I am implementing some algorithm on haskell. This algorithm requires generating some data.
I have a function of an algorithm which takes generation function as a parameter. For example, algorithm is just multiplying input data by n:
algo :: a -> ??? -> [a]
algo n dgf = map (\x -> x * n) $ dgf
dgf is used to generate data. How to write function header correctly, as dgf can be any function with any number of parameters?
Another variant is accepting not the generation function but already generated data.
algo :: a -> [b] -> [a]
algo n d = (\x -> n*x) d
So, now let's imagine I'm generation data with stdGen, which uses IO. How can I make function more generic, so that it could accept both IO instance and plain values like just [1,2,3]. This also relates to variant with function, as it can also produce IO.
All in all, which solution is better - having a generation function or a pre-generated data?
Thanks in advance.
One option is to take a stream rather than a list. If generating the values involves performing IO, and there may be many many values, this is often the best approach. There are several packages that offer streams of some sort, but I'll use the streaming package in this example.
import qualified Streaming.Prelude as S
import Streaming
algo :: Monad m => a -> Stream (Of a) m r -> Stream (Of a) m r
algo a = S.map (a +)
You can read Stream (Of a) m r as "a way to use operations in m to produce successive values of type a and finally a result of type r". This algo function doesn't commit to any particular way of generating the data; they can be created purely:
algo a (S.each [these, are, my, elements])
or within IO,
algo a $ S.takeWhile (> 3) (S.readLn :: Stream (Of Int) IO ())
or using a randomness monad, or whatever you like.
For contrast, I'm going to take the opposite approach as dfeuer's answer.
Just use lists.
Consider your first example:
algo :: a -> ??? -> [a]
algo n dgf = map (\x -> x * n) $ dgf
You ask "How to write function header correctly, as dgf can be any function with any number of parameters?"
Well, one way is to use uncurrying.
Normally, Haskell functions are curried. If we have a function like
add :: Int -> Int -> Int
add x y = x + y
And we want a function that adds two to its input we can just use add 2.
>>> map (add 2) [1..10]
[3,4,5,6,7,8,9,10,11,12]
Because add is not actually a function that takes two arguments,
it's a function of one argument that returns a function of one argument.
We could have added parentheses to the argument of add above to make this more clear:
add :: Int -> (Int -> Int)
In Haskell, all functions are functions of one argument.
However, we can also go the other way - uncurry a function
that returns a function to get a function that takes a pair:
>>> :t uncurry
uncurry :: (a -> b -> c) -> (a, b) -> c
>>> :t uncurry add
uncurry add :: (Int, Int) -> Int
This can also be useful, say if we want to find the sum of each pair in a list:
>>> map (uncurry add) [ (1,2), (3,4), (5,6), (7,8), (9,10) ]
[3,7,11,15,19]
In general, we can uncurry any function of type a0-> a1 -> ... -> aN -> b
into a function (a0, a1, ..., aN) -> b, though there might not be
a cute library function to do it for us.
With that in mind, we could implement algo by passing it an uncurried
function and a tuple of values:
algo :: Num a => a -> (t -> [a]) -> t -> [a]
algo n f t = map (\x -> x * n) $ f t
And then use anonymous functions to uncurry our argument functions:
>>> algo 2 (\(lo,hi) -> enumFromTo lo hi) (5, 10)
[10,12,14,16,18,20]
>>> algo 3 (\(a,b,c,d) -> zipWith (+) [a..b] [c..d]) (1, 5, 10, 14)
[33,39,45,51,57]
Now we could do it this way, but we don't need to. As implemented above,
algo is only using f and t once. So why not pass it the list directly?
algo' :: Num a => a -> [a] -> [a]
algo' n ns = map (\x -> x * n) ns
It calculates the same results:
>>> algo' 2 $ (\(lo,hi) -> enumFromTo lo hi) (5, 10)
[10,12,14,16,18,20]
>>> algo' 2 $ enumFromTo 5 10
[10,12,14,16,18,20]
>>> algo' 3 $ (\(a,b,c,d) -> zipWith (+) [a..b] [c..d]) (1, 5, 10, 14)
[33,39,45,51,57]
>>> algo' 3 $ zipWith (+) [1..5] [10..14]
[33,39,45,51,57]
Furthermore, since haskell is non-strict, the argument to algo' isn't evaluated
until it's actually used, so we don't have to worry about "wasting" time computing
arguments that won't actually be used:
algo'' :: Num a => a -> [a] -> [a]
algo'' n ns = [n,n,n,n]
algo'' doesn't use the list passed to it, so it's never forced, so whatever
computation is used to calculate it never runs:
>>> let isPrime n = n > 2 && null [ i | i <- [2..n-1], n `rem` i == 0 ]
>>> :set +s
>>> isPrime 10000019
True
(6.18 secs, 2,000,067,648 bytes)
>>> algo'' 5 (filter isPrime [1..999999999999999])
[5,5,5,5]
(0.01 secs, 68,936 bytes)
Now to the second part of your question - what if your data is being generated within some monad?
