The Permutation Pigeon-hole Principle

Question

In the game of sudoku, many players like to "pencil in" possible numbers that can go in each square:

The above row can be represented as an array:

[[1,2,9], [6], [5], [7], [1,2,9], [1,2,9], [3], [1,2,4], [8]]

Now, notice that there is only 1 place where a 4 can go. This effectively lets us simplify the above list to:

[[1,2,9], [6], [5], [7], [1,2,9], [1,2,9], [3], [4], [8]]

The goal of this challenge is to take a list of possible numbers in a permutation, and deduce which possibilities can be eliminated.

As another example, lets say you have the following array of possibilities:

[[0,1,3], [0,2,3], [1,2], [1,2]]

The last two places must be filled with 1 and 2. Therefore, we can remove those possibilities from the first two elements in the array:

[[0,3], [0,3], [1,2], [1,2]]

As another example:

[[0,1,2,3], [0,2], [0,2], [0,2]]

Its impossible to construct a permutation from the above possibilities, as there's only 1 location for both 1 and 3, and you would want to return an empty array.

You need to input a list of possibilities and output the remaining possibilities after the maximum number of possibilities have been eliminated.

• If a particular array is impossible, you either need to return an empty array, or an array where one of the subarrays is empty.
• You may assume that the array will be well-formed, and have at least 1 element.
• Given an array of size N, you can assume the numbers in the subarray will always be in the range [0:N), and that N <= 10
• You may not assume that every number from 0 to N-1 will be present
• You may assume that numbers within a single subarray are unique.
• If a subarray contains only a single possibility, you can either represent the possibility in an array or by itself. [[1],[2],[0]], [1,2,0], [[1,2],0,[1,2]] are all valid.
• You may accept the array either in a reasonable string format or in list/array format.
• Subarrays can be in any order.
• Instead of dealing with ragged arrays, you can pad empty places with -1.

Test cases

[[0]]                                         -> [[0]]
[[1],[0]]                                     -> [[1],[0]]
[[1],[1]]                                     -> []
[[1],[0,1]]                                   -> [[1],[0]]
[[0,1,2],[1,2],[1,2]]                         -> [[0],[1,2],[1,2]]
[[0,1],[1,2],[0,2]]                           -> [[0,1],[1,2],[0,2]]
[[2,1],[1,2],[1,2]]                           -> []
[[0,3],[2,1],[3,0],[3,2]]                     -> [[0,3],[1],[0,3],[2]]
[[0,1],[0,1],[2,3],[2,3,0]]                   -> [[0,1],[0,1],[2,3],[2,3]]
[[0,1],[0,3],[3,2],[0]]                       -> [[1],[3],[2],[0]]
[[3,5,2],[0,2,4],[4,0],[0,1,3,5],[2,1],[2,4]] -> [[3,5],[0,2,4],[4,0],[3,5],[1],[2,4]]
[[6,9,8,4],[4,5],[5,3,6],[3,8,6,1,4],[3,1,9,6],[3,7,0,2,4,5],[9,5,6,8],[6,5,8,1,3,7],[8],[8,0,6,2,5,6,3]] -> [[6,9,4],[4,5],[5,3,6],[3,6,1,4],[3,1,9,6],[0,2],[9,5,6],[7],[8],[0,2]]
[[3,5,0],[5,7],[5,1,2],[1,3,0],[5,3],[5,0],[5,3,7,8,0,6],[7,5,0,1,8],[1,0,8],[0,6]] -> []
[[9,0,2,3,7],[0,7,6,5],[6,9,4,7],[9,1,2,3,0,5],[2,8,5,7,4,6],[6,5,7,1],[5,9,4],[5,9,3,8,1],[5,0,6,4],[0,7,2,1,3,4,8]] -> [[9,0,2,3,7],[0,7,6,5],[6,9,4,7],[9,1,2,3,0,5],[2,8,5,7,4,6],[6,5,7,1],[5,9,4],[5,9,3,8,1],[5,0,6,4],[0,7,2,1,3,4,8]]
[[2,6,0],[0,4,3],[0,6,2],[0,7],[0,9,2,3,6,1,4],[1,7,2],[2,7,8],[8,6,7],[6,5,2,8,0],[5,8,1,4]] -> [[2,6,0],[3],[0,6,2],[0,7],[9],[1],[2,7,8],[8,6,7],[5],[4]]
[[8],[8,0,6,5,7,2,4,1],[8,6,9,3,5,0,7],[3,9,1,0],[9],[9,2,6],[2,8,3],[3,1,6,8,2],[6],[6,4,5,3,0,7]] -> [[8],[5,7,4],[5,7],[0],[9],[2],[3],[1],[6],[4,5,7]]
[[8,1,0],[5,8,7,6,2,0],[6,8,2],[2,4,0,9],[4,1,7,3,6,8],[8,1],[8,0,3],[0,8,2],[0,8,3],[1,8,0]] -> []

