I'm just learning haskell (on my own, for fun) and I've come up against a wall.
My Question:
How can I define a function
flrt = (floor . sqrt)
When I try it in a file and compile, GCHi complains with the following:
AKS.hs:11:9:
No instance for (RealFrac Integer)
arising from a use of `floor'
Possible fix: add an instance declaration for (RealFrac Integer)
In the first argument of `(.)', namely `floor'
In the expression: (floor . sqrt)
In an equation for `flrt': flrt = (floor . sqrt)
AKS.hs:11:17:
No instance for (Floating Integer)
arising from a use of `sqrt'
Possible fix: add an instance declaration for (Floating Integer)
In the second argument of `(.)', namely `sqrt'
In the expression: (floor . sqrt)
In an equation for `flrt': flrt = (floor . sqrt)
I don't understand why the resulting function isn't just Int -> Int.
I've just finished my second year of CS and done a basic PL course. I've heard of, but don't quite get types yet. I tried reading through a few haskell tutorials but it's all going above my head.
P.S. - I also don't understand what a monad is. (a lot of the other questions that my search turned up talked about these)
P.P.S. - My full source
bar = \a b -> if (2^a) > b
then (a-1)
else bar (a+1) b
foo = bar 1
flrt :: Integer -> Integer
flrt = (floor . sqrt)
aks target = if (target < 2)
then putStr "Not a Prime.\n\n"
else if elem (mod target 10) [0,2,4,5,6,8]
then putStr "Composite\n\n"
else if (elem target) [a^b | a <- [3,5..(flrt target)], b <- [1.. (foo target)]]
then putStr "Composite\n\n"--}
else
putStr "filler"
The problem is that you're trying to use Integer as the input. Haskell is strongly typed, which means there are no implicit coercions or conversions of any kind. Look at signatures of functions you're trying to compose:
sqrt :: Floating a => a -> a
floor :: (RealFrac a, Integral b) => a -> b
And at the signature of your function inferred by GHC:
> :t floor . sqrt
floor . sqrt :: (RealFrac b, Integral c, Floating b) => b -> c
So, to have a function that goes from Integer (which doesn't have a Floating instance) to Integer, you have to first convert your argument to Floating, which can be done by using fromIntegral:
> :t floor . sqrt . fromIntegral
floor . sqrt . fromIntegral :: (Integral a, Integral c) => a -> c
As copumpkin remarked, it might actually be a bad idea to convert to floating point here, because this comes with loss of precision and therefore might, even with rounding, yield incorrect results for sufficiently large integer inputs.
I assume all numbers you're dealing with will at least be small enough that there is some floating-point representation for them, e.g. all are < 10300. But, for instance
Prelude> round(sqrt.fromInteger$10^60 :: Double) ^ 2
1000000000000000039769249677312000395398304974095154031886336
Prelude> {- and not -} 10^60 {- == (10^30)^2 == (sqrt$10^60) ^ 2 -}
1000000000000000000000000000000000000000000000000000000000000
Which is way off, in terms of absolute difference. Still it's certainly a rather good approximation relative to the numbers themselves, so you can use it as a quickly determined starting point for an algorithm to find the exact result. You can implement Newton/Raphson (in this case AKA Heron) with Integers:
flrt :: Integer -> Integer -- flrt x ≈ √x, with flrt x^2 ≤ x < flrt(x+1)^2
flrt x = approx (round . (sqrt::Double->Double) . fromInteger $ x)
where approx r
| ctrl <= x, (r+1)^2 > x = r
| otherwise = approx $ r - diff
where ctrl = r^2
diff = (ctrl - x) // (2*r) -- ∂/∂x x² = 2x
a//b = a`div`b + if (a>0)==(b>0) then 1 else 0 -- always away from 0
This now works as desired:
*IntegerSqrt> (flrt $ 10^60) ^ 2
1000000000000000000000000000000000000000000000000000000000000
The division always away from 0 in the Newton-Raphson correction is here necessary to prevent getting stuck in an infinite recursion.
Related
Ugh. The following code fails to compile:
factorsOf number = [(x, quot number x) | x <- [2..toInteger $ floor $ sqrt number], number `mod` x == 0]
The following error is thrown:
"No instance for (Floating Integer) arising from a use of `sqrt'"
Please help? I'm clearly not grokking Haskell coercion.