Rather than convince algo to operate on monadic values, you could take the stream
based approach as dfeuer explains. Or you could just use a list.
Just because you're in a monad, doesn't mean that your values suddenly become strict.
For example, want a infinite list of random numbers? No problem.
newRandoms :: Num a -> IO [a]
newRandoms = unfoldr (\g -> Just (random g)) <$> newStdGen
Now I can just pass those to some algorithm:
>>> rints <- newRandoms :: IO [Int]
(0.00 secs, 60,624 bytes)
>>> algo'' 5 rints
[5,5,5,5]
(0.00 secs, 68,920 bytes)
For a small program which is just reading input from a file or two, there's no problem
with just using readFile and lazy I/O to get a list to operate on.
For example
>>> let grep pat lines = [ line | line <- lines, pat `isInfixOf` line ]
>>> :set +s
>>> dict <- lines <$> readFile "/usr/share/dict/words"
(0.01 secs, 81,504 bytes)
>>> grep "poop" dict
["apoop","epoophoron","nincompoop","nincompoopery","nincompoophood","nincompoopish","poop","pooped","poophyte","poophytic","whisterpoop"]
(0.72 secs, 423,650,152 bytes)
Related
Given a list of integers xs, let:
count :: [Integer] -> Integer -> Integer
count xs n = length . filter (==n) $ xs
count the number of times the integer n occurs in the list.
Now, given a "list" (some sort of array of integers, can be something besides a List) of length n, write a function
countSequence :: [Integer] -> Integer -> Integer -> Integer
countSequence xs n m = [count xs x | x <- [0..m]]
that outputs the "list of counts" (0th index contains number of times 0 occurs in the list, 1st index contains number of times 1 occurs in the list, etc) that has time compleity o(m*n)
The above implementation I've given has complexity O(m*n). In Python (which I'm more familiar with), it's easy to do this in O(m + n) time --- iterate through the list, and each element increment a counter in some other list, which is initialized to be all zeros and length (m+1).
How could I get a better implementation in Haskell? I'd prefer if it wasn't some trivial way to implement the Python solution (such as adding another input to the function to keep the "list of counts" in and then interating through it).
In O(n+m) (sort of, I think, maybe):
import Data.Ix (inRange)
import qualified Data.IntMap.Strict as IM
countSequence m =
foldl' count IM.empty . filter (inRange (0,m))
where count a b = IM.insertWith (+) b 1 a
gives
> countSequence 2 [1,2,3,1,2,-1]
fromList [(1,2),(2,2)]
I haven't used n because you also didn't use n and I'm not sure what it's supposed to be. I also moved the list to the last argument to put it in a position to be eta reduced.
I think you should use your Python intuition -- iterate through the one list and increment a counter in another list. Here's an implementation with O(n+m) runtime:
import Data.Array
countSequence xs m = accumArray (+) 0 (0,m) [(x, 1) | x <- xs, inRange (0,m) x]
(This use case is even the motivating example for the existence of accumArray in the documentation!) In ghci:
> countSequence ([1..5] ++ [1,3..5] ++ [1,4..5] ++ [1,5]) 3
array (0,3) [(0,0),(1,4),(2,1),(3,2)]
I guess using Data.IntMap would be as efficient as it gets for this job. One foldr pass is done to establish the IntMap (cm) and a map to construct a new list holding the counts of elements at corresponding positions.
import qualified Data.IntMap.Lazy as IM
countSequence :: [Int] -> [Int]
countSequence xs = map (\x -> let cm = foldr (\x m -> IM.alter (\mx -> if mx == Nothing then Just 1 else fmap (+1) mx) x m) IM.empty xs
in IM.findWithDefault 0 x cm) xs
*Main> countSequence [1,2,5,1,3,7,8,5,6,4,1,2,3,7,9,3,4,8]
[3,2,2,3,3,2,2,2,1,2,3,2,3,2,1,3,2,2]
*Main> countSequence [4,5,4]
[2,1,2]
*Main> *Main> countSequence [9,8,7,6,5]
[1,1,1,1,1]
I've been solving a few combinatoric problems on Haskell, so I wrote down those 2 functions:
permutations :: (Eq a) => [a] -> [[a]]
permutations [] = [[]]
permutations list = do
x <- list
xs <- permutations (filter (/= x) list)
return (x : xs)
combinations :: (Eq a, Ord a) => Int -> [a] -> [[a]]
combinations 0 _ = [[]]
combinations n list = do
x <- list
xs <- combinations (n-1) (filter (> x) list)
return (x : xs)
Which works as follows:
*Main> permutations [1,2,3]
[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
*Main> combinations 2 [1,2,3,4]
[[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]]
Those were uncomfortably similar, so I had to abstract it. I wrote the following abstraction:
combinatoric next [] = [[]]
combinatoric next list = do
x <- list
xs <- combinatoric next (next x list)
return (x : xs)
Which receives a function that controls how to filter the elements of the list. It can be used to easily define permutations:
permutations :: (Eq a) => [a] -> [[a]]
permutations = combinatoric (\ x ls -> filter (/= x) ls)
But I couldn't define combinations this way since it carries an state (n). I could extend the combinatoric with an additional state argument, but that'd become too clunky and I remember such approach was not necessary in a somewhat similar situation. Thus, I wonder: is it possible to define combinations using combinatorics? If not, what is a better abstraction of combinatorics which successfully subsumes both functions?
