I am creating a function to factorize any given number in haskell. And so i have created this:
primes :: [Integer]
primes = 2:(sieve [3,5..])
where
sieve (p:xs) = p : sieve [x |x <- xs, x `mod` ((p+1) `div` 2 ) > 0]
factorize 2 = [2]
factorize x
| divisible x = w:(factorize (x `div` w))
| otherwise = [x]
where
divisible :: Integer -> Bool
divisible y = [x |x <- (2:[3,5..y]), y `mod` x == 0] /= []
firstfactor :: Integer -> [Integer] -> Integer
firstfactor a (x:xs)
| a `ifdiv` x = x
| otherwise = firstfactor a xs
firstfactor a _ = a
ifdiv x y = mod x y == 0
w = firstfactor x primes
The function works fine, but appends 1 to the end of the list, for example factorize 80 would give this list: [2,2,2,2,5,1] My question is, why does this happen?
This comes up from two parts of the code. Firstly, factorize 1 is [1]. Secondly, since x is always divisible by itself, your very final call will always have w == x, so the recursion will be w:(factorize (w `div` w)) which is always w:(factorize 1).
To solve this, you can just add an extra base case to throw out the factors of 1:
factorize 1 = []
factorize ...
Also, you can drop the factorize 2 = [2] case because it gets subsumed by the otherwise case you already have.
factorize 1 = [] makes sense mathematically, because 1 has no prime factors (remember that 1 is not a prime number itself!). This follows the same logic behind product [] = 1—1 is the identity for multiplication, which makes it the "default" value when you have nothing to multiply.
Related
Create a divisors :: Int -> [Int] function that returns the divisors
of x.
I've successfully made a divides function that returns True if y is a divisor of x.
divides x y | mod x y == 0 = True
| mod x y /= 0 = False
I've tried to use it to filter numbers from [1..n] , but can't exactly get a grasp on how the filter function works. Can anyone set me in the right direction?
divisors n = filter divides n [1..n]
You are in the right track. Only that your problem is not exactly how filter works, but how Haskell works.
divisors n = filter (divides n) [1..n]
The above will do the trick. See, filter takes two arguments, so does divisors. But you are giving it three arguments at filter divides n [1..n].
Btw,
divides x y | mod x y == 0 = True
| mod x y /= 0 = False
is semantically equivalent to
divides x y = mod x y == 0
and operationally it doesn't repeat the mod calculation and the test mod x y /= 0.
filter takes as its second argument a list to filter - here your [1..n] - and as first argument a function which takes one of those list elements as input and returns a boolean.
So think about what this function would be, in terms of your pre-defined function divides. Since, as you have it, divides x y actually means "y is a divisor of x", you can easily express this with a lambda function:
divisors n = filter (\m -> divides n m) [1..n]
which can be further simplified by "eta-reduction" (or coloquially, "cancelling the m") to
divisors n = filter (divides n) [1..n]
This question already has answers here:
Finite comprehension of an infinite list
(3 answers)
Haskell - Why does this list comprehension return an infinite list?
(1 answer)
How to filter an infinite list in Haskell [duplicate]
(4 answers)
Closed 1 year ago.
I have a predefined function "primes" that returns an infinite list of all prime numbers. I wrote the following code for a function "prime_factors n" to return all the factors of n that are also prime.
prime_factors n = [x | x<-primes, n `mod` x == 0]
On executing prime_factors 12 it gives me the correct output [2,3 but keeps executing after that with no output. Why is this happening and how can I put a stop it?
For reference, here's the entire code segment:
primes :: [Integer]
primes = f [2..] where f (p:xs) = p: f [x | x <- xs, x mod p/=0] f [] = []
prime_factors n = [x | x<-primes, n mod x == 0, x*2 <= n]
primes :: Int -> [Integer]
primes n = take n $ f [2..]
where f (p:xs) = p: f [x | x <- xs, x `mod` p/=0]
prime_factors :: Int -> [Integer]
prime_factors n = [x | x<-primes n, n `mod` fromIntegral x == 0]
This is not a beautiful solution (am relatively new to Haskell myself), but it stops the program from running indefinitely by using the take function to ensure that you never get more entries than a finite number of entries up to n, which by definition will be far more than you need because there will always be fewer primes than natural numbers up to a given threshold, i.e., length [2,3] < length [1,2,3] etc.
Hopefully, someone else can give you a more elegant solution. But this one works.
