I'm trying to solve problem #26 from Project Euler in Haskell, but I have encountered some problems.
I've managed to figure out that the recurring cycle of the reciprocal is only related to its prime divisors, so I think I just need to find out a reciprocal of a prime number with the longest recurring cycle. So I wrote a algorithm in Haskell:
isPrime :: Int -> Bool
isPrime k
| k <= 1 = error "Seriously?"
| otherwise = null [ x | x <- [2..floor(sqrt(fromIntegral k))], k `mod` x == 0]
lp = [x | x <- [7..1000], isPrime x]
s = map (\n -> head [x | x <- [ceiling(logBase 10 (fromIntegral n))..], 10^x `mod` n == 1]) lp
main::IO()
main = print $ maximum s
However, it cannot produce an answer. I have tried using the lamda, which could produce the digits of recurring cycle, with a few prime numbers and I manage to get the right count of digits (I hope there would be no problem with the algorithm). I have also checked the output of the list s, which produced [6,2,6,16,18,45,23,15,3,5,63, without an end. I have no idea why it is the case, since if I manually apply the function to each prime number I can get the right output.
Can anybody tell me what is wrong with my code, or is my way of solving it turns out to be wrong? Thanks.
Int is not a good choice here, since you operate pretty big numbers, with the 10^x. Int is Bounded, so wraps around past its upper bound:
> maxBound :: Int
9223372036854775807
> (maxBound :: Int) + 1
-9223372036854775808
Omitting the signature for isPrime altogether, we get
> :t lp
lp :: Integral b => [b]
Trying
> map (\n -> (n, head [x | x <- [ceiling(logBase 10 (fromIntegral n))..],
10^x `mod` n == 1]))
(lp :: [Int])
[(7,6),(11,2),(13,6),(17,16),(19,18),(23,45),(29,23),(31,15),(37,3),(41,5),(43,63),
(47,Interrupted.
we see that your calculation gets stuck on 47. But using [Integer] (or nothing at all, so it defaults to Integer by itself), we successfully get the full result. You just misinterpret it. Re-read the problem statement, and you'll see.
(also, the answer for 43 in the above snippet is incorrect, while those for 7, 11, 13 are correct. Getting wrong results for bigger numbers is a strong signal that we have some integer wrap-around arithmetic errors going on; and that's how I found it out).
Related
--for number divisible by 15 we can get it easily
take 10 [x | x <- [1..] , x `mod` 15 == 0 ]
--but for all how do I use the all option
take 10 [x | x <- [1..] , x `mod` [2..15] == 0 ]
take 10 [x | x <- [1..] , all x `mod` [2..15] == 0 ]
I want to understand how to use all in this particular case.
I have read Haskell documentation but I am new to this language coming from Python so I am unable to figure the logic.
First you can have a function to check if a number is mod by all [2..15].
modByNumbers x ns = all (\n -> x `mod` n == 0) ns
Then you can use it like the mod function:
take 10 [x | x <- [1..] , x `modByNumbers` [2..15] ]
Alternatively, using math, we know that the smallest number divible by all numbers less than n is the product of all of the prime numbers x less than n raised to the floor of the result of logBase x n.
A basic isPrime function:
isPrime n = length [ x | x <- [2..n], n `mod` x == 0] == 1
Using that to get all of the primes less than 15:
p = [fromIntegral x :: Float | x <- [2..15], isPrime x]
-- [2.0,3.0,5.0,7.0,11.0,13.0]
Now we can get the exponents:
e = [fromIntegral (floor $ logBase x 15) :: Float | x <- p']
-- [3.0,2.0,1.0,1.0,1.0,1.0]
If we zip these together.
z = zipWith (**) p e
-- [8.0,9.0,5.0,7.0,11.0,13.0]
And then find the product of these we get the smallest number divisible by all numbers between 2 and 15.
smallest = product z
-- 360360.0
And now to get the rest we just need to multiply that by the numbers from 1 to 15.
map round $ take 10 [smallest * x | x <- [1..15]]
-- [360360,720720,1081080,1441440,1801800,2162160,2522520,2882880,3243240,3603600]
This has the advantage of running substantially faster.
Decompose the problem.
