There are several similar questions but none quite solve it for me, for example this question explains how to remove borders from fullscreen floating windows.
Using XMonad.Layout.NoBorders you can do lots of cool stuff like remove borders from certain windows or if it is the only window or only fullscreen floating windows.
I couldn't find anything for all floating windows, however if someone could just point me to some tool that I could use to check if a window is floating or not, I am sure I could try hack up a solution.
Any suggestions are welcome
I'll be using the source code of XMonad.Layout.NoBorders as a reference, since I can't find anything more fitting that already exists. We want to see how it implements "remove borders on fullscreen floating windows", to see if it can be easily relaxed to "remove borders on floating windows" (without the fullscreen constraint).
According to the answer on the question you linked:
layoutHook = lessBorders OnlyFloat $ avoidStruts $ myLayout
OnlyFloat seems to be the specifier for "remove borders on fullscreen floating windows", so let's check the definition of that:
data Ambiguity = Combine With Ambiguity Ambiguity
| OnlyFloat
| Never
| EmptyScreen
| OtherIndicated
| Screen
deriving (Read, Show)
Not too helpful on its own. We should look elsewhere to see how the code treats these values.
It's a pretty safe bet that the first function to check is lessBorders:
lessBorders :: (SetsAmbiguous p, Read p, Show p, LayoutClass l a) =>
p -> l a -> ModifiedLayout (ConfigurableBorder p) l a
lessBorders amb = ModifiedLayout (ConfigurableBorder amb [])
From the type signature of lessBorders, we can see that:
OnlyFloat :: (SetsAmbiguous p, Read p, Show p) => p
This is a good sign, as it means lessBorders doesn't explicitly expect an Ambiguity: we can extend the functionality here by implementing our own SetsAmbiguous and passing it to the existing lessBorders. Let's now look at SetsAmbiguous, and Ambiguity's implementation of it:
class SetsAmbiguous p where
hiddens :: p -> WindowSet -> Maybe (W.Stack Window) -> [(Window, Rectangle)] -> [Window]
instance SetsAmbiguous Ambiguity where
hiddens amb wset mst wrs
| Combine Union a b <- amb = on union next a b
| Combine Difference a b <- amb = on (\\) next a b
| Combine Intersection a b <- amb = on intersect next a b
| otherwise = tiled ms ++ floating
where next p = hiddens p wset mst wrs
nonzerorect (Rectangle _ _ 0 0) = False
nonzerorect _ = True
screens =
[ scr | scr <- W.screens wset,
case amb of
Never -> True
_ -> not $ null $ integrate scr,
nonzerorect . screenRect $ W.screenDetail scr]
floating = [ w |
(w, W.RationalRect px py wx wy) <- M.toList . W.floating $ wset,
px <= 0, py <= 0,
wx + px >= 1, wy + py >= 1]
ms = filter (`elem` W.integrate' mst) $ map fst wrs
tiled [w]
| Screen <- amb = [w]
| OnlyFloat <- amb = []
| OtherIndicated <- amb
, let nonF = map integrate $ W.current wset : W.visible wset
, length (concat nonF) > length wrs
, singleton $ filter (1==) $ map length nonF = [w]
| singleton screens = [w]
tiled _ = []
integrate y = W.integrate' . W.stack $ W.workspace y
hiddens is the only method here that we need to implement. Its arguments are our SetsAmbiguous value, a WindowSet, and some other things, and it returns a list of windows that should not show borders. There's a lot of logic for the combining operations and other Ambiguity values, but those don't matter to us right now. What we care about is this snippet:
floating = [ w |
(w, W.RationalRect px py wx wy) <- M.toList . W.floating $ wset,
px <= 0, py <= 0,
wx + px >= 1, wy + py >= 1]
This is very promising. It defines a set of floating windows by extracting all windows from the floating section of the WindowSet, converting it to a list (initially it's a Data.Map), and filtering out all the windows that don't cover the entire screen. All we need to do is remove the filter.