This isn't a direct answer to your question (sorry), but I don't think your code is correct. The Eq and Ord constraints tipped me off - they shouldn't be necessary - so I wrote a couple of QuickCheck properties.
prop_numberOfPermutations xs = length (permutations xs) === factorial (length xs)
where _ = (xs :: [Int]) -- force xs to be instantiated to [Int]
prop_numberOfCombinations (Positive n) (NonEmpty xs) = n <= length xs ==>
length (combinations n xs) === choose (length xs) n
where _ = (xs :: [Int])
factorial :: Int -> Int
factorial x = foldr (*) 1 [1..x]
choose :: Int -> Int -> Int
choose n 0 = 1
choose 0 r = 0
choose n r = choose (n-1) (r-1) * n `div` r
The first property checks that the number of permutations of a list of length n is n!. The second checks that the number of r-combinations of a list of length n is C(n, r). Both of these properties fail when I run them against your definitions:
ghci> quickCheck prop_numberOfPermutations
*** Failed! Falsifiable (after 5 tests and 4 shrinks):
[0,0,0]
3 /= 6
ghci> quickCheck prop_numberOfCombinations
*** Failed! Falsifiable (after 4 tests and 1 shrink):
Positive {getPositive = 2}
NonEmpty {getNonEmpty = [3,3]}
0 /= 1
It looks like your functions fail when the input list contains duplicate elements. Writing an abstraction for an incorrect implementation isn't a good idea - don't try and run before you can walk! You might find it helpful to read the source code for the standard library's definition of permutations, which does not have an Eq constraint.
First let's improve the original functions. You assume that all elements are distinct wrt their equality for permutations, and that they're distinct and have an ordering for combinations. These constraints aren't necessary and as described in the other answer, the code can produce wrong results. Following the robustness principle, let's accept just unconstrained lists. For this we'll need a helper function that produces all possible splits of a list:
split :: [a] -> [([a], a, [a])]
split = loop []
where
loop _ [] = []
loop rs (x:xs) = (rs, x, xs) : loop (x:rs) xs
Note that the implementation causes prefixes returned by this function to be reversed, but it's nothing we require.
This allows us to write generic permutations and combinations.
permutations :: [a] -> [[a]]
permutations [] = [[]]
permutations list = do
(pre, x, post) <- split list
-- reversing 'pre' isn't really necessary, but makes the output
-- order natural
xs <- permutations (reverse pre ++ post)
return (x : xs)
combinations :: Int -> [a] -> [[a]]
combinations 0 _ = [[]]
combinations n list = do
(_, x, post) <- split list
xs <- combinations (n-1) post
return (x : xs)
Now what they have in common:
At each step they pick an element to output,
update the list of elements to pick from and
stop after some condition is met.
The last point is a bit problematic, as for permutations we end once the list to choose from is empty, while for combinations we have a counter. This is probably the reason why it was difficult to generalize. We can work around this by realizing that for permutations the number of steps is equal to the length of the input list, so we can express the condition in the number of repetitions.
For such problems it's often very convenient to express them using StateT s [] monad, where s is the state we're working with. In our case it'll be the list of elements to choose from. The core of our combinatorial functions can be then expressed with StateT [a] [] a: pick an element from the state and update the state for the next step. Since the stateful computations all happen in the [] monad, we automatically branch all possibilities. With that, we can define a generic function:
import Control.Monad.State
combinatoric :: Int -> StateT [a] [] b -> [a] -> [[b]]
combinatoric n k = evalStateT $ replicateM n k
And then define permutations and combinations by specifying the appropriate number of repetitions and what's the core StateT [a] [] a function:
permutations' :: [a] -> [[a]]
permutations' xs = combinatoric (length xs) f xs
where
f = StateT $ map (\(pre, x, post) -> (x, reverse pre ++ post)) . split
combinations' :: Int -> [a] -> [[a]]
combinations' n xs = combinatoric n f xs
where
f = StateT $ map (\(_, x, post) -> (x, post)) . split
I'm trying to resolve problem 14 of Project Euler (http://projecteuler.net/problem=14) and I hit a dead end using Haskell.
Now, I know that the numbers may be small enough and I could do a brute force, but that isn't the purpose of my exercise.
