I have this self-defined IO() Triangle, which has each row of a triangle as an element in a list:
import Data.List ( intersperse )
type Triangle = Char -> Int -> [String]
centeredTriangle :: Triangle
centeredTriangle c n = [replicate (n-i) ' ' ++ intersperse ' ' (replicate i c) | i <- [0 .. n]]
Output:
Ok, one module loaded.
ghci> centeredTriangle '*' 6
[" "," *"," * *"," * * *"," * * * *"," * * * * *","* * * * * *"]
When I run my main function, I use unlines to print out the triangle like so:
triangles :: Int -> Int -> Int -> IO()
triangles a b c = do
putStr $ unlines $ centeredTriangle '*' a
putStr $ unlines $ centeredTriangle '*' b
putStr $ unlines $ centeredTriangle '*' c
Output:
ghci> triangles 1 2 3
*
*
* *
*
* *
* * *
I want to print the triangles on the same line, like so:
ghci> triangles 1 2 3
*
* * *
* * * * * *
I realise it may be a bigger task than I first anticipated, but my first thought was to use centeredTriangle to get the bottom line (last element of the list) of each triangle, and put them toghether in a new string which I then put as the last element in a new list of strings. I thought If I do this for each element (starting from the last) and up to the top I can print the triangles in the same line using my main funtion. How do I achieve this?
Using type Triangle = [String],
Write a function boundingBox :: Triangle -> (Int,Int).
Write a function maxBoundingBox :: [Triangle] -> (Int,Int).
Write a function putInBox :: Triangle -> (Int,Int) -> Triangle.
Write a function makeSameSize :: [Triangle] -> [Triangle].
Write a function sideToSide :: Triangle -> [String] -> [String].
Use it in foldr to conjoin all the resized Triangles together.
Some of the above functions aren't needed if you are the one creating these triangles in the first place. Then you just know the bounding box of the triangle you would create from the given parameters.
Related
I am trying to understand why my code doesn't work:
writeRow :: Int -> IO()
writeRow x = putStr(concat (replicate x "* "))
triangle :: Int -> IO()
triangle x = do
writeRow x
putStr ""
triangle x-1
My thinking is, that writeRow creates a row of x times " * ", so if x = 4, triangle 4 will write
* * * *
As triangle is called recursivly with x -1, the next line will be:
* * *
And so on ... until:
* * * *
* * *
* *
*
But it just outputs it all on one line:
* * * * * * * *
What seems to be the issue? :D
putStr only prints the string, it does not write a new line, you should use putStrLn. Furthermore you should use triangle (x-1) so with parenthesis around the x-1 part, and specify a base case for the recursion with the triangle:
writeRow :: Int -> IO()
writeRow x = putStr(concat (replicate x "* "))
triangle :: Int -> IO()
triangle x | x <= 0 = pure ()
triangle x = do
writeRow x
putStrLn ""
triangle (x-1)
With these modifications, we obtain:
Prelude> triangle 4
* * * *
* * *
* *
*
Inspired by this question, I have made this code which prints out triangles:
type TriangleMaker = Char -> Int -> [String]
topLeftTriangle :: TriangleMaker
topLeftTriangle c n = [replicate i c | i <- [1 .. n]]
centeredTriangle :: TriangleMaker
centeredTriangle c n = [replicate (n-i) ' ' ++ replicate i c | i <- [0 .. n]]
getType :: IO TriangleMaker
getType = do
let menu = [topLeftTriangle, centeredTriangle]
putStr $ unlines [
"What type of triangle do you want to print? (type 1 and then type the int size)",
"1) Top Left",
"2) Centered"]
line <- getLine
return (menu !! ((read line :: Int) - 1))
trekant :: IO()
trekant = do
triangle <- getType
size <- getLine
putStr $ unlines $ triangle '*' (read size :: Int)
It gives me this output in ghci:
Ok, one module loaded.
