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Imagine an $n \times n \times n$ Rubik's cube, where we can transition the state of the cube using Singmaster moves under either the face- or quarter-turn metrics. We call a face of this cube "solved" if all of the symbols on the face are of the same color (there are six total colors).

How many total cube states are there when $k = {0, 1, 2, 3, 4}$ of the cube faces are solved, and what "distribution" are these states drawn from? Namely, what is the probability that they contain the fully solved state of the cube? Is this probability what one would expect for a uniform random sampling from the set of all possible cube states?

Note: This is (hopefully) a simplification of an earlier question asking for the probability that a greedy algorithm solves a Rubik's cube, where once a face is solved, the algorithm cannot backtrack and perturb the face (though rotations of the face are allowed).

Note 2: I know there are only going to be a small number of possible cube states when we freeze a face. Exact counts would be fantastic, but my goal here is to understand what distribution they are being drawn from.

Imagine an $n \times n \times n$ Rubik's cube, where we can transition the state of the cube using Singmaster moves under either the face- or quarter-turn metrics. We call a face of this cube "solved" if all of the symbols on the face are of the same color (there are six total colors).

How many total cube states are there when $k = {0, 1, 2, 3, 4}$ of the cube faces are solved?

Note: This is (hopefully) a simplification of an earlier question asking for the probability that a greedy algorithm solves a Rubik's cube, where once a face is solved, the algorithm cannot backtrack and perturb the face (though rotations of the face are allowed).

Imagine an $n \times n \times n$ Rubik's cube, where we can transition the state of the cube using Singmaster moves under either the face- or quarter-turn metrics. We call a face of this cube "solved" if all of the symbols on the face are of the same color (there are six total colors).

How many total cube states are there when $k = {0, 1, 2, 3, 4}$ of the cube faces are solved, and what "distribution" are these states drawn from? Namely, what is the probability that they contain the fully solved state of the cube? Is this probability what one would expect for a uniform random sampling from the set of all possible cube states?

Note: This is (hopefully) a simplification of an earlier question asking for the probability that a greedy algorithm solves a Rubik's cube, where once a face is solved, the algorithm cannot backtrack and perturb the face (though rotations of the face are allowed).

Note 2: I know there are only going to be a small number of possible cube states when we freeze a face. Exact counts would be fantastic, but my goal here is to understand what distribution they are being drawn from.

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Imagine an $n \times n \times n$ Rubik's cube, where we can transition the state of the cube using Singmaster moves under either the face- or quarter-turn metrics. We call a face of this cube "solved" if all of the symbols on the face are of the same color (there are six total colors).

How many total cube states are there when $k = {0, 1, 2, 3, 4}$ of the cube faces are solved?

Note: This is (hopefully) a simplification of an earlier question asking for the probability that a greedy algorithm solves a Rubik's cube, where once a face is solved, the algorithm cannot backtrack and perturb the face (though rotations of the face are allowed).

Imagine an $n \times n \times n$ Rubik's cube, where we can transition the state of the cube using Singmaster moves under either the face- or quarter-turn metrics. We call a face of this cube "solved" if all of the symbols on the face are of the same color (there are six total colors).

How many total cube states are there when $k = {0, 1, 2, 3, 4}$ of the cube faces are solved?

Note: This is (hopefully) a simplification of an earlier question asking for the probability that a greedy algorithm solves a Rubik's cube, where once a face is solved, the algorithm cannot backtrack and perturb the face.

Imagine an $n \times n \times n$ Rubik's cube, where we can transition the state of the cube using Singmaster moves under either the face- or quarter-turn metrics. We call a face of this cube "solved" if all of the symbols on the face are of the same color (there are six total colors).

How many total cube states are there when $k = {0, 1, 2, 3, 4}$ of the cube faces are solved?

Note: This is (hopefully) a simplification of an earlier question asking for the probability that a greedy algorithm solves a Rubik's cube, where once a face is solved, the algorithm cannot backtrack and perturb the face (though rotations of the face are allowed).

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Counting the number of Rubik's cube states in which k = 0 to 64 of the faces are "solved"

Imagine an $n \times n \times n$ Rubik's cube, where we can transition the state of the cube using Singmaster moves under either the face- or quarter-turn metrics. We call a face of this cube "solved" if all of the symbols on the face are of the same color (there are six total colors).

How many total cube states are there when $k = {0, 1, 2, 3, 4, 5, 6}$$k = {0, 1, 2, 3, 4}$ of the cube faces are solved?

Note: This is (hopefully) a simplification of an earlier question asking for the probability that a greedy algorithm solves a Rubik's cube, where once a face is solved, the algorithm cannot backtrack and perturb the face.

Counting the number of Rubik's cube states in which k = 0 to 6 of the faces are "solved"

Imagine an $n \times n \times n$ Rubik's cube, where we can transition the state of the cube using Singmaster moves under either the face- or quarter-turn metrics. We call a face of this cube "solved" if all of the symbols on the face are of the same color (there are six total colors).

How many total cube states are there when $k = {0, 1, 2, 3, 4, 5, 6}$ of the cube faces are solved?

Note: This is (hopefully) a simplification of an earlier question asking for the probability that a greedy algorithm solves a Rubik's cube, where once a face is solved, the algorithm cannot backtrack and perturb the face.

Counting the number of Rubik's cube states in which k = 0 to 4 of the faces are "solved"

Imagine an $n \times n \times n$ Rubik's cube, where we can transition the state of the cube using Singmaster moves under either the face- or quarter-turn metrics. We call a face of this cube "solved" if all of the symbols on the face are of the same color (there are six total colors).

How many total cube states are there when $k = {0, 1, 2, 3, 4}$ of the cube faces are solved?

Note: This is (hopefully) a simplification of an earlier question asking for the probability that a greedy algorithm solves a Rubik's cube, where once a face is solved, the algorithm cannot backtrack and perturb the face.

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