Ordered partitions

Ordered partitions
You are encouraged to solve this task according to the task description, using any language you may know.

In this task we want to find the ordered partitions into fixed-size blocks. This task is related to [Combinations] in that it has to do with discrete mathematics and moreover a helper function to compute combinations is (probably) needed to solve this task.

partitions(arg1,arg2,...,argn) should generate all distributions of the elements in {1,...,sum{arg1,arg2,...,argn}} into #{arg1,arg2,...,argn} blocks of respective size arg1,arg2,...,argn.

E.g. partitions(2,0,2) would create:

{({1, 2}, {}, {3, 4}), ({1, 3}, {}, {2, 4}), ({1, 4}, {}, {2, 3}), 
 ({2, 3}, {}, {1, 4}), ({2, 4}, {}, {1, 3}), ({3, 4}, {}, {1, 2})}

E.g. partitions(1,1,1) would create:

{({1}, {2}, {3}), ({1}, {3}, {2}), ({2}, {1}, {3}), ({2}, {3}, {1}), 
 ({3}, {1}, {2}), ({3}, {2}, {1})}

Note that the number of elements in the list is

(arg1+arg2+...+argn choose arg1) * (arg2+arg3+...+argn choose arg2) * ... * (argn choose argn)

(i.e. the multinomial coefficient). Also, partitions(1,1,1) creates the permutations of {1,2,3} and thus there would be 3! = 6 elements in the list.

Note: Do not use functions that are not in the standard library of the programming language you use. Your file should be written so that it can be executed on the command line and by default outputs the result of partitions(2,0,2). If the programming language does not support polyvariadic functions pass a list as an argument.

Notation

Remarks on the used notation for the task in order to understand it easierly.

{1,...,n} denotes the set of consecutive numbers from 1 to n, e.g. {1,2,3} if n = 3. sum is the function that takes as its sole argument a set of natural numbers and computes the sum of the numbers, e.g. sum{1,2,3} = 6. arg1,arg2,...,argn are the arguments - natural numbers - that the sought function receives. The operator # returns the number of elements in a set, e.g. #{1,2,3} = 3. () is a tuple, e.g. (1,2,3).

Haskell

<lang haskell> comb :: Int -> [a] -> a comb 0 _ = [[]] comb _ [] = [] comb k (x:xs) = [ x:cs | cs <- comb (k-1) xs ] ++ comb k xs

partitions :: [Int] -> [[[Int]]] partitions xs = p [1..sum xs] xs

   where p _ []      = [[]]
         p xs (k:ks) = [ cs:rs | cs <- comb k xs, rs <- p (xs `minus` cs) ks ]
         minus xs ys = [ x | x <- xs, not $ x `elem` ys ]

main = do putStrLn $ show $ partitions [2,0,2] </lang>

An alternative where minus is not needed anymore because comb now not only keeps the chosen elements but also the not chosen elements together in a tuple.

<lang haskell> comb :: Int -> [a] -> [([a],[a])] comb 0 xs = [([],xs)] comb _ [] = [] comb k (x:xs) = [ (x:cs,zs) | (cs,zs) <- comb (k-1) xs ] ++

               [ (cs,x:zs) | (cs,zs) <- comb  k    xs ]

partitions :: [Int] -> [[[Int]]] partitions xs = p [1..sum xs] xs

   where p _ []      = [[]]
         p xs (k:ks) = [ cs:rs | (cs,zs) <- comb k xs, rs <- p zs ks ]

main = do putStrLn $ show $ partitions [2,0,2] </lang>