When trying to solve this challenge I stumbled upon something I was not able to explain myself.
First I generate an infinite list of prime numbers as follows:
primes = [n|n<-[2..],product[1..n-1]`rem`n==n-1]
This has the inferred type [Integer] so Int-overflow should not be a problem.
Then I try to make 2-tuples of subsequent primes (goal: [(2,3),(5,7),...]). To achieve this I wrote another function:
listtotuples l=[ (l!!i, l!!(i+1) ) |i<-[0,2..]]
Strangely this listtotuples function seems to work fine on e.g. [0..], but it just stops working when I apply it to primes, the output is just (after interrupting)
[(2,3),(5,7),(11,13),(Interrupted.
I do not understand why this happens, can anyone explain?
EDIT: This does not only happen when trying to output the infinite list, but also e.g. using take 10 $ listtotuples primes in Prelude after having loaded a file with the two lines from above. It does get stuck at the exact same point.
I am using Windows 7 with GHCi 7.10.2.
EDIT2: The full contents of my file are as follows:
order p m=head[n-1|n<-[0..],mod m (p^n)>0]
primes = [n|n<-[2..],product[1..n-1]`rem`n==n-1]
listtotuples l=[ (l!!i, l!!(i+1) ) |i<-[0,2..]]
p=listtotuples primes
f n=product[r^(order s n) * s^(order r n)|(r,s)<-take n p]
The problem disappears as soon as I comment/remove the last line (the function f, but I still think this is very strange, as f is not called and does not have anything todo with the functions above. Also if I replace take n p in the function f with [(2,3)] everything works as defined.
The addition of f forces your primes to have type Int, which does overflow during the factorial operation. The reasoning goes like this:
take :: Int -> [a] -> [a]
In f n, the use of take n p forces n :: Int.
Because the arguments to order must have the same type, the calls order r n and order s n force r, s :: Int.
(r, s) <- take n p forces p :: [(Int, Int)]
p = listtotuples primes forces primes :: [Int]
Simple fixes include breaking steps 2 or 3 above; use take (fromInteger n) p to break step 2 or order s (fromIntegral n) and order r (fromIntegral n) to break step 3.
...and now you know why adding top-level type signatures is considered a best practice. =)
Related
I'm learning Haskell and there's a lot of type-checking that seems completely nonsensical to me. I have written a simple function to count the number of occurrences of a given element in a given list, as such:
-- Count the number of occurrences of an element in a list.
countOcc :: (Eq a) => [a] -> a -> Int
countOcc xs x = length $ filter (== x) xs
Now, using this explicitly with calls such as:
countOcc "str" 's'
This executes fine, and returns correctly. However, this causes an error:
countOcc "str" "str"!!0
I haven't the foggiest why this should cause an error. "str"!!0 gives 's', a Char, which is exactly the same type passed in the second parameter of the first call.
I'm sure there are some nuances to Haskell's type system that I'm overlooking, or haven't broached yet. Ideally, I'd like to know why this is erroneous and furthermore, I'd like to know, according to Haskell's ideology, why it should be erroneous.
The following works fine:
countOcc :: (Eq a) => [a] -> a -> Int
countOcc xs x = length $ filter (== x) xs
main = print $ countOcc "str" ("str"!!0) -- 1
As far as I know, function applictaion has the highest precedence; although !! has precedence level of 9, it is still lower than function application.
I have a recursive Haskell function that takes a number n and generates a list that is n long of squared numbers from 1 up.
The code works but instead of the list for n=3 being [9,4,1] I want it to be [1,4,9].
sqNList :: Int -> [Int]
sqNList 0 = []
sqNList n = n*n : sqNList (n-1)
I've tried swappig the : for a ++ but then the pattern matching doesn't work. I have only just started Haskell today so may be obvious
The best approach is to implement sqNList in terms of a helper function that counts from 1 up to n. That way you can just generate the list in the correct order in the first place.
sqNList :: Int -> [Int]
sqNList n = sqNListHelper 1 n
sqNListHelper :: Int -> Int -> [Int]
sqNListHelper current end = ...
There are a wide variety of more sophisticated approaches, but this is the easiest approach to debug/test interactively that also has you figuring out how to do everything yourself.
More sophisticated approaches can include list comprehensions or composing functions like map and enumFromTo to build up the logic from simpler pieces.
