Haskell Programming Tutorial: Recursive Functions on Lists
Computing with lists
 There are two approaches to working with lists:
 Write functions to do what you want, using recursive definitions that traverse the list structure.
 Write combinations of the standard list processing functions.
 The second approach is preferred, but the standard list processing functions do need to be defined, and those definitions use the first approach (recursive definitions).
 We’ll cover both methods.
Recursion on lists

 A list is built from the empty list \([]\) and the function \(cons\; :: \; a\rightarrow [a] \rightarrow [a]\). In Haskell, the function \(cons\) is actually written as the operator \((:)\) , in other words : is pronounced as
cons
.
 A list is built from the empty list \([]\) and the function \(cons\; :: \; a\rightarrow [a] \rightarrow [a]\). In Haskell, the function \(cons\) is actually written as the operator \((:)\) , in other words : is pronounced as

 Every list must be either

 \([]\) or

 \((x : xs)\) for some \(x\) (the head of the list) and \(xs\) (the tail)

 Every list must be either
4.8

 The recursive definition follows the structure of the data:

 Base case of the recursion is \([]\).

 Recursion (or induction) case is \((x : xs)\).

 The recursive definition follows the structure of the data:
Some examples of recursion on lists
Recursive definition of length
The length of a list can be computed recursively as follows:length :: [a] > Int  function type
length [] = 0  base case
length (x:xs) = 1 + length xs  recursion case
Recursive definition of filter

 filter is given a predicate (a function that gives a Boolean result) and a list, and returns a list of the elements that satisfy the predicate.
filter :: (a>Bool) > [a] > [a]
filter (<5) [3,9,2,12,6,4]  > [3,2,4]
filter
is shown below. This relies on guards.
filter :: (a > Bool) > [a] > [a]
filter pred [] = []
filter pred (x:xs) pred x = x : filter pred xs otherwise = filter pred xs
Computations over lists

 Many computations that would be for/while loops in an imperative language are naturally expressed as list computations in a functional language.

 There are some common cases:

 Perform a computation on each element of a list: \(map\)

 Iterate over a list, from left to right: \(foldl\)

 Iterate over a list, from right to left: \(foldr\)

 There are some common cases:

 It’s good practice to use these three functions when applicable

 And there are some related functions that we’ll see later
Function composition

 We can express a large computation by “chaining together” a sequence of functions that perform smaller computations

 Start with an argument of type \(a\)

 Apply a function \(g :: a \to b\) to it, getting an intermediate result of type \(b\)

 Then apply a function \(f :: b \to c\) to the intermediate result, getting the final result of type \(c\)

 The entire computation (first \(g\), then \(f\)) is written as \(f \circ g\).

 This is traditional mathematical notation; just remember that in \(f \circ g\), the functions are used in right to left order.

 Haskell uses
.
as the function composition operator(.) :: (b>c) > (a>b) > a > c (f . g) x = f (g x)
 Haskell uses
Performing an operation on every element of a list: map

 map applies a function to every element of a list
map f [x0,x1,x2]  > [f x0, f x1, f x2]
 map applies a function to every element of a list
Composition of maps

 map is one of the most commonly used tools in your functional toolkit

 A common style is to define a set of simple computations using map, and to compose them.
map f (map g xs) = map (f . g) xs
 A common style is to define a set of simple computations using map, and to compose them.
Recursive definition of map
map :: (a > b) > [a] > [b]
map _ [] = []
map f (x:xs) = f x : map f xs
Folding a list (reduction)

 An iteration over a list to produce a singleton value is called a fold

 There are several variations: folding from the left, folding from the right, several variations having to do with “initialisation”, and some more advanced variations.

 Folds may look tricky at first, but they are extremely powerful, and they are used a lot! And they aren’t actually very complicated.
Left fold: foldl

 foldl is fold from the left

 Think of it as an iteration across a list, going left to right.

 A typical application is \(foldl\, f\, z\, xs\)

 The \(z :: b\) is an initial value

 The \(xs :: [a]\) argument is a list of values which we combine systematically using the supplied function \(f\)

 A useful intuition: think of the \(z :: b\) argument as an “accumulator”.

 The function \(f\) takes the current value of the accumulator and a list element, and gives the new value of the accumulator.
foldl :: (b>a>b) > b > [a] > b
 The function \(f\) takes the current value of the accumulator and a list element, and gives the new value of the accumulator.
Examples of foldl with function notation
\[\begin{aligned}\mathtt{foldl\,f\,z\,[]} &\rightsquigarrow & z\\
\mathtt{foldl\,f\,z\,[x0]} & \rightsquigarrow & f\,z\,x0\\
\mathtt{foldl\,f\,z\,[x0,x1]} & \rightsquigarrow & f\,(f\,z\,x0)\,x1\\
\mathtt{foldl\,f\,z\,[x0,x1,x2]} & \rightsquigarrow & f\,(f\,(f\,z\,x0)\,x1)\, x2\end{aligned}\]
Examples of foldl with infix notation
In this example, + denotes an arbitrary operator for f; it isn’t supposed to mean specifically addition.foldl (+) z []  > z
foldl (+) z [x0]  > z + x0
foldl (+) z [x0,x1]  > (z + x0) + x1
foldl (+) z [x0,x1,x2]  > ((z + x0) + x1) + x2
Recursive definition of foldl
foldl :: (b > a > b) > b > [a] > b
foldl f z0 xs0 = lgo z0 xs0wherelgo z [] = zlgo z (x:xs) = lgo (f z x) xs
Right fold: foldr

 Similar to \(foldl\), but it works from right to left
foldr :: (a > b > b) > b > [a] > b
Examples of foldr with function notation
\[\begin{aligned}\mathtt{foldr\,f\, z\, [] } & \rightsquigarrow & z\\
\mathtt{foldr\, f\, z\, [x0] } & \rightsquigarrow & f\, x0\, z\\
\mathtt{foldr\, f\, z\, [x0,x1] } & \rightsquigarrow & f\, x0\, (f\, x1\, z)\\
\mathtt{foldr\, f\, z\, [x0,x1,x2] } & \rightsquigarrow & f\, x0\, (f\, x1\, (f\, x2\, z))\end{aligned}\]
Examples of foldr with operator notation
foldr (+) z []  > z
foldr (+) z [x0]  > x0 + z
foldr (+) z [x0,x1]  > x0 + (x1 + z)
foldr (+) z [x0,x1,x2]  > x0 + (x1 + (x2 + z))
Recursive definition of foldr
foldr :: (a > b > b) > b > [a] > b
foldr k z = gowherego [] = zgo (y:ys) = y `k` go ys
Relationship between foldr and list structure
We have seen that a list[x0,x1,x2]
can also be written as
x0 : x1 : x2 : []
foldr (:) [] [x0,x1,x2] >x0 : x1 : x2 : []
Some applications of folds
sum xs = foldr (+) 0 xs
product xs = foldr (*) 1 xs
sum = foldr (+) 0
product = foldr (*) 1
Functional Programming in Haskell: Supercharge Your Coding
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