# [Haskell-beginners] CSES programming problems at https://cses.fi/problemset/

Julian Ong julian_ong at yahoo.com
Sun Jun 28 23:45:18 UTC 2020

``` I've simplified and optimized it slightly (no need to use a monad for moveKnightUR) but overall it's still not fast enough to pass the CSES test. I'm wondering if the recursion is somehow inefficient because of two instances of solveK (k-1)...?---- main :: IO ()main = do    line <- getLine    let n = read line :: Integer    putStr \$ unlines \$ map show \$ reverse \$ solveK n

solveK :: Integer -> [Integer]solveK k    | k == 1 =     | otherwise = (solveFrameK k + head (solveK (k-1))) : solveK (k-1)
-- Returns list of knight moves in the upper right (k-1) x (k-1) portion of the board excluding the first column and first rowmoveKnightUR :: Integer -> (Integer, Integer) -> [(Integer, Integer)]moveKnightUR k (c, r) = filter (\(c', r') -> c' `elem` [2..k] && r' `elem` [2..k]) [(c-1, r+2), (c+1, r+2), (c+2, r+1), (c+2, r-1), (c+1, r-2), (c-2, r+1)]    -- Returns list of left and bottom border squares for k x k board in (col, row) format with (1, 1) being the lower left squaregenBorder :: Integer -> [(Integer, Integer)]genBorder k = [(1, a) | a <- [1..k]] ++ [(a, 1) | a <- [2..k]]
-- Formula for combinations C(n, r)combinations :: Integer -> Integer -> Integercombinations n r = product [1..n] `div` (product [1..(n-r)] * product [1..r])
-- Calculates additional number of two knight placements along the left and bottom border and from that border into the upper right (k-1) x (k-1) regionsolveFrameK :: Integer -> IntegersolveFrameK k    | k == 1 = 0    | k == 2 = 6    | otherwise = ((combinations (2*k-1) 2) - 2) + (k-1) * (k-1) * (2*k-1) - sum (map (toInteger . length) (map (moveKnightUR k) (genBorder k)))----
Julian
On Sunday, June 28, 2020, 04:27:07 PM PDT, Julian Ong <julian_ong at yahoo.com> wrote:

I realized I did not answer the question Doug posed, but the algorithm as originally presented works correctly and calculates correctly the number of possible knight pairings for each k x k board and generates the correct output requested by the problem.
The issue is still that, as I have implemented it in Haskell, it doesn't run fast enough to pass the automated CSES testing for n=10000. I am very curious whether it's possible to pass the speed testing for this problem using Haskell and if so how.
On Sunday, June 28, 2020, 09:49:02 AM PDT, Irfon-Kim Ahmad <irfon at ambienautica.com> wrote:

On 2020-06-28 11:26 a.m., Doug McIlroy wrote:

I'm currently stuck on the Two Knights problem.

Having placed one knight on the board, in how many
places can you put the other?

If you check the website indicated, it's a slight variation on that:

"Your task is to count for k=1,2,…,n the number of ways two knights can be placed on a k×k chessboard so that they do not attack each other."

The input is n (an integer that can range from 1 to 10000), the output is a single integer for each value from 1 to n, one per line, the memory limit is 512MB, and the maximum runtime is 1.00 seconds.

_______________________________________________
Beginners mailing list