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Today's Topics:
1. What's an idiomatic Haskell solution to solve the "Maximum
Subarray Problem"? (Dominik Bollmann)
2. Re: What's an idiomatic Haskell solution to solve the
"Maximum Subarray Problem"? (Theodore Lief Gannon)
3. Re: What's an idiomatic Haskell solution to solve the
"Maximum Subarray Problem"? (Theodore Lief Gannon)
----------------------------------------------------------------------
Message: 1
Date: Sun, 17 Jul 2016 00:40:42 +0200
From: Dominik Bollmann <[email protected]>
To: [email protected]
Subject: [Haskell-beginners] What's an idiomatic Haskell solution to
solve the "Maximum Subarray Problem"?
Message-ID:
<87vb05qk9h.fsf@t450s.i-did-not-set--mail-host-address--so-tickle-me>
Content-Type: text/plain
Hi all,
I've recently been trying to implement the "maximum subarray problem"
from [1] in Haskell. My first, naive solution looked like this:
maxSubArray :: [Int] -> [Int]
maxSubArray [] = []
maxSubArray [x] = [x]
maxSubArray xs@(_:_:_) = maxArr (maxArr maxHd maxTl) (maxCrossingArray hd tl)
where
(hd,tl) = splitAt (length xs `div` 2) xs
maxHd = maxSubArray hd
maxTl = maxSubArray tl
maxCrossingArray :: [Int] -> [Int] -> [Int]
maxCrossingArray hd tl
| null hd || null tl = error "maxArrayBetween: hd/tl empty!"
maxCrossingArray hd tl = maxHd ++ maxTl
where
maxHd = reverse . foldr1 maxArr . tail $ inits (reverse hd)
-- we need to go from the center leftwards, which is why we
-- reverse the list `hd'.
maxTl = foldr1 maxArr . tail $ inits tl
maxArr :: [Int] -> [Int] -> [Int]
maxArr xs ys
| sum xs > sum ys = xs
| otherwise = ys
While I originally thought that this should run in O(n*log n), a closer
examination revealed that the (++) as well as maxHd and maxTl
computations inside function `maxCrossingArray` are O(n^2), which makes
solving one of the provided test cases in [1] infeasible.
Hence, I rewrote the above code using Data.Array into the following:
data ArraySum = ArraySum {
from :: Int
, to :: Int
, value :: Int
} deriving (Eq, Show)
instance Ord ArraySum where
ArraySum _ _ v1 <= ArraySum _ _ v2 = v1 <= v2
maxSubList :: [Int] -> [Int]
maxSubList xs = take (to-from+1) . drop (from-1) $ xs
where
arr = array (1, length xs) [(i,v) | (i,v) <- zip [1..] xs]
ArraySum from to val = findMaxArr (1, length xs) arr
findMaxArr :: (Int, Int) -> Array Int Int -> ArraySum
findMaxArr (start, end) arr
| start > end = error "findMaxArr: start > end"
| start == end = ArraySum start end (arr ! start)
| otherwise = max (max hd tl) (ArraySum leftIdx rightIdx
(leftVal+rightVal))
where
mid = (start + end) `div` 2
hd = findMaxArr (start, mid) arr
tl = findMaxArr (mid+1, end) arr
(leftIdx, leftVal) = snd $ findMax mid [mid-1,mid-2..start]
(rightIdx, rightVal) = snd $ findMax (mid+1) [mid+2,mid+3..end]
findMax pos = foldl' go ((pos, arr ! pos), (pos, arr ! pos))
go ((currIdx, currSum), (maxIdx, maxSum)) idx
| newSum >= maxSum = ((idx, newSum), (idx, newSum))
| otherwise = ((idx, newSum), (maxIdx, maxSum))
where newSum = currSum + (arr ! idx)
I believe this runs in O(n*log n) now and is fast enough for the purpose
of solving the Hackerrank challenge [1].
However, I feel this second solution is not very idiomatic Haskell code
and I would prefer the clarity of the first solution over the second, if
somehow I could make it more efficient.
Therefore my question: What would be an efficient, yet idiomatic
solution to solving the "maximum subarray problem" in Haskell? (Note:
I'm aware that this problem can be solved in O(n), but I'm also happy with
idiomatic Haskell solutions running in O(n*log n))
Thanks, Dominik.
[1] https://www.hackerrank.com/challenges/maxsubarray
------------------------------
Message: 2
Date: Sat, 16 Jul 2016 22:14:23 -0700
From: Theodore Lief Gannon <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] What's an idiomatic Haskell solution
to solve the "Maximum Subarray Problem"?
