I have the following two type signatures in Haskell:
foo :: (a -> (a,b)) -> a -> [b]
bar :: (a -> b) -> (a -> b -> c) -> a -> c
I want to write a concrete implementation of these two functions but I'm really struggling to understand where to start.
I understand that foo takes a function (a -> (a,b)) and returns a and a list containing b.
And bar takes a function (b -> c) which returns a function (a -> b -> c) which finally returns a and c.
Can anyone show me an example of a concrete implementation?
How do I know where to start with something like this and what goes on the left side of the definition?
You have some misunderstandings there:
I understand that foo takes a function (a -> (a,b)) and returns a and a list containing b.
No, it doesn't return a. It expects it as another argument, in addition to that function.
And bar takes a function (b -> c) which returns a function (a -> b -> c) which finally returns a and c.
Same here. Given g :: a -> b, bar returns a function bar g :: (a -> b -> c) -> a -> c. This function, in turn, given a function h :: (a -> b -> c), returns a function of type a -> c. And so it goes.
It's just like playing with pieces of a puzzle:
foo :: (a -> (a,b)) -> a -> [b]
-- g :: a -> (a,b)
-- x :: a
-- g x :: (a,b)
foo g x = [b] where
(a,b) = g x
bar :: (a -> b) -> (a -> b -> c) -> a -> c
-- g :: a -> b
-- x :: a
-- g x :: b
-- h :: a -> b -> c
-- h x :: b -> c
-- h x (g x) :: c
bar g h x = c where
c = ....
There's not much free choice for us here. Although, there are more ways to get more values of type b, for foo. Instead of ignoring that a in (a,b) = g x, we can use it in more applications of g, so there actually are many more possibilities there, like
foo2 :: (a -> (a,b)) -> a -> [b]
foo2 g x = [b1,b2] where
(a1,b1) = g x
(a2,b2) = g a1
and many more. Still, the types guide the possible implementations. foo can even make use of foo in its implementation, according to the types:
foo3 :: (a -> (a,b)) -> a -> [b]
foo3 g x = b : bs where
(a,b) = g x
bs = ...
So now, with this implementation, the previous two become its special cases: foo g x === take 1 (foo3 g x) and foo2 g x === take 2 (foo3 g x). Having the most general definition is probably best.
In addition to #will-nes's answer, it will be useful to treat (->) as a right-associative infix operator. So something like f: a -> b -> c is the same as f: a -> (b -> c). So this is saying f is a function that takes a value of type a and returns you a value of type b -> c, which is, another function, one that takes a value of type b and returns you a value of type c.
So the types in your example can be re-written as follows
foo :: (a -> (a,b)) -> (a -> [b])
bar :: (a -> b) -> ((a -> (b -> c)) -> (a -> c))
Similarly, you can think of arguments to a function in pieces as well, as being left-associative (like + and -), though there's no explicit operator in this case. foo a b c d e is the same as ((((foo a) b) c) d) e. For example, let's say we have a function f: Int -> Int -> Int (which is the same as f: Int -> (Int -> Int)). You don't have to provide both arguments at once. So you can write g = f 1, which has the type (Int -> Int). And then you can provide an argument to g, like g 2, which has the type Int. f 1 2 and let g = f 1 in g 2 are more or less the same. Here's a more concrete example of how this works:
Prelude> f = (+)
Prelude> g = f 1
Prelude> g 2
3
Prelude> :t f
f :: Num a => a -> a -> a
Prelude> :t g
g :: Num a => a -> a
Prelude> :t g 2
g 2 :: Num a => a
In #will-nes's sample implementation examples, he defines the functions with all of the arguments up front, but you don't have to think of them that way. Just think of f: a -> b -> c as taking a value of type a and returning another function. While most of the methods you encounter will use all of their arguments up-front, there might be cases in which you don't want to do that. Here's an example:
veryExpensive :: A -> B
unstagedFun :: A -> (B -> C) -> C
unstagedFun a f = f (veryExpensive a)
stagedFun :: A -> (B -> C) -> C
stagedFun a = let b = veryExpensive a in \f -> f b
(You can also rewrite the latter as let b = veryExpensive a in ($ b))
Of course, with compiler optimizations, I wouldn't be surprised if the unstaged version staged automatically, but hopefully this offers some motivation for thinking of functions as not having multiple arguments, but rather, as a single argument, but they may return other functions that may themselves return functions (but also only take a single argument).
