fixing parity earlier

[mccartym]

I have been solving the 3x3x3 for a few years now, and have recently started learning the 4x4x4. I can solve it easily unless I run into parity I have seen the algorithms to fix parity, and at one point I had them memorized. What I was wondering, is if there is simpler algorithm for fixing parity if you do it earlier in the method. I use the petrus method , and so I notice the parity before I fix the whole bottom layer. Is there some sort of shortcut? thanks!

 

[deco122392]

ok first how do you solve the "bottem layer"? second can you explain which step of the Pet' method? im guessing the "correct bad edges" step? is that correct?

 

[mccartym]

I solve the bottom layer just using the way I learned with the petrus method. Permute corners, rotate corners, permute edges. And your assumption was correct, I see that I have a parity problem in the correct bad edges step.

 

[AvGalen]

You should search for 4x4x4 FMC. By adding 1 move you can remove (or create) the "OLL-parity". After centers are done it is "to late" to efficiently solve parity

 

[MistArts]

You can find a "OLL-parity" yourself and use it if you have parity after correcting "edges" to "flip" one "dedge" which might be shorter as you are not preserving as much.

 

[Stefan]

After centers are done it is "to late" to efficiently solve parity

Depends on your definition of "efficient". Can still easily be done in seven moves at that point.

 

[AvGalen]

Not so easily for a mere mortal like me.

And 7 isn't efficient compared to the 1 it takes when you have only 3 centers done (that is the point Lars vdB and I determined the best to solve parity with a highly experimental speedsolving technique we never fully finished)

 

[Stefan]

Oops. Made a mistake, can't do it in seven, need the standard nine. Sorry.

How do you do it in one move after three centers?

 

[AvGalen]

FMC-style: You do 3 centers, calculated whether you have parity or not and solve the 4th center with 1 extra "slice"-move if you had parity

 

[jazzthief81]

FMC-style: You do 3 centers, calculated whether you have parity or not and solve the 4th center with 1 extra "slice"-move if you had parity

Did you mean to say: "solve the 3rd center with 1 extra slice move"?

Once you have solved 3 centers, everything you do after that consists of:

• Slice
• Move outer layer
• Slice back

So the parity remains unchanged after solving the 3rd center.

