Storing values in an N x N grid Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)mapM for both keys and values of Data.MapStoring hierarchical data in a databaseCapture the notion of invertible functionsSquare grid collision detectionStoring array inside closureA function determining intervals of values greater than thresholdJavaScript 2D Grid WrapperCreating and storing binary arraysBuilding an interval-based gridStoring computed values in an array
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Storing values in an N x N grid
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)mapM for both keys and values of Data.MapStoring hierarchical data in a databaseCapture the notion of invertible functionsSquare grid collision detectionStoring array inside closureA function determining intervals of values greater than thresholdJavaScript 2D Grid WrapperCreating and storing binary arraysBuilding an interval-based gridStoring computed values in an array
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I am trying to write a program to store/display a grid of size N x N with cells containing either a 1 or a 0 in preparation for further computation:
module Board where
import Data.List as List
data CellState = One | Zero deriving (Eq, Ord)
data Cell = Cell cellPos :: (Int, Int), cellState :: CellState deriving (Eq, Ord)
type Board = [Cell]
instance Show CellState where
show One = "1"
show Zero = "0"
instance Show Cell where
show (Cell c x) = show (c,x)
genPositions :: Int -> [(Int, Int)]
genPositions x = [ (a,b) | a <- [0..(x-1)], b <- [0..(x-1)] ]
genCellState :: Int -> CellState
genCellState 0 = Zero
genCellState 1 = One
newBoard :: Int -> [Int] -> Board
newBoard i [x] = [Cell (round $ sqrt(fromIntegral i), round $ sqrt(fromIntegral i)) (genCellState x)]
where positions = genPositions $ round $ sqrt(fromIntegral i)
newBoard i (x : xs) = [Cell (positions!!(i - 1 - length xs)) (genCellState x)] ++ newBoard i xs
where positions = genPositions $ round $ sqrt(fromIntegral i)
What improvements can I make to this, both in terms of good practises and performance? I don't like Board being a list of Cells. I posted this on Stack Overflow by accident (have since deleted it) and someone recommended changing type Board = [Cell]
to newtype Board = Board getBoard :: Array (Int,Int) CellState
but I'm not too familiar with arrays in Haskell so not sure how this would work exactly. From what I have read they apply a function to the elements in the range [Int..Int]
and return a list of tuples with the input and output.
array haskell
$endgroup$
add a comment |
$begingroup$
I am trying to write a program to store/display a grid of size N x N with cells containing either a 1 or a 0 in preparation for further computation:
module Board where
import Data.List as List
data CellState = One | Zero deriving (Eq, Ord)
data Cell = Cell cellPos :: (Int, Int), cellState :: CellState deriving (Eq, Ord)
type Board = [Cell]
instance Show CellState where
show One = "1"
show Zero = "0"
instance Show Cell where
show (Cell c x) = show (c,x)
genPositions :: Int -> [(Int, Int)]
genPositions x = [ (a,b) | a <- [0..(x-1)], b <- [0..(x-1)] ]
genCellState :: Int -> CellState
genCellState 0 = Zero
genCellState 1 = One
newBoard :: Int -> [Int] -> Board
newBoard i [x] = [Cell (round $ sqrt(fromIntegral i), round $ sqrt(fromIntegral i)) (genCellState x)]
where positions = genPositions $ round $ sqrt(fromIntegral i)
newBoard i (x : xs) = [Cell (positions!!(i - 1 - length xs)) (genCellState x)] ++ newBoard i xs
where positions = genPositions $ round $ sqrt(fromIntegral i)
What improvements can I make to this, both in terms of good practises and performance? I don't like Board being a list of Cells. I posted this on Stack Overflow by accident (have since deleted it) and someone recommended changing type Board = [Cell]
to newtype Board = Board getBoard :: Array (Int,Int) CellState
but I'm not too familiar with arrays in Haskell so not sure how this would work exactly. From what I have read they apply a function to the elements in the range [Int..Int]
and return a list of tuples with the input and output.
array haskell
$endgroup$
add a comment |
$begingroup$
I am trying to write a program to store/display a grid of size N x N with cells containing either a 1 or a 0 in preparation for further computation:
module Board where
import Data.List as List
data CellState = One | Zero deriving (Eq, Ord)
data Cell = Cell cellPos :: (Int, Int), cellState :: CellState deriving (Eq, Ord)
type Board = [Cell]
instance Show CellState where
show One = "1"
show Zero = "0"
instance Show Cell where
show (Cell c x) = show (c,x)
genPositions :: Int -> [(Int, Int)]
genPositions x = [ (a,b) | a <- [0..(x-1)], b <- [0..(x-1)] ]
genCellState :: Int -> CellState
genCellState 0 = Zero
genCellState 1 = One
newBoard :: Int -> [Int] -> Board
newBoard i [x] = [Cell (round $ sqrt(fromIntegral i), round $ sqrt(fromIntegral i)) (genCellState x)]
where positions = genPositions $ round $ sqrt(fromIntegral i)
newBoard i (x : xs) = [Cell (positions!!(i - 1 - length xs)) (genCellState x)] ++ newBoard i xs
where positions = genPositions $ round $ sqrt(fromIntegral i)
What improvements can I make to this, both in terms of good practises and performance? I don't like Board being a list of Cells. I posted this on Stack Overflow by accident (have since deleted it) and someone recommended changing type Board = [Cell]
to newtype Board = Board getBoard :: Array (Int,Int) CellState
but I'm not too familiar with arrays in Haskell so not sure how this would work exactly. From what I have read they apply a function to the elements in the range [Int..Int]
and return a list of tuples with the input and output.
