This Haskell program prints "1.0" How can I get it to print "1"?
fact 0 = 1
fact x = x * fact (x-1)
place m n = (fact m) / (fact n) * (fact (m-n))
main = do
print (place 0 0)
By using the / operation, you are asking haskell to use a fractional data type. You probably don't want that in this case. It is preferable to use an integral type such as Int or Integer. So I suggest to do the following:
1. Add a type declaration for the fact function, something like fact :: Integer -> Integer
2. Use quot instead of /.
So your code should look like this:
fact :: Integer -> Integer
fact 0 = 1
fact x = x * fact (x-1)
place :: Integer -> Integer -> Integer
place m n = (fact m) `quot` (fact n) * (fact (m-n))
main = do
print (place 0 0)
Also, as #leftaroundabout pointed out, you probably want to use a better algorithm for computing those binomial numbers.
You could just use round:
print (round $ place 0 0)
This changes the formatting to the one you want. redneb's answer is, however, the right approach.
Related
Good morning everyone!
I'm using the following function as a fitting example of a function that needs to have a simple input and output. In this case it's a function that converts a number from decimal to binary form, as a list of digits no less, just because it is convenient later on.
I chose to write it like this, because even though a number goes in and a list comes out, another structure is needed as an intermediate step, that will hold the digits found so far and hold the quotient of the division, as input for the next step of the loop. I will clean up the necessary mess before outputing anything, though, by selecting the part of the structure that I'm interested in, in this case the second one , and not counters or other stuff, that I'm done with. (As I mentioned this is an example only, and it's not unusual in other cases to initialize the until loop with a triplet like (a,b,c), only to pick one of them at the end, as I see fit, with the help of additional function, like pickXof3.)
So there,
dec2Bin :: Int -> [Int]
dec2Bin num = snd $ until
(\(n,l) -> n <=0) -- test
(\(n,l) -> (fst $ division n, (snd $ division n):l)) -- main function
(num,[]) -- initialization
where division a = divMod a 2
I find it very convenient that Haskell, although lacking traditional for/while loops has a function like until, which reminds me very much of Mathematica's NextWhile, that I'm familiar with.
In the past I would write sth even uglier, like two functions, a "helper" one and a "main" one, like so
dec2BinHelper :: (Int,[Int]) -> (Int,[Int])
dec2BinHelper (n,l)
| n <= 0 = (n,l)
| otherwise = dec2BinHelper (fst $ division n, (snd $ division n):l)
where division a = divMod a 2
-- a function with the sole purpose to act as a front-end to the helper function, initializing its call parameters and picking up its output
dec2Bin :: Int -> [Int]
dec2Bin n = snd $ dec2BinHelper (n,[])
which I think is unnecessarily bloated.
Still, while the use of until allows me to define just one function, I get the feeling that it could be done even simpler/easier to read, perhaps in a way more fitting to functional programming. Is that so? How would you write such a function differently, while keeping the input and output at the absolutely essential values?
I strongly prefer your second solution. I'd start a clean-up with two things: use pattern matching, and use where to hide your helper functions. So:
dec2Bin :: Int -> [Int]
dec2Bin n = snd $ dec2BinHelper (n, []) where
dec2BinHelper (n, l)
| n <= 0 = (n, l)
| otherwise = dec2BinHelper (d, m:l)
where (d, m) = divMod n 2
Now, in the base case, you return a tuple; but then immediately call snd on it. Why not fuse the two?
dec2Bin :: Int -> [Int]
dec2Bin n = dec2BinHelper (n, []) where
dec2BinHelper (n, l)
| n <= 0 = l
| otherwise = dec2BinHelper (d, m:l)
where (d, m) = divMod n 2
There's no obvious reason why you should pass these arguments in a tuple, rather than as separate arguments, which is more idiomatic and saves some allocation/deallocation noise besides.
dec2Bin :: Int -> [Int]
dec2Bin n = dec2BinHelper n [] where
dec2BinHelper n l
| n <= 0 = l
| otherwise = dec2BinHelper d (m:l)
where (d, m) = divMod n 2
You can swap the arguments to dec2BinHelper and eta-reduce; that way, you will not be shadowing the definition of n.
