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Is it possible to "solve" a Rubik's Cube (the original with 3x3 elements per side and six sides) to make it look like a die (each side showing a different die pattern from 1 to 6, opposite sides sum up to 7)?

(dice patterns are: [·] [.'] [⋰] [: :] [:·:] [:::])

With "solve" I mean to use one of the well known solutions to solve the cube level by level.

Looking closer I thought it would be easier to "make" the 1 last, i.e. on the last level of the cube. And because it might be impossible to make a side "uni-colored", it may consist of many different colors per side as long as the dice pattern is still visible. As you can see in this picture Example Picture
(this picture shows the [:·:] in orange, because there aren't enough white panels left)

Any ideas? Happy solving! ;-)

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    $\begingroup$ What counts as the dice pattern being "visible" exactly? It's a bit unclear, and I'm not sure how to tell if something looks like the correct number. Is it something like "each face is only interpretable as one of the six standard layouts ([·] [.'] [⋰] [: :] [:·:] [:::])? $\endgroup$
    – Deusovi
    Aug 27 at 6:14
  • $\begingroup$ @Deusovi - Yes exactly, recognizable as one of the standard patterns will be sufficient no matter which colors it consists of. And even the patterns may be mixed with multiple colors, e.g yellow orange and red. As long as you can "see" the pattern. $\endgroup$
    – Matt
    Aug 27 at 6:23
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    $\begingroup$ Right, what counts as being able to "see" it? If I have a face with [RYO / GBG / OYR], does that mean you can "see" a 6 on it (even though you can also see a 4 if you just look at red and orange, or a 2 if you just look at red)? $\endgroup$
    – Deusovi
    Aug 27 at 6:36
  • $\begingroup$ @Deusovi - Good question. I would say, in the example you gave, it is a 4 because GBG "interrupts" the 6 pattern (if you think of a rainbow, then green and blue are too far away from yellow and red). Regarding the two, it is more difficult to decide, but the rule "sum of opposite sides is 7" makes it unique again. So as long as the sum is correct, you can interpret this rule in a more relaxed way. ;-) $\endgroup$
    – Matt
    Aug 27 at 8:03
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I will assume that the pips of a number do have to be the same colour, with the background an arbitrary mix of the other 5 colours, and that the six numbers must be on six different faces (e.g. you cannot combine the 1 and 4 on one face).

Then this is not possible.

Proof:

The 4,5,6 have their corners the same colour.

Suppose those three faces are adjacent around a cube corner like on a normal die. Any corner piece of the cube that has a square in two (or three) of those faces is uniquely determined by those two (or three) colours, so this fixes four of the corner pieces. Now each of the three faces has three of its corners already determined, leaving only one piece left of the correct colour to fill its fourth corner. In this way 7 of the cube's 8 corner pieces are forced to be in their correct relative positions. It is then impossible to make the diagonal corner pattern of the 2 and 3.

If the 4,5,6 are not adjacent around a cube corner, then there is one opposite pair with the third adjacent to both. The corner pieces of that third face will be in their correct relative positions (again because each corner piece lies in two of the faces and and is therefore uniquely determined by the two colours). This makes it impossible to create the diagonal corner pattern for the 2 or 3 on any of the four adjacent faces. There is no room left on the cube for both 2 and 3, so here too it is impossible to make all six numbers.

Suppose instead that we make the background a single colour, and allow the pips to be a mix of any of the other 5 colours.

This too is impossible.

The numbers 1,2,3,4,5 all must have their four edges the same colour, so the 6 which needs at least 2 colours for the edges is not possible.

So there will have to be faces where the pips have two or more colours, and faces where the background has two or more colours.

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  • $\begingroup$ Yes I think that having the dice patterns in a unique color is impossible, but how about mixing them with similar colors as discussed with @Deusovi (comments below the question)? It is even allowed to have different background colors as long as you can still clearly recognize the dice patterns. $\endgroup$
    – Matt
    Aug 27 at 8:09
  • $\begingroup$ So if you allow more than 2 colors for the pips and/or for the background, it is solveable. Right? $\endgroup$
    – Matt
    Aug 27 at 11:51
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    $\begingroup$ "The 4,5,6 have their corners the same colour. If those three faces are adjacent around a cube corner like on a normal die, then this forces the 7 of the cube's 8 corner pieces to be in their correct relative positions." Could you please expand on that assertion? I don't think I understand its meaning or why it's true. $\endgroup$
    – Stef
    Aug 27 at 13:54
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    $\begingroup$ @Stef I have expanded my answer a bit. I hope this makes it clearer. $\endgroup$ Aug 27 at 14:31
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    $\begingroup$ To summarize, you have described the preconditions so far. If we take the "multicolored" version, how would such a solution look like? And if the solution requires to disregard the "opposite faces sum up to 7" rule, how would the setup look like? $\endgroup$
    – Matt
    Aug 27 at 15:13
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Piece of cake...if we are "a bit lenient" (cough) about the "dice look"... (hard cough) ...and, additionally, employ, uh, "selective blindness"... (hard persistent cough)

Yes, Jaap Scherphuis already answered that a proper solution is impossible (and, of course, the first thing I did was to upvote his answer; his answer should be the accepted answer). But workarounds are, in this case, kind of easy. First I though of some sort of binary coding...but that's too complicated.

So I chose white, went to some online Rubik's Cube simulator and did five moves:

Five moves

On top, you see one (1) white square. This is, obviously, the (1) side of your die. The (6) side is, of course, following the "7 rule", on the bottom (not shown).

To the right, you see three (3) white squares. This is, obviously, the (3) side of your die. The (4) side is, of course, following the "7 rule", on the opposite side (not shown).

To the left, you see five (5) white squares. This is, obviously, the (5) side of your die. The (2) side is, of course, ...ah, you guessed it already.

Now, there is, of course, some minor "cosmetic issue". If you happen to look at the (2), (4) or (6) side, you won't see any white pieces (so "x" is unknown). But don't despair. As you happen to know the "7 rule", you can just lift the cube, look at bottom (which will yield value "y"), and, with a bit of math, you can solve the x=7-y equation to yield "x".

Well, technically, you don't need to lift the cube, you can just look at the five exposed sides to determine what's on the bottom.

Important notice: this is meant to be a "fun answer", as it does not meet the "look like a die" criteria. But it might help you if you want to play "Schocken" and lack the proper three dice to do so...but, luckily, find three guys in the pub who happen to carry their Rubik's cube with them all the time.

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  • $\begingroup$ Wow, lol ... and how do you show up the answer of all questions (42) on a standard Rubik Cube? If you don't know, ask the guys at the pub the world's end ... :-D $\endgroup$
    – Matt
    Aug 27 at 19:11
  • $\begingroup$ Or you could use a black permanent Edding pen (black works best even on blue and green) and draw on each side one of the patterns [·] [.'] [⋰] [: :] [:·:] [:::] - this works for each setup of the cube. If you want to make it difficult, make a random pattern on the cube, then draw the dice patterns, then solve it (you know showing the right colors on each side) and give it to some guy at the pub near you, and ask him to restore the dice patterns. $\endgroup$
    – Matt
    Aug 27 at 19:46
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    $\begingroup$ You don't need to look at the other side of the cube to find what even number you've thrown. Yellow is opposite white on a solved cube (and in this case this is conserved even after the moves you've done) so subtract the number of yellows you see from 7 to get your score. $\endgroup$ Aug 28 at 16:25
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    $\begingroup$ +1 for the humoristic answer :-D $\endgroup$
    – Matt
    Aug 29 at 8:10

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