That's expected after a 66 turn Little Chops solve, which I still have no idea how anyone would do something like that. As for the Tetra... I need 4 more. The Edge-Turning guys are killing me - especially those with movable centers.

Yea, that little chop solve gets me everytime I see it.
The best I've ever done is 113 / 9:38.92
And i felt like THAT was extremely lucky.

I went ahead and did 4.1.5. I didn't do 4.1.2 yet, lack of shift click on that puzzle is going to drive me to figure out a different method. And who knows when I'll get around to that, I'm about to go out of town for quite a while on friday.

Was that on a different program like UMC (Ultimate Magic Cube)? Cuz that time isn't listed on the scoreboards, nor does the applet measure to hundredths of a second.


I guess GB could add shift-click..., but I would never use it. I solve the 12 edges first. I get a lot of flexibility here because there are no centers to align to. The group of 3 stickers on whatever face then designates the face color.

I pick 4 non-adjacent faces and solve the 3 additional pieces on each to from a solved hexagon on each. This I do using pyraminx-type turns and setup turns. next I pair 2 pieces to each of the 8 corners so that it's solvable using only commutative turns of the 4 chosen faces.

tetra_e5 / 5.2.6
There seems to be a minor code bug on this one... When making shift-click turns, you can see through the internal surfaces. GB should probably fix this.

Anyhow I just solved 5.2.5 earning 1FM while beating fusion's time by a tiny bit. So I have 3 more to go.

Well, I'm dethroned.

I was too angry with my mom, so I've been gone for about a half a week.

Same here.


By two people now, so you better catch back up.

I have 4 left to tie Campbell at 116. (I'm really at 112, cuz my last four 5.2.x solves are not yet in.)

My To-Do list is fairly straightforward now. I need 2.2.8d, 2.2.9abc, 2.2.10abcd, 3.2.3, 3.3.3, and 3.3.5. So the Custom Dogic stuff and 3 others.

Well, I'm happy because I just beat 3.4.8!



4fde47b0227beff2684e2a63ace952d3
4902eeca58e56ecb401219cc32a7584f
b095623dc486682fd0866406c84b613e
c27689ec135ea1bc436e91dc237a85f4

Your time is better than mine, but I bet you didn't encounter every possible parity problem like I did. What did you run into while solving this?

I just solved 2.2.8d, and have 7 more left to finish off the 2.2.9abc and 2.2.10abcd. There are 25 puzzles under 2.2.x. I have solved 14 of them and have 11 remaining (3,4,5,7,9,10). After the 2.2.9 and 2.2.10, then only 4 more. It would be awesome to kill off all of 2.2.x, but the deep-cut ones (3,4,5,7) look really scary to me. I haven't tried them though, and not one has solved any of those 4 except Campbell who solved 2.2.4 and 2.2.5 recently. I'm sure he will get 2.2.3 and 2.2.7 soon.

Well when I was pairing up the edges, I first got one flipped. Once I flipped it, I realize that I need to switch two edges. But those problems took altogether only 3 short 3-cycles. That was the only parity I encountered.

Btw I solve in this order:

1. centers
2. edges
3. solve like a 3x3
4. little triangle bits

I don't know how you would flip a single edge-group, and I can't think of a way that could be possible so I'm curious as to how that could happen.
Switching two edges is easy to do, it is like coming up with a commutator for 4x4 edge-pairs. You put one half of each edge-group together, flip it in place, and then join the other two halves and flip that in place. I think 20 turns should be optimal for that.

I solve it differently, I put together entire edge-groups including the triangle bits first. I solve centers after solving 6 of these edge-groups and then I solve the other 6. Then I just solve like a 3x3, except possible EP-parity.

Hem, I just noticed my 5.2.6 solve didn't get entered in to the scoreboards. I'll have to e-mail GB.

You wouldn't really flip an edge but you would do a 3-cycle of the edge parts with two of the pieces being identical.

For the triangular bits yes, but there aren't any identical wing pieces. You have to do two 2-swaps, it's probably the most optimal way.

ah, I think we are having a miscommunication. I was calling the 3-sticker pieces the triangle bits and the 1-sticker pieces the edges.