After making that change, and removing all unnecessary code pertaining to tiled windows and set operations (which is most of the implementation), we end up with simply:
import XMonad.Layout.NoBorders
import qualified XMonad.StackSet as W
import qualified Data.Map as M
data AllFloats = AllFloats deriving (Read, Show)
instance SetsAmbiguous AllFloats where
hiddens _ wset _ _ = M.keys $ W.floating wset
We can then say:
layoutHook = lessBorders AllFloats $ myLayout...
Related
I'm trying to make what I think is called an Ulam spiral using Haskell.
It needs to go outwards in a clockwise rotation:
6 - 7 - 8 - 9
| |
5 0 - 1 10
| | |
4 - 3 - 2 11
|
..15- 14- 13- 12
For each step I'm trying to create coordinates, the function would be given a number and return spiral coordinates to the length of input number eg:
mkSpiral 9
> [(0,0),(1,0),(1,-1),(0,-1),(-1,-1),(-1,0),(-1,1),(0,1),(1,1)]
(-1, 1) - (0, 1) - (1, 1)
|
(-1, 0) (0, 0) - (1, 0)
| |
(-1,-1) - (0,-1) - (1,-1)
I've seen Looping in a spiral solution, but this goes counter-clockwise and it's inputs need to the size of the matrix.
I also found this code which does what I need but it seems to go counterclock-wise, stepping up rather than stepping right then clockwise :(
type Spiral = Int
type Coordinate = (Int, Int)
-- number of squares on each side of the spiral
sideSquares :: Spiral -> Int
sideSquares sp = (sp * 2) - 1
-- the coordinates for all squares in the given spiral
coordinatesForSpiral :: Spiral -> [Coordinate]
coordinatesForSpiral 1 = [(0, 0)]
coordinatesForSpiral sp = [(0, 0)] ++ right ++ top ++ left ++ bottom
where fixed = sp - 1
sides = sideSquares sp - 1
right = [(x, y) | x <- [fixed], y <- take sides [-1*(fixed-1)..]]
top = [(x, y) | x <- reverse (take sides [-1*fixed..]), y <- [fixed]]
left = [(x, y) | x <- [-1*fixed], y <- reverse(take sides [-1*fixed..])]
bottom = [(x, y) | x <- take sides [-1*fixed+1..], y <- [-1*fixed]]
-- an endless list of coordinates (the complete spiral)
mkSpiral :: Int -> [Coordinate]
mkSpiral x = take x endlessSpiral
endlessSpiral :: [Coordinate]
endlessSpiral = endlessSpiral' 1
endlessSpiral' start = coordinatesForSpiral start ++ endlessSpiral' (start + 1)
After much experimentation I can't seem to change the rotation or starting step direction, could someone point me in the right way or a solution that doesn't use list comprehension as I find them tricky to decode?
Let us first take a look at how the directions of a spiral are looking:
R D L L U U R R R D D D L L L L U U U U ....
We can split this in sequences like:
n times n+1 times
_^_ __^__
/ \ / \
R … R D … D L L … L U U … U
\_ _/ \__ __/
v v
n times n+1 times
We can repeat that, each time incrementing n by two, like:
data Dir = R | D | L | U
spiralSeq :: Int -> [Dir]
spiralSeq n = rn R ++ rn D ++ rn1 L ++ rn1 U
where rn = replicate n
rn1 = replicate (n + 1)
spiral :: [Dir]
spiral = concatMap spiralSeq [1, 3..]