ghci> trekant
What type of triangle do you want to print? (type 1 and then type the int size)
1) Top Left
2) Centered
1
6
*
**
***
****
*****
******
ghci> trekant
What type of triangle do you want to print? (type 1 and then type the int size)
1) Top Left
2) Centered
2
6
*
**
***
****
*****
******
I want to make it so that I can use a string as input instead of a char, like so:
trekant :: IO()
trekant = do
triangle <- getType
size <- getLine
putStr $ unlines $ triangle " *" (read size :: Int)
That way, (I think) I'll get a centered triangle as output:
ghci> trekant
What type of triangle do you want to print? (type 1 and then type the int size)
1) Top Left
2) Centered
2
6
*
* *
* * *
* * * *
* * * * *
* * * * * *
Or am I way off here? How can I re-write this so that the triangle is centered?
In case you want to generate a triangle in the center, you should add spaces between two stars, this thus means that the string looks like:
centeredTriangle :: TriangleMaker
centeredTriangle c n = [replicate (n-i) ' ' ++ concat (replicate i [c, ' ']) | i <- [0 .. n]]
We thus generate a string where we have n-i spaces followed by n times the "* " string.
Perhaps it is more elegant to work with intersperse :: a -> [a] -> [a] where we intersperse a list of '*' characters with spaces:
import Data.List(intersperse)
centeredTriangle :: TriangleMaker
centeredTriangle c n = [replicate (n-i) ' ' ++ intersperse ' ' (replicate i c) | i <- [0 .. n]]
This then produces:
ghci> trekant
What type of triangle do you want to print? (type 1 and then type the int size)
1) Top Left
2) Centered
2
6
*
* *
* * *
* * * *
* * * * *
* * * * * *
ghci> trekant
What type of triangle do you want to print? (type 1 and then type the int size)
1) Top Left
2) Centered
2
10
*
* *
* * *
* * * *
* * * * *
* * * * * *
* * * * * * *
* * * * * * * *
* * * * * * * * *
* * * * * * * * * *
Below I have defined a function that converts a list of base-3 digits to the corresponding numerical value. For example:
f "201" = (2 * 3^2) + (0 * 3^1) + (1 * 3^0) = 19
f "12" = 5
f "1202" = 47
f "120221" = 430
Here is a definition using comprehension:
f :: String -> Int
f str = sum (listToFinal (stringToTuples str))
Helper functions:
-- 1) converts "201" to "102"
reverse "str"
-- 2) converts "102" to 102
stringToInt :: String -> Int
stringToInt str = read str :: Int
-- 3) converts 102 to ['1','0','2']
intToList :: Int -> [Int]
intToList 0 = []
intToList x = intToList (x `div` 10) ++ [x `mod` 10]
-- 4) converts "201" to [(1,0),(0,1),(2,2)] using reverse, stringToInt, intToList
stringToTuples :: String -> [(Int,Int)]
stringToTuples str = zip (intToList (stringToInt (reverse str))) [0..]
-- 5) converts [(1,0),(0,1),(2,2)] to [1*3^0, 0*3^1, 2*3^2]
listToFinal :: [(Int,Int)] -> [Int]
listToFinal list = [ x * (3^y) | (x,y) <- list ]
Now I'd like to do it with recursion only (well, using basic & library functions too, of course).
An idea: I was thinking of taking the head of each element in the list and simply multiplying it with 3^(length of string - 1). The only problem is, with each recursive call the power of three would have to decrease by 1, e.g. given:
recursive_version "201" = (2 * 3^2) + (0 * 3^1) + (1 * 3^0)
How to implement this?
