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WANTED: New Dedge Flip Algorithm!

You are clearly not familiar with Dan Cohen's parity song.


OK. I am cyclily stumped. He uses K4 hybrid, and is working towards a Dr. of Music. This is a parity paradoxical to me. Please post your link to this familiar song.

reThinker

surprisingly those are both almost correct.

The song is just a little ditty used to remember the moves of lucasparity, which is very simple. My little brother cubes about two times a year and can remember the OLL parity due to the little song. As for my method, I do a form of reduction that utilizes the 4x4's ability to be flexible, and directly solve a large portion of the cube before fulling reducing it to a 3x3. I still get OLL parity just like any reduction user would, let alone the fact that you can still get the OLL parity case when using K4.
 
Never mind, got 19q. This one costs $2.

EDIT: Comes with the 21q alg for free.

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?

I am waiting for you to post your advertisements in the "buy/sell/trade" section. This is not puzzle theory and is unacceptable from a "super moderator".

And, besides, I don't think the person who wrote the cube solver program you're using to find these algs will be very happy with you trying to gain profit!
 

It is very good to actually have this here -

r' D2 l B2 r' U2 r U2 l' B2 E2 l U2 l' (Pochman)

Nice alg ends up flipping the cube around spatially by relocating all of the pieces like doing a y2, and also a 4-cycle of edges to end up with a pure dbl parity fixer. Really neat use of cold, pure math!

reThinker
 
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You know, guys, you don't go "hey, I bet there's a better alg for <case> out there, just WAITING to be found! You guys are smart, I bet you can do it! Won't it be so cool?". You find the alg and then you post it. I mean, sure, I understand that you wish there was a better alg, but a better one isn't known and people aren't gonna make one appear out of thin air just because you think they should be able to. We already know the OLL parity alg is long and if we knew a nice shorter one we'd obviously tell everyone (and use it). This is just not a good way to discover algs. If you have an idea or an alg, tell it, but if you don't have any ideas and you just think there might be a great one we haven't found yet, well, go find it or I don't believe you.
 
This is just not a good way to discover algs

Sorry - previous ways were not getting good results.

Yeh, I am not just sitting around here banging on a pot, and expecting to be fed. Look!, I just cooked up another nice meal of parity stew, and shared it with everyone. Check out my earlier post on even-cycle ideas http://www.speedsolving.com/forum/showthread.php?p=292339#post292339, and give me your take on that, but don't bother trying to dissuade me. I'm on a mission, and I can't be stopped. ;)

reThinker
 
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Here is a possible speed version of Grandmaster P's earlier alg:

ColdHardMath™ - [(Rr)' D2 (Ll) B2 (Rr)' U2 (Rr) U2 (Ll)' B2 E2 (Ll) U2 (Ll)']

and this speed dedge flip for comparison:

FrederickBadie -[(Rr)' U2 (Ll) F2 (Ll)' F2 (Rr)2 U2 (Rr) U2 (Rr)' U2 F2 (Rr)2 F2]


reThinker
 
The song is just a little ditty used to remember the moves of lucasparity, which is very simple. My little brother cubes about two times a year and can remember the OLL parity due to the little song.


OLDlucasparity™ = (Rr) U2 x (Rr) U2 (Rr) U2 (Rr)' U2 (Ll) U2 (Rr)' U2 (Rr) U2 (Rr)' U2 (Rr)'

Correct me if this isn't it.

reThinker
 
I'm interested also to see it. But more than that (it's not a breaking alg anyway) I'm also interested in how you found this alg.
 
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.....

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.

No, only even cycles of odd permutations.

For example, here is a 10 cycle that is an even permutation:
R2 D R' M' B L' R B' M B L R' B' R D' R2
and any 3 composite edge cycle on a 4X4X4 is really a 6 cycle of even permutation. We naturally think of it as a 3 cycle, but it is really a 6 cycle (6 pieces are being moved, not just 3).

The most common 4 cycle performed by speedcubers, r2 U2 r2 (Uu)2 r2 u2 is an even permutation cycle as well.
 
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For example, here is a 10 cycle that is an even permutation:

and any 3 composite edge cycle on a 4X4X4 is really a 6 cycle of even permutation. We naturally think of it as a 3 cycle, but it is really a 6 cycle (6 pieces are being moved, not just 3).

The most common 4 cycle performed by speedcubers, r2 U2 r2 (Uu)2 r2 u2 is an even permutation cycle as well.

The permutation produced by r2 U2 r2 (Uu)2 r2 u2 is not a 4-cycle. It consists of two 2-cycles. The permutation produced by R2 D R' M' B L' R B' M B L R' B' R D' R2 is not a 10-cycle, but rather is two 5-cycles.

An even permutation always consists of an even number of even-length cycles, and an odd permutation always consists of an odd number of even-length cycles. The number of odd-length cycles does not matter in either case.
 
The permutation produced by r2 U2 r2 (Uu)2 r2 u2 is not a 4-cycle. It consists of two 2-cycles. The permutation produced by R2 D R' M' B L' R B' M B L R' B' R D' R2 is not a 10-cycle, but rather is two 5-cycles.

An even permutation always consists of an even number of even-length cycles, and an odd permutation always consists of an odd number of even-length cycles. The number of odd-length cycles does not matter in either case.

The even parity can also be interpreted as 2 3 cycles as well. So your statement is incorrect. A 5 cycle can be made by 1 3 cycle and 1 2 cycle or 2 3 cycles...
 
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The permutation produced by r2 U2 r2 (Uu)2 r2 u2 is not a 4-cycle. It consists of two 2-cycles. The permutation produced by R2 D R' M' B L' R B' M B L R' B' R D' R2 is not a 10-cycle, but rather is two 5-cycles.

An even permutation always consists of an even number of even-length cycles, and an odd permutation always consists of an odd number of even-length cycles. The number of odd-length cycles does not matter in either case.

The even parity can also be interpreted as 2 3 cycles as well. So your statement is incorrect. A 5 cycle can be made by 1 3 cycle and 1 2 cycle or 2 3 cycles...

No, I think you are confusing cycles and permutations. A 5-cycle can not be made by a 3-cycle and a 2-cycle.
 
What about the other ones I mentioned? The ones that you mentioned before, don't leave those out. I had a feeling you were only going to mention the 5 cycle being composed of 1 3 cycle and 1 2 cyle....
 
What about the other ones I mentioned? The ones that you mentioned before, don't leave those out. I had a feeling you were only going to mention the 5 cycle being composed of 1 3 cycle and 1 2 cyle....
Please... learn proper style like Bruce did it: "a 3-cycle and a 2-cycle". Improves readability a lot. Thank you. And:

- "The even parity can also be interpreted as 2 3 cycles as well"
Huh?

- "A 5 cycle can be made by [...] 2 3 cycles"
And your point is?

Or which ones are you referring to?
 
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