In today’s Programming Praxis exercise, our goal is to determine whether four given points in a 2D plane make up a square. Let’s get started, shall we?

An import:

import Data.List

The provided solution calculates the squares distances between the first and the three other points and considers the points a square when the smallest two are equal and the largest is the sum of the two smallest, i.e. a^{2} + b^{2} = c^{2}. This solution is incorrect, however. If we take the points (0, 0), (1, 0), (0, 1) and (-1, -1), both conditions are satisfied despite the fact that the four points do not form a square. To handle this case correctly, we can use the same basic approach, albeit expanded a bit. Instead of just three distances, we calculate the distances between all six pairs. Of these six, the four smallest ones should be equal, as should the two larger ones. Pythagoras’ theorem can remain the same.

isSquare :: (Num a, Ord a) => (a, a) -> (a, a) -> (a, a) -> (a, a) -> Bool
isSquare p q r s = and [a == b, b == c, c == d, e == f, a + b == e] where
[a,b,c,d,e,f] = sort [l | (h:t) <- tails [p,q,r,s], l <- map (dist h) t]
dist (x1,y1) (x2,y2) = (x2-x1)^2 + (y2-y1)^2

Testing reveals that this version handles the old test cases correctly, as well as the example described above.

main :: IO ()
main = do print $ isSquare (0,0) (0,1) (1,1) (1,0)
print $ isSquare (0,0) (2,1) (3,-1) (1, -2)
print $ isSquare (0,0) (1,1) (0,1) (1,0)
print . not $ isSquare (0,0) (0,2) (3,2) (3,0)
print . not $ isSquare (0,0) (3,4) (8,4) (5,0)
print . not $ isSquare (0,0) (0,0) (1,1) (0,0)
print . not $ isSquare (0,0) (0,0) (1,0) (0,1)
print . not $ isSquare (0,0) (1,0) (0,1) (-1,-1)

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Tags: bonsai, code, Haskell, kata, points, praxis, programming, square

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