I want to create a function as mentioned in the title. The specific is that it adds the digits in reversed order, you can see that in the test cases: 12 -> 1; 852369 -> 628; 1714 -> 11; 12345 -> 42; 891 -> 9; 448575 -> 784; 4214 -> 14
The main idea is that when the number is bigger than 99 it enters the helper function which has i - indicator if the the digit is on an even position, and res which stores the result. Helper begins to cycle n as it checks whether or not the current digit is on even position and adds it to the result.
So far I've tried the following code:
everyOther :: Int -> Int
everyOther n
| n < 10 = error "n must be bigger than 10 or equal"
| n < 100 = div n 10
| otherwise = helper n 0 0
where
helper :: Int -> Int -> Int -> Int
helper n i res
| n < 100 = res
| i == 1 = helper (div n 10) (i - 1) (res + (mod n 10)*10)
| otherwise = helper (div n 10) i res
Any help would be appreciated!
You can obtain the one but last digit of x with mod (div x 10) 10. You can use this with an accumulator that accumulates the value by each time multiplying with 10, so:
everyOther :: Int -> Int
everyOther = go 0
where go a v
| v < 10 = a
| otherwise = go (10*a + mod (div v 10) 10) (div v 100)
If v is thus less than 10, we can return the accumulator, since there is no "other digit" anymore. If that is not the case, we multiply a with 10, and add mod (div v 10) 10 to add the other digit to it, and recurse with the value divided by 100 to move it two places to the right.
We can improve this, as #Daniel Wagner says, by making use of quotRem :: Integral a => a -> a -> (a, a):
everyOther :: Int -> Int
everyOther = go 0
where go a v
| v < 10 = a
| otherwise = let (q, r) = v `quotRem` 100 in go (10*a + r `quot` 10) q
here we thus work with the remainder of a division by 100, and this thus avoids an extra modulo.
Related
I'm solving a practice problem in Haskell where I'm trying to count the palindrome numbers between 2 given integers. Single-digit numbers are palindromes. I've tried solving it with a helper function but I can't make it take the smaller number from the main function. Any help would be appreciated!
So far I typed this:
main :: IO()
main = do
print $ countPalindromes 5 13 == 5 -- 6 7 8 9 11
print $ countPalindromes 13 5 == 5 -- 6 7 8 9 11
rev :: Int -> Int
rev n = helper n 0
where
helper :: Int -> Int -> Int
helper 0 result = result
helper n result = helper (div n 10) (result * 10 + mod n 10)
isPalindrome :: Int -> Bool
isPalindrome x = rev x == x
countPalindromes :: Int -> Int -> Int
countPalindromes a b
| a > b = helper b a 0
| otherwise = helper a b 0
where
helper :: Int -> Int -> Int -> Int
helper a b count
| a <= b && isPalindrome (a - 1) = count + 1
| otherwise = helper (a - 1) b count
That's not your problem. The problem is that helper a b count only returns count + 1 if a is a palindrome, without ever checking if a + 1, a + 2, etc, are palindromes as well. When the first number is a palindrome, it returns 0 + 1 == 1 and done. (Your definition of helper is also counting the wrong way; it's decrementing a instead of incrementing as you need to do if you ever want a <= b to be false.)
helper needs to recurse whether or not a is a palindrome; the only difference is in the value of its third argument.
helper a b count | a > b = count -- base
| isPalindrome a = helper (a + 1) b (count + 1)
| otherwise = helper (a + 1) b count
Note that b never changes; it doesn't need to be an argument to helper. Instead, you can make a recursive call to countPalindromes to ensure a < b:
countPalindromes :: Int -> Int -> Int
countPalindromes a b
| a > b = countPalindromes b a
| otherwise = helper a 0
where
helper :: Int -> Int -> Int
helper a count
| a > b = count -- base case
| isPalindrom a = helper (a + 1) (count + 1)
| otherwise = helper (a + 1) count
Tail recursion also isn't terribly important in Haskell. You can write helper more naturally
helper a | a > b = 0
| isPalindrome a = 1 + helper (a + 1)
| otherwise = helper (a + 1)
Note, too, that the only difference between isPalindrome returning True or False is whether you add 1 or 0 to the recursive return value. You can capture that with fromEnum:
helper a | a > b = 0
| otherwise = (fromEnum (isPalindrome a)) + helper (a + 1)
As an exercise, note that you don't need explicit recursion at all. You can use filter to get the values in range that are palindromes, then simply count the number of values in the resulting list.
I am writing some code to work with arbitrary radix numbers in haskell. They will be stored as lists of integers representing the digits.