For 3.4.8, I would say that the only 3-sticker pieces are the corners.... I would call the 2-sticker edge pieces, the "wings" to correspond to common 4x4 naming. Not sure of a good name for the 1-sticker edge pieces yet. But we need one because these things come up on several of the puzzles, including the 6.2.x spheres.

As a sidenote, I *just* solved 3.2.3 finally, but even with careful turning, could not beat Campbell's turn-count. 432 puts me in 2nd place on it. I thought I had figured out a good turn-optimized method, but it didn't hold up very well.

I need 2 more puzzles to tie Campbell at 116 now.

-.- Ya, I meant the 2-sticker pieces. So what I really meant was I was calling the 2-sticker pieces the triangle bits and the 1-sticker pieces the edges. I agree with naming the 2-sticker pieces wings as that makes sense. I don't have a good name for the 1-sticker pieces either though.

But I just solved 3.4.1! I don't know why I solved the 3x3 + skewb before the 2x2 + skewb but I did haha

I mean, we really, really need a name for those pieces. And I for one don't think it's a good idea to call them "edges"... functionally 4 of them can occur on the same edge-group on certain puzzles like 4.1.8 and 6.2.x. These pieces are incredibly important to talk about. Eventually I want to write up a tutorial on dealing with these pieces. I could suggest "diamond pieces" regardless of their actual 2d geometry (on 4.1.x they are actually shaped like a diamond, in the mathematical sense of the word).

I guess it'd be nice to have a name for the tip-adjacent pieces on the 4.1.x stuff too. It can get complicated when there are two layers of those pieces like on 4.1.6. Oh how about "tippy pieces", so then on 4.1.6 you can have "inner-tippies" and "outer-tippies". Although it's highly ambiguous if I mean closer to the tip or closer to the center of the face. But I am basing on the tips, as if they are "a column of 5x5 edges" (well 2 intersecting in this case).


Notice my ridiculous turn-count on that one. (Hope you didn't beat it...) I was afraid of it for the longest time too.

lol ya right! I got like 370 turns or so


And I just solved 3.4.2 as well but my time and move count both sucked. Especially because I ran into parity. So I'm gonna redo it now.

EDIT: well my second attempt wasn't too much better although I did cut both time and move count

The never before solved 2.2.3:

This is *SO* unfair!!!

I just solved 2.2.3. I get to be the first (unless someone was working on it today too)! I believe I now tie Campbell for leader, at 116 puzzles.

So here is how I solved the CO-parity that I had posted an image of on my last post. With the triangle in the center of the view "pointed up" and the corner we need to twist at that top point, we do turns of D and that top point in such a way that we cycle the 5 edges (that circle the bad corner) to their correct relative placements using a chain of simple commutators (11 turns or so). Then I used a sort of (RF'R'F)^9 (length 36) alg to do another commutative-chain to cycle the 5 centers to where the belong (the ones adjacent to the bad corner - which by now is not even really there anymore, so I mean 'the slot where it was'). Then I use whatever means to 5-cycle the 5 immediately adjacent edges to where they need to be. (I used 4 turns to get one, 4 turns to get a second and then a ton of setup moves to set up a perfect 3-cycle to finish and then undo setup moves.)

Now I have all of centers, edges, and corners done. I just have to fix the "tippies". And I do this using a commutator alg of length 10.

====
The CO-parity problem is not something I should have run in to, especially not *that* late in the game. What I did was solve like:
Centers, Edges, Tippies, then Corners.

Okay now here's what I should have done, which I realized half way though that my alg for Tippies was entirely preserving corners:
1. Solve Centers just like 1.1.12 (Impossiball), ignoring CO. Thinking about it this way makes it very easy, but in speed and turn-minimization.

2. Solve the Edges using the various 4-turn commutators with setups. Think Pyraminx Crystal...

3. [Try and] solve Corners - both orientation and permutation at the same time. So like if a corner is in but has wrong CO, then that is bad... try to not have that happen in the first place, but if it does then it needs to be taken out at some point, so just do it sooner than later by placing it in one of the incidental-slots/crud-locations prior to an alg execution. Here we can abuse a 12-turn alg that does 2-2-cycle very similar to (RF'R'F)^3 on cubes. As a sidenote, an alg that does this in 'pure form' (preserving tippies) would be *that* composed 7 times (i.e. length 84, yikes!).