Here is a much simpler approach; note that, through the use of foldl, it's only "implicitly" recursive, though. For information, digitToInt is exported by Data.Char.
import Data.Char
import Data.List ( foldl' )
--- horner x xs : the value of polynomial 'xs' at point 'x'
horner :: Int -> [Int] -> Int
horner x = foldl' (\c1 c0 -> c1 * x + c0) 0
-- f s : the integer whose representation in base 3 is string 's'
f :: String -> Int
f = horner 3 . map digitToInt
When you define it recursively, the natural way to decrement the length is trimming the array from the head. For example:
base3 x = base3' x 0 where
base3' (d:ds) v = base3' ds $ v + d * 3 ^ length ds
base3' [] v = v
I hope this works by just pasting and running it with "runghc euler4.hs 1000". Since I am having a hard time learning Haskell, can someone perhaps tell me how I could improve here? Especially all those "fromIntegral" are a mess.
module Main where
import System.Environment
main :: IO ()
main = do
args <- getArgs
let
hBound = read (args !! 0)::Int
squarePal = pal hBound
lBound = floor $ fromIntegral squarePal /
(fromIntegral hBound / fromIntegral squarePal)
euler = maximum $ takeWhile (>squarePal) [ x | y <- [lBound..hBound],
z <- [y..hBound],
let x = y * z,
let s = show x,
s == reverse s ]
putStrLn $ show euler
pal :: Int -> Int
pal n
| show pow == reverse (show pow) = n
| otherwise = pal (n-1)
where
pow = n^2
If what you want is integer division, you should use div instead of converting back and forth to Integral in order to use ordinary /.
module Main where
import System.Environment
main :: IO ()
main = do
(arg:_) <- getArgs
let
hBound = read arg :: Int
squarePal = pal hBound
lBound = squarePal * squarePal `div` hBound
euler = maximum $ takeWhile (>squarePal) [ x | y <- [lBound..hBound],
z <- [y..hBound],
let x = y * z,
let s = show x,
s == reverse s ]
print euler
pal :: Int -> Int
pal n
| show pow == reverse (show pow) = n
| otherwise = pal (n - 1)
where
pow = n * n
(I've re-written the lbound expression, that used two /, and fixed some styling issues highlighted by hlint.)
Okay, couple of things:
First, it might be better to pass in a lower bound and an upper bound for this question, it makes it a little bit more expandable.
If you're only going to use the first two (one in your previous case) arguments from the CL, we can handle this with pattern matching easily and avoid yucky statements like (args !! 0):
(arg0:arg1:_) <- getArgs
Let's convert these to Ints:
let [a, b] = map (\x -> read x :: Int) [arg0,arg1]
Now we can reference a and b, our upper and lower bounds.
Next, let's make a function that runs through all of the numbers between an upper and lower bound and gets a list of their products:
products a b = [x*y | x <- [a..b], y <- [x..b]]
We do not have to run over each number twice, so we start x at our current y to get all of the different products.
from here, we'll want to make a method that filters out non-palindromes in some data set:
palindromes xs = filter palindrome xs
where palindrome x = show x == reverse $ show x
finally, in our main function:
print . maximum . palindromes $ products a b
Here's the full code if you would like to review it:
import System.Environment
main = do
(arg0:arg1:_) <- getArgs
let [a, b] = map (\x -> read x :: Int) [arg0,arg1]
print . maximum . palindromes $ products a b
products a b = [x*y | x <- [a..b], y <- [x..b]]
palindromes = filter palindrome
where palindrome x = (show x) == (reverse $ show x)
Here is my first Haskell program. What parts would you write in a better way?
