Please help, I've been trying to get this code to work but I can't find the errors. Below is my code
sumToN f x 1 = f (x 1)
sumToN f x n = f x n + f x (n-1)
facOfN 0 = 1
facOfN n = n * facOfN (n-1) sgfr
sineApprox x n = ((-1) ^ n) * ((x ** (2*n+1))/facOfN(2*n+1)
sine x n = sumToN (sineApprox x n)
When I try to load the file I get the following error.
ERROR file:F:\sine.hs:8 - Syntax error in expression (unexpected `;', possibly due to bad layout)
Any assistance would be greatly appreciated.
As already said in the comments, you've forgotten to close a paren. It'll work like that:
sineApprox x n = ((-1) ^ n) * ((x ** (2*n+1))/facOfN(2*n+1))
Note that this problem would have been obvious with a better text editor. Being a beginner, I suggest you switch to iHaskell, which has a very simple interface and yet reasonably powerful editor features.
The problem would also have been obvious if you hadn't used so many unnecessary parens. The following can be omitted just like that, some can be replaced with $. While we're at style...
sumToN f x n -- checking ==1 is not safe in general
| n<=1 = f $ x 1
| otherwise = f x n + f x (n-1)
facOfN = product [1..n]
sineApprox x n = (-1)^n * x**(2*n+1) / facOfN (2*n+1)
sine x = sumToN . sineApprox x
On another note: in general, you should always use type signatures. This code actually has problems because all the counter variables are automaticall floating point (like everything else). They should really be Ints, which requires a conversions in the factorial†:
sumToN :: Num n => (Int -> n) -> Int -> n
sumToN f x n
| n<1 = 0
| otherwise = f x n + f x (n-1)
facOfN :: Num n => Int -> n
facOfN = product [1 .. fromIntegral n]
sineApprox :: Fractional n => n -> Int -> n
sineApprox x n = (-1)^n * x^(2*n+1) / facOfN (2*n+1)
sine
sine x = sumToN . sineApprox x
†BTW, explicitly using factorials is almost always a bad idea, as the numbers quickly get intractibly huge. Also, you're doing a lot of duplicate work. Better multiply as you add along!
Related
I am trying to solve the problems from Project Euler using Haskell, but I got sucked at #24
I'm trying to use factorials to solve problem but just can't work for the last three digits, here is my code:
import Data.List
fact n = product [n, n-1 .. 1]
recur :: Int -> Int -> [Int] -> [Int]
recur x y arr
| y > 1 = arr !! d : recur r (y-1) (delete (arr !! d) arr)
| otherwise = arr
where d = x `div` fact y
r = x `mod` fact y
main::IO()
main = print(recur 1000000 9 [0..9])
(I know it is now not really "functional")
I managed to get result [2,7,8,3,9,1,4,5,0,6], while the right answer I accidently figured out by hand is 2783915460.
I just want to know why this algorithm doesn't work for the last three digits. Thanks.
Unadulterated divMod is wrong for this algorithm. You need
dvm x facty | r == 0 = (d-1, facty)
| otherwise = (d, r)
where
(d, r) = divMod x facty
instead:
recur x y arr
.......
.......
where (d, r) = x `dvm` fact y
We cannot have zero combinations to do left. Zero means none.
Also the pattern guard condition should be changed to y > 0. Only when the length of the remaining choices list is 1 (at which point y is 0) there's no more choices to be made and we just use the last available digit left.
This code either returns the first factor of an Integer starting from 2 or returns nothing if it's a prime.
Example: firstFactorOf 24 returns "Just 2"
Example: firstFactorOf 11 returns "Nothing"
My question is, how would I return the value 2 rather than "Just 2" if there is a factor or return the value x if there is no factor.
firstFactorOf x
| m == Nothing = m
| otherwise = m
where m =(find p [2..x-1])
p y = mod x y == 0
//RETURNS:
ghci> firstFactorOf 24
Just 2
ghci> firstFactorOf 11
Nothing
Haskell is statically typed, meaning that you can define a function Maybe a -> a, but the question is what to do with the Nothing case.
