cage/cage variants sub minute average potential?

[cubacca1972]

I did a test run of 15 solves to check the frequency of needing only 3 steps to solve the last 4 centers. 12/15 solves were 3 steps or less. 3/15 required 4 steps or less.

The sample size is too small. I quit due to using a relatively stiff Rubik's brand cube, and the subsequent MCP joint pain in my right thumb.

That being said, the actual frequency seems to be tending towards 80%.

 

[cubacca1972]

Another alg:

r2 U r' u2 r U' r2 u2

And its cousin:

r2 U2 r' u2 r U2 r2 u2

And its 2nd cousin:

r2 U2 r' E2 r U2 r2 E2

And its 3rd cousin:

r2 U r' E2 r U' r2 E2

 

[Poke]

Sorry if this sounds Noobish, but... what type of cube, 4x4x4 or 5x5x5

 

[rachmaninovian]

cubacca: lol the 2x2 CLL can mess up edges =P
so nah, i'd still fridrich for 3x3.
and the algs you gave are like...complicated. =P

poke: you should read the earlier parts of this topic and find that this is a thread more for a cage variant on a 4x4.

 

[cubacca1972]

cubacca: lol the 2x2 CLL can mess up edges =P
so nah, i'd still fridrich for 3x3.
and the algs you gave are like...complicated. =P

poke: you should read the earlier parts of this topic and find that this is a thread more for a cage variant on a 4x4.

Complicated? Could you elaborate? Do you know shorter algs that do the same thing?

Regarding Roux and CLL- If you mean solving the corners of the last layer in one algorithm, and are worried about ruining edges in the first 2 layers, then just learn the algorithms which don't ruin the edges.

 

[rachmaninovian]

i suck at blockbuilding, so roux is out of the question =P =P
i'd just do fridrich...almost sub 20 already with it...and i need it for r5 anyway.

complicated as in they do too much stuff at the same time...i see 3 swaps?

 

[cubacca1972]

i suck at blockbuilding, so roux is out of the question =P =P
i'd just do fridrich...almost sub 20 already with it...and i need it for r5 anyway.

complicated as in they do too much stuff at the same time...i see 3 swaps?

These are step 1 and step 2 algorithms for my system. Do the reverse of each algorithm. In my system, only the U face center will matter, not the adjacent centers. Watch the U center cubies only.

Roux is funky, but looks to be the most viable alternate system to Fridrich. Still, it needs to mesh with your personal style.

 

[cubacca1972]

In some nasty cases, the alg l E' L2 E l' E' L2 E (which 3-cycles bars) may be useful too.

I love this algorithm, and its full center swap variant. I do the mirror of it, as I prefer moves with my right hand.

 

[rachmaninovian]

i suck at blockbuilding, so roux is out of the question =P =P
i'd just do fridrich...almost sub 20 already with it...and i need it for r5 anyway.

complicated as in they do too much stuff at the same time...i see 3 swaps?

These are step 1 and step 2 algorithms for my system. Do the reverse of each algorithm. In my system, only the U face center will matter, not the adjacent centers. Watch the U center cubies only.

Roux is funky, but looks to be the most viable alternate system to Fridrich. Still, it needs to mesh with your personal style.

ah okay. do post more algs =P i do see potential in this system because the algs aren't as bad I thought they would be..=P

However I would think your initial system would be better: direct solving the first centre, then the opposite, then the last 2 in one alg.
in k4, ELL is done this way: 1st dedge in one alg, 2nd dedge in one alg, last 2 dedges in another alg. It is an efficient way to solve the last 2 dedges, and I would very much like a system like this =)

since I have already all the cases for the last set of algs, what is left is the first 2 sets...=P
to shorten the number of algs, I would only generate/find cases for stuff with 2 centres already done, for those that have only 1 centre done, I would just cheat with a niklas.
The current system I'm using is more like solving 1 entire centre, followed by another, and the last 2 be it adjacent or opposite. personally I prefer finishing with adjacent centres because they are faster and easier to control.

 

[cubacca1972]

ah okay. do post more algs =P i do see potential in this system because the algs aren't as bad I thought they would be..=P

I am editing post 42 of this thread to stick in algorithms for each case.

