trying-to-speedcube...
Member
@cmowla: Do you know you can also just do an edge 3-cycle in 9 moves?
-
Welcome to the Speedsolving.com, home of the web's largest puzzle community!
You are currently viewing our forum as a guest which gives you limited access to join discussions and access our other features.
Registration is fast, simple and absolutely free so please, join our community of 40,000+ people from around the world today!
If you are already a member, simply login to hide this message and begin participating in the community!
WANTED: New Dedge Flip Algorithm!
- Thread starter reThinking the Cube
- Start date
reThinking the Cube
Member
- Joined
- Dec 11, 2009
- Messages
- 294
cmowla ,
OK. Friend. I know (and everyone else knows as well) that you can solve this parity problem using "whatever" 2 edge swap as base).
But you know, and I know (and everyone else may NOT know) that there are better algs to get this done!
You can quote me on this: "pure swap cannot be less moves than unpure swap"!!!!!!!!!!!!!
You can do better, and you know it!
reThinker
OK. Friend. I know (and everyone else knows as well) that you can solve this parity problem using "whatever" 2 edge swap as base).
But you know, and I know (and everyone else may NOT know) that there are better algs to get this done!
You can quote me on this: "pure swap cannot be less moves than unpure swap"!!!!!!!!!!!!!
You can do better, and you know it!
reThinker
KwS Pall
Member
I use for 4x4 double parity fix if I see an oll parity - usually i have 1 look oll (only 8 cases out of 7x are 2 look)
setup corners for non-pure parity fix in order to speed up
setup corners for non-pure parity fix in order to speed up
Christopher Mowla
Premium Member
If you are taking about 9 block quarter turn moves, then no. The minimum that I know is 10 bqtm: I just used the standard 12qtm here.
I know that these algs I gave were longer than what the optimal alg is, but, as I said, they are for a start. And, just because I have found the briefest pure edge flip in the world doesn't mean that I can find the briefest non-pure edge flip that meets your request.
Last edited:
Stefan
Member
Proof, please.
Lucas Garron
Administrator
You officially do not have a brain.
By the way, I have a 21bqtm alg that takes a 3x3x3 state "with a flipped edge" (that is, a reduced 3x3x3 on a 4x4x4 with odd edge parity) to solved.
Last edited:
Stefan
Member
Is that true, or are you just acting like him?
Lucas Garron
Administrator
It's actually true. Want me to PM it to you?
qqwref
Member
- Joined
- Dec 18, 2007
- Messages
- 7,834
- Location
- a <script> tag near you
- WCA
- 2006GOTT01
- YouTube
- Visit Channel
I'd be interested in seeing it.
Cyrus C.
Member
+1, although he might not want to release it incase he writes a book on it.
PHPJaguar
Member
- Joined
- Dec 5, 2009
- Messages
- 142
...
Lucas Garron
Administrator
Nice. Now, can you find one with 21 outer block turns?
Lucas Garron
Administrator
I'll sell it to you, or anyone, for a $1 no-distribution license. Free for German competitors, though, including Stefan.
PM me for a Paypal account or to make other arrangements.
Stefan
Member
Ugh. Maybe. But not now and not since I can get yours for free. PM, please? And hey, consider that qqwref's real name is more German than yours
Lucas Garron
Administrator
Never mind, got 19q. This one costs $2.
EDIT: Comes with the 21q alg for free.
EDIT: Comes with the 21q alg for free.
Last edited:
reThinking the Cube
Member
- Joined
- Dec 11, 2009
- Messages
- 294
I can smell smoke!
The double parity that Jakub mentioned he uses to fix this dedge flip is actually a 4-cycle of 4x4x4 edges.
IMO - the main reason that the pure alg approach is running into trouble, is that it is forced to do a very specific 2-cycle that will also place the swapped edge pieces back into the original dedge. By not constraining this to just a 2-cycle (i.e. returning the edge pieces back to opposite sides of the same dedge location) should give an easier way.