I am trying to memorize the intermediate results in a Map of type Map Integer (Bool, Integer) with the meaning of:
- the first Integer (the key) holds the number
- the Tuple (Bool, Interger) holds either (True, Length) or (False, Number)
where Length = length of the chain
Number = the number before him
Ex:
for 13: the chain is 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
My map should contain :
13 - (True, 10)
40 - (False, 13)
20 - (False, 40)
10 - (False, 20)
5 - (False, 10)
16 - (False, 5)
8 - (False, 16)
4 - (False, 8)
2 - (False, 4)
1 - (False, 2)
Now when I search for another number like 40 i know that the chain has (10 - 1) length and so on.
I want now, if I search for 10, not only to tell me that length of 10 is (10 - 3) length and update the map, but also I want to update 20, 40 in case they are still (False, _)
My code:
import Data.Map as Map
solve :: [Integer] -> Map Integer (Bool, Integer)
solve xs = solve' xs Map.empty
where
solve' :: [Integer] -> Map Integer (Bool, Integer) -> Map Integer (Bool, Integer)
solve' [] table = table
solve' (x:xs) table =
case Map.lookup x table of
Nothing -> countF x 1 (x:xs) table
Just (b, _) ->
case b of
True -> solve' xs table
False -> {-WRONG-} solve' xs table
f :: Integer -> Integer
f x
| x `mod` 2 == 0 = x `quot` 2
| otherwise = 3 * x + 1
countF :: Integer -> Integer -> [Integer] -> Map Integer (Bool, Integer) -> Map Integer (Bool, Integer)
countF n cnt (x:xs) table
| n == 1 = solve' xs (Map.insert x (True, cnt) table)
| otherwise = countF (f n) (cnt + 1) (x:xs) $ checkMap (f n) n table
checkMap :: Integer -> Integer -> Map Integer (Bool, Integer) -> Map Integer (Bool, Integer)
checkMap n rez table =
case Map.lookup n table of
Nothing -> Map.insert n (False, rez) table
Just _ -> table
At the {-WRONG-} part we should update all the values like in the following example:
--We are looking for 10:
10 - (False, 20)
|
V {-finally-} update 10 => (True, 10 - 1 - 1 - 1)
20 - (False, 40) ^
| |
V update 20 => 20 - (True, 10 - 1 - 1)
40 - (False, 13) ^
| |
V update 40 => 40 - (True, 10 - 1)
13 - (True, 10) ^
| |
---------------------------
The problem is that I don't know if its possible to do 2 things in a function like updating a number and continue the recurence. In a C like language I may do something like (pseudocode):
void f(int n, tuple(b,nr), int &length, table)
{
if(b == False) f (nr, (table lookup nr), 0, table);
// the bool is true so we got a length
else
{
length = nr;
return;
}
// Since this is a recurence it would work as a stack, producing the right output
table update(n, --cnt);
}
The last instruction would work since we are sending cnt by reference. Also we always know that it will finish at some point and cnt should not be < 1.
The easiest optimization (as you have identified) is memoization. You have attempted create a memoization system yourself, however have come across issues on how to store the memoized values. There are solutions to doing this in a maintainable way, such as using a State monad or a STArray. However, there is a much simpler solution to your problem - use haskell's existing memoization. Haskell by default remembers constant values, so if you create a value that stores the collatz values, it will be automatically memoized!
A simple example of this is the following fibonacci definition:
fib :: Int -> Integer
fib n = fibValues !! n where
fibValues = 1 : 1 : zipWith (+) fibValues (tail fibValues)
The fibValues is a [Integer], and as it is just a constant value, it is memoized. However, that doesn't mean it is all memoized at once, since as it is an infinte list, this would never finish. Instead, the values are only calculated when needed, as haskell is lazy.
So if you do something similar with your problem, you will get memoization without a lot of the work. However, using a list like above won't work well in your solution. This is because the collatz algorithm uses many different values to get the result for a given number, so the container used will require random access to be efficient. The obvious choice is an array.
collatzMemoized :: Array Integer Int
Next, we need to fill up the array with the correct values. I'll write this function pretending a collatz function exists that calculates the collatz value for any n. Also, note that arrays are fixed size, so a value needs to be used to determine the maximum number to memoize. I'll use a million, but any value can be used (it is a memory/speed tradeoff).
collatzMemoized = listArray (1, maxNumberToMemoize) $ map collatz [1..maxNumberToMemoize] where
maxNumberToMemroize = 1000000
That is pretty straightforward, the listArray is given bounds, and the a list of all the collatz values in that range is given to it. Remember that this won't calculate all the collatz values straight away, as the values are lazy.