The easiest approach that consumes probably the least amount of memory is to work with an accumulator: a parameter you pass and update through every recursive call.
Right now you use n as an accumulator, and decrement that, but we can decide to use an accumulator i instead that starts at 1, and keeps incrementing:
helper :: Int -> Int -> [Int]
helper i n | i > n = []
| otherwise = i*i : helper (i+1) n
Now we of course have to call it with i = 1 ourselves, which is not ideal, but we can use a where clause that scopes the helper in the sqNList:
sqNList :: Int -> [Int]
sqNList n = helper 1 n
where helper :: Int -> Int -> [Int]
helper i n | i > n = []
| otherwise = i*i : helper (i+1) n
Now we can improve this a bit. For instanc there is no need to pass n through the helper calls, since it does not change, and is defined at the top level:
sqNList :: Int -> [Int]
sqNList n = helper 1
where helper :: Int -> [Int]
helper i | i > n = []
| otherwise = i*i : helper (i+1)
There is furthermore no need to only work with Ints: we can use any type a for which Num a and Ord a:
sqNList :: (Num a, Ord a) => a -> [a]
sqNList n = helper 1
where helper i | i > n = []
| otherwise = i*i : helper (i+1)
This is a more advanced technique, so you may want to just save this for future study.
Most recursive functions tend to use one of the same general forms of recursion, so knowing which higher-order function to use saves you the effort of reimplementing (possibly incorrectly) the recursion, and lets you focus on the work that is unique to your problem.
This particular type of recursion is captured by the unfoldr function, defined in Data.List, which lets you generate a (possibly infinite) list using a generator function and an initial seed value.
Its definition can be as simple as
unfoldr f x = case f x of
Nothing -> []
Just (new, next) = new : unfoldr f next
although the actual definition is slightly more complicated for efficiency.
Essentially, you just call the generator f on the seed value x. If the result is Nothing, return an empty list. Otherwise, build a list containing a result and the list produced by a recursive call with a new seed value.
Applying this to your problem, we produce a square and the next integer as long as the input is still small enough.
import Data.List (unfoldr)
sqNList n = unfoldr generator 1
where generator x = if x > n then Nothing else Just (x*x, x+1)
So for a simple example like sqNList 3, it proceeds as follows:
Call generator 1 and get back Just (1, 2); the accumulator is now [1].
Call generator 2 and get back Just (4, 3); the accumulator is now
[1,4].
Call generator 3 and get back Just (9, 4); the accumulator is now [1,4,9].
Call generator 4 and get back Nothing. Return the accumulator [1,4,9] as the result.
Note that you generate the infinite list of all squares simply by never returning Nothing:
allSquares = unfoldr (\x -> Just (x*x, x+1)) 1
Haskell, being lazy, only generates elements in the list as you need them, so you could also define sqNList by taking only the first n items of the infinite list.
sqNList n = take n (unfoldr (\x -> Just (x*x, x+1)) 1)
I've written the following Haskell code to produce a list where the nth element is the number of 1s in writing 1..n as binary numbers (it's related to euler 391, incidentally):
buildList :: a -> (a -> a) -> [a]
buildList start f = start : buildList (f start) f
differences :: [[Int]]
differences = buildList [0] (\x -> x ++ map (+1) x)
sequenceK' :: Int -> [Int]
sequenceK' n = tail $ scanl (+) 0 (last $ take n differences)
which results in sequenceK' n giving a list of 2^(n-1) elements.
This question has two parts:
a) Why does the time taken to compute head $ sequenceK' n increase with n? -- due to ghc's laziness, I would expect the time to remain more or less constant.
b) Is it possible to define an infinite version of this list so that I can do things like take and takeWhile without having to worry about the value of the parameter passed to sequenceK'?
a) Because you're calling last $ take n differences, which has to do more work the bigger n is.
b) Yep, it's possible. The least-thinking solution is to just take the earliest element we see at each particular depth:
*Main> take 20 . map head . transpose $ differences
[0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,1,2,2,3]
The better solution is to generate only the meaningful bits. We can do this by observing the following equality:
differences' = 1 : (differences' >>= \x -> [x, x+1])
Actually, this is slightly off, as you can probably guess:
*Main> take 20 differences'
[1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,1,2,2,3,2,3]
But it's easily fixed by just tacking a 0 on front.
