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11

2

This is a question about Rubik's Cube and generalizations of this puzzle, such as Rubik's Revenge, Professor's cube or in general the $n \times n \times n$ cube.

Let $g(n)$ be the smallest number $m$, such that every realizable arrangement of the $n \times n \times n$ cube can be solved with $m$ moves. In other words, this is the "radius" of the Cayley graph of the $n \times n \times n$ cube group with respect to the canonical generating system.

We have $g(1)=0$, $g(2)=11$ and - quite recently - in 2010 it was proven that $g(3)=20$: God's number is 20.

Question. Is anything known about $g(4)$ or $g(5)$?

I expect that the precise number is unkwown since the calculation for Rubik's cube already took three decades. Nevertheless, is there any work in progress? Are any lower or upper bounds known?

I would like to ask the same question about $g(n)$ for $n>5$, or rather:

Question. Is anything known about the asymptotic value of $g(n)$?

edited Jan 14 2012 at 20:01

Goldstern
5,250●1●12●27

asked Oct 11 2011 at 16:24

Martin Brandenburg
29.1k●2●45●129

2

Perhaps it should be mentioned that this result is for the half-turn metric. In light of the comments at matthewkahle.wordpress.com/tag/gods-number, it appears that the situation for the quarter-turn metric will prove to be more elegant mathematically, where the answer is expected to be 26 with a unique position requiring 26 moves to solve. (Also, why restrict to three dimensions? Let's ask for the asymptotics for God's number for the n^k cube). – Peter McNamara Oct 11 2011 at 20:19

2

Thanks Peter; I didn't know that 20 refers to the half-turn metric. And now I'm rather disappointed that the number for the quater-turn metric (which seems much more natural to me) is still unknown! – Martin Brandenburg Oct 12 2011 at 12:29

Theo Johnson-Freyd · Answer 1 · 2011-10-11 16:55:02Z UTC

10

I am not an expert, but I remember seeing this press release from MIT not too long ago. The corresponding arXiv article should answer your second question.

answered Oct 11 2011 at 16:55

Theo Johnson-Freyd
22.3k●2●27●110

9

Just to make the answer to Q2 explicit: $g(n)= \Theta(n^2 / \log n)$. – Joseph O'Rourke Oct 11 2011 at 17:36

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This is fantastic! I will try to read the paper. This indeed answers Q2. I should have asked Q1 separatedly. – Martin Brandenburg Oct 11 2011 at 19:56

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Erik Demaine mentioned this result last week at his public lecture "Geometric puzzles: Algorithms and complexity" in the Joint Meetings. math.harvard.edu/~elkies/Pix/jmm12/demaine5.jpg – Noam D. Elkies Jan 14 2012 at 20:09

God’s number for the $n \times n \times n$-cube

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