You already know how to take the first 10 elements of a list, so set that aside and forget about it. There are infinitely many numbers divisible by all of [2,15], your remaining task is to list them all.
There are infinitely many natural numbers (unconstrained), and you already know how to list them all ([1..]), so your remaining task is to transform that list into the "sub-list" who's elements are divisible by all of [2,15].
You already know how to transform a list into the "sub-list" satisfying some constraint (predicate :: X -> Bool). You're using a list comprehension in your posted code, but I think the rest of this is going to be easier if you use filter instead. Either way, your remaining task is to represent "is divisible by all of [2,15]" as a predicate..
You already know how to check if a number x is divisible by another number y. Now for something new: you want to abstract that as a predicate on x, and you want to parameterize that predicate by y. I'm sure you could get this part on your own if asked:
divisibleBy :: Int -> (Int -> Bool)
divisibleBy y x = 0 == (x `mod` y)
You already know how to represent [2,15] as [2..15]; we can turn that into a list of predicates using fmap divisibleBy. (Or map, worry about that difference tomorrow.) Your remaining task is to turn a list of predicates into a predicate.
You have a couple of options, but you already found all :: (a -> Bool) -> [a] -> Bool, so I'll suggest all ($ x). (note)
Once you've put all these pieces together into something that works, you'll probably be able to boil it back down into something that looks a little bit like what you first wrote.
I have been trying to learn haskell by trying to do some simple problems.
The Problem
Currently, I am trying to implement a function primeFactorization :: Integer -> [(Integer, Integer)] such that the output is a list of tuples containing the prime factor and the power it is raise to in the number.
Example Output
> primeFactorization 120
[(2,3), (3,1), (5,1)] since 120 = 2^3 * 3^1 * 5^1
My (Partial) Solution
primeFactorization :: Integer -> [Integer]
primeFactorization n =
let
factors :: Integer -> [Integer]
factors n = [x | x <- [2..n-1], n `mod` x == 0]
isPrime :: Integer -> Bool
isPrime n
| n `elem` [0, 1] = False
| n == 2 = True
| n > 2 = null [ x | x <- [2..(ceiling . sqrt . fromIntegral) n], n `mod` x == 0]
| otherwise = False
in
filter isPrime $ (factors n)
This is a working implementation to get the prime factors of a number. However as seen it only outputs the prime factors. I am not sure on how to store the number of times in haskell. Also, considering it is un-idiomatic to iterate in haskell I don't know how I would implement the solution. In python, I would do:
def pf(number):
factors=[]
d=2
while(number>1):
while(number%d==0):
factors.append(d)
number=number/d
d+=1
return factors
So, the question: How to implement the powers of the prime factors?
NOTE:
I already saw: Prime factorization of a factorial however that does not answer my question.
This is NOT a homework problem, I am learning independently.
You can always replace imperative-language loops (as long as they don't meddle with any global state) with recursion. That may not be the most elegant approach, but in this case it seems perfectly appropriate to imitate your inner Python loop with a recursive function:
dividerPower :: Integer -> Integer -> Int
dividerPower n d
| n`rem`d == 0 = 1 + dividerPower (n`quot`d) d
| otherwise = 0
(This counts “backwards” compared to the Python loop. You could also make it tail-recursive with a helper function and count forwards over an accumulator variable, but that's more awkward and I don't think there's a memory/performance benefit that would justify it in this case.)
You can either use that together with your Haskell code (for each of the factors you've already found, check how often it occurs), or extend it so the whole thing works like the Python solution (which is actually a lot more efficient, because it avoids for every number checking whether it's prime). For that you just need to give back the final n in the result. Let's use a where block for handling the pattern matching, and also make the rem and:
dividePower :: Integer -> Integer -> (Integer, Int)
dividePower n d
| r == 0 = (nfin, p'+1)
| otherwise = (n, 0)
where (n', r) = n `quotRem` d
(nfin, p') = dividePower n' d
Then the equivalent to your Python code is
pf :: Integer -> Integer -> [(Integer, Int)]
pf = go 2
where go d n
| n>1 = (d, p) : go (d+1) n'
| otherwise = []
where (n', p) = dividePower n d
This actually gives you, like in Python, the list including also non-dividers (with power 0). To avoid that, change the list-building to
| n>1 = (if p>0 then ((d,p):) else id) $ go (d+1) n'
This is a spoiler to task #3 of Project Euler! Don't continue to read, if you want to solve it by yourself.