Haskell has two functions that can be helpful here: fromMaybe and fromJust:
fromMaybe :: a -> Maybe a -> a
fromJust :: Maybe a -> a
fromJust simply assumes that you will always provide it a Just x, and return x, in the other case, it will throw an exception.
fromMaybe on the other hand expects two parameters, the first - an a is the "default case" the value that should be returned in case of Nothing. Next it is given a Maybe a and in case it is a Just x, x is returned. In the other case (Nothing) as said before the default is returned.
In your comment you say x should be returned in case no such factor exists. So I propose you define a new function:
firstFactorOfJust :: Integral a => a -> a
firstFactorOfJust x = fromMaybe x $ firstFactorOf x
So this function firstFactorOfJust calls your firstFactorOf function and if the result is Nothing, x will be returned. In the other case, the outcome of firstFactorOf will be returned (but only the Integral part, not the Just ... part).
EDIT (simplified)
Based on your own answer that had the intend to simplify things a bit, I had the idea that you can simplify it a bit more:
firstFactorOf x | Just z <- find ((0 ==) . mod x) [2..x-1] = z
| otherwise = x
and since we are all fan of optimization, you can already stop after sqrt(x) iterations (a well known optimization in prime checking):
isqrt :: Int -> Int
isqrt = floor . sqrt . fromIntegral
firstFactorOf x | Just z <- find ((0 ==) . mod x) [2..isqrt x] = z
| otherwise = x
Simplified question
For some reason there was some peculiarly complicated aspect in your question:
firstFactorOf x
| m == Nothing = m
| otherwise = m
where m =(find p [2..x-1])
p y = mod x y == 0
Why do you use guards to make a distinction between two cases that generate the exact same output? You can fold this into:
firstFactorOf x = m
where m = (find p [2..x-1])
p y = mod x y == 0
and even further:
firstFactorOf x = find p [2..x-1]
where p y = mod x y == 0
If you want it to return the first factor of x, or x, then this should work:
firstFactorOf x =
let
p y = mod x y == 0
m = (find p [2..x-1])
in
fromMaybe x m
import Data.List
import Data.Maybe
firstFactorOf x
| m == Nothing = x
| otherwise = fromJust m
where m =(find p [2..x-1])
p y = mod x y == 0
This was what I was after. Not sure why you guys made this so complicated.
I was reading the Haskell Prelude and finding it pretty understandable, then I stumbled upon the exponention definition:
(^) :: (Num a, Integral b) => a -> b -> a
x ^ 0 = 1
x ^ n | n > 0 = f x (n-1) x
where f _ 0 y = y
f x n y = g x n where
g x n | even n = g (x*x) (n `quot` 2)
| otherwise = f x (n-1) (x*y)
_ ^ _ = error "Prelude.^: negative exponent"
I do not understand the need for two nested wheres.
What I understood so far:
(^) :: (Num a, Integral b) => a -> b -> a
The base must be a number and the exponent intege, ok.
x ^ 0 = 1
Base case, easy.
g x n | even n = g (x*x) (n `quot` 2)
| otherwise = f x (n-1) (x*y)
Exponention by squaring... kind of ... Why is the f helper needed? Why are f and g given single letter names? Is it just optimization, am I missing something obvious?
_ ^ _ = error "Prelude.^: negative exponent"
N > 0 was checked before, N is negative if we arrived here, so error.
My implementation would be a direct translation to code of:
Function exp-by-squaring(x, n )
if n < 0 then return exp-by-squaring(1 / x, - n );
else if n = 0 then return 1; else if n = 1 then return x ;
else if n is even then return exp-by-squaring(x * x, n / 2);
else if n is odd then return x * exp-by-squaring(x * x, (n - 1) / 2).
Pseudocode from wikipedia.
To illustrate what #dfeuer is saying, note that the way f is written it either:
f returns a value
or, f calls itself with new arguments
Hence f is tail recursive and therefore can easily be transformed into a loop.
On the other hand, consider this alternate implementation of exponentiation by squaring:
-- assume n >= 0
exp x 0 = 1
exp x n | even n = exp (x*x) (n `quot` 2)
| otherwise = x * exp x (n-1)
The problem here is that in the otherwise clause the last operation performed is a multiplication. So exp either:
returns 1
calls itself with new arguments
calls itself with some new arguments and multiplies the result by x.
exp is not tail recursive and therefore cannot by transformed into a loop.
f is indeed an optimization. The naive approach would be "top down", calculating x^(n `div` 2) and then squaring the result. The downside of this approach is that it builds a stack of intermediate computations. What f lets this implementation do is to first square x (a single multiplication) and then raise the result to the reduced exponent, tail recursively. The end result is that the function will likely operate entirely in machine registers. g seems to help avoid checking for the end of the loop when the exponent is even, but I'm not really sure if it's a good idea.
