Haskell's expressiveness enables us to rather easily define a powerset function:
import Control.Monad (filterM)
powerset :: [a] -> [[a]]
powerset = filterM (const [True, False])
To be able to perform my task it is crucial for said powerset to be sorted by a specific function, so my implementation kind of looks like this:
import Data.List (sortBy)
import Data.Ord (comparing)
powersetBy :: Ord b => ([a] -> b) -> [a] -> [[a]]
powersetBy f = sortBy (comparing f) . powerset
Now my question is whether there is a way to only generate a subset of the powerset given a specific start and endpoint, where f(start) < f(end) and |start| < |end|. For example, my parameter is a list of integers ([1,2,3,4,5]) and they are sorted by their sum. Now I want to extract only the subsets in a given range, lets say 3 to 7. One way to achieve this would be to filter the powerset to only include my range but this seems (and is) ineffective when dealing with larger subsets:
badFunction :: Ord b => b -> b -> ([a] -> b) -> [a] -> [[a]]
badFunction start end f = filter (\x -> f x >= start && f x <= end) . powersetBy f
badFunction 3 7 sum [1,2,3,4,5] produces [[1,2],[3],[1,3],[4],[1,4],[2,3],[5],[1,2,3],[1,5],[2,4],[1,2,4],[2,5],[3,4]].
Now my question is whether there is a way to generate this list directly, without having to generate all 2^n subsets first, since it will improve performance drastically by not having to check all elements but rather generating them "on the fly".
If you want to allow for completely general ordering-functions, then there can't be a way around checking all elements of the powerset. (After all, how would you know the isn't a special clause built in that gives, say, the particular set [6,8,34,42] a completely different ranking from its neighbours?)
However, you could make the algorithm already drastically faster by
Only sorting after filtering: sorting is O (n · log n), so you want keep n low here; for the O (n) filtering step it matters less. (And anyway, number of elements doesn't change through sorting.)
Apply the ordering-function only once to each subset.
So
import Control.Arrow ((&&&))
lessBadFunction :: Ord b => (b,b) -> ([a]->b) -> [a] -> [[a]]
lessBadFunction (start,end) f
= map snd . sortBy (comparing fst)
. filter (\(k,_) -> k>=start && k<=end)
. map (f &&& id)
. powerset
Basically, let's face it, powersets of anything but a very small basis are infeasible. The particular application “sum in a certain range” is pretty much a packaging problem; there are quite efficient ways to do that kind of thing, but you'll have to give up the idea of perfect generality and of quantification over general subsets.
Since your problem is essentially a constraint satisfaction problem, using an external SMT solver might be the better alternative here; assuming you can afford the extra IO in the type and the need for such a solver to be installed. The SBV library allows construction of such problems. Here's one encoding:
import Data.SBV
-- c is the cost type
-- e is the element type
pick :: (Num e, SymWord e, SymWord c) => c -> c -> ([SBV e] -> SBV c) -> [e] -> IO [[e]]
pick begin end cost xs = do
solutions <- allSat constraints
return $ map extract $ extractModels solutions
where extract ts = [x | (t, x) <- zip ts xs, t]
constraints = do tags <- mapM (const free_) xs
let tagged = zip tags xs
finalCost = cost [ite t (literal x) 0 | (t, x) <- tagged]
solve [finalCost .>= literal begin, finalCost .<= literal end]
test :: IO [[Integer]]
test = pick 3 7 sum [1,2,3,4,5]
We get:
Main> test
[[1,2],[1,3],[1,2,3],[1,4],[1,2,4],[1,5],[2,5],[2,3],[2,4],[3,4],[3],[4],[5]]
For large lists, this technique will beat out generating all subsets and filtering; assuming the cost function generates reasonable constraints. (Addition will be typically OK, if you've multiplications, the backend solver will have a harder time.)
(As a side note, you should never use filterM (const [True, False]) to generate power-sets to start with! While that expression is cute and fun, it is extremely inefficient!)
