Here's a video for the Tripod method:
http://www.youtube.com/watch?v=1na2FsB7ThI
Sorry for the long wait!
http://www.youtube.com/watch?v=1na2FsB7ThI
Sorry for the long wait!
Supercritical Method
- Thread starter ostracod
- Start date
krazedkat
Regular
Sounds kind of like my combo method (Froux) that I came up with awhile ago....
1. Build 1x2x3 block on LEFT
2. Build 1x2x3 block on RIGHT
3. Finish bottom two edges
4. Solve last layer...
1. Build 1x2x3 block on LEFT
2. Build 1x2x3 block on RIGHT
3. Finish bottom two edges
4. Solve last layer...
fanwuq
Legend
Bad idea and definitely not original and nothing similar except naming to the above Fridrus method.
It was discussed and shot down a while ago. Pure Roux or Pure Fridrich are more efficient [insert number smaller, but very close to 100]% of the time.
krazedkat
Regular
I know... I don't use that method.. I use Roux... 
...
...
qqwref
Legend
- Joined
- Dec 18, 2007
- Messages
- 7,834
- Location
- a <script> tag near you
- WCA ID
- 2006GOTT01
- YouTube
- Visit Channel
Hi, I've been working on tripod a bit and I thought of a good way to do the LL. It's more efficient than your approach, I think, and it does it in two looks maximum, which is pretty good if you don't want to learn every alg. (By the way I'm not really fond of your "three edges" start, it is a very inefficient way to do a 2x2x3 block. Just do it Petrus style, or even all at once if you can see it.)
First you have to learn the clockwise and counterclockwise "supertwistflip", a pretty short algorithm which twists the three corners in the same direction and flips both edges (of the LL). The ccw one is U' R U R2' F R F2' U F U' and the cw one is the inverse. You can learn fingertricks to make these really fast later. So now when you get to LL you recognize the edges first:
* Case 1: edges solved. You either have a commutator or some kind of corner twist case, so do it intuitively in one step.
* Case 2: edges permuted, but flipped. Use either supertwistflip to go to Case 1.
* Case 3: edges swapped, and maybe flipped. One corner will be correctly permuted, so find it, and if it's twisted apply the correct supertwistflip to fix it. So now you have two unsolved edges (possibly flipped) and two unsolved corners (possibly twisted). This is relatively easy to set up into a PLL, which should be J, R, V, or Y. Here's what you do: set up the corners first (do nothing if they are oriented, otherwise move one of them [without messing up the edges] so it is oriented and still on the LL), and then fix the edges with one move of the M or S slice. Then do your PLL (making sure to swap the pieces' new positions, not the old ones) and undo it all.
Here's an example: set up with D' R U R' U' D R' U R' U' R' U R2 U' y. This is Case 3, and the UBR edge is twisted, so apply the clockwise supertwistflip. Now you have something which looks like a J perm except that all four pieces are twisted. The setup is (corners) L'B' (edges) M', to a J perm. So the entire thing is:
U F' U' F2 R' F' R2 U' R' U
L' B' M' (R' U2 R U R' U2 L U' R U L') M B L.
First you have to learn the clockwise and counterclockwise "supertwistflip", a pretty short algorithm which twists the three corners in the same direction and flips both edges (of the LL). The ccw one is U' R U R2' F R F2' U F U' and the cw one is the inverse. You can learn fingertricks to make these really fast later. So now when you get to LL you recognize the edges first:
* Case 1: edges solved. You either have a commutator or some kind of corner twist case, so do it intuitively in one step.
* Case 2: edges permuted, but flipped. Use either supertwistflip to go to Case 1.
* Case 3: edges swapped, and maybe flipped. One corner will be correctly permuted, so find it, and if it's twisted apply the correct supertwistflip to fix it. So now you have two unsolved edges (possibly flipped) and two unsolved corners (possibly twisted). This is relatively easy to set up into a PLL, which should be J, R, V, or Y. Here's what you do: set up the corners first (do nothing if they are oriented, otherwise move one of them [without messing up the edges] so it is oriented and still on the LL), and then fix the edges with one move of the M or S slice. Then do your PLL (making sure to swap the pieces' new positions, not the old ones) and undo it all.