PS: Leaving off toInteger compiles but throws a type-ambiguity error at runtime.
It is highly advisable to always start design of a Haskell function with the type signature, and only then write the implementation. In this case, you probably want
factorsOf :: Integer -> [(Integer, Integer)]
So, within factorsOf n = ..., the variable n will have type Integer. That's the problem: you're trying to take the square root of an integer, but sqrt is only defined on floating numbers. So you need to convert to such a number before taking the root. After the root, you'll then want to truncate back to an integer, but floor already does that. toInteger is not needed.
factorsOf :: Integer -> [(Integer, Integer)]
factorsOf n
= [ (x, n`quot`x)
| x <- [2 .. floor . sqrt $ fromIntegral n]
, n `mod` x == 0
]
I am teaching myself Haskell, and playing with list comprehensions. I wrote this list comprehension:
[ c | a <- [1..3], b <- [1..4], let c = hyp a b, c == round c]
I hoped it would produce the list of values c where c is an integer (c == round c), which would be only 5, but it does not compile. Playing around some more, I've found I can't really embed any functions in a list comprehension, I'm sure there is a way, I just don't know how.
Here's the error code:
<interactive>:158:1: error:
• Ambiguous type variable ‘t0’ arising from a use of ‘it’
prevents the constraint ‘(Floating t0)’ from being solved.
Probable fix: use a type annotation to specify what ‘t0’ should be.
These potential instances exist:
instance Floating Double -- Defined in ‘GHC.Float’
instance Floating Float -- Defined in ‘GHC.Float’
• In the first argument of ‘print’, namely ‘it’
In a stmt of an interactive GHCi command: print it
Thanks!
First of all, include necessary definitions in a question like this.
hyp :: Floating a => a -> a -> a
hyp a b = sqrt $ a^2 + b^2
Now. You can “embed” functions all right in list comprehensions. Apparently you just chose some unfortunate ones! round has the following type:
GHCi, version 7.10.2: http://www.haskell.org/ghc/ :? for help
Prelude> :t round
round :: (Integral b, RealFrac a) => a -> b
So, for round c == c to make sense, you'd need to have a number type that's both an instance of Integral and of RealFrac. In other words, this type would contain fractions, yet all its elements would be integers†. Well, you can't have your cake and eat it too!
This problem would have been much more evident, like so often in Haskell, if you had actually written out some type signatures. While fooling around like that, it often helps to just select some simple example type. Perhaps the most reasonable thing would seem:
Prelude> [ c | a <- [1..3], b <- [1..4], let c = hyp a b, c == round c] :: [Integer]
<interactive>:6:41:
No instance for (Floating Integer) arising from a use of ‘hyp’
In the expression: hyp a b
In an equation for ‘c’: c = hyp a b
In the expression:
[c |
a <- [1 .. 3], b <- [1 .. 4], let c = hyp a b, c == round c] ::
[Integer]
Ok, so Integer doesn't work because you're trying to do real arithmetic, with those square roots in hyp. This is not possible with Integer, you need a Floating type like Double. Let's try that:
Prelude> [ c | a <- [1..3], b <- [1..4], let c = hyp a b, c == round c] :: [Double]
<interactive>:8:55:
No instance for (Integral Double) arising from a use of ‘round’
In the second argument of ‘(==)’, namely ‘round c’
In the expression: c == round c
In a stmt of a list comprehension: c == round c
Ok, this is because as I said, round always gives integral-type results. However, you can always convert such an integral type to a Double again:
Prelude> [ c | a <- [1..3], b <- [1..4], let c = hyp a b, c == fromIntegral (round c)] :: [Double]
[5.0]
Note that this is not really a good solution though: you don't really want the result to be floating point if you go already check that the elements are really integral! I'd recommend in this case not just evaluating hyp as such at all. Better use this comprehension:
Prelude> [ c | a <- [1..3], b <- [1..4], let c² = a^2 + b^2; c = round . sqrt $ fromIntegral c², c^2==c²] :: [Integer]
[5]
One big argument for this version is that it does the comparison in Integer, not in Double. Floating-point equality comparison is something you should best stay away from entirely if you can help it; in this case it's mostly harmless because the interesting thing is the integer subset, which can in fact be represented exactly (unlike decimal fractions like 0.1). Still you ca get wrong results this way: in particular, for sufficiently large floating-point numbers, c == fromInteger (round c) will always be true, because above a certain threshold all values are integral after all.