array haskell
$endgroup$
I am trying to write a program to store/display a grid of size N x N with cells containing either a 1 or a 0 in preparation for further computation:
module Board where
import Data.List as List
data CellState = One | Zero deriving (Eq, Ord)
data Cell = Cell cellPos :: (Int, Int), cellState :: CellState deriving (Eq, Ord)
type Board = [Cell]
instance Show CellState where
show One = "1"
show Zero = "0"
instance Show Cell where
show (Cell c x) = show (c,x)
genPositions :: Int -> [(Int, Int)]
genPositions x = [ (a,b) | a <- [0..(x-1)], b <- [0..(x-1)] ]
genCellState :: Int -> CellState
genCellState 0 = Zero
genCellState 1 = One
newBoard :: Int -> [Int] -> Board
newBoard i [x] = [Cell (round $ sqrt(fromIntegral i), round $ sqrt(fromIntegral i)) (genCellState x)]
where positions = genPositions $ round $ sqrt(fromIntegral i)
newBoard i (x : xs) = [Cell (positions!!(i - 1 - length xs)) (genCellState x)] ++ newBoard i xs
where positions = genPositions $ round $ sqrt(fromIntegral i)
What improvements can I make to this, both in terms of good practises and performance? I don't like Board being a list of Cells. I posted this on Stack Overflow by accident (have since deleted it) and someone recommended changing type Board = [Cell]
to newtype Board = Board getBoard :: Array (Int,Int) CellState
but I'm not too familiar with arrays in Haskell so not sure how this would work exactly. From what I have read they apply a function to the elements in the range [Int..Int]
and return a list of tuples with the input and output.
array haskell
array haskell
edited Nov 22 '17 at 1:00
Jamal♦
30.6k11121227
30.6k11121227
asked Nov 22 '17 at 0:54
user6731064user6731064
111
111
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
By defining r = round . sqrt . fromIntegral
, newBoard
fits on the screen. positions
doesn't appear to be used in newboard
's first case.
Half the code disappears if we interpret Cell
as ((Int, Int), Int)
.
newBoard
's first case can be pushed one recursion call deeper, mapping []
to []
instead of [x]
to the current right hand side.
[_] ++ _
should be simplified as _ : _
.
positions
is only used once, therefore I inline it.
type Cell = ((Int, Int), Int) -- (position, state)
genPositions :: Int -> [(Int, Int)]
genPositions x = [ (a,b) | a <- [0..(x-1)], b <- [0..(x-1)] ]
r = round . sqrt . fromIntegral
newBoard :: Int -> [Int] -> [Cell]
newBoard i [] = []
newBoard i (x : xs) = (genPositions (r i)!!(i - 1 - length xs), x) : newBoard i xs
Successive elements of the list returned by genPositions
and xs
are zipped together; zip
captures this pattern. i
is now not needed in its non-rooted form and I recommend changing the interface to take N
as an argument instead. Non-square arguments can currently crash !!
anyway. genPositions
is only used once, therefore I inline it.
type Cell = ((Int, Int), Int) -- (position, state)
newBoard :: Int -> [Int] -> [Cell]
newBoard n = zip $ liftA2 (,) [0..n-1] [0..n-1]
For type Board = Array (Int, Int) Int
, Data.Array allows newBoard n = listArray ((0,0),(n-1,n-1))
.
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By defining r = round . sqrt . fromIntegral
, newBoard
fits on the screen. positions
doesn't appear to be used in newboard
's first case.
Half the code disappears if we interpret Cell
as ((Int, Int), Int)
.
newBoard
's first case can be pushed one recursion call deeper, mapping []
to []
instead of [x]
to the current right hand side.
[_] ++ _
should be simplified as _ : _
.
positions
is only used once, therefore I inline it.
type Cell = ((Int, Int), Int) -- (position, state)
genPositions :: Int -> [(Int, Int)]
genPositions x = [ (a,b) | a <- [0..(x-1)], b <- [0..(x-1)] ]
r = round . sqrt . fromIntegral
newBoard :: Int -> [Int] -> [Cell]
newBoard i [] = []
newBoard i (x : xs) = (genPositions (r i)!!(i - 1 - length xs), x) : newBoard i xs
Successive elements of the list returned by genPositions
and xs
are zipped together; zip
captures this pattern. i
is now not needed in its non-rooted form and I recommend changing the interface to take N
as an argument instead. Non-square arguments can currently crash !!
anyway. genPositions
is only used once, therefore I inline it.
type Cell = ((Int, Int), Int) -- (position, state)
newBoard :: Int -> [Int] -> [Cell]
newBoard n = zip $ liftA2 (,) [0..n-1] [0..n-1]
For type Board = Array (Int, Int) Int
, Data.Array allows newBoard n = listArray ((0,0),(n-1,n-1))
.
