# Random Cubing Discussion

#### Kit Clement

This is clearly false. Repeat just F R U R' U' F twice and you'll see that two edges are unoriented, regardless of which of the 3 orientations you choose. That alg does affect 2 edges' EO state, but it also cycles pieces, so it doesn't switch the EO state of the same pieces each time.

#### Alex Davison

##### Member
This is clearly false. Repeat just F R U R' U' F twice and you'll see that two edges are unoriented, regardless of which of the 3 orientations you choose. That alg does affect 2 edges' EO state, but it also cycles pieces, so it doesn't switch the EO state of the same pieces each time.
ahh true, i didn't take into account permutation

#### abunickabhi

##### Member
I doubt this will have any benefit to speedsolving... But a few days ago i realized that any algorithm repeated an even number of times will not affect EO. If you think of an algorithm as a "toggle" for the some of the edges (by which I mean it switches some edges from bad->good or good->bad) it becomes pretty clear that if you repeat the algorithm twice, all of the "toggled" edges are toggled back... meaning that EO isn't affected.

I know this is stupid but I thought it was interesting nonetheless. there's probably there's some other post that already mentioned this but whatever lol
Every algorithm affects EO differently. It depends on the number of F and B turns the alg has, and if it has rotations too. As Kit said, the refutation of the even conjecture, is F R U R' U' F.

#### xyzzy

##### Member
A better counterexample would be something that truly leaves edges flipped in place after an even number of iterations, like M' U M U' M' U M U M' U2 M (flips all four last layer edges after two iterations).

#### qwr

##### Member
Brian Sun, well known for his "pls" vids where he shows someone's video then shows a "better" alg, apparently got copyright striked. I think this use of copyright striking is malicious and the clips are used in fair use (specifically for use in scholarly criticism), though youtube's copyright system is widely known to be abusable.
That aside, some people in discords apparently view his "pls" vids as derogatory or condescending which might be the case.

I have no idea what this is about

#### xyzzy

##### Member
Random observation: Any scramble with (at least) 10 edge pieces in the correct location but flipped must take at least 9 moves to solve.

Not hard to prove by computer search (fire up Cube Explorer and manually check the twenty or so different such scrambles), but the fun thing is that this can be proved without computer search or massive case exhaustion.

The ten flipped edges will be flipped for any of the three EO orientations. For, say, the F-B axis, there will be either 10 or 12 bad edges, and since each F/B quarter turn will affect at most 4 bad edges, this means that at least 3 F/B quarter turns must be used. Same reasoning applies to the other two axes. Therefore at least 3+3+3 = 9 moves must be used to solve any such scramble.

A slight modification of the above argument leads to a stronger conclusion, one that you can't prove just by using Cube Explorer: any solution must use 9 moves that are not half turns.

#### MJS Cubing

##### Member
I like this thread because I can come and pretend that all the math makes sense to me. who says you need to be good at math to solve a cube?

#### qwr

##### Member
I like this thread because I can come and pretend that all the math makes sense to me. who says you need to be good at math to solve a cube?
if you're talking about xyzzy's most recent post, all you need to know is the concept of good edges and bad edges, which you can probably understand if you know roux basics

#### Cubing Forever

##### Member
Random observation: Any scramble with (at least) 10 edge pieces in the correct location but flipped must take at least 9 moves to solve.

Not hard to prove by computer search (fire up Cube Explorer and manually check the twenty or so different such scrambles), but the fun thing is that this can be proved without computer search or massive case exhaustion.

The ten flipped edges will be flipped for any of the three EO orientations. For, say, the F-B axis, there will be either 10 or 12 bad edges, and since each F/B quarter turn will affect at most 4 bad edges, this means that at least 3 F/B quarter turns must be used. Same reasoning applies to the other two axes. Therefore at least 3+3+3 = 9 moves must be used to solve any such scramble.

A slight modification of the above argument leads to a stronger conclusion, one that you can't prove just by using Cube Explorer: any solution must use 9 moves that are not half turns.
This is just a modification to "superflip takes at least 20 moves to solve" thing bc the edges are superflipped but we do not take into account the corner permutation thus leading to shorter solutions (9 being the lower bound). Am I right?
(TLDR, what you say is, a superflip with corners unsolved needs at least 9 moves to be solved right?)

#### xyzzy

##### Member
This is just a modification to "superflip takes at least 20 moves to solve" thing bc the edges are superflipped but we do not take into account the corner permutation thus leading to shorter solutions (9 being the lower bound). Am I right?
(TLDR, what you say is, a superflip with corners unsolved needs at least 9 moves to be solved right?)
Yeah, pretty much, except showing that superflip needs 20 moves requires a relatively long computer search, whereas this is a three-line proof where the most complicated calculation is 10/4 = 2.5 . (Obviously, calculating more lets you get better bounds.)