Prelude> [ c | a <- take 4 [1000000000000..], b <- take 5 [1000000000000..], let c = hyp a b, c == fromIntegral (round c)] :: [Float]
[1.4142135e12,1.4142135e12,1.4142135e12,1.4142135e12,1.4142135e12,1.4142135e12,1.4142135e12,1.4142135e12,1.4142135e12,1.4142135e12,1.4142135e12,1.4142135e12,1.4142135e12,1.4142135e12,1.4142135e12,1.4142135e12,1.4142135e12,1.4142135e12,1.4142135e12,1.4142135e12]
But none of these are actually correct integral hypothenuses, as you can see with the version that does the comparison in Integer:
Prelude> [ c | a <- take 4 [1000000000000..], b <- take 5 [1000000000000..], let c² = a^2 + b^2; c = round . sqrt $ fromIntegral c², c^2==c²] :: [Integer]
[]
Strictly speaking, this improved version is also not safe though – it doesn't give false positives, but may not find actual Pythagorean triples because the lossy floating-point steps can already destroy equality. To do it real properly, you need an all-integral
intSqrt :: Integral a => a -> Maybe a
This could probably done quite efficiently by taking the float sqrt as a starting value for a few rounds of pseudo Newton-Raphson in integer arithmetic.
†In principle, the round function could also have a more relaxed signature.
round' :: (Num b, RealFrac a) => a -> b
round' = fromInteger . round
With that version, your original code would work.
I'm new to Haskell and I'm writing a program that calculates the limit of a function. So given two lists a and b, a delta dx = 0.001, and the limits of integration l and r, I want to recursively compute the area under the curve with equation:
a1(x)^b1 + a2(x)^b2 + ... + an(x)bn where x is all the values between l an r with an increment of dx between each value. The technical part isn't that important I guess but it helps to read the code:
import Text.Printf (printf)
-- This function should return a list [area].
solve :: Int -> Int -> [Int] -> [Int] -> [Double]
solve l r x y = [area l r x y]
area l r a b = if (l < r)
then (calc l a b) * 0.001 + (area (l + 1) r a b)
else (calc r a b) * 0.001
calc n (a:arest) (b:brest) = (fromIntegral(n) ^^ b) * fromIntegral(a) + (calc n arest brest)
calc n [] [] = 0
--Input/Output.
main :: IO ()
main = getContents >>= mapM_ (printf "%.1f\n"). (\[a, b, [l, r]] -> solve l r a b). map (map read. words). lines
I get no error with the above code but as soon as I change area (l + 1) r a b to area (l + 0.001) r a b I get the following error message:
No instance for (Fractional Int) arising from a use of `area'
I tried making a new class and having a be an abstract type but that didn't work, any other ideas?
So the problem is that Int is not a Fractional type. In other words, it does not have a value called 0.001 [note 1], but you have requested Haskell to give you such a value in your code.
You are making this request because 0.001 is fed to the (+) function with another argument (in this case l) which is of type Int. This is a problem because the function has type (+) :: (Num a) => a -> a -> a: in other words, there are a lot of different functions (+) all having the type a -> a -> a; one of these functions exists for every type a in the Num type class.
Since we know that one argument to the function is an Int, it follows that we're using the specific function (+) :: Int -> Int -> Int. That is why l + 0.001 gets weird.
As for solving the problem: You probably wanted l and r to be of type Double (they're left and right bounds on where a number can be?) but if you're sure that they must be Ints then you probably meant to write fromIntegral l + 0.001.
Side note on style: parentheses in Haskell are always just grouping/precedence, functions are higher precedence than operators which are higher precedence than special forms (let, case, if, do), and function application is always left-associative or "greedy nom": a function eats whatever is immediately in front of it. What you have written:
(fromIntegral(n) ^^ b) * fromIntegral(a) + (calc n arest brest)
is probably better written as:
fromIntegral a * fromIntegral n ^^ b + calc n arest brest
The parentheses around calc are not necessary (because operators like + have lower precedence than function applications), nor are the parentheses around n and a (because those sub-expressions are indivisible chunks; fromIntegral(n) is identical to fromIntegral (n) is identical to fromIntegral n).