4. If you encounter CO-parity then fix it now. It is so much less painful to fix without worrying about tippies.

5. Solve the tippies using 3-cycles. Find a commutator alg that does this. I'm not revealing mine yet.
====

On my solve I did the lengthy length 84 alg about 13-14 times for solving corners after tippies to preserve them. That was pretty gross. Using up 2391 turns for my solve would be easily and greatly beat, using what I know now. A way to improve the turn-count in my method would be to somehow compute the CO-parity right after step 2. Although I have never tried this, nor do I know of a way to do so off the top of my head. Fixing things at that juncture would give a *MASSIVE* turn-count bonus.


-Doug

I just finished solving 3.3.4. It is only a combination of 3.3.3 and 3.3.5.

I also had to redo my 2x2+Skewb solve because it sucked and I thought of a better way to do it.

It looks like you solved a whole lot of puzzles in the last 24 hours. Dozens! You are about to pass Michael up and you've seriously lowered his 1T points. You've also lowered my 1FM points a little.

I just solved 2.2.4 and it took me 2 hours. Last night when I solved 2.2.3 (which is actually harder because the only difference is that it has centers), it turns out I wasn't the first, cuz Campbell solved it that day too. So I am guessing he now has a 2 point lead on me again (at 119).

I'm not sure how to solve 2.2.5, the centers move in a strange way that I can't deal with. It's probably best that I gather points through 2.2.9 and 2.2.10 now, and skip 2.2.5.

So I solved 2.2.4, by first solving an entire turning-layer. Next I solved the "ring of 10 edges" under that without regard to tippies. I then tried to solve some of the corners a bit until I had 3 left on the same layer and tried to count the CO, and then fix it early. So after I did that, CO-parity was still there (but different... it started out -2 I think, so I did +2 but then it was still 1 off for some reason). So I had to fix that once again. For the final 3-edge cycle to get to all edges done, I got lazy and did it in away that not only messed up a lot of corners but messed up a lot of solved tippies on the other side. So I had to go though and solve a few tippies, then a corner or two, and a few more tippies and so forth. I did have to use my 84-turn alg about 3 times, which was pretty awful on turn-count.

Still I had a slightly lower turn-count than Campbell on it, but I'm sure he can easily beat it now seeing his turn-count on 2.2.3 which is the same but with centers.

Yes it is 179 and here's the *current* break down:
12-sided: 37
20-sided: 32
6-sided: 44
8-sided: 32
4-sided: 27
Sphere: 7

It's easy to count by multiplying things on the page - I usually count some rows and multiple by 5 and then adjust as needed. So where are these extra 8 puzzles hiding? GB could add a 2x2 Arrow Cube, a sort of 5.2.0 that is edge-turning but only has 4 corners and 4 centers (always solvable in like 4 turns I bet), a Super-Gigaminx, a Teraminx (ctrl-click?), a face-turning octagonal-prism, a Square-1, a Ruibk's UFO, and why not a 1x2x2 while we are at it.

I have been hoping to see more frequent posts here, but looks like I'll give it another go...

For icosa_f3 (2.1.1), I just found a length 29 alg that does this:

So I could solve edges, then face-pieces, then corners (somehow, not sure how yet), I would be left with almost 60 of those tippy pieces to do using as much as about 29*60=1740 turns plus setup moves. If I can score at least 2 tippies per 3-cycle then that cuts it nearly in half, but adds setup moves.

In conclusion, I just proved that this guy is feasible to solve. The only danger being that on such an extra-long sitting, I'd run into a power outage (since it's the season for such things where I am at).

I just did 5.1.1 to 5.1.5. Got stuck on centre tips of 5.1.6, still havn't managed them.

Good. The 5.1.x series was pretty easy for me. (The 5.2.x stuff was really hard in contrast.)

For 5.1.6, I solved the those pieces first. I call them "face-corners". Next I pair the edges like doing edges on 5x5-reduction. And then I solve like Halpern-Meier Pyramid.

