How do you make an anonymous recursive function (something simple for example factorial n?) I have heard it is possible but no idea how to make it work in OCaml.
let a =
fun x -> ....
I just don't know how to keep it going...
Here is a definition of factorial using only anonymous functions:
let fact =
(fun f -> (fun x a -> f (x x) a) (fun x a -> f (x x) a))
(fun f n -> if n < 2 then 1 else n * f (n - 1))
It requires the use of the -rectypes flag.
Here's a session showing that it works:
$ rlwrap ocaml -rectypes
OCaml version 4.03.0
let fact =
(fun f -> (fun x a -> f (x x) a) (fun x a -> f (x x) a))
(fun f n -> if n < 2 then 1 else n * f (n - 1));;
val fact : int -> int = <fun>
# fact 8;;
- : int = 40320
I cheated somewhat by looking up the Y Combinator here: Rosetta Code: Y Combinator
Update
Disclaimer: you would do better to read up on lambda calculus, fixed points, and the Y Combinator than to get your info from me. I'm not a theorist, just a humble practitioner.
Following the actual computation is almost impossible (but definitely worth doing I'm sure). But at a high level the ideas are like this.
The first line of the definition is the Y Combinator, which in general calculates the fixed point of a function. It so happens that a recursive function is the fixed point of a function from functions to functions.
So the first goal is to find the function whose fixed point is the factorial function. That's the second line of the definition. If you give it a function of type int -> int, it gives you back another function of type int -> int. And if you give it the factorial function, it gives you back the factorial function. This means that the factorial function is its fixed point.
So then when you apply the Y Combinator to this function, you do indeed get the factorial function.
Let me try to expand a bit on Jeffrey Scofield's answer. A non-anonymous recursive definition of the factorial function could be
let rec fact n =
if n < 2 then 1 else n * fact (n - 1)
The first problem you encounter when you try to define an anonymous recursive function is how to do the actual recursive call (fact (n - 1) in our case). For a call we need a name and we do not have a name for an anonymous function. The solution is to use a temporary name. With the temporary name f, the definition body is just
fun n -> if n < 2 then 1 else n * f (n - 1)
This term does not have a type, because the "temporary name" f is unbound. But we can turn it into a value that does have a type by bounding f as well. Let us call the result g:
let g = fun f n -> if n < 2 then 1 else n * f (n - 1)
g is not yet anonymous at the moment, but only because I want to refer to it again.
Observe that g has type (int -> int) -> (int -> int). What we want (the factorial function) will have type (int -> int). So g takes something of the type we want (a function type in this case) and produces something of the same type. The intuition is that g takes an approximation of the factorial function, namely a function f which works for all n up to some limit N and returns a better approximation, namely a function that works for all n up to N+1.
Finally we need something that turns g into an actual recursive definition.
Doing so is a very generic task. Recall that g improves the approximation quality. The final factorial function fact is one which cannot be further improved. So applying g to fact should be the same as just fact. (Actually that is only true from a value point of view. The actual computation inherent in g fact n for some n is different from that of just fact n. But the returned values are the same.) In other words, fact is a fixed point of g. So what we need is something that computes fixed points.
Luckily, there is a single function that does so: The Y combinator. From a value point of view, the Y combinator (let us use y in OCaml, as uppercase is reserved for constructors) is defined by the fact that y g = g (y g) for all g: given some function g, the combinator returns one of its fixed points.
Consequently,
y : (`a -> `a) -> `a
In our case the type variable is instantiated by (int -> int).
One possible way to define y would be
let y = fun g -> (fun x -> g (x x)) (fun x -> g (x x))
but this works only with lazy evaluation (as, I believe, Haskell has). As OCaml has eager evaluation, it produces a stack overflow when used. The reason is that OCaml tries to turn something like y g 8 into
g (y g) 8
g (g (y g)) 8
g (g (g (y g))) 8
...
without ever getting to call g.
The solution is to use deferred computation inside of y:
let y = fun g -> (fun x a -> g (x x) a) (fun x a -> g (x x) a)
One drawback is that y does not work for arbitrary types any more. It only works for function types.
y : ((`b -> `c) -> (`b -> `c)) -> (`b -> `c)
But you asked for recursive definitions of functions anyway, not for recursive definitions of other values. So, our definition of the factorial function is y g with y and g defined as above. Neither y nor g are anonymous yet, but that can be remedied easily:
(fun g -> (fun x a -> g (x x) a) (fun x a -> g (x x) a))
(fun f n -> if n < 2 then 1 else n * f (n - 1))
UPDATE:
Defining y only works with the -rectypes option. The reason is that we apply x to itself.
