Renslay
Legend
There is a link in the bottom which says "alg.cubing.net". You can see there the solution in an animation.
For example Alexander Lau's 6.83:
http://cubesolv.es/solve/3920
Show it as an animation:
alg.cubing.net
[Help Thread] Roux Discussion and Help
- Thread starter LamentConfiguration
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There is a link in the bottom which says "alg.cubing.net". You can see there the solution in an animation.
For example Alexander Lau's 6.83:
http://cubesolv.es/solve/3920
Show it as an animation:
alg.cubing.net
Kirjava
Colourful
Hello All
I am making a video about what method beginners should choose and I know Roux a little and I need help...
Can anyone give me a Scramble and reconstruction that is working ??'
I searched this : http://cubesolv.es/?solver=&competition=&puzzle=&time-specifier=&time-value=&tags=roux
and I did not find a single scramble that solves the cube (I might be doing it wrong)
So, can anyone give me a scramble and reconstruction that you have solved with for Roux from a list there or any scramble and reconstruction
Thanks in advance
Please don't make youtube videos if you're completely inept.
DeeDubb
Veteran
How fast should I get before I start worrying about memorizing CLLs?
TheOneOnTheLeft
tOOtl
It's up to you, basically whenever you think CMLL is the part of your solve that you can most improve at. I think most people start when they average somewhere between 25-30 seconds, although I'm now sub-20 and still don't know them all. If you enjoy learning algs, or if you are very slow at learning algs, maybe start earlier. If you particularly dislike learning algs, just focus on other parts of your solve and you'll still improve.
5BLD's and Waffle's guides both recommend starting to learn 1-look CMLL when you're sub-30.
CriticalCubing
Veteran
No Sir, you got me wrong. I am making a video where I show them all 4 major speedcubing method (basic steps, etc) and in the end an example solve. I know till CMLL but the after cycling etc goes over my head. SO I thought I will practice a Roux reconstruction, and show it after I can do it at a fast speed. Already know till CMLL and know all CMLL algs. I hope you got me...
I hope you got me...
Kirjava
Colourful
No Sir, you got me wrong. I am making a video where I show them all 4 major speedcubing method (basic steps, etc) and in the end an example solve. I know till CMLL but the after cycling etc goes over my head. SO I thought I will practice a Roux reconstruction, and show it after I can do it at a fast speed. Already know till CMLL and know all CMLL algs.
it's more that I feel that if you struggle finding and executing a reconstruction, the quality of your videos will likely be poor
Renslay
Legend
I agree. However, for example, I started to learn CMLLs when I was around sub-50. Although you won't see much improvements on your time (you can easily improve more with practicing F2B anf focusing on look-ahead), it's good if you get familiar with them in an earlier stage.
DeeDubb
Veteran
No Sir, you got me wrong. I am making a video where I show them all 4 major speedcubing method (basic steps, etc) and in the end an example solve. I know till CMLL but the after cycling etc goes over my head. SO I thought I will practice a Roux reconstruction, and show it after I can do it at a fast speed. Already know till CMLL and know all CMLL algs.
Watch Donovan's tutorials on cycling. They explain it really well.
http://www.youtube.com/watch?v=MwH9_lCJFSU
http://www.youtube.com/watch?v=ac6L6x-NEm8
TheOneOnTheLeft
tOOtl
I think this is a really good idea for starting to pay attention to non-LL stickers (ie not just the colour of the top face), and starting to recognise the CMLL even before you know the alg for it. At the very least it means you know when you'll get a skip, and once I start to pay attention to that, I fairly quickly learn where the bar is going to be for an adjacent swap so I don't pause between orientation and permutation.
DeeDubb
Veteran
Good idea. I guess I should at least start recognizing the cases.
I have another question. I was watching Alex Lau's Roux Example Solves and he talks about how sometimes he can solve the LU/RU edges while he's fixing the bad edges. He does one at 4:10 of the video. Is there a list of algorithms somewhere for these cases? I looked on Waffle's, Rouxtorials, and Donovan's, and none of them mention this possibility.
He didn't solve 4b during 4a at all there. I think you're asking for a list of how to finish the "3 on top, 1 on bottom" case so that you end up with ideal 4b edge placement, or even a skip. Unfortunately there isn't much advanced Roux information out there.
TheNextFeliks
Veteran
There's lots of Roux info out there. Not a lot on this though.
I went through all 4a + 4b possibilities by hand and figured out what to look for and how to solve them by working backwards from ideal "3 on top, 1 on bottom" setups. I never wrote it down. I'll use tonight to compile what I know and post it here.
