I have this code for sorting list of even number of items:
sort [] = []
sort l = sortN l (length l)
sortN l 0 = l
sortN l n = sortN (swap l) (n - 1)
swap [] = []
swap (a:b:t) | a <= b = a : b : (swap t)
swap (a:b:t) | b < a = b : a : (swap t)
and I am stuck. For some reason, which I don't understand it returns results as if only one swap was always called.
Also, please do not post here better and more efficient ways how to sort. I am aware of them. I want to know why this is wrong, not other solutions.
Thank you.
The problem is you are only considering the same pairs of element on each pass. Consider:
swap (a:b:t) | a <= b = a : b : (swap t)
swap (a:b:t) | b < a = b : a : (swap t)
What if we have a list [4,1,1,1]?
swap [4,1,1,1] = 1 : 4 : swap [1,1] = 1:4:1:1:[]
Ok, now lets look at the next iteration of sortN and its call to swap:
swap [1,4,1,1] = 1 : 4 : swap [1,1]
So you see, swap implicitly assumes every pair of elements is ordered within the list. Instead consider the implementation a: swap (b:t) and b : swap (a : t).
Let me offer you a working version (there was the one-element case missing from swap so as far as I see you will run into trouble else:
swap [] = []
swap [a] = [a]
swap (a:b:t)
| a <= b = a : swap (b:t)
| otherwise = b : swap (a:t)
I will not be tempted into giving you a different algorithm but still you don't need to run swap xs length xs times - you just have to run it till the output is not changing anymore:
sort :: Ord a => [a] -> [a]
sort = fix swap
swap :: Ord a => [a] -> [a]
swap [] = []
swap [a] = [a]
swap (a:b:t)
| a <= b = a : swap (b:t)
| otherwise = b : swap (a:t)
fix :: Eq a => (a -> a) -> a -> a
fix f x = let x' = f x
in if x' /= x then fix f x' else x
fix here will do exactly that - it applies f to x till the result is not changing anymore.
for example:
sort [3,6,1,3,2] will call swap only 4 times
sort [1..10] it will be called only once
testing your implementations
maybe you are interested in how you can easily check your implementations - here is a QuickCheck test (I used to verify my solution too) that does this:
import Test.QuickCheck
isSorted :: Ord a => [a] -> Bool
isSorted [] = True
isSorted [_] = True
isSorted (a:b:t) = a<=b && isSorted (b:t)
checkSortAlg :: ([Int] -> [Int]) -> IO ()
checkSortAlg sortAlg = quickCheck test
where
test xs = isSorted $ sortAlg xs
you just have to run it like this (for example in ghci):
checkSortAlg sort
it should print +++ OK, passed 100 tests.
here is what it does for for the version with the missing swap [a] = [a] case:
(after 3 tests and 1 shrink):
Exception:
SimpleSort.hs:(9,1)-(12,30): Non-exhaustive patterns in function swap
[0]
this tells you that after 3 tests it used some list that after shrinking to [0] still threw the Non-exhaustive patterns exception - so it basically tells you to look for the single element case here ;) - of course the compiler should have too
Related
Write the recursive function adjuster. Given a list of type
x, an int and an element of type x, either remove from the front of the
list until it is the same length as int, or append to the end of the list
until it is the same length as the value specified by the int.
expected:
adjuster [1..10] (-2) 2 -> *** Exception: Invalid Size
adjuster [1..10] 0 2 -> []
adjuster "apple" 10 ’b’ -> "applebbbbb"
adjuster "apple" 5 ’b’ -> "apple"
adjuster "apple" 2 ’b’ -> "le"
adjuster [] 3 (7,4) -> [(7,4),(7,4),(7,4)]
What i did:
adjuster (x:xs) count b
| count < 0 = error "Invalid Size"
| count == 0 = []
| count < length xs = adjuster xs (count-1) b
| otherwise = (adjuster xs (count-1) b):b
the error that I'm getting:
* Occurs check: cannot construct the infinite type: t ~ [t]
Expected type: [t]
Actual type: [[t]]
* In the expression: (adjuster xs (count - 1) b) : b
In an equation for `adjuster':
adjuster (x : xs) count b
| count < 0 = error "Invalid Size"
| count == 0 = []
| count < length xs = adjuster xs (count - 1) b
| otherwise = (adjuster xs (count - 1) b) : b
* Relevant bindings include
b :: [[t]] (bound at code01.hs:21:23)
adjuster :: [a] -> Int -> [[t]] -> [t] (bound at code01.hs:21:1)
I'm new in haskell.I'll really appreciate some help.