Here is the code I am trying to use: This should generate all primes up to 100
sieve_primes = [x | x<-[2..100], y<-[2..50], z <-[2..25], (x*z) `mod` y /= 0]
The code
isPrime n = length [x | x<-[2..n], n `mod` x == 0] == 1
computes all the factors just to count them. You don't need to count them: as soon as the second factor is found you can stop your search without checking for further ones.
So, either replace length ... == 1 with a custom predicate, or take 2 elements from the list comprehension before checking its length.
What you had in mind was probably
Prelude> [x| x<-[2..100], not $ elem x [y*z| y<-[2..50], z<-[2..25]]]
[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97]
This is very slow. At least we can rearrange the pieces,
Prelude> [x| let sieve=[y*z| y<-[2..50], z<-[2..25]],
x<-[2..100], not $ elem x sieve]
[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97]
This is still very slow, for any number much above even 1000 (where you'd use 500 and 250). Then again, why the 25 (250) limit? Your code follows the
primes = [x| x<-[2..], not $ elem x
[y*z| y<-[2..x`div`2], z<-[2..min y (x`div`y)]]]
idea, i.e. y*z = 2*y .. min (y*y) x, so with the known top limit (x <= n) it should be
primesTo n = [x| let sieve=[y*z| y<-[2..n`div`2], z<-[2..min y (n`div`y)]],
x<-[2..n], not $ elem x sieve]
(incidentally, max (min y (n/y)) {y=2..n/2} = sqrt n, so we could've used 10 instead of 25, (and 31 instead of 250, for the 1000)).
Now 1000 is not a problem, only above ~ 10,000 we again begin to see that it's slow (still), running at n2.05..2.10 empirical orders of growth (quick testing interpreted code in GHCi at n = 5000, 10000, 15000).
As for your second (now deleted) function, it can be rewritten, step by step improving its speed, as
isPrime n = length [x | x<-[2..n], n `mod` x == 0] == 1
= take 1 [x | x<-[2..n], n `mod` x == 0] == [n]
= [x | x<- takeWhile ((<=n).(^2)) [2..n], n `mod` x == 0] == []
= and [n `mod` x > 0 | x<- takeWhile ((<=n).(^2)) (2:[3,5..n])]
now, compiled, it can get first 10,000 primes in few tenths of a second.
So I'm totally new to Haskell, hope it doesn't show too much. By any means, I was trying to create a function to determine if a number is prime. The basic idea is this: Given a number, see if it is divisible by any other number less than it. If so, return false. If not, it's prime, return true. The code so far (that is known to be working) is this:
divisible :: Integer -> Integer -> Bool
divisible x 2 = even x
divisible x y = ((x `mod` y) /= 0) && not (divisible x (y-1))
isPrime :: Integer -> Bool
isPrime x = not (even x) && not (divisible x (x-1))
Produces:
ghci> isPrime 9
False
ghci> isPrime 13
True
What I'd like to do is optimize this a bit, since I only need to check values less than or equal to sqrt(x). The problem is, when I try and implement this, stuff gets crazy:
isPrime x = not (even x) && not (divisible x (ceiling(sqrt(fromIntegral(x-1)))))
Besides the fact that it looks terrible (I said I was new), it doesn't give the correct result:
ghci> isPrime 9
False
ghci> isPrime 13
False
I'm trying to figure out what changed, because:
ghci> ceiling(sqrt(13))
4
Seems to be giving me the correct number. I know this is a small problem, but I'm seriously confused...
You got your conditions mixed up:
divisible x y = ((x `mod` y) /= 0) && not (divisible x (y-1))
should be
divisible x y = (x `mod` y) == 0 || divisible x (y-1)
for the test to work.
As is, your divisible function expands e.g.
divisible 21 5 = (21 `mod` 5 /= 0) && not (divisible 21 4)
= (21 `mod` 5 /= 0) && not ((21 `mod` 4 /= 0) && not (divisible 21 3))
= not ((21 `mod` 4 /= 0) && not ((21 `mod` 3 /= 0) && not (divisible 21 2)))
= not (True && not (False && not (divisible 21 3)))
= not (not False)
= False
since 21 `mod` 3 == 0, and isPrime 21 evaluates to True with the square root bound.
However, I get
*StrangePrime> isPrime 9
True
*StrangePrime> isPrime 13
True
with your code using the square root.