dec2Bin :: Int -> [Int]
dec2Bin = dec2BinHelper [] where
dec2BinHelper l n
| n <= 0 = l
| otherwise = dec2BinHelper (m:l) d
where (d, m) = divMod n 2
Since you know that n > 0 in the recursive call, you can use the slightly faster quotRem in place of divMod. You could also consider using bitwise operations like (.&. 1) and shiftR 1; they may be even better, but you should benchmark to know for sure.
dec2Bin :: Int -> [Int]
dec2Bin = dec2BinHelper [] where
dec2BinHelper l n
| n <= 0 = l
| otherwise = dec2BinHelper (r:l) q
where (q, r) = quotRem n 2
When you don't have a descriptive name for your helper function, it's traditional to name it go or loop.
dec2Bin :: Int -> [Int]
dec2Bin = go [] where
go l n
| n <= 0 = l
| otherwise = go (r:l) q
where (q, r) = quotRem n 2
At this point, the two sides of the conditional are short enough that I'd be tempted to put them on their own line, though this is something of an aesthetic choice.
dec2Bin :: Int -> [Int]
dec2Bin = go [] where
go l n = if n <= 0 then l else go (r:l) q
where (q, r) = quotRem n 2
Finally, a comment on the name: the input isn't really in decimal in any meaningful sense. (Indeed, it's much more physically accurate to think of the input as already being in binary!) Perhaps int2Bin or something like that would be more accurate. Or let the type speak for itself, and just call it toBin.
toBin :: Int -> [Int]
toBin = go [] where
go l n = if n <= 0 then l else go (r:l) q
where (q, r) = quotRem n 2
At this point I'd consider this code quite idiomatic.
I am trying to build a function that converts a Decimal(Int) into a Binary number.
Unfortunately other than in java it is not possible to divide an int by two in haskell.
I am very new to functional programming so the problem could be something trivial.
So far I could not find another solution to this problem but
here is my first try :
fromDecimal :: Int -> [Int]
fromDecimal 0 = [0]
fromDecimal n = if (mod n 2 == 0) then
do
0:fromDecimal(n/2)
else
do
1:fromDecimal(n/2)
I got an java implementation here which I did before :
public void fromDecimal(int decimal){
for (int i=0;i<values.length;i++){
if(decimal % 2 = 0)
values[i]=true ;
decimal = decimal/ 2;
else {values[i]= false;
} }
}
Hopefully this is going to help to find a solution!
There are some problems with your solution. First of all, I advise not to use do at all, until you understand what do does. Here we do not need do at all.
Unfortunately other than in java it is not possible to divide an int by two in haskell.
It actually is, but the / operator (which is in fact the (/) function), has type (/) :: Fractional a => a -> a -> a. An Int is not Fractional. You can perform integer division with div :: Integral a => a -> a -> a.
So then the code looks like:
fromDecimal :: Int -> [Int]
fromDecimal 0 = [0]
fromDecimal n = if (mod n 2 == 0) then 0:fromDecimal (div n 2) else 1:fromDecimal (div n 2)
But we can definitely make this more elegant. mod n 2 can only result in two outcomes: 0 and 1, and these are exactly the ones that we use at the left side of the (:) operator.
So we do not need to use an if-then-else at all:
fromDecimal :: Int -> [Int]
fromDecimal 0 = [0]
fromDecimal n = mod n 2 : fromDecimal (div n 2)
Likely this is still not exactly what you want: here we write the binary value such that the last element, is the most significant one. This function will add a tailing zero, which does not make a semantical difference (due to that order), but it is not elegant either.
We can define an function go that omits this zero, if the given value is not zero, like:
fromDecimal :: Int -> [Int]
fromDecimal 0 = [0]
fromDecimal n = go n
where go 0 = []
go k = mod k 2 : go (div k 2)
If we however want to write the most significant bit first (so in the same order as we write decimal numbers), then we have to reverse the outcome. We can do this by making use of an accumulator:
fromDecimal :: Int -> [Int]
fromDecimal 0 = [0]
fromDecimal n = go n []
where go 0 r = r
go k rs = go (div k 2) (mod k 2:rs)
You cannot / integers in Haskell – division is not defined in terms of integral numbers! For integral division use div function, but in your case more suitable would be divMod that comes with mod gratis.