I'm writing a program in Haskell that generates random triads of ints in order to play them as chords using Euterpea. Here's what my list of lists looks like (obviously, it's infinite, but I'll take 10 $ chop 3 $ randInts, where chop is just a function that splits a list up into (ordered) groupings of three and randPitch simply generates ints over the interval 24 - 84):
[[27,33,48],[53,59,82],[31,49,62],[50,60,63],[51,56,79],[57,59,60],[52,63,69],[52,77,79],[32,32,37],[39,77,84]]
To learn a bit about music theory, I'm applying various filters to these triads. The one that's giving me trouble (at least to implement in Haskell, which is a new language for me) is what Dmitri Tymoczko would call "conjunct melodic notation"--i.e., melodies (in this case the top 'note' or pitch of the 3-tuple) should only move short distances.
What I want to do is write a function that recursively filters the endless list of triads such that what results is a list of triads whose top note only moves by <= n between each chord: essentially, a kind of random walk where you are only allowed to move n semitones for any one step. Here are the functions I have so far:
jumpSize :: (Num a, Ord a) => [[a]] -> [[a]]
jumpSize (_:[]) = []
jumpSize (x:y:[]) = (abs (maximum x - maximum y) : x) : []
jumpSize (x:y:zs) = (abs (maximum x - maximum y) : x) : jumpSize (y:zs)
This will prepend the distance from triad x to y at the beginning of list x. So take 5 $ jumpSize $ chop 3 $ randPitch gives me:
[[34,27,33,48],[20,53,59,82],[1,31,49,62],[16,50,60,63],[19,51,56,79]]
I tried to write a recursive function that filters out all the chords with jumps of great than n, recalculates the new jumpSizes and then applies itself again. However, I'm having trouble because this function is either crashing GHCi or giving me all sorts of problems. Structurally, I would like to still produce the chords by filtering over the random array, but I think I'm missing something about how to best do this in a functional language. Maybe I need to supply an initial generator chord? Thanks!
jumpRecur :: (Num a, Ord a) => a -> [[a]] -> [[a]]
jumpRecur n (xs)
| [x | x <- xs, head x > n] == [] = xs
| otherwise = jumpRecur n $ jumpSize $ filter (\x -> head x <= n) xs
Here's a sample output:
*Main> jumpRecur 5 $ jumpSize $ chop 3 $ take 1000 $ randPitch
[[2,2,2,2,2,2,2,2,70,70,76],[1,1,1,1,1,1,1,1,55,74,74],[1,1,1,1,1,1,1,5,26,28,73], [3,3,3,3,3,3,0,1,26,69,74],[5,5,5,5,5,5,5,5,33,43,77],[0,0,0,0,0,0,0,0,47,67,82],[2,2,2,2,2,2,4,4,37,66,82],[3,3,3,3,3,3,3,3,59,69,84],[4,4,4,4,4,4,4,4,59,79,81],[4,4,5,5,5,5,5,5,28,69,77],[5,5,5,5,5,5,5,5,54,68,73],[0,0,0,0,0,0,0,0,32,73,78],[5,5,5,5,5,5,1,4,52,62,78],[0,0,0,0,0,0,0,0,58,71,73],[3,3,3,3,3,3,4,4,25,64,73],[1,1,1,1,1,3,3,3,35,42,76],[4,4,4,4,4,4,4,4,35,39,77]]
It's probably good Haskell idiom to generalize the filter, as follows:
jumpBy :: (a -> a -> Bool) -> [a] -> [a]
jumpBy ok (x:y:zs) | ok x y = x:jumpBy ok (y:zs) -- accept y & step
| otherwise = jumpBy ok (x:zs) -- reject y & retry
jumpBy _ xs = xs
If you are sure all input will be infinite lists, you don't really need that last line which handles the ends of finite lists, but it's good practice to handle all cases.
Then, use your general filter with your specific test:
jumpSize = jumpBy (\x y -> abs(maximum x - maximum y) < n)
I want to iterate 2 (or 3) infinite lists and find the "smallest" pair that satisfies a condition, like so:
until pred [(a,b,c) | a<-as, b<-bs, c<-cs]
where pred (a,b,c) = a*a + b*b == c*c
as = [1..]
bs = [1..]
cs = [1..]