However I would think your initial system would be better: direct solving the first centre, then the opposite, then the last 2 in one alg.

My goal is to first make 4SCS (4 Step Center Solve) viable. Next, I want to get an acurate probability of needing 3 or less steps to finish the centers with this system.

After that, it will become clear if adding the rest of the algorithms to make it a 3SCS is worth it. There are 15 algorithms to know to draw cubies from adjacent faces to sort/solve the first face, and the same 15 for the opposite face. Evolving to 3SCS would mean making the jump to 39 algorithms to direct solve the first center, so adding 24 algorithms. You could replace the first 15 algorithms from 4SCS with faster ones (since you wouldn't need to preserve the opposite face) for step 1, but that would get silly. In any case, that's at least 24 new algorithms which will only be used in step 1.

That is why I want an accurate probability of getting a 3 step solve for 4SCS: Is it worth it to find/learn the extra algorithms? Plus, consider the much-hated setup moves required if involving all 4 centers.

The current system I'm using is more like solving 1 entire centre, followed by another, and the last 2 be it adjacent or opposite. personally I prefer finishing with adjacent centres because they are faster and easier to control.

Finish with adjacent centers every time....that should be doable. The 39 algorithm set for the first center, a new set of 15 algorithms to solve an adjacent center (new set which preserves an adjacent solved center instead of the opposite center), then 6 for the last 2 centers. The nice feature would be that you could see both centers at the same time.

btw, if you have any algorithms to add to post 42, post them, and I will add to the list.

 

[rachmaninovian]

well after thinking a lot, i'm more suited to freestyling, though i know that if there is a more rigid system, i might be better of with that..
i'd say that i'll just learn some of the useful stuff here and there, either from your list of algs, or tmoy's.

I would say that your system is more like TMOY's, but split into 2. first "orientation", then "permutation", thus it's not exactly direct solving =P ah wells...the number of centre algs i know as of this moment, excluding mirrors or similar stuff, are 6 for adjacent faces, and 6 for opposite faces.

pretty pathetic, huh? =P =P
i'd probably learn some of your cases to cover some of the bad cases i really do hate.

 

[cubacca1972]

well after thinking a lot, i'm more suited to freestyling, though i know that if there is a more rigid system, i might be better of with that..
i'd say that i'll just learn some of the useful stuff here and there, either from your list of algs, or tmoy's.

I would say that your system is more like TMOY's, but split into 2. first "orientation", then "permutation", thus it's not exactly direct solving =P ah wells...the number of centre algs i know as of this moment, excluding mirrors or similar stuff, are 6 for adjacent faces, and 6 for opposite faces.

pretty pathetic, huh? =P =P
i'd probably learn some of your cases to cover some of the bad cases i really do hate.

Yeah, it would have to be 3SCS in order to make it a pure direct solve. On the other hand, it seems like there would be plenty of direct solves under the 4SCS method.

 

[cubacca1972]

Actually, 4SCS can be made into a direct solve system:

Step 1 (if necessary): For the first center, direct solve any of its center cubies which are in the opposite center (e.g., if solving blue center first, solve any blue center cubies that are in the green face).

Step 2: Direct solve the first center (any remaining center cubies on the adjacent faces).

Step 3: Solve the opposite center

Step 4: Solve the remaining centers

So staying true to the philosophy of direct solving is possible.

Regarding the finish, I was thinking about which two centers would be the best to finish with, opposite or adjacent. I think that opposite might be the best, because I suspect that the algs for solving the second center are shorter if you solve the opposite center from the first. I have no proof of this, just a gut feeling. I haven't really studied the harder of the algorithms to solve the last two faces. The easy cases seem to be much faster for the easier cases though--well, at least the swap of full centers and the swap of 2 adjacent for 2 adjacent.

 

[rachmaninovian]

technically the move count for opposite centre finish is lower on average.

and the new description is what I kinda do =P just that step 3 can be either opposite or adjacent centres, so the finish depends as well.

the advantage of freestyling allows space for more freedom, like 3 cycling 3 different color centre pieces. =P a lot of my sub 55s solves had cases as such.