Also, you are going to get reversed output, so you can reverse manually it after that, or use more memory-efficient version with accumulator:
decToBin :: Int -> [Int]
decToBin = go [] where
go acc 0 = acc
go acc n = let (d, m) = n `divMod` 2 in go (m : acc) d
go will give you an empty list for 0. You may add it manually if the list is empty:
decToBin = (\l -> if null l then [0] else l) . go [] where ...
Think through how your algorithm will work. It starts from 2⁰, so it will generate bits backward from how we ordinarily think of them, i.e., least-significant bit first. Your algorithm can represent non-negative binary integers only.
fromDecimal :: Int -> [Int]
fromDecimal d | d < 0 = error "Must be non-negative"
| d == 0 = [0]
| otherwise = reverse (go d)
where go 0 = []
go d = d `rem` 2 : go (d `div` 2)
In Haskell, when we generate a list in reverse, go ahead and do so but then reverse the result at the end. The reason for this is consing up a list (gluing new items at the head with :) has a constant cost and the reverse at the end has a linear cost — but appending with ++ has a quadratic cost.
Common Haskell style is to have a private inner loop named go that the outer function applies when it’s happy with its arguments. The base case is to terminate with the empty list when d reaches zero. Otherwise, we take the current remainder modulo 2 and then proceed with d halved and truncated.
Without the special case for zero, fromDecimal 0 would be the empty list rather than [0].
The binary numbers are usually strings and not really used in calculations.
Strings are also less complicated.
The pattern of binary numbers is like any other. It repeats but at a faster clip.
Only a small set is necessary to generate up to 256 (0-255) binary numbers.
The pattern can systematically be expanded for more.
The starting pattern is 4, 0-3
bd = ["00","01","10","11"]
The function to combine them into larger numbers is
d2b n = head.drop n $ [ d++e++f++g | d <- bd, e <- bd, f <- bd, g <- bd]
d2b 125
"01111101"
If it's not obvious how to expand, then
bd = ["000","001","010","011","100","101","110","111"]
Will give you up to 4096 binary digits (0-4095). All else stays the same.
If it's not obvious, the db2 function uses 4 pairs of binary numbers so 4 of the set. (2^8) - 1 or (2^12) - 1 is how many you get.
By the way, list comprehension are sugar coated do structures.
Generate the above patterns with
[ a++b | a <- ["0","1"], b <- ["0","1"] ]
["00","01","10","11"]
and
[ a++b++c | a <- ["0","1"], b <- ["0","1"], c <- ["0","1"] ]
["000","001","010","011","100","101","110","111"]
More generally, one pattern and one function may serve the purpose
b2 = ["0","1"]
b4 = [ a++b++c++d | a <- b2, b <- b2, c <- b2, d <- b2]
b4
["0000","0001","0010","0011","0100","0101","0110","0111","1000","1001","1010","1011","1100","1101","1110","1111"]
bb n = head.drop n $ [ a++b++c++d | a <- b4, b <- b4, c <- b4, d <- b4]
bb 32768
"1000000000000000"
bb 65535
"1111111111111111"
To calculate binary from decimal directly in Haskell using subtraction
cvtd n (x:xs) | x>n = 0:(cvtd n xs)
| n>x = 1:(cvtd (n-x) xs)
| True = 1:[0|f<-xs]
Use any number of bits you want, for example 10 bits.