I almost managed to get it working, but I have run into the problem of converting a list of tuples [(a_1,b_1),...,(a_n,b_n)] into a single list which is defined as follows:
for all i, L(a_i) = b_i.
if there is no i such that a_i = k, a(k)=0
In other words, this is a list of (position,value) pairs for values in an array. If a position does not have a corresponding value, it should be set to zero.
I have read this (https://wiki.haskell.org/How_to_work_on_lists) but I don't think any of these methods are suitable for this task.
baseN :: Integer -> Integer -> [Integer]
baseN n b = convert_digits (baseN_digits n b)
chunk :: (Integer, Integer) -> [Integer]
chunk (e,m) = m : (take (fromIntegral e) (repeat 0))
-- This is broken because the exponents don't count for each other's zeroes
convert_digits :: [(Integer,Integer)] -> [Integer]
convert_digits ((e,m):rest) = m : (take (fromIntegral (e)) (repeat 0))
convert_digits [] = []
-- Converts n to base b array form, where a tuple represents (exponent,digit).
-- This works, except it ignores digits which are zero. thus, I converted it to return (exponent, digit) pairs.
baseN_digits :: Integer -> Integer -> [(Integer,Integer)]
baseN_digits n b | n <= 0 = [] -- we're done.
| b <= 0 = [] -- garbage input.
| True = (e,m) : (baseN_digits (n-((b^e)*m)) b)
where e = (greedy n b 0) -- Exponent of highest digit
m = (get_coef n b e 1) -- the highest digit
-- Returns the exponent of the highest digit.
greedy :: Integer -> Integer -> Integer -> Integer
greedy n b e | n-(b^e) < 0 = (e-1) -- We have overshot so decrement.
| n-(b^e) == 0 = e -- We nailed it. No need to decrement.
| n-(b^e) > 0 = (greedy n b (e+1)) -- Not there yet.
-- Finds the multiplicity of the highest digit
get_coef :: Integer -> Integer -> Integer -> Integer -> Integer
get_coef n b e m | n - ((b^e)*m) < 0 = (m-1) -- We overshot so decrement.
| n - ((b^e)*m) == 0 = m -- Nailed it, no need to decrement.
| n - ((b^e)*m) > 0 = get_coef n b e (m+1) -- Not there yet.
You can call "baseN_digits n base" and it will give you the corresponding array of tuples which needs to be converted to the correct output
Here's something I threw together.
f = snd . foldr (\(e,n) (i,l') -> ( e , (n : replicate (e-i-1) 0) ++ l')) (-1,[])
f . map (fromIntegral *** fromIntegral) $ baseN_digits 50301020 10 = [5,0,3,0,1,0,2,0]
I think I understood your requirements (?)
EDIT:
Perhaps more naturally,
f xs = foldr (\(e,n) fl' i -> (replicate (i-e) 0) ++ (n : fl' (e-1))) (\i -> replicate (i+1) 0) xs 0
I need to determine a recursive function crosssum :: Int -> Int in Haskell to calculate the cross sum of positive numbers. I am not allowed to use any functions from the hierarchical library besides (:), (>), (++), (<), (>=), (<=), div, mod, not (&&), max, min, etc.
crosssum :: Int -> Int
cross sum x = if x > 0
then x `mod` 10
+ x `div` 10 + crosssum x
else 0
so whenever I fill in e.g. crosssum 12 it says 'thread killed'. I do not understand how to get this right. I would appreciate any ideas. Thx
One of the problems with your code is that x is not reduced (or changed somehow) when it's passed as an argument to the recursive call of crosssum. That's why your program never stops.
The modified code:
crosssum :: Int -> Int
crosssum x = if x > 0
then x `mod` 10 + crosssum (x `div` 10)
else 0
is going to have the following logic
crosssum 12 = 2 + (crosssum 1) = 2 + (1 + (crosssum 0)) = 2 + 1 + 0
By the way, Haskell will help you to avoid if condition by using pattern-matching to receive more readable code:
crosssum :: Int -> Int
crosssum 0 = 0
crosssum x =
(mod x 10) + (crosssum (div x 10))
divMod in Prelude is very handy, too. It's one operation for both div and mod, In fact for all 2 digit numbers dm n = sum.sequence [fst,snd] $ divMod n 10
cs 0 = 0; cs n = m+ cs d where (d,m) = divMod n 10
cs will do any size number.