The total number of cases excluding mirrors/reflections for adjacent centre finish is 6. number in parentheses are the move counts..
1. niklas (8)
2. 1x2 blocks on both U and F face (8)
3. 1x2 block on U face and diagonal on F face (12)
4. diagonal on both F and U (right now I use 2 niklas without saving moves because it has the fastest execution...a better alg would be good..) (16)
5. 3 centres needed to be swapped on both F and U faces (16)
6. entire F and U centres swopped. (10)

for opposites, change the F to D face and the move counts are
1. 8
2. 4
3.12
4.10
5.10
6. 4

but the recognition for adjacent is faster...so it really just cancels out =P

last notes: direct solving is still fastest by itself as it is very clear what you are doing at the moment.

the cases of centres i need to find are 4 cycle centres, and also the direct algs for the more recent algs you released a few posts before this.
and of course, a better alg for case 5 of adjacent centres if possible...right now its just 2 niklas' withOUT cancellations..lol

EDIT: just for fun, an average of 5 for niklas (without AUFING): (2.41), 1.71, 1.52, 1.50, (1.27) ==> 1.57...maybe slightly higher during a solve..

 

[TMOY]

You can perform the case 3 for opposite centers in 10 moves only:
Lw' U Lw u2 r2 u2 r2 Lw' U' Lw
That's what I'm usually doing.

 

[cubacca1972]

technically the move count for opposite centre finish is lower on average.

and the new description is what I kinda do =P just that step 3 can be either opposite or adjacent centres, so the finish depends as well.

the advantage of freestyling allows space for more freedom, like 3 cycling 3 different color centre pieces. =P a lot of my sub 55s solves had cases as such.

The total number of cases excluding mirrors/reflections for adjacent centre finish is 6. number in parentheses are the move counts..
1. niklas (8)
2. 1x2 blocks on both U and F face (8)
3. 1x2 block on U face and diagonal on F face (12)
4. diagonal on both F and U (right now I use 2 niklas without saving moves because it has the fastest execution...a better alg would be good..) (16)
5. 3 centres needed to be swapped on both F and U faces (16)
6. entire F and U centres swopped. (10)

for opposites, change the F to D face and the move counts are
1. 8
2. 4
3.12
4.10
5.10
6. 4

but the recognition for adjacent is faster...so it really just cancels out =P

last notes: direct solving is still fastest by itself as it is very clear what you are doing at the moment.

the cases of centres i need to find are 4 cycle centres, and also the direct algs for the more recent algs you released a few posts before this.
and of course, a better alg for case 5 of adjacent centres if possible...right now its just 2 niklas' withOUT cancellations..lol

EDIT: just for fun, an average of 5 for niklas (without AUFING): (2.41), 1.71, 1.52, 1.50, (1.27) ==> 1.57...maybe slightly higher during a solve..

Man, I love this thread. The analysis just gets deeper and deeper.

I think that adjacent centers for the final step is going to prove viable, but will need to have a lower average move count for the second center algorithms. Otherwise, you would be confronted with two sets of algorithms which have a larger average move count.

Just an experiment, since you have the fast hands:

For each case to be solved in the last step, do the appropriate algorithm 10 times in a row, non-stop. Compare times between adjacent centers and opposite centers.

Which one wins for each case (adjacent vs opposite)? By how much?

Or, do the gauntlet for adjacent, and compare it with the gauntlet for opposite.

 

[rachmaninovian]

just gonna do average of 5 for each...i'm quite tired now. 10 is abit too much; and also this would have only a very small percentage error if you are consistent..

without AUFing i can save like 1 move so i will do it. so niklas is 7 moves now =)
rolling averages:
1. adjacent: (1.41), 1.55, (1.58), 1.52, 1.47 =>1.51
opposite: (2.03), 1.93, (1.63), 1.90, 1.84 =>1.89
comments: opposite took a while to warm up.

2. A: 3.11, 2.94, 3.06, (2.81), (4.84) => 3.04
O: (2.40), 1.94, (1.59), 2.16, 2.00 => 2.03
comments: you all know that opposites take more time to recognize so yea. for this case it would be only slightly faster for opposite.

