ruby while loop translated to haskell

Question

I've just started learning a bit of Haskell and functional programming, but I find it very difficult getting a hang of it :)

I am trying to translate a small piece of ruby code to Haskell (because I like the concept functional programming and Haskell proposes and even more because I come from a mathematics field and Haskell seems very mathematical):

class Integer
  def factorial
    f = 1; for i in 1..self; f *= i; end; f
  end
end

boundary = 1000
m = 0

# Brown Numbers - pair of integers (m,n) where n factorial is equal with square root of m

while m <= boundary

    n = 0

    while n <= boundary
        puts "(#{m},#{n})" if ((n.factorial + 1) == (m ** 2)) 
        n += 1
    end

    m += 1
end

I could only figure out how to do factorials:

let factorial n = product [1..n]

I cannot figure out how to do the while loops or equivalent in Haskell, even though I found some examples that were far to confusing for me.

The idea is that the loops start from 0 (or 1) and continue (with an increment of 1) until it reaches a boundary (in my code is 1000). The reason there is a boundary is because I was thinking of starting parallel tasks that do the same operation but on different intervals so the results that I expect are returned faster (one operation would be done on 1 to 10000, another on 10000 to 100000, etc.).

I would really appreciate it if anyone could help out with this :)

On the very basic level [1..n] generates a list of indices that a normal loop would give you. If each iteration produces on element, use a map. If you need to combine the iterations into one element, use a fold — Bartek Banachewicz
– Bartek Banachewicz, Commented Jul 21, 2014 at 10:44
@bartekbanachewicz ~ I confused [1..n] with a range of numbers in ruby, or is it the same thing but it can also be used instead of loop? — Roland
– Roland, Commented Jul 21, 2014 at 11:32
It's a range of numbers alright, but there are tools to apply some operation on every element of a range. I just pointed out that [1..n] are values of index in a typical for loop. — Bartek Banachewicz
– Bartek Banachewicz, Commented Jul 21, 2014 at 11:34

shree.pat18 · Accepted Answer · 2014-07-21 14:36:45Z

8

Try this:

let results =  [(x,y) | x <- [1..1000], y <- [1..1000] ,1 + fac x == y*y]
               where fac n = product [1..n]

This is a list comprehension. More on that here.

To map it to your Ruby code,

The nested loops in m and n are replaced with x and y. Basically there is iteration over the values of x and y in the specified ranges (1 to 1000 inclusive in this case).
The check at the end is your filter condition for getting Brown numbers.
where allows us to create a helper function to calculate the factorial.

Note that instead of a separate function, we could have computed the factorial in place, like so:

(1 + product[1..x]) == y * y

Ultimately, the (x,y) on the left side means that it returns a list of tuples (x,y) which are your Brown numbers.

OK, this should work in your .hs file:

results :: [(Integer, Integer)] --Use instead of `Int` to fix overflow issue
results =  [(x,y) | x <- [1..1000], y <- [1..1000] , fac x == y*y]
        where fac n = product [1..n]

edited Jul 21, 2014 at 14:36

answered Jul 21, 2014 at 10:42

shree.pat18

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11 Comments

ThreeFx Over a year ago

fac n = product does not compile. Change it to fac n = product[1..n]

shree.pat18 Over a year ago

My bad. Missed it earlier.

Roland Over a year ago

Holy moly, this is so much less code than any other languages I tried :)

Ørjan Johansen Over a year ago

@rolandjitsu You are getting integer overflow. Replace Int by the unbounded Integer throughout.

Ørjan Johansen Over a year ago

@rolandjitsu Int is a type that is limited in size to a machine word, for speed and space efficiency, so it overflows when you get over that limit. Integer is a type that's based (usually) on the GMP bignum library, it can grow to several megabytes in size.

|

Luis Casillas · Accepted Answer · 2014-07-23 22:42:48Z

0

To add to shree.pat18's answer, maybe an exercise you could try is to translate the Haskell solution back into Ruby. It should be possible, because Ruby has ranges, Enumerator::Lazy and Enumerable#flat_map. The following rewritten Haskell solution should perhaps help:

import Data.List (concatMap)

results :: [(Integer, Integer)]
results = concatMap (\x -> concatMap (\y -> test x y) [1..1000]) [1..1000]
    where test x y = if fac x == y*y then [(x,y)] else []
          fac n = product [1..n]

Note that Haskell concatMap is more or less the same as Ruby Enumerable#flat_map.

answered Jul 23, 2014 at 22:42

Luis Casillas

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