# Communication via a Rubik's cube

I watched The Promised Neverland recently and after exploring a little, I found out that Norman used a 5x5x5 Rubik's cube (aka a professor's cube) to communicate when he was held at the $$\Lambda\text{-}7214$$ research facility. The image below describes the same.

I began wondering how one could possibly use a cube for communication. One idea struck me which involved using one row to encode one character. Here's the method I devised:

The image above shows a row of a 5x5x5 cube as seen from the front.
Here are the numbers associated with the six colours that constitute the cube:

• White: 1
• Blue: 2
• Orange: 3
• Green: 4
• Red: 5
• Yellow: 6

It's hard for me to explain how I came up with this or why it is what it is but I think stating some examples will help you understand it just fine.

Let's say that the character that we need to encode is at the $$x^{\mathrm{th}}$$ position in the alphabet. Here's what we need to do to encode it in a row.

First, we need to see if $$x\leq24$$. If yes, we follow algorithm $$A$$. If not, we follow algorithm $$B$$.

Algorithm $$A$$:

• Find $$\left\lceil\dfrac x6\right\rceil \overset{\mathrm{def}}{=} y$$ where $$\lceil k\rceil$$ gives the smallest integer greater than or equal to $$k$$.
• Find $$x-\left\lfloor\dfrac x6\right\rfloor\overset{\mathrm{def}}{=} z$$ where $$\lfloor k\rfloor$$ gives the greatest integer smaller than or equal to $$k$$. If $$z=0$$, make it $$6$$.
• Fill the $$y^{\mathrm{th}}$$ square from the beginning with the color corresponding to $$z$$ and fill the last square with the colour corresponding to $$y$$

We have devised a way to represent the first $$24$$ alphabets in this way. The remaining two will be covered via Algorithm $$B$$.

Algorithm $$B$$:

• Paint the last square with the colour corresponding to $$4+(x-24)=x-20$$. So, if $$x=25$$, paint the last square with the color red, if it's $$26$$, paint it with the color green.

Examples:

If we need to encode $$S$$, we first find the value of $$x$$ which comes out to be $$19$$. Also, $$y=4$$ and $$z=1$$. $$S$$ is denoted by:

(Black: not relevant)

Similarly, the word TEST will be denoted by:

To decrypt a row, let $$x$$ be the number corresponding to the color of the rightmost square and for $$x\leq4$$, let $$y$$ be the number corresponding to the $$y^{\mathrm{th}}$$ square from the left. If $$x>4$$, then for $$x=5$$, the alphabet is $$Y$$ and for $$x=6$$, the alphabet is $$Z$$.

For $$x\leq4$$, the alphabet is the one at the position $$6(x-1)+y$$

What I want to know about is the potential drawbacks of this method. One is, obviously that three out of the five "squares" in each row are wasted. Another drawback is that it's hard to show more than 5 words on the whole cube but that doesn't seem to be a problem with the method.

Also, what are some alternatives to this? What are some other ways that you have heard of to accomplish this and how do they compare with this method?

PS: This is my first question on Puzzling SE. So, if I was unable to make my method clear, please let me know so that I can edit the question for the same. I don't know if what I talk about is something on-topic here. If it is and it interests people and making the question longer is fine, I would very much like to elaborate. Edits on the tags are very welcome.

• To be clear, do you not care about encoding the rotation? That is, If you pass me the cube, I would need to examine each side and every rotation, until I find the one that works (that can be decoded into something that makes sense) ? Commented Dec 22, 2020 at 20:12
• Maybe you could use something like the Nihilist cipher but without the part of adding the numbers. Then, you have a way to convert letters into pairs of numbers from 1 to 5. This allows you to use 5 colors to write 12 letters on one side of the cube, leaving you one extra color and one extra square to encode the rotation. Commented Dec 22, 2020 at 20:20
• Each face can encode $6^25$ possibilities, which is equivalent to 13 English characters. Commented Dec 23, 2020 at 2:04
• @DmitryKamenetsky That sounds interesting! How did you calculate the number of possibilities, by the way? As far as I know, if we were simply finding out the number of ways to arrange the colours on a face, the number of arrangements would be much more... Commented Dec 23, 2020 at 15:01
• @RajdeepSindhu Ciphers that use 5 by 5 matrices like nihilist, playfair, collon, ... usually remove the letter J to be able to fit all letters, i.e. declare J=I as same letters. If someone wants to turn my comment into an answer, I don't mind. (Is every face of possible color permutations actually reachable by cube rotations is for what I haven't found a reference yet.) Commented Dec 23, 2020 at 15:16