Now we can use Dir here to calculate the next coordinate, like:
move :: (Int, Int) -> Dir -> (Int, Int)
move (x, y) = go
where go R = (x+1, y)
go D = (x, y-1)
go L = (x-1, y)
go U = (x, y+1)
We can use scanl :: (a -> b -> a) -> a -> [b] -> [a] to generate the points, like:
spiralPos :: [(Int, Int)]
spiralPos = scanl move (0,0) spiral
This will yield an infinite list of coordinates for the clockwise spiral. We can use take :: Int -> [a] -> [a] to take the first k items:
Prelude> take 9 spiralPos
[(0,0),(1,0),(1,-1),(0,-1),(-1,-1),(-1,0),(-1,1),(0,1),(1,1)]
The idea with the following solution is that instead of trying to generate the coordinates directly, we’ll look at the directions from one point to the next. If you do that, you’ll notice that starting from the first point, we go 1× right, 1× down, 2× left, 2× up, 3× right, 3× down, 4× left… These can then be seperated into the direction and the number of times repeated:
direction: > v < ^ > v < …
# reps: 1 1 2 2 3 3 4 …
And this actually gives us two really straightforward patterns! The directions just rotate > to v to < to ^ to >, while the # of reps goes up by 1 every 2 times. Once we’ve made two infinite lists with these patterns, they can be combined together to get an overall list of directions >v<<^^>>>vvv<<<<…, which can then be iterated over to get the coordinate values.
Now, I’ve always thought that just giving someone a bunch of code as the solution is not the best way to learn, so I would highly encourage you to try implementing the above idea yourself before looking at my solution below.
Welcome back (if you did try to implement it yourself). Now: onto my own solution. First I define a Stream data type for an infinite stream:
data Stream a = Stream a (Stream a) deriving (Show)
Strictly speaking, I don’t need streams for this; Haskell’s predefined lists are perfectly adequate for this task. But I happen to like streams, and they make some of the pattern matching a bit easier (because I don’t have to deal with the empty list).
Next, I define a type for directions, as well as a function specifying how they interact with points:
-- Note: I can’t use plain Left and Right
-- since they conflict with constructors
-- of the ‘Either’ data type
data Dir = LeftDir | RightDir | Up | Down deriving (Show)
type Point = (Int, Int)
move :: Dir -> Point -> Point
move LeftDir (x,y) = (x-1,y)
move RightDir (x,y) = (x+1, y)
move Up (x,y) = (x,y+1)
move Down (x,y) = (x,y-1)
Now I go on to the problem itself. I’ll define two streams — one for the directions, and one for the number of repetitions of each direction:
dirStream :: Stream Dir
dirStream = Stream RightDir $ Stream Down $ Stream LeftDir $ Stream Up dirVals
numRepsStream :: Stream Int
numRepsStream = go 1
where
go n = Stream n $ Stream n $ go (n+1)
At this point we’ll need a function for replicating each element of a stream a specific number of times:
replicateS :: Stream Int -> Stream a -> Stream a
replicateS (Stream n ns) (Stream a as) = conss (replicate n a) $ replicateS ns as
where
-- add more than one element to the beginning of a stream
conss :: [a] -> Stream a -> Stream a
conss [] s = s
conss (x:xs) s = Stream x $ appends xs s
This gives replicateS dirStream numRepsStream for the stream of directions. Now we just need a function to convert those directions to coordinates, and we’ve solved the problem:
integrate :: Stream Dir -> Stream Point
integrate = go (0,0)
where
go p (Stream d ds) = Stream p (go (move d p) ds)
spiral :: Stream Point
spiral = integrate $ replicateS numRepsStream dirStream
Unfortunately, it’s somewhat inconvenient to print an infinite stream, so the following function is useful for debugging and printing purposes:
takeS :: Int -> Stream a -> [a]
takeS 0 _ = []; takeS n (Stream x xs) = x : (takeS (n-1) xs)
I'm trying to print out my 2d array in game of life, but i'm not quite sure how to go on with it. So I need some help with my printArray function I'm not quite sure how to proceed. Her is the code below, everything is working.. Except printing it out in the right manner.
