The Rubik's Twist has been a fun time sink. From the wiki page,

[It] is a toy with twenty-four wedges that are right isosceles triangular prisms. The wedges are connected by spring bolts, so that they can be twisted, but not separated. By being twisted, the Rubik's Snake can be made to resemble a wide variety of objects, animals, or geometric shapes. Its "ball" shape in its packaging is a non-uniform concave rhombicuboctahedron.

This page has an applet to show configurations in 3D and play around with it, which will give you an idea of what the puzzle is like (requires Wolfram CDF player plugin.)

A friend asked what shape would minimize surface area and we thought the standard 'ball' shape would be the solution:

We denoted the square side of the wedge as $1u^2$ and thus the long rectangular side of the wedge is $\sqrt{2}u^2$ and a triangle is $\frac{1}{2}u^2$.

In the ball configuration, we counted 12 rectangle panels showing (blue, orange, and red in the above image) 6 "squares" (12 triangles) and 8 niches each made of 3 triangles giving:

  • 12 + 24 = 36 triangles ($18u^2$)
  • 12 rectangle panels ($12\sqrt{2}u^2$)

$18u^2 + 12\sqrt{2}u^2$ surface area, which is just under $35u^2$.

However, we found that there are configurations with smaller surface areas like:

We calculated the surface area as just under $32u^2$ (12 rectangular panels, 2 squares and 26 triangles.)

The definition of surface we are using is any visible face - the area which would have to be painted if someone were to hold it and look all around it, no area would appear unpainted. (See the comments for extra clarification.)

In the ball shape (see first image) there is a 1x1x1 void in the center which is not counted as surface area as it is not visible. I'm not sure if that makes the problem much more difficult, however, I'd be interested to know of a solution in either case.

I calculate the surface area by counting up the number of visible rectangular panels, ($\sqrt{2}u^2$) triangles ($\frac{1}{2}u^2$) and the two square ends of the snake ($1u^2$ each.) By adding these units up I get the surface area.

So, for the ball we are excluding the $6u^2$ on the inside. Additionally, partial turns that expose the "inside" of the snake are not allowed (see below) as these are not valid configurations.

(Partial turn exposing inside grey face)

The ball shape had a lot of outcroppings which add up. We thought that another way of minimizing surface area is by maximizing touching faces. What configuration has the lowest visible surface area, and can it be derived or proved mathematically?

Or, can the configuration above be proven to exhibit the lowest surface area?

Or, can a computer program be written to iterate over (or in some way consider) all combinations (of which there can be no more than $4^{23}$ simply by observing each of the 23 twists can exist in 4 positions, although some positions are physically impossible as it overlaps itself) and find the solution?

  • $\begingroup$ This likely depends on what you are considering "surface". For example, do you consider the outer envelope of the resulting figure, i.e. the empty faces in this octahedron, or only the "empty-facing" faces of the Twist? $\endgroup$ – A.P. Apr 27 '15 at 19:17
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    $\begingroup$ This still isn't compltely clear: in the "ball" picture you posted how do you consider the grey space between, say, the blue, orange, and red faces? Do you count it as three triangles (thus considering the "hole") or do you count it as one triangle (thus considering the surface of a rhombicuboctahedron)? $\endgroup$ – A.P. Apr 27 '15 at 20:53
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    $\begingroup$ I still don't know how to answer your question, but I can tell you that there are $6,770,518,220,623$ ($\approx 4^{21.81}$) different configurations. Further, there are programs that simulate the Rubik Twist, so with enough time it may be possible to run an exhaustive search. $\endgroup$ – A.P. May 9 '15 at 11:14
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    $\begingroup$ For the 'extended' definition (where internal holes are taken into account), that computation relatively easy: the total surface area of all faces, minus twice the surface area of any faces 'shared' in the configuration. The goal should then be to maximize the number of shared faces. $\endgroup$ – Steven Stadnicki Mar 31 '16 at 18:54
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    $\begingroup$ I found another NOT closed loop solution of $(15+12\sqrt{2})u^2 = 31.97 u^2$. Solution: (003)2(10203)2(31)2(123)2(1003)2(3). It seems to be as close to a $(2 \times 2 \times 2) u^3$ cube as possible. $\endgroup$ – Cuc Jun 20 '17 at 1:09

I do not know the Rubix's Twist well enough to determine possibilities and what not. However, I do know how to compute the minimum surface area of a cubic region.

You have two equations (note that $V$ is a volume constant selected beforehand):

$V = L\cdot W \cdot H \rightarrow \frac{V}{WH} = L$

$2LW + 2LH + 2WH = P \rightarrow 2V/H + 2V/W + 2WH = P$

$P$ is now a variable defined by the above two-arguments function of $H$ and $W$. Using multi variate calculus techniques, one can find the minima of the function. Those minima will give the ideal length, width, and volume (of the rectangular prismic form) such that the surface area is minimal. Of course, there might be shapes other than rectangles that have less surface area, but I doubt it.

There are dozens of max min tests in $3$-space but in general the minimums occur where the gradient is the $(0,0)$ vector.

$F_w(W,H) = -4V/W^2 + 2H$

$F_h(W,H) = -4V/H^2 + 2W$

Places where the gradient is $0$ aren't apparent to me, but I'm sure this is trivial to look at. I hope this helps. Note that this won't give you the combination of the Rubix's Twist, but it will instead give the ideal cubic to try and build from the shape.


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