Haskell Function that calculates e^x - haskell

Implement a function that calculates the value of e ^ x, x is a parameter of the function, an integer.
To do this, use the Taylor series expansion to calculate the potency of e.
The function will receives as a parameter, in addition to the exponent x, the number of terms of the series, which will operate as a maximum value of n.
For the resolution of this function, recursion must be used.
I made this:
factorial 0 = 1
factorial n = n * factorial (n-1)
consigna3::Int->Int->Float
consigna3 _ 0 = 1
consigna3 x n = (fromIntegral(x^n) / fromIntegral(factorial n)) + consigna3 x (n-1)
But some results are wrong, this is what I expected:
Ejemplo 1: Main> funcion3 1 1
2.0
Ejemplo 2: Main> funcion3 1 10
2.718282
Ejemplo 3: Main> funcion3 2 10
7.388997
Ejemplo 4: Main> funcion3 10 20
21991.48
Ejemplo 5: Main> funcion3 10 30
22026.46
Ejemplo 6: Main> funcion3 0 30
1.0
The results (10 20) and (10 30) do not match what the function I did returns.
What I am doing wrong? Thanks and sorry for my English.

You are using Int for calculations that will overflow an Int. Instead, convert to Float right away, and then using Float for everything. So:
consigna3 x n = ((fromIntegral x)^n / factorial (fromIntegral n)) + consigna3 x (n-1)
There are two critical changes from Int to Float here: first, you do x^n where x :: Int, but I do fromIntegral x^n where fromIntegral x :: Float; second, you do factorial n where n :: Int, but I do factorial (fromIntegral n) where fromIntegral n :: Float.

