Recently I am trying to figure out how to do some programming in Haskell.
I'm trying to do some simple operations. Right now I'm stuck with an operation like in this example:
input = [1,2,3,4]
output = [1,2,2,3,3,3,4,4,4,4]
That is, for each element x in input, produce x elements of x in output. So, for element 1 in input, append [1] to output. Then, for element 2 in input, append elements [2,2] to output. Then, for element 3, append [3,3,3], etc. The algorithm should work only on standard numbers.
I know it's very easy, and it's trivial to perform it in "normal" imperative programming, but as Haskell's functions are stateless, I'm having a problem in how to approach this.
Could anyone please give me some hint how can an absolute Haskell beginner cope with this?
You've just discovered monads!
Here's the general idea of what you're doing:
For each a-element in the input (which is a container-type M a, here [a]), you specify an entire new container M b. But as a final result, you want just a single "flat" container M b.
Well, let's take a look at the definition of the Monad type class:
class (Applicative m) => Monad m where
return :: a -> m a
(>>=) :: m a -> (a -> m b) -> m b
which is exactly what you need. And lists are an instance of Monad, so you can write
replicates :: [Int] -> [Int]
replicates l = l >>= \n -> replicate n n
Alternatively, this can be written
replicates l = do
n <- l
replicate n n
It might be interesting to know that the, perhaps easier to understand, list comprehension
replicates l = [ n | n <- l, _ <- [1..n] ]
as suggested by chi, is actually just syntactic sugar for another monad expression:
[ n | n <- l, _ <- [1..n] ] ≡ l >>= \n -> [1..n] >>= \_ -> return n
... or least it used to be in some old version of GHC, I think it now uses a more optimised implementation of list comprehensions. You can still turn on that de-sugaring variant with the -XMonadComprehensions flag.
Yet another solution, exploiting list comprehensions:
output = [ n | n <- input , m <- [1..n] ]
Compare the above with the imperative Python code:
for n in input: -- n <- input
for m in range(1,n+1): -- m <- [1..n] (in Python the second extreme is excluded, hence +1)
print n -- the n in [ n | ... ]
Note that m is unused -- in Haskell it is customary to can call it _ to express this:
output = [ n | n <- input , _ <- [1..n] ]
As a beginner, I more easily understand something like this:
concat $ map (\x -> take x $ repeat x) [1,2,3,4]
For "list as monads" it is important to know that there is also "concat" operation under the hood (in bind definition), IMO
A simple solution:
rep (x:xs) = replicate x x ++ rep xs
rep [] = []
Hints:
replicate 5 "a" gives ["a","a","a","a","a"], and it works the same way for any type in the second argument, but first argument must be of type Int
the operator ++ concatenates two lists
so the inferred type of rep is [Int] -> [Int], if you need to use other types you should use conversion functions
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.
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 ...) )).