Is it possible to know the entire configuration of a Rubik's cube looking at only two sides and not rotating the cube? In other words: what is the minimum information required to create a two-dimensional map of a $3\times 3\times 3$ Rubik's cube?
[Math] Can a Rubik’s cube be mapped knowing only two sides
recreational-mathematicsrubiks-cube
Related Solutions
Yes, the result is very mathematical, though computers were used for many of the calculations. Tomas Rokicki, Herbert Kociemba, Morley Davidson, and John Dethridge proved this result in July 2010 and put up a very nice webpage at cube20.org, with some history of the problem and their basic methods.
The team consisted of a programmer, a math teacher, a mathematician, and an engineer. There are two parts to the problem: find a method that lets you solve any cube in M moves or less, and find a really hard starting position, that takes at least N moves to solve. These are upper and lower bounds for the problem-- the goal is to keep finding better methods for solving cubes, or to keep finding tougher starting positions, until you can show M=N.
In 1981, it was known that some positions took at least 18 moves, and Morwen Thistlethwaite had a method to solve any cube in 52 moves or less. The website lists the progress over the last few decades, narrowing in on the number 20. Part of the progress comes from having more powerful computers to work with, but a lot of it is having a better idea (math!) to simplify the problem.
Here's a cool table (copied straight from cube20.org) showing how many starting positions require 20 moves, 19 moves, 18 moves, etc. You can see there are many, many starting positions; and it's actually really tough to find one that takes 20 moves to solve.
Distance Count of Positions
0 1
1 18
2 243
3 3,240
4 43,239
5 574,908
6 7,618,438
7 100,803,036
8 1,332,343,288
9 17,596,479,795
10 232,248,063,316
11 3,063,288,809,012
12 40,374,425,656,248
13 531,653,418,284,628
14 6,989,320,578,825,358
15 91,365,146,187,124,313
16 about 1,100,000,000,000,000,000
17 about 12,000,000,000,000,000,000
18 about 29,000,000,000,000,000,000
19 about 1,500,000,000,000,000,000
20 about 300,000,000
As for bigger cubes, there are many methods for solving bigger cubes already. It's actually not that much tougher than solving the 3x3x3 cube; it just takes longer. But I don't think many people will work on finding the optimal number of moves for bigger cubes-- it will be much harder (probably not possible with current computers) and is not as interesting (because everyone cares most about the 3x3x3 cube).
You can certainly achieve a rotation of side as a composition of moves of the other sides only. The most instructive way to see this is to take your favorite algorithm for solving the cube from an arbitrary position and figure out how to modify it such that it never turns the first side you solve. (Whether this is easy depends on what your favorite algorithm is, of course. My own usual one is almost there except for the initial-cross stage and the final turning of the top corners, both of which are easily replaced). Then turn your chosen side by one quarter, and use the modified algorithm to solve from that position. Whatever moves you make during the solve will add up to a "turn this side without turning it" combination.
Also, 5 is the minimum number of single-side turns that generate the group, because restricting movement to any set of 4 sides will either leave one edge unmovable or be unable to flip an edge to the opposite orientation.
A full set of relations between 5 basic quarter-turns would be absolute horrible, though.
If the generators are not required to be simple moves, we don't need as many as 5 of them. It's easy to get down to 3 quite complex operations:
- $\alpha$, a cyclic permutation of 11 of the 12 edges simultaneously with 7 of the 8 corners.
- $\beta$, swaps the edge that $\alpha$ doesn't touch with another one, and swaps the corner that $\alpha$ doesn't touch with another one.
- $\gamma$, turns two corners and flips two edges.
Here there's some hope of being able to write down some reasonably natural relations, but I haven't tried to see whether the details work out.
I'm not sure if we can get down to two generators -- could the role of $\gamma$ be folded into $\beta$, perhaps? Edit: The next time the question came up (and I had forgotten this one existed) I tried anew and came up with a blueprint for a two-generator solution: Minimal generating set of Rubik's Cube group
One generator is of course not possible, since the group is not abelian.
Best Answer
Take a solved cube and flip all four edges with a red face. Now you can permute those edges freely without the effect being visible on any other side than the one with the red center. So even seeing five sides of the cube is not enough to reconstruct the last one.
(And by "freely" I mean that any even permutation of the edges can be reached by legal cube moves, of course).