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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
$endgroup$
bumped to the homepage by Community♦ 5 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
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
$endgroup$
bumped to the homepage by Community♦ 5 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
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
61
61
bumped to the homepage by Community♦ 5 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 5 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
add a comment |
1 Answer
1
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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. [_] ++ _ ~> _ : _. 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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$begingroup$
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. [_] ++ _ ~> _ : _. 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. [_] ++ _ ~> _ : _. 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. [_] ++ _ ~> _ : _. 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. [_] ++ _ ~> _ : _. 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))
.
answered Nov 22 '17 at 23:49
GurkenglasGurkenglas
2,939512
2,939512
add a comment |
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