Haskell/Fix and recursion< Haskell
fix function is a particularly weird-looking 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.
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 yield an infinite application stream of
f (f (f (... )))? The resolution to this is our good friend, lazy evaluation. Essentially, this sequence of applications of
f will converge to a value if (and only if)
f is a lazy function. Let's see some 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). Then we try some examples. Since the definition of
fix is so simple, let's expand our examples to explain what happens:
fix (2+) = 2 + (fix (2+)) = 2 + (2 + fix (2+)) = 2 + (2 + (2 + fix (2+))) = 2 + (2 + (2 + (2 + fix (2+)))) = ...
It's clear that this will never converge to any value. Let's expand the next example:
fix (const "hello") = const "hello" (fix (const "hello")) = "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:) = 1 : fix (1:) = 1 : (1 : fix (1:)) = 1 : (1 : (1 : fix (1:)))
Although this similarly looks like it'll never converge to a value, keep in mind that when you type
fix (1:) into GHCi, what it's really doing is applying
show to that. So we should look at how
show (fix (1:)) evaluates (for simplicity, we'll pretend
show on lists doesn't put commas between items):
show (fix (1:)) = "[" ++ map show (fix (1:)) ++ "]" = "[" ++ map show (1 : fix (1:)) ++ "]" = "[" ++ "1" ++ map show (fix (1:)) ++ "]" = "[" ++ "1" ++ "1" ++ map show (fix (1:)) ++ "]"
So although the
map show (fix (1:)) will never terminate, it does produce output: GHCi can print the beginning of the string,
"[" ++ "1" ++ "1", and continue to print more as
map show (fix (1:)) produces more. 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 (\next guess tol val -> if abs(guess^2-val) < tol then guess else next ((guess + val / guess) / 2.0) tol val) 2.0 0.0001 25.0 = let f next guess tol val = if abs(guess^2-val) < tol then guess else next ((guess + val / guess) / 2.0) tol val in fix f 2.0 0.0001 25.0 = let f ... = ... in f (fix f) 2.0 0.0001 25.0 -- next = fix f = f (fix f) = f next ... = 5.000000000016778
What, if anything, will the following expressions converge to?
fix and fixed pointsEdit
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 least-defined 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.
|For each of the functions
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
Example: Fixed points of
Prelude> (2+) undefined *** Exception: Prelude.undefined
(2+) gives us
undefined back. So
⊥ is a fixed point of
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 "least-defined" clause comes in. Types in Haskell have a partial order on them called definedness. In any type,
⊥ is the least-defined 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 non-bottom
n. Similar comments apply to other simple types like
(). For "layered" values such as lists or
Maybe, the picture is more complicated, and we refer to the chapter on denotational semantics.
⊥ is the least-defined value for all types and
fix finds the least-defined 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.
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
Prelude> let fact n = if n == 0 then 1 else n * fact (n-1) in fact 5 120 Prelude> fix (\rec n -> if n == 0 then 1 else n * rec (n-1)) 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:
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 (n-1) + rec (n-2))) [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 (n-1)
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 (n-1)
All we did was substitute
f in the definition of
fact'. But this looks exactly like a recursive definition of a factorial function!
fact' itself as its first parameter in order to create a recursive function out of a higher-order function.
We can also consider things from a more operational point of view. Let's actually expand the definition of
fix fact' = fact' (fix fact') = (\rec n -> if n == 0 then 1 else n * rec (n-1)) (fix fact') = \n -> if n == 0 then 1 else n * fix fact' (n-1) = \n -> if n == 0 then 1 else n * fact' (fix fact') (n-1) = \n -> if n == 0 then 1 else n * (\rec n' -> if n' == 0 then 1 else n' * rec (n'-1)) (fix fact') (n-1) = \n -> if n == 0 then 1 else n * (if n-1 == 0 then 1 else (n-1) * fix fact' (n-2)) = \n -> if n == 0 then 1 else n * (if n-1 == 0 then 1 else (n-1) * (if n-2 == 0 then 1 else (n-2) * fix fact' (n-3))) = ...
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.
The typed lambda calculusEdit
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 top-level 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> (n-1)
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 (n-1)) (λm:Nat. if iszero m then 1 else m * <blank> (m-1))
This expands to:
λn:Nat. if iszero n then 1 else n * (if iszero n-1 then 1 else (n-1) * <blank> (n-2))
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 (n-1) ((λg:Nat→Nat. λm:Nat. if iszero n' then 1 else n' * g (m-1)) (λp:Nat. if iszero p then 1 else p * <blank> (p-1)))) -> λn:Nat. if iszero n then 1 else n * (if iszero n-1 then 1 else (n-1) * (if iszero n-2 then 1 else (n-2) * <blank> (n-3)))
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 (n-1))
This is a perfect factorial function in the typed lambda calculus plus
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
fix (λx:T. x)
This, in Haskell-speak, 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 well-typed term reduces to a value is lost.
Fix as a data typeEdit
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 a|Three a a a data FingerTree a=U a|Up (FingerTree (Node a))
It is not easy to implement this using Fix.