Back to basics

How many values can we construct from the following type?

data Bool = False | True

Note: in this discussion, we're explicitly omitting well-typed but non-terminating constructs such as the following:

loop :: Bool
loop = loop

wtf :: Bool
wtf = undefined

crash :: Bool
crash = error "fnord"

Ordering

Another well-known type:

data Ordering = LT | EQ | GT

Clearly we can construct three different values of this type.

A zero-valued type

In Haskell 2010, we can create types from which no values can be constructed:

data Empty

This type has no value constructors (and we can't use deriving syntax on it).

Zero, one, two...

So big deal, we can create types with zero or more constructors:

data Empty

data One = One

data Bool = False | True

Adding some parameters

Another type to ponder.

data A = A Bool
       | B Ordering

We can construct five values of this type:

A False
A True
B LT
B EQ
B GT

A different tack

How many values can this type represent?

data Fnord = Fnord Bool Ordering

What about this one?

data Quaternion = Point Double Double Double Double

Switching the notation: sums

Let's take a different perspective for a moment, and do some arithmetic.

data Sum = A Bool
         | B Ordering

If we exhaustively enumerate the possible values of this type, we see that there are as many values as:

Values of Bool...
...added to...
Values of Ordering

Let's write that number as Bool + Ordering

Switching the notation: products

From reading this type:

data Product = Product Bool Ordering

Following the previous example, it's pretty clear that we can create as many values of type Ordering as there are:

Values of Bool...
...multiplied by...
Values of TrafficLight

Let's write that number as Bool × Ordering

From arithmetic to algebra: sums

Now let's introduce polymorphism into the mix.

data Either a b = Left a | Right b

We don't know how many values there are of this type, since neither a nor b is specified.

But we can still write an algebraic expression that will compute the right number, once we plug concrete types in:

a + b

From arithmetic to algebra: products

This should be a no-brainer:

data Triple a b c = Triple a b c

The algebraic expression that describes the number of values of this type is no surprise:

a × b × c

Mixing sums and products

How many values are there of this type?

type Foo a b c = Foo a b
               | Bar b c

Clarity on naming: sums

Consider a type that consists only of zero-parameter constructors:

data Rainbow = Red | Orange | Yellow | Green {- etc -}

These are often referred to as sum types.

Clarity on naming: products

If a type has only one constructor, and that constructor takes parameters:

data Point = Point Int Int

We refer to this as a product type.

Algebraic data types

Haskell's type system admits sum types, product types, and types that are a mixture of both.

This should make it clear why the word "algebraic" appears in the phrase.

The frob merchant

This is a frob.

                 ___             ____
                /__/\     ______/___/\
                \  \ \   /          /\\
                 \  \ \_/__        /  \
                 _\  \ \  /\______/__  \
                // \__\/ /  \       /\  \
        _______//_______/    \     / _\_/_____
       /      / \       \    /    / /        /\
    __/      /   \       \  /    / /        / _\__
   / /      /     \_______\/    / /        / /   /\
  /_/______/___________________/ /________/ /___/  \
  \ \      \    ___________    \ \        \ \   \  /
   \_\      \  /          /\    \ \        \ \___\/
      \      \/          /  \    \ \        \  /
       \_____/          /    \    \ \________\/
            /__________/      \    \  /
            \   _____  \      /_____\//
             \ /    /\  \    /    \  /
              /____/  \  \  /______\/\
              \    \  /___\/     \  \ \
               \____\/            \__\/

The frob merchant's web store

Suppose we're building a web app, where we want to send frobs to customers of our web site.

data Customer = Customer {
      custID :: Int
    , custName :: String
    , custAddress :: Address
    }

newtype Zip = Zip Int

data Address = Address {
      addrStreet :: String
    , addrCity :: String
    , addrState :: String
    , addrZip :: Zip
    }

Oh noes!

A customer has made a mistake in entering their shipping zip code. They've called us up, irate that we've been unable to fulfil their urgent frob order.

So. We need to change their zip code.

In a C-like language, this would be easy:

struct Customer *cust;

/* ... */

cust->custAddress->addrZip = 94043;

Getting at a zip code

Haskell's record syntax automatically defines "accessor" or "getter" functions for us:

custAddress :: Customer -> Address
addrZip :: Address -> Zip

Given a Customer, we can obviously use function composition to get their zip:

custZip :: Customer -> Zip
custZip = addrZip . custAddress

Unfortunately, we lack a "good" facility for updating records. Let's see what that means.

Setting a zip code

We need to modify a zip code, but we're working in a pure language, so clearly a "zip code setter" is going to be a function that returns a new value that is identical to the previous value except for the zip.

setAddrZip :: Zip -> Address -> Address

If we have a new Address and we want to "modify" a Customer, we need a similar function:

setCustAddress :: Address -> Customer -> Customer

Ultimately, our goal is actually to write this function:

setCustZip :: Zip -> Customer -> Customer

Record update syntax

Along with record syntax, Haskell provides an "update" syntax:

setAddrZip :: Zip -> Address -> Address
setAddrZip zip addr = addr { addrZip = zip }

The expression on the right means this:

Make a copy of addr
All fields in the new value should be the same as in addr...
...except for addrZip, which should have the value zip

Does this solve our problem?