Now, the collatz function can be written. The most important part is to only check the collatzMemoized array if the number being checked is within its bounds:
collatz :: Integer -> Int
collatz 1 = 1
collatz n
| inRange (bounds collatzMemoized) nextValue = 1 + collatzMemoized ! nextValue
| otherwise = 1 + collatz nextValue
where
nextValue = case n of
1 -> 1
n | even n -> n `div` 2
| otherwise -> 3 * n + 1
In ghci, you can now see the effectiveness of the memoization. Try collatz 200000. It will take about 2 seconds to finish. However, if you run it again, it will complete instantly.
Finally, the solution can be found:
maxCollatzUpTo :: Integer -> (Integer, Int)
maxCollatzUpTo n = maximumBy (compare `on` snd) $ zip [1..n] (map collatz [1..n]) where
and then printed:
main = print $ maxCollatzUpTo 1000000
If you run main, the result will be printed in about 10 seconds.
Now, a small problem with this approach is it uses a lot of stack space. It will work fine in ghci (which seems to use be more flexible with regards to stack space). However, if you compile it and try to run the executable, it will crash (with a stack space overflow). So to run the program, you have to specify more when you compile it. This can be done by adding -with-rtsopts='K64m' to the compile options. This increases the stack to 64mb.
Now the program can be compiled and ran:
> ghc -O3 --make -with-rtsopts='-K6m' problem.hs
Running ./problem will give the result in less than a second.
You are going about memoization the hard way, trying to write an imperative program in Haskell. Borrowing from David Eisenstat's solution, we'll solve it as j_random_hacker suggested:
collatzLength :: Integer -> Integer
collatzLength n
| n == 1 = 1
| even n = 1 + collatzLength (n `div` 2)
| otherwise = 1 + collatzLength (3*n + 1)
The dynamic programming solution for this is to replace the recursion with looking things up in a table. Let's make a function where we can replace the recursive call:
collatzLengthDef :: (Integer -> Integer) -> Integer -> Integer
collatzLengthDef r n
| n == 1 = 1
| even n = 1 + r (n `div` 2)
| otherwise = 1 + r (3*n + 1)
Now we could define the recursive algorithm as
collatzLength :: Integer -> Integer
collatzLength = collatzLengthDef collatzLength
Now we could also make a tabled version of this (it takes a number for the table size, and returns a collatzLength function that is calculated using a table of that size):
-- A utility function that makes memoizing things easier
buildTable :: (Ix i) => (i, i) -> (i -> e) -> Array i e
buildTable bounds f = array $ map (\x -> (x, f x)) $ range bounds
collatzLengthTabled :: Integer -> Integer -> Integer
collatzLengthTabled n = collatzLengthTableLookup
where
bounds = (1, n)
table = buildTable bounds (collatzLengthDef collatzLengthTableLookup)
collatzLengthTableLookup =
\x -> Case inRange bounds x of
True -> table ! x
_ -> (collatzLengthDef collatzLengthTableLookup) x
This works by defining the collatzLength to be a table lookup, with the table being the definition of the function, but with recursive calls replaced by table lookup. The table lookup function checks to see if the argument to the function is in the range that is tabled, and falls back on the definition of the function. We can even make this work for tabling any function like this:
tableRange :: (Ix a) => (a, a) -> ((a -> b) -> a -> b) -> a -> b
tableRange bounds definition = tableLookup
where
table = buildTable bounds (definition tableLookup)
tableLookup =
\x -> Case inRange bounds x of
True -> table ! x
_ -> (definition tableLookup) x
collatzLengthTabled n = tableRange (1, n) collatzLengthDef
You just need to make sure that you
let memoized = collatzLengthTabled 10000000
... memoized ...
So that only one table is built in memory.
I remember finding memoisation of dynamic programming algorithms very counterintuitive in Haskell, and it's been a while since I've done it, but hopefully the following trick works for you.
But first, I don't quite understand your current DP scheme, though I suspect it may be quite inefficient as it seems like it will need to update many entries for each answer. (a) I don't know how to do this in Haskell, and (b) you don't need to do this to solve the problem efficiently ;-)
I suggest the following approach instead: first build an ordinary recursive function that computes the right answer for an input number. (Hint: it will have a signature like collatzLength :: Int -> Int.) When you have this function working, just replace its definition with the definition of an array whose elements are defined lazily with the array function using an association list, and replace all recursive calls to the function to array lookups (e.g. collatzLength 42 would become collatzLength ! 42). This will automagically populate the array in the necessary order! So your "top-level" collatzLength object will now actually be an array, rather than a function.
As I suggested above, I would use an array instead of a map datatype to hold the DP table, since you will need to store values for all integer indices from 1 up to 1,000,000.
I don't have a Haskell compiler handy, so I apologize for any broken code.
Without memoization, there's a function
collatzLength :: Integer -> Integer
collatzLength n
| n == 1 = 1
| even n = 1 + collatzLength (n `div` 2)
| otherwise = 1 + collatzLength (3*n + 1)
With memoization, the type signature is
memoCL :: Map Integer Integer -> Integer -> (Map Integer Integer, Integer)
since memoCL receives a table as input and gives the updated table as output. What memoCL needs to do is intercept the return of the recursive call with a let form and insert the new result.