You are trying to construct a list within lists within lists and so on and so forth …
Why is this?
(:) :: a -> [a] -> [a]
The colon operator takes an element and a list of such elements as an argument and constructs a list from that (by prepending that element).
In your case if (adjuster ...) had type [a] then b must be of type [[a]], by line 4 which is the same as the end result, but line 3 says the type is [a] - which is different. This is what GHC tries to tell you.
How to fix it?
First of all, it is always a good advice to add a type signature to every top level function:
adjuster :: [a] -> Int -> a -> [a]
which should clean up your error-message and keep you honest, when implementing your function.
So how to fix this: - you could use b:adjuster xs (count-1) b but this would yield a result in the wrong order - so
choose a different operator: (++) and wrap the b inside a list.
| otherwise = (adjuster xs (count-1) b)++[b]
Now a few more hints:
turn on -Wall when you compile your file - this will show you that you missed the case of adjuster [] ...
using length is a relatively expensive operation - as it needs to traverse the full list to be calculated.
As an exercise - try to modify your function to not use length but only work with the base cases [] for list and 0 for count (here the function replicate might be helpful).
Here is another approach, without error handling
adjuster xs n v = tnr n $ (++) (replicate n v) $ tnr n xs
where tnr n r = take n $ reverse r
if you play with the signature, perhaps cleaner this way
adjuster n v = tnr . (++) (replicate n v) . tnr
where tnr = take n . reverse
Right now I'm working on a problem in Haskell in which I'm trying to check a list for a particular pair of values and return True/False depending on whether they are present in said list. The question goes as follows:
Define a function called after which takes a list of integers and two integers as parameters. after numbers num1 num2 should return true if num1 occurs in the list and num2 occurs after num1. If not it must return false.
My plan is to check the head of the list for num1 and drop it, then recursively go through until I 'hit' it. Then, I'll take the head of the tail and check that against num2 until I hit or reach the end of the list.
I've gotten stuck pretty early, as this is what I have so far:
after :: [Int] -> Int -> Int -> Bool
after x y z
| y /= head x = after (drop 1 x) y z
However when I try to run something such as after [1,4,2,6,5] 4 5 I get a format error. I'm really not sure how to properly word the line such that haskell will understand what I'm telling it to do.
Any help is greatly appreciated! Thanks :)
Edit 1: This is the error in question:
Program error: pattern match failure: after [3,Num_fromInt instNum_v30 4] 3 (Num_fromInt instNum_v30 2)
Try something like this:
after :: [Int] -> Int -> Int -> Bool
after (n:ns) a b | n == a = ns `elem` b
| otherwise = after ns a b
after _ _ _ = False
Basically, the function steps through the list, element by element. If at any point it encounters a (the first number), then it checks to see if b is in the remainder of the list. If it is, it returns True, otherwise it returns False. Also, if it hits the end of the list without ever seeing a, it returns False.
after :: Eq a => [a] -> a -> a -> Bool
after ns a b =
case dropWhile (/= a) ns of
[] -> False
_:xs -> b `elem` xs
http://hackage.haskell.org/package/base-4.8.2.0/docs/src/GHC.List.html#dropWhile
after xs p1 p2 = [p1, p2] `isSubsequenceOf` xs
So how can we define that? Fill in the blanks below!
isSubsequenceOf :: Eq a => [a] -> [a] -> Bool
[] `isSubsequenceOf` _ = ?
(_ : _) `isSubsequenceOf` [] = ?
xss#(x : xs) `isSubsequenceOf` (y:ys)
| x == y = ?
| otherwise = ?
after :: [Int] -> Int -> Int -> Bool
Prelude> let after xs a b = elem b . tail $ dropWhile (/=a) xs
Examples:
Prelude> after [1,2,3,4,3] 88 7
*** Exception: Prelude.tail: empty list
It raises an exception because of tail. It's easy to write tail' such that it won't raise that exception. Otherwise it works pretty well.