Message-ID:
<CAJoPsuAu1u4=ozakqpseewibxt9h4fzzfpdicdqirtwg89d...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
-- Not beautifully idiomatic, but not too bad, and O(n):
data SolutionState = SSInitial | SS Int Int Int
solve :: [Int] -> SolutionState
solve = foldr go SSInitial where
go x (SS dense best sparse) =
let dense' = max x (dense + x)
best' = max best dense'
sparse' = max (sparse + x) (max sparse x)
in SS dense' best' sparse'
go x SSInitial = SS x x x
On Sat, Jul 16, 2016 at 3:40 PM, Dominik Bollmann <[email protected]
> wrote:
>
> Hi all,
>
> I've recently been trying to implement the "maximum subarray problem"
> from [1] in Haskell. My first, naive solution looked like this:
>
> maxSubArray :: [Int] -> [Int]
> maxSubArray [] = []
> maxSubArray [x] = [x]
> maxSubArray xs@(_:_:_) = maxArr (maxArr maxHd maxTl) (maxCrossingArray hd
> tl)
> where
> (hd,tl) = splitAt (length xs `div` 2) xs
> maxHd = maxSubArray hd
> maxTl = maxSubArray tl
>
> maxCrossingArray :: [Int] -> [Int] -> [Int]
> maxCrossingArray hd tl
> | null hd || null tl = error "maxArrayBetween: hd/tl empty!"
> maxCrossingArray hd tl = maxHd ++ maxTl
> where
> maxHd = reverse . foldr1 maxArr . tail $ inits (reverse hd)
> -- we need to go from the center leftwards, which is why we
> -- reverse the list `hd'.
> maxTl = foldr1 maxArr . tail $ inits tl
>
> maxArr :: [Int] -> [Int] -> [Int]
> maxArr xs ys
> | sum xs > sum ys = xs
> | otherwise = ys
>
> While I originally thought that this should run in O(n*log n), a closer
> examination revealed that the (++) as well as maxHd and maxTl
> computations inside function `maxCrossingArray` are O(n^2), which makes
> solving one of the provided test cases in [1] infeasible.
>
> Hence, I rewrote the above code using Data.Array into the following:
>
> data ArraySum = ArraySum {
> from :: Int
> , to :: Int
> , value :: Int
> } deriving (Eq, Show)
>
> instance Ord ArraySum where
> ArraySum _ _ v1 <= ArraySum _ _ v2 = v1 <= v2
>
> maxSubList :: [Int] -> [Int]
> maxSubList xs = take (to-from+1) . drop (from-1) $ xs
> where
> arr = array (1, length xs) [(i,v) | (i,v) <- zip [1..] xs]
> ArraySum from to val = findMaxArr (1, length xs) arr
>
> findMaxArr :: (Int, Int) -> Array Int Int -> ArraySum
> findMaxArr (start, end) arr
> | start > end = error "findMaxArr: start > end"
> | start == end = ArraySum start end (arr ! start)
> | otherwise = max (max hd tl) (ArraySum leftIdx rightIdx
> (leftVal+rightVal))
> where
> mid = (start + end) `div` 2
> hd = findMaxArr (start, mid) arr
> tl = findMaxArr (mid+1, end) arr
> (leftIdx, leftVal) = snd $ findMax mid [mid-1,mid-2..start]
> (rightIdx, rightVal) = snd $ findMax (mid+1) [mid+2,mid+3..end]
> findMax pos = foldl' go ((pos, arr ! pos), (pos, arr ! pos))
> go ((currIdx, currSum), (maxIdx, maxSum)) idx
> | newSum >= maxSum = ((idx, newSum), (idx, newSum))
> | otherwise = ((idx, newSum), (maxIdx, maxSum))
> where newSum = currSum + (arr ! idx)
>
> I believe this runs in O(n*log n) now and is fast enough for the purpose
> of solving the Hackerrank challenge [1].
>
> However, I feel this second solution is not very idiomatic Haskell code
> and I would prefer the clarity of the first solution over the second, if
> somehow I could make it more efficient.
>
> Therefore my question: What would be an efficient, yet idiomatic
> solution to solving the "maximum subarray problem" in Haskell? (Note:
> I'm aware that this problem can be solved in O(n), but I'm also happy with
> idiomatic Haskell solutions running in O(n*log n))
>
> Thanks, Dominik.
>
> [1] https://www.hackerrank.com/challenges/maxsubarray
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
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Message: 3
Date: Sat, 16 Jul 2016 22:23:52 -0700
From: Theodore Lief Gannon <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] What's an idiomatic Haskell solution
to solve the "Maximum Subarray Problem"?
