CODEWITHSUNDEEP
haskellmathfunctional-programminglist-comprehensionequation
vytaute
vytaute

Reputation: 1480

Equation solving with Haskell

I want to find all numbers up to N in Haskell, that satisfies the equation (Fermat theorem):

enter image description here

Some solutions could be:

enter image description here

So I try to do it like so in Haskell:

main :: IO ()
main =  do
    let arr = [ z * z == x * x + y * y | x <- [1..13], y <- [1..13], z <- [1..13]] in print arr

I get bool values list:

[False,False,False,False,False,False,False,False,False,False,False,False,False,False,False,False,False,...]

There are 4 True values in total in that list. My questons are:

  1. How do I get all tuples (x, y, z) that satisfies the given equation?
  2. How can I count True values in that array?

UPDATE: One more question. How can I count how many numbers do not belong to ANY tuple? I thought it would logically go like this:

solutions2 = [
    (z)
  | x <- [1..13]
  , y <- [x..13]
  , z <- [y..13]
  , z * z /= x * x + y * y, 
  z * z /= x * x - y * y, 
  z * z /= y * y - x * x
  ]

But that returned way too many numbers. Any ideas about how to find numbers that do not belong to any triples?

UPDATE2: I have been experimenting and diging more. How can I pass number n, that I read from input to a previously efined function (calculation method)? I would like to be able to get solution array, print it and its lenght.

sol n = solution = [
    (x, y, z)
  | x <- [1..n]
  , y <- [x..n]
  , z <- [y..n]
  , z * z == x * x + y * y
  ]

main :: IO ()
main = do 
    putStrLn "Enter n:"
    n <- getLine
    let mySol = sol n
    print (mySol)

But i gives me error:

error: parse error on input `='

and also:

Failed, no modules loaded.

Upvotes: 1

Views: 378

Answers (1)

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 476813

You can work with a filter and yield an (x, y, z) tuple in case the filter is satisfied:

solutions = [
    (x, y, z)
  | x <- [1..13]
  , y <- [1..13]
  , z <- [1..13]
  , z * z == x * x + y * y
  ]

main :: IO ()
main =  do
    print solutions

This gives us the six solutions:

Prelude> solutions 
[(3,4,5),(4,3,5),(5,12,13),(6,8,10),(8,6,10),(12,5,13)]

You can also print print (length solutions) to obtain the number of solutions.

Upvotes: 4

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