module GameOfLife where
import Data.List
import System.IO
import Text.Show
import Data.Array
import System.Random
width :: Int
width = 5
height :: Int
height = 5
data State = Alive | Dead deriving (Eq, Show)
type Pos = (Int,Int)
type Board = Array Pos State
startBoard :: Pos -> Board
startBoard (width,height) =
let bounds = ((0,0),(width - 1,height - 1))
in array bounds $ zip (range bounds) (repeat Dead)
set :: Board -> [(Pos,State)] -> Board
set = (//)
get :: Board -> [Pos] -> [State]
get board pos = map (board!) pos
neighbours :: Board -> Pos -> [Pos]
neighbours board c#(x,y) =
filter (/= c) $ filter (inRange (bounds board)) [(x',y') | x' <- [x -
1..x + 1], y' <- [y - 1..y + 1]]
nextGen :: Board -> Board
nextGen board =(irrelevant code for the question..)
printArray :: Board -> String
printArray arr =
unlines [unwords [show (arr ! (x, y)) | x <- [1..5]] | y <- [1..5]]
My output:
[((0,0),Dead),((0,1),Dead),((0,2),Dead),((0,3),Dead),((1,0),Dead),
((1,1),Dead),((1,2),Dead),((1,3),Dead),((2,0),Dead),((2,1),Dead),
((2,2),Dead)2,3),Dead)]
My preferable output:
1 2 3 4 5
1 . . . . .
2 n n n . .
3 n X n . .
4 n n n . .
5 . . . . .
To start to answer your question, I suggest breaking the problem into several pieces:
Print out the numbers across the top.
Number each row as you print them.
Decide what symbol to print in each cell.
Tackle each of these pieces one at a time. If it helps, rather than think in terms of "printing" just build up a String object. Once you have a String, printing is pretty trivial.
Making tree like data structures is relatively easy in Haskell. However, what if I want a structure like the following:
A (root)
/ \
B C
/ \ / \
D E F
So if I traverse down the structure through B to update E, the returned new updated structure also has E updated if I traverse through C.
Could someone give me some hints about how to achieve this? You can assume there are no loops.
I would flatten the data structure to an array, and operate on this instead:
import Data.Array
type Tree = Array Int -- Bounds should start at (1) and go to sum [1..n]
data TreeTraverse = TLeft TreeTraverse | TRight TreeTraverse | TStop
Given some traverse directions (left, right, stop), it's easy to see that if we go left, we simply add the current level to our position, and if we go right, we also add the current position plus one:
getPosition :: TreeTraverse -> Int
getPosition = getPosition' 1 1
where
getPosition' level pos (TLeft ts) = getPosition' (level+1) (pos+level) ts
getPosition' level pos (TRight ts) = getPosition' (level+1) (pos+level + 1) ts
getPosition' _ pos (TStop) = pos
In your case, you want to traverse either ABE or ACE:
traverseABE = TLeft $ TRight TStop
traverseACE = TRight $ TLeft TStop
Since we already now how to get the position of your element, and Data.Array provides some functions to set/get specific elements, we can use the following functions to get/set tree values:
getElem :: TreeTraverse -> Tree a -> a
getElem tt t = t ! getPosition tt
setElem :: TreeTraverse -> Tree a -> a -> Tree a
setElem tt t x = t // [(getPosition tt, x)]
To complete the code, lets use your example:
example = "ABCDEF"
exampleTree :: Tree Char
exampleTree = listArray (1, length example) example
And put everything to action:
main :: IO ()
main = do
putStrLn $ "Traversing from A -> B -> E: " ++ [getElem traverseABE exampleTree]
putStrLn $ "Traversing from A -> C -> E: " ++ [getElem traverseACE exampleTree]
putStrLn $ "exampleTree: " ++ show exampleTree ++ "\n"
putStrLn $ "Setting element from A -> B -> E to 'X', "
let newTree = setElem traverseABE exampleTree 'X'
putStrLn $ "but show via A -> C -> E: " ++ [getElem traverseACE newTree]
putStrLn $ "newTree: " ++ show newTree ++ "\n"
Note that this is most-likely not the best way to do this, but the first thing that I had in mind.