I am trying to learn Haskell by writing programs for Project Euler. At the moment I'm trying to solve task #3 which asks for the largest prime factor of the number 600851475143.
To do this, I create a list liste which contains all numbers, which are divisors of this number (up to its squareroot). My strategy is now, to count the divisors of these numbers, to decide, if they are prime.
number = 600851475143
-- sn = sqrt number
sn = 775146
liste = [x | x <- [1..sn], (mod number x == 0)]
-- liste = [1,71,839,1471,6857,59569,104441,486847]
primelist :: Int -> [Int]
primelist z = [y | y <- [1..z], mod z y == 0]
main = print [primelist x | x <- liste]
The result, which should appear here, should be a list containing 8 lists with the divisors of the elements of liste. Instead, the list
[[1],[1,3],[1,29],[1,3,29,87]]
is printed.
How is this behaviour to be explained?
The Problem is the type declaration primelist :: Int -> [Int]. It forces Haskell to use native integers, i.e. 32-Bit integers on a 32-Bit platform. However, if you leave it out, Haskell will infer the function type to be Integer -> [Integer]. Integers allow computations with arbitrary precision, but are a little bit slower than native types.
To quote from "What's the difference between Integer and Int" in the Haskell FAQ:
Operations on Int can be much faster than operations on Integer, but
overflow and underflow can cause weird bugs.
Now isn't that the truth.
I am not sure if this will help you, but I also am working through Project Euler to help teach myself Haskell, and I devised the following solution:
defacto :: Integer -> Integer -> Integer
defacto x p | x == p = 1
| x`mod`p==0 = defacto (x`div`p) p
| otherwise = x
gpf :: Integer -> Integer
gpf = \x -> prim (x,primes)
prim :: (Integer,[Integer]) -> Integer
prim (x,(p:ps)) | p > x = 1
| (defacto x p) == 1 = p
| otherwise = prim((defacto x p),ps)
n :: Integer
n = 600851475143
Here, defacto de-factors a prime out of a number, so defacto 2 12 returns 4 and defacto 5 14 returns 14. gpf is a function to find the greatest prime factor, though it requires a list of primes up to x to be in scope. The key component is prim, which either returns 1 if the number is smaller than the next prime, returns the first prime in its prime list if x is a perfect power of that prime (i.e. if all other primes smaller than p have been factored out of x), and otherwise performs a recursive call on the defactored x and the truncated prime list. This has the effect of continuously shrinking x while linearly traversing the prime list, so that we need not test any primes that cannot factor into x, and we don't need to keep retesting the same primes on the reduced value of x. Hope this helps you.
I'm new in Haskell and try to solve 3 problem from http://projecteuler.net/.
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
My solution:
import Data.List
getD :: Int -> Int
getD x =
-- find deviders
let deriveList = filter (\y -> (x `mod` y) == 0) [1 .. x]
filteredList = filter isSimpleNumber deriveList
in maximum filteredList
-- Check is nmber simple
isSimpleNumber :: Int -> Bool
isSimpleNumber x = let deriveList = map (\y -> (x `mod` y)) [1 .. x]
filterLength = length ( filter (\z -> z == 0) deriveList)
in
case filterLength of
2 -> True
_ -> False
I try to run for example:
getD 13195
> 29
But when i try:
getD 600851475143
I get error Exception: Prelude.maximum: empty list Why?
Thank you #Barry Brown, I think i must use:
getD :: Integer -> Integer
But i get error:
Couldn't match expected type `Int' with actual type `Integer'
Expected type: [Int]
Actual type: [Integer]
In the second argument of `filter', namely `deriveList'
In the expression: filter isSimpleNumber deriveList
Thank you.
Your type signature limits the integer values to about 2^29. Try changing Int to Integer.
Edit:
I see that you already realised that you need to use Integer instead of Int. You need to change the types of both getD and isSimpleNumber otherwise you will get a type mismatch.
Also in general, if you are having trouble with types, simply remove the type declarations and let Haskell tell you the correct types.
Main> :t getD
getD :: Integral a => a -> a
Main> :t isSimpleNumber
isSimpleNumber :: Integral a => a -> Bool
After you found the error, may I point out that your solution is quite verbose? In this case a very simple implementation using brute force is good enough:
getD n = getD' n 2 where
getD' n f | n == f = f
| n `mod` f == 0 = getD' (n `div` f) f
| otherwise = getD' n (succ f)
this question is easy enough for brute-force solution, but it is a bad idea to do so because the whole idea of project euler is problems you need to really think of to solve (see end of answer)
so here are some of your program's flaws:
first, use rem instead of mod. it is more efficient.
some mathematical thinking should have told you that you don't need to check all numbers from 1 to x in the isprime function and the getD function, but checking all numbers from the squareroot to one (or reversed) should be sufficient. note that in getD you will actually need to filter numbers between x and the square root, because you search for the biggest one.
why do you use the maximum function in getD? you know the list is monotonically growing, so you may as well get the last one.
despite you only need the biggest divisor (which is prime) you compute the divisors list from small to big making the computer check for each value if it is a divisor or not although discarding the result once a bigger divisor is found. it should be fixed by filtering the list of numbers from x to 1, not from 1 to x. this will cause the computer to check divisibility (how should I say that?) for the biggest possible divisor, not throwing to the trash the knowledge of previous checks. note that this optimization takes effect only if the previous point is optimized, because otherwise the computer will compute all divisors anyway.
with the previous points mixed, you should have filtered all numbers [x,x-1 .. squareroot x] and taken the first.
you don't use an efficient isPrime function. if I were you, I would have searched for an isprime library function, which is guaranteed to be efficient.
and there are more..
with this kind of code you will never be able to solve harder project euler problems. they are designed to need extra thinking about the problem (for instance noticing you don't have to check numbers greater from the square root) and writing fast and efficient code. this is the purpose of project euler; being smart about programming. so don't skip it.
I started learning haskell yesterday and I got stuck on a problem. After a bit of trying different things I thought I'd finally come here and ask how to fix this. Also, feel free to criticize the way I have done things so far so I can know what direction to go. Thanks.
module Main where
main = putStrLn lastPrime
where
lastPrime :: String
lastPrime = show(last(take 10001 primes))
primes :: [Int]
primes = [x| x <- [1..],length [a| a <- [1..lessen(x)], mod x a /= 0] == x - 2]
lessen :: Int -> Int
lessen a = ceiling(sqrt(a))
To fix your type error, you want this:
lessen :: Int -> Int
lessen a = ceiling (sqrt (fromIntegral a))
a has type Int, but sqrt is expecting a floating point type, and the easiest way to convert an integral type to another numeric type is fromIntegral.
In addition to the type error in lessen you have a logic error in primes:
length [a| a <- [1..lessen(x)], mod x a /= 0] == x - 2
You're (rightly) only considering elements up to lessen x. This has the consequence that the list will almost never have exactly x - 2 elements. As a consequence you'll get into an infinite loop when trying to get more than two elements out of that list (because there is no 3rd element for which the condition is true, but haskell doesn't know that so it iterates to infinity trying to find it).
Also note that taking the length of a list is an O(n) operation and there's almost always a better way to achieve what you want.
As a style note, I would recommend defining a separate method isPrime. This will make your code look like this:
module Main where
main = putStrLn lastPrime
where
lastPrime :: String
lastPrime = show(last(take 10001 primes))
isPrime x = length [a| a <- [1..lessen(x)], mod x a /= 0] == x - 2]
primes :: [Int]
primes = [x| x <- [1..], isPrime x]
lessen :: Int -> Int
lessen a = ceiling(sqrt (fromIntegral a))
This IMHO makes the list comprehension much more readable. However it still contains the bug. To get rid of the bug, I'd suggest defining isPrime using a different approach. Going up to lessen x is fine (except you should start from 2 because 1 cleanly divides everything), but instead of building a new list with all the divisors, you should just check whether any of the numbers in the range divides x. For this we can use the higher order function any, so we get this:
isPrime x = not (any (\a -> mod x a == 0) [2 .. lessen x])