x)

treeElement :: (a -> b) -> BinTree a -> BinTree b
treeElement = fmap

treeLeft :: (BinTree a -> BinTree a) -> BinTree a -> BinTree a
treeLeft f (Tree l r e) = Tree (f l) r e
treeLeft f r = r

treeRight :: (BinTree a -> BinTree a) -> BinTree a -> BinTree a
treeRight f (Tree l r e) = Tree l (f r) e
treeRight f r = r

treeNonLeaf :: (a -> a) -> BinTree a -> BinTree a
treeNonLeaf f (Tree l r e) = Tree (treeNonLeaf f l) (treeNonLeaf f r) (f e)
treeNonLeaf f r = r

treeLeaf :: (a -> a) -> BinTree a -> BinTree a
treeLeaf f (Node x) = Node (f x)
treeLeaf f (Tree l r x) = Tree (treeLeaf f l) (treeLeaf f r) x

Here treeElement walks into the data part of each node to do the modification. Since treeElement = fmap, we can generalize element = fmap, to make it work for not only lists and BinTrees, but also any other Functors.

treeLeft and treeRight walks into left subtree and right subtree, respectively. But to target at the value of a BinTree, we need treeV, which takes care of walking into the data field of a tree.

treeNonLeaf and treeLeaf are just like treeElement but works merely on non-leaf nodes and leaf nodes, respectively.

See them in action:

λ> let d1 = Tree (Tree (Node (1,1)) (Node (1,2)) (2,2)) (Node (3,4)) (5,6)
λ> (treeElement . first) even d1 -- test for each node, if the `fst` part is an even?
Tree (Tree (Node (False,1)) (Node (False,2)) (True,2)) (Node (False,4)) (False,6)
λ> (treeLeft . treeV . first) (* 4) d1 -- walk into left subtree, multiple `fst` part of its value by 4
Tree (Tree (Node (1,1)) (Node (1,2)) (2,2)) (Node (3,4)) (5,6)
λ> (treeLeft . treeRight . treeV . first) (* 4) d1 -- walk into the right subtree of the left subtree, ..
Tree (Tree (Node (1,1)) (Node (4,2)) (2,2)) (Node (3,4)) (5,6)
λ> (treeNonLeaf . first) (const 100) d1 -- for each value of the `fst` part of each non-leaf node
Tree (Tree (Node (1,1)) (Node (1,2)) (100,2)) (Node (3,4)) (100,6)
λ> (treeLeaf . second) (const 100) d1 -- for each value of the `snd` part of each leaf node
Tree (Tree (Node (1,100)) (Node (1,100)) (2,2)) (Node (3,100)) (5,6)

# Combining semantic editor combinators

The semantic editor combinators can also be chained together.

Suppose we have some combinators:

λ> let c13 = (inPos 1) (const 3)
λ> let c05 = (inPos 0) (const 5)
λ> let c06 = (inPos 0) (const 6)
λ> let cf1 = first (const 1)

For a given list, we want to apply c13 and c05:

λ> let d = replicate 5 0
λ> c06 (c13 d)
[6,3,0,0,0]
λ> (c06 . c13) d
[6,3,0,0,0]

So we can use function composition to compose semantic editors:

λ> let d2 = (10,replicate 5 0)
λ> (cf1 . (second c13)) d2
(1,[0,3,0,0,0])
λ> (cf1 . (second c13) . (second c05)) d2
(1,[5,3,0,0,0])
λ> (cf1 . (second (c13 . c05))) d2
(1,[5,3,0,0,0])

The last example says “apply 1, (back to the top level and then) walk into the snd part, apply c05 and then c13”. I think this “distributive property” will make things extremely flexible.

Since the function composition happens from right to left, if some combinators are trying to modify the same field, the “leftmost” one take effect:

λ> (c05 . c06) [0,1]
[5,1]
λ> (c06 . c05) [0,1]
[6,1]