I am working on a program to get the closest prime number by the exponent of 2, this is between an interval.
module Main where
import Data.Char
import System.IO
import Control.Monad (liftM)
data PGetal = G Bool | P Int
instance Show PGetal where
show (P n) = show n
show (G False) = "GEEN PRIEMGETAL GEVONDEN"
mPriem::(Int, Int) -> PGetal
mPriem (x,y) | (x > y) = G False
| (x > 1000000) = G False
| (y > 1000000) = G False
| (null (getAllPriem(x,y))) = G False
| otherwise = P (kleinsteVerschilF(getAllPriem(x,y),1000000,1))
kleinsteVerschilF:: ([Int], Int , Int) -> Int
kleinsteVerschilF ([],_, priemGetal) = priemGetal
kleinsteVerschilF (priem1:priemcss, kleinsteVerschil,priemGetal)=
if(kleinsteVerschil <= kleinsteVerschilMetLijst (priem1,(getMachtenVanTwee(0)),1000000))then kleinsteVerschilF(priemcss, kleinsteVerschil,priemGetal)
else kleinsteVerschilF (priemcss,kleinsteVerschilMetLijst(priem1,(getMachtenVanTwee(0)),1000000), priem1)
kleinsteVerschilMetLijst :: (Int,[Int],Int) -> Int
kleinsteVerschilMetLijst ( _,[],kleinsteVerschil) = kleinsteVerschil
kleinsteVerschilMetLijst (x,tweeMachten1:tweeMachtencss,kleinsteverschil)=
if((abs(x-tweeMachten1)) < kleinsteverschil)
then kleinsteVerschilMetLijst(x,tweeMachtencss, (abs(x-tweeMachten1)))
else kleinsteVerschilMetLijst(x,tweeMachtencss, kleinsteverschil)
getAllPriem :: (Int, Int) ->[Int]
getAllPriem (x,y) = filter isPriem [x..y]
getMachtenVanTwee ::(Int) -> [Int]
getMachtenVanTwee (macht)
|(functieMachtTwee(macht)< 1000000) = (functieMachtTwee(macht)) : (getMachtenVanTwee ((macht+1)))
| otherwise = []
functieMachtTwee:: (Int) -> Int
functieMachtTwee (x) = 2^x
isPriem n = (aantalDelers n)==2
aantalDelers n = telAantalDelersVanaf n 1
telAantalDelersVanaf n kandidaatDeler
| n == kandidaatDeler = 1
| mod n kandidaatDeler == 0
= 1 + telAantalDelersVanaf n (kandidaatDeler+1)
| otherwise
= telAantalDelersVanaf n (kandidaatDeler+1)
aantalDelers2 getal = telDelers getal 1 0
where telDelers n kandidaat teller
| n == kandidaat = 1+teller
| mod n kandidaat == 0
= telDelers n (kandidaat+1) (teller+1)
| otherwise
= telDelers n (kandidaat+1) teller
transform :: [String] -> [PGetal]
transform [] = []
transform (cs:css) =
let (a : b: _ ) = words cs
in (mPriem ((read(a)),(read(b))): transform css)
main :: IO ()
main = do
n <- read `liftM` getLine :: IO Int
lss <- lines `liftM` getContents
let cases = take n lss
let vs = (transform (lss))
putStr $ unlines $ map show vs
When I use the mPriem function, it works fine.
But it needs to work with an input txt file, so I made a .exe file with the ghc command. I also added this .txt file in the folder.
10
1 1
1 3
1 100
200 250
14 16
5 10
20 31
16 50
100 120
5200 7341
When I use in command line this command, it does nothing. There is no output. I can't CTRL+C to stop the program, so I think it crashes. But I don't know what's wrong.
type invoer.txt | programma.exe
Your program works, but is not that efficient and personally I find it not that elegant (sorry :S) because you introduce a lot of "noise". As a result it takes a lot of time before output is written.
If I understand the problem statement correctly, each line (except the first), contains two integers, and you need to count the amount of prime numbers between these two numbers (bounds inclusive?)
First of all, you can do this more elegantly by defining a function: cPrime :: Int -> Int -> Int that takes as input the two numbers and returns the amount of prime numbers:
cPrime :: Int -> Int -> Int
cPrime a b = count $ filter isPrime [a .. b]
You can improve performance by improving your prime checking algorithm. First of all, you do not need to check whether 1 is a divisor, since 1 is always a divisor. Furthermore, you can prove mathematically that there is no divisor greater than sqrt(n) (except for n) that divides n; unless there is another divider that is smaller than sqrt(n). So that means that you can simply enumerate all numbers between 2 and sqrt n and from the moment one of these is a divisor, you can stop: you have proven the number is not prime:
isPrime :: Int -> Bool
isPrime 1 = False
isPrime 2 = True
isPrime n = all ((0 /=) . mod n) (2:[3,5..m])
where m = floor $ sqrt $ fromIntegral n
Now I'm not sure what you aim to do with kleinsteVerschilF.