Anyways I was starting to feel like I'm talking to myself here. I really hope people would post more about these puzzles. As for me, I've plateaued. It'd be really time consuming for me to try and maintain the pace with Campbell now that he's at 120 and I'm 3 behind.

I'm getting close to working out a full 2.1.1 method (that is, something that takes less than 3 hours). I refuse to just chip away at it over *several* days. I'll probably solve all the edges in one sitting, go eat, solve all the corners, go take a shower, and then solve like 30 of the 60 tippy-pieces, sleep, and then finish the rest the next day. Although I'm not sure what would be more tedious: corners or tippies. Edges will be fun, and just like solving the dual-edges for 1.2.1... or actually no, because although each has 30 of them, 2.1.1 has 2 faces of every color to give some extra freedom. So it's actually most similar to solving edges on the 4.2.4 (the way it turns and the lack of centers to frame things).

A better way I just thought of, is to solve them in corner-groups. Solve like 3 or 4 tippies on the same group, and then bring in the actual corner, and then "close". Repeat 11 more times!

In the meanwhile, I think I can attack 2.2.9a for *time*, and worked out a good LBL-ish method for it that should be bad on turn-count.

After your first post about 2.1.1, I decided to try it, because I knew I culd do it. It took 40 minutes to do the sides and the big one stickered pieces. The corners took another half hour. I started to do the tippies and I just gave up because I could tell it would take a very very long time. So good luck lol. By the way, I found a three-cycle of tippies that is 20 moves, but the three tippies are really spread out and difficult to set up.

A few days ago, I finished off the 2.2.x section except for one: 2.2.7. I came close once while trying it, but couldn't seam to get it. What I did was group side pieces together like on a 5x5 so that it turned into a 2.2.3. It was a little different from a 2.2.3 obviously so I couldn't quite get that part. I got lucky on 2.2.3 and didn't have the corner parity .

I also solved the 3.6.3 yesterday (the mix between 3.3.3, a master skewb, and a 3x3x3). It was even easier than the 3.3.3 because less set-up moves can be used. The only part that's different than it is that the corners are broken up into different pieces that can be easily grouped together first.

Campbell: A 2 turn, 5 second solve on 4.1.1 (Skewb Diamond).... SERIOUSLY!? Awh man!

How did that that happen? I thought my 5 turn, 7 second solve was incredible. The chances of even encountering something that is solvable in 2 turns or less is 57/138240 ~= .04123%, or worst than 1 out of 2425.

haha after I saw the 2 move solve, I tried for like 30 minutes to get a 1 move solve. I just kept pushing scramble over and over again but i never got it =(

Well ya... *sarcastically*, I list the odds on my last post to understand the rarity of that and detour people from attempting such a thing, cuz it's a waste of time. What I'm curious of, is how HE managed to pull it off.

Anyhow, I just solved 2.2.9a and 2.2.9b so I'm 1 away from tying Campbell again (as in 'again' I am 1 away, not that I was ever actually tied with him at any point). So I have 5 sort of freebies with 2.2.9c-2.2.10d, and then maybe do a 3.3.x or two, or maybe the 2.1.1.

So I have a bit of a problem with solving the longer puzzles... My computer crashes all the time now (probably from the heat), and it makes it impossible to keep from frustration when it happens on a big puzzle.

Hem, I think activity here, on other boards, and on GB's site has dwindled down during this US Nationals weekend in Atlanta, GA. (Which is where I should be! BLARG!)

I'm sure you are going to pass me soon. Unless I figure out the spheres or something, you're going to have the lead for a while I think.

About the skewb diamond solve, yeah I got really lucky. I just kept hitting scramble until something that was a little done already like I do with all the really easy, quick puzzles and got one that could be solve in two moves. It was kinda hilarious.

I need to go to a competition someday. I wasn't at a competition though over the weekends. I was on vacation at Six Flags in Dallas, Texas. It was a lot of fun but it's good to be home.

Good job Doug on the Icosahedra. It will be a while before I move onto those. I have still got a lot of catching up to do solving these guys.

I solved 4.3.1, which took forever.