Prelude> after [1,2,3,4,3] 2 7
False
Prelude> after [1,2,3,4,3] 2 4
True
There's a series of examples I'm trying to do to practice Haskell. I'm currently learning about continuation passing, but I'm a bit confused as to how to implement a function like find index of element in list that works like this:
index 3 [1,2,3] id = 2
Examples like factorial made sense since there wasn't really any processing of the data other than multiplication, but in the case of the index function, I need to compare the element I'm looking at with the element I'm looking for, and I just can't seem to figure out how to do that with the function parameter.
Any help would be great.
first let me show you a possible implementation:
index :: Eq a => a -> [a] -> (Int -> Int) -> Int
index _ [] _ = error "not found"
index x (x':xs) cont
| x == x' = cont 0
| otherwise = index x xs (\ind -> cont $ ind + 1)
if you prefer point-free style:
index :: Eq a => a -> [a] -> (Int -> Int) -> Int
index _ [] _ = error "not found"
index x (x':xs) cont
| x == x' = cont 0
| otherwise = index x xs (cont . (+1))
how it works
The trick is to use the continuations to count up the indices - those continuations will get the index to the right and just increment it.
As you see this will cause an error if it cannot find the element.
examples:
λ> index 1 [1,2,3] id
0
λ> index 2 [1,2,3] id
1
λ> index 3 [1,2,3] id
2
λ> index 4 [1,2,3] id
*** Exception: not found
how I figured it out
A good way to figure out stuff like this is by first writing down the recursive call with the continuation:
useCont a (x:xs) cont = useCont a xs (\valFromXs -> cont $ ??)
And now you have to think about what you want valFromXs to be (as a type and as a value) - but remember your typical start (as here) will be to make the first continuation id, so the type can only be Int -> Int. So it should be clear that we are talking about of index-transformation here. As useCont will only know about the tail xs in the next call it seems natural to see this index as relative to xs and from here the rest should follow rather quickly.
IMO this is just another instance of
Let the types guide you Luke
;)
remarks
I don't think that this is a typical use of continuations in Haskell.
For once you can use an accumulator argument for this as well (which is conceptional simpler):
index :: Eq a => a -> [a] -> Int -> Int
index _ [] _ = error "not found"
index x (x':xs) ind
| x == x' = ind
| otherwise = index x xs (ind+1)
or (see List.elemIndex) you can use Haskells laziness/list-comprehensions to make it look even nicer:
index :: Eq a => a -> [a] -> Int
index x xs = head [ i | (x',i) <- zip xs [0..], x'== x ]
If you have a value a then to convert it to CPS style you replace it with something like (a -> r) -> r for some unspecified r. In your case, the base function is index :: Eq a => a -> [a] -> Maybe Int and so the CPS form is
index :: Eq a => a -> [a] -> (Maybe Int -> r) -> r
or even
index :: Eq a => a -> [a] -> (Int -> r) -> r -> r
Let's implement the latter.
index x as success failure =
Notably, there are two continuations, one for the successful result and one for a failing one. We'll apply them as necessary and induct on the structure of the list just like usual. First, clearly, if the as list is empty then this is a failure
case as of
[] -> failure
(a:as') -> ...
In the success case, we're, as normal, interested in whether x == a. When it is true we pass the success continuation the index 0, since, after all, we found a match at the 0th index of our input list.
case as of
...
(a:as') | x == a -> success 0
| otherwise -> ...
So what happens when we don't yet have a match? If we were to pass the success continuation in unchanged then it would, assuming a match is found, eventually be called with 0 as an argument. This loses information about the fact that we've attempted to call it once already, though. We can rectify that by modifying the continuation
case as of
...