$endgroup$
add a comment |
$begingroup$
By defining r = round . sqrt . fromIntegral
, newBoard
fits on the screen. positions
doesn't appear to be used in newboard
's first case.
Half the code disappears if we interpret Cell
as ((Int, Int), Int)
.
newBoard
's first case can be pushed one recursion call deeper, mapping []
to []
instead of [x]
to the current right hand side.
[_] ++ _
should be simplified as _ : _
.
positions
is only used once, therefore I inline it.
type Cell = ((Int, Int), Int) -- (position, state)
genPositions :: Int -> [(Int, Int)]
genPositions x = [ (a,b) | a <- [0..(x-1)], b <- [0..(x-1)] ]
r = round . sqrt . fromIntegral
newBoard :: Int -> [Int] -> [Cell]
newBoard i [] = []
newBoard i (x : xs) = (genPositions (r i)!!(i - 1 - length xs), x) : newBoard i xs
Successive elements of the list returned by genPositions
and xs
are zipped together; zip
captures this pattern. i
is now not needed in its non-rooted form and I recommend changing the interface to take N
as an argument instead. Non-square arguments can currently crash !!
anyway. genPositions
is only used once, therefore I inline it.
type Cell = ((Int, Int), Int) -- (position, state)
newBoard :: Int -> [Int] -> [Cell]
newBoard n = zip $ liftA2 (,) [0..n-1] [0..n-1]
For type Board = Array (Int, Int) Int
, Data.Array allows newBoard n = listArray ((0,0),(n-1,n-1))
.
$endgroup$
add a comment |
$begingroup$
By defining r = round . sqrt . fromIntegral
, newBoard
fits on the screen. positions
doesn't appear to be used in newboard
's first case.
Half the code disappears if we interpret Cell
as ((Int, Int), Int)
.
newBoard
's first case can be pushed one recursion call deeper, mapping []
to []
instead of [x]
to the current right hand side.
[_] ++ _
should be simplified as _ : _
.
positions
is only used once, therefore I inline it.
type Cell = ((Int, Int), Int) -- (position, state)
genPositions :: Int -> [(Int, Int)]
genPositions x = [ (a,b) | a <- [0..(x-1)], b <- [0..(x-1)] ]
r = round . sqrt . fromIntegral
newBoard :: Int -> [Int] -> [Cell]
newBoard i [] = []
newBoard i (x : xs) = (genPositions (r i)!!(i - 1 - length xs), x) : newBoard i xs
Successive elements of the list returned by genPositions
and xs
are zipped together; zip
captures this pattern. i
is now not needed in its non-rooted form and I recommend changing the interface to take N
as an argument instead. Non-square arguments can currently crash !!
anyway. genPositions
is only used once, therefore I inline it.
type Cell = ((Int, Int), Int) -- (position, state)
newBoard :: Int -> [Int] -> [Cell]
newBoard n = zip $ liftA2 (,) [0..n-1] [0..n-1]
For type Board = Array (Int, Int) Int
, Data.Array allows newBoard n = listArray ((0,0),(n-1,n-1))
.
$endgroup$
By defining r = round . sqrt . fromIntegral
, newBoard
fits on the screen. positions
doesn't appear to be used in newboard
's first case.
Half the code disappears if we interpret Cell
as ((Int, Int), Int)
.
newBoard
's first case can be pushed one recursion call deeper, mapping []
to []
instead of [x]
to the current right hand side.
[_] ++ _
should be simplified as _ : _
.
positions
is only used once, therefore I inline it.
type Cell = ((Int, Int), Int) -- (position, state)
genPositions :: Int -> [(Int, Int)]
genPositions x = [ (a,b) | a <- [0..(x-1)], b <- [0..(x-1)] ]
r = round . sqrt . fromIntegral
newBoard :: Int -> [Int] -> [Cell]
newBoard i [] = []
newBoard i (x : xs) = (genPositions (r i)!!(i - 1 - length xs), x) : newBoard i xs
Successive elements of the list returned by genPositions
and xs
are zipped together; zip
captures this pattern. i
is now not needed in its non-rooted form and I recommend changing the interface to take N
as an argument instead. Non-square arguments can currently crash !!
anyway. genPositions
is only used once, therefore I inline it.
type Cell = ((Int, Int), Int) -- (position, state)
newBoard :: Int -> [Int] -> [Cell]
newBoard n = zip $ liftA2 (,) [0..n-1] [0..n-1]
For type Board = Array (Int, Int) Int
, Data.Array allows newBoard n = listArray ((0,0),(n-1,n-1))
.
edited 9 mins ago
200_success
131k17157422
131k17157422
answered Nov 22 '17 at 23:49
GurkenglasGurkenglas
2,949512
2,949512
add a comment |
add a comment |
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