There is also an "intuitive" way to accomplish anonymous recursion without resorting to Y combinators.
It makes use of a let binding to store the value of a lambda that accepts itself as an argument, so that it can call itself with itself as the first parameter, like so:
let fact = (let fact0 = (fun self n -> if n < 2 then 1 else n * self self (n - 1)) in (fun n -> fact0 fact0 n));;
It's anonymous only to the extent that it is not defined with let rec.
Related
I'm looking for a function with this signature:
chainTraversal :: k -> (k -> a -> Maybe (k, b)) -> Map k a -> Map k b
You give it an initial key to start at, a function and a map.
It will extract the element at the position k in the Map, and feed that element to the function. Based on this, the function will return another key to look at next.
It's some mix between a filter and a traversal, with the elements themselves giving the next position to open. The result is the list of elements that has been traversed. It can be shorter than the original map.
Edit: taking into account a comment.
Since all the lookups are done in the original Map:
foo :: k -> (k -> a -> Maybe (k, b)) -> Map k a -> Map k b
foo k f m = fromList $ unfoldr g k
where
g k = (\(k', b) -> (k', (k, b))) -- k ? k' ? you decide
<$> (f' k =<< (m `at` k))
f' k (k', a) = f k a -- or: f k' a ? you decide
or something like that.
You'll have to implement the at function in terms of one of the lookupNN functions of your choosing.
It's not a filter since it must stop on the first Nothing produced by f.
There is no existing function with that signature and behavior. You'll have to write it yourself.
Usually test which contain question about SML have questions that ask you to find the signature/type of a function.
For example - What is the type of the following function:
fun foo f g x y = f (f x (g x) y) y;
Solution:
val foo = fn : ('a -> 'b -> 'b -> 'a) -> ('a -> 'b) -> 'a -> 'b -> 'b -> 'a
I was wondering if there is a good algorithm I could follow in order to solve those kind of questions. Every time I try to solve one of those, I get confused and fail.
Start with what you know, then figure out a bit here and a bit there until there are no unknowns.
Here is one possibility:
Call the unknown types FOO, F, G, X, and Y, respectively.
Then look for something small and easy and start assigning types.
(g x)
is clearly an application of a function to one argument.
Set X = a and G = a -> b.
Then look at the enclosing expression:
(f x (g x) y)
| |
v v
a b
So far, we know that F = a -> b -> Y -> C, for some C.
Go outwards again:
f (f x (g x) y) y
Since both x and (f x (g x) y) are first arguments to f, they must be the same type a, and the same idea applies to y and (g x), giving them the type b.
So, F = a -> b -> b -> a and, since the outer f is only given two arguments, the type of the right-hand side must be b -> a.
Thus
X = a
Y = b
G = a -> b
F = a -> b -> b -> a
FOO = (a -> b -> b -> a) -> (a -> b) -> a -> b -> (b -> a)
And, since arrows associate to the right, FOO is equivalent to
(a -> b -> b -> a) -> (a -> b) -> a -> b -> b -> a
There are several ways to derive the type of a function depending on how close to the compiler's algorithm you want to go and how much you want to cut corners with intuition, which can come handy in practice and perhaps in exams, depending on the focus of the exam.
An example by Ionuț G. Stan cuts very few corners, and has a quite verbose notation. This mechanical approach is very safe, spells out everything and takes some time.
This current example by molbdnilo takes a middle ground and does some equational reasoning, but also relies on some level of intuition. I think this is generally the way you want to be able to do it, since it takes less time and space by hand.
An example by me links to various other examples for a diversity in practical approaches.
How would I go about applying a function n-times in OCaml if I don't
know the functions argument?
I want the call iter(n, fun x -> 2+x) 0 to evaluate to 2*n since
that would be the same as 2+2+2+2.... Also, if n=0 it should
return the identity function.
My attempt:
let rec iter : int * (int -> int) -> (int -> int)
= fun (n,f) ->
if n = 0 then f
else iter((n-1), f( f () ))
Possible duplicate: OCaml recursive function to apply a function n times but this question has an argument for the anonymous function so the answers does not help me.
You may not “have an argument” right now, but since the result is a function you can always just bring an argument in scope by returning a lambda:
let rec iter : int * (int -> int) -> (int -> int)
= fun (n,f) ->
if n = 0 then f
else fun x -> iter(n-1, f) (f x);;
Try it online!
Note that, as Willem remarks, your base case is probably wrong: for n=0, you want to return the identity function regardless of what function is passed in. Otherwise you get strange behaviour, for instance such a function should generally fulfill iter (n, fun x -> x+1) 0 ≡ n, but with your base case it gives n+1.
I would write the function thus:
let rec iter : int -> ('a -> 'a) -> ('a -> 'a)
= fun n f x -> if n = 0
then x
else iter (n-1) f (f x);;
Try it online!
Here I've not explicitly mentioned the identity function, but because I just return x when n is zero, that's what the identity function does. Alternatively, you can return another lambda which just passes the argument through:
let rec iter : int -> ('a -> 'a) -> 'a -> 'a
= fun n f ->
if n = 0
then fun x -> x (* identity function *)
else fun x -> iter (n-1) f (f x);;
I am not really sure if this is what you want - a small modification of the answer you linked to seems to do the job though:
*Edit: identity function depends on the nature of the function you pass to iter (right?), so I am not really sure how you can get it just from looking at f. That's why I am only returning f for now. And repeat fun x -> x + 2 n times - wouldn't that give you x + 2 * n?
let iter n f =
let chain_func f1 f2 arg = f1 (f2 arg) in
let rec aux n f newf =
if n <= 0 then newf else aux (n - 1) f ( chain_func f newf ) in
aux (n - 1) f f;;
*Edit 2: identity function is fun x -> x so the last line needs to be fixed to: aux n f (fun x -> x)
In pure functional languages like Haskell, is there an algorithm to get the inverse of a function, (edit) when it is bijective? And is there a specific way to program your function so it is?
In some cases, yes! There's a beautiful paper called Bidirectionalization for Free! which discusses a few cases -- when your function is sufficiently polymorphic -- where it is possible, completely automatically to derive an inverse function. (It also discusses what makes the problem hard when the functions are not polymorphic.)
What you get out in the case your function is invertible is the inverse (with a spurious input); in other cases, you get a function which tries to "merge" an old input value and a new output value.
No, it's not possible in general.
Proof: consider bijective functions of type
type F = [Bit] -> [Bit]
with
data Bit = B0 | B1
Assume we have an inverter inv :: F -> F such that inv f . f ≡ id. Say we have tested it for the function f = id, by confirming that
inv f (repeat B0) -> (B0 : ls)
Since this first B0 in the output must have come after some finite time, we have an upper bound n on both the depth to which inv had actually evaluated our test input to obtain this result, as well as the number of times it can have called f. Define now a family of functions
g j (B1 : B0 : ... (n+j times) ... B0 : ls)
= B0 : ... (n+j times) ... B0 : B1 : ls
g j (B0 : ... (n+j times) ... B0 : B1 : ls)
= B1 : B0 : ... (n+j times) ... B0 : ls
g j l = l
Clearly, for all 0<j≤n, g j is a bijection, in fact self-inverse. So we should be able to confirm
inv (g j) (replicate (n+j) B0 ++ B1 : repeat B0) -> (B1 : ls)
but to fulfill this, inv (g j) would have needed to either
evaluate g j (B1 : repeat B0) to a depth of n+j > n
evaluate head $ g j l for at least n different lists matching replicate (n+j) B0 ++ B1 : ls
Up to that point, at least one of the g j is indistinguishable from f, and since inv f hadn't done either of these evaluations, inv could not possibly have told it apart – short of doing some runtime-measurements on its own, which is only possible in the IO Monad.
⬜
You can look it up on wikipedia, it's called Reversible Computing.
In general you can't do it though and none of the functional languages have that option. For example:
f :: a -> Int
f _ = 1
This function does not have an inverse.
Not in most functional languages, but in logic programming or relational programming, most functions you define are in fact not functions but "relations", and these can be used in both directions. See for example prolog or kanren.
Tasks like this are almost always undecidable. You can have a solution for some specific functions, but not in general.
Here, you cannot even recognize which functions have an inverse. Quoting Barendregt, H. P. The Lambda Calculus: Its Syntax and Semantics. North Holland, Amsterdam (1984):
A set of lambda-terms is nontrivial if it is neither the empty nor the full set. If A and B are two nontrivial, disjoint sets of lambda-terms closed under (beta) equality, then A and B are recursively inseparable.
Let's take A to be the set of lambda terms that represent invertible functions and B the rest. Both are non-empty and closed under beta equality. So it's not possible to decide whether a function is invertible or not.
(This applies to the untyped lambda calculus. TBH I don't know if the argument can be directly adapted to a typed lambda calculus when we know the type of a function that we want to invert. But I'm pretty sure it will be similar.)
If you can enumerate the domain of the function and can compare elements of the range for equality, you can - in a rather straightforward way. By enumerate I mean having a list of all the elements available. I'll stick to Haskell, since I don't know Ocaml (or even how to capitalise it properly ;-)
What you want to do is run through the elements of the domain and see if they're equal to the element of the range you're trying to invert, and take the first one that works:
inv :: Eq b => [a] -> (a -> b) -> (b -> a)
inv domain f b = head [ a | a <- domain, f a == b ]
Since you've stated that f is a bijection, there's bound to be one and only one such element. The trick, of course, is to ensure that your enumeration of the domain actually reaches all the elements in a finite time. If you're trying to invert a bijection from Integer to Integer, using [0,1 ..] ++ [-1,-2 ..] won't work as you'll never get to the negative numbers. Concretely, inv ([0,1 ..] ++ [-1,-2 ..]) (+1) (-3) will never yield a value.
However, 0 : concatMap (\x -> [x,-x]) [1..] will work, as this runs through the integers in the following order [0,1,-1,2,-2,3,-3, and so on]. Indeed inv (0 : concatMap (\x -> [x,-x]) [1..]) (+1) (-3) promptly returns -4!
The Control.Monad.Omega package can help you run through lists of tuples etcetera in a good way; I'm sure there's more packages like that - but I don't know them.
Of course, this approach is rather low-brow and brute-force, not to mention ugly and inefficient! So I'll end with a few remarks on the last part of your question, on how to 'write' bijections. The type system of Haskell isn't up to proving that a function is a bijection - you really want something like Agda for that - but it is willing to trust you.
(Warning: untested code follows)
So can you define a datatype of Bijection s between types a and b:
data Bi a b = Bi {
apply :: a -> b,
invert :: b -> a
}
along with as many constants (where you can say 'I know they're bijections!') as you like, such as:
notBi :: Bi Bool Bool
notBi = Bi not not
add1Bi :: Bi Integer Integer
add1Bi = Bi (+1) (subtract 1)
and a couple of smart combinators, such as:
idBi :: Bi a a
idBi = Bi id id
invertBi :: Bi a b -> Bi b a
invertBi (Bi a i) = (Bi i a)
composeBi :: Bi a b -> Bi b c -> Bi a c
composeBi (Bi a1 i1) (Bi a2 i2) = Bi (a2 . a1) (i1 . i2)
mapBi :: Bi a b -> Bi [a] [b]
mapBi (Bi a i) = Bi (map a) (map i)
bruteForceBi :: Eq b => [a] -> (a -> b) -> Bi a b
bruteForceBi domain f = Bi f (inv domain f)
I think you could then do invert (mapBi add1Bi) [1,5,6] and get [0,4,5]. If you pick your combinators in a smart way, I think the number of times you'll have to write a Bi constant by hand could be quite limited.
After all, if you know a function is a bijection, you'll hopefully have a proof-sketch of that fact in your head, which the Curry-Howard isomorphism should be able to turn into a program :-)
I've recently been dealing with issues like this, and no, I'd say that (a) it's not difficult in many case, but (b) it's not efficient at all.
Basically, suppose you have f :: a -> b, and that f is indeed a bjiection. You can compute the inverse f' :: b -> a in a really dumb way:
import Data.List
-- | Class for types whose values are recursively enumerable.
class Enumerable a where
-- | Produce the list of all values of type #a#.
enumerate :: [a]
-- | Note, this is only guaranteed to terminate if #f# is a bijection!
invert :: (Enumerable a, Eq b) => (a -> b) -> b -> Maybe a
invert f b = find (\a -> f a == b) enumerate
If f is a bijection and enumerate truly produces all values of a, then you will eventually hit an a such that f a == b.
Types that have a Bounded and an Enum instance can be trivially made RecursivelyEnumerable. Pairs of Enumerable types can also be made Enumerable:
instance (Enumerable a, Enumerable b) => Enumerable (a, b) where
enumerate = crossWith (,) enumerate enumerate
crossWith :: (a -> b -> c) -> [a] -> [b] -> [c]
crossWith f _ [] = []
crossWith f [] _ = []
crossWith f (x0:xs) (y0:ys) =
f x0 y0 : interleave (map (f x0) ys)
(interleave (map (flip f y0) xs)
(crossWith f xs ys))
interleave :: [a] -> [a] -> [a]
interleave xs [] = xs
interleave [] ys = []
interleave (x:xs) ys = x : interleave ys xs
Same goes for disjunctions of Enumerable types:
instance (Enumerable a, Enumerable b) => Enumerable (Either a b) where
enumerate = enumerateEither enumerate enumerate
enumerateEither :: [a] -> [b] -> [Either a b]
enumerateEither [] ys = map Right ys
enumerateEither xs [] = map Left xs
enumerateEither (x:xs) (y:ys) = Left x : Right y : enumerateEither xs ys
The fact that we can do this both for (,) and Either probably means that we can do it for any algebraic data type.
Not every function has an inverse. If you limit the discussion to one-to-one functions, the ability to invert an arbitrary function grants the ability to crack any cryptosystem. We kind of have to hope this isn't feasible, even in theory!
In some cases, it is possible to find the inverse of a bijective function by converting it into a symbolic representation. Based on this example, I wrote this Haskell program to find inverses of some simple polynomial functions:
bijective_function x = x*2+1
main = do
print $ bijective_function 3
print $ inverse_function bijective_function (bijective_function 3)
data Expr = X | Const Double |
Plus Expr Expr | Subtract Expr Expr | Mult Expr Expr | Div Expr Expr |
Negate Expr | Inverse Expr |
Exp Expr | Log Expr | Sin Expr | Atanh Expr | Sinh Expr | Acosh Expr | Cosh Expr | Tan Expr | Cos Expr |Asinh Expr|Atan Expr|Acos Expr|Asin Expr|Abs Expr|Signum Expr|Integer
deriving (Show, Eq)
instance Num Expr where
(+) = Plus
(-) = Subtract
(*) = Mult
abs = Abs
signum = Signum
negate = Negate
fromInteger a = Const $ fromIntegral a
instance Fractional Expr where
recip = Inverse
fromRational a = Const $ realToFrac a
(/) = Div
instance Floating Expr where
pi = Const pi
exp = Exp
log = Log
sin = Sin
atanh = Atanh
sinh = Sinh
cosh = Cosh
acosh = Acosh
cos = Cos
tan = Tan
asin = Asin
acos = Acos
atan = Atan
asinh = Asinh
fromFunction f = f X
toFunction :: Expr -> (Double -> Double)
toFunction X = \x -> x
toFunction (Negate a) = \a -> (negate a)
toFunction (Const a) = const a
toFunction (Plus a b) = \x -> (toFunction a x) + (toFunction b x)
toFunction (Subtract a b) = \x -> (toFunction a x) - (toFunction b x)
toFunction (Mult a b) = \x -> (toFunction a x) * (toFunction b x)
toFunction (Div a b) = \x -> (toFunction a x) / (toFunction b x)
with_function func x = toFunction $ func $ fromFunction x
simplify X = X
simplify (Div (Const a) (Const b)) = Const (a/b)
simplify (Mult (Const a) (Const b)) | a == 0 || b == 0 = 0 | otherwise = Const (a*b)
simplify (Negate (Negate a)) = simplify a
simplify (Subtract a b) = simplify ( Plus (simplify a) (Negate (simplify b)) )
simplify (Div a b) | a == b = Const 1.0 | otherwise = simplify (Div (simplify a) (simplify b))
simplify (Mult a b) = simplify (Mult (simplify a) (simplify b))
simplify (Const a) = Const a
simplify (Plus (Const a) (Const b)) = Const (a+b)
simplify (Plus a (Const b)) = simplify (Plus (Const b) (simplify a))
simplify (Plus (Mult (Const a) X) (Mult (Const b) X)) = (simplify (Mult (Const (a+b)) X))
simplify (Plus (Const a) b) = simplify (Plus (simplify b) (Const a))
simplify (Plus X a) = simplify (Plus (Mult 1 X) (simplify a))
simplify (Plus a X) = simplify (Plus (Mult 1 X) (simplify a))
simplify (Plus a b) = (simplify (Plus (simplify a) (simplify b)))
simplify a = a
inverse X = X
inverse (Const a) = simplify (Const a)
inverse (Mult (Const a) (Const b)) = Const (a * b)
inverse (Mult (Const a) X) = (Div X (Const a))
inverse (Plus X (Const a)) = (Subtract X (Const a))
inverse (Negate x) = Negate (inverse x)
inverse a = inverse (simplify a)
inverse_function x = with_function inverse x
This example only works with arithmetic expressions, but it could probably be generalized to work with lists as well. There are also several implementations of computer algebra systems in Haskell that may be used to find the inverse of a bijective function.
No, not all functions even have inverses. For instance, what would the inverse of this function be?
f x = 1
I have a university course about functional programming, where I use SML. As a preparation for the exam, I am working on some of the older exam sets without solutions.
One of the only questions I really have problems with is the following question using foldl:
Consider the program skeleton: fun
addGt k xs = List.foldl (...) ... xs;
Fill in the two missing pieces
(represented by the dots ...), so that
addGt k xs is the sum of those
elements in xs, which are greater than
k. For example, addGt 4 [1, 5, 2, 7,
4, 8] = 5 + 7 + 8 = 20
I am sure this is really easy, but I have a very hard time understanding the foldl and foldr functions.
What I have now is the following (which seems to be very wrong if you ask my compiler!):
fun addGt(k,xs) = List.foldl ( fn x => if x > k then op+ else 0) 0 xs;
I would really appreciate some help with this question, and maybe a very short comment which would cast some light on the foldl and foldr functions!
A solution that I just though of is the following:
fun addGt(k, xs) = List.foldl (fn (x, y) => if x >= 5 then x + y else y) 0 xs;
But let me explain. First of all check the type of the List.foldl function, it's:
('a * 'b -> 'b) -> 'b -> 'a list -> 'b
So List.foldl is a curried function that takes as first parameter another function of type ('a * 'b -> 'b). You used (fn x => if x > k then op+ else 0) which has type int -> int. You should instead provide List.foldl with a function that takes a tuple of type int * int and returns an int, so something like this: (fn (x, y) => do stuff). That's why your code didn't compile, you passed a wrong type of function in foldl.
Now you can think of foldl this way:
foldl f b [x_1, x_2, ..., x_(n - 1), x_n] = f(x_n, f(x_(n - 1), ..., f(x2, f(x1, b)) ...)) where f is a function of type ('a * 'b -> 'b), b is something of type 'b and the list [x_1, x_2, ..., x_(n - 1), x_n] is of type 'a list.
And similar for foldr you can think it in this way:
foldr f b [x_1, x_2, ..., x_(n - 1), x_n] = f(x_1, f(x_2, ..., f(x_(n - 1), f(x_ n, b))
If you call foldl f s ls on a list, ls = [x1, x2, ..., xn], then you get the result:
f(xn, ... f(x2, f(x1, s)))
That is, it starts by finding
a1 = f(x1, s)
Then
a2 = f(x2, a1)
and so on, until it's through the list.
When it's done, it returns an.
You can think of the a-values as being a sort of accumulator, that is, ai is the result as it would be if the list was only [x1, x2, ..., xi] (or rather, the first i elements of the list).
Your function will usually have the form:
fn (x, a) => ...
What you then need to do is think: Okay, if I have the next element in the list, x(i+1), and the value ai, which is the result for the list [x1, x2, ..., xi], what do I need to do to find the value a(i+1), which is the result for the list [x1, x2, ..., xi, x(i+1)].
s can be thought of as the value given to the empty list.
foldr works the same way, only you start from the back of the list instead of from the front.