Related

Summing a finite prefix of an infinite series

The number π can be calculated with the following infinite series sum:
I want to define a Haskell function roughlyPI that, given a natural number k, calculates the series sum from 0 to the k value.
Example: roughlyPi 1000 (or whatever) => 3.1415926535897922
What I did was this (in VS Code):
roughlyPI :: Double -> Double
roughlyPI 0 = 2
roughlyPI n = e1/e2 + (roughlyPI (n-1))
where
e1 = 2**(n+1)*(factorial n)**2
e2 = factorial (2*n +1)
factorial 0 = 1
factorial n = n * factorial (n-1)
but it doesn't really work....
*Main> roughlyPI 100
NaN
I don't know what's wrong. I'm new to Haskell, by the way.
All I really want is to be able to type in a number that will give me PI at the end. It can't be that hard...
As mentioned in the comments, we need to avoid large divisions and instead intersperse smaller divisions within the factorials. We use Double for representing PI but even Double has its limits. For instance 1 / 0 == Infinity and (1 / 0) / (1 / 0) == Infinity / Infinity == NaN.
Luckily, we can use algebra to simplify the formula and hopefully delay the blowup of our Doubles. By dividing within our factorial the numbers don't grow too unwieldy too quickly.
This solution will calculate roughlyPI 1000, but it fails on 1023 with NaN because 2 ^ 1024 :: Double == Infinity. Note how each iteration of fac has a division as well as a multiplication to help keep the numbers from blowing up. If you are trying to approximate PI with a computer, I believe there are better algorithms, but I tried to keep it as conceptually close to your attempt as possible.
roughlyPI :: Integer -> Double
roughlyPI 0 = 2
roughlyPI k = e + roughlyPI (k - 1)
where
k' = fromIntegral k
e = 2 ** (k' + 1) * fac k / (2 * k' + 1)
where
fac 1 = 1 / (k' + 1)
fac p = (fromIntegral p / (k' + fromIntegral p)) * fac (p - 1)
We can do better than having a blowup of Double after 1000 by doing computations with Rationals then converting to Double with realToFrac (credit to #leftaroundabout):
roughlyPI' :: Integer -> Double
roughlyPI' = realToFrac . go
where
go 0 = 2
go k = e + go (k - 1)
where
e = 2 ^ (k + 1) * fac k / (2 * fromIntegral k + 1)
where
fac 1 = 1 % (k + 1)
fac p = (p % (k + p)) * fac (p - 1)
For further reference see Wikipedia page on approximations of PI
P.S. Sorry for the bulky equations, stackoverflow does not support LaTex
First note that your code actually works:
*Main> roughlyPI 91
3.1415926535897922
The problem, as was already said, is that when you try to make the approximation better, the factorial terms become too big to be representable in double-precision floats. The simplest – albeit somewhat brute-force – way to fix that is to do all the computation in rational arithmetic instead. Because numerical operations in Haskell are polymorphic, this works with almost the same code as you have, only the ** operator can't be used since that allows fractional exponents (which are in general irrational). Instead, you should use integer exponents, which is anyway the conceptually right thing. That requires a few fromIntegral:
roughlyPI :: Integer -> Rational
roughlyPI 0 = 2
roughlyPI n = e1/e2 + (roughlyPI (n-1))
where
e1 = 2^(n+1)*fromIntegral (factorial n^2)
e2 = fromIntegral . factorial $ 2*n + 1
factorial 0 = 1
factorial n = n * factorial (n-1)
This now works also for much higher degrees of approximation, although it takes a long time to carry around the giant fractions involved:
*Main> realToFrac $ roughlyPI 1000
3.141592653589793
The way to go in such cases is to calculate the ratio of consecutive terms and calculate the terms by rolling multiplications of the ratios:
-- 1. -------------
pi1 n = Sum { k = 0 .. n } T(k)
where
T(k) = 2^(k+1)(k!)^2 / (2k+1)!
-- 2. -------------
ts2 = [ 2^(k+1)*(k!)^2 / (2k+1)! | k <- [0..] ]
pis2 = scanl1 (+) ts2
pi2 n = pis2 !! n
-- 3. -------------
T(k) = 2^(k+1)(k!)^2 / (2k+1)!
T(k+1) = 2^(k+2)((k+1)!)^2 / (2(k+1)+1)!
= T(k) 2 (k+1)^2 / (2k+2) (2k+3)
= T(k) (k+1)^2 / ( k+1) (2k+3)
= T(k) (k+1) / (k+1 + k+2)
= T(k) / (1 + (k+2)/(k+1))
= T(k) / (2 + 1 /(k+1))
-- 4. -------------
ts4 = scanl (/) 2 [ 2 + 1/(k+1) | k <- [0..]] :: [Double]
pis4 = scanl1 (+) ts4
pi4 n = pis4 !! n
This way we share and reuse the calculations as much as possible. This leads to the most efficient code, hopefully leading to the smallest cumulative numerical error. The formula also turned out to be exceptionally simple, and could even be simplified further as ts5 = scanl (/) 2 [ 2 + recip k | k <- [1..]].
Trying it out:
> pis2 = scanl1 (+) $ [ fromIntegral (2^(k+1))*fromIntegral (product[1..k])^2 /
fromIntegral (product[1..(2*k+1)]) | k <- [0..] ] :: [Double]
> take 8 $ drop 30 pis2
[3.1415926533011587,3.141592653447635,3.141592653519746,3.1415926535552634,
3.141592653572765,3.1415926535813923,3.141592653585647,3.141592653587746]
> take 8 $ drop 90 pis2
[3.1415926535897922,3.1415926535897922,NaN,NaN,NaN,NaN,NaN,NaN]
> take 8 $ drop 30 pis4
[3.1415926533011587,3.141592653447635,3.141592653519746,3.1415926535552634,
3.141592653572765,3.1415926535813923,3.141592653585647,3.141592653587746]
> take 8 $ drop 90 pis4
[3.1415926535897922,3.1415926535897922,3.1415926535897922,3.1415926535897922,
3.1415926535897922,3.1415926535897922,3.1415926535897922,3.1415926535897922]
> pis4 !! 1000
3.1415926535897922

cross sum operation in Haskell

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.

