Not sure why you'd want a K4 solve from me, but I just recorded a QTPI solve. It's slow and filled with pauses, and I was going to delete until I realized it's actually my PB single.
Not sure why you'd want a K4 solve from me, but I just recorded a QTPI solve. It's slow and filled with pauses, and I was going to delete until I realized it's actually my PB single.
Because Thom Barlow ( @Kirjava ) is very particular himself about what is considered K4, I'm pretty sure he would not consider Jules' solves K4. In this video at least (edit. And also in his most-recent average of 5 video), Jules completed all centers first and then made the 1x3x4 block.
And since most "methods" are just minor variations of each other (this could also explain why Thom is very particular about the order of steps when performing his method), we can safely say that Jules uses his own variant of K4.
So Jules' times cannot be taken as a testament to the speed potential of K4.
@abunickabhi , here is a video of Thom's PB of 50.44 using his 4x4x4 method. (At the time he first published the method, I think he had a PB of 1:05. So 4 years after knowing the method, he decreased his time by about 15 seconds. I know he was a decent speedcuber, but he was also method inventor and software tool creator. So speed improvement potential can be better for those who just speedcube "full time". Combining that with the improved hardware...)
I actually like the concept of this method. However, I'm not sure that I would ever use it. Since OLL parity is created by the number of rotations used when reducing the centers, it seems as though You're just adding extra steps to reducing a 4x4 to a 2x2. Lots of 3-cycle commutators with some extra double checking. The corners of a 4x4 are always solvable, because they're a 2x2. If you reduce a 4x4 to a 2x2, you won't have parity, just a bunch of 3-cycles of the center pieces.
it seems as though You're just adding extra steps to reducing a 4x4 to a 2x2. Lots of 3-cycle commutators with some extra double checking. The corners of a 4x4 are always solvable, because they're a 2x2.
You will certainly encounter some form of parity in the process of doing that. Maybe not the single edge flip variation, but some form of it. For example, what if I have this case?
"In short", a method like this (or any Cage variant, 2x2x2 reduction, etc.) should be advertised because it's an alternate way to solve the 4x4x4. It should be elegant in its own right. But making people put in 3 times the effort to solve the 4x4x4 just because it avoids just one manifestation of parity is, I'm sorry to say, nonsensical.
If the single edge flip parity algs bother them that much, I think it's much faster for them to learn how they work from someone who already knows. ("Face the problem" rather than "avoid" it.) 9 minute video that explains how 2 different parity algs work. And just 1 commutator (versus dozens).
Because Thom Barlow ( @Kirjava ) is very particular himself about what is considered K4, I'm pretty sure he would not consider Jules' solves K4. In this video at least (edit. And also in his most-recent average of 5 video), Jules completed all centers first and then made the 1x3x4 block.
And since most "methods" are just minor variations of each other (this could also explain why Thom is very particular about the order of steps when performing his method), we can safely say that Jules uses his own variant of K4.
So Jules' times cannot be taken as a testament to the speed potential of K4.
@abunickabhi , here is a video of Thom's PB of 50.44 using his 4x4x4 method. (At the time he first published the method, I think he had a PB of 1:05. So 4 years after knowing the method, he decreased his time by about 15 seconds. I know he was a decent speedcuber, but he was also method inventor and software tool creator. So speed improvement potential can be better for those who just speedcube "full time". Combining that with the improved hardware...)
Not sure why you'd want a K4 solve from me, but I just recorded a QTPI solve. It's slow and filled with pauses, and I was going to delete until I realized it's actually my PB single.
Quarter Turn Parity Intuitive is a super cool idea. I would love to give it a try this weekend. The solving is a bit more conscious compared to say Yau method.
Sorry for the necro, but you've linked this thread in a few other places, so I figure I might as well post here.
It's great to see that we use the same method! I even wrote a post about it back when I came up with it (https://www.speedsolving.com/threads/new-4x4x4-method.695/), but you've done a much better job explaining it and drumming up interest. I have a couple of questions and comments. How do you always solve exactly 7 of the 10 edges? I leave a slice for last, and that could be 0 edges left (yay, skip!), a 3-cycle of edges left (what you address), or a pair of swaps. I used to do this last case with two 3-cycles, but I recently figured out a commutator to do it in one step: z' R F U F' U2 R' u2 R U2 F U' F' R' u2. I hadn't even thought of moving centers with the edges. That's clever.
When solving the edges, it's possible to also solve some of the last layer centers at the same time (the ones not in the last slice). This makes more of an impact on bigger cubes, but it may make that step too cumbersome. I usually skip doing this.
For centers, I usually do 8-move commutators with 0-2 setup moves (rarely 2), so I can spot them pretty quickly. Sometimes I permute blocks of centers, especially on bigger cubes. I also use r2 u2 r2 u2 to swap two connected centers on opposite faces. There are more variations of this alg.
For the 1x3x4 blocks, I sometimes switch between blocks, and I sometimes pair up some edges to use later. Basically I use what pieces I see, so I don't have to hunt all over the cube for one specific piece. Flexibility in this step helps prevent long awkward pauses.
When extending to 2x3x4, sometimes I'll screw up and have the centers in the wrong order. If I'm solving with white on top and green in front and I make a slice with blue in front, yellow on bottom, and green in back, I'll just do M2 Uw2 M2 Uw2. This puts the slice on the other side of the cube and corrects it. I also use this alg when I try to use reduction or K4 and screw up my cross.
Whoa, amazing that someone else does this! There are not many intuitive cubers around. I just happened to get back into cubing recently, so even though this is a very old post it is very timely for me.
The "7 of 10 edges" step can indeed end up with 4 unsolved edges requiring 2 commutators. But the first commutator solves a single edge, thus leaving you with 7 edges solved and 3 unsolved for the final edge commutator. I didn't mean to suggest that you could always avoid the 2-commutator cases. That's just part of the step. I will have to try out your method for solving the 4-edge case in one step!
Those 4-move center commutators are certainly very satisfying! One of my favorite parts of this method is finding efficient center commutators.
I am concentrating on 3x3 Roux these days but I do some QTPI solves every now and then. At this point, I'm using a permute-first 2-look CMLL method for corners so I don't use exclusively commutators anymore (well all of the algs I use are commutators but...) I have a guide written up for it, which I will post some time. It's significantly more move efficient than orient-first.
I wish I had found your explanation years ago when I was first learning 4x4, and stubbornly refusing to learn any algorithms (well I am still pretty stubborn about that). I share your sentiment that I dislike 3x3 reduction solves - I understand that it's good for speed, but it's just so much less interesting than direct solving for me. I like the unique challenges that 4x4 poses, and I want to accentuate rather than avoid them.
I'm sure that someone else will invent this method every 10 years or so.
I'm curious, have you had any success using this method for speed, or is it just for fun? I'm solidly in the "just for fun" camp myself.
Edit: Great video, that's much faster than my PB, which I think was 1:44 if I remember correctly. You are very fluent with those center comms!
Edit: Actually, maybe it is easier to force exactly 3 edge pieces unsolved if you don't always leave the unsolved edges in the same slice. If you solve the UL/UR edges, that's 6 pieces left in the free slice and UF/UB. So you would just need to solve exactly 3 of those 6, which seems easy enough ... maybe. This will give me something to investigate!
I made another version of this method by borrowing a bit from Triforce. My new version builds half-centers like Triforce and rearranges the steps a bit so that the last step where you solve the centers with commutators is simplified and more consistent. This method can be still done intuitively with commutators, but there's also more room to flex your alg knowledge, or even learn some algs specifically for this method. Hopefully that makes this version more attractive for speedsolving. Here are the steps:
1: First block and second square
Similar to original QTPI, but don't finish the second block. I build two opposite centers, then build the first block on the left, then second square on the right, but sometimes I'll do simultaneous block building or simply save a paired edge to use later.
2: Extend first block to a 2x3x4
Similar to step 3 of the original method, but we're putting off solving the corners. Also, there's a bit more freedom in solving this slice layer, since you can do R U* R' type stuff to change the alignment of a wing with the center pieces of the U face.
3: Pair remaining L/R edges and make center bars (half-centers)
This step is borrowed from Triforce. We can start out by pairing one more edge to store on the R layer next to the second square. Then use r slices and U moves to pair up the half centers. Now pair the remaining two edges with Redux/Yau/Meyer type inserts into the M-slice to preserve our half-centers. Note that U2 moves are perfectly fine because scrambling up the order of the half centers is fine in this method. Also, I like to make sure my last F2L/second block edge is oriented. If you know ZBLL but not ZBLS, you might want to orient all three remaining L/R edges. Once we're done with this step, solve the U centers. This will make the next step a lot easier; if you skip solving the U centers you'll have to worry about center rotation, like solving a picture cube.
4: Finish second block and solve corners and L/R LL edges
Finish second block F2L-style. This is why I like to have that edge oriented, so I don't have to rotate to solve it and rotate back afterwards. If you know ZBLS, you can do that too. Next is solving last layer, but the two unpaired edges are like wild cards. If you do OLL/PLL, you can choose whether to regard the unpaired edges as oriented or not, so you can automatically avoid dot OLLs. For PLL, the recognition is different and a bit weird, but the fact that the unpaired edges can be swapped or not swapped avoids PLL parity. (ZBLL is also possible, but with similarly weird recognition.) Also, there are a small number of algs that rotate the F, D, or B centers, which we don't want to happen. In PLL, it's just the MU EPLLs that do this, so use the RU algs instead. In general, algs with M's and Rw's are usually the ones that screw up our centers, so test them and look for alternative algs if they do mess up the centers.
5: Solve the Ul line intuitively
Next up is solving the line of wing-center bar-wing on Ul to leave just the r slice remaining. This is similar to step 4 of the original method, but you pair up one of the wings with a center bar first, before connecting with the other wing. You could also solve the Ur line instead, or leave out the center bar to solve in the next step.
6: Solve the last slice
You can do this with steps 5 and 6 of the original method. Or if you want to learn some algs, solving the last slice looks a lot like PLLs (including parity PLLs), so the full last slice can be solved with 43 algs. I call it PLS, for Permutation of Last Slice and I have some algs here. If you want even more algs, you could leave out solving the Ul center in the previous step, and you'd get something that looks a bit like HKPLL. I suggest learning the centers only cases, since they are nice algs and there aren't many cases.
