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 !
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.