It's now known that to reach the solved position of the rubik cube from any other position you need at most $20$ moves (a $180^{\circ}$ of a face is counted as one move , not two ) .

See for example here : http://www.cube20.org/

Now I'll define $d(A,B)$ be the distance between the two positions of the cube $A$ and $B$.That is the minimal number of moves to reach $B$ from $A$ .

Let's denote with $P$ the set of all the possible positions and with $\phi$ the solved position .

From the above :

$$\max_{A \in P} d(A,\phi)=20$$

Now my question is :

What is $$\max_{A,B \in P} d(A,B)$$ That is : How far two positions can be from each other ?

What I have :

It's obvious that $d(A,B) \leq d(A,\phi)+d(B,\phi)$ (triangle inequality :))

So $$d(A,B) \leq 20+20=40$$ but I kind of doubt that $40$ is the maximum .

My intuition is that it's in the $20$-$30$ range .

What do you think?

Thanks in advance for all the help !

  • 2
    $\begingroup$ What is so special in the ''solved'' position? It seems to me that any position can be assumed as a solution so the maximum distance from any two positions is $20$. I'm wrong? $\endgroup$ – Emilio Novati Oct 16 '15 at 14:42

Let $G$ the Rubik's cube group of legal moves of the Rubik's cube under concatenation. Then there is a one-to-one correspondence between the set $P$ of legal positions of the Rubik's cube, and the group $G$, by associating to every position $X\in P$ a concatenation of moves $\varphi(X)\in G$ leading from the starting position to $X$. This immediately yields a distance function $d'$ on $G$ defined by $$d'(g,h):=d(\varphi^{-1}(g),\varphi^{-1}(h)),$$ for all $g,h\in G$. Now it is clear that for all $g_1,g_2,h\in G$ we have $$d'(g_1,g_2)=d'(g_1h,g_2h),$$ because a series of moves leading from $\varphi^{-1}(g_1)$ to $\varphi^{-1}(g_2)$ also leads from $\varphi^{-1}(g_1h)$ to $\varphi^{-1}(g_2h)$, and vice versa, as such series of moves represent $(g_2h)(g_1h)^{-1}=g_2g_1^{-1}\in G$. Hence for all $A,B\in P$ we have \begin{eqnarray*} d(A,B)&=&d'(\varphi(A),\varphi(B))=d'(\varphi(A)\varphi(B)^{-1},\varphi(B)\varphi(B)^{-1})\\ &=&d'(\varphi(A)\varphi(B)^{-1},\varphi(\operatorname{id}_G))=d(AB^{-1},\phi), \end{eqnarray*} which shows that $d(A,B)\leq20$ holds for all $A,B\in P$.

On a more conceptual note; considering the 'shifted' distance function $$d''(A_1,A_2):=d'(\varphi(A_1)\varphi(B)^{-1},\varphi(A_2)\varphi(B)^{-1}),$$ which I implicitly use in the last part of my answer, comes down to considering $B$ as the 'solved state' of the Rubik's cube in stead of $\phi$. The rest f the argument then says that a series of moves leads from $A$ to $B$ if and only if it leads from "$AB^{-1}$" to $\phi$. This is also the basic idea of the other answer by Mike Haskel.


By symmetry, no two positions can be more than 20 steps apart. There's nothing special about the "solved" configuration: the same argument that shows you can get from any starting $A$ to the solved state in no more than 20 steps shows that you can get from any starting $A$ to any fixed $B$ in no more than 20 steps.

  • $\begingroup$ I like your answer. Simple, and straight to the point! No rigor needed. $\endgroup$ – Christopher Mowla Oct 17 '15 at 12:52

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