Here's the other "setter" function we need, which follows the same pattern:

setCustAddress :: Address -> Customer -> Customer
setCustAddress addr cust = cust { custAddress = addr }

Now we can write that setCustZip function we wanted:

setCustZip :: Zip -> Customer -> Customer
setCustZip zip cust =
    setCustAddress (setAddrZip zip (custAddress cust)) cust

Trouble is, the above looks much uglier to me than the corresponding C:

cust->custAddress->addrZip = 94043;

Worse, we have to write each of our Haskell "setter" functions by hand. Ugh.

Is the situation hopeless?

Here are our desiderata:

We want to be able to access fields within records.
We want to be able to compose accesses, so that we can inspect fields within records that are themselves fields of records.
We want to be able to update fields within records.
We want to be able to compose updates, so that we can modify fields within records that are themselves fields of records.

With Haskell's record syntax, we get #1 and #2, sort of #3 (if we squint), and definitely not #4.

Lenses

What we want is a type that behaves something like this:

data Lens rec fld = Lens {
      get :: rec -> fld
    , set :: fld -> rec -> rec
    }

This "bundles together" a record type rec with a field type fld, so that we know:

how to get a field out of a record, and
how to update a field within a record.

(Why the name "lens"? Because it lets us focus on a field within a record.)

How should lenses behave?

We need three laws to hold for lenses.

If we put something into a record, we can get it back out.

get l (put l b a) == b

If we get something out of a record, and put it back in, the result is identical to the original record.

put l (get l a) a == a

Two successive put operations must give the same result as a single put of the second value:

put l b1 (put l b2 a) == put l b1 a

(We call these properties "laws" because they must hold in order for us to be able to reason about lenses.)

What does a real lens look like?

The following definitions correspond to those in the data-lens package.

newtype Lens rec fld = Lens (rec -> Store fld rec)

where

data Store fld rec = Store (fld -> rec) fld

That's hard to follow, so let's dig in and try to understand. First, we'll get rid of the name Store, to give the tuple:

(fld -> rec, fld)

Then we'll substitute this into the definition of Lens:

newtype Lens rec fld = Lens (rec -> (fld -> rec, fld))

Simplifying further

If we ignore all the newtype noise, we're left with a very simple type:

rec -> (fld -> rec, fld)

That is, a Lens is:

A function that accepts a record type rec as its argument
It returns a pair
The first element is a setter: give it a field value of type fld, and it will return a new record
The second element is the current value of the field

Why the coupling?

Why does a lens give us both the value of a field and a function for setting a new value of that field?

Suppose that computing the path to the right place in the record for the getter is expensive.
This representation allows the setter to reuse that computation.

We can also reduce the number of laws that a lens must obey from 3 to 2 (but that's beyond our scope).

The get operator

Here is our getter:

(^.) :: rec -> Lens rec fld -> fld
a ^. (Lens f) = pos (f a)
infixr 9 ^.

-- internal
pos :: Store fld rec -> fld
pos (Store _ s) = s

The set operator

And here is our setter:

(^=) :: Lens rec fld -> fld -> rec -> rec
(Lens f) ^= b = peek b . f
infixr 4 ^=

-- internal
peek :: fld -> Store fld rec -> rec
peek s (Store g _) = g s

Constructing a lens

Given a getter and a setter, we can build a lens:

lens :: (rec -> fld) -> (fld -> rec -> rec) -> Lens rec fld
lens get set = Lens $ \a -> Store (\b -> set b a) (get a)

Alternatively, we can construct a lens from an isomorphism between record and field types:

iso :: (rec -> fld) -> (fld -> rec) -> Lens rec fld
iso f g = Lens (Store g . f)

A lens for points

Consider our venerable Point type:

data Point = Point {
      ptX :: Int
    , ptY :: Int
    } deriving (Show)

We need to define two lenses for this type, one to focus on the x coordinate, and another for y:

x, y :: Lens Point Int
x = lens ptX (\x pt -> pt {ptX = x})
y = lens ptY (\y pt -> pt {ptY = y})

Using our lens on points

The getter:

>> let pt = Point 1 1
>> pt ^. x
1

The setter:

>> (x ^= 2) pt
Point {ptX = 2, ptY = 1}

Revisiting nested data

Let's define a line type, with lenses for its beginning and end points:

data Line = Line {
      lnBeg :: Point
    , lnEnd :: Point
    } deriving (Show)

beg, end :: Lens Line Point
beg = lens lnBeg (\b l -> l {lnBeg = b})
end = lens lnEnd (\e l -> l {lnEnd = e})

Suppose we want to access the x coordinate of the end of the line.