3. A: (3.19), 2.94, 2.75, (2.46), 2.79 => 2.83
O: 3.06, (3.83), 3.06, (2.81), 3.13 => 3.08
again, opposite case will take slightly longer to inspect. adjacent might face more lockups..but that can be prevented..>.<

4. A: 4.44, 4.08, 4.03, (4.61), (3.75) => 4.16
O: (3.22), 4.28, (4.30), 4.09, 4.09 => 4.15
comments: draw. recognition for opposite would be slightly slower though.

5. A: 3.71, 4.50, 4.09, (3.63), (4.58) => 4.10
O: 4.77, 4.31, (3.65), (5.81), 4.25 => 4.44
I hate the opposite case A LOT.

6. A: 3.84, (4.34), (5.22), 4.02, 3.71 => 4.07
O: 2.50, (2.55), 1.81, (1.78), 2.47 => 2.26
quite obvious...opposite wins in this case.

so the final verdict is
1. adjacent
2. opposite
3. adjacent
4. draw...while adjacent is easy to spam the move count is high, lock ups are easy to happen, while the opposite's move count is lower, but the execution is not as slick.
5. adjacent
6. opposite

case 6 happens very rarely? for both cases.
overall winner would be very much be adjacent cases because many of the moves are restricted to <r l U> which are very fingertrickable.

and also, in general adjacents are easier to recognize and set up than opposite. that gives adjacents a boosted advantage.

lower move count on average does not necessarily mean faster times. =P
interesting result though.

 

[cubacca1972]

just gonna do average of 5 for each...i'm quite tired now. 10 is abit too much; and also this would have only a very small percentage error if you are consistent..

without AUFing i can save like 1 move so i will do it. so niklas is 7 moves now =)
rolling averages:
1. adjacent: (1.41), 1.55, (1.58), 1.52, 1.47 =>1.51
opposite: (2.03), 1.93, (1.63), 1.90, 1.84 =>1.89
comments: opposite took a while to warm up.

2. A: 3.11, 2.94, 3.06, (2.81), (4.84) => 3.04
O: (2.40), 1.94, (1.59), 2.16, 2.00 => 2.03
comments: you all know that opposites take more time to recognize so yea. for this case it would be only slightly faster for opposite.

3. A: (3.19), 2.94, 2.75, (2.46), 2.79 => 2.83
O: 3.06, (3.83), 3.06, (2.81), 3.13 => 3.08
again, opposite case will take slightly longer to inspect. adjacent might face more lockups..but that can be prevented..>.<

4. A: 4.44, 4.08, 4.03, (4.61), (3.75) => 4.16
O: (3.22), 4.28, (4.30), 4.09, 4.09 => 4.15
comments: draw. recognition for opposite would be slightly slower though.

5. A: 3.71, 4.50, 4.09, (3.63), (4.58) => 4.10
O: 4.77, 4.31, (3.65), (5.81), 4.25 => 4.44
I hate the opposite case A LOT.

6. A: 3.84, (4.34), (5.22), 4.02, 3.71 => 4.07
O: 2.50, (2.55), 1.81, (1.78), 2.47 => 2.26
quite obvious...opposite wins in this case.

so the final verdict is
1. adjacent
2. opposite
3. adjacent
4. draw...while adjacent is easy to spam the move count is high, lock ups are easy to happen, while the opposite's move count is lower, but the execution is not as slick.
5. adjacent
6. opposite

case 6 happens very rarely? for both cases.
overall winner would be very much be adjacent cases because many of the moves are restricted to <r l U> which are very fingertrickable.

and also, in general adjacents are easier to recognize and set up than opposite. that gives adjacents a boosted advantage.

lower move count on average does not necessarily mean faster times. =P
interesting result though.

Very interesting results. I have the frequencies for each case scribbled out somewhere. Weighing your results against the actual frequencies might come in to play.

I will continue to find the algs as I have been (i.e., solve the opposite center to the first solved center).

If you are going to search the algorithms for solving an adjacent center after the first one is solved, it would be nice if you could structure it so that you solve a center, get the solved center to the D face, then solve the B center. That way, you will already be looking at the two remaining centers for the final step.