(a:as') ...
| otherwise -> index x as' (fun idx -> success (idx + 1)) failure
Another way to think about it is that we have the collect "post" actions in the continuation since ultimately the result of the computation will pass through that code
-- looking for the value 5, we begin by recursing
1 :
2 :
3 :
4 :
5 : _ -- match at index 0; push it through the continuation
0 -- lines from here down live in the continuation
+1
+1
+1
+1
This might be even more clear if we write the recursive branch in pointfree style
| otherwise -> index x as' (success . (+1)) failure
which shows how we're modifying the continuation to include one more increment for each recursive call. All together the code is
index :: Eq a => a -> [a] -> (Int -> r) -> r -> r
index x as success failure
case as of
[] -> failure
(a:as') | x == a -> success 0
| otherwise -> index x as' (success . (+1)) failure
So I'm trying to define a function in Haskell that if given an integer and a list of integers will give a 'true' or 'false' whether the integer occurs only once or not.
So far I've got:
let once :: Eq a => a -> [a] -> Bool; once x l =
But I haven't finished writing the code yet. I'm very new to Haskell as you may be able to tell.
Start off by using pattern matching:
once x [] =
once x (y:ys) =
This won't give you a good program immediately, but it will lead you in the right direction.
Here's a solution that doesn't use pattern matching explicitly. Instead, it keeps track of a Bool which represents if a occurance has already been found.
As others have pointed out, this is probably a homework problem, so I've intentionally left the then and else branches blank. I encourage user3482534 to experiment with this code and fill them in themselves.
once :: Eq a => a -> [a] -> Bool
once a = foldr f False
where f x b = if x == a then ??? else ???
Edit: The naive implementation I was originally thinking of was:
once :: Eq a => a -> [a] -> Bool
once a = foldr f False
where f x b = if x == a then b /= True else b
but this is incorrect as,
λ. once 'x' "xxx"
True
which should, of course, be False as 'x' occurs more than exactly once.
However, to show that it is possible to write once using a fold, here's a revised version that uses a custom monoid to keep track of how many times the element has occured:
import Data.List
import Data.Foldable
import Data.Monoid
data Occur = Zero | Once | Many
deriving Eq
instance Monoid Occur where
mempty = Zero
Zero `mappend` x = x
x `mappend` Zero = x
_ `mappend` _ = Many
once :: Eq a => a -> [a] -> Bool
once a = (==) Once . foldMap f
where f x = if x == a then Once else Zero
main = do
let xss = inits "xxxxx"
print $ map (once 'x') xss
which prints
[False,True,False,False,False]
as expected.
The structure of once is similar, but not identical, to the original.
I'll answer this as if it were a homework question since it looks like one.
Read about pattern matching in function declarations, especially when they give an example of processing a list. You'll use tools from Data.List later, but probably your professor is teaching about pattern matching.
Think about a function that maps values to a 1 or 0 depending on whethere there is a match ...
match :: a -> [a] -> [Int]
match x xs = map -- fill in the thing here such that
-- match 3 [1,2,3,4,5] == [0,0,1,0,0]
Note that there is the sum function that takes a list of numbers and returns the sum of the numbers in the list. So to count the matches a function can take the match function and return the counts.
countN :: a -> [a] -> Int
countN x xs = ? $ match x xs
And finally a function that exploits the countN function to check for a count of only 1. (==1).
Hope you can figure out the rest ...
You can filter the list and then check the length of the resulting list. If length == 1, you have only one occurrence of the given Integer:
once :: Eq a => a -> [a] -> Bool
once x = (== 1) . length . filter (== x)
For counting generally, with import Data.List (foldl'), pointfree
count pred = foldl' (\ n x -> if pred x then n + 1 else n) 0
applicable like
count (< 10) [1 .. 10] == 9
count (== 'l') "Hello" == 2
gives
once pred xs = count pred xs == 1
Efficient O(n) short-circuit predicated form, testing whether the predicate is satisfied exactly once:
once :: (a -> Bool) -> [a] -> Bool
once pred list = one list 0
where
one [] 1 = True
one [] _ = False
one _ 2 = False
one (x : xs) n | pred x = one xs (n + 1)
| otherwise = one xs n
Or, using any:
none pred = not . any pred
once :: (a -> Bool) -> [a] -> Bool
once _ [] = False
once pred (x : xs) | pred x = none pred xs
| otherwise = one pred xs
gives
elemOnce y = once (== y)
which
elemOnce 47 [1,1,2] == False
elemOnce 2 [1,1,2] == True
elemOnce 81 [81,81,2] == False
I am doing another Project Euler problem and I need to find when the result of these 3 lists is equal (we are given 40755 as the first time they are equal, I need to find the next:
hexag n = [ n*(2*n-1) | n <- [40755..]]
penta n = [ n*(3*n-1)/2 | n <- [40755..]]
trian n = [ n*(n+1)/2 | n <- [40755..]]
