p028.hs

{- 
 - Solution to Project Euler problem 28
 - Copyright (c) Project Nayuki. All rights reserved.
 - 
 - https://www.nayuki.io/page/project-euler-solutions
 - https://github.com/nayuki/Project-Euler-solutions
 -}


{- 
 - From the diagram, let's observe the four corners of an n * n square (where n is odd).
 - It's not hard to convince yourself that the top right corner always has the value n^2.
 - Working counterclockwise (backwards), the top left corner has the value n^2 - (n - 1),
 - the bottom left corner has the value n^2 - 2(n - 1), and the bottom right is n^2 - 3(n - 1).
 - Putting it all together, this outermost ring contributes 4n^2 - 6(n - 1) to the final sum.
 - 
 - Incidentally, the closed form of this sum is (4m^3 + 3m^2 + 8m - 9) / 6, where m = size.
 -}

size = 1001  -- Must be odd
main = putStrLn (show ans)
ans = 1 + sum [4 * n * n - 6 * (n - 1) | n <- [3, 5 .. size]]