Basically I found that there are fifteen "3 on top, 1 on bottom" cases when combined with all possible 4b edge permutations (not including Z-axis mirrors), and only five of them are ideal. I worked backwards to the other three 4a cases to find the ones that are worth learning. I say "three" 4a cases because I don't count the 6-flip as I just use an algorithm for it, and one of the others is the "2 adjacent on top, 2 on bottom" case, which is just an M2 from the "3 on top, 1 on bottom" case, so it was trivial. The remaining six can be grouped in M2 pairs as well, so there are really only three that take effort to learn. One of those pairs, the "2 opposite on top, 2 on bottom" and "4 on top" cases, is fairly easy as well, because none of the edges change orientation while cycling to the "3 on top, 1 on bottom" case. So those two can be learned very intuitively. That leaves two pairs that require a fair amount of study. Combine that with there only being five ideal "3 on top, 1 on bottom" cases, and that leaves only 10 cases to learn. It's really not that much.
Last edited:
GuRoux
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Most people do this with look ahead because it might be a bit hard to recognize each case.
DeeDubb
Veteran
That would be awesome. Thanks.
I'm still in my infant stages, and I'm sure some of these things will be intuitively picked up, but seeing some info would be really great for my development.
Just to make sure I'm following, the ideal "3 top 1 bottom" is when you get the 3 bad edges on top, and 1 bad edge on bottom, then you do the MUM'?
Also what is the 6-flip? Is that 6 bad edges? Which algorithm do you use for that?
Renslay
Legend
Yes, that is the case.
If you have UL,UR,UF,DF as wrong edges, you can choose between M' U M, M' U M', M' U' M and M' U' M'. Choose the one which can help at Step 4b).
Similarly, if you have UL,UR,UB,DB as wrong edges, you can do M U* M*, where U* can be U or U' and M* can be M or M' too (4 solutions). Usually one of them can make Step 4b) easier then the others.
6-flip is when all the 6 edges are wrong. A quick algorithm to fix that is R U' r' U' M' U r U r'.
The alg I use is r' F R U M' U' R' F' R.
I have already covered the 6-flip in my LSE write up, so I'll paste that part here. Unfortunately this is all specific to the alg I use for 6-flip, but that's how it is. I can't write about every 6-flip alg, obviously. Also, clearly this will take longer than one night now that I've decided to cover all of LSE. It has turned into a full advanced LSE write up because LSE tends to be a "mind-flow" where everything connects to the next part, and so when covering the 4b part I find that I'm tending to link it to 4c. I removed the 4c stuff from the paste below because it was not complete. "Mind-flow" may be a corny term, but I think it fits.
After using what I've written here for a while now, I am able to do my 6-flip + 4b with my eyes closed. This does not come at the cost of a pause for recognition, because I react only to a small amount of information on the U stickers before starting the alg, and then finish the 4b based on what I saw just as I started it. Think of it similarly to predicting the AUF of a PLL, same mind-flow. Except pretend there are only four PLLs and they're recognized using only U stickers. It feels the same as in CFOP how just after I start a PLL, I can look away and finish the solve because I jsut "know" the AUF based on a color relationship I saw just as I started the alg, except instead of just AUF, it's the entirety of 4b.
Please let me know if something is unclear or just long-winded, because I'm using the same approach to the rest of the LSE. Let me know of any errors you find as well, even just spelling errors.
Okay enough, here it is:
When all of the edges on U are bad, all of the stickers on U are side stickers. This means that if your 4b edges are on U, then the 4b color stickers will be plainly visible on the U face.
First of all, it's useful to know how the edges move in pairs to have a better understanding of what is going on.
The edges in UB and UR move to UL and UF, UL and UF move to DB and DF, and lastly DB and DF move to UB and UR. To get a feel for it, put your right middle finger on UB and right thumb on UR, then "hop" them across the top of the cube together diagonally so that your middle finger is UL, and your thumb is on UF. Then move them down so that you have your middle finger on DB and your thumb on DF, then go back up to UB and UR. Go through that motion several times and now you know how the edges move around in pairs. Do the alg to a solved cube and look where the edges moved to. The permutation is a 3-cycle of pairs, and the orientation is a toggled state, so you need to do the alg 2 * 3 = 6 times to get back to the cube's original state (solved in this case).
The main thing to see from this is that if you do the alg and your 4b edges end up adjacent to one another on U, then you have essentially the worst 4b case possible, second only to a pop during the 6-flip and your 4b edges flying out of the window of a moving car. So, if your 4b edges are at UB and UR, and you do this alg, they end up adjacent at UL and UF. If they are both on D, they end up adjacent at UB and UR. Getting this case nullifies the benefit of using a non-<M, U> solution in the first place. This can be avoided 100% of the time, and even influenced to get benefit.
There are ways to set up the 4b edges so that they get solved completely every time during the 6-flip, but that takes too much recognition and extra moves, so it's not worth it. All the stuff below is realistic because you're reacting to almost nothing, and in the same way every time. You only have to react immediately to which of the top stickers are 4b colors. The 4b is predicted entirely from a simple color relationship as you start the alg.
If you see both of your 4b colors on top:
- If they are adjacent, move them to UL and UF, do the 6-flip, AUF ("adjustment of the U face" with U, U' or U2) if necessary, and finish with M2.
For example, start with a solved cube and apply: r' F R U M' U' R' F' R U' M U2 M' U'. As you start the alg, look at the cube from the front left. This will be similar to predicting AUF for PLL in CFOP, where you look for a color relationship (same, opposite, or adjacent colors) just as you start the alg, which tells you what the AUF will be after the alg. In the case of this 6-flip alg, the corners are not moved at all so predicting the AUF is very simple. If you see that UL is the same as the left corner stickers, the AUF is U'. If they are opposite, it is U. If UF is the same as the front corner stickers, the AUF is U2, and if they are opposite there is no AUF. When there is no AUF, integrate an M2 into the end of the alg, making it r' F R U M' U' R' F' (r M'). The solution to this example is then: r' F R U M' U' R' F' R U' M2 (U M2 U2 M U2 M').
The initial adjustment to make sure the 4b edges are at UL and UF cannot be considered an extra move, because it saves moves in all cases. If the edges were to remain at UB and UR, you end up with the worst 4b case possible at UL and UF, taking either five or six moves depending on the U corners to solve 4b. If the edges were to remain at UL and UB, it would take four or six moves to solve 4b. If the edges were to remain at UR and UF, it would take four or six moves to solve 4b. It takes zero or one moves to place the edges at UL and UF, and takes one or two to solve 4b afterward, making one, two, or three for all combinations. So, in all cases, it is less moves than the other possibilities, which were four, five, or six moves.
- If they are opposite, put them in UL and UR. One 4b edge will stay on top, the other will move to the bottom. Two out of three possibilities have an easily predicted cancel.
For example, start with a solved cube and apply: r' F R U M' U' R' F' R M U2 M' U. This one is easy. Just as you start the 6-flip alg, See the color relationship between UR and the front corner stickers. The UR edge will end up between the front corners, and the UL edge will end up at DB, so the 4b case is not only predicted, it's ready to go. Notice that on the current set up, the UR sticker matches the front corner stickers. This is a trigger to end the alg with a cancel and go straight into the 4b. Solve this example with: r' F R U M' U' R' F' r U2 M' (U' M' U2 M' U2). This cancel works for when it is the adjacent color 4b case as well. See that whether you do U or U' to move the 4b edge from UB and UF to UL and UR makes no difference, because the UR sticker has the same-color relationship to the front U corner sticker as the UL sticker has to the back U corner stickers.
The initial adjustment to make sure the 4b edges are at UL and UR cannot be considered an extra move, because in all cases the number of moves is either the same or one less. If the edges were to remain at UB and UF, they would move to UL and DF, requiring a U turn to be ready for the 4b. This means that a cancel is not possible, and so that move is never saved. In the one other case, the U turn is being done after instead of before, resulting in the same number of moves. This means that adjusting the 4b edges to UL and UR from UB and UF before the 6-flip alg results in an easily predictable cancel 3/4 of the time on top of making the 4b much easier to predict overall at no loss.
If only one 4b color is on top:
- Unfortunately, check bottom (if not looked ahead to during CMLL). If 4b edge on D is at DF, place 4b on U at UB. If 4b edge on D is at DB, place 4b edge on U at UR. Do 6-flip, possibly with integrated M2 at end, solve 4b if not a complete skip already. Not as complex as it sounds.
