My Haskell solution to problem at http://beust.com/weblog/2011/10/30/a-new-coding-challenge/

solve.hs

-- Solution to http://beust.com/weblog/2011/10/30/a-new-coding-challenge/

import Data.List(isInfixOf)

partitions n 1             = [[n]]
partitions n k | n > 0     = [m:ms | m <- [1..n `div` 2], ms <- partitions (n-m) (k-1)]
partitions n k | otherwise = []

weighings []     = [0]
weighings (w:ws) = [m + n | m <- weighings ws, n <- [-w, 0, w]]

-- solve 40 4 returns [[1,3,9,27]]
solve n k = [p | p <- partitions n k, [1..n] `isInfixOf` (weighings p)]