Once you've established identity, it can be done.
But first you must establish identity.
In many languages, values can be distinct from each other, but equal. In Python, for example:
>>> a = [1]
>>> b = [1]
>>> a == b
True
>>> a is b
False
You want to update E in one branch of the tree, and also update all other elements for which that element is E. But Haskell is referentially transparent: it has no notion of things being the same object; only equality, and even that is not applicable for every object.
One way you could do this is equality. Say this was your tree:
__A__
/ \
B C
/ \ / \
1 2 2 3
Then we could go through the tree and update all the 2s to, say, four. But this isn't exactly what you want in some cases.
In Haskell, if you want to update one thing in multiple places, you'll have to be explicit about what is and isn't the same thing. Another way you could deal with this is to tag each different value with a unique integer, and use that integer to determine identity:
____________A___________
/ \
B C
/ \ / \
(id=1)"foo" (id=2)"bar" (id=2)"bar" (id=3)"baz"
Then we could update all values with an identity of 2. Accidental collisions cannot be a problem, as there can be no collisions except those that are intentional.
This is essentially what STRef and IORef do, except they hoist the actual value into the monad's state and hide the identities from you. The only downside of using these is you'll need to make much of your code monadic, but you're probably not going to get away from that easily whatever you do. (Modifying values rather than replacing them is an inherently effectful thing to do.)
The structure you gave was not specified in much detail so it's impossible to tailor an example to your use case, but here's a simple example using the ST monad and a Tree:
import Control.Monad
import Control.Monad.ST
import Data.Tree
import Data.Traversable (traverse)
import Data.STRef
createInitialTree :: ST s (Tree (STRef s String))
createInitialTree = do
[a, b, c, d, e, f] <- mapM newSTRef ["A", "B", "C", "D", "E", "F"]
return $ Node a [ Node b [Node d [], Node e []]
, Node c [Node e [], Node f []]
]
dereferenceTree :: Tree (STRef s a) -> ST s (Tree a)
dereferenceTree = traverse readSTRef
test :: ST s (Tree String, Tree String)
test = do
tree <- createInitialTree
before <- dereferenceTree tree
let leftE = subForest (subForest tree !! 0) !! 1
writeSTRef (rootLabel leftE) "new" -- look ma, single update!
after <- dereferenceTree tree
return (before, after)
main = do
let (before, after) = runST test
putStrLn $ drawTree before
putStrLn $ drawTree after
Observe that although we only explicitly modified the value of the left E value, it changed on the right side, too, as desired.
I should note that these are not the only ways. There are probably many other solutions to this same problem, but they all require you to define identity sensibly. Only once that has been done can one begin the next step.
I'm not sure if our assignment was presented in the most functional-enabling of ways, but I have to work with it. I have a "map" that represents a pacman game state:
B B B B
B P _ B
B . . B
B B B B
where B is a border tile, P is pacman, _ is an empty space, and . is a food pellet. There are many rules when moving pacman, but consider one:
When pacman moves into a tile occupied by a food pellet, replace the pacman tile with an empty space and the food pellet with pacman. This function would have the definition:
move:: [[Char]] -> [[Char]]
Right now I've got functions that give me the (x,y) coordinate tuple of pacman and his new location, and I was planning to use the !! function to "overwrite" the tiles. However, I know a little of list operations such as :. Could I use : to accomplish this task?
Rather than modify the string, I would instead define a function of type:
type Position = (Int, Int)
type Board = [[Char]]
renderBoard :: Position -> Board
Then I would just modify pacman's position and re-render the board:
move :: Position -> Position
Edit: To answer your specific question, you can do this easily using the lens library:
import Control.Lens
move :: Position -> Position -> Board -> Board
move (oldX, oldY) (newX, newY) = (ix oldX.ix oldY .~ '_') . (ix newX.ix newY .~ 'P')
Below is a replace function that you can use to replace a particular position character in the 2d array [[Char]]
replace :: [[Char]] -> (Int,Int) -> Char -> [[Char]]
replace chars (x',y') c = do
(x,row) <- zip [0..] chars
return [if x == x' && y == y' then c else r | (y,r) <- zip [0..] row]
The second argument is the position which needs to be updated with the Char value in 3rd position.