-- table must have an initial entry for 1
memoCL table n = case Map.lookup n table of
Just m -> (table, m)
Nothing -> let (table', m) = memoCL table (collatzStep n) in (Map.insert n (1 + m) table', 1 + m)
collatzStep :: Integer -> Integer
collatzStep n = if even n then n `div` 2 else 3*n + 1
At some point you'll get sick of the above idiom. Then it's time for monads.
I eventually modify the {-WRONG-} part to do what it should with a call to mark x (b, n) [] xs table where
mark :: Integer -> (Bool, Integer) -> [Integer] -> [Integer] -> Map Integer (Bool, Integer) -> Map Integer (Bool, Integer)
mark crtElem (b, n) list xs table
| b == False = mark n (findElem n table) (crtElem:list) xs table
| otherwise = continueWith n list xs table
continueWith :: Integer -> [Integer] -> [Integer] -> Map Integer (Bool, Integer) -> Map Integer (Bool, Integer)
continueWith _ [] xs table = solve' xs table
continueWith cnt (y:ys) xs table = continueWith (cnt - 1) ys xs (Map.insert y (True, cnt - 1) table)
findElem :: Integer -> Map Integer (Bool, Integer) -> (Bool, Integer)
findElem n table =
case Map.lookup n table of
Nothing -> (False, 0)
Just (b, nr) -> (b, nr)
But it seams that there are better (and far less verbose) answers than this 1
Maybe you might find interesting how I solved the problem. Its is pretty functional though it might be not the most efficient thing on earth :)
You can find the code here: https://github.com/fmancinelli/project-euler/blob/master/haskell/project-euler/Problem014.hs
P.S.: Disclaimer: I was doing Project Euler exercises in order to learn Haskell, so the quality of the solution could be debatable.
Since we are studying recursion schemes, here's one for you.
Let's consider functor N(A,B,X)=A+B*X, which is a stream of Bs with the last element being A.
{-# LANGUAGE DeriveFunctor
, TypeFamilies
, TupleSections #-}
import Data.Functor.Foldable
import qualified Data.Map as M
import Data.List
import Data.Function
import Data.Int
data N a b x = Z a | S b x deriving (Functor)
This stream is handy for several kinds of iterations. For one, we can use it to represent a chain of Ints in a Collatz sequence:
type instance Base Int64 = N Int Int64
instance Foldable Int64 where
project 1 = Z 1
project x | odd x = S x $ 3*x+1
project x = S x $ x `div` 2
This is just a algebra, not a initial one, because the transformation is not a isomorphism (same chain of Ints is part of a chain for 2*x and (x-1)/3), but this is sufficient to represent the fixpoint Base Int64 Int64.
With this definition, cata is going to feed the chain to the algebra given to it, and you can use it to construct a memo Map of integers to the chain length. Finally, anamorphism can use it to generate a stream of solutions to the problem of different sizes:
problems = ana (uncurry $ cata . phi) (M.empty, 1) where
phi :: M.Map Int64 Int ->
Base Int64 (Prim [(Int64, Int)] (M.Map Int64 Int, Int64)) ->
Prim [(Int64, Int)] (M.Map Int64 Int, Int64)
phi m (Z v) = found m 1 v
phi m (S x ~(Cons (_, v') (m', _))) = maybe (notFound m' x v') (found m x) $
M.lookup x m
The ~ before (Cons ...) means lazy pattern matching. We don't touch the pattern until the values are needed. If not for lazy pattern matching, it would always construct the whole chain, and using the map would be useless. With lazy pattern matching we only construct the values v' and m' if the chain length for x was not in the map.
Helper functions construct the stream of (Int, chain length) pairs:
found m x v = Cons (x, v) (m, x+1)
notFound m x v = Cons (x, 1+v) (M.insert x (1+v) m, x+1)
Now just take the first 999999 problems, and figure out the one that has the longest chain:
main = print $ maximumBy (compare `on` snd) $ take 999999 problems
This works slower than array-based solution, because Map lookup is logarithmic of map size, but this solution is not fixed size. Still, it finishes in about 5 seconds.
While solving a problem, I had to calculate the divisors of a number. I have two implementations that produce all divisors > 1 for a given number.
The first is using simple recursion:
divisors :: Int64 -> [Int64]
divisors k = divisors' 2 k
where
divisors' n k | n*n > k = [k]
| n*n == k = [n, k]
| k `mod` n == 0 = (n:(k `div` n):result)
| otherwise = result
where result = divisors' (n+1) k
The second one uses list processing functions from the Prelude:
divisors2 :: Int64 -> [Int64]
divisors2 k = k : (concatMap (\x -> [x, k `div` x]) $!
filter (\x -> k `mod` x == 0) $!
takeWhile (\x -> x*x <= k) [2..])
