Haskell: recursion with array arguments - haskell

Disclaimer: I'm new to Haskell and I don't remember a lot about FP from university, so there may be more than one or two errors in my code. This is also my code for Euler Problem 3.
I'm trying to recursively call a function with two arrays as arguments and an array as a result.
The goal:
assume n is 10 for this question
create a list of all natural numbers from 1 to n (variable is 'allNumbers' is code)
create another list of all natural numbers from 1 to n (variable is 'allFactors' is code)
take the first element in 'allFactors' and multiply the rest of the numbers of 'allFactors' by this number. (this generates an array of numbers)
remove all these numbers from 'allNumbers'
continue from 1 to n until 'allFactors' is empty.
Here is my code:
mkList :: Int -> [Int]
mkList n = [1..n-1]
modArray :: Int -> Int -> [Int]
modArray a b = [ x*b | x <- [1..a], x `mod` b == 0]
modArrayAll :: [Int] -> [Int] -> [Int]
modArrayAll [] [] = []
modArrayAll (x:xs) (y:ys) = (e)
where
m = head( ys)
n = length( xs)
e = (modArrayAll xs ys ) \\ modArray n m
(in main)
let allNumbers = mkList (first + 1)
let allFactors = mkList (first + 1)
let mainList2 = modArrayAll allNumbers allFactors
This results in a null list. However, if I have:
e = xs \\ modArray n m --WORKS for one iteration
I get all the odd numbers from 1 to 10.
My Question: Why isn't this working the way I would expect it? I would expect that the recursive stack would hit the empty array condition and just return an empty array which would not be removed from the calling array and it would continue on returning just the prime numbers?


I copied your goal notes:
-- assume n is 10 for this question
n=10
-- create a list of all natural numbers from 1 to n (variable is 'allNumbers' is code)
allNumbers = [1..n]
-- create another list of all natural numbers from 1 to n (variable is 'allFactors' is code)
allFactors = [2..n] -- i suspect you really wanted this rather than [1..n]
-- take the first element in 'allFactors' and
-- multiply the rest of the numbers of 'allFactors' by this number.
-- (this generates an array of numbers)
-- continue from 1 to n until 'allFactors' is empty
factorProducts = [ x*y | x <- allFactors, y <- allFactors]
-- remove all these numbers from 'allNumbers'
whatYouWanted = allNumbers \\ factorProducts
At the moment you seem to still be thinking in a fairly imperative mindset. Try thinking more about what you want, not how to get it :)


modArray n m creates a list of multiples of m, which you then remove from the "main list" of integers. But modArray n m includes 1*m, so each number is removed because it is a "multiple" of itself. In your test case you get only the odd numbers as a result, while you would want 2 to still be in the resulting list. Additionally 1 is included in your list of factors, which will eliminate all numbers, since they are all multiples of 1.
The terminating case of the recursion is modArrayAll [] [] = [], so there an empty list is returned. Then in the surrounding recursive calls this return value is used here:
(modArrayAll xs ys) \\ modArray n m
This tries to remove further elements (those returned by modArray n m) from the already empty list returned by modArrayAll xs ys. No new elements are added anywhere and the result list stays empty. With your algorithm you want the []-case to return the whole list of numbers, not an empty one. Then the \\ modArray n m in the surrounding recursive function calls can filter out more and more of the non-prime factors.

Related

Why does my function not work with an infinite list?

I'm trying to learn haskell and implemented a function conseq that would return a list of consecutive elements of size n.
conseq :: Int -> [Int] -> [[Int]]
conseq n x
| n == length(x) = [x]
| n > length(x) = [x]
| otherwise = [take n x] ++ (conseq n (drop 1 x))
This works correctly.
> take 5 $ conseq 2 [1..10]
[[1,2],[2,3],[3,4],[4,5],[5,6]]
However, if I pass [1..] instead of [1..10], the program gets stuck in an infinite loop.
As I understood it, haskell has lazy evaluation so I should still be able to get the same result right? Is it length? Shouldn't the first two conditions evaluate to false as soon as the length becomes greater than n?
What did I misunderstand?