As #dfeuer mentions below: secretly, when you write 0.001 it does not have a definite type; rather it is translated to fromRational 0.001 internally, where the latter 0.001 is a definite value of the definite type Rational, just as when you write 4 it is translated to fromInteger 4 where the latter 4 is a definite value of the definite type Integer. The problem is really that there is no fromRational function for Int, because Int is not part of the Fractional typeclass which defines fromRational. And it's not part of that typeclass because the language designers preferred an error to a silent rounding/dropping of a fraction.
I'm trying to write a program in Haskell
that gets a list (of integer) and prints out the number of elements that are bigger than the list's average
So far I tried
getAVG::[Integer]->Double
getAVG x = (fromIntegral (sum x)) / (fromIntegral (length x))
smallerThanAVG:: [Integer]->Integer
smallerThanAVG x = (map (\y -> (if (getAVG x > y) then 1 else 0)) x)
For some reason I'm getting this error
Couldn't match expected type `Double'
against inferred type `Integer'
Expected type: [Double]
Inferred type: [Integer]
In the second argument of `map', namely `x'
It could be that I haven't written the logic correctly, although I think I did..
Ideas?
These errors are the best kind, because they pinpoint where you have made a type error.
So let's do some manual type inference. Let's consider the expression:
map (\y -> (if (getAvg x > y) then 1 else 0)) x
There are a few constraints we know off the bat:
map :: (a -> b) -> [a] -> [b] -- from definition
(>) :: Num a => a -> a -> Bool -- from definition
getAvg :: [Integer] -> Double -- from type declaration
1, 0 :: Num a => a -- that's how Haskell works
x :: [Integer] -- from type declaration of smallerThanAVG
Now let's look at the larger expressions.
expr1 = getAvg x
expr2 = (expr1 > y)
expr3 = (if expr2 then 1 else 0)
expr4 = (\y -> expr3)
expr5 = map expr4 x
Now let's work backwards. expr5 is the same as the RHS of smallerThanAVG, so that means it has the same result type as what you've declared.
expr5 :: Integer -- wrong
However, this doesn't match our other constraint: the result of map must be [b] for some b. Integer is definitely not a list (although if you get facetious, it could be coerced into a list of bits). You probably meant to sum that list up.
expr6 = sum expr5
sum :: Num a => [a] -> a
Now let's work forwards.
expr1 :: Double -- result type of getAvg
y :: Double -- (>) in expr2 requires both inputs to have the same type
expr4 :: (Integer -> [a]) -- because for `map foo xs` (expr5)
-- where xs :: [a], foo must accept input a
y :: Integer -- y must have the input type of expr4
Herein lies the conflict: y cannot be both a Double and an Integer. I could equivalently restate this as: x cannot be both a [Double] and [Integer], which is what the compiler is saying. So tl;dr, the kicker is that (>) doesn't compare different types of Nums. The meme for this sort of problem is: "needs more fromIntegral".
(getAvg x > fromIntegral y)
Your code has two errors.
Although the type signature in the code declares that smallerThanAVG x evaluates to an Integer, its code is map ... x, which clearly evaluates to a list instead of a single Integer.
In the code getAVG x > y, you are comparing a Double to an Integer. In Haskell, you can only compare two values of the same type. Therefore, you have to use fromIntegral (or fromInteger) to convert an Integer to a Double. (This is essentially what caused the error message in the question, but you have to get used to it to figure it out.)
There are several ways to fix item 1 above, and I will not write them (because doing so would take away all the fun). However, if I am not mistaken, the code you are aiming at seems like counting the number of elements that are smaller than the average, in spite of what you write before the code.
Styling tips:
You have many superfluous parentheses in your code. For example, you do not have to parenthesize the condition in an if expression in Haskell (unlike if statement in C or Java). If you want to enjoy Haskell, you should learn Haskell in a good style instead of the Haskell which just works.
You call the variable which represents a list “x” in your code. It is conventional to use a variable name such as xs to represent a list.
Others have explained the errors beautifully.
I'm still learning Haskell but I'd like to provide an alternative version of this function:
greaterThanAvg :: [Int] -> [Int]
greaterThanAvg xs = filter (>avg) xs
where avg = sum xs `div` length xs
cnt = length $ greaterThanAvg [1,2,3,4,5]
This is not a duplicate question. Read below...