Also, "almost superflips" (e.g. this) are within scope for my argument.

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#### Cubing Forever

##### Member
Random observation (could be false)
Diag PLLs can never be less than 13 moves long. The shortest diag PLLs are:
Nb (R2 U2 F2 U' R2 U2 R2 F2 U' F2 U2 F2 R2) and Na (L2 U2 F2 U L2 U2 L2 F2 U F2 U2 F2 L2) both of which are 13 moves long(in HTM).
(Ikr the words "shortest" and "N perms" shouldn't be in one sentence lol)

By definition, 2GLL means solving the LL with a 2 Gen moveset (RU, LU or MU). With this definition, we can include the ELL cases as 2GLL cases too since ELL is just cycling or flipping edges which can be done with MU. With this, the total number of 2GLLs is 84+29=113 cases.
However, if we define 2GLL as a subset of ZBLL, then we cannot include ELL since the edges need not be oriented for ELL but they need to be oriented for ZBLL. So, defining 2GLL as a subset of 1LLL(rather than ZBLL), we can assume that the ELLs can be classified as "2GLLs" right?

#### xyzzy

##### Member
Diag PLLs can never be less than 13 moves long. The shortest diag PLLs are:
Nb (R2 U2 F2 U' R2 U2 R2 F2 U' F2 U2 F2 R2) and Na (L2 U2 F2 U L2 U2 L2 F2 U F2 U2 F2 L2) both of which are 13 moves long(in HTM).
Correct (for face turn metric); optimal Y perm is 13 moves too, and optimal E perm and V perm are both 14 moves.

2GLL means solving the LL with a 2 Gen moveset
[…]
we can assume that the ELLs can be classified as "2GLLs" right?
"2GLL" already specifically means the last layer cases that can be solved using only R and U moves. It's not a catch-all term for all 2-gen-solvable last layer subsets using any two generators you want.

If, say, you choose Rw and U as your generators, then you get the 1LLL subset with corners permuted and no other restriction. If you choose more esoteric generators (like R U R' and F' U F or something; nobody said the generators had to be single moves) then you might get all the last layer cases.

(Also: just because it's called "2-gen last layer" doesn't mean that you have to use 2-gen algs to solve those cases. I use RUL/RUS/RrU/RBL/RUD for a handful of 2GLL cases.)

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#### qwr

##### Member
I came up with a silly J perm executions

R U2 R' U' R U2 r' F R' F' r
L' U2 L U L' U2 l F' L F l'

#### Cubing Forever

##### Member
any solution must use 9 moves that are not half turns.
Part of the reason is because half turns preserve EO and quarter turns break it right?

(Take this with a pinch of salt because I'm not as smart as you lol)

#### H0BB3

##### Member
What do you think about the roxenda magic cube megaminx /pyraminx/random cube box off of amazon

#### rubik2005

##### Member
What do you think about the roxenda magic cube megaminx /pyraminx/random cube box off of amazon
Roxenda isn't a cube brand, but rather the name of the seller on the Amazon paga. Same thing with D-Fantix. The actual brands are Qiyi, Moyu, Cyclone Boys, ShengShou, YJ, GAN, and maybe a few others that I forgot. Amazon is generally not the best place for you to order cubes, for sometimes they charge you more and don't always offer benefits like discount codes and rewards as seen in stores like TheCubicle, SpeedCubeShop, DailyPuzzles, etc.

As for budget cubes, the Moyu MF3RS 2020 is conside the best performing cube (and only comes in at around $9 USD), the YJ MGC is one of the best big-cube lines (4x4 - 7x7), and there are other options for WCA events such as the YLM M square-one and the Cyclone Boys skewb which people seem to reccomend. #### SH03L4C3 ##### Member Roxenda isn't a cube brand, but rather the name of the seller on the Amazon paga. Same thing with D-Fantix. The actual brands are Qiyi, Moyu, Cyclone Boys, ShengShou, YJ, GAN, and maybe a few others that I forgot. Amazon is generally not the best place for you to order cubes, for sometimes they charge you more and don't always offer benefits like discount codes and rewards as seen in stores like TheCubicle, SpeedCubeShop, DailyPuzzles, etc. As for budget cubes, the Moyu MF3RS 2020 is conside the best performing cube (and only comes in at around$9 USD), the YJ MGC is one of the best big-cube lines (4x4 - 7x7), and there are other options for WCA events such as the YLM M square-one and the Cyclone Boys skewb which people seem to reccomend.
yes. On amazon, a meilong M is $10, and on TC it is$6 (not including reward points and discount codes).