I have the following list (it’s a length 2 list, but in my assignment I have a length +n list)
xxs = [(11,22,[(33,33,33),(44,44,44)]),(55,66,[(77,77,77),(88,88,88)])]
I’m trying to “replace” one 3-tuple (p1 or p2 or p3 or p4 from the image bellow) by list index (n) and by sub-list index (p).
The function, at the end, should be like:
fooo newtuple n p = (…)
For example: (replace p3 for (98,98,98):
fooo (98,98,98) 2 1
[(11, 22, [(33,33,33) , (44,44,44)]) , (55, 66, [(98,98,98),(88,88,88)])]
I planned the code like following this steps:
Access the pn that I want to change. I manage to achieve it by:
fob n p = ((aux2 xxs)!!n)!!p
where aux2 [] = []
aux2 ((_,_,c):xs) = c:aux2 xs
“replace” the 3-tuple. I really need some help here. I’m stuck. the best code (in my head it makes some sense) that I’ve done: (remember: please don’t be too bad on my code, I’ve only been studying Haskell only for 5 weeks)
foo n p newtuple = fooAux newtuple fob
where fooAux _ [] = []
fooAux m ((_):ds) = m:ds
fob n p = ((aux2 xxs)!!n)!!p
where aux2 [] = []
aux2 ((_,_,c):xs) = c:aux2 xs
Finally I will put all back together, using splitAt.
Is my approach to the problem correct? I really would appreciate some help on step 2.
I'm a bit new to Haskell too, but lets see if we can't come up with a decent way of doing this.
So, fundamentally what we're trying to do is modify something in a list. Using functional programming I'd like to keep it a bit general, so lets make a function update.
update :: Int -> (a -> a) -> [a] -> [a]
update n f xs = pre ++ (f val) : post
where (pre, val:post) = splitAt n xs
That will now take an index, a function and a list and replace the nth element in the list with the result of the function being applied to it.
In our bigger problem, however, we need to update in a nested context. Luckily our update function takes a function as an argument, so we can call update within that one, too!
type Triple a = (a,a,a)
type Item = (Int, Int, [Triple Int])
fooo :: Triple Int -> Int -> Int -> [Item] -> [Item]
fooo new n p = update (n-1) upFn
where upFn (x,y,ps) = (x,y, update (p-1) objFn ps)
objFn _ = new
All fooo has to do is call update twice (once within the other call) and do a little "housekeeping" work (putting the result in the tuple correctly). The (n-1) and (p-1) were because you seem to be indexing starting at 1, whereas Haskell starts at 0.
Lets just see if that works with our test case:
*Main> fooo (98,98,98) 2 1 [(11,22,[(33,33,33),(44,44,44)]),(55,66,[(77,77,77),(88,88,88)])]
[(11,22,[(33,33,33),(44,44,44)]),(55,66,[(98,98,98),(88,88,88)])]
First, we need a general function to map a certain element of a list, e.g.:
mapN :: (a -> a) -> Int -> [a] -> [a]
mapN f index list = zipWith replace list [1..] where
replace x i | i == index = f x
| otherwise = x
We can use this function twice, for the outer list and the inner lists. There is a little complication as the inner list is part of a tuple, so we need another helper function:
mapTuple3 :: (c -> c) -> (a,b,c) -> (a,b,c)
mapTuple3 f (x,y,z) = (x,y,f z)
Now we have everything we need to apply the replace function to our use case:
fooo :: Int -> Int -> (Int,Int,Int) -> [(Int,Int,[(Int,Int,Int)])]
fooo n p newTuple = mapN (mapTuple3 (mapN (const newTuple) p)) n xxs
Of course in the inner list, we don't need to consider the old value, so we can use const :: a -> (b -> a) to ignore that argument.
So you've tried using some ready-made function, (!!). It could access an item in a list for you, but forgot its place there, so couldn't update. You've got a solution offered, using another ready-made function split, that tears a list into two pieces, and (++) which glues them back into one.
But to get a real feel for it, what I suspect your assignment was aiming at in the first place (it's easy to forget a function name, and it's equally easy to write yourself a new one instead), you could try to write the first one, (!!), yourself. Then you'd see it's real easy to modify it so it's able to update the list too.
To write your function, best think of it as an equivalence equation:
myAt 1 (x:xs) = x
myAt n (x:xs) | n > 1 = ...
when n is zero, we just take away the head element. What do we do when it's not? We try to get nearer towards the zero. You can fill in the blanks.