Here's an example: set up with D' R U R' U' D R' U R' U' R' U R2 U' y. This is Case 3, and the UBR edge is twisted, so apply the clockwise supertwistflip. Now you have something which looks like a J perm except that all four pieces are twisted. The setup is (corners) L'B' (edges) M', to a J perm. So the entire thing is:
U F' U' F2 R' F' R2 U' R' U
L' B' M' (R' U2 R U R' U2 L U' R U L') M B L.
And yes, qqwref, I probably should get into the habit of doing a petrus style 2 by 2 by 3 block solve. I don't use the 3 edge cross anymore, but I've still been doing it by placing corners first then edges. I think I have an easier time recognizing what to do fast (it takes me 15 seconds to build a block in this way), but I know petrus style would be much faster. New years resolution...
qqwref
Legend
- Joined
- Dec 18, 2007
- Messages
- 7,834
- Location
- a <script> tag near you
- WCA ID
- 2006GOTT01
- YouTube
- Visit Channel
Corners first? Ewwwwwww... Just try the Petrus way. It'll be hard at first, but if you practice it becomes easier and easier. I just did an average of 5 of 7.81 seconds to build it and I'm not even good at Petrus. I'm sure it's way more efficient than what you're doing anyway. If you want to be a slow intuitive solver you should concentrate on efficiency
About this "one look last layer system"... I know it's doable, and I think Fridrich's website already has algorithms for all cases (and if she doesn't I *know* Helmstetter's site does), but there's no way I would learn them all. My LL proposal above was for doing it in as few steps as possible without learning any new algorithms.
Yes, I know it's despicable that I do a corners first block; it's a remnant from when I started using the Petrus method as my second method. As a more experienced cuber I should break such habits.
A 1 alg last layer is considerably easier than Fridrich's site would suggest, because with our nice last layer (<-this should get some sort of abbreviation...) it is very easy to apply single move conjugates to form new nice last layers. For instance, the simple 3 corner cycle algorithm can normally cover only 2 cases (clockwise and counterclockwise), but when conjugates are used, this number is increased to 6 cases. I would estimate that one would need to learn maybe 30 algorithms, but that is a rough guess. The down side is you have to remember a lot of cases and which conjugates you should use, but this is a lot easier than remembering a lot of algorithms.
Here is an example code + solution:
"FUR:cycle-cB"
The F means that the left corner's top sticker is in front. The U means the right corner's top sticker is on top. The R means that the corner which is normally on the left side is now on the right. A colon separates the case code from the solution code. The name of the algorithm comes next, in this case "cycle" is the simple 3 corner cycle. Lastly, c stands for conjugate, and B means that the solver will turn the back side to form a new nice last layer. After turning the back side, the solver applies the cycle algorithm to the new LL, then turns the back side... well, back. You can try it for yourself if you want.
I will find out the specific number of algorithms when I find a solution to each case. I have already done 40 cases, all of which can be solved with one of 10 algorithms. I have devised a special case identification code, solution code, and a super fast recognition system, which lets the solver look at only 3 pieces in the nice last layer (the corner + edge + corner in front). This is going to be VERY fast!
PS: I think I am a slow solver for 3 reasons:
1. I have the dexterity of a horseshoe crab. I don't know any finger tricks.
2. Intuitive solving requires a lot of practice to master.
3. A 1 look last layer would do me good. XD It would actually shave off around 8 seconds.
A 1 alg last layer is considerably easier than Fridrich's site would suggest, because with our nice last layer (<-this should get some sort of abbreviation...) it is very easy to apply single move conjugates to form new nice last layers. For instance, the simple 3 corner cycle algorithm can normally cover only 2 cases (clockwise and counterclockwise), but when conjugates are used, this number is increased to 6 cases. I would estimate that one would need to learn maybe 30 algorithms, but that is a rough guess. The down side is you have to remember a lot of cases and which conjugates you should use, but this is a lot easier than remembering a lot of algorithms.
Here is an example code + solution:
"FUR:cycle-cB"
The F means that the left corner's top sticker is in front. The U means the right corner's top sticker is on top. The R means that the corner which is normally on the left side is now on the right. A colon separates the case code from the solution code. The name of the algorithm comes next, in this case "cycle" is the simple 3 corner cycle. Lastly, c stands for conjugate, and B means that the solver will turn the back side to form a new nice last layer. After turning the back side, the solver applies the cycle algorithm to the new LL, then turns the back side... well, back. You can try it for yourself if you want.