The above wouldn't get very far, as a == b == 1 throughout the run of the program.
Is there a nice way to dovetail the problem, e.g. build the infinite sequence [(1,1,1),(1,2,1),(2,1,1),(2,1,2),(2,2,1),(2,2,2),(2,2,3),(2,3,2),..] ?
Bonus: is it possible to generalize to n-tuples?
There's a monad for that, Omega.
Prelude> let as = each [1..]
Prelude> let x = liftA3 (,,) as as as
Prelude> let x' = mfilter (\(a,b,c) -> a*a + b*b == c*c) x
Prelude> take 10 $ runOmega x'
[(3,4,5),(4,3,5),(6,8,10),(8,6,10),(5,12,13),(12,5,13),(9,12,15),(12,9,15),(8,15,17),(15,8,17)]
Using it's applicative features, you can generalize to arbitrary tuples:
quadrupels = (,,,) <$> as <*> as <*> as <*> as -- or call it liftA4
But: this alone does not eliminate duplication, of course. It only gives you proper diagonalization. Maybe you could use monad comprehensions together with an approach like Thomas's, or just another mfilter pass (restricting to b /= c, in this case).
List comprehensions are great (and concise) ways to solve such problems. First, you know you want all combinations of (a,b,c) that might satisfy a^2 + b^2 = c^2 - a helpful observation is that (considering only positive numbers) it will always be the case that a <= c && b <= c.
To generate our list of candidates we can thus say c ranges from 1 to infinity while a and b range from one to c.
[(a,b,c) | c <- [1..], a <- [1..c], b <- [1..c]]
To get to the solution we just need to add your desired equation as a guard:
[(a,b,c) | c <- [1..], a <- [1..c], b <- [1..c], a*a+b*b == c*c]
This is inefficient, but the output is correct:
[(3,4,5),(4,3,5),(6,8,10),(8,6,10),(5,12,13),(12,5,13),(9,12,15)...
There are more principled methods than blind testing that can solve this problem.
{- It depends on what is "smallest". But here is a solution for a concept of "smallest" if tuples were compared first by their max. number and then by their total sum. (You can just copy and paste my whole answer into a file as I write the text in comments.)
We will need nub later. -}
import Data.List (nub)
{- Just for illustration: the easy case with 2-tuples. -}
-- all the two-tuples where 'snd' is 'n'
tuples n = [(i, n) | i <- [1..n]]
-- all the two-tuples where 'snd' is in '1..n'
tuplesUpTo n = concat [tuples i | i <- [1..n]]
{-
To get all results, you will need to insert the flip of each tuple into the stream. But let's do that later and generalize first.
Building tuples of arbitrary length is somewhat difficult, so we will work on lists. I call them 'kList's, if they have a length 'k'.
-}
-- just copied from the tuples case, only we need a base case for k=1 and
-- we can combine all results utilizing the list monad.
kLists 1 n = [[n]]
kLists k n = do
rest <- kLists (k-1) n
add <- [1..head rest]
return (add:rest)
-- same as above. all the klists with length k and max number of n
kListsUpTo k n = concat [kLists k i | i <- [1..n]]
-- we can do that unbounded as well, creating an infinite list.
kListsInf k = concat [kLists k i | i <- [1..]]
{-
The next step is rotating these lists around, because until now the largest number is always in the last place. So we just look at all rotations to get all the results. Using nub here is admittedly awkward, you can improve that. But without it, lists where all elements are the same are repeated k times.