Using normal Haskell machinery, we know we can just use composition:

>> let l = Line (Point 1 2) (Point 3 4)
>> (ptX . lnEnd) l
3

Function composition: not gnar enough

By now, we are familiar with (and love) function composition:

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

However, we can make composition more abstract:

import Prelude hiding (id, (.))

class Category cat where
  id :: cat a a
  (.) :: cat b c -> cat a b -> cat a c

Now we can recast function composition as just an instance of this more general Category class:

instance Category (->) where
    id a = a
    f . g = \x -> f (g x)

Category? Composition? Abstraction? Huh?

We care about the Category class because it turns out we can compose lenses!

import Control.Category

instance Category Lens where
    id = Lens (Store id)

    Lens f . Lens g = Lens $ \a -> case g a of
      Store wba b -> case f b of
    Store wcb c -> Store (wba . wcb) c

How do we do this in practice?

Just as we compose two functions to get another function, when we compose two lenses, we get another lens.

Composition of lenses

Access a nested field:

>> let l = Line (Point 1 2) (Point 3 4)
>> l ^. (x . beg)

Modify a nested field:

>> ((y . end) ^= 7) l
Line {lnBeg = Point {ptX = 1, ptY = 2},
      lnEnd = Point {ptX = 3, ptY = 7}}

A map as a lens

Lenses are not restricted to use solely with algebraic data types.

They're just as applicable to container types, for instance:

import qualified Data.Map as Map
import Data.Map (Map)

mapLens :: (Ord k) => k -> Lens (Map k v) (Maybe v)
mapLens k = Lens $ \m ->
            let set Nothing  = Map.delete k m
                set (Just v) = Map.insert k v m
                get          = Map.lookup k m
            in Store set get

It's all about focus

We now know how the "algebraic" got into "algebraic data type", and why (and how) we'd want to focus on an element within a type.

What's next?

Lenses and triples

Suppose we have this type.

(Int,Int,Int)

How many lenses must we define in order to be able to work with all of its fields?

Values in a triple

In our type:

(Int,Int,Int)

We can create the following number of values:

Int × Int × Int

Which of course we can shorten as Int³ (raised to the 3rd power).

Suppose we want to update the first field (using a lens, manual update, or whatever - the mechanism doesn't matter).

Let's poke a hole in that field:

(_,Int,Int)

Clearly we're now representing just Int × Int (or Int²) values, because we no longer care what value used to be present in the first field.

Poking more holes

There are in fact three different ways we could poke holes in our triple:

(_,Int,Int)
(Int,_,Int)
(Int,Int,_)

And each one can express Int² values, for a total of:

3 × Int²

Getting a little more abstract

Let's parameterise our triple, so we no longer know or care what the type in each field is:

(x,x,x)

This can hold x³ values.

When we poke holes in each field, we find that the total number of values expressible is:

3x²

This ought to remind you of differential calculus.

Isn't that remarkable?

Lists - again?

We're so very familiar with the list type by now.

data List x = Null
            | Cons x (List x)

From the symbolic shenanigans we saw earlier, let's compute n(x), the number of values expressible in a list of type x:

n(x) = 1 + x × n(x)

Just as the list is a recursive data type, this is a recurrence relation.

Let's perform a symbolic differentiation on this expression:

nʹ(x) = 0 + 1 × n(x) + x × nʹ(x)

(The last part is from the Leibniz rule.)

More algebraic crunching

We now have:

nʹ(x) = 0 + 1 × n(x) + x × nʹ(x)

Or more simply:

nʹ(x) = n(x) + x × nʹ(x)

Rearranging:

nʹ(x)(1 - x) = n(x)

And again:

nʹ(x) = n(x) / (1 - x)

And finally:

nʹ(x) = n(x)²

In other words, the derivative of a list is the product of two lists.

Wow! But what does this mean?

It's quite amazing that symbolic differentiation works on recursive data types. This discovery was made by McBride.

But what can we do with this knowledge?

Recall our earlier phrasing of "poking a hole" in a triple. Clearly there's a correspondence between "poking a hole" and modifying data.

We can use these ideas to both modify a list and move around in it.

Introducing the zipper

Here is the derivative of a list, expressed as a data type:

data Zipper a = Zipper [a] a [a]
              deriving (Show)

The unadorned a in the middle is our current focus point. (It's not required, just a detail of this particular implementation.)

In a regular list, we can only move in one direction: from the head to the tail.

From a list to a zipper

This function constructs a zipper from a list:

fromList :: [a] -> Zipper a
fromList (x:xs) = Zipper [] x xs
fromList _      = error "empty!"

List-like iteration

Here's iteration in the normal "towards the tail" direction:

next :: Zipper a -> Zipper a
next (Zipper ys y (x:xs)) = Zipper (y:ys) x xs
next z                    = z

Notice that we save "where we've been" in our other list. This is critically important.

Going backwards

Since we have saved where we've been in the list, we can step back there again!

prev :: Zipper a -> Zipper a
prev (Zipper (y:ys) x xs) = Zipper ys y (x:xs)
prev z                    = z

We can use the fact that we can pattern match against nearby elements on both sides of our current focus to perform useful operations that need local context, e.g. sliding window algorithms, convolutions, etc.

Zippers and such

The purpose of this lecture