11

Suppose, Initial state Rubik's Cube 6x6x6

444444
444444
444444
444444
444444
444444

000000 111111 222222 333333
000000 111111 222222 333333
000000 111111 222222 333333
000000 111111 222222 333333
000000 111111 222222 333333
000000 111111 222222 333333

555555
555555
555555
555555
555555
555555

I want to find a final state using basic slice and face moves, where in each face 6x6 the columns and rows don't have repeating colors (numbers), for example:

0 2 3 4 5 1
1 5 0 2 4 3
5 0 4 3 1 2
2 3 5 1 0 4
4 1 2 5 3 0
3 4 1 0 2 5

I found a final state very close to the desired one (only one piece out of place) Using the following algorithm

bucket is rubik cube 6x6x6 empty
fill_board()
    if length available positions is 0
        return True
    position = pop available positions
    for each N in random.sample([0,1,2,3,4,5,6],6)
        bucket[position.i][position.j] = N
        if bucked is not valid
            continue
        if fill_board                           //Recursion call
            return True
        bucket[position.i][position.j] = None   //backtracking
    append position in available positions      //backtracking
    return False

Nearby state found:

015324
451230
120453
342501
203145
534012

320154 135042 315024 315240
254013 513420 420153 052134
413202 351204 153240 143502 ---- check first 2
045231 204153 531402 524013
102345 042315 042531 401325
531420 420531 204315 230451

042351
215403
153024
530142
421530
304215

My hypothesis is that there is no sudoku state for the rubik's cube 6x6x6, I am agricultural engineer and I do not know how to validate or refute it, with my algorithm if I find a sudoku state would refute it

Is there someone who can help me?

  • Might there be some relationship to the 36 officers problem? https://en.m.wikipedia.org/wiki/Thirty-six_officers_problem – Oscar Lanzi Jul 31 '17 at 19:52
  • This is similar, but not the same problem .... P.S. I didn't know about "36 officers problem" Euler :) – Jose Ricardo Bustos M. Jul 31 '17 at 20:01
  • If you ran the backtracking algorithm, and the algorithm is correct, and it didn't find any solutions, then isn't that a proof that there aren't any? Oh but it looks like your algorithm is not backtracking, but trying random solutions… so there is no guarantee that it will explore the entire space of possibilities. – ShreevatsaR Aug 01 '17 at 22:32
  • 1
    @ShreevatsaR cube 6x6x6 has 1.572 * 10^116 combinations (states), I cann't explore the entire space of possibilities – Jose Ricardo Bustos M. Aug 01 '17 at 22:45
  • @JoseRicardoBustosM. Thanks, that makes sense :-) – ShreevatsaR Aug 01 '17 at 23:00

1 Answers1

6

Yes, there is such a state.

At least http://www.randelshofer.ch/rubik/vcube6/X900.01.html claims to present two sequences of moves that each will create one.

Here they are (but I'm not sure how to read the notation):

Algorithm Michael Z. R. Gottlieb 2008
F R' B' D2 · F' B2 · D2 L D' F R' F' U' F' R' B R U2 TF TR' TB' TD2 · TF' TB2 · TD2 TL TD' TF TR' TF' TU' TF' TR' TB TR TU2 T3B T3L T3U2 T3B2 T3L' T3B T3U T3L' T3U2 T3L' T3B' (47 btm)

First

part1

Second

part2

Final

part3

Algorithm Walter Randelshofer 2014
U SF SR SD' F2 SR2 B U SF2 D2 WR2 U WR2 TU S2F S2R S2D' TF2 S2R2 TB TU S2F2 TD2 M2R2 TU M2R2 T3U2 T3B T3L2 T3U' T3L2 T3U T3B2 T3L' (34 btm)