Bottling Computation Patterns

Polymorphism and Equational Abstractions are the Secret Sauce

Refactor arbitrary repeated code patterns …

… into precisely specified and reusable functions

EXERCISE: Iteration

Write a function that squares a list of Int

squares :: [Int] -> [Int]
squares ns = ???

When you are done you should see

>>> squares [1,2,3,4,5]
[1,4,9,16,25]

Pattern: Iteration

Next, lets write a function that converts a String to uppercase.

>>> shout "hello"
"HELLO"

Recall that in Haskell, a String is just a [Char].

shout :: [Char] -> [Char]
shout = ???

Hoogle to see how to transform an individual Char

Iteration

Common strategy: iteratively transform each element of input list

Like humans and monkeys, shout and squares share 93% of their DNA

Super common computation pattern!

Abstract Iteration “Pattern” into Function

Remember D.R.Y. (Don’t repeat yourself)

Step 1 Rename all variables to remove accidental differences

-- rename 'squares' to 'foo'
foo []     = []
foo (x:xs) = (x * x)     : foo xs

-- rename 'shout' to 'foo'
foo []     = []
foo (x:xs) = (toUpper x) : foo xs

Step 2 Identify what is different

• In squares we transform x to x * x

• In shout we transform x to Data.Char.toUpper x

Step 3 Make differences a parameter

• Make transform a parameter f
foo f []     = []
foo f (x:xs) = (f x) : foo f xs

Done We have bottled the computation pattern as foo (aka map)

map f []     = []
map f (x:xs) = (f x) : map f xs

map bottles the common pattern of iteratively transforming a list:

QUIZ

What is the type of map ?

map :: ???
map f []     = []
map f (x:xs) = (f x) : map f xs

A. (Int -> Int) -> [Int] -> [Int]

B. (a -> a) -> [a] -> [a]

C. [a] -> [b]

D. (a -> b) -> [a] -> [b]

E. (a -> b) -> [a] -> [a]

The type precisely describes map

>>> :type map
map :: (a -> b) -> [a] -> [b]

That is, map takes two inputs

• a transformer of type a -> b
• a list of values [a]

and it returns as output

• a list of values [b]

that can only come by applying f to each element of the input list.

Reusing the Pattern

Lets reuse the pattern by instantiating the transformer

shout

-- OLD with recursion
shout :: [Char] -> [Char]
shout []     = []
shout (x:xs) = Char.toUpper x : shout xs

-- NEW with map
shout :: [Char] -> [Char]
shout xs = map (???) xs

squares

-- OLD with recursion
squares :: [Int] -> [Int]
squares []     = []
squares (x:xs) = (x * x) : squares xs

-- NEW with map
squares :: [Int] -> [Int]
squares xs = map (???) xs

EXERCISE

Suppose I have the following type

type Score = (Int, Int) -- pair of scores for Hw0, Hw1

Use map to write a function

total :: [Score] -> [Int]
total xs = map (???) xs

such that

>>> total [(10, 20), (15, 5), (21, 22), (14, 16)]
[30, 20, 43, 30]

The Case of the Missing Parameter

Note that we can write shout like this

shout :: [Char] -> [Char]
shout = map Char.toUpper

Huh. No parameters? Can someone explain?

The Case of the Missing Parameter

In Haskell, the following all mean the same thing

Suppose we define a function

add :: Int -> Int -> Int
add x y = x + y

Now the following all mean the same thing

plus x y = add x y

Why? equational reasoning! In general

foo x = e x

-- is equivalent to

foo   = e

as long as x doesn’t appear in e.

Thus, to save some typing, we omit the extra parameter.

Pattern: Reduction

sumList :: [Int] -> Int
sumList []     = 0
sumList (x:xs) = x + sumList xs

Next, a function that concatenates the Strings in a list

catList :: [String] -> String
catList []     = ""
catList (x:xs) = x ++ (catList xs)

Lets spot the pattern!

Step 1 Rename

foo []     = 0
foo (x:xs) = x + foo xs

foo []     = ""
foo (x:xs) = x ++ foo xs

Step 2 Identify what is different

1. ???

2. ???

Step 3 Make differences a parameter

foo p1 p2 []     = ???
foo p1 p2 (x:xs) = ???

EXERCISE: Reduction/Folding

This pattern is commonly called reducing or folding

foldr :: (a -> b -> b) -> b -> [a] -> b
foldr op base []     = base
foldr op base (x:xs) = op x (foldr op base xs)

Can you figure out how sumList and catList are just instances of foldr?

sumList :: [Int] -> Int
sumList xs = foldr (?op) (?base) xs

catList :: [String] -> String
catList xs = foldr (?op) (?base) xs

Executing foldr

To develop some intuition about foldr lets “run” it a few times by hand.

foldr op b (a1:a2:a3:a4:[])
==>
a1 `op` (foldr op b (a2:a3:a4:[]))
==>
a1 `op` (a2 `op` (foldr op b (a3:a4:[])))
==>
a1 `op` (a2 `op` (a3 `op` (foldr op b (a4:[]))))
==>
a1 `op` (a2 `op` (a3 `op` (a4 `op` foldr op b [])))
==>
a1 `op` (a2 `op` (a3 `op` (a4 `op` b)))

Look how it mirrors the structure of lists!

