Haskell/Fix and recursion
The fix
function is a particularly weirdlooking function when you first see it. However, it is useful for one main theoretical reason: introducing it into the (typed) lambda calculus as a primitive allows you to define recursive functions.
Introducing fix
[edit]
Let's have the definition of fix
before we go any further:
fix :: (a > a) > a fix f = let {x = f x} in x
This immediately seems quite magical. Surely fix f
will cause an infinite series of nested applications of f
s: x = f x = f (f x) = f (f (f ( ... )))
? The resolution to this is our good friend, lazy evaluation. Essentially, this infinite sequence of applications of f
will be avoided if (and only if) f
is a lazy function. Let's see some examples:
Example: fix
examples
Prelude> :m Control.Monad.Fix Prelude Control.Monad.Fix> fix (2+) *** Exception: stack overflow Prelude Control.Monad.Fix> fix (const "hello") "hello" Prelude Control.Monad.Fix> fix (1:) [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...
We first import the Control.Monad.Fix
module to bring fix
into scope (this is also available in the Data.Function
module). Then we try some examples. Since the definition of fix
is so simple, let's expand our examples to explain what happens:
 fix f = let {x = f x} in x fix (2+) = let {x = 2 + x} in x = let {x = 2 + x} in 2 + x = let {x = 2 + x} in 2 + (2 + x) = let {x = 2 + x} in 2 + (2 + (2 + x)) = let {x = 2 + x} in 2 + (2 + (2 + (2 + x))) = ...
(+)
is an eager function, so it needs full value of x
before it can compute ... the value of x
. It's clear that this calculation will never stop. Let's expand the next example:
fix (const "hello") = let {x = const "hello" x} in x = "hello"
This is quite different: we can see after one expansion of the definition of fix
that because const
ignores its second argument, the evaluation concludes. The evaluation for the last example is a little different, but we can proceed similarly:
fix (1:) = let {x = 1 : x} in x = let {x = 1 : x} in 1 : x
Here 1 : x
is already in weak head normal form ((:)
is a lazy data constructor), so the expansion stops. This actually creates a cyclic structure. Each time new element of this list is requested by a consumer function, x
's definition is consulted, and it is already known to be 1 : x
.
Thus, requesting elements from this list one by one will return 1
s one after another while requested, without limit. So take 10 (fix (1:))
will produce a list of ten 1
s, but trying to print them all by typing fix (1:)
into GHCi causes the infinite stream of 1
s to be printed.
Well actually it causes the evaluation of show (fix (1:)) = "[" ++ intercalate "," (map show (fix (1:))) ++ "]"
, and although the map show (fix (1:))
will never terminate, it does produce its output incrementally, piece by piece, one "1"
at a time:
"[" ++ intercalate "," (map show (fix (1:))) ++ "]" = "[" ++ intercalate "," (map show (let {x = 1 : x} in x)) ++ "]" = "[" ++ intercalate "," (map show (let {x = 1 : x} in 1 : x)) ++ "]" = "[" ++ "1" ++ "," ++ intercalate "," (map show (let {x = 1 : x} in x)) ++ "]" = "[1," ++ intercalate "," (map show (let {x = 1 : x} in 1 : x)) ++ "]" = "[1," ++ "1" ++ "," ++ intercalate "," (map show (let {x = 1 : x} in x)) ++ "]" = "[1,1," ++ intercalate "," (map show (let {x = 1 : x} in 1 : x)) ++ "]" = "[1,1," ++ "1" ++ "," ++ intercalate "," (map show (let {x = 1 : x} in x)) ++ "]" = "[1,1,1," ++ intercalate "," (map show (let {x = 1 : x} in 1 : x)) ++ "]" = ...
This is lazy evaluation at work: the printing function doesn't need to consume its entire input string before beginning to print, it does so as soon as it can start.
Lastly, iteratively calculating an approximation of a square root of a number,
 fix f = let {x = f x} in x fix (\next guess tol val > if abs(guess^2val) < tol then guess else next ((guess + val / guess) / 2.0) tol val) 2.0 0.0001 25.0 = (let {next = (\next guess tol val > if abs(guess^2val) < tol then guess else next ((guess + val / guess) / 2.0) tol val) next} in next) 2.0 0.0001 25.0 = let {next guess tol val = if abs(guess^2val) < tol then guess else next ((guess + val / guess) / 2.0) tol val} in next 2.0 0.0001 25.0 = 5.000000000016778
Exercises 

What, if anything, will the following expressions converge to?