In fact, any EVEN cycle (2,4,6,8,10) of the 4x4x4 edge pieces that make up the last 5 unsolved dedges will end up solving the flip parity, as long as the edge piece cycling ends up with paired dedges. This dedge pairing would normally be broken by these cycles, but here there is a solution because it is possible to target the piece cycles so that the dedges are reformed into their newly cycled locations.
The double parity is doing just that. By cycling 4 edge pieces (UFl-->UBl-->UFr-->UBr) the edge parity changes, and the dedge pieces also end up swapped as the natural consequence of this 4-cycle of edges. If the alg for this is nicer than the single parity one derived from 2-cycle edges, then the double parity alg should be the one used for fixing single parity as well.
This idea will work for other even cycles of 4x4x4 edges too. For example, the 6-cycle (UFl-->UBl-->URb-->UFr-->UBr-->URf) will fix edge parity, and at the same time move the dedges in a 3-cycle (UF-->UB-->UR). Note that the dedge parity in this case remains unchanged. It makes no difference though for this alg, since in practice, it takes too long to figure out if the dedge (PLL) parity is even or odd at this point in the solve. To just go ahead and change the dedge parity like the double parity does, or leave it the same like this 6-cycle, will end up being correct half of the time either way.
There are also 5 corners that can move, and do not have to be returned at the end of the alg. I don't even care what the corner parity is yet, since it will usually take longer to determine the corner parity here, than it will take to execute the alg that is used to fix it later on.
Should be no more excuses now. Somebody WILL get this.
reThinker
The double parity that Jakub mentioned he uses to fix this dedge flip is actually a 4-cycle of 4x4x4 edges.
IMO - the main reason that the pure alg approach is running into trouble, is that it is forced to do a very specific 2-cycle that will also place the swapped edge pieces back into the original dedge. By not constraining this to just a 2-cycle (i.e. returning the edge pieces back to opposite sides of the same dedge location) should give an easier way.
In fact, any EVEN cycle (2,4,6,8,10) of the 4x4x4 edge pieces that make up the last 5 unsolved dedges will end up solving the flip parity, as long as the edge piece cycling ends up with paired dedges. This dedge pairing would normally be broken by these cycles, but here there is a solution because it is possible to target the piece cycles so that the dedges are reformed into their newly cycled locations.
The double parity is doing just that. By cycling 4 edge pieces (UFl-->UBl-->UFr-->UBr) the edge parity changes, and the dedge pieces also end up swapped as the natural consequence of this 4-cycle of edges. If the alg for this is nicer than the single parity one derived from 2-cycle edges, then the double parity alg should be the one used for fixing single parity as well.
This idea will work for other even cycles of 4x4x4 edges too. For example, the 6-cycle (UFl-->UBl-->URb-->UFr-->UBr-->URf) will fix edge parity, and at the same time move the dedges in a 3-cycle (UF-->UB-->UR). Note that the dedge parity in this case remains unchanged. It makes no difference though for this alg, since in practice, it takes too long to figure out if the dedge (PLL) parity is even or odd at this point in the solve. To just go ahead and change the dedge parity like the double parity does, or leave it the same like this 6-cycle, will end up being correct half of the time either way.
There are also 5 corners that can move, and do not have to be returned at the end of the alg. I don't even care what the corner parity is yet, since it will usually take longer to determine the corner parity here, than it will take to execute the alg that is used to fix it later on.
Should be no more excuses now. Somebody WILL get this.
reThinker
Last edited:
Christopher Mowla
Premium Member
Did you use your brain, or a computer solver?
Stefan
Member
What exactly is "this"?
Lucas Garron
Administrator
UPDATE:
As a holiday special offer, I will now actually sell anyone the 19 outer block turn algorithm for the price of $1. That's right, for only $1 (and agreeing not to redistribute) you get to see the world's shortest known alg for OLL parity on 4x4x4. What a deal!
And of course, the 21q is still included! What are you waiting for?
Trade for any hand-found 23q pure OLL parity algorithms also accepted.
Last edited:
