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I've been working on a problem related to the 3x3x3 Rubik's Cube where you allow faces to be turned by $45^\circ$ instead of just the usual $90^\circ$. We know for the standard 3x3x3 the cube is sliced up into 27 regions and 26 of them are the external pieces of the puzzle (8 corners + 12 edges + 6 face centers).

For the $45^\circ$ Rubik's Cube, when you turn a face by $45^\circ$ there have to be additional cuts added to enable the adjacent faces to turn. One interesting property though is that no matter how many cuts you add, it seems like there are still configurations where faces are be blocked unless more cuts are added. The twisty puzzle community calls this property jumbling when an infinite number of cuts are required. I'm trying to prove that the $45^\circ$ Rubik's Cube jumbles.

Here is an example of a Rubik's Cube where a few cuts have been added to enable some $45^\circ$ turns before things get blocked. In this photo the left face with the yellow octagon center was twisted $45^\circ$ and then the right face with the red octagon center was twisted $45^\circ$. Now the yellow face can not be twisted any further without adding more cuts through the pieces that are blocking it: 45 degree Rubik's Cube after two twists

I call the above sequence [45F 45R].

To go about trying to prove that an infinite number of cuts are require, I set out to show something simpler that implies that. Namely, that a point on the puzzle gets rotated around some axis by an irrational angle if you repeat [45F 45R]. If repeating these two moves over and over rotates points by an irrational angle, then the order of the sequence (the period) is infinite and an infinite number of cuts must be added, implying that the puzzle jumbles. Here is an animation showing how the corner point moves when you do [45F 45R] repeatedly: corner point rotating animation

So I build the rotation matrix for 45F and 45R and found the combined rotation matrix:

$$\left( \begin{array}{ccc} \sqrt{2}/2 & \sqrt{2}/2 & 0 \\ -1/2 & 1/2 & -\sqrt{2}/2 \\ -1/2 & 1/2 & \sqrt{2}/2 \end{array} \right)$$

If you do some additional work on that matrix you'll find that the axis points are being rotated around is $$[-1,\ -\sqrt{2} + 1,\ 1]$$ and the angle points are rotated through is $$\arccos ( \frac{\sqrt{2}}{2} - \frac{1}{4} )$$

Here is a plot of several different points through a few hundred applications of [45F 45R] and a plot of the axis computed from the rotation matrix: the orbit of several points about the axis

All that is left now is to prove that $\theta = \arccos ( \frac{\sqrt{2}}{2} - \frac{1}{4} )$ is an irrational angle. By irrational angle I mean that $\theta$ is not of the form $\frac{m}{n}\pi$.

The strategy I've tried is to assume that $\theta = \frac{m}{n}\pi$ and then show that leads to a contradiction. This seems like a good way to go because $\cos(n\theta) = T_n(\cos(\theta))$ where $T_n$ is the $n^{\mathrm{th}}$ Chebyshev polynomial. This is simple because $n\theta = n\frac{m}{n}\pi = m\pi$ where $m$ is an integer. $\cos(m\pi) = \pm 1$

Since we know $\cos(\theta) = \frac{\sqrt{2}}{2} - \frac{1}{4}$ if we can show $T_n(\frac{\sqrt{2}}{2} - \frac{1}{4}) \ne \pm 1$ for any $T_n$ then $\theta$ can't be of the form $\frac{m}{n}\pi$ which is a contradiction and would prove that the angle is irrational.

Unfortunately I'm stuck. I've tried expanding out various Chebyshev polynomials and substituting $y$ for $\sqrt{2}$ plugging in $\frac{y}{2} - \frac{1}{4}$ and it certainly seems like the odd powers of $y$ could never fully cancel out but I can't work out the details.

So, is there a way to prove $\arccos ( \frac{\sqrt{2}}{2} - \frac{1}{4} )$ is an irrational angle?

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  • $\begingroup$ Nice question! +1 $\endgroup$ – Rogelio Molina Jul 6 '15 at 23:49
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    $\begingroup$ There's a proof at arxiv.org/abs/1006.2938. Someone might want to write it up as an answer... $\endgroup$ – Chris Culter Jul 7 '15 at 0:54
  • $\begingroup$ @ChrisCulter thanks for the link. I actually read through this paper a few times yesterday while trying to answer the question myself. There was a lot of content I didn't understand and algebraic integers was one of those things. I didn't realize they weren't just an integer. Did you have some aspect of that paper in mind other than the algebraic integer bit Will Jagy used in his answer? $\endgroup$ – Brandon Enright Jul 7 '15 at 2:19
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book http://www.amazon.com/Irrational-Numbers-Carus-Mathematical-Monographs/dp/0883850389

page 37, Theorem 3.9, due to Lehmer: given integers $n > 2$ and $k,$ then $$ 2 \cos \left( \frac{2k \pi}{n} \right) $$ is an algebraic integer of degree $\phi(n)/2.$

Twice your cosine is $$ t = \sqrt 2 - \frac{1}{2}. $$ This satisfies $$ 4 t^2 + 4 t - 7 = 0 $$ and is an algebraic number but not an algebraic integer. See https://en.wikipedia.org/wiki/Algebraic_integer#Non-example

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  • $\begingroup$ Out of curiosity, did you know of this property before or did you search it out after reading the question? $\endgroup$ – gabe Jul 7 '15 at 1:03
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    $\begingroup$ The point is that $2cos(2kπ/n)$ is the sum of two roots of $1$, which are algebraic integers. $\endgroup$ – Lev Borisov Jul 7 '15 at 1:16
  • $\begingroup$ I'd read through quite a bit of Niven's book trying to answer this myself. Most of it was over my head though. I didn't realize an algebraic integer wasn't just "an integer". One question though: you showed $\sqrt(2) - \frac{1}{2}$ is a root of that polynomial but is that sufficient to show that it isn't an algebraic integer. Is there some reason why it couldn't also be the root of some other monic polynomial? $\endgroup$ – Brandon Enright Jul 7 '15 at 2:16
  • $\begingroup$ @BrandonEnright editted in at the end $\endgroup$ – Will Jagy Jul 7 '15 at 2:21
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    $\begingroup$ @WillJagy indeed. Thanks again. I'm glad your proof is so simple and easy to understand! You've made a small subset of the nerdy puzzle community quite happy :-) $\endgroup$ – Brandon Enright Jul 7 '15 at 3:14

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