fix
and fixed points[edit]
A fixed point of a function f
is a value a
such that f a == a
. For example, 0
is a fixed point of the function (* 3)
since 0 * 3 == 0
. This is where the name of fix
comes from: it finds the leastdefined fixed point of a function. (We'll come to what "least defined" means in a minute.) Notice that for both of our examples above that converge, this is readily seen:
const "hello" "hello" > "hello" (1:) [1,1,..] > [1,1,...]
And since there's no number x
such that 2+x == x
, it also makes sense that fix (2+)
diverges.
Exercises 

For each of the functions f in the above exercises for which you decided that fix f converges, verify that fix f finds a fixed point. 
In fact, it's obvious from the definition of fix
that it finds a fixed point. All we need to do is write the equation for fix
the other way around:
f (fix f) = fix f
Which is precisely the definition of a fixed point! So it seems that fix
should always find a fixed point. But sometimes fix
seems to fail at this, as sometimes it diverges. We can repair this property, however, if we bring in some denotational semantics. Every Haskell type actually includes a special value called bottom, written ⊥
. So the values with type, for example, Int
include, in fact, ⊥
as well as 1, 2, 3
etc.. Divergent computations are denoted by a value of ⊥
, i.e., we have that fix (2+) = ⊥
.
The special value undefined
is also denoted by this ⊥
. Now we can understand how fix
finds fixed points of functions like (2+)
:
Example: Fixed points of (2+)
Prelude> (2+) undefined *** Exception: Prelude.undefined
So feeding undefined
(i.e., ⊥
) to (2+)
gives us undefined
back. So ⊥
is a fixed point of (2+)
!
In the case of (2+)
, it is the only fixed point. However, there are other functions f
with several fixed points for which fix f
still diverges: fix (*3)
diverges, but we remarked above that 0
is a fixed point of that function. This is where the "leastdefined" clause comes in. Types in Haskell have a partial order on them called definedness. In any type, ⊥
is the leastdefined value (hence the name "bottom"). For simple types like Int
, the only pairs in the partial order are ⊥ ≤ 1
, ⊥ ≤ 2
and so on. We do not have m ≤ n
for any nonbottom Int
s m
, n
. Similar comments apply to other simple types like Bool
and ()
. For "layered" values such as lists or Maybe
, the picture is more complicated, and we refer to the chapter on denotational semantics.
So since ⊥
is the leastdefined value for all types and fix
finds the leastdefined fixed point, if f ⊥ = ⊥
, we will have fix f = ⊥
(and the converse is also true). If you've read the denotational semantics article, you will recognise this as the criterion for a strict function: fix f
diverges if and only if f
is strict.
Recursion[edit]
If you have already come across examples of fix
, chances are they were examples involving fix
and recursion. Here's a classic example:
Example: Encoding recursion with fix
Prelude> let fact n = if n == 0 then 1 else n * fact (n1) in fact 5 120 Prelude> fix (\rec n > if n == 0 then 1 else n * rec (n1)) 5 120
Here we have used fix
to "encode" the factorial function: note that (if we regard fix
as a language primitive) our second definition of fact
doesn't involve recursion at all. In a language like the typed lambda calculus that doesn't feature recursion, we can introduce fix
in to write recursive functions in this way. Here are some more examples:
Example: More fix
examples
Prelude> fix (\rec f l > if null l then [] else f (head l) : rec f (tail l)) (+1) [1..3] [2,3,4] Prelude> map (fix (\rec n > if n == 1  n == 2 then 1 else rec (n1) + rec (n2))) [1..10] [1,1,2,3,5,8,13,21,34,55]
So how does this work? Let's first approach it from a denotational point of view with our fact
function. For brevity let's define:
fact' rec n = if n == 0 then 1 else n * rec (n1)
This is the same function as in the first example above, except that we gave a name to the anonymous function so that we're computing fix fact' 5
now. fix
will find a fixed point of fact'
, i.e. the function f
such that f == fact' f
. But let's expand what this means:
f = fact' f = \n > if n == 0 then 1 else n * f (n1)
All we did was substitute rec
for f
in the definition of fact'
. But this looks exactly like a recursive definition of a factorial function! fix
feeds fact'
itself as its first parameter in order to create a recursive function out of a higherorder function.