One of the main reasons why using length is not a good idea is because when it has to be evaluated on an infinite list, it will get stuck in an infinite loop.
The good news is however, we don't need length. It would also make the time complexity worse. We can work with two enumerators, one is n-1 places ahead of the other. If this enumerator reaches the end of the list, then we know that the first enumerator still has n-1 elements, and thus we can stop yielding values:
conseq :: Int -> [a] -> [[a]]
conseq n ys = go (drop (n-1) ys) ys
where go [] _ = []
go (_:as) ba#(~(_:bs)) = take n ba : go as bs
This gives us thus:
Prelude> conseq 3 [1 ..]
[[1,2,3],[2,3,4],[3,4,5],[4,5,6],[5,6,7],[6,7,8],[7,8,9],[8,9,10],[9,10,11],[10,11,12],[11,12,13],[12,13,14],[13,14,15],[14,15,16],[15,16,17],[16,17,18],[17,18,19],[18,19,20],[19,20,21],[20,21,22],[21,22,23],[22,23,24],[23,24,25],[24,25,26],[25,26,27],…
Prelude> conseq 3 [1 .. 4]
[[1,2,3],[2,3,4]]

The first thing your function does is calculate length(x), so it knows whether it should return [x], [x], or [take n x] ++ (conseq n (drop 1 x))
length counts the number of elements in the list - all the elements. If you ask for the length of an infinite list, it never finishes counting.

What does split in Haskell do?

I'm new to Haskell and I came into the code below and don't actually understand what it does.
I know that the toDigits function is declared to take an Integer and return an array of Integers. If the argument n equals 0 or less, we return an empty array, otherwise...? That's the mystery!
toDigits :: Integer -> [Integer]
toDigits n
| n < 1 = []
| otherwise = reverse $ split [] n
where split _ 0 = []
split acc m = lastDigit m : split acc (dropLastDigit m)
Could you please explain to me?

reverse $ split [] n
Is the same as
reverse (split [] n)
It reverses the return value of split [] n and returns the result.
Split is defined in the next line.
It takes a list (an accumulator) and an Integer and does this:
Note that I assume that split is defined as follows (current implementation does not use acc). I also assume lastDigit and dropLastDigit do as their names would suggest:
split acc 0 = acc
split acc m = split (lastDigit m : acc) (dropLastDigit m)
Now, split returns acc if m is zero, otherwise it recursively prepends last digit of m to acc and pass it as the first argument to split, and removes the last digit from m and pass it as the second argument to split. In other words, this function eventually splits a number to its digits and returns the result as a list of Integers. Having this said calling split [] 1234 would return [1, 2, 3, 4]. You probably don't need to reverse the result of calling split.

Here, it takes a number and builds a list of its digits, one by one. At each moment it first checks whether the number argument is zero (then the returned value is an empty list), and, if not, a) takes its last digit; b) prepends the digit to the result of a recursive call to itself, now with number with the last digit dropped. The acc argument is never actually used, since the recursion was finally settled to be modulo cons.
Note that list is finally built in least-endian order: the first element to the list is the least significant digit of the number.

Double elements in a list if values are over certain threshold

Trying to double elements of a list that are greater than 5 for example.
to double every element in a list i would do this
doubleAll n = [2*x| x <-n]
now i want to double all elements in list that are greater than 5 using list comprehensions.
So if i do this
doubleAll n = [2*x| x <-n, x>5]
My list [1,2,3,4,5] would result in to [10]. But i want my list to show [1,2,3,4,10]
Can anyone explain what i did wrong and how can this be fixed?

An interpretation of [2*x| x <-n, x>5] is:
Take the next element from n and call it x
Proceed if x>5, otherwise go to step 1
Return the value 2*x as the next element of the list.
Repeat
From this it is clear that the x>5 filters out elements of n. The expression is equivalent to:
map (\x -> 2*x) ( filter (\x -> x>5) n )
As Arthur mentioned, you want something like:
[ if x > 5 then 2*x else x | x <- n ]
It's interpretation is:
Take the next value of n and call it x
Return the value if x > 5 then 2*x else x as then next value of the list.

This is clearer to understand if you don't use list comprehensions and use the map and filter operations instead:
-- Apply a function to every element of the list.
map :: (a -> b) -> [a] -> [b]
-- Throw out list elements that don't pass the test.
filter :: (a -> Bool) -> [a] -> [a]
Your original doubleAll is equivalent this:
-- Multiply every element of the list by two.
doubleAll xs = map (*2) xs
The version where you double only if x > 5 would be this:
-- Apply the `step` function to every element of the list. This step function
-- multiplies by two if x >= 5, otherwise it just returns its argument.
doubleAll xs = map step xs
where step x | x >= 5 = 2*x
| otherwise = x
The problem with the list comprehension version that you wrote is that it's instead equivalent to this:
-- Filter out list elements that are smaller than 5, then double the remaining ones.
doubleAll xs = map (*2) (filter (>=5) xs)
The list comprehension solution that produces the result you want would instead be this:
doubleAll xs = [if x >= 5 then x*2 else x | x <- xs]
As a more general remark, I always recommend to newcomers to stay away from list comprehensions and learn the higher-order list functions, which are more general and less magical.