Looking at the bottom of the cube first may seem terrible, but not only does it guarantee that you don't end up with the even worse situation of both 4b edges on top adjacent to each other, it also has potential to skip the 4b completely or possibly have a two move 4b. The worst case is a four move 4b, five if you include the initial AUF before the 6-flip. Remember that all of this stuff results in being able to do 6-flip + 4b with your eyes closed so no relying on awkward look ahead during a crazy alg. As you improve at looking for 4b edges during CMLL, you will have to look at the bottom for this case less and less. An alternative is to watch the right side of the U layer as you do the "U M' U'", which will either reveal the 4b edge that was at DB, or it won't. Reacting to that means you are having to make a decision immediately based on a barely visible sticker that disappears two turns later. I recommend dealing with looking at the bottom of the cube for now, and rely on CMLL look ahead as you improve. It's not like a 6-flip, let alone this particular case, is very common anyway, and with future CMLL look ahead it doesn't even matter.
The idea here is that if the 4b edge on D is at DB, then it will move to UB. So, if you move the 4b edge that is on U to the UR position, you will end up with 4b edges opposite one another at UB and UF. If the UR edge has a same-color relationship with the front corner stickers, you will get a complete 4b skip. If they have either of the other color relationships, finish the alg with an M2 during the R, preparing for an AUF before finishing the 4b with M2. So it becomes: r' F R U M' U' R' F' (r M'). I do the F' at the end by pulling down with my left index finger, which happens to make the M2 easy to do along-side the R. An alternative to this is to place the 4b edge that is on top at UF, do an M2, and place it back at UR. This sets up a skip, but requires more thought and is the same number of moves anyway. I don't bother.
For example, start with a solved cube and apply: r' F R U M' U' R' F' R U'. The 4b edge on bottom is at DB, so we moved the 4b edge on U to UR. See that the AUF will be a U2 because when looking at the cube from the front right, we see that UR has an opposite-color relationship with the front corner stickers. If the relationship was that UR is the same as the front corner stickers, then we would have a 4b skip. If UR is the same as the right corner stickers, the AUF is U. If they are opposite, the AUF is U'. In this example they are opposite, so we get have an AUF of U2: r' F R U M' U' R' F' (r M') U2 M2 (U' M U2 M U2 M2).
Now an example of when the 4b edge on D is at DF. Start with a solved cube and apply: r' F R U M' U' R' F' R U2 M2 U2. The last U2 places the 4b edge on U at UB, so that it moves to UL as the DF 4b edge moves to UR. Look at the front left of the cube while doing this. The UB edge has a same-color relationship with the left corner stickers, so we are about to get a 4b skip. If they are opposite, just do an M2 and you have yourself another skip after the 6-flip. If UB has a same-color relationship with the front corner stickers, then the AUF will be U, regardless of whether you do U M2 after the 6-flip or U' M2. If the relationship is opposite, the AUF will be U', again regardless of the initial U direction. These last two color relationships are essentially the worst case scenarios possible with this 6-flip.
If none are visible:
- Do M2 (U/U') to convert the case from a terrible one to a great one. This adds two moves, but makes the 4b very predictable as described above (in fact it's the best case overall), and if the 4b edges are left on the bottom then they are guaranteed to end up at UL and UF, which is terrible, taking either five or 6 moves to solve the 4b. You will save several moves and improve look ahead 100% of the time by converting cases here.
For example, start with a solved cube and apply: r' F R U M' U' R' F' r U2 M. No 4b colors are on U, so do M2 U. Now it's like the case described above. We see that UR is opposite the front corner stickers, so we can predict our 4b case by imagining the UR edge between the front corners. The resulting solution is: r' F R U M' U' R' F' R M U2 M' (U' M2).
To practice the 6-flip, first of course learn the alg, then do an <M, U> scramble and look for good edges. Pretend they're bad and do what you would usually do for 4a as if they were bad. This sets up a 6-flip case.
DeeDubb
Veteran
Wow, Dominate, what a great write up!
So, just adding some AUFs to your algorithm to either place the edges or set up predictable edges. Very nice.
So, just adding some AUFs to your algorithm to either place the edges or set up predictable edges. Very nice.
DeeDubb
Veteran
Ok, another block building question. I get in this situation a lot in 2, where I'm working on pairing edge corners, and I get one reversed, which requires me to spin the corner to the wrong side, plug in the edge, then spin it to the correct side to fill it in. I'm wondering what alternatives there are:
http://alg.garron.us/?alg=M_U2_M2_U-_R_U2_R-&ini=R_U_R-_M_U2_M2_U2&stage=F2L
http://alg.garron.us/?alg=M_U2_M2_U-_R_U2_R-&ini=R_U_R-_M_U2_M2_U2&stage=F2L
I use (r M') U2 M U R' or r' R2 U2 M2 U R' randomly.
Even M' U' R U R' U R U' R' U R U' R' can be fast, but it's certainly not a pretty move-count.
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