I'm declaring the following function:
divisors x = [(a, x/a) | a <- [2..(sqrt x)], x `mod` a == 0]
What I want to obtain is the divisors of x: A list of tuples that will contain (n, k) such as n * k = x
Example:
> divisors x
[(1,10), (2, 5)]
Why the above code isn't working?
It gives me the error:
*Main> divisors 10
<interactive>:1:0:
Ambiguous type variable `t' in the constraints:
`Floating t'
arising from a use of `divisors' at <interactive>:1:0-10
`Integral t'
arising from a use of `divisors' at <interactive>:1:0-10
Probable fix: add a type signature that fixes these type variable(s)
I've tried manually setting the signature of the function without success...
The problem is sqrt returns a Floating a, and you really just want integers when finding divisors. You can turn a Floating a into an Integral a with ceiling, floor or round. I will use ceiling, as I'm not sure if using floor or average won't skip a divisor.
The sqrt function also only accepts a floating number, so you will have to convert an integer into a floating before giving it to it (this can be done with fromIntegral).
Also, you use /, which also works with floating numbers. Using div is better as it works with integral numbers (rounding when necessary).
divisors x = [(a, x `div` a) | a <- [2..(ceiling $ sqrt $ fromIntegral x)], x `mod` a == 0]
With this, divisors 10 will give [(2,5)] (your code stops the (1,10) case from happening - I'm guessing this was intentional). Unfortunately you will get duplicates, eg divisors 12 will return [(2,6),(3,4),(4,3)], but that shouldn't be too hard to fix if it is a problem.
You can see the problem if you ask for the type:
divisors :: (Integral t, Floating t) => t -> [(t, t)]
and then check what things are both Integral and Floating:
Prelude> :info Floating
class Fractional a => Floating a where
instance Floating Float -- Defined in GHC.Float
instance Floating Double -- Defined in GHC.Float
and
Prelude> :info Integral
class (Real a, Enum a) => Integral a where
instance Integral Integer -- Defined in GHC.Real
instance Integral Int -- Defined in GHC.Real
so, it can be neither Int, Integer, Float or Double. You're in trouble...
Thankfully, we can convert between types, so that while sqrt needs a Floating, and mod needs an Integral (btw, rem is faster), we can either, e.g., do away with floating point division:
divisors :: Integer -> [(Integer, Integer)]
divisors x = [(a, x `div` a) | a <- [2..ceiling (sqrt (fromIntegral x))], x `rem` a == 0]
> divisors 100
[(2,0),(4,0),(5,0),(10,0)]
However, you need to think hard about what you really mean to do when converting integer types to floating point, via sqrt...
In Haskell, integer division and fractional division are different operations, and have different names. The slash operator, /, is for fractional division. Integer division is accomplished with div or quot (the difference between the two having to do with the behavior when there are negative numbers involved).
Try replacing x/a with
x `quot` a
instead.
The compiler error tells you exactly this: that you're treating a type sometimes as an integral number (by using mod), and sometimes as a fractional number (by using /), and it's not sure how to pick a type that acts like both of those.
You'll have a similar issue with sqrt, once that's sorted, though. There again, you need to be consistent about whether your types are integers or (in that case) floating point. For the purpose of finding possible divisors, it should suffice to range up to the greatest integer less that the floating point, so consider using floor (sqrt (fromIntegral x))). The fromIntegral converts x (which must have an integral type) to a different type -- in this case, it will default to Double. The floor then converts the Double result back into an integral type.
Instead of taking the square-root to bound the search, you can allow the comprehension to range over an infinite list, and use takeWhile to stop the search when the remainder is greater than the divisor:
divisors x = takeWhile (uncurry (<=)) [(a, x `div` a) | a <- [1..], x `mod` a == 0]
> divisors 100
[(1,100),(2,50),(4,25),(5,20),(10,10)]
Note: your original example shows (1,10) as one of the divisors of 10, so I started the comprehension from 1 instead of 2.
Hmm, this does search beyond the square-root until it hits the next factor above.
How about this:
divisors x = [(a, x `div` a) | a <- takeWhile ((<= x) . (^2)) [1..], x `mod` a == 0]