I found a 3-cycle algorithm that was 32 turns. The way I did it was to solve the centers first, then corners, and last the 2-colored edge pieces. The edge pieces was where I had used the 3-cycle at. I might do it over but I might have to find a few more algorithms to speed it up.

That's a breakthrough. I played with 4.3.5 for a while and couldn't get anything. Also try the Little Chop duel, 4.3.3. I would assume you can apply your method from Little Chop on it.

I just solved 2.2.10b in 394 turns, which takes 1FM by a huge gap. So now I am 22/25 on the 2.2.x. Campbell is 24/25 and Noah is 21/25. We all need 2.2.7! It turns very strangely. There is no shift-click, and it's a little bit like Pentultimate... maybe it's similar to the Master-Pentultimate (1.1.6). It also don't function like the 1.1.8 either because you can't turn tips by themselves. I suspect that any solution will be over 1500 turns. I am missing an alg to 3-cycle the tippy pieces. The one I use on 2.2.3 doesn't work here because I would need the tip to turn with it.

I begin to wonder if it was GB's intention to have a puzzle that turns like this. There doesn't seem to be an intuitive way of solving it.

Here's a possible order: wings (no idea), corners (easy enough), middle edges (40 turn alg but messes up tippies), tippies (not sure sure yet).
This method is missing a lot of key algs. The two main problems is "how do we pair the wings with the corners?" and "how do we 3-cycle the tippies?".

Oh wow, I just found a short alg to split up wing pairs:


Because you can also pair wings across, then solve those groups, and then cycle corners around.

Btw, solving the center-pieces is easy because there you can just twist 2 adj corners commutatively for 36 turns and get the usual 2-2-swap. Thus, I would solve those last.

So I finally got around to solving 3.3.3, but I did so in the most inefficient way possible... I solved the corners, tried to solve the centers but then had to fix the parity there, then solved edge permutation, then solved the pieces adjacent to the edges (edge-thingies), then solved edge orientation, then was left with the pieces adjacent to centers (face-edges). At that point I had no idea how to proceed.

I ended up having to solve those face-edges while messing up edges and edge-thingies (but preserving corners and centers). And then going back and solving EP, then EO, then the edge-thingies. It's the way I should have solved it initially. The setup moves got very complicated at the end. But I don't think I would have gotten 1T or 1FM, if I hit scramble to start all over. I just wanted to check it off my list.

I am now somewhat convinced I had not solved 3.3.3 in the past (over a year ago was when I tried).

After that, I noticed that 3.6.3 (no one had solved before), looks very similar. I solved it in about 55 minutes using 659 turns, which should hold up as 1FM for a long time. My method was this:

1. Solve the induced 3.4.1 (2x2+Skewb) using whatever method. (You get one if you ignore the segmented-3x3's edges and centers.)
2. Solve centers.
3. Solve all the face-edges. It's optional, but I used some 3x3 stuff to solve some of the face-edges. Apply algs from 3.3.3, and by using 3x3 turns for setting up, things go much faster.
4. Solve EP using 3.3.3 stuff. Use 3x3 turns for setup to lower turn count a bit.
5. Solve EO. (It would have been smart to combine step 3 and 4 as much as possible of course.)
6. Solve the edge-thingies using 3.3.3's 3-cycle alg but you can use a variety of setup moves. I mostly used 3x3 ones, and kept forgetting obvious single edge-turns. You can get creative and use Skewb turns if you are comfortable with that.

Of course this outline only makes sense if you already know how to solve 3.3.3, and I purposely omitted algs (the notation for full algs would be atrocious because it is triple-turning/triple-cut). If I write a 3.3.3 solution up, I will include algs... because after step 1, it's essentially the steps to solving a 3.3.3.

Ops, it seems that someone did solve 3.6.3 before I did. Awh...
At least I did it in much fewer moves.

I noticed 3.6.3 a while ago and tried it. I tried it after I noticed that Campbell had solved it and I got an EO Parity (flip one edge). I had no idea how to fix that so I was going to try it again. I will probably end up doing it today and probably won't get fewest moves on it. I am working on solving a few more to pass Michael. I might redo 4.3.1 also because I think I should have taken less time on it.

EDIT: I got 00:34:42 1090 turns on 4.3.1. I'm satisfied with that.

I wasn't aware that was possible. I tried to get there on a solved one just now. My first idea was to flip the UR edge-group using: (UR{edge-turn} F2{face-turn})^3, but it leaves the U and R centers swapped and also the UFL and DFR corner-groups swapped.


That'll be another 1T point for you. You have passed Michael. You have 2 more than him, but 26 left to tie Noah now. What a gap! With all your 1T solves, I thought you could come in first on that ranking - that that would be a good goal for you, but looking at it, Campbell's 2'nd and got such a huge lead. Although his lead is on puzzles where there are 3 or less solvers (usually him, Noah, and me). Noah doesn't try for time, and I rarely do. I might do separate time-attacks on later passes at stuff. I would probably use a different method and turn off animation in those cases. So once you attempt those you have a good shot at 1T. This is because on many of the ones Michael has done, you were able to steal 1T from him and he's the current leader for 1T points.

Anyhow I think I'll re-solve 2.2.3 for 1FM, and solve 2.2.10 c&d today. It's bizarre to see that I now have a 5pt lead on Campbell somehow. He blames it on the Spheres but if I can pull off a +7pt lead then that excuse won't hold up anymore . (Since there are 7 spheres, and Noah can solve them too.)

Btw it looks like Michael performed really well at the US Nationals last weekend. First in 4x4 at that level of competition is incredible! He got 2nd place in 5x5 over FrankM as well. Fifth in One-Handed, s6th in Fewest-Moves, 2nd on Magic, 4th in Master Magic, 4th in Megaminx, 3rd in Pyraminx... he really racked up there.

Cohen, who participates on GB puzzles, broke 5x5 WRs (single attempt and average) of course, and everybody already knows that I'm sure.

Update
I solved all of the 4.2.x puzzles, I solved 3.4.10 in 00:44:45 with 1019 turns, and I passed Michael.

The centers on 3.4.10 were hard to figure out, everything else was easy to solve. For my first try on this puzzle, I think it went better than I thought.
The way I solved 3.4.10 was -
Solve edges like Super X
Solve centers
Solve inside edges
Doug, how did you solve 3.4.10?

I found a better 14 turn 3-cycle for 4.3.1 also. I still need to solve 3.6.3 and all of the 4.1.x puzzles.

First of all, I just solved 2.2.10c in under 1hr using 392 turns (compare to the others: 1021 and 1268). I've been solving about 1 new puzzle a day on average, although I did lower my Impossiball time significantly today (after holding 1FM, I thought I needed a less than pathetic time).

Okay. So you beat me on time on 3.4.10, but I also only attempted it once. I forget exactly how, it was one of the less memorable solves for me. Playing around with it again, I probably converted it to a 3.2.6 as my main strategy. I did this by "pairing the edges" first using mostly 2x2 turns. [Read the following carefully:] Then for each of 4 non-adjacent vertices, I formed a solved-group of 3 edges onto it. Then using mainly turns of those vertices, I solved 3 long 1-sticker pieces to each of the 4 groups. Next I formed groups of the remaining 6-piece-slivers. Then I fix those 4 vertex-groups' CO and solved the rest like 'centers of a Skewb'.

Forming those groups is the tricky part, which I omit because that secret takes all the fun out of it!

Btw, 'vertices' isn't in the default Firefox dictionary... but neither is 'okay'. How annoying. I just started using a different computer cuz my old one kept crashing, and just added all the common puzzle names .

EDIT: I just noticed that that last solve to take me to 126 puts me, and me alone, past the 70% completion mark (unless GB adds new puzzles at a rate proportionately faster than I can solve them).

Doug
That was really how I solved 3.4.10 also. I converted it to 3.2.6 as much as possible and solved from there.

I solved 3.6.3 in 00:35:11 in 1172 turns. I also solved 4.1.5 and got third place. Now I just have to solve that last five 4.1.x puzzles to get 90.

Well from your description, that's not what I got. But I think converting it to 3.2.6 is the way to go. But in forming the six groups of 6 pieces near the end, there are lots of ways you can minimize turns I bet. Your turn-count was kinda of high. Although, I'm not sure how Michael got it in about 100 turns lass than me. I wasn't super careful, but if i was I don't think I could have taken it down by more than 60.

As a minor update, I just re-solved Icosa-Pentultimate (2.2.6) for 26:29 and a few less turns. This is in preparation for solving 2.2.7 (Not Yet Solved by anyone).

So do we have a plan of attack on that one? Noah said 2.2.7 was similar to one of the 1.1.x (1.1.6 or 1.1.8 or both, I forget). Both of which I can't solve and he and Campbell can. I will try to find the magical 3-cycle on tippies today and solve it. I think I have a way now of pairing the 3-pieces edges like on 5x5 edge reduction. Then solve edge-groups and corners together in a LBL way. Then solve the somewhat trivial centers using 2-2-swaps. Then 3-cycle the tippes for the the finish.

EDIT: I just found a length 24 alg that does the perfect 3-cycle of tippies. So now I (theoretically) can solve it. Just a matter of time...

I just solved 2.2.7 in 1456 turns.

I decided early on that the 36-turn alg for doing 2-2 pure swap of centers was going to cost me a lot near the end. So I solved Centers-First, which is a bit like solving a 3x3 Corners-First. Well more like solving a Megaminx Corners-First...

So for this first step it's just like solving an Impossiball without CO and 2-fold indistinguishably. This is easy but confusing at times. And it's also something that needs to be preserved throughout the rest of the solve, by strictly using commutators between non-adjacent corners. Moreover, it's important to check it occasionally during the next step: edge-pairing!

If on 5x5, our "working layer" for edge-pairing is the "u,e,d slices" then here it will be any arbitrary layer. I tend to stick with it being on top tip of the screen (with the triangle face in the center of the view pointing up). We setup two things we can pair in at least 2 of those 5 working-slots. We do the turn to pair, and then use a 4-turn commutator replacement scheme to grab the a *useful* crud edge-triple from the other places (including the ring of 5 slots which move during the pairing-slice, although really only the nearest/front 2 are easy to get at).

This step goes on for a while since there are 30 edge-groups, you'll need to do this about 30 times (because you form 2 semi-pairs each time using an Arnaud approach).

Next you solve the induced 2.2.3 and pray you don't run into the nasty CO-parity. Although with the way I solved it's a bit easier.

For the induced 2.2.3, what I did was start with solving a 'cross' of 3 edge-groups onto a center. Remember that since centers are already solved (relative to each other) you may only use commutators (hopefully 4-turn...). Then I solve a bit like LBL. I solve the 6 additional edge-groups nearest my Cross-Face, which forms 3 distinct c/e-pairs. I solve the c/e-pairs using a keyhole-method. First I insert a corner (proper orientation, but don't worry about tippies yet), then I move it out of the way and insert the edge and move the corner back. I can do this for the 2nd c/e-pair, but for the 3rd, I solved the edge first and then I have an arsenal of fairly short algs to place the corner (like 17 turns for a bad case).

Then we do the usual Megaminx-ish stuff and attach pieces of 'cross' to one of the other faces (choice between 3) and solve like before. And then once again to a 3rd face (choice between 2 now). At this juncture we have 10 edge-groups and 6 corners remaining. I chose to solve the ring of 5 edges next, because those can each be done with 4-turn algs and up to 1 setup turn. Next I solved one of the 5 c/e-pairs. Now I have 5 corners and 4 edge-groups remaining. After a miracle, I have all the edges solved. Then I solve the 5 corners or using some brute force. I managed to cut some moves by applying the keyhole idea again.

It should now look something like this:


Now I assert that it is possible to 3-cycle tippies using a 24-turn alg, which I might reveal at a later date. But I found that the setup moves were a bit nicer with the tip not affected on turns of the vertex. Well I'll toss out a bone: it is of the form X Y Z Y' X' Y Z' Y', where X is a length 4 commutator, Y is length 2, and Z is a length 4 commutator. Oh that is a commutator too when interpreted as X (Y Z Y') X' (Y Z Y')'.

Anyhow, I 3-cycled my way to victory. Now I have one more left (2.2.10d) to finish off all of 2.2.x.

@GB
I would request a variation on 2.2.7, where the tips also turn. It becomes a different puzzle. (A significantly less annoying puzzle...)

