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patternMinor

Better or more idiomatic way to apply a sequence of functions in Haskell

Submitted by: @import:stackexchange-codereview··
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Problem

This is one of my Haskell solutions to a variation of the N-Queens problem, where the queens can move like knights in addition to their normal moves. It has been optimized somewhat to avoid investigating redundant combinations.

place :: Int -> [[Int]]
place 0 = [[]]
place n = go $ take n $ repeat [1..n] 
    where go [] = [[]]
          go (row:rest) = do
                q  filter (\y -> abs(y - q) /= x) r)) . (zip [1..])
          notK q = map (\(f, r) -> filter f r) . (zip (kFilters q))
          kFilters q = (\x -> abs (x - q) /= 2) : (\x -> abs (x - q) > 1) : (repeat (const True))

solutions = length . place

main = do
    n <- readLn
    putStrLn $ show $ solutions n


I am satisfied with the performance, but I feel there must be a more elegant way to apply a series of functions (in this case, filters) to a list.

In each iteration of the recursive function go, the top row of the board is selected and then for each position in sequence a queen is positioned and the function recurs with a filtered copy of the rest of the board, so that each iteration has only safe squares to choose from. The safe function applies three filters to the board:

  • notC removes all spaces in the same column as the new queen.



  • notD removes any spaces on a diagonal from the new queen.



  • notK removes knight moves from the next two lines.



I feel that notK in particular could be implemented more cleanly and idiomatically but I couldn't see a better way to apply one function to the first item of a list, another to the second and something else to the rest. And using zip does save me from having to check for the end of the board.

I wouldn't be surprised if there is a better way to write notD. So I am looking for more expressive ways to apply a sequence of varying functions to successive list items.

UPDATE:

I realise that I can use uncurry to clean up notK...

notK q = map (uncurry filter) . (zip (kFilters q))


and the two filters in Kfilters ca

Solution

notK q = map (uncurry filter) . (zip (kFilters q))


is equivalent to

notK q rows = zipWith filter (kFilters q) rows


Onto making above Haskell less idiomatic and less pointfree and more readable.

notC q = map (filter (/= q))
notD q = (map (\(x, r) -> filter (\y -> abs(y - q) /= x) r)) . (zip [1..])
notK q = map (\(f, r) -> filter f r) . (zip (kFilters q))


These three are logically similar but structurally dissimilar.

First let's factor out common components, give them meaningful names, and some comments to help those readers without the Developer's Manual.

-- a new queen placed in column q
-- would take a piece placed in column x
-- n rows down
-- because (r, q) and (r+n, x)

-- are on the same column
sameColumn q n x = x == q

-- are on the same diagonal
sameDiagonal q n x = abs(x - q) == n

-- or (n, |q - x|) is in [(1,2),(2,1)]
onKnightMove :: Int -> Int -> Int -> Bool
onKnightMove q n x = (n, abs(x - q)) `elem` [(1,2),(2,1)]


The last one needs a better name still, oh well. At least they are of the same form, with parameter name use consistent among them (and elsewhere).

rowFilterList :: Int -> (Int -> Int -> Int -> Bool) -> [Int -> Bool]
rowFilterList q pred = map (\n x -> not $ pred q n x) [1..]


such that kFilters q = rowFilterList q onKnightMove

filterRows :: Int -> (Int -> Int -> Int -> Bool) -> [[Int]] -> [[Int]]
filterRows q pred rows = zipWith filter (rowFilterList q pred) rows


This way notC q = filterRows q sameColumn and notD q = filterRows q sameDiagonal

notK q . notD q . notC q a repetition is evident. And we now can factor it out easily.

filterWithAll :: Int -> [[Int]] -> [Int -> Int -> Int -> Bool] -> [[Int]]
filterWithAll q rows preds = foldl (\rows' pred -> filterRows q pred rows') rows preds


such that safe q rows = filterWithAll q rows [sameColumn, sameDiagonal, onKnightMove]

By factoring out [sameColumn, sameDiagonal, onKnightMove] as a parameter to place n one could easily generalize place from n super-queens to n queens or n rooks.

Code Snippets

notK q = map (uncurry filter) . (zip (kFilters q))
notK q rows = zipWith filter (kFilters q) rows
notC q = map (filter (/= q))
notD q = (map (\(x, r) -> filter (\y -> abs(y - q) /= x) r)) . (zip [1..])
notK q = map (\(f, r) -> filter f r) . (zip (kFilters q))
-- a new queen placed in column q
-- would take a piece placed in column x
-- n rows down
-- because (r, q) and (r+n, x)

-- are on the same column
sameColumn q n x = x == q

-- are on the same diagonal
sameDiagonal q n x = abs(x - q) == n

-- or (n, |q - x|) is in [(1,2),(2,1)]
onKnightMove :: Int -> Int -> Int -> Bool
onKnightMove q n x = (n, abs(x - q)) `elem` [(1,2),(2,1)]
rowFilterList :: Int -> (Int -> Int -> Int -> Bool) -> [Int -> Bool]
rowFilterList q pred = map (\n x -> not $ pred q n x) [1..]

Context

StackExchange Code Review Q#37081, answer score: 6

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