Dovetail iteration over infinite lists in Haskell

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.

Iterating a function and analysing the result in haskell

Ok, referring back to my previous question, I am still working on learning haskell and solving the current problem of finding the longest chain from the following iteration:
chain n | n == 0 = error "What are you on about?"
| n == 1 = [1]
| rem n 2 == 0 = n : chain (n `div` 2)
| otherwise = n : chain (3 * n + 1)
I have this bit sorted, but I need to find the longest chain from a starting number below 1,000,000. So how do I make it do each starting number up to 1,000,000 and then print the one with the longest chain length.
I can do it for one example with:
Main> length (chain n)
I assume I need the output as an array and then use the maximum function to find the value largest chain length and then see how far along it is in the array of answers.
Is this a good way to go about finding a solution or is there a better way (perhaps with better efficiency)?

You are right about the maximum part. To get the list (that's what Haskell's []s are, arrays are different structures) you need to use the map higher-order function, like this:
chainLength n = length (chain n)
lengths = map chainLength [1..1000000]
Essentially, map takes as arguments a function and a list. It applies the function to each element in the list and returns the list of the results.
Since you will be needing the number whose chain has that length, you may want change the chainLength function to return the number as well, like this:
chainLength n = (n, length (chain n))
That way you will have an array of pairs, with each number and its chain length.
Now you need to get the pair with the largest second component. That's where the maximumBy function comes in. It works just like maximum but takes a function as a parameter to select how to compare the values. In this case, the second component of the pair. This comparison function takes two numbers and returns a value of type Ordering. This type has only three possible values: LT, EQ, GT, for less than, equal, and greater than, respectively.
So, we need a function that given two pairs tells us how the second components compare to each other:
compareSnd (_, y1) (_, y2) = compare y1 y2
-- Or, if you import Data.Function, you can write it like this (thanks alexey_r):
compareSnd = compare `on` snd -- reads nicely
I used the default compare function that compares numbers (well, not just numbers).
Now we only need to get the maximum using this function:
longestChain = maximumBy compareSnd lengths
That gets you a pair of the number with the longest chain and the corresponding length. Feel free to apply fst and snd as you please.
Note that this could be more much more concisely using zip and composition, but since you tagged the question as newbie, I thought it better to break it down like this.

SPOILER (solving the problem for positive integers under 100):
module Test where
import Data.List -- this contains maximumBy
chain n
| n == 0 = error "What are you on about?"
| n == 1 = [1]
| rem n 2 == 0 = n : chain (n `div` 2)
| otherwise = n : chain (3 * n + 1)
chains = map (\x -> (x,chain x)) [1..100]
cmpSnd (a,b) (c,d)
| length b > length d = GT
| length b == length d = EQ
| otherwise = LT
solve = (fst . maximumBy cmpSnd) chains
The chains function makes use of map. It applies a function to every element of a list of a values, so
map succ [1,2]
is the same as
[succ 1,succ 2]
The cmpSnd function is a comparison function that probably exists somewhere deep in the Hierarchical Libraries, but I could not find it, so I created it. GT means "the first value is greater than the second", the rest is trivial.
Solve takes the maximum (by utilizing the comparison function we defined earlier) of the list. This will be a pair of an integer and a list. It will return the integer only (because of the fst).
A comment: Your chain function is not tail-recursive. This means that large chains will inevitably result in a Stack Overflow. You shall add an explicit accumulator variable and make it tail-recursive.

Something like
fst $ maximumBy (length . snd) $ zip [1..1000000] $ map chain [1..1000000]
(untested)
i.e. don't work out how far along the longest chain is in the list of longest chains, but carry around the seed values with the chains instead.

I studied Haskell years ago, so I don't remember it that well. On the other hand I've tested this code and it works. You will get the max chain and the number that generates it. But as fiships has stated before, it will overflow for big values.
chain :: Int -> [Int]
chain n
| n == 0 = []
| n == 1 = [1]
| rem n 2 == 0 = n : chain (n `div` 2)
| otherwise = n : chain (3 * n + 1)
length_chain :: Int -> Int
length_chain n = length (chain n)
max_pos :: (Int,Int) -> Int -> [Int] -> (Int,Int)
max_pos (m,p) _ [] = (m,p)
max_pos (m,p) a (x:xs)
| x > m = max_pos (x,a) (a+1) xs
| otherwise = max_pos (m,p) (a+1) xs
The instruction will be
Main> max_pos (0,0) 1 (map length_chain [1..10000])
(262,6171)

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