We can also consider things from a more operational point of view. Let's actually expand the definition of fix fact'
:
fix fact' = fact' (fix fact') = (\rec n > if n == 0 then 1 else n * rec (n1)) (fix fact') = \n > if n == 0 then 1 else n * fix fact' (n1) = \n > if n == 0 then 1 else n * fact' (fix fact') (n1) = \n > if n == 0 then 1 else n * (\rec n' > if n' == 0 then 1 else n' * rec (n'1)) (fix fact') (n1) = \n > if n == 0 then 1 else n * (if n1 == 0 then 1 else (n1) * fix fact' (n2)) = \n > if n == 0 then 1 else n * (if n1 == 0 then 1 else (n1) * (if n2 == 0 then 1 else (n2) * fix fact' (n3))) = ...
Notice that the use of fix
allows us to keep "unravelling" the definition of fact'
: every time we hit the else
clause, we product another copy of fact'
via the evaluation rule fix fact' = fact' (fix fact')
, which functions as the next call in the recursion chain. Eventually we hit the then
clause and bottom out of this chain.
Exercises 


The typed lambda calculus[edit]
In this section we'll expand upon a point mentioned a few times in the previous section: how fix
allows us to encode recursion in the typed lambda calculus. It presumes you've already met the typed lambda calculus. Recall that in the lambda calculus, there is no let
clause or toplevel bindings. Every program is a simple tree of lambda abstractions, applications and literals. Let's say we want to write a fact
function. Assuming we have a type called Nat
for the natural numbers, we'd start out something like the following:
λn:Nat. if iszero n then 1 else n * <blank> (n1)
The problem is, how do we fill in the <blank>
? We don't have a name for our function, so we can't call it recursively. The only way to bind names to terms is to use a lambda abstraction, so let's give that a go:
(λf:Nat→Nat. λn:Nat. if iszero n then 1 else n * f (n1)) (λm:Nat. if iszero m then 1 else m * <blank> (m1))
This expands to:
λn:Nat. if iszero n then 1 else n * (if iszero n1 then 1 else (n1) * <blank> (n2))
We still have a <blank>
. We could try to add one more layer in:
(λf:Nat→Nat. λn:Nat. if iszero n then 1 else n * f (n1) ((λg:Nat→Nat. λm:Nat. if iszero n' then 1 else n' * g (m1)) (λp:Nat. if iszero p then 1 else p * <blank> (p1)))) > λn:Nat. if iszero n then 1 else n * (if iszero n1 then 1 else (n1) * (if iszero n2 then 1 else (n2) * <blank> (n3)))
It's pretty clear we're never going to be able to get rid of this <blank>
, no matter how many levels of naming we add in. Never, that is, unless we use fix
, which, in essence, provides an object from which we can always unravel one more layer of recursion and still have what we started off:
fix (λf:Nat→Nat. λn:Nat. if iszero n then 1 else n * f (n1))
This is a perfect factorial function in the typed lambda calculus plus fix
.
fix
is actually slightly more interesting than that in the context of the typed lambda calculus: if we introduce it into the language, then every type becomes inhabited, because given some concrete type T
, the following expression has type T
:
fix (λx:T. x)
This, in Haskellspeak, is fix id
, which is denotationally ⊥
. So we see that as soon as we introduce fix
to the typed lambda calculus, the property that every welltyped term reduces to a value is lost.
Fix as a data type[edit]
It is also possible to make a fix data type in Haskell.
There are three ways of defining it.
newtype Fix f = Fix (f (Fix f))
or using the RankNTypes extension
newtype Mu f=Mu (forall a.(f a>a)>a) data Nu f=forall a.Nu a (a>f a)
Mu and Nu help generalize folds, unfolds and refolds.
fold :: (f a > a) > Mu f > a fold g (Mu f)=f g unfold :: (a > f a) > a > Nu f unfold f x=Nu x f refold :: (a > f a) > (g a> a) > Mu f > Nu g refold f g=unfold g . fold f
Mu and Nu are restricted versions of Fix. Mu is used for making inductive noninfinite data and Nu is used for making coinductive infinite data. Eg)
newpoint Stream a=Stream (Nu ((,) a))  forsome b. (b,b>(a,b)) newpoint Void a=Void (Mu ((,) a))  forall b.((a,b)>b)>b
Unlike the fix point function the fix point types do not lead to bottom. In the following code Bot is perfectly defined. It is equivalent to the unit type ().
newtype Id a=Id a newtype Bot=Bot (Fix Id)  equals newtype Bot=Bot Bot  There is only one allowable term. Bot $ Bot $ Bot $ Bot ..,
The Fix data type cannot model all forms of recursion. Take for instance this nonregular data type.
data Node a=Two a aThree a a a data FingerTree a=U aUp (FingerTree (Node a))
It is not easy to implement this using Fix.