As far as I understand it exponentiation is solved by squaring as long as the exponent is even.
This leads to the answer why f is needed in case of an odd number - we use f to return the result in the case of g x 1, in every other odd case we use f to get back in the g-routine.
You can see it best I think if you look at an example:
x ^ n | n > 0 = f x (n-1) x
where f _ 0 y = y
f x n y = g x n
where g x n | even n = g (x*x) (n `quot` 2)
| otherwise = f x (n-1) (x*y)
2^6 = -- x = 2, n = 6, 6 > 0 thus we can use the definition
f 2 (6-1) 2 = f 2 5 2 -- (*)
= g 2 5 -- 5 is odd we are in the "otherwise" branch
= f 2 4 (2*2) -- note that the second '2' is still in scope from (*)
= f 2 4 (4) -- (**) for reasons of better readability evaluate the expressions, be aware that haskell is lazy and wouldn't do that
= g 2 4
= g (2*2) (4 `quot` 2) = g 4 2
= g (4*4) (2 `quot` 2) = g 16 1
= f 16 0 (16*4) -- note that the 4 comes from the line marked with (**)
= f 16 0 64 -- which is the base case for f
= 64
Now to your question of using single letter function names - that's the kind of thing you have to get used to it is a way most people in the community write. It has no effect on the compiler how you name your functions - as long as they start with a lower case letter.
As others noted, the function is written using tail-recursion for efficiency.
However, note that one could remove the innermost where while preserving tail-recursion as follows: instead of
x ^ n | n > 0 = f x (n-1) x
where f _ 0 y = y
f x n y = g x n
where g x n | even n = g (x*x) (n `quot` 2)
| otherwise = f x (n-1) (x*y)
we can use
x ^ n | n > 0 = f x (n-1) x
where f _ 0 y = y
f x n y | even n = f (x*x) (n `quot` 2) y
| otherwise = f x (n-1) (x*y)
which is also arguably more readable.
I have however no idea why the authors of the Prelude chose their variant.
I'm writing a bit of code to help me with some math stuff. I'm trying to implement the Miller test, not Miller-Rabin, and I need to make a list of a bunch of exponents. Here's the code so far. It inserts the last result twice for some reason, and I don't know why. I must not understand how the until function works.
import Math.NumberTheory.Powers
divides::Integer->Integer->Bool
divides x y = y `mod` x == 0
factorcarmichael::Integer->(Integer,Integer)
factorcarmichael n = until (\(_, s) -> not $ divides 2 s)
(\(r, s) -> (r+1, div s 2))
(0, n-1)
second::((Integer,Integer),[Integer])->[Integer]
second (x,xs) = xs
millerlist::Integer->Integer->[Integer]
millerlist a n = second $ until (\((r,s), xs) -> r<0)
(\((r,s), xs) -> ((r-1,s), (powerMod a ((2^r)*s) n):xs))
(factoredcarmichael, [])
where
factoredcarmichael = factorcarmichael n
Also, the millerlist function is a little kludgy. If someone can suggest an alternative, that would be nice.
The output I'm getting for
millerlist 8888 9746347772161
repeats the last element twice.
That is because
7974284540860^2 ≡ 7974284540860 (mod 9746347772161)
so the number appears twice in the list. But your list is one too long, I believe. I think you only want the remainder of a^(2^k*s) modulo n for 0 <= k < r.
As for alternatives, is there a particular reason why you're not using Math.NumberTheory.Primes.isStrongFermatPP? If you're only interested in the outcome, that's less work coding.
If you want to generate the list, what about
millerlist a n = go r u
where
(r,s) = factorcarmichael n
u = powerMod a s n
go 0 m = []
go k m = m : go (k-1) ((m*m) `mod` n)
or
millerlist a n = take (fromInteger r) $ iterate (\m -> (m*m) `mod` n) u
where
(r,s) = factorcarmichael n
u = powerMod a s n
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