Assuming two different (valid) configurations for a Rubik's Cube - A and B - does going from A to B directly have the same complexity as solving the cube?

Considering solving a cube can be done in 20 moves, we would never need more than 40 moves, considering: A -> [solved] <- B ...but what about A -> B?

If there is a difference, what makes the "solved" state special?

Bonus Round: Is there a(n online) solver where you can input two custom configurations? (My Google-skills didn't find one.)


2 Answers 2


There is nothing special about the solved state. Solving the cube means achieving a particular permutation of the cubelets' faces; that's true whether what you're trying to end up with is the state we call "solved" or some other state; God's number is just the maximum number of operations you need to do to achieve any achievable permutation of cubelet faces.

  • $\begingroup$ I was thinking about this, but it's not so clear-cut, in my opinion. To solve a Rubik's cube, the only restriction is that one of many permutations is achieved (as long as an entire face is colored the same, we're safe). This is not quite the same, is there a good way of thinking about it? (also upvoted) $\endgroup$
    – sunfishho
    Commented Jul 10, 2019 at 16:51
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    $\begingroup$ I think it is the same: if you know the colour of every cubelet face then you know exactly where everything is (other than rotating the centres of the faces, which can't be detected unless you have pictures on the faces or something). $\endgroup$
    – Gareth McCaughan
    Commented Jul 10, 2019 at 16:54
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    $\begingroup$ That's my issue as well ... I'm pretty certain it's the same. But I couldn't find anything on somehow re-mapping one of the solvers so that it would "think" it's solving the cube, but actually moving it to the intended configuration B. $\endgroup$ Commented Jul 10, 2019 at 16:59
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    $\begingroup$ @Orangenhain Since each permutation decides the position of each cubelet (and the rotation of each cubelet except the centers), which face goes in which position is decided uniquely by the target configuration. So, write T on all the faces you want to go on the top, B on all the faces you want to go on the bottom, etc., and then solve the cube looking at the letters (ignoring the actual colors). This will produce the desired configuration. $\endgroup$
    – Deusovi
    Commented Jul 10, 2019 at 17:10
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    $\begingroup$ @im_so_meta_even_this_acronym Imagine you had a Rubiks cube where you want to reach an end state. For that end state, paste little tiny stickers on the respective faces, such that the end state has the stickers representing a regularly solved rubiks cube. The puzzle is then equivalent to solving the cube based on the stickers, thus going from A to B has the same God Number as solving the cube. $\endgroup$
    – skyeriding
    Commented Jul 12, 2019 at 8:44

Corroborating the answer already given with more precise terminology: Let $G$ be the free group on the letters $R$, $U$, $F$, $B$, $R$, $D$ (so, just words consisting of these letters and their formal inverses, with concatenation + reduction being the group operation), and let $S$ be the set of all valid cube states. $G$ acts on $S$ in the natural way, producing a Cayley graph (adjacent nodes being those achievable from one another by a single move). Then God's number is just the maximum path-distance away from the solved state, on this graph.

The question posed in the OP is: What if we consider God's number as being relative to a general state, rather than just the solved state? This is the same notion, since any one node on the Cayley graph can be carried to another one arbitrarily specified, via a graph automorphism (namely, by an appropriate permutation, i.e. element of $G$).

Altogether it's like the (lack of) difference between asking for the furthest distance from $(1,0)$ on the unit circle, versus the furthest distance between two arbitrary points on the unit circle; there is a family of symmetries to allow us to fix one of our parameters.


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