I solved centers first, and then corners, and then edges using a 6-turn 2-2-swap alg, and then the 1-sticker pieces using some sort of 3-cycle that took 26-turns.

I got it in 1174 turns and 3hr 40min, which does not beat fusion in either way. You say it "took forever" but your solve was only 34 minutes.... mine took forever.

So there is an incredibly subtle parity phenomenon here. The only way to avoid completely is to solve the 24 edges first. Study this:

Notice how it's now impossible to finish solving by 3-cycling those 1-sticker pieces. This is because they form 4 split-orbitals of 6 pieces each. Since there are 8 colors total, you could have used process of elimination to find the two colors not on the orbital and make sure they are faces the sides of the 'rings of 6'. So to be clear, despite floating-centers there is a forced color-orientation due to these pieces. I ran into this parity and had to basically cycle everything (2 sets of 3 faces) about an axis perpendicular to another face. Figuring out which direction and what needed to be cycled was quite tricky - something I spent like 10 minutes on (not turning). The actual parity fix I used was like 10 turns long, but it messed up a edges. That's what took me so long and so many turns more. Having to cycle centers and everything, kinda sucks.

I noticed a problem with the applet:

The time stops and the "Game Over" box appears when I finish but the 'certificate' button is disabled... I wonder why this is if it wasn't a bug.

Good job figuring out 4.3.1. I redid 4.3.1 because I thought my one hour time took forever. Those one-sticker pieces you have in your first picture are part of what I solved first before moving on to the corners then the edges. That way you don't run into any parity problems.

I'm still working on solving all of the 4.1.x puzzles and some of the 2.2.x puzzles.

I also noticed the Certificate button as well. I don't know why it has been disabled.

I solved all of the 4.1.x puzzles and with times that I am happy with. I am now working on the 2.2.x puzzles to help break the 100 mark. I solved the Pentultimate using what algorithms I found and the ones Noah provided. I can solve half of the Pentultimate in a minute or less, but the last half is what takes forever. I found my own center 3-cycle also.

Campbell
How did you solve the Pentultimate corners so fast? Do you have better algorithms or was it a lucky solve?

Small update from me... I wanted to redo 3x3 for time, and on just one single attempt, I got really lucky and got it in 33 turns. Pretty wierd - after cross, the first 3 c/e-pairs were already paired. I followed a basic 6-turn ZBF2L case, then an easy anti-sune (7 turns).

Just now I redid Super-X for FM, and got 59 turns, which is 2nd. I was devising a method to combat the parity issues. I experimented with converting to "Dino-Cube with Parities", and converted one in *15* turns! But things didn't work out. I even had one case land perfectly at 53 turns (which would tie 1st place), but had a single flipped edge-pair which is a bad parity...

I also re-did 2.2.8b for 184 turns to take 1FM.


I noticed the 4.1.x stuff. This is especialyl since you knocked me off on 1FM on a few. I almost have all the 2.2.x, just one remaining.


I've been assuming he solves in a totally different way than I do. Probably uses a commutator approach, with some LBL stuff, to get both low turn-count and low times. I'm sure he has better algs with such low-turn counts. But I'm curious for his answer.

To me, any sort of block building method on a deep cut puzzle would not work very well.

Well, there does exist a length 8 alg that 3-cycles groups of pieces. (I'd give it, but it's it's too hard to come up with notation for it.) As in corner/face/corner groups. Using the idea of coming up with a commutator for solving J-Perm PLL on 3x3 by 3-cycling c/e-blocks.

So what is the general consensus now? Solve Corners-First or Centers-First? (to minimize turns)

For the pentultimate, I think it is quicker to go for corners first. The first 5 can be done intuitively around one face and so can 4 of the ones in the next layer. If centers were done first, you would be forced to use algorithms to get those 9 out of 20 corners and that would take a while and use a lot of moves.

After the other 11 corners are solved with the double corner swap that has been mentioned before. I do not use a different algorithm for flipping the corners. Then I solve the 11 (or less) remaining centers with a 20 move alg that swaps two pairs of centers. It's long, but it only needs to be done 2-3 times.