I will find out the specific number of algorithms when I find a solution to each case. I have already done 40 cases, all of which can be solved with one of 10 algorithms. I have devised a special case identification code, solution code, and a super fast recognition system, which lets the solver look at only 3 pieces in the nice last layer (the corner + edge + corner in front). This is going to be VERY fast!
PS: I think I am a slow solver for 3 reasons:
1. I have the dexterity of a horseshoe crab. I don't know any finger tricks.
2. Intuitive solving requires a lot of practice to master.
3. A 1 look last layer would do me good. XD It would actually shave off around 8 seconds.
PSS: qqwref, is this you?!
http://www.worldcubeassociation.org/results/p.php?i=2006GOTT01
I don't know how many people have the same name, and if you haven't updated the times in your signature recently... but if it is, I bow down to you! XD
http://www.worldcubeassociation.org/results/p.php?i=2006GOTT01
I don't know how many people have the same name, and if you haven't updated the times in your signature recently... but if it is, I bow down to you! XD
Last edited:
krazedkat
Regular
That IS super critical!
I should really start a new thread, because we are not talking about the Supercritical method now, but the Tripod method. :| The methods are vaguely similar, though, with nice last layers and using "Frifri" to solve corner edge pairs. (Supercritical relies on it too much though!)
qqwref
Legend
- Joined
- Dec 18, 2007
- Messages
- 7,834
- Location
- a <script> tag near you
- WCA ID
- 2006GOTT01
- YouTube
- Visit Channel
1. If you get a good cube and learn a bunch of finger tricks, I think you'd be surprised at how fast you could go. If you just want to practice turnspeed, try to do scrambles as fast as possible. 10 seconds or so is a good time
2. That's not an obstacle, that's a challenge
3. 8 seconds? Wow. Have you tried my method for doing the last layer?
Yeah. The times in my signature are unofficial averages of 12, so they're going to differ from the official averages of 5 by a bit (in either direction
).A challenge would still require more work nonetheless, but I agree, it's less of an "obstacle" as it is a way of gaining a better understanding for the cube while feeling good about "actually" solving it... I know that the Tripod method doesn't completely use intuition (because of the last step), but it's very close.
I timed myself for doing a 2 corner orienting algorithm (one which is somewhat long, but I am very good at doing), and I got about 7 seconds. Add 1 second for recognition, and that's how much longer it would take to do a 2 look last layer. The first algorithm might require a second longer for recognition, but not much.
Here is how long it takes me (right now) to do each step:
Block: 15-17 seconds
2 by 2 by 1 blocks: 15 seconds
Frifri: 7 seconds
Last layer (2 algs): 15 seconds
That adds up to around 53 seconds.
I timed myself for doing a 2 corner orienting algorithm (one which is somewhat long, but I am very good at doing), and I got about 7 seconds. Add 1 second for recognition, and that's how much longer it would take to do a 2 look last layer. The first algorithm might require a second longer for recognition, but not much.
Here is how long it takes me (right now) to do each step:
Block: 15-17 seconds
2 by 2 by 1 blocks: 15 seconds
Frifri: 7 seconds
Last layer (2 algs): 15 seconds
That adds up to around 53 seconds.
I went on a little tangent this weekend:
http://www.youtube.com/watch?v=Tm-8ydC2wcc
It is an edges first method. I have no idea how efficient it is. Here are the steps:
1. Solve 5 edges in a pseudo block.
2. Fix "bad" edges (like in the Petrus method).
3. Solve 2 pairs of adjacent edges, one on each unsolved face.
4. Solve the remaining 3 edges, fixing a parity if there is one.
5. Solve the 4 corners on one side, using 1 of 2 algorithms for each.
6. Solve the last 4 corners, using 1 algorithm to orient and 1 to permute.
http://www.youtube.com/watch?v=Tm-8ydC2wcc
It is an edges first method. I have no idea how efficient it is. Here are the steps:
1. Solve 5 edges in a pseudo block.
2. Fix "bad" edges (like in the Petrus method).
3. Solve 2 pairs of adjacent edges, one on each unsolved face.
4. Solve the remaining 3 edges, fixing a parity if there is one.
5. Solve the 4 corners on one side, using 1 of 2 algorithms for each.
6. Solve the last 4 corners, using 1 algorithm to orient and 1 to permute.
Similar threads
