cubernya
Premium Member
Just perform all the CMLLs. It won't matter if the cube ends up solved or not, as long as you did all 42 algs.
The One-Answer Question Thread [closed]
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I don't remember the details, but I do remember specifically that it was limited to 10^21 different scrambles due to it being 70 moves; more moves meant more different scrambles. By not remembering details, though, for instance, I don't remember whether there is some number of moves where all scrambles are reachable or not. There was a big discussion about this somewhere when Stefan first proposed it; it seems to me it may have been over on Yahoo! Groups, not here.
Mal
Veteran
Can someone tell me the edge parity 6x6 algorithim where there is only one edge left?
Pete the Geek
Regular
I'd really enjoy seeing the background for these numbers. I've done some searching to no avail. The source code linked on Stefan's page says that it analyses "all 2^70 possible scrambles" That big number is 10^21. I'd also like to see the calculation of the number of possible states of the megaminx, particularly what assumptions are made.
tasguitar7
Regular
Combinations on megaminx
30 edges - 2 orientations
20 corners - 3 orientations
only even permutations for edges or corners (they must both be even)
30 edges - 30!/2 permutations, 2^29 orientations (last orientation is based on the orientation of the other edges)
20 corners - 20!/2 permutations, 3^19 orientations (same parenthetical as above but for corners)
so 30!/2*2^29*20!/2*3^19 = 1.01 x 10^68
http://en.wikipedia.org/wiki/Megaminx#Number_of_combinations
Noahaha
blindmod
Is it possible/efficient to be CN ZZ?
Do you need to use your left hand, or can you rotate the cube?
Do you need to use your left hand, or can you rotate the cube?
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I think the thing I was thinking of on the Yahoo! group was the discussion about the validity of megaminx scrambles. It turns out there was a pretty good long discussion about scramble quality and ideas for alternate methods here. Also, the wiki states that generating every possible scramble with Stefan's method requires over 200 moves, implying it eventually covers all possibilities, but I have no idea what the source of that claim is. (Perhaps it's in the thread I quoted - I admit I never read through the entire thing.)
There are 70 moves in the megaminx scramble, each of which can take 2 directions. It's like a combination lock: there are n digits (let's say n=5) and each digit is in base x (let's say x=10). So that number of different combinations is x^n (10^5=10000, each number from 00000 to 99999). The megaminx scrambles are the same: n=70 moves, x=2 directions, so the number of combinations (scrambles) is 2^70 ~= 10^21.
Of course, some of those scrambles may reach the same positions, so the actual number of positions reachable in 70 move scrambles would be lower than 2^70.
tasguitar7 correctly explained the total number of positions.
tasguitar7
Regular
I guess its possible but ZZ seems like it would be the hardest method to achieve full CN with, and the benefits of it probably wouldn't out weigh the enormity of the task of learning it/relearning the cube and adjusting to strange EOline. I'm not really sure how CN would help much with ZZ.
Also, because ZZ is so <R,U,L> intensive you should already be using your left hand anyway. Oh, and also, cube rotations are something to be avoided in almost any method; although, I still rotate too much.
Noahaha
blindmod
Thanks. I'm practicing other methods to help me with my Petrus, and I find I really like ZZ and Roux. I think those three are a good combination since Petrus trains R and U, ZZ R U and L, while Roux trains M. I guess I'll be good at all the turns by the end.
Kingsman08
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wit 2 assembly?
hi guys,
i dont know how to assemble my wit-2 :/ could you guys lead me to a good tutorial? thanks a bunch
hi guys,
i dont know how to assemble my wit-2 :/ could you guys lead me to a good tutorial? thanks a bunch
cubingawsumness
Veteran
My PB is 28 seconds, and average is around 40 secs.
I first had a rubik's brand cube, which gave me solves of around 50 secs to a minute, and now, after very little research (bad choice), I have a dayan zhanchi stickerless. Is this not a good idea? Should I get like a "transition" cube, or something so that I can improve more?
Thank you for your help.
I first had a rubik's brand cube, which gave me solves of around 50 secs to a minute, and now, after very little research (bad choice), I have a dayan zhanchi stickerless. Is this not a good idea? Should I get like a "transition" cube, or something so that I can improve more?
Thank you for your help.
cubecraze1
Veteran
stickerless cubes are good unless you want to go to a competition,
you must have a sticker/tiled cube or you can't use them in competition
tasguitar7
Regular
The cube you have is just fine. A common quote around here is "it's the cuber not the cube", and it's true. The only way you will get better on any cube is by practicing, while cubing is more enjoyable with a "good" cube you don't technically need one to be fast. And Zhanchis are one of the most popular cubes around, you can't really go wrong with a Dayan. So practice! Also you may want to lube your cube to make it smoother.
cubingawsumness
Veteran
thank you tasguitar7 and cubecraze1 for your answers
i thought i read somewhere that you shouldnt use a dayan zhanchi unless u average like sub 20 or something
i guess whoever said that was wrong then
thanks again
i thought i read somewhere that you shouldnt use a dayan zhanchi unless u average like sub 20 or something
i guess whoever said that was wrong then
thanks again
cubecraze1
Veteran
not wrong just some peoples opinion
tasguitar7
Regular
Anytime, and well on here there is a private forum.
Only members that apply and are approved may post there in an attempt to make it among the best info on the site. As you can see the general consensus among the more experienced members is that good cubes promote good cubing habits.
http://www.speedsolving.com/forum/s...-quot-beginner-quot-cubes&p=699468#post699468
cubingawsumness
Veteran
i see
well then sorry that my question was repetitive as i didnt know and didnt take the time to seriously search
and thanks
well since i already have a zhanchi ill just go by ur opinion
and people will always have different opinions but this one seems more like a general consensus it seems
thanks
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NullPointer
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I've been using Frank Morris's method for the 5x5 when solving the 4x4, and I've been wondering if I switch to a different one for the 4x4. Is there a faster way I should be solving the 4x4?
cubelover111
Regular
what is a size of a sticker in "X" x "Y" i know its O.6 inch x 0.6 inch. I cant explain it very well but like there is resolution of a picture what is the resolution of the sticker?
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