• (:) is replaced by op
• [] is replaced by base

So

foldr (+) 0 (x1:x2:x3:x4:[])
==> x1 + (x2 + (x3 + (x4 + 0))

Typing foldr

foldr :: (a -> b -> b) -> b -> [a] -> b
foldr op base []     = base
foldr op base (x:xs) = op x (foldr op base xs)

foldr takes as input

• a reducer function of type a -> b -> b
• a base value of type b
• a list of values to reduce [a]

and returns as output

• a reduced value b

QUIZ

Recall the function to compute the len of a list

len :: [a] -> Int
len []     = 0
len (x:xs) = 1 + len xs

Which of these is a valid implementation of Len

A. len = foldr (\n -> n + 1) 0

B. len = foldr (\n m -> n + m) 0

C. len = foldr (\_ n -> n + 1) 0

D. len = foldr (\x xs -> 1 + len xs) 0

E. All of the above

The Missing Parameter Revisited

We wrote foldr as

foldr :: (a -> b -> b) -> b -> [a] -> b
foldr op base []     = base
foldr op base (x:xs) = op x (foldr op base xs)

but can also write this

foldr :: (a -> b -> b) -> b -> [a] -> b
foldr op base  = go
where
go []     = base
go (x:xs) = op x (go xs)

Can someone explain where the xs went missing ?

Trees

Recall the Tree a type from last time

data Tree a
= Leaf
| Node a (Tree a) (Tree a)

For example here’s a tree

tree2 :: Tree Int
tree2 = Node 2 Leaf Leaf

tree3 :: Tree Int
tree3 = Node 3 Leaf Leaf

tree123 :: Tree Int
tree123 = Node 1 tree2 tree3

Some Functions on Trees

Lets write a function to compute the height of a tree

height :: Tree a -> Int
height Leaf         = 0
height (Node x l r) = 1 + max (height l) (height l)

Here’s another to sum the leaves of a tree:

sumTree :: Tree Int -> Int
sumTree Leaf         = ???
sumTree (Node x l r) = ???

Gathers all the elements that occur as leaves of the tree:

toList :: Tree a -> [a]
toList Leaf         = ???
toList (Node x l r) = ???

Lets give it a whirl

>>> height tree123
2

>>> sumTree tree123
6

>>> toList tree123
[1,2,3]

Pattern: Tree Fold

Can you spot the pattern? Those three functions are almost the same!

Step 1: Rename to maximize similarity

-- height
foo Leaf         = 0
foo (Node x l r) = 1 + max (foo l) (foo l)

-- sumTree
foo Leaf         = 0
foo (Node x l r) = foo l + foo r

-- toList
foo Leaf         = []
foo (Node x l r) = x : foo l ++ foo r

Step 2: Identify the differences

1. ???
2. ???

Step 3 Make differences a parameter

foo p1 p2 Leaf         = ???
foo p1 p2 (Node x l r) = ???

Pattern: Folding on Trees

tFold op b Leaf         = b
tFold op b (Node x l r) = op x (tFold op b l) (tFold op b r)

Lets try to work out the type of tFold!

tFold :: t_op -> t_b -> Tree a -> t_out

QUIZ

Suppose that t :: Tree Int.

What does tFold (\x y z -> y + z) 1 t return?

a. 0

b. the largest element in the tree t

c. the height of the tree t

d. the number-of-leaves of the tree t

e. type error

EXERCISE

Write a function to compute the largest element in a tree or 0 if tree is empty or all negative.

treeMax :: Tree Int -> Int
treeMax t = tFold f b t
where
f    = ???
b    = ???

Map over Trees

We can also write a tmap equivalent of map for Trees

treeMap :: (a -> b) -> Tree a -> Tree b
treeMap f (Leaf x)   = Leaf (f x)
treeMap f (Node l r) = Node (treeMap f l) (treeMap f r)

which gives

>>> treeMap (\n -> n * n) tree123     -- square all elements of tree
Node 1 (Node 4 Leaf Leaf) (Node 9 Leaf Leaf)

EXERCISE

Recursion is HARD TO READ do we really have to use it ?

Lets rewrite treeMap using tFold !

treeMap :: (a -> b) -> Tree a -> Tree b
treeMap f t = tFold op base t
where
op     = ???
base   = ???

When you are done, we should get

>>> animals = Node "cow" (Node "piglet" Leaf Leaf) (Leaf "hippo" Leaf Leaf)
>>> treeMap reverse animals
Node "woc" (Node "telgip" Leaf Leaf) (Leaf "oppih" Leaf Leaf)

Examples: foldDir

data Dir a
= Fil a             -- ^ A single file named `a`
| Sub a [Dir a]     -- ^ A sub-directory name `a` with contents `[Dir a]`

data DirElem a
= SubDir a          -- ^ A single Sub-Directory named `a`
| File a            -- ^ A single File named `a`

foldDir :: ([a] -> r -> DirElem a -> r) -> r -> Dir a -> r
foldDir f r0 dir = go [] r0 dir
where
go stk r (Fil a)    = f stk r (File a)
go stk r (Sub a ds) = L.foldl' (go stk') r' ds
where
r'   = f stk r (SubDir a)
stk' = a:stk

foldDir takes as input

• an accumulator f of type [a] -> r -> DirElem a -> r

• takes as input the path [a] , the current result r, the next DirElem [a]

• and returns as output the new result r

• an initial value of the result r0 and

• directory to fold over dir

And returns the result of running the accumulator over the whole dir.

Examples: Spotting Patterns In The “Real” World

These patterns in “toy” functions appear regularly in “real” code

2. Spot the patterns and eliminate recursion using HOFs.

3. Finally refactor the code to “swizzle” and “unswizzle” without duplication.

Try it yourself

• Rewrite the code that swizzles Char to use the Map k v type in Data.Map

Which is more readable? HOFs or Recursion

At first, recursive versions of shout and squares are easier to follow

• fold takes a bit of getting used to!

With practice, the higher-order versions become easier

• only have to understand specific operations

• recursion is lower-level & have to see “loop” structure

• worse, potential for making silly off-by-one errors

Indeed, HOFs were the basis of map/reduce and the big-data revolution

• Can parallelize and distribute computation patterns just once

• Reuse across hundreds or thousands of instances!