Are all Megaminx states possible with WCA scrambles?

[Spencer131]

This is a hard theoretical question if anyone likes proving cube-related stuff

I have no idea how you would even go about proving or disproving this

 

[Deleted member 36445]

If the number of moves for a WCA Megaminx scrambler is larger or equal to the God number for megaminx then yes

 

[Deleted member 36445]

If the number of moves for a WCA Megaminx scrambler is larger or equal to the God number for megaminx then yes

Now I was thinking that the WCA scrambler uses only R++, R--, D--, D++, U and U', then I really don't know because that's independent of the God number for Megaminx if you use every face for the solution

 

[Deleted member 36445]

According to this: https://pastebin.com/JUfcdSjv

The upper bound for Megaminx is 194 htm

 

[THE_V01DL0RD]

I feel like it would be the same as just using wide moves on a 3x3 and as long if the amount of moves is approximate to the gods number than I think yes.

 

[THE_V01DL0RD]

According to this: https://pastebin.com/JUfcdSjv

The upper bound for Megaminx is 194 htm

What's the lower bound?

 

[Deleted member 36445]

What's the lower bound?

48

 

[Spencer131]

If the number of moves for a WCA Megaminx scrambler is larger or equal to the God number for megaminx then yes

What if some scrambles lead to the same state? I'm almost certain some scrambles would be the same

 

[1973486]

The answer is no and it's not even close. WCA scrambles can generate 2^70 states (the first move can be either R++ or R--, the second D++ or D--, etc for 70 moves). That's about 10^21.

The total number of Megaminx states is about 10^68.

 

[xyzzy]

70 moves is nowhere near enough, but it is possible to reach every state with enough moves.

You can verify this for yourself with GAP (which presumably uses Schreier-Sims internally). You can use Twizzle to produce GAP files for megaminx (and other puzzles supported by Twizzle).

Or, here's a less computation-heavy proof.

By repeating any move sequence enough times, we will eventually return to the initial state; stopping at just one iteration before thus lets us effectively do the inverse sequence. We will use this trick to build up simpler moves.

(R−− D++ R++ D++ R++ D++ R++ D++ R++ D++ U) (R++ D++ R++ D++ R++ D++ R++ D++ R++ D++ U)' = R+
(Note that "(R++ D++ R++ D++ R++ D++ R++ D++ R++ D++ U)' " really means (R++ D++ R++ D++ R++ D++ R++ D++ R++ D++ U)1679, which is possible with long enough scrambles.)

Now that we have isolated R+ (and by the same logic, R−), we can get:
R− R− (R++ D++ R++ D++ R++ D++ R++ D++ R++ D++ U) = D++ R++ D++ R++ D++ R++ D++ R++ D++ U
and
R− R− (R++ D−− R++ D++ R++ D++ R++ D++ R++ D++ U) = D−− R++ D++ R++ D++ R++ D++ R++ D++ U

From these we can also isolate D+ and D− with the same process:
(D−− R++ D++ R++ D++ R++ D++ R++ D++ U) (D++ R++ D++ R++ D++ R++ D++ R++ D++ U)' = D+

Then we can isolate U:
(D− D− R− R−)5 (R++ D++ R++ D++ R++ D++ R++ D++ R++ D++ U) = U

Then we have y rotations:
y = U D−
y' = U' D+

From which we create a rotation around a different face:
R+ y2 R− y2 R− y2 R+ y2' = Fv'

Fv and y generate all rotations of the dodecahedron, so ⟨U, Fv, y⟩ = ⟨all face turns, all rotations⟩.

The U, Fv and y moves created by this process are like a million moves long if you fully expand them in WCA scramble notation, so this is a very poor (but finite) upper bound on God's number for WCA-style scrambles.

If the number of moves for a WCA Megaminx scrambler is larger or equal to the God number for megaminx then yes

This is true only if the scrambler can do turns on every side, which is not the case with WCA scrambles.

 

[Bingus Ringus McDingus]

70 moves is nowhere near enough, but it is possible to reach every state with enough moves.

You can verify this for yourself with GAP (which presumably uses Schreier-Sims internally). You can use Twizzle to produce GAP files for megaminx (and other puzzles supported by Twizzle).

Or, here's a less computation-heavy proof.

By repeating any move sequence enough times, we will eventually return to the initial state; stopping at just one iteration before thus lets us effectively do the inverse sequence. We will use this trick to build up simpler moves.

(R−− D++ R++ D++ R++ D++ R++ D++ R++ D++ U) (R++ D++ R++ D++ R++ D++ R++ D++ R++ D++ U)' = R+
(Note that "(R++ D++ R++ D++ R++ D++ R++ D++ R++ D++ U)' " really means (R++ D++ R++ D++ R++ D++ R++ D++ R++ D++ U)1259, which is possible with long enough scrambles.)

Now that we have isolated R+ (and by the same logic, R−), we can get:
R− R− (R++ D++ R++ D++ R++ D++ R++ D++ R++ D++ U) = D++ R++ D++ R++ D++ R++ D++ R++ D++ U
and
R− R− (R++ D−− R++ D++ R++ D++ R++ D++ R++ D++ U) = D−− R++ D++ R++ D++ R++ D++ R++ D++ U

From these we can also isolate D+ and D− with the same process:
(D−− R++ D++ R++ D++ R++ D++ R++ D++ U) (D++ R++ D++ R++ D++ R++ D++ R++ D++ U)' = D+

Then we can isolate U:
(D− D− R− R−)5 (R++ D++ R++ D++ R++ D++ R++ D++ R++ D++ U) = U

Then we have y rotations:
y = U D−
y' = U' D+

From which we create a rotation around a different face:
R+ y2 R− y2 R− y2 R+ y2' = Fv'

Fv and y generate all rotations of the dodecahedron, so ⟨U, Fv, y⟩ = ⟨all face turns, all rotations⟩.