Without the square root, it happened to work for odd numbers, because the difference between an odd composite and any of its divisors is always even. Unfolding divisible a few steps for n = p*m, where p is the smallest prime factor of the odd composite n, we see
divisible n (n-1) = n `mod` (n-1) /= 0 && not (divisible n (n-2))
= not (divisible n (n-2))
= not (n `mod` (n-2) /= 0 && not (divisible n (n-3)))
= not (not (divisible n (n-3)))
= not^2 (divisible n (n-3))
and inductively
divisible n (n-1) = not^(k-1) (divisible n (n-k))
if there are no divisors of n larger than n-k. Now, in the above situation, the largest divisor of n is m = n - (p-1)*m, so we obtain
divisible n (n-1) = not^((p-1)*m-1) (divisible n m)
= not^((p-1)*m-1) (n `mod` m /= 0 && not (...))
But n `mod` m == 0, so we have
divisible n (n-1) = not^((p-1)*m-1) False
Since p is odd, p-1 is even, and hence so is (p-1)*m, so altogether we have an odd number of nots, which is the same as one not, giving
divisible n (n-1) = True
isPrime n = not (even n) && not (divisible n (n-1)) = True && not True = False
If p is an odd prime, the unfolding reaches divisible p (p-1) = not^(p-3) (divisible p (p - (p-2))). p-3 is even, divisible p 2 is even p, which is False.
Generally, consider divisible n s for an odd n, and let d be the largest divisor of n not exceeding s, if n is composite, or d = 2 if n is prime. The unfolding of divisible n s still proceeds the same way
divisible n s = not^k (divisible n (s-k))
while no divisor has been found and s-k > 2. So in the case of a composite n, we find
divisible n s = not^(s-d) (divisible n d)
= not^(s-d) (n `mod` d /= 0 && not (...))
= not^(s-d) False
= odd (s-d)
= even s -- since d is odd, as a divisor of an odd number
and in the case of an odd prime n,
divisible n s = not^(s-2) (divisible n 2)
= not^(s-2) (even n)
= not^(s-2) False
= odd s
So divisible n s measures the parity of the distance of s to the next smaller divisor of n or to 2, whichever is larger. When s was n-1, the starting point was always even, so it worked out correctly, but ceiling (sqrt (fromIntegral (n-1))) can be odd, in which case the results are flipped and composites are declared prime and vice versa.
You can make your divisible function work for the primality test of odd numbers with a square root bound if you make sure that the first call gets an even second argument (so if ceiling (sqrt (fromIntegral (n-1))) is odd, start at ceiling (sqrt (fromIntegral (n-1))) + 1), but the logic of that function is confusing, and its name doesn't correctly describe its results.
A more idiomatic way to write it would be
isPrime n = and [n `mod` k /= 0 | k <- [2 .. ceiling (sqrt $ fromIntegral n)]]
The test becomes more efficient when one skips candidate divisors that are already known to be nondivisors from prior tests, easy is skipping all even numbers except 2,
isPrime 2 = True
isPrime n = all ((/= 0) . (n `mod`)) (2 : [3, 5 .. ceiling (sqrt (fromIntegral n))])
slightly more involved, but still more efficient is also skipping multiples of 3
isPrime n = all ((/= 0) . (n `mod`)) (takeWhile (<= bound) (2:3:scanl (+) 5 (cycle [2,4])))
where
bound = ceiling (sqrt (fromIntegral (n-1)))
In the same vein one can eliminate the multiples of more small primes from the trial divisors, each gaining a bit of efficiency, but at the cost of a more complicated wheel, e.g. also eliminating multiples of 5 leads to
isPrime n = all ((/= 0) . (n `mod`)) (takeWhile (<= bound) (2:3:5: scanl (+) 7 (cycle [4,2,4,2,4,6,2,6])))
where
bound = ceiling (sqrt (fromIntegral (n-1)))
Here's how I'd do it:
divisibleBy :: (Integral a) => a -> a -> Bool
divisibleBy x y = mod x y == 0
isPrime :: (Integral a) => a -> Bool
isPrime x = or $ map (divisibleBy x) [2..(x-1)]
divisibleBy is a simple test of divisibility. isPrime performs this test on all integers between 1 and x, returning true if x is divisible by any of those integers. You might change the upper bound to root x, as you've done in your code, but otherwise this works.