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've defined a function f1 and f2,so that I can use at the and the function composition (fkomp), which is supposed to use f1 and f2 to calculate 2^x by every element in a given List.
f1 :: Int -> Int
f1 x = product (replicate x 2)
f2 :: (a -> b) -> [a] -> [b]
f2 f xs = [f x | x <- xs]
fkomp :: [Int] -> [Int]
fkomp xs = f2 f1 $ xs
It works,but the problem is,that i can't write my code with composition:
fkomp xs = f2.f1 $ xs
I've been typing every single combination but it doesn't work with composition.
Could someone lighten my path ?
Thanks a lot
Ok, let's look just at the types (it's like a puzzle - the types have to fit):
f1 :: Int -> Int
f2 :: (a -> b) -> [a] -> [b] = (a -> b) -> ([a] -> [b])
in order to compose the both you need ones co-domain to be the same as the others domain.
This is because the composition has type:
(.) :: (b -> c) -> (a -> b) -> a -> c
See the b has to fit ;)
So for your f1 and f2 you would need either Int ~ (a -> b) or Int ~ ([a] -> [b]) both of which are not working well (as you found out).
BUT you kind of have the ability to apply f1 to f2 as f1 just fits f2 first argument (as you have seen too) - so I am a bit confused why you even want to use composition here.
remarks
your functions are a bit strange - I think the usual way to write them would be
f1 x = 2 ^ x
f2 = map
or even
fkomp :: [Int] -> [Int]
fkomp = map (2^)
note that the last one is not function-composition but (just as your case) function-application: I apply the function (2^) :: Int -> Int to map :: (Int -> Int) -> [Int] -> [Int] and get a function of type [Int] -> [Int] as the result (if you check the types in GHCi you will see a more generic versions but I think this is a bit more clear)
I have this code:
esprimo :: Int->Bool
esPrimo x = if length (div x x) == 2 then True else False
But I pulled the error is above
In addition to what sibi said, I think what you are trying to do is this:
isPrime :: Int -> Bool
isPrime x = if length [d | d <- [1..x], x `mod` d == 0] == 2 then True else False
this is basically the direct translation of the mathematical concept of beeing prime into Haskell.
As you don't need the if as it checks the same == already returns a bit more readable might be:
isPrime :: Int -> Bool
isPrime x = length divisors == 2
where divisors = [d | d <- [1..x], x `isMultipleOf` d]
isMultipleOf m n = m `mod` n == 0
Please note that this is of course not the most performant prime-test.
The exact reason for your error is because of the different cases you have used in the type signature and the type definition:
esprimo :: Int -> Bool -- p should be capital here to work.
esPrimo x = if length (div x x) == 2 then True else False
Haskell is case sensitive, so esprimo and esPrimo are different. That being said there is other type error in your code: the type of div is div :: Integral a => a -> a -> a, so it returns a and you are applying length function on it. But length function only accepts list i.e [a] and not a which will produce you type error.
I have a funciton say foo :: [Integer] -> Bool , but it only works if the incoming argument is valid for some criteria, otherwise it should terminate immediately.
foo x | not $ isSorted x = False
| otherwise = some_recursive_stuff_here
where
isSorted ax = ax == sort ax
etc.
But I don't want to check invariant every time if it is sorted or not. Is there a good way to deal with that other then introducing another internal function?
You can carry around "proof" that your invariant holds by creating a newtype.
newtype Sorted a = Sorted { fromSorted :: [a] }
sorted :: Ord a => [a] -> Sorted a
sorted = Sorted . sort
foo :: Sorted Integer -> Bool
foo (Sorted as) -> some_recursive_stuff_here
If you hide the Sorted constructor in a separate module then users of your code will be unable to use foo without creating proof of sorting first. They also won't be able to sort a Sorted so you can be sure it's only occurred once.
You can even support proof-maintaining operations if you like.
instance Monoid (Sorted a) where
mempty = Sorted mempty
mappend (Sorted as) (Sorted bs) = Sorted (go as bs) where
-- lazy stable sort
go :: Ord a => [a] -> [a] -> [a]
go [] xs = xs
go xs [] = xs
go (x:xs) (y:ys) | x == y = x : y : go xs ys
| x < y = x : go xs (y:ys)
| x > y = y : go (x:xs) ys
(This code is available now on Hackage: http://hackage.haskell.org/package/sorted)