-- Multiplication table
-- Returns n*n multiplication table in base b
import Text.Printf
import Data.List
import Data.Char
-- Returns n*n multiplication table in base b
mulTable :: Int -> Int -> String
mulTable n b = foldl (++) (verticalHeader n b w) (map (line n b w) [0..n])
where
lo = 2* (logBase (fromIntegral b) (fromIntegral n))
w = 1+fromInteger (floor lo)
verticalHeader :: Int -> Int -> Int -> String
verticalHeader n b w = (foldl (++) tableHeader columnHeaders)
++ "\n"
++ minusSignLine
++ "\n"
where
tableHeader = replicate (w+2) ' '
columnHeaders = map (horizontalHeader b w) [0..n]
minusSignLine = concat ( replicate ((w+1)* (n+2)) "-" )
horizontalHeader :: Int -> Int -> Int -> String
horizontalHeader b w i = format i b w
line :: Int -> Int -> Int -> Int -> String
line n b w y = (foldl (++) ((format y b w) ++ "|" )
(map (element b w y) [0..n])) ++ "\n"
element :: Int -> Int -> Int -> Int -> String
element b w y x = format (y * x) b w
toBase :: Int -> Int -> [Int]
toBase b v = toBase' [] v where
toBase' a 0 = a
toBase' a v = toBase' (r:a) q where (q,r) = v `divMod` b
toAlphaDigits :: [Int] -> String
toAlphaDigits = map convert where
convert n | n < 10 = chr (n + ord '0')
| otherwise = chr (n + ord 'a' - 10)
format :: Int -> Int -> Int -> String
format v b w = concat spaces ++ digits ++ " "
where
digits = if v == 0
then "0"
else toAlphaDigits ( toBase b v )
l = length digits
spaceCount = if (l > w) then 0 else (w-l)
spaces = replicate spaceCount " "
Here are some suggestions:
To make the tabularity of the computation more obvious, I would pass the list [0..n] to the line function rather than passing n.
I would further split out the computation of the horizontal and vertical axes so that they are passed as arguments to mulTable rather than computed there.
Haskell is higher-order, and almost none of the computation has to do with multiplication. So I would change the name of mulTable to binopTable and pass the actual multiplication in as a parameter.
Finally, the formatting of individual numbers is repetitious. Why not pass \x -> format x b w as a parameter, eliminating the need for b and w?
The net effect of the changes I am suggesting is that you build a general higher-order function for creating tables for binary operators. Its type becomes something like
binopTable :: (i -> String) -> (i -> i -> i) -> [i] -> [i] -> String
and you wind up with a much more reusable function—for example, Boolean truth tables should be a piece of cake.
Higher-order and reusable is the Haskell Way.
You don't use anything from import Text.Printf.
Stylistically, you use more parentheses than necessary. Haskellers tend to find code more readable when it's cleaned of extraneous stuff like that. Instead of something like h x = f (g x), write h = f . g.
Nothing here really requires Int; (Integral a) => a ought to do.
foldl (++) x xs == concat $ x : xs: I trust the built-in concat to work better than your implementation.
Also, you should prefer foldr when the function is lazy in its second argument, as (++) is – because Haskell is lazy, this reduces stack space (and also works on infinite lists).
Also, unwords and unlines are shortcuts for intercalate " " and concat . map (++ "\n") respectively, i.e. "join with spaces" and "join with newlines (plus trailing newline)"; you can replace a couple things by those.
Unless you use big numbers, w = length $ takeWhile (<= n) $ iterate (* b) 1 is probably faster. Or, in the case of a lazy programmer, let w = length $ toBase b n.
concat ( (replicate ((w+1)* (n+2)) "-" ) == replicate ((w+1) * (n+2)) '-' – not sure how you missed this one, you got it right just a couple lines up.
You do the same thing with concat spaces, too. However, wouldn't it be easier to actually use the Text.Printf import and write printf "%*s " w digits?
Norman Ramsey gave excellent high-level (design) suggestions; Below are some low-level ones:
First, consult with HLint. HLint is a friendly program that gives you rudimentary advice on how to improve your Haskell code!
In your case HLint gives 7 suggestions. (mostly about redundant brackets)
Modify your code according to HLint's suggestions until it likes what you feed it.
More HLint-like stuff:
concat (replicate i "-"). Why not replicate i '-'?
Consult with Hoogle whenever there is reason to believe that a function you need is already available in Haskell's libraries. Haskell comes with tons of useful functions so Hoogle should come in handy quite often.
