Combining two list comprehensions Haskell

Question

I'm trying to calculate the first triangle number with 501 divisorsin Haskell. I already made two list comprehensions, one listing all the triangle numbers and one listing all divisors of a given number. Now I want to make one big list with all values divisors of each triangle number. (eg [[1],[1,3],[1,2,3,6],[1,2,5,10] etc..)

How can I use my triangleNumbers list in my divisors list? My code is below.

triangleNumbers = [t | a <- [0..], let t = a*(a+1)/2] 
divisors triangleNumbers = [d | d <- [1..triangleNumbers], triangleNumbers 
`mod` d == 0]
numDivisors = takeWhile(<501) length . divisors
answer = divisors !! numDivisors

Answer 1

First of all, your triangleNumbers is a bit weird: it contains Fractional s, instead of Integrals s. This makes it more troublesome to perform accurate division calculations. So we better modify this:

triangleNumbers = [ div (a*(a+1)) 2 | a <- [0..]]

Note that we can write the expression in the head of the list comprehension. We also use div over / since div is an integer division. We know for sure that we will not lose data by performing an integer division, since either a or a+1 is even, and the multiplication of a number with an even number is always even. This results in the following list:

Prelude> take 10 triangleNumbers 
[0,1,3,6,10,15,21,28,36,45]

Now we want a function that maps numbers on the divisors. We can make a generic function:

divisors x = [d | d <- [1..x], mod x d == 0]

Now we can use map , to map a list of numbers to a list of lists where each list contains the divisors of the original number. So:

Prelude> map divisors [1,2,3,5,8,13,21]
[[1],[1,2],[1,3],[1,5],[1,2,4,8],[1,13],[1,3,7,21]]

We can however also give the map divisors the (infinite list) of triangleNumbers . For instance for the first 10 triangleNumbers :

Prelude> take 10 $ map divisors $ triangleNumbers 
[[],[1],[1,3],[1,2,3,6],[1,2,5,10],[1,3,5,15],[1,3,7,21],[1,2,4,7,14,28],[1,2,3,4,6,9,12,18,36],[1,3,5,9,15,45]]

Now we only need to filter the numbers that have 501 elements or more. We can do this by checking that if we drop 500 elements, we still have a list that contains at least one element. So with:

hasAtLeastLength :: Int -> [a] -> Bool
hasAtLeastLength n = not . null . drop (n-1)

So now we can filter all the elements where the hasAtLeastLengh 501 (divisors x) for a number x . This will thus produce the list of all these numbers:

filter (hasAtLeastLength 501 . divisors) triangleNumbers

This will produce an infinite list of all triangleNumbers that have at least 501 divisors. We can use head to finally obtain the first element:

head $ filter (hasAtLeastLengh 501 . divisors) triangleNumbers

This takes a large amount of time. The code works however quite fast if we work with 10 divisors:

Prelude> filter (hasAtLeastLength 10 . divisors) triangleNumbers
[120,210,276,300,378,496,528,630,666,780,820,990,1035,1128,1176,1275,1326,1485,1540,1596,1770,1830,1953,2016,2080,2145,2346,2415,2556,2628,2775,2850,2926,3003,3160,3240,3321,3486,3570,3828,3916,4005,4095,4186,4278,4560,4656,4851,4950,5050,5356,5460,5565,5778,5886,6105,6216,6328,6555,6670,6786,6903,7140,7260,7626,7750,7875,8001,8128,8256,8385,8646,8778,9045,9180,9316,9730,9870,10296,10440,10731,10878,11175,11325,11476,11628,11781,11935,12090,12246,12720,12880,13041,13203,13530,13695,14028,14196,14365,14535,14706,15225,15400,15576,16110,16290,16653,16836,17020,17205,17391,17578,17766,17955,18336,18528,18915,19110,19306,19503,19701,19900,20100,20706,20910,21321,21528,21736,21945,22155,22578,23220,23436,24090,24310,24531,24976,25200,25425,25878,26106,26565,26796,27028,27261,27495,27730,27966,28203,28680,28920,29403,29646,29890,30135,30381,30628,30876,31125,31626,31878,32385,32640,32896,33411,33670,33930,34191,34716,34980,35245,35511,35778,36315,36585,36856,37128,37401,37675,37950,38226,38781,39060,39340,40470,40755,41041,41328,41616,41905,42195,42486,43071,43365,43660,43956,44253,44850,45150,46056,46360,46665,46971,47278,47586,47895,48516,48828,49455,49770,50721,51040,51360,51681,52003,52326,52650,...

which means that it produces an answer. This means however that you will have to come up with something smarter than simply enumerating.