-}
rotate n l = let (init, end) = splitAt n l
in end ++ init
rotations k l = nub [rotate i l | i <- [0..k-1]]
rotatedKListsInf k = concatMap (rotations k) $ kListsInf k
{- What remains is to convert these lists into tuples. This is a bit awkward, because every n-tuple is a separate type. But it's straightforward, of course. -}
kListToTuple2 [x,y] = (x,y)
kListToTuple3 [x,y,z] = (x,y,z)
kListToTuple4 [x,y,z,t] = (x,y,z,t)
kListToTuple5 [x,y,z,t,u] = (x,y,z,t,u)
kListToTuple6 [x,y,z,t,u,v] = (x,y,z,t,u,v)
{- Some tests:
*Main> take 30 . map kListToTuple2 $ rotatedKListsInf 2
[(1,1),(1,2),(2,1),(2,2),(1,3),(3,1),(2,3),(3,2),(3,3),(1,4),(4,1),(2,4),(4,2),(3,4),
(4,3),(4,4),(1,5),(5,1),(2,5),(5,2),(3,5),(5,3),(4,5),(5,4),(5,5),(1,6),(6,1),
(2,6), (6,2), (3,6)]
*Main> take 30 . map kListToTuple3 $ rotatedKListsInf 3
[(1,1,1),(1,1,2),(1,2,1),(2,1,1),(1,2,2),(2,2,1),(2,1,2),(2,2,2),(1,1,3),(1,3,1),
(3,1,1),(1,2,3),(2,3,1),(3,1,2),(2,2,3),(2,3,2),(3,2,2),(1,3,3),(3,3,1),(3,1,3),
(2,3,3),(3,3,2),(3,2,3),(3,3,3),(1,1,4),(1,4,1),(4,1,1),(1,2,4),(2,4,1),(4,1,2)]
Edit:
I realized there is a bug: Just rotating the ordered lists isn't enough of course. The solution must be somewhere along the lines of having
rest <- concat . map (rotations (k-1)) $ kLists (k-1) n
in kLists, but then some issues with repeated outputs arise. You can figure that out, I guess. ;-)
-}
It really depends on what you mean by "smallest", but I assume you want to find a tuple of numbers with respect to its maximal element - so (2,2) is less than (1,3) (while standard Haskell ordering is lexicographic).
There is package data-ordlist, which is aimed precisely at working with ordered lists. It's function mergeAll (and mergeAllBy) allows you to combine a 2-dimensional matrix ordered in each direction into an ordered list.
First let's create a desired comparing function on tuples:
import Data.List (find)
import Data.List.Ordered
compare2 :: (Ord a) => (a, a) -> (a, a) -> Ordering
compare2 x y = compare (max2 x, x) (max2 y, y)
where
max2 :: Ord a => (a, a) -> a
max2 (x, y) = max x y
Then using mergeAll we create a function that takes a comparator, a combining function (which must be monotonic in both arguments) and two sorted lists. It combines all possible elements from the two lists using the function and produces a result sorted list:
mergeWith :: (b -> b -> Ordering) -> (a -> a -> b) -> [a] -> [a] -> [b]
mergeWith cmp f xs ys = mergeAllBy cmp $ map (\x -> map (f x) xs) ys
With this function, it's very simple to produce tuples ordered according to their maximum:
incPairs :: [(Int,Int)]
incPairs = mergeWith compare2 (,) [1..] [1..]
Its first 10 elements are:
> take 10 incPairs
[(1,1),(1,2),(2,1),(2,2),(1,3),(2,3),(3,1),(3,2),(3,3),(1,4)]
and when we (for example) look for the first pair whose sum of squares is equal to 65:
find (\(x,y) -> x^2+y^2 == 65) incPairs
we get the correct result (4,7) (as opposed to (1,8) if lexicographic ordering were used).
This answer is for a more general problem for a unknown predicate. If the predicate is known, more efficient solutions are possible, like others have listed solutions based on knowledge that you don't need to iterate for all Ints for a given c.
When dealing with infinite lists, you need to perform breadth-first search for solution. The list comprehension only affords depth-first search, that is why you never arrive at a solution in your original code.
counters 0 xs = [[]]
counters n xs = concat $ foldr f [] gens where
gens = [[x:t | t <- counters (n-1) xs] | x <- xs]
f ys n = cat ys ([]:n)
cat (y:ys) (x:xs) = (y:x): cat ys xs
cat [] xs = xs
cat xs [] = [xs]
main = print $ take 10 $ filter p $ counters 3 [1..] where
p [a,b,c] = a*a + b*b == c*c
counters generates all possible counters for values from the specified range of digits, including a infinite range.