I find that the first implementation is faster (I printed the whole list returned, so that no part of the result remains unevaluated due to laziness). The two implementations produce differently ordered divisors, but that is not a problem for me. (In fact, if k is a perfect square, the square root is output twice in the second implementation - again not a problem).
In general are such recursive implementations faster in Haskell? Also, I would appreciate any pointers to make either of these codes faster. Thanks!
EDIT:
Here is the code I am using to compare these two implementations for performance: https://gist.github.com/3414372
Here are my timing measurements:
Using divisor2 with strict evaluation ($!)
$ ghc --make -O2 div.hs
[1 of 1] Compiling Main ( div.hs, div.o )
Linking div ...
$ time ./div > /tmp/out1
real 0m7.651s
user 0m7.604s
sys 0m0.012s
Using divisors2 with lazy evaluation ($):
$ ghc --make -O2 div.hs
[1 of 1] Compiling Main ( div.hs, div.o )
Linking div ...
$ time ./div > /tmp/out1
real 0m7.461s
user 0m7.444s
sys 0m0.012s
Using function divisors
$ ghc --make -O2 div.hs
[1 of 1] Compiling Main ( div.hs, div.o )
Linking div ...
$ time ./div > /tmp/out1
real 0m7.058s
user 0m7.036s
sys 0m0.020s
The recursive version is not in general faster than the list-based version. This is because the GHC compiler employs List fusion optimizations when a computation follows a certain pattern. This means that list generators and "list transformers" might be fused into one big generator instead.
However, when you use $!, you basically tell the compiler to "Please produce the first cons of this list before performing the next step." This means that GHC is forced to at least compute one intermediate list element, which disables the whole fusion optimization entirely.
So, the second algorithm is slower, because you produce intermediate lists that have to be constructed and destructed, while the recursive algorithm simply produces a single list straight away.
Since you asked, to make it faster a different algorithm should be used. Simple and straightforward is to find a prime factorization first, then construct the divisors from it somehow.
Standard prime factorization by trial division is:
factorize :: Integral a => a -> [a]
factorize n = go n (2:[3,5..]) -- or: `go n primes`
where
go n ds#(d:t)
| d*d > n = [n]
| r == 0 = d : go q ds
| otherwise = go n t
where (q,r) = quotRem n d
-- factorize 12348 ==> [2,2,3,3,7,7,7]
Equal prime factors can be grouped and counted:
import Data.List (group)
primePowers :: Integral a => a -> [(a, Int)]
primePowers n = [(head x, length x) | x <- group $ factorize n]
-- primePowers = map (head &&& length) . group . factorize
-- primePowers 12348 ==> [(2,2),(3,2),(7,3)]
Divisors are usually constructed, though out of order, with:
divisors :: Integral a => a -> [a]
divisors n = map product $ sequence
[take (k+1) $ iterate (p*) 1 | (p,k) <- primePowers n]
Hence, we have
numDivisors :: Integral a => a -> Int
numDivisors n = product [ k+1 | (_,k) <- primePowers n]
The product here comes from the sequence in the definition above it, because sequence :: Monad m => [m a] -> m [a] for list monad m ~ [] constructs lists of all possible combinations of elements picked by one from each member list, sequence_lists = foldr (\xs rs -> [x:r | x <- xs, r <- rs]) [[]], so that length . sequence_lists === product . map length, and or course length . take n === n for infinite argument lists.
In-order generation is possible, too:
ordDivisors :: Integral a => a -> [a]
ordDivisors n = foldr (\(p,k)-> foldi merge [] . take (k+1) . iterate (map (p*)))
[1] $ reverse $ primePowers n
foldi :: (a -> a -> a) -> a -> [a] -> a
foldi f z (x:xs) = f x (foldi f z (pairs xs)) where
pairs (x:y:xs) = f x y:pairs xs
pairs xs = xs
foldi f z [] = z
merge :: Ord a => [a] -> [a] -> [a]
merge (x:xs) (y:ys) = case (compare y x) of
LT -> y : merge (x:xs) ys
_ -> x : merge xs (y:ys)
merge xs [] = xs
merge [] ys = ys
{- ordDivisors 12348 ==>
[1,2,3,4,6,7,9,12,14,18,21,28,36,42,49,63,84,98,126,147,196,252,294,343,441,588,
686,882,1029,1372,1764,2058,3087,4116,6174,12348] -}
This definition is productive, too, i.e. it starts producing the divisors right away, without noticeable delay:
{- take 20 $ ordDivisors $ product $ concat $ replicate 5 $ take 11 primes
==> [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]
(0.00 secs, 525068 bytes)
numDivisors $ product $ concat $ replicate 5 $ take 11 primes
==> 362797056 -}
Suppose someone makes a program to play chess, or solve sudoku. In this kind of program it makes sense to have a tree structure representing game states.