Using this function you should be able to implement your move function.
Here's a simple approach that doesn't rely on libraries. First, we define a function .~ that allows you to set the index of a list -
set n x xs = take n xs ++ (x : drop (n+1) xs)
and give it a convenient alias so that we can use it in infix form.
n .~ x = \xs -> set n x xs
This allows you to do things like
>> let list = [1,2,3,4]
>> 1 .~ 10 $ list
[1,10,3,4]
Now it's a simple matter to extend that to a function that modifies two-dimensional lists
(n,m) .= x = \xs -> n .~ (m .~ x $ xs!!n) $ xs
so that you can do things like
>> let listOfList = [[1,2,3],[4,5,6],[7,8,9]]
>> (1,1) .= 100 $ listOfList
[[1,2,3],[4,100,6],[7,8,9]]
Now you can easily write a function move that takes an old position, a new position and the current board, and modifies the board in the way you want
type Pos = (Int,Int)
type Board = [[Char]]
move :: Pos -> Pos -> Board -> Board
move (x,y) (x',y') board = board''
where
board' = (x, y ) .= '_' $ board
board'' = (x',y') .= 'P' $ board'
That is, the first line in the where clause modifies the board to replace PacMan's old position with an empty space, and the second line modifies the board that's reutrned form that, to put PacMan in the new position.
I am attempting to implement the following algorithm, as detailed here.
Start with a flat terrain (initialize all height values to zero).
Pick a random point on or near the terrain, and a random radius
between some predetermined minimum and maximum. Carefully choosing
this min and max will make a terrain rough and rocky or smooth and
rolling.
Raise a hill on the terrain centered at the point, having the given
radius.
Go back to step 2, and repeat as many times as necessary. The number
of iterations chosen will affect the appearance of the terrain.
However, I start to struggle once I get to the point where I have to select a random point on the terrain. This random point is wrapped in an IO monad, which is then passed up my chain of functions.
Can I cut the IO off at a certain point and, if so, how do I find that point?
The following is my (broken) code. I would appreciate any suggestions on improving it / stopping the randomness from infecting everything.
type Point = (GLfloat, GLfloat, GLfloat)
type Terrain = [Point]
flatTerrain :: Double -> Double -> Double -> Double -> Terrain
flatTerrain width length height spacing =
[(realToFrac x, realToFrac y, realToFrac z)
| x <- [-width,-1+spacing..width], y <- [height], z <- [-length,-1+spacing..length]]
hill :: Terrain -> Terrain
hill terrain = hill' terrain 100
where hill' terrain 0 = terrain
hill' terrain iterations = do
raised <- raise terrain
hill' (raise terrain) (iterations - 1)
raise terrain = do
point <- pick terrain
map (raisePoint 0.1 point) terrain
raisePoint r (cx,cy,cz) (px,py,pz) =
(px, r^2 - ((cx - px)^2 + (cz - pz)^2), pz)
pick :: [a] -> IO a
pick xs = randomRIO (0, (length xs - 1)) >>= return . (xs !!)
The algorithm says that you need to iterate and in each iteration select a random number and update the terrain which can be viewed as generate a list of random points and use this list to update the terrain i.e iteration to generate random numbers == list of random numbers.