I tried adding in the other lists as predicates of the first list, but that didn't work:
hexag n = [ n*(2*n-1) | n <- [40755..], penta n == n, trian n == n]
I am stuck as to where to to go from here.
I tried graphing the function and even calculus but to no avail, so I must resort to a Haskell solution.
Your functions are weird. They get n and then ignore it?
You also have a confusion between function's inputs and outputs. The 40755th hexagonal number is 3321899295, not 40755.
If you really want a spoiler to the problem (but doesn't that miss the point?):
binarySearch :: Integral a => (a -> Bool) -> a -> a -> a
binarySearch func low high
| low == high = low
| func mid = search low mid
| otherwise = search (mid + 1) high
where
search = binarySearch func
mid = (low+high) `div` 2
infiniteBinarySearch :: Integral a => (a -> Bool) -> a
infiniteBinarySearch func =
binarySearch func ((lim+1) `div` 2) lim
where
lim = head . filter func . lims $ 0
lims x = x:lims (2*x+1)
inIncreasingSerie :: (Ord a, Integral i) => (i -> a) -> a -> Bool
inIncreasingSerie func val =
val == func (infiniteBinarySearch ((>= val) . func))
figureNum :: Integer -> Integer -> Integer
figureNum shape index = (index*((shape-2)*index+4-shape)) `div` 2
main :: IO ()
main =
print . head . filter r $ map (figureNum 6) [144..]
where
r x = inIncreasingSerie (figureNum 5) x && inIncreasingSerie (figureNum 3) x
Here's a simple, direct answer to exactly the question you gave:
*Main> take 1 $ filter (\(x,y,z) -> (x == y) && (y == z)) $ zip3 [1,2,3] [4,2,6] [8,2,9]
[(2,2,2)]
Of course, yairchu's answer might be more useful in actually solving the Euler question :)
There's at least a couple ways you can do this.
You could look at the first item, and compare the rest of the items to it:
Prelude> (\x -> all (== (head x)) $ tail x) [ [1,2,3], [1,2,3], [4,5,6] ]
False
Prelude> (\x -> all (== (head x)) $ tail x) [ [1,2,3], [1,2,3], [1,2,3] ]
True
Or you could make an explicitly recursive function similar to the previous:
-- test.hs
f [] = True
f (x:xs) = f' x xs where
f' orig (y:ys) = if orig == y then f' orig ys else False
f' _ [] = True
Prelude> :l test.hs
[1 of 1] Compiling Main ( test.hs, interpreted )
Ok, modules loaded: Main.
*Main> f [ [1,2,3], [1,2,3], [1,2,3] ]
True
*Main> f [ [1,2,3], [1,2,3], [4,5,6] ]
False
You could also do a takeWhile and compare the length of the returned list, but that would be neither efficient nor typically Haskell.
Oops, just saw that didn't answer your question at all. Marking this as CW in case anyone stumbles upon your question via Google.
The easiest way is to respecify your problem slightly
Rather than deal with three lists (note the removal of the superfluous n argument):
hexag = [ n*(2*n-1) | n <- [40755..]]
penta = [ n*(3*n-1)/2 | n <- [40755..]]
trian = [ n*(n+1)/2 | n <- [40755..]]
You could, for instance generate one list:
matches :: [Int]
matches = matches' 40755
matches' :: Int -> [Int]
matches' n
| hex == pen && pen == tri = n : matches (n + 1)
| otherwise = matches (n + 1) where
hex = n*(2*n-1)
pen = n*(3*n-1)/2
tri = n*(n+1)/2
Now, you could then try to optimize this for performance by noticing recurrences. For instance when computing the next match at (n + 1):
(n+1)*(n+2)/2 - n*(n+1)/2 = n + 1
so you could just add (n + 1) to the previous tri to obtain the new tri value.
Similar algebraic simplifications can be applied to the other two functions, and you can carry all of them in accumulating parameters to the function matches'.
That said, there are more efficient ways to tackle this problem.