First, we obtain a list of generators of valid combinations of counters - for each permitted digit, combine it with all permitted combinations for counters of smaller size. This may result in a generator that produces a infinite number of combinations. So, we need to borrow from each generator evenly.
So gens is a list of generators. Think of this as a list of all counters starting with one digit: gens !! 0 is a list of all counters starting with 1, gens !! 1 is a list of all counters starting with 2, etc.
In order to borrow from each generator evenly, we could transpose the list of generators - that way we would get a list of first elements of the generators, followed by a list of second elements of the generators, etc.
Since the list of generators may be infinite, we cannot afford to transpose the list of generators, because we may never get to look at the second element of any generator (for a infinite number of digits we'd have a infinite number of generators). So, we enumerate the elements from the generators "diagonally" - take first element from the first generator; then take the second element from the first generator and the first from the second generator; then take the third element from the first generator, the second from the second, and the first element from the third generator, etc. This can be done by folding the list of generators with a function f, which zips together two lists - one list is the generator, the other is the already-zipped generators -, the beginning of one of them being offset by one step by adding []: to the head. This is almost zipWith (:) ys ([]:n) - the difference is that if n or ys is shorter than the other one, we don't drop the remainder of the other list. Note that folding with zipWith (:) ys n would be a transpose.
For this answer I will take "smallest" to refer to the sum of the numbers in the tuple.
To list all possible pairs in order, you can first list all of the pairs with a sum of 2, then all pairs with a sum of 3 and so on. In code
pairsWithSum n = [(i, n-i) | i <- [1..n-1]]
xs = concatMap pairsWithSum [2..]
Haskell doesn't have facilities for dealing with n-tuples without using Template Haskell, so to generalize this you will have to switch to lists.
ntuplesWithSum 1 s = [[s]]
ntuplesWithSum n s = concatMap (\i -> map (i:) (ntuplesWithSum (n-1) (s-i))) [1..s-n+1]
nums n = concatMap (ntuplesWithSum n) [n..]
Here's another solution, with probably another slightly different idea of "smallest". My order is just "all tuples with max element N come before all tuples with max element N+1". I wrote the versions for pairs and triples:
gen2_step :: Int -> [(Int, Int)]
gen2_step s = [(x, y) | x <- [1..s], y <- [1..s], (x == s || y == s)]
gen2 :: Int -> [(Int, Int)]
gen2 n = concatMap gen2_step [1..n]
gen2inf :: [(Int, Int)]
gen2inf = concatMap gen2_step [1..]
gen3_step :: Int -> [(Int, Int, Int)]
gen3_step s = [(x, y, z) | x <- [1..s], y <- [1..s], z <- [1..s], (x == s || y == s || z == s)]
gen3 :: Int -> [(Int, Int, Int)]
gen3 n = concatMap gen3_step [1..n]
gen3inf :: [(Int, Int, Int)]
gen3inf = concatMap gen3_step [1..]
You can't really generalize it to N-tuples, though as long as you stay homogeneous, you may be able to generalize it if you use arrays. But I don't want to tie my brain into that knot.
I think this is the simplest solution if "smallest" is defined as x+y+z because after you find your first solution in the space of Integral valued pythagorean triangles, your next solutions from the infinite list are bigger.
take 1 [(x,y,z) | y <- [1..], x <- [1..y], z <- [1..x], z*z + x*x == y*y]
-> [(4,5,3)]
It has the nice property that it returns each symmetrically unique solution only once. x and z are also infinite, because y is infinite.
This does not work, because the sequence for x never finishes, and thus you never get a value for y, not to mention z. The rightmost generator is the innermost loop.
take 1 [(z,y,x)|z <- [1..],y <- [1..],x <- [1..],x*x + y*y == z*z]
Sry, it's quite a while since I did haskell, so I'm going to describe it with words.
As I pointed out in my comment. It is not possible to find the smallest anything in an infinite list, since there could always be a smaller one.
What you can do is, have a stream based approach that takes the lists and returns a list with only 'valid' elements, i. e. where the condition is met. Lets call this function triangle
You can then compute the triangle list to some extent with take n (triangle ...) and from this n elements you can find the minium.