Easier Way?

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Easier Way?

Postby orionshunter » Thu Aug 12, 2010 7:14 pm

On a 5x5x5, when combining the final wings to the final edge pieces, is there an easy method of doing this?  I can get down to 3 wings left to place and I have a difficult time placing them.  There is no parity error as far as I know since there are 3 wings left and all the algorithms are 3 moves each.

Any ideas?

Thanks!
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Re: Easier Way?

Postby cuber969 » Thu Aug 12, 2010 9:27 pm

Yes.  Commutators.  Commutators, pieces of commutators, and combinations of commutators (maybe nested commutators, pieces of commutators with other moves, repeated commutators, etc.) cover a very big portion (may be all) of all algorithms there are which swap very few pieces on any puzzle, not just cubes.

Check out Thom Barlow's K4 Method ELL page.  It consists of '3-cycle' cases which probably match your description.  So scroll down to where it says '3-cycles' and view the 28 possible cases if all 3 wings are concentrated in one face (if not, you can do set-up moves to put all three in the same face).  I believe once you understand the concept of cycles (in an n-cycle, n pieces are swapped with each other simultaneously), swapping 3 wings in any fashion will come easy to you.  (For 3-cycles, 3 pieces can be cycled in two ways:  clockwise or counter-clockwise).
http://snk.digibase.ca/k4/7.htm

And yes, only 3 wings needing to be cycled is not parity.  An even number of wings which need to be cycled all at once are parities (2-cycles, 4-cycles, 6-cycles, etc.).  Also, reduction users may consider EVEN multiples of even cycles to be parity as well (ODD multiples of even cycles are obviously parity).  Such as two 2-cycles (also on Thom's page).  The two 2-cycle most well-known is 'PLL Parity'.
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Re: Easier Way?

Postby orionshunter » Fri Aug 13, 2010 7:06 am

I'm using this solution to solve the 5x5x5:
http://www.alchemistmatt.com/cube/5by5cube.html

I'm trying to keep the number of algorithms down to a minimum.  Once I get down to 3-6 wings, I'm looking for one or two maximum algorithms to complete them.  Is there a general strategy to prevent having nothing to work with when down to 3-6 wings in Step 6 of the above method?

Thanks!
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Re: Easier Way?

Postby cuber969 » Fri Aug 13, 2010 8:32 am



(Ignore the code above.  There apparently is some kind of bug in this
forum.)


'Is there a general strategy to prevent having nothing
to work with when down to 3-6 wings in Step 6 of the above method?'

I assume you are saying that you are wondering if there is a way to complete
the other 18-21 wings in such a manner that the algorithms to fix the remaining
3-6 wings is a minimum.




Unfortunately, every scramble is different, so in general, using a few types of
cycling algorithms to tackle all scrambles will not yield an efficient
solve.  In general, to solve wings in the fewer moves, you will need to
know a variety of algorithms to do:  2-cycles, 3-cycles, 4-cycles,
5-cycles, 6-cycles, 2 2-cycles and 2 3-cycles.



In the solution you are using, it appears that the only cycles which are one
3-cycle, one 2-cycle, and one 4-cycle.



The algorithm for 6a (3-cycle)

Changing the algorithm for 6a from R'r' F R' F' Rr F'  to r' F R' F' r F R F', it is clearly a
3-cycle of wings (just not 'pure' like the adjusted one).



Algorithms 6b and 6c are obviously not wing-involved algorithms, but just
set-up for algorithm 6a.



The algorithm for 6d (4-cycle)

Changing the algorithm from  (Rr U2)5 to (r U2)4 r, it is clearly a 4-cycle.



The algorithm for 6e (2-cycle)

Changing the algorithm from Rr Rr B2 U2 Ll U2 R'r' U2 Rr U2 F2 Rr F2 L'l' B2 Rr Rr
to

r r B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r r
, it is clearly a 2-cycle.



There are definitely many more kinds of 3-cycles, 2-cycles, and 4-cycle
algorithms not covered in that guide, and, like I said, there are other types
of cycles not covered at all.



But with all of that said, here is what I suggest you do:



Probably the most practical strategy you can use is to complete all composite
edges so that the only remaining wings to complete all have one of their two
stickers to be the same color.  So, for example, solve all wings which do
not have yellow in them.  Then you can do a few outer-layer turns to bring
the 4 incomplete composite edges (there will always be a maximum of 4 of them:
but 1-3 are also possibilities) into the same face.  Then use a
combination of the algorithms on Thom Barlow's page (which I previously gave a
link to) to solve the remaining 2-8 wings.  Thom said that between 2-cycles,
3-cycles, 4-cycles, and 2 2-cycles, only 3 algorithms are required to solve 2-8
wings which all have one of their two stickers as the same color (e.g. yellow).



If there are 2 composite edges incomplete, another good resource which does
this (besides Thom Barlow's page, which is all you really need) is this: http://rachmaninovian.webs.com/step4b.htm

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