So here we returned the element found. What if we wanted to replace it? Replace it with what? - this calls another parameter into existence,
myRepl 1 (x:xs) y = (y:xs)
myRepl n (x:xs) y | n > 1 = x : myRepl ...
Now you can complete the rest, I think.
Lastly, Haskell is a lazy language. That means it only calls into existence the elements of a list that are needed, eventually. What if you replace the 7-th element, but only first 3 are later asked for? The code using split will actually demand the 7 elements, so it can return the first 3 when later asked for them.
Now in your case you want to replace in a nested fashion, and the value to replace the old one with is dependent on the old value: newVal = let (a,b,ls)=oldVal in (a,b,myRepl p ls newtuple). So indeed you need to re-write using functions instead of values (so that where y was used before, const y would go):
myUpd 1 (x:xs) f = (f x:xs)
myUpd n ... = ...
and your whole call becomes myUpd n xxs (\(a,b,c)->(a,b,myUpd ... (const ...) )).
I'm new in Haskell and try to solve 3 problem from http://projecteuler.net/.
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
My solution:
import Data.List
getD :: Int -> Int
getD x =
-- find deviders
let deriveList = filter (\y -> (x `mod` y) == 0) [1 .. x]
filteredList = filter isSimpleNumber deriveList
in maximum filteredList
-- Check is nmber simple
isSimpleNumber :: Int -> Bool
isSimpleNumber x = let deriveList = map (\y -> (x `mod` y)) [1 .. x]
filterLength = length ( filter (\z -> z == 0) deriveList)
in
case filterLength of
2 -> True
_ -> False
I try to run for example:
getD 13195
> 29
But when i try:
getD 600851475143
I get error Exception: Prelude.maximum: empty list Why?
Thank you #Barry Brown, I think i must use:
getD :: Integer -> Integer
But i get error:
Couldn't match expected type `Int' with actual type `Integer'
Expected type: [Int]
Actual type: [Integer]
In the second argument of `filter', namely `deriveList'
In the expression: filter isSimpleNumber deriveList
Thank you.
Your type signature limits the integer values to about 2^29. Try changing Int to Integer.
Edit:
I see that you already realised that you need to use Integer instead of Int. You need to change the types of both getD and isSimpleNumber otherwise you will get a type mismatch.
Also in general, if you are having trouble with types, simply remove the type declarations and let Haskell tell you the correct types.
Main> :t getD
getD :: Integral a => a -> a
Main> :t isSimpleNumber
isSimpleNumber :: Integral a => a -> Bool
After you found the error, may I point out that your solution is quite verbose? In this case a very simple implementation using brute force is good enough:
getD n = getD' n 2 where
getD' n f | n == f = f
| n `mod` f == 0 = getD' (n `div` f) f
| otherwise = getD' n (succ f)
this question is easy enough for brute-force solution, but it is a bad idea to do so because the whole idea of project euler is problems you need to really think of to solve (see end of answer)
so here are some of your program's flaws:
first, use rem instead of mod. it is more efficient.
some mathematical thinking should have told you that you don't need to check all numbers from 1 to x in the isprime function and the getD function, but checking all numbers from the squareroot to one (or reversed) should be sufficient. note that in getD you will actually need to filter numbers between x and the square root, because you search for the biggest one.
why do you use the maximum function in getD? you know the list is monotonically growing, so you may as well get the last one.
despite you only need the biggest divisor (which is prime) you compute the divisors list from small to big making the computer check for each value if it is a divisor or not although discarding the result once a bigger divisor is found. it should be fixed by filtering the list of numbers from x to 1, not from 1 to x. this will cause the computer to check divisibility (how should I say that?) for the biggest possible divisor, not throwing to the trash the knowledge of previous checks. note that this optimization takes effect only if the previous point is optimized, because otherwise the computer will compute all divisors anyway.
with the previous points mixed, you should have filtered all numbers [x,x-1 .. squareroot x] and taken the first.
you don't use an efficient isPrime function. if I were you, I would have searched for an isprime library function, which is guaranteed to be efficient.
and there are more..
with this kind of code you will never be able to solve harder project euler problems. they are designed to need extra thinking about the problem (for instance noticing you don't have to check numbers greater from the square root) and writing fast and efficient code. this is the purpose of project euler; being smart about programming. so don't skip it.