So the complete conditions for this scrambled cube can be described alternatively as the following:

  1. Each face has less than 3 squares of the same color, which leads to

1a. Each face has 5 or 6 different colors.

(2. It's a huge bonus if the same-color squares are not adjacent to each other. In other words, they don't share an edge.

2a. It's even better if they don't share a vertex!)

If it's possible, could you please provide pictures or sequence algorithms? Thank you!


Yes it is possible!

Do you know how to get it into a checkered state? Looking at any face you then do as follows....

Keep the same face towards you the entire time.

Turn left side clock-wise

Turn right side clock-wise

Turn top side clock-wise

Turn bottom side clock-wise

Turn left side clock-wise

I don't have enough reputation to post more than 2 pictures.

I now have enough reputation for more then 2 images

enter image description here enter image description here

There are a couple colors that are sharing edges but there are no more then 2 of the same color per side

Let me know if you need more information.


I had some more time to mess around with it and have the answer for Q2 but not Q2a. There is only one difference

The last step.. Instead of doing Left side clock-wise do left side counter clock-wise

enter image description here enter image description here

  • $\begingroup$ Checkered meaning this i.ytimg.com/vi/yzEv4aQBb7I/hqdefault.jpg $\endgroup$
    – Saloaty
    Jun 29 '17 at 20:33
  • $\begingroup$ This is cool! I never knew... $\endgroup$
    – undo
    Jun 29 '17 at 20:56
  • $\begingroup$ Thank you! For now conditions 1 & 1a are fulfilled. Can you or anyone solve condition 2? My uneducated guess is that condition 2a is even several times tougher than 2! $\endgroup$
    – 12tn2
    Jun 30 '17 at 8:02
  • 1
    $\begingroup$ Yes! I've found the steps from a solved cube as: D R' L' U D R2 L2 F2 B2. Now we only have the obstacle of 2a left ;) $\endgroup$
    – 12tn2
    Jul 1 '17 at 3:00
  • $\begingroup$ Oh nice! That one will for sure will be more difficult. $\endgroup$
    – Saloaty
    Jul 1 '17 at 3:08

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