So you can do something like:
selectRandomPoints :: [Points] -> Int -> IO [Points] -- generate Int times random points
updateTerrain :: Terrain -> [Points] -> Terrain
-- somewhere in IO
do
pts <- selectRandomPoints allPts iterationCount
let newTerrain = updateTerrain t pts
One of the most useful features of haskell is to know a function is deterministic just based on its type - it makes testing much easier. For this reason, I would base my design on limiting randomness as much as possible, and wrapping the core non random functions with a random variant. This is easily done with the MonadRandom type class, which is the best way of writing code in haskell that requires random values.
For fun, I wrote a console version of that hill generator. It is pretty basic, with a lot of hard coded constants. However, it does provide a pretty cool ascii terrain generator :)
Note with my solution all of the calculations are isolated in pure, non random functions. This could then be tested easily, as the result is deterministic. As little as possible occurs in the IO monad.
import Control.Monad
import Control.Monad.Random
import Data.List
import Data.Function (on)
type Point = (Double, Double, Double)
type Terrain = [Point]
-- Non random code
flatTerrain :: Double -> Double -> Double -> Double -> Terrain
flatTerrain width length height spacing = [(realToFrac x, realToFrac y, realToFrac z)
| x <- [-width,-width+spacing..width], y <- [height], z <- [-length,-length+spacing..length]]
-- simple terrain displayer, uses ascii to render the area.
-- assumes the terrain points are all separated by the same amount
showTerrain :: Terrain -> String
showTerrain terrain = unlines $ map (concat . map showPoint) pointsByZ where
pointsByZ = groupBy ((==) `on` getZ) $ sortBy (compare `on` getZ) terrain
getZ (_, _, z) = z
getY (_, y, _) = y
largest = getY $ maximumBy (compare `on` getY) terrain
smallest = getY $ minimumBy (compare `on` getY) terrain
atPC percent = (largest - smallest) * percent + smallest
showPoint (_, y, _)
| y < atPC (1/5) = " "
| y < atPC (2/5) = "."
| y < atPC (3/5) = "*"
| y < atPC (4/5) = "^"
| otherwise = "#"
addHill :: Double -- Radius of hill
-> Point -- Position of hill
-> Terrain -> Terrain
addHill radius point = map (raisePoint radius point) where
raisePoint :: Double -> Point -> Point -> Point
-- I had to add max py here, otherwise new hills destroyed the
-- old hills with negative values.
raisePoint r (cx,cy,cz) (px,py,pz) = (px, max py (r^2 - ((cx - px)^2 + (cz - pz)^2)), pz)
-- Some random variants. IO is an instance of MonadRandom, so these function can be run in IO. They
-- can also be run in any other monad that has a MonadRandom instance, so they are pretty flexible.
-- creates a random point. Note that the ranges are hardcoded - an improvement would
-- be to be able to specify them, either through parameters, or through reading from a Reader
-- monad or similar
randomPoint :: (MonadRandom m) => m Point
randomPoint = do
x <- getRandomR (-30, 30)
y <- getRandomR (0,10)
z <- getRandomR (-30, 30)
return (x, y, z)
addRandomHill :: (MonadRandom m) => Terrain -> m Terrain
addRandomHill terrain = do
radius <- getRandomR (0, 8) -- hardcoded again
position <- randomPoint
return $ addHill radius position terrain
-- Add many random hills to the Terrain
addRandomHills :: (MonadRandom m) => Int -> Terrain -> m Terrain
addRandomHills count = foldr (>=>) return $ replicate count addRandomHill
-- testing code
test hillCount = do
let terrain = flatTerrain 30 30 0 2
withHills <- addRandomHills hillCount terrain
-- let oneHill = addHill 8 (0, 3, 0) terrain
-- putStrLn $ showTerrain oneHill
putStrLn $ showTerrain withHills
main = test 200
Example output:
... .. ..*. .***^^^***.
... ... .***. .***^^^*^^*.
... .. .*^**......*^*^^^^.
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Nope, you can't escape IO. Perhaps you can do all your randomness up front and rewrite your functions to take that randomness as a parameter; if not, you can use MonadRandom or similar to track a random seed or just put everything in IO.