 

[rachmaninovian]

looking at the k4 model, it seems to imply that you know an arsenal of algs, and use them in as little number of steps as possible.

besides these 12 algs, I really only know another one alg, which cycles 3 centres of different colour:
r U l U' r' U l'
which has been one of my main tricks which saved me from some bad times.
I have yet to find a nice alg to 3-cycle blocks between 3 colours...actually from the previous pages i believed that you (cubebacca1972) has already had such an alg..but it might require setup moves of 3 faces.

sure enough, I can cycle such cases, but I do it intuitively by setting up and doing one of the 2 algs...=P another one of my tricks.
for example i can do l2 U y' l' E l U2 l' E' l U' y l2 or something along that line.

trying to find such an adjacent solving centre system would deny me from have opposite cases =P not all my finishes have adjacent finishes, its more like a mix of the 2 depending on the easiest centre to solve after the first one.

once again, 3-cycling between 3 colours has made me not achieve such a finish...a lot of my solves end with 3cycling of 3 different coloured cubies?

I would really just come up with more algs by hook or by crook, and use them intuitively...=P

Also, I think your new description of the 4CSC is more original; since TMOY had already had the "orientation+permutation" approach.
I actually do this sometimes, without knowing it. =) when I see opposite cases, I immediately do it, especially for cases 2,4,6. Case 4 I would say once again that it is on par with its adjacent counterpart..=P
yea and most likely I would solve the first centre as such, with another 1/2 algs. that makes the maximum number of algs 3. normally if 3 algs are used, more centres are solved on the way, so most likely the solve would have a maximum 5 algs used, or if you are unlucky, 6, which is the maximum number of algs I would use. still, not to bad for my previous analysis of 4.08 (?) number of algs on average used for 60 solves. I would really allow +/- 1 number of algs, so technically anything below than 3 is either the result of luck, or by mocktricks like 3cycling between 3 colors.
still, such a system would probably yield a 4.x average number of algs for me. If I'm feeling bored enough i'd do 10 such solves to try..
whichever way seems fastest to do atm, I would do it.

EDIT: I shall just do an average...xD but using the easiest centre to solve next approach..
4, 5, 3, 3, 4, 4, 2, 5, 3, 5
comments: 7th solve was with heavy centre control, domino style. 8th solve would be better off if i tried adjacent centres. 10th solve would be the same with adjacent centres approach, but adjacent centres approach would be faster.
the average is 3.8, which is around the average of 4.08. This is not a very accurate average because I was trying some new centre control technique, which produced 2 of the 3's, and of course the 2.
All in all i would be using around 4 algs on average to solve it in this manner...
I should start using this centre control system =P it is pwning...
EDIT: after much thought, centre control on average gives easier centres, though not necessarily decreasing the number of algs. sometimes it gives way too lucky centres like that 2 alg finish (and very easy ones, case 1 opposite, case 2 adjacent finish). without that centre control I think I would use 4 algs...=P

 

[cubacca1972]

I am not too sure how old TMOY's method is, but the first version of separating centers then sollving center pairs that I have seen was on Jaap's puzzle page, from which I first learned the corners first/cage method.

I do like my revised 4SCS steps better than the original as well. I think it also flows a bit nicer. Plus, it would make an easier transition to 3SCS, since you wouldn't have to unlearn sorting behavior.

Here was that 3 cycle of 1x2 blocks on 3 faces:

r' U2 r E2 r' U2 r E2

Regarding freestyle solves:

It reminds me a lot of the orientation step in Roux. The finish in that system is rather organic vs. the machine-like finish of Fridrich and other 2 loook LL systems.

Adopting 4SCS or something like it would mean abandoning that sense of freedom that you get from freestyle. The only real "freedom" that you would get from 4SCS would be picking which center you start with; after that, you would pretty much be locked in for the rest of the solve.

A complete algorithm set is still needed. Unless the vast majority of the algorithms are really good/fast, there won't be much reason to change what is already working for you.

I do find it interesting that you have been getting many more sub 60 solves since the start of this thread.