I want to reverse an Integer in Haskell with recursion. I have a small issue.
Here is the code :
reverseInt :: Integer -> Integer
reverseInt n
| n>0 = (mod n 10)*10 + reverseInt(div n 10)
| otherwise = 0
Example 345
I use as input 345 and I want to output 543
In my program it will do....
reverseInt 345
345>0
mod 345 10 -> 5
reverseInt 34
34
34>0
mod 34 10 -> 4
reverseInt 3
3>0
mod 3 10 -> 3
reverseInt 0
0=0 (ends)
And at the end it returns the sum of them... 5+4+3 = 12.
So I want each time before it sums them, to multiple the sum * 10. So it will go...
5
5*10 + 4
54*10 + 3
543
Here's a relatively simple one:
reverseInt :: Int -> Int
reverseInt 0 = 0
reverseInt n = firstDigit + 10 * (reverseInt $ n - firstDigit * 10^place)
where
n' = fromIntegral n
place = (floor . logBase 10) n'
firstDigit = n `div` 10^place
Basically,
You take the logBase 10 of your input integer, to give you in what place it is (10s, 100s, 1000s...)
Because the previous calculation gives you a floating point number, of which we do not need the decimals, we use the floor function to truncate everything after the decimal.
We determine the first digit of the number by doing n 'div' 10^place. For example, if we had 543, we'd find place to be 2, so firstDigit = 543/100 = 5 (integer division)
We use this value, and add it to 10 * the reverse of the 'rest' of the integer, in this case, 43.
Edit: Perhaps an even more concise and understandable version might be:
reverseInt :: Int -> Int
reverseInt 0 = 0
reverseInt n = mod n 10 * 10^place + reverseInt (div n 10)
where
n' = fromIntegral n
place = (floor . logBase 10) n'
This time, instead of recursing through the first digit, we're recursing through the last one and using place to give it the right number of zeroes.
reverseInt :: Integer -> Integer
reverseInt n = snd $ rev n
where
rev x
| x>0 = let (a,b) = rev(div x 10)
in ((a*10), (mod x 10)*a + b)
| otherwise = (1,0)
Explanation left to reader :)
I don't know convenient way to found how many times you should multiply (mod n 10) on 10 in your 3rd line. I like solution with unfoldr more:
import Data.List
listify = unfoldr (\ x -> case x of
_ | x <= 0 -> Nothing
_ -> Just(mod x 10, div x 10) )
reverse_n n = foldl (\ acc x -> acc*10+x) 0 (listify n)
In listify function we generate list of numbers from integer in reverse order and after that we build result simple folding a list.
Or just convert it to a string, reverse it and convert it back to an integer:
reverseInt :: Integer -> Integer
reverseInt = read . reverse . show
More (not necessarily recursion based) answers for great good!
reverseInt 0 = 0
reverseInt x = foldl (\x y -> 10*x + y) 0 $ numToList x
where
numToList x = if x == 0 then [] else (x `rem` 10) : numToList (x `div` 10)
This is basically the concatenation of two functions : numToList (convert a given integer to a list 123 -> [1,2,3]) and listToNum (do the opposite).
The numToList function works by repeatedly getting the lowest unit of the number (using rem, Haskell's remainder function), and then chops it off (using div, Haskell's integer division function). Once the number is 0, the empty list is returned and the result concatenates into the final list. Keep in mind that this list is in reverse order!
The listToNum function (not seen) is quite a sexy piece of code:
foldl (\x y -> 10*x + y) 0 xs
This starts from the left and moves to the right, multiplying the current value at each step by 10 and then adding the next number to it.
I know the answer has already been given, but it's always nice to see alternative solutions :)
The first function is recursive to convert the integer to a list. It was originally reversing but the re-conversion function reversed easier so I took it out of the first. The functions can be run separately. The first outputs a tuple pair. The second takes a tuple pair. The second is not recursive nor did it need to be.
di 0 ls = (ls,sum ls); di n ls = di nn $ d:ls where (nn,d) = divMod n 10
di 3456789 []
([3,4,5,6,7,8,9],42)
rec (ls,n) = (sum [y*(10^x)|(x,y) <- zip [0..] ls ],n)
Run both as
rec $ di 3456789 []
(9876543,42)