Can some one tell me, why i am getting the following error:
error: [-Wincomplete-patterns, -Werror=incomplete-patterns]
Pattern match(es) are non-exhaustive
In a case alternative: Patterns not matched: []
|
54 | case list of
| ^^^^^^^^^^^^...
Thats my test:
testMinBy :: Test
testMinBy = TestCase $ do
assertEqual "test1" (minBy (\x -> -x) [1,2,3,4,5]) 5
assertEqual "test2" (minBy length ["a", "abcd", "xx"]) "a"
minBy :: Ord b => (a -> b) -> [a] -> a
minBy measure list =
case list of
(x:y:xs) -> minBy measure (if measure x > measure y then y:xs else x:xs)
[x] -> x
Your pattern does not matches with the empty list. Indeed, that is what the error is saying. You can match the empty list, for example with:
minBy :: Ord b => (a -> b) -> [a] -> a
minBy measure list =
case list of
(x:y:xs) -> minBy measure (if measure x > measure y then y:xs else x:xs)
[x] -> x
[] -> error "Empty list"
Your function however is not very efficient: it will recalculate measure multiple times if an item is the current minimum, and will also pack and unpack lists. You can work with an accumulator here, like:
minBy :: Ord b => (a -> b) -> [a] -> a
minBy _ [] = error "Empty list"
minBy f (x:xs) = go xs x (f x)
where go [] y _ = y
go (y₁:ys) y₀ fy₀
| fy₁ < fy₀ = go ys y₁ fy₁
| otherwise = go ys y₀ fy₀
where fy₁ = f y₁
This means it only once has to check for an empty list, and then knows for sure that this is a non-empty list if it enumerates. It also will determine the f of each item exactly once, and uses accumulators to avoid packing and unpacking a "cons".
I want to write a function that returns the longest prefix of a list, where applying a function to every item in that prefix produces a strictly ascending list.
For example:
longestAscendingPrefix (`mod` 5) [1..10] == [1,2,3,4]
longestAscendingPrefix odd [1,4,2,6,8,9,3,2,1] == [1]
longestAscendingPrefix :: Ord b => (a -> b) -> [a] -> [a]
longestAscendingPrefix _ [] = []
longestAscendingPrefix f (x:xs) = takeWhile (\y z -> f y <= f z) (x:xs)
This code snippet produces the error message in the title. It seems the problem lies within that lambda function.
takeWhile has type takeWhile :: (a -> Bool) -> [a] -> [a]. The first parameter is thus a function that maps an element of the list to a Bool. Your lambda expression has type Ord b => a -> a -> Bool, which does not make much sense.
You can work with explicit recursion with:
longestAscendingPrefix :: Ord b => (a -> b) -> [a] -> [a]
longestAscendingPrefix f = go
where go [] = []
go [x] = …
go (x1:x2:xs) = …
where you need to fill in the … parts the last one makes a recursive call to go.
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 want to merge two functions into one.
The functions are
data Dice = Stone Char deriving (Show, Eq)
calculateScore :: [Dobbelsteen] -> Int
calculateScore xs = sum[giveValueDice x | x <- xs]
giveValueDice :: Dice -> Int
giveValueDice (Stone d) = if d == 'W' then 5 else digitToInt d
Normally, I would just copy one line into the first function en change a little bit to make the syntax correct. But here I am kinda lost how to do this
As you've already noticed, you can't just directly inline here; the reason for this is that you're pattern matching on Stone in giveValueDice. This pattern matching can be moved inside the right side of the list comprehension:
calculateScore :: [Dobbelsteen] -> Int
calculateScore xs = sum[if d == 'W' then 5 else digitToInt d | (Stone d) <- xs]
Another method for merging these two functions is using a where clause, which 'merges' one function into another while keeping them distince:
calculateScore :: [Dobbelsteen] -> Int
calculateScore xs = sum[giveValueDice x | x <- xs]
where
giveValueDice :: Dice -> Int
giveValueDice (Stone d) = if d == 'W' then 5 else digitToInt d
I have to derive the type of this function:
func x = map -1 x
And I've already found a way, using a tip to change it to a lambda expression:
func = \x -> (map) - (1 x)
If I express it like that, its fine and I get the same type as the original, but I'm not sure why its grouped like this. Could someone explain it?
For example, why isn't it like this:
func = \x -> (map - 1) x
or something similar.
I know it's a useless function etc. but I can't change the function, I just have to derive its type.
If you write this function in a file, eg:
test.hs has func x = map -1 x
and use :t func in the interpreter, it will reply:
func :: (Num (t -> (a -> b) -> [a] -> [b]),
Num ((a -> b) -> [a] -> [b])) =>
t -> (a -> b) -> [a] -> [b]
I now believe you meant to ask why
func x = map -1 x
has the type (Num (t -> (a -> b) -> [a] -> [b]), Num ((a -> b) -> [a] -> [b])) => t -> (a -> b) -> [a] -> [b], and how you can bracket the expression to make it have that type.
First, you have to recognise that the space is an operator in haskell, and has the highest precedence of all.
Let's use # instead of space, with highest precedence we can:
infixl 9 #
f # x = f x
We can replace and space without an operator with #:
func x = map - 1 # x
because the space between 1 and x was the only one without an operator (- is between map and 1).
Since # has higher precedence than -, we get
func x = map - (1 # x)
or equivalently
func x = map - (1 x)
Another example
func2 x = map (-1) x
> :t func2
func2 :: Num (a -> b) => [a] -> [b]
This translates as
func2' x = map # (-1) # x
but why isn't there a # between the - and the 1? In this case, - in front of a numeric literal like 1 means negate:
> (-1)
-1
> (negate 1)
-1
> (subtract 1)
<interactive>:73:1:
No instance for (Show (a0 -> a0))
arising from a use of `print'
Possible fix: add an instance declaration for (Show (a0 -> a0))
In a stmt of an interactive GHCi command: print it
So this function is trying to map the negative of 1 over a list. For that to work, it would need negative 1 to be a function, which is why it needs a numeric instance for functions (the Num (a->b) => at the start of the type).
but i'm not sure why its grouped like this. Could someone explain it? In example, why its not like that:
func = \x -> (map - 1) x
Precedence. The language definition specifies that the precedence of (prefix) function application is higher than that of any infix operator, so
map -1 x
is parsed as the application of the infix operator (-) to the two operands map and 1 x, like 3 + 4 * 5 is parsed 3 + (4 * 5) due to the higher precedence of (*) compared to that of (+).
Although the interpreter has assigned a type to the expression, it's not a sensible one. Let's see what the function should be
func x = map -1 x
looks like we want to bracket that like this
func x = map (-1) x
in the hope that it subtracts one from each element of a list, but unfortunately, the - is considered to be negation when it's in front of a numeric literal, so we need to bracket it to change it into the subtraction function:
func x = map ((-) 1) x
Now this function subtracts each number in the list from 1:
func [1,2,3]
=[(-) 1 1, (-) 1 2, (-) 1 3]
=[ 1-1, 1-2, 1-3]
=[ 0, -1, -2]
The type is
func :: Num a => [a] -> [a]
If you wanted to subtract one from each element of the list, rather than subtracting each element of the list from 1, you could use func x = map (subtract 1) x. As hammar points out, the subtract function exists exactly for the purpose of allowing this.
Your alternative
func = \x -> (map - 1) x
This can't work because (-) has type Num a => a -> a -> a, whereas map has type (a -> b) -> [a] -> [b]. You can't subtract one from a function, because a function isn't a numeric value.