Number of 1 bits in a decimal number

How can I write a function to return the number of 1 bits corresponding a decimal number? Maybe a square root function? The type should be this:
bits :: Int -> Int
EDIT: SOlved
uns :: Int -> Int
uns 0 = 0
uns 1 = 1
uns x | mod x 2 == 1 = 1 + uns (div x 2)
| otherwise = uns (div x 2)
What about this old trick?
import Data.Bits
countOnes 0 = 0
countOnes x = 1 + countOnes (x .&. (x-1))
It only recurses n times where n is the number of one bits in x.
Of course, if you are going to import Data.Bits, then you might as well use popCount as #Cirdec suggested.
Use popCount from Data.Bits, also known as the Hamming weight of a number. The advantage is that some CPUs have instructions specifically for calculating this, resulting in high performance.
λ> import Data.Bits
λ> :t popCount
popCount :: Bits a => a -> Int
λ> popCount 255
8
λ> popCount 0xa5
4
sum . map (`mod` 2) . takeWhile ( /= 0) . iterate (`quot` 2)
I guess Luka has the better version but maybe you struggle a bit with understanding it, so here is what I hinted at (which is the same Luka did only with naive recursion):
bits :: Int -> Int
bits 0 = 0
bits n = let (d,r) = n `divMod` 2
in r + bits d
The idea is to get continous divide by 2 and look at the remainder - if it's 1 then you found a set bit, if not (it has to be 0 then) then there was none.

Haskell reverse Integer with recursion

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)

Mod Haskell Homework

My homework was to provide a function that computes 'x^y mod n' -for any n < (sqrt maxint32)
So I started by writing doing this:
modPow :: Int -> Int -> Int -> Int
modPow x y n = (x `mod` n) ^ (y `mod` n) `mod` n
Which seemed to work fine, for any number of n, although my next homework question involved using x^n mod n = x (Camichael numbers) and I could never get modPow to work.
So I made another modPow using pseudocode for mod exponentiation, -from wikipedia:
modPow2 :: Int -> Int -> Int -> Int
modPow2 x y n
= loopmod 1 1
where
loopmod count total = if count > y
then total
else loopmod (count+1) ((total*x) `mod` n)
Which now correctly produces the right answer for my next question, (x^n mod n = x) -for checking for Camichael numbers.
ALTHOUGH, modPow2 does not work for big numbers of 'y' (STACK-OVERFLOW!!)
How could I adjust modPow2 so it no longer gets a stackoverflow in the cases where y > 10,000 (but still less than sqrt of maxint 32 -which is around 46,000)
Or is there a fix on my original modPow so it works with x^n mod n = x? (I always do 560 561 561 as inputs and it gives me back 1 not 560 (561 is a carmichael number so should give 560 back)
Thanks alot.
Your formula for modPow is wrong, you can't just use y mod n as the exponent, it will lead to wrong results. For example:
Prelude> 2^10
1024
Prelude> 2^10 `mod` 10
4
Prelude> 2^(10 `mod` 10) `mod` 10
1
For a better modPow function you could use that x2n+1 = x2n ⋅ x and x2n = xn ⋅ xn and that for multiplication you actually can simply use the mod of the factors.
Where did you get your formula for modPow from?
(x ^ y) `mod` n = ((x `mod` n) ^ (y `mod` φ n)) `mod` n where φ is Euler's totient function.
This is probably because the argument total is computed lazily.
If you use GHC, you can make loopmod strict in total by placing a ! in frontof the argument, i.e.
loopmod count !total = ...
Another way would be to force evaluation of total like so: Replace the last line with
else if total == 0 then 0 else loopmod (count+1) ((total*x) `mod` n)
This does not change semantics (because 0*xis 0 anyway, so the reminder must be 0 also) and it forces hugs to evaluate total in every recursion.
If you are looking for implementation ( a^d mod n ) then
powM::Integer->Integer->Integer->Integer
powM a d n
| d == 0 = 1
| d == 1 = mod a n
| otherwise = mod q n where
p = powM ( mod ( a^2 ) n ) ( shiftR d 1 ) n
q = if (.&.) d 1 == 1 then mod ( a * p ) n else p

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