This tree would be very large, "practically infinite". Which isn't by itself a problem as Haskell supports infinite data structures.
An familiar example of an infinite data structure:
fibs = 0 : 1 : zipWith (+) fibs (tail fibs)
Nodes are only allocated when first used, so the list takes finite memory. One may also iterate over an infinite list if they don't keep references to its head, allowing the garbage collector to collect its parts which are not needed anymore.
Back to the tree example - suppose one does some iteration over the tree, the tree nodes iterated over may not be freed if the root of the tree is still needed (for example in an iterative deepening search, the tree would be iterated over several times and so the root needs to be kept).
One possible solution for this problem that I thought of is using an "unmemo-monad".
I'll try to demonstrate what this monad is supposed to do using monadic lists:
import Control.Monad.ListT (ListT) -- cabal install List
import Data.Copointed -- cabal install pointed
import Data.List.Class
import Prelude hiding (enumFromTo)
nums :: ListT Unmemo Int -- What is Unmemo?
nums = enumFromTo 0 1000000
main = print $ div (copoint (foldlL (+) 0 nums)) (copoint (lengthL nums))
Using nums :: [Int], the program would take a lot of memory as a reference to nums is needed by lengthL nums while it is being iterated over foldlL (+) 0 nums.
The purpose of Unmemo is to make the runtime not keep the nodes iterated over.
I attempted using ((->) ()) as Unmemo, but it yields the same results as nums :: [Int] does - the program uses a lot of memory, as evident by running it with +RTS -s.
Is there anyway to implement Unmemo that does what I want?
Same trick as with a stream -- don't capture the remainder directly, but instead capture a value and a function which yields a remainder. You can add memoization on top of this as necessary.
data UTree a = Leaf a | Branch a (a -> [UTree a])
I'm not in the mood to figure it out precisely at the moment, but this structure arises, I'm sure, naturally as the cofree comonad over a fairly straightforward functor.
Edit
Found it: http://hackage.haskell.org/packages/archive/comonad-transformers/1.6.3/doc/html/Control-Comonad-Trans-Stream.html
Or this is perhaps simpler to understand: http://hackage.haskell.org/packages/archive/streams/0.7.2/doc/html/Data-Stream-Branching.html
In either case, the trick is that your f can be chosen to be something like data N s a = N (s -> (s,[a])) for an appropriate s (s being the type of your state parameter of the stream -- the seed of your unfold, if you will). That might not be exactly correct, but something close should do...
But of course for real work, you can scrap all this and just write the datatype directly as above.
Edit 2
The below code illustrates how this can prevent sharing. Note that even in the version without sharing, there are humps in the profile indicating that the sum and length calls aren't running in constant space. I'd imagine that we'd need an explicit strict accumulation to knock those down.
{-# LANGUAGE DeriveFunctor #-}
import Data.Stream.Branching(Stream(..))
import qualified Data.Stream.Branching as S
import Control.Arrow
import Control.Applicative
import Data.List
data UM s a = UM (s -> Maybe a) deriving Functor
type UStream s a = Stream (UM s) a
runUM s (UM f) = f s
liftUM x = UM $ const (Just x)
nullUM = UM $ const Nothing
buildUStream :: Int -> Int -> Stream (UM ()) Int
buildUStream start end = S.unfold (\x -> (x, go x)) start
where go x
| x < end = liftUM (x + 1)
| otherwise = nullUM
sumUS :: Stream (UM ()) Int -> Int
sumUS x = S.head $ S.scanr (\x us -> maybe 0 id (runUM () us) + x) x
lengthUS :: Stream (UM ()) Int -> Int
lengthUS x = S.head $ S.scanr (\x us -> maybe 0 id (runUM () us) + 1) x
sumUS' :: Stream (UM ()) Int -> Int
sumUS' x = last $ usToList $ liftUM $ S.scanl (+) 0 x
lengthUS' :: Stream (UM ()) Int -> Int
lengthUS' x = last $ usToList $ liftUM $ S.scanl (\acc _ -> acc + 1) 0 x
usToList x = unfoldr (\um -> (S.head &&& S.tail) <$> runUM () um) x
maxNum = 1000000
nums = buildUStream 0 maxNum
numsL :: [Int]
numsL = [0..maxNum]
-- All these need to be run with increased stack to avoid an overflow.
-- This generates an hp file with two humps (i.e. the list is not shared)
main = print $ div (fromIntegral $ sumUS' nums) (fromIntegral $ lengthUS' nums)
-- This generates an hp file as above, and uses somewhat less memory, at the cost of
-- an increased number of GCs. -H helps a lot with that.
-- main = print $ div (fromIntegral $ sumUS nums) (fromIntegral $ lengthUS nums)
-- This generates an hp file with one hump (i.e. the list is shared)
-- main = print $ div (fromIntegral $ sum $ numsL) (fromIntegral $ length $ numsL)