Message-ID:
<cajopsuchgu_y8eonnmfcqwdd4zkvntub9kdh6k-et3mduoz...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
Hmm, I clipped out all the boilerplate for interacting with HackerRank's
expected I/O formats, but probably should have left this in for clarity:
prettySS :: SolutionState -> String
prettySS SSInitial = "Whoops! Didn't I filter nulls?"
prettySS (SS _ dense sparse) = unwords $ map show [dense, sparse]
On Sat, Jul 16, 2016 at 10:14 PM, Theodore Lief Gannon <[email protected]>
wrote:
> -- Not beautifully idiomatic, but not too bad, and O(n):
>
> data SolutionState = SSInitial | SS Int Int Int
>
> solve :: [Int] -> SolutionState
> solve = foldr go SSInitial where
> go x (SS dense best sparse) =
> let dense' = max x (dense + x)
> best' = max best dense'
> sparse' = max (sparse + x) (max sparse x)
> in SS dense' best' sparse'
> go x SSInitial = SS x x x
>
>
> On Sat, Jul 16, 2016 at 3:40 PM, Dominik Bollmann <
> [email protected]> wrote:
>
>>
>> Hi all,
>>
>> I've recently been trying to implement the "maximum subarray problem"
>> from [1] in Haskell. My first, naive solution looked like this:
>>
>> maxSubArray :: [Int] -> [Int]
>> maxSubArray [] = []
>> maxSubArray [x] = [x]
>> maxSubArray xs@(_:_:_) = maxArr (maxArr maxHd maxTl) (maxCrossingArray
>> hd tl)
>> where
>> (hd,tl) = splitAt (length xs `div` 2) xs
>> maxHd = maxSubArray hd
>> maxTl = maxSubArray tl
>>
>> maxCrossingArray :: [Int] -> [Int] -> [Int]
>> maxCrossingArray hd tl
>> | null hd || null tl = error "maxArrayBetween: hd/tl empty!"
>> maxCrossingArray hd tl = maxHd ++ maxTl
>> where
>> maxHd = reverse . foldr1 maxArr . tail $ inits (reverse hd)
>> -- we need to go from the center leftwards, which is why we
>> -- reverse the list `hd'.
>> maxTl = foldr1 maxArr . tail $ inits tl
>>
>> maxArr :: [Int] -> [Int] -> [Int]
>> maxArr xs ys
>> | sum xs > sum ys = xs
>> | otherwise = ys
>>
>> While I originally thought that this should run in O(n*log n), a closer
>> examination revealed that the (++) as well as maxHd and maxTl
>> computations inside function `maxCrossingArray` are O(n^2), which makes
>> solving one of the provided test cases in [1] infeasible.
>>
>> Hence, I rewrote the above code using Data.Array into the following:
>>
>> data ArraySum = ArraySum {
>> from :: Int
>> , to :: Int
>> , value :: Int
>> } deriving (Eq, Show)
>>
>> instance Ord ArraySum where
>> ArraySum _ _ v1 <= ArraySum _ _ v2 = v1 <= v2
>>
>> maxSubList :: [Int] -> [Int]
>> maxSubList xs = take (to-from+1) . drop (from-1) $ xs
>> where
>> arr = array (1, length xs) [(i,v) | (i,v) <- zip [1..] xs]
>> ArraySum from to val = findMaxArr (1, length xs) arr
>>
>> findMaxArr :: (Int, Int) -> Array Int Int -> ArraySum
>> findMaxArr (start, end) arr
>> | start > end = error "findMaxArr: start > end"
>> | start == end = ArraySum start end (arr ! start)
>> | otherwise = max (max hd tl) (ArraySum leftIdx rightIdx
>> (leftVal+rightVal))
>> where
>> mid = (start + end) `div` 2
>> hd = findMaxArr (start, mid) arr
>> tl = findMaxArr (mid+1, end) arr
>> (leftIdx, leftVal) = snd $ findMax mid [mid-1,mid-2..start]
>> (rightIdx, rightVal) = snd $ findMax (mid+1) [mid+2,mid+3..end]
>> findMax pos = foldl' go ((pos, arr ! pos), (pos, arr ! pos))
>> go ((currIdx, currSum), (maxIdx, maxSum)) idx
>> | newSum >= maxSum = ((idx, newSum), (idx, newSum))
>> | otherwise = ((idx, newSum), (maxIdx, maxSum))
>> where newSum = currSum + (arr ! idx)
>>
>> I believe this runs in O(n*log n) now and is fast enough for the purpose
>> of solving the Hackerrank challenge [1].
>>
>> However, I feel this second solution is not very idiomatic Haskell code
>> and I would prefer the clarity of the first solution over the second, if
>> somehow I could make it more efficient.
>>
>> Therefore my question: What would be an efficient, yet idiomatic
>> solution to solving the "maximum subarray problem" in Haskell? (Note:
>> I'm aware that this problem can be solved in O(n), but I'm also happy with
>> idiomatic Haskell solutions running in O(n*log n))
>>
>> Thanks, Dominik.
>>
>> [1] https://www.hackerrank.com/challenges/maxsubarray
>> _______________________________________________
>> Beginners mailing list
>> [email protected]
>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>>
>
>
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