I'm new to Haskell, and I'm trying a bit:
isPrime :: Integer->Bool
isPrime x = ([] == [y | y<-[2..floor (sqrt x)], mod x y == 0])
I have a few questions.
Why when I try to load the .hs, WinHugs say: Instances of (Floating Integer, RealFrac Integer) required for definition of isPrime?
When the interpreter finds one element in the right set, it immediately stops or it computes all the set? I think you know what I mean.
Sorry about my english.
1) The problem is that sqrt has the type (Floating a) => a -> a, but you try to use an Integer as argument. So you have to convert your Integer first to a Floating, e.g. by writing sqrt (fromIntegral x)
2) I see no reason why == shouldn't be lazy, but for testing for an empty collection you can use the null function (which is definitely lazy, as it works on infinite lists):
isPrime :: Integer->Bool
isPrime x = null [y | y<-[2..floor (sqrt (fromIntegral x))], x `mod` y == 0]
But in order to get an more idiomatic solution, break the problem into smaller sub-problems. First, we need a list of all elements y with y*y <= x:
takeWhile (\y -> y*y <= x) [2..]
Then we need only the elements that divide x:
filter (\y -> x `mod`y == 0) (takeWhile (\y -> y*y <= x) [2..])
Then we need to check if that list is empty:
isPrime x = null (filter (\y -> x `mod`y == 0) (takeWhile (\y -> y*y <= x) [2..]))
And if this looks to lispy to you, replace some of the parens with $
isPrime x = null $ filter (\y -> x `mod` y == 0) $ takeWhile (\y -> y*y <= x) [2..]
For additional clarity you can "outsource" the lambdas:
isPrime x = null $ filter divisible $ takeWhile notTooBig [2..] where
divisible y = x `mod`y == 0
notTooBig y = y*y <= x
You can make it almost "human readable" by replacing null $ filter with not $ any:
isPrime x = not $ any divisible $ takeWhile notTooBig [2..] where
divisible y = x `mod`y == 0
notTooBig y = y*y <= x
Because sqrt has the type Floating a => a -> a. This means the input has to be a Floating type and the output will be the same type. In other words x needs to be a Floating type. However you declared x to be of type Integer, which is not a Floating type. In addition floor needs a RealFrac type, so x needs to be that as well.
The error message suggests that you fix that by making Integer a Floating type (by defining an instance Floating Integer (and the same for RealFrac).
Of course this is not the correct approach in this case. Rather you should use fromIntegral to convert x to a Real (which is an instance of Floating and RealFrac) and then give that to sqrt.
Yes. As soon as == sees that the right operand has at least one element, it knows it is not equal to [] and thus returns False.
That being said, null is a more idiomatic way to check whether a list is empty than [] ==.
Regarding the second point, it stops, for example:
[] == [x | x <- [1..]]
Returns False
Landei's solution is great, however, if you want a more efficient¹ implementation we have (thanks to BMeph):
-- list of all primes
primes :: [Integer]
primes = sieve (2 : 3 : possible [1..]) where
sieve (p : xs) = p : sieve [x | x <- xs, x `mod` p > 0]
possible (x:xs) = 6*x-1 : 6*x+1 : possible xs
isPrime :: Integer -> Bool
isPrime n = shortCircuit || (not $ any divisible $ takeWhile inRangeOf primes) where
shortCircuit = elem n [2,3] || (n < 25 && ((n-1) `mod` 6 == 0 || (n+1) `mod` 6 == 0))
divisible y = n `mod` y == 0
inRangeOf y = y * y <= n
The 'efficiency' comes from the use of constant primes. It improves the search in two ways:
The Haskell runtime could cache the results so subsequent invocations are not evaluated
It eliminates a range of numbers by logic
note that the sieve value is simply a recursive table, where says the head of
the list is prime, and adds it to it. For the rest of the lists if there is no
other value already in the list that composes the number then its also prime
possible is list of all possible primes, since all possible primes are in the
form 6*k-1 or 6*k-1 except 2 and 3
The same rule is applied for shortCircuit too to quickly bail out of calculations
Footnote by D.F.
¹ It's still a terribly inefficient way to find primes. Don't use trial division if you need primes larger than a few thousand, use a sieve instead. There are several far more efficient implementations on hackage.
I think WinHugs needs to import a module for Integer and etc... Try Int
The interpreter will not compute anything until you call e.g. isPrime 32 then it will lazily compute the expression.
PS your isPrime implementation is not the best implementation!