cvtd 639 [2^e|e<-[9,8..0]]
[1,0,0,1,1,1,1,1,1,1]
import Data.List
dec2bin x =
reverse $ binstr $ unfoldr ndiv x
where
binstr = map (\x -> "01" !! x)
exch (a,b) = (b,a)
ndiv n =
case n of
0 -> Nothing
_ -> Just $ exch $ divMod n 2
Hi i programing function in Haskell, that convert binary number to decimal
i have function remainder
--return last bit
portion :: Integer -> Integer
portion10 n = n `div` 10
remainder10 :: Integer -> Integer
remaindern10 n = n `mod` 10
now i wanna binToDec with use recursive
binToDec 0 = 0
binToDec binary = (remainder10 binary) * 2^x++ + binToDec (portion10 binary)
i need know how i can implement 2^x++ in me binToDec function :/ please help :)
First off, you can't change variables in Haskell. Pretty much eveything is immutable. If you want a counter, the simplest way to get one is to add another argument to a function:
binToDec binary = binToDec binary 0 where
binToDec' 0 _ = 0
binToDec' binary counter = (remainder10 binary) * 2^counter + binToDec' (portion10 binary) (counter+1)
Secondly, however, you can actually do rather better in this case. Note that calculating 2^n requires squaring 2 several times, and doing some extra multiplications. You can actually do without that.
Or without explicit recursion, with just playing with list functions
import Data.List
import Data.Tuple
binToDec x = sum $ zipWith (*) (unfoldr remainder x)
(iterate (*2) 1) where
remainder 0 = Nothing
remainder x = Just . swap $ divMod x 10
iterate (*2) 1 is the answer to your question , how to "implement 2^x++".
Or you can do it like this, just another version:
binToDec :: Int -> Int -> Int
binToDec 0 _ = 0
binToDec b c = mod b 10 * 2^c + binToDec (div b 10) (c + 1)
I have this code to work out the sum of squares of integers in the range of m:n
sumsquares :: Integral a=> Int -> Int -> Int -> Int
sumsquares m n middle
| m > n = error "First number cannot be bigger than second number"
|m==n = m*m
|otherwise = m*m + sumsquares (m+1)n
How would i redefine the function sumsquares for this purpose?
If there is more than one number in the range m:n, compute the middle of the range and add the sum of the squares of (m:middle) to sum of the squares (middle+1:n),
otherwise there is only one number in the range m:n, so m = = n, and the solution is just the square of m. (Note that with this approach the recursion combines two half- solutions: each sub-problem is approximately half in size of the overall problem).
In your original function, the class constraint Integral a in the type signature is obsolete (a is not mentioned anywhere else in the signature, is it?). Furthermore, the third parameter of the function (middle) remains unused. Hence, you could have written it as
sumsquares :: Int -> Int -> Int
sumsquares m n
| m > n = error "First number cannot be bigger than second number"
| m == n = m * m
| otherwise = m * m + sumsquares (m + 1) n
Rewriting it to move from a decrease-and-conquer scheme to a strict divide-and-conquer scheme then just involves adapting the recursive case accordingly:
sumsquares :: Int -> Int -> Int
sumsquares m n
| m > n = error "First number cannot be bigger than second number"
| m == n = m * m
| otherwise = let middle = (m + n) `div` 2
in sumsquares m middle + sumsquares (middle + 1) n
The question remains, of course, why you would want to make this change. One reason could be that you are preparing your algorithm to be adapted for parallelisation: then, indeed, divide-and-conquer is often a better fit than decrease-and-conquer.
The Following code:
unSum :: Float -> Float
unSum x = (y + y`mod`2 + 2) / 2
where
y = x*(x+1) / 2
gives me this error when I try to load it into WinHugs 98:
Hugs> :load "D:\\kram\\unSumme2.hs"
ERROR file:.\unSumme2.hs:2 - Instance of Integral Float required for definition of unSumme2
What's the essence of this and how am I to do it? Anyway, I dont now if it serves my porpuse, I want to calculate the sum of the uneven numbers until x without recursivity.
It's because you're using mod. The definition of mod, from the Standard Prelude:
class (Real a, Enum a) => Integral a where
[...]
div, mod :: a -> a -> a
[...]
n `mod` d = r where (q,r) = divMod n d
In other words, it expects a to be of the numeric typeclass Integral, which includes only whole numbers.
If you want to stick with Floats, try using mod', from Data.Fixed, per this answer.
Alternatively, if you just want to do integer division, you could change your function's signature to
unSum :: Int -> Int
or similar.