This is a code-golf so make your answers as short as possible!

Answer 1

Brachylog, 21 bytes

:1fz:da|,[]
:2a#d
:Am

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Predicate 0 (main predicate)

:1fz:da|,[]
:1f            Find all solutions of Predicate 1 using Input as Input.
   z           Transpose
    :da        Deduplicate each.
       |,[]    If there is no solution, return [] instead.

Predicate 1 (auxiliary predicate 1)

:2a#d
:2a     Each element of Output satisfies Predicate 2 with each element of Input as Input
   #d   Each element is different

Predicate 2 (auxiliary predicate 2)

:Am     Output is member of Input

Answer 2

Jelly, 10 bytes

Œp⁼Q$ÐfZQ€

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Œp⁼Q$ÐfZQ€   Main chain, argument: z

Œp           Cartesian product
  ⁼Q$Ðf      Filter for those that remain unchanged when uniquified
       Z     Transpose
        Q€   Uniquify each subarray

Answer 3

Pyth, 11 bytes

{MC{I#.nM*F

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{MC{I#.nM*F
         *F  reduce by Cartesian product
             produces nested arrays
      .nM    flatten each
   {I#       filter for invariant under deduplication
  C          transpose
{M           deduplicate each

Answer 4

Haskell, 100 bytes

import Data.List
p z=map nub$transpose$filter(and.(flip$zipWith elem)z)$permutations[0..length z-1]

Answer 5

Actually, 27 bytes

Rd╗R`╜∙"♂i"£M╗`MX╜`;╔=`░┬♂╔

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Answer 6

Python 3, 101 99 bytes

Thanks to @TLW for -2 bytes

from itertools import*
lambda x:list(map(set,zip(*[i for i in product(*x)if len(i)==len(set(i))])))

An anonymous function that takes input via argument of a list of lists and returns a list of sets.

How it works

from itertools import*        Import Python's library for iterator generation
lambda x                      Anonymous function with input possibilities x as a
                              list of lists
...for i in product(*x)...    For i in the Cartesian product of x, ie all candidate
                              arrangements:
[...if len(i)==len(set(i))]    Filter into list by non-duplicity (set removes
                               duplicates, so there are no duplicates if the length
                               of i is the same as the length of the set of
                               the elements of i)
zip(*...)                     Unpack and take the transpose, leaving the modified
                              possibilities with duplicates
map(set,...)                  Remove duplicates
:list(...)                    Return the modified possibilities as a list of sets

Try it on Ideone

Answer 7

Mathematica, 46 bytes

Union/@Thread@Select[Tuples@#,DuplicateFreeQ]&

Answer 8

PHP, 245 231 bytes

131 117 for the cartesian product function, 114 for the other stuff

function c($a){if ($a){if($u=array_pop($a))foreach(c($a)as$p)foreach($u as$v)yield $p+[count($p)=>$v];}else yield[];}
function p($a){foreach(c($a)as$i)if(max(array_count_values($i))<2)foreach($i as$k=>$v)$r[$k][$v]=$v;return$r?:[];}

I ran into memory issues on some of the test cases, with a recursive function for the cartesian product. Worked better with this generator class and function c($a){$b=[];foreach($a as$i)$b[]=new \ArrayIterator($i);return new CartesianProductIterator($b);}.
But my generator is shorter and does the same job.

The larger examples, however, result in an Internal Server Error (both with the iterator and the generator) after a while on my machine. Currently no time to check the server logs, unfortunately.