# Haskell Quiz/Splitting the Loot/Solution TomPlick

## Examples

Using the examples from http://www.rubyquiz.com/quiz65.html:

```
> splitLoot 3 [1..4]
[]
> splitLoot 2 [9, 12, 14, 17, 23, 32, 34, 40, 42, 49]
[[[9,12,32,34,49],[14,17,23,40,42]],[[14,17,23,40,42],[9,12,32,34,49]]]
> head $ splitLoot 3 [1..4]
*** Exception: Prelude.head: empty list
> head $ splitLoot 2 [9, 12, 14, 17, 23, 32, 34, 40, 42, 49]
[[9,12,32,34,49],[14,17,23,40,42]]
```

## Solution

```
import Control.Monad.List
partitionSet :: [a] -> [([a], [a])]
partitionSet [] = [([], [])]
partitionSet (x:xs) =
concat [[(x:p1, p2), (p1, x:p2)] | (p1, p2) <- partitionSet xs]
type Distribution = [[Integer]]
splitLoot :: Integer -> [Integer] -> [Distribution]
splitLoot 0 [] = return []
splitLoot 0 _ = mzero
splitLoot numPeople jewels =
let (share, leftovers) = (sum jewels) `quotRem` numPeople
in if leftovers /= 0
then []
else do (mine, restOfJewels) <- partitionSet jewels
if (sum mine) == share
then do restOfDivision <- splitLoot (numPeople - 1) restOfJewels
return (mine : restOfDivision)
else mzero
```

## Explanation

`partitionSet`

takes a list and returns a list of ways to divide the set into two. For example, `partitionSet [1, 2]`

returns `[([1,2],[]),([2],[1]),([1],[2]),([],[1,2])]`

.

`splitLoot`

operates by trying to take one person's loot from the jewels, then dispersing the remainder to the rest of the people. First, care must be taken that an equal distribution is possible (hence `leftovers /= 0`

). Then, we examine each partition of the jewels into two sets `mine`

and `restOfJewels`

. For each `mine`

that adds up to a fair share, we attempt division of `restOfJewels`

among the rest of the people.

`splitLoot`

uses the list monad; under the list monad, a binding of the form `x <- y`

means roughly that `x`

is to take on each member of the list `y`

, one at a time.

`splitLoot`

returns all solutions; if only one is desired, call `head`

on the result.