Need to concatenate strings? Search for [String] -> String, and voila you found concat. Now go replace all those folds.
The previous search also suggested unlines. Actually, this even better suits your needs. It's magic!
Optional: pause and thank in your heart to Neil M for making Hoogle and HLint, and thank others for making other good stuff like Haskell, bridges, tennis balls, and sanitation.
Now, for every function that takes several arguments of the same type, make it clear which means what, by giving them descriptive names. This is better than comments, but you can still use both.
So
-- Returns n*n multiplication table in base b
mulTable :: Int -> Int -> String
mulTable n b =
becomes
mulTable :: Int -> Int -> String
mulTable size base =
To soften the extra characters blow of the previous suggestion: When a function is only used once, and is not very useful by itself, put it inside its caller's scope in its where clause, where it could use the callers' variables, saving you the need to pass everything to it.
So
line :: Int -> Int -> Int -> Int -> String
line n b w y =
concat
$ format y b w
: "|"
: map (element b w y) [0 .. n]
element :: Int -> Int -> Int -> Int -> String
element b w y x = format (y * x) b w
becomes
line :: Int -> Int -> Int -> Int -> String
line n b w y =
concat
$ format y b w
: "|"
: map element [0 .. n]
where
element x = format (y * x) b w
You can even move line into mulTable's where clause; imho, you should.
If you find a where clause nested inside another where clause troubling, then I suggest to change your indentation habits. My recommendation is to use consistent indentation of always 2 or always 4 spaces. Then you can easily see, everywhere, where the where in the other where is at. ok
Below's what it looks like (with a few other changes in style):
import Data.List
import Data.Char
mulTable :: Int -> Int -> String
mulTable size base =
unlines $
[ vertHeaders
, minusSignsLine
] ++ map line [0 .. size]
where
vertHeaders =
concat
$ replicate (cellWidth + 2) ' '
: map horizontalHeader [0 .. size]
horizontalHeader i = format i base cellWidth
minusSignsLine = replicate ((cellWidth + 1) * (size + 2)) '-'
cellWidth = length $ toBase base (size * size)
line y =
concat
$ format y base cellWidth
: "|"
: map element [0 .. size]
where
element x = format (y * x) base cellWidth
toBase :: Integral i => i -> i -> [i]
toBase base
= reverse
. map (`mod` base)
. takeWhile (> 0)
. iterate (`div` base)
toAlphaDigit :: Int -> Char
toAlphaDigit n
| n < 10 = chr (n + ord '0')
| otherwise = chr (n + ord 'a' - 10)
format :: Int -> Int -> Int -> String
format v b w =
spaces ++ digits ++ " "
where
digits
| v == 0 = "0"
| otherwise = map toAlphaDigit (toBase b v)
spaces = replicate (w - length digits) ' '
0) add a main function :-) at least rudimentary
import System.Environment (getArgs)
import Control.Monad (liftM)
main :: IO ()
main = do
args <- liftM (map read) $ getArgs
case args of
(n:b:_) -> putStrLn $ mulTable n b
_ -> putStrLn "usage: nntable n base"
1) run ghc or runhaskell with -Wall; run through hlint.
While hlint doesn't suggest anything special here (only some redundant brackets), ghc will tell you that you don't actually need Text.Printf here...
2) try running it with base = 1 or base = 0 or base = -1
If you want multiline comments use:
{- A multiline
comment -}
Also, never use foldl, use foldl' instead, in cases where you are dealing with large lists which must be folded. It is more memory efficient.
A brief comments saying what each function does, its arguments and return value, is always good. I had to read the code pretty carefully to fully make sense of it.
Some would say if you do that, explicit type signatures may not be required. That's an aesthetic question, I don't have a strong opinion on it.
One minor caveat: if you do remove the type signatures, you'll automatically get the polymorphic Integral support ephemient mentioned, but you will still need one around toAlphaDigits because of the infamous "monomorphism restriction."