8
$\begingroup$

This puzzle is a follow-up for Does The Rubik’s Cube In This Painting Have A Solved State?

What is the most complicated illegal “partial state” possible for a Rubik’s Cube?

To be more specific: a partial state is a configuration of 27 colours for three adjacent faces, such as shown below. A partial state is legal if you can call the colours for all stickers on the three unseen faces and solve the resulting cube. Solving means all faces are a single colour, but not necessarily the standard permutation of colours (therefore White is not necessarily opposite Yellow etc). If this is not possible then the partial state is illegal.

Given an illegal partial state, the complexity is the minimum number of stickers you must consider to prove it is indeed illegal. In the diagram below we need only consider the five green edge stickers to prove illegality because … here’s a knowledge bomb from Puzzling Stack Exchange … every colour in a Rubik’s Cube has exactly four edge pieces 😊 Therefore the complexity is 5.

Your task is to find an illegal partial state with the highest possible complexity as defined above.

enter image description here

$\endgroup$
4
  • $\begingroup$ Cool puzzle. Do you have a way to measure complexity of an arbitrary partial state? $\endgroup$ – Dmitry Kamenetsky Mar 22 at 10:06
  • 1
    $\begingroup$ @DmitryKamenetsky an arbitrary partial state is either legal or illegal. If illegal then the above definition works. If it is legal then it doesn't make sense to ask what is the minimum number of stickers needed to prove it's illegal. In that case I would define the complexity as NaN :) $\endgroup$ – happystar Mar 23 at 10:49
  • $\begingroup$ One problem with this question is that every answer basically needs to be a separate question of finding the complexity of that particular configuration. Both answers that have been found so far have been found to have less complexity than the poster originally thought, and there's no good way to guarantee that we've found the minimum number. $\endgroup$ – Rob Watts Mar 23 at 15:56
  • $\begingroup$ Rubik's Cubes may have 4 edge pieces per colour, but do they add to 45? $\endgroup$ – user253751 Mar 23 at 21:22
1
$\begingroup$

I'm afraid I have to answer your question with a question of my own: how would you score this impossible cube:

enter image description here

$\endgroup$
2
  • $\begingroup$ [spoiler]Looks like 8 to me. There are four yellow edges and four orange edges, yet one of those should be the same piece, an orange/yellow for a total of 7 edges with orange OR yellow on.[/spoiler] $\endgroup$ – AdamV Mar 23 at 13:16
  • 1
    $\begingroup$ @AdamV you've considered 10 stickers - it's only a problem because we can see that the orange and yellow faces are adjacent. $\endgroup$ – Rob Watts Mar 23 at 15:59
1
$\begingroup$

I believe this one

enter image description here

scores

26 Currently 19, see @DiscreteLizarad's comment

Intended proof:

Once we know that red and yellow faces are adjacent the presence of 4 yellow and 4 red separate edge pieces will prevent a solution. If we fix the orientation suggested by the three mid pieces then the central corner shows the bottom must be blue. if the yellow face were opposite the red the right (orange blue) corner would have a hidden yellow. The two white-green corners have another hidden yellow. Plus three visibly yellow corners is one too many. Contradiction, meaning that yellow and red faces cannot be opposite. This has used all pieces except the back right green edge.

$\endgroup$
2
  • 2
    $\begingroup$ I think this one has complexity at most 19. Inspecting the 4 red and 4 yellow faces on separate edge cubes implies red is opposite to yellow. This implies red does not share a corner with yellow. (8 faces so far) Inspecting the 2 green-white and 2 blue-orange corner pairs implies there are at most 2 red faces on those 4 corners. (16 faces so far) Inspecting the 3 yellow corner faces implies these corners have no red face. (19 faces inspected) Hence, at most 2 of the (partially) visible corners have a red face, so the cube has at most 3 corners with a red face, so the cube must be invalid. $\endgroup$ – Discrete lizard Mar 23 at 9:56
  • $\begingroup$ @Discretelizard Yep, you found a shortcut. Let me see whether I can fix it. $\endgroup$ – loopy walt Mar 23 at 10:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.