The U, Fv and y moves created by this process are like a million moves long if you fully expand them in WCA scramble notation, so this is a very poor (but finite) upper bound on God's number for WCA-style scrambles.


This is true only if the scrambler can do turns on every side, which is not the case with WCA scrambles.

I'm scared of your brain.

 

[unirox13]

With the WCA scramble... Will you see all of the potential scramble positions? Probably not, obviously proven, the numbers don't exactly line up. However, will you always see a new scrambled position? More than likely. That scramble technique was developed for speed and the simplicity by which it creates a scrambled puzzle. In order to see every scrambled position of the mega, you would literally need to live and stay awake for years longer than possible. I think that the WCA scramble works fantastically. I will admit that during at home solves, I usually do a few extra moves to separate stuff in order to make it look more scrambled.

 

[bldcuber314]

The answer is no and it's not even close. WCA scrambles can generate 2^70 states (the first move can be either R++ or R--, the second D++ or D--, etc for 70 moves). That's about 10^21.

The total number of Megaminx states is about 10^68.

I don't think 2^70 is a correct number for the number of WCA scrambles as that is the number of possible sequences, but some of those could be the same scramble(correct me if I'm wrong, it could be that WCA Mega scrambles are all individual)

 

[d---]

This is a hard theoretical question if anyone likes proving cube-related stuff

I have no idea how you would even go about proving or disproving this

The short answer, no.
Pretty sure the megaminx scramble can't reach a solved state, but idk.

 

[Jorian Meeuse]

I don't think 2^70 is a correct number for the number of WCA scrambles as that is the number of possible sequences, but some of those could be the same scramble(correct me if I'm wrong, it could be that WCA Mega scrambles are all individual)

Well no, technically it's 2^77 because of the U/U' at the end of each line. Whether some of these are the same or not is irrelevant here (although it seems unlikely for such short scrambles). All that matters is that there are more megaminx states than scramble sequences, meaning if you could reach each state with WCA scrambles, there would have to be scrambles that achieve multiple states at the same time, which is clearly impossible.

 

[THE_V01DL0RD]

70 moves is nowhere near enough, but it is possible to reach every state with enough moves.

You can verify this for yourself with GAP (which presumably uses Schreier-Sims internally). You can use Twizzle to produce GAP files for megaminx (and other puzzles supported by Twizzle).

Or, here's a less computation-heavy proof.

By repeating any move sequence enough times, we will eventually return to the initial state; stopping at just one iteration before thus lets us effectively do the inverse sequence. We will use this trick to build up simpler moves.

(R−− D++ R++ D++ R++ D++ R++ D++ R++ D++ U) (R++ D++ R++ D++ R++ D++ R++ D++ R++ D++ U)' = R+
(Note that "(R++ D++ R++ D++ R++ D++ R++ D++ R++ D++ U)' " really means (R++ D++ R++ D++ R++ D++ R++ D++ R++ D++ U)1259, which is possible with long enough scrambles.)

Now that we have isolated R+ (and by the same logic, R−), we can get:
R− R− (R++ D++ R++ D++ R++ D++ R++ D++ R++ D++ U) = D++ R++ D++ R++ D++ R++ D++ R++ D++ U
and
R− R− (R++ D−− R++ D++ R++ D++ R++ D++ R++ D++ U) = D−− R++ D++ R++ D++ R++ D++ R++ D++ U

From these we can also isolate D+ and D− with the same process:
(D−− R++ D++ R++ D++ R++ D++ R++ D++ U) (D++ R++ D++ R++ D++ R++ D++ R++ D++ U)' = D+

Then we can isolate U:
(D− D− R− R−)5 (R++ D++ R++ D++ R++ D++ R++ D++ R++ D++ U) = U

Then we have y rotations:
y = U D−
y' = U' D+

From which we create a rotation around a different face:
R+ y2 R− y2 R− y2 R+ y2' = Fv'

Fv and y generate all rotations of the dodecahedron, so ⟨U, Fv, y⟩ = ⟨all face turns, all rotations⟩.

The U, Fv and y moves created by this process are like a million moves long if you fully expand them in WCA scramble notation, so this is a very poor (but finite) upper bound on God's number for WCA-style scrambles.


This is true only if the scrambler can do turns on every side, which is not the case with WCA scrambles.

Your IQ is the amount of possible megaminx scrambles

 

[1973486]

Well no, technically it's 2^77 because of the U/U' at the end of each line. Whether some of these are the same or not is irrelevant here (although it seems unlikely for such short scrambles). All that matters is that there are more megaminx states than scramble sequences, meaning if you could reach each state with WCA scrambles, there would have to be scrambles that achieve multiple states at the same time, which is clearly impossible.

The U/U' is dependent on the previous move

 

[Jorian Meeuse]

The U/U' is dependent on the previous move

Oh wow I never realised that

 

[Deleted member 36445]

if not all the states can be achieves, doesn't that mean that eventually a new scrambler will have to be used?

 

[PapaSmurf]

Probably after the heat death of the universe.

Also, I could give you a scrambler that goes to beyond God's number, but if it was <RU> then it wouldn't reach all states, nowhere near.

@xyzzy's proof is a good one.