I'm writing a combs function in haskell
what it needs to do is, when I provide it with a deck of cards, give me every combination of hands possible from that deck of size x
This is the relevant code
combs :: Int -> [a] -> [[a]]
combs 0 _ = [[ ]]
combs i (x:xs) = (filter (isLength i) y)
where y = subs (x:xs)
combs _ _ = [ ]
isLength :: Int -> [a] -> Bool
isLength i x
| length x == i = True
| otherwise = False
subs :: [a] -> [[a]]
subs [ ] = [[ ]]
subs (x : xs) = map (x:) ys ++ ys
where ys = subs xs
However, when I ask it to compute a combs 5 [1..52], e.g. a hand of 5 out of a full deck, it does not provide a result, and keeps running for a really long time
Does anyone know what the problem is and how to speed up this algorithm?
To extract i items from x:xs you can proceed in two ways:
you keep the x, and extract only i-1 elements from xs
you discard x, and extract all the i elements from xs
Hence, a solution is:
comb :: Int -> [a] -> [[a]]
comb 0 _ = [[]] -- only the empty list has 0 elements
comb _ [] = [] -- can not extract > 0 elements from []
comb i (x:xs) = [ x:ys | ys <- comb (i-1) xs ] -- keep x case
++ comb i xs -- discard x case
By the way, the above code also "proves" a well-known recursive formula for the binomial coefficients. You might already have met this formula if you attended a calculus class.
Letting B(k,n) = length (comb k [1..n]), we have
B(k+1,n+1) == B(k,n) + B(k+1,n)
which is just a direct consequence of the last line of the code above.
Right now it's a bit hard to see what you are trying to do - but I guess the problems you have is that you gonna filter and map a lot.
I think a simple way to get what you need is this:
module Combinations where
import Data.List (delete)
combs :: Eq a => Int -> [a] -> [[a]]
combs 0 _ = [[]]
combs i xs = [ y:ys | y <- xs, ys <- combs (i-1) (delete y xs) ]
which uses delete from Data.List
It should be lazy enough to find you combinations quick - of course all will take a while ;)
λ> take 5 $ combs 5 [1..52]
[[1,2,3,4,5],[1,2,3,4,6],[1,2,3,4,7],[1,2,3,4,8],[1,2,3,4,9]]
how does it work
it's one of those recursive combinatorial algorithm that works by selecting a first card y from all the cards xs, and then recursivley gets the rest of the handysfrom the deck without the selected carddelete a xsand then putting it back togethery:ys` inside the list-monad (here using list-comprehensions).
BTW: ther are 311,875,200 such decks ;)
version without list-comprehensions
here is a version without comprehensions in case your system has issues here:
combs :: Eq a => Int -> [a] -> [[a]]
combs 0 _ = [[]]
combs i xs = do
y <- xs
ys <- combs (i-1) (delete y xs)
return $ y:ys
version that will remove permutations
this one uses Ord to get sort the items in ascending order and in doing so removing duplciates in respect to permutaion - for this to work xs is expected to be pre-sorted!
Note chi's version is working with fewer constraints and might be more preformant too - but I thougt this is nice and readable and goes well with the version before so maybe it's of interest to you.
I know it's not a thing often done in Haskell/FP where you strife for the most general and abstract cases but I come form an environment where most strive for readability and understanding (coding for the programmer not only for the compiler) - so be gentle ;)
combs' :: Ord a => Int -> [a] -> [[a]]
combs' 0 _ = [[]]
combs' i xs = [ y:ys | y <- xs, ys <- combs' (i-1) (filter (> y) xs) ]
Related
Here is the list of lists: [[1,2,3],[1,2,3,4],[1,2,3]]
How can I increment each element of the second list by the length of the first list, and increment the third list by the length of the first list + second list? The first list should remain unchanged.
Intended output: [[1,2,3],[4,5,6,7],[8,9,10]]
Since the first list has length 3, the second list is generated by [1+3, 2+3, 3+3, 4+3].
Since the first list + second list combined have length 7, the third list is generated by [1+7, 2+7, 3+7].
Ideally it should work with any number of lists.
So far, I've had slight sucess using this:
scanl1 (\xs ys -> [y + length xs | y <- ys]) [[1,2,3],[1,2,3,4],[1,2,3]]
which outputs: [[1,2,3],[4,5,6,7],[5,6,7]]
scanl1 is a good idea, but it's not quite right, because you don't want your accumulator to be a list, but rather to be an integer. So you really want scanl, not scanl1. I'll leave it as an exercise for you to see how to adjust your solution - given that you managed to write something almost-right with scanl1, I don't think you'll find it too hard once you have the right function.
In the comments, jpmariner suggests mapAccumL :: (s -> a -> (s, b)) -> s -> [a] -> (s, [b])). That's perfectly typed for what we want to do, so let's see how it would look.
import Data.Traversable (mapAccumL)
addPreviousLengths :: [[Int]] -> [[Int]]
addPreviousLengths = snd . mapAccumL go 0
where go n xs = (n + length xs, map (+ n) xs)
λ> addPreviousLengths [[1,2,3],[1,2,3,4],[1,2,3]]
[[1,2,3],[4,5,6,7],[8,9,10]]
mapAccumL really is the best tool for this job - there's not much unnecessary complexity involved in using it. But if you're trying to implement this from scratch, you might try the recursive approach Francis King suggested. I'd suggest a lazy algorithm instead of the tail-recursive algorithm, though:
incrLength :: [[Int]] -> [[Int]]
incrLength = go 0
where go _ [] = []
go amount (x:xs) =
map (+ amount) x : go (amount + length x) xs
It works the same as the mapAccumL version. Note that both versions are lazy: they consume only as much of the input list as necessary. This is an advantage not shared by a tail-recursive approach.
λ> take 3 . incrLength $ repeat [1]
[[1],[2],[3]]
λ> take 3 . addPreviousLengths $ repeat [1]
[[1],[2],[3]]
There are many ways to solve this. A simple recursion is one approach:
lst :: [[Int]]
lst = [[1,2,3],[1,2,3,4],[1,2,3]]
incrLength :: [[Int]] -> Int -> [[Int]] -> [[Int]]
incrLength [] _ result = result
incrLength (x:xs) amount result =
incrLength xs (amount + length x) (result ++ [map (+amount) x])
(Edit: it is more efficient to use (:) in this function. See #amalloy comment below. The result then has to be reversed.
incrLength :: [[Int]] -> Int -> [[Int]] -> [[Int]]
incrLength [] _ result = reverse result
incrLength (x:xs) amount result =
incrLength xs (amount + length x) (map (+amount) x : result)
End Edit)
Another approach is to use scanl. We use length to get the length of the inner lists, then accumulate using scanl.
map length lst -- [3,4,3]
scanl (+) 0 $ map length lst -- [0,3,7,10]
init $ scanl (+) 0 $ map length lst -- [0,3,7]
Then we zip the lst and the accumulated value together, and map one over the other.
incrLength' :: [[Int]] -> [[Int]]
incrLength' lst =
[map (+ snd y) (fst y) | y <- zip lst addlst]
where
addlst =init $scanl (+) 0 $ map length lst
main = do
print $ incrLength lst 0 [] -- [[1,2,3],[4,5,6,7],[8,9,10]]
I know it is impossible to sort infinite lists, but I am trying to write a definition of the infinite increasing list of multiples of n numbers.
I already have the function
multiples :: Integer -> [Integer]
multiples n = map (*n) [1..]
that returns the infinite list of multiples of n. But now I want to build a function that given a list of Integers returns the increasing infinite list of the multiples of all the numbers in the list. So the function multiplesList :: [Integer] -> [Integer] given the input [3,5] should yield [3,5,6,9,10,12,15,18,20,....].
I'm new at Haskell, and I'm struggling with this. I think I should use foldr or map since I have to apply multiples to all the numbers in the input, but I don't know how. I can't achieve to mix all the lists into one.
I would really appreciate it if someone could help me.
Thank you!
You are in the right path. following the comments here is a template you can complete.
multiples :: Integer -> [Integer]
multiples n = map (*n) [1..]
-- This is plain old gold recursion.
mergeSortedList :: [Integer] -> [Integer] -> [Integer]
mergeSortedList [] xs = undefined
mergeSortedList xs [] = undefined
mergeSortedList (x:xs) (y:ys)
| x < y = x:mergeSortedList xs (y:ys) -- Just a hint ;)
| x == y = undefined
| x > y = undefined
multiplesList :: [Integer] -> [Integer]
multiplesList ms = undefined -- Hint: foldX mergeSortedList initial xs
-- Which are initial and xs?
-- should you foldr or foldl?
We can easily weave two infinite lists together positionally, taking one element from each list at each step,
weave (x:xs) ys = x : weave ys xs
or we could take longer prefixes each time,
-- warning: expository code only
weaveN n xs ys = take n xs ++ weaveN n ys (drop n xs)
but assuming both lists are not only infinite but also strictly increasing (i.e. there are no duplicates in the lists), we can guide the taking of prefixes by the head value of the opposite list:
umerge :: Ord a => [a] -> [a] -> [a]
-- warning: only works with infinite lists
umerge xs (y:ys) = a ++ [y | head b > y ] ++ umerge ys b
where
(a,b) = span (< y) xs
This is thus a possible encoding of the unique merge operation ("unique" meaning, there won't be any duplicates in its output).
Testing, it seems to work as intended:
> take 20 $ umerge [3,6..] [5,10..]
[3,5,6,9,10,12,15,18,20,21,24,25,27,30,33,35,36,39,40,42]
> [3,6..42] ++ [5,10..42] & sort & nub
[3,5,6,9,10,12,15,18,20,21,24,25,27,30,33,35,36,39,40,42]
> [ p | let { ms :: [Integer] ; ms = takeWhile (< 25^2) $
foldl1 umerge [[p*p,p*p+p..] | p <- [2..25]] },
p <- [2..545], not $ elem p ms ]
[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,
97,101,...........,499,503,509,521,523,541]
> length it
100
And with an ingenious little tweak (due to Richard Bird as seen in the JFP article by Melissa O'Neill) it can even be used to fold an infinite list of ascending lists, provided that it is sorted in ascending order of their head elements, so the head of the first argument is guaranteed to be the first in the output and can thus be produced without testing:
umerge1 :: Ord a => [a] -> [a] -> [a]
-- warning: only works with infinite lists
-- assumes x < y
umerge1 (x:xs) ~(y:ys) = x : a ++ [y | head b > y ] ++ umerge ys b
where
(a,b) = span (< y) xs
Now
> take 100 [ p | let { ms :: [Integer] ;
ms = foldr1 umerge1 [[p*p,p*p+p..] | p <- [2..]] },
p <- [2..], not $ elem p $ takeWhile (<= p) ms ]
[2,3,5,7,11,13, ...... 523,541]
the same calculation works indefinitely.
to the literalists in the audience: yes, calling elem here is Very Bad Thing. The OP hopefully should have recognized this on their own, (*) but unfortunately I felt compelled to make this statement, thus inadvertently revealing this to them, depriving them of their would-be well-earned a-ha moment, unfortunately.
Also, umerge1's definition can be radically simplified. Again, this is left to the OP to discover on their own. (which would, again, be much better for them if I wasn't compelled to make this remark revealing it to them --- finding something on your own is that much more powerful and fulfilling)
(*) and search for ways to replace it with something more efficient, on their own. No, this code is not presented as The Best Solution to Their Problem.
I have a function which was written in list comprehension. As a learning I decided to try to convert this function into a functional application using map, zip, fold, etc. I am having a really hard time converting this particular one.
It might seem unreasonable for what it is doing, but it is part of a bigger function and I want to get this piece working first.
combination :: Int -> [a] -> [([a],[a])]
combination 0 xs = [([],xs)]
combination n (x:xs) = [ (x:ys,zs) | (ys,zs) <- combination (n-1) xs ]
It's just a map:
combination :: Int -> [a] -> [([a],[a])]
combination 0 xs = [([],xs)]
combination n (x:xs) = map (\(ys, zs) -> (x:ys,zs)) (combination (n-1) xs)
I'm really new to programming and Haskell in particular (so new that I actually don't know if this is a stupid question or not). But I was watching the lecture given by Eric Meijer (http://channel9.msdn.com/Series/C9-Lectures-Erik-Meijer-Functional-Programming-Fundamentals) and i was fascinated by the program written by Dr. Graham Hutton in lecture 11; The countdown problem.
My question is:
Is there a way of "filtering" the list of solutions by the length (number of elements), so that the list of solutions are restricted to the solutions that only uses (for example) three of the source numbers? In other words, I would like to change the question from "given the numbers [1,2,3,4,5,6,8,9] construct 18 using the operators..." to "given the numbers [..] which three numbers can be used to construct..."
In my futile attempts, I've been trying to put a kind restriction on his function subbags (which returns all permutations and subsequences of a list)
subbags :: [a] -> [[a]]
subbags xs = [zs | ys <- subs xs, zs <- perms ys]
So that I get all the permutations and subsequences that only contain three of the source numbers. Is this possible? If so, how?
Like I said, I have no idea if this is even a legitimate question - but I have gone from curious to obsessed, so any form of help or hint would be greatly appreciated!
The simplest way would be to just select from the candidates three times
[ (x, y, z) | x <- xs, y <- xs, z <- xs ]
although this assumes that repeat use of a single number is OK.
If it's not, we'll have to get smarter. In a simpler scenario we'd like to pick just two candidates:
[ (x, y) | x <- xs, y <- ys, aboveDiagonal (x, y) ]
in other words, if we think of this as a cartesian product turning a list into a grid of possibilities, we'd like to only consider the values "above the diagonal", where repeats don't happen. We can express this by zipping the coordinates along with the values
[ (x, y) | (i, x) <- zip [1..] xs
, (j, y) <- zip [1..] xs
, i < j
]
which can be extended back out to the n=3 scenario
[ (x, y, z) | (i, x) <- zip [1..] xs
, (j, y) <- zip [1..] xs
, (k, z) <- zip [1..] xs
, i < j
, j < k
]
Ultimately, however, this method is inefficient since it still has to scan through all of the possible pairs and then prune the repeats. We can be a bit smarter by only enumerating the above diagonal values to begin with. Returning to n=2 we'll write this as
choose2 :: [a] -> [(a, a)]
choose2 [] = []
choose2 (a:as) = map (a,) as ++ choose2 as
In other words, we pick first all of the pairs where the head of the list comes first and a value in the tail of the list comes second—this captures one edge of the upper triangle—and then we recurse by adding all of the upper diagonal values of the list of candidates sans the head.
This method can be straightforwardly extended to the n=3 case by using the n=2 case as a building block
choose3 :: [a] -> [(a, a, a)]
choose3 [] = []
choose3 (a:as) = map (\(y, z) -> (a, y, z)) (choose2 as) ++ choose3 as
which also provides a direct generalization to the fully general n dimensional solution
choose :: Int -> [a] -> [[a]]
choose 0 as = [[]] -- there's one way to choose 0 elements
choose _ [] = [] -- there are 0 ways to choose (n>0) elements of none
choose 1 as = map (:[]) as -- there are n ways to choose 1 element of n
choose n (a:as) = map (a:) (choose (n-1) as) ++ choose n as
I like this solution, which does not require the list elements to be an instance of Eq:
import Data.List (tails)
triples ls = [[x,y,z] | (x:xs) <- tails ls,
(y:ys) <- tails xs,
z <- ys]
This returns only subsequences, not permutations, though.
How can I split a list into 2 sublists, where first sublist includes elements from begin of initial list and equals to first element, and second sublist contains others elements? I have to resolve this without using Prelude functions.
My base solution is:
partSameElems :: [a] -> ([a],[a])
partSameElems [] = ([],[])
partSameElems (x:xs) = fstList (x:xs) scdList (x:xs)
where
fstList (x:y:xs) = if x == y then x:y:fstList xs {- I need to do Nothing in else section? -}
scdList (x:xs) = x:scdList xs
For example:
[3,3,3,3,2,1,3,3,6,3] -> ([3,3,3,3], [2,1,3,3,6,3])
Now I can offer my version of solution:
partSameElems :: Eq a => [a] -> ([a],[a])
partSameElems [] = ([],[])
partSameElems (x:xs) = (fstList (x:xs), scdList (x:xs))
where
fstList [] _ = []
fstList (x:xs) el = if x == el then x:fstList xs el else []
scdList [] _ = []
scdList (x:xs) el = if x /= el then (x:xs) else scdList xs el
This is easier if you don't try to do it in two passes.
parSameElems [] = ([], [])
parSameElems lst = (reverse revxs, ys)
where (revxs, ys) = accum [] lst
accum xs [y] = ((y:xs), [])
accum xs (y1:y2:ys) | y1 == y2 = accum (y1:xs) (y2:ys)
| otherwise = ((y1:xs), (y2:ys))
Not sure you can use guard syntax in where clauses. You will also have to implement reverse yourself since you can't use Prelude, but that's easy.
Note: I haven't actually run this. Make sure you try and debug it.
Also, don't write the type signature yourself. Let ghci tell you. You got it wrong in your first try.
Another implementation can be
partition [] = ([],[])
partition xa#(x:xs) = (f,s)
where
f = takeWhile (==x) xa
s = drop (length f) xa
should be clear what it does.
> partition [3,3,3,3,2,1,3,3,6,3]
([3,3,3,3],[2,1,3,3,6,3])
I assume the "without resorting to Prelude function" means it's educational. Probably aimed at working on recursion, given it's manipulation of List data. So let's emphasize this
Recursive algorithms are simpler to express when input and output types are identical.
Let's rather say that instead of a list [3,3,3,3,2,1,3,3,6,3], your input data is composed of
the front list, but at this stage it's empty
the remainder, at this stage equals to input [3,3,3,2,1,3,3,6,3]
recursion input is then ([],[3,3,3,2,1,3,3,6,3])
The type of the central function will be ([a],[a]) -> ([a],[a])
Now, each recursion step will take the front element of the remainder and either put if in the front list or stop recursion (you reached the final state and can return the result)
module SimpleRecursion where
moveInFront :: (Eq a) => ([a],[a]) -> ([a],[a])
moveInFront (xs , [] ) = ( xs , [])
moveInFront ([] , y:ys ) = moveInFront ( y:[] , ys)
moveInFront (x:xs , y:ys ) = if x == y then moveInFront ( y:x:xs , ys)
else (x:xs, y:ys)
partSameElems :: (Eq a) => [a] -> ([a],[a])
partSameElems a = moveInFront ([],a)
What we have here is a classical recursion scheme, with
- stop condition (x /= y)
- recursion clause
- coverage of trivial cases
Notes :
- writing y:x:xs actually reverses the front list but since all values are equal the result is ok
Please don't do that kind of trick in the code of an actual program, it would come back to bite you eventually
- the function only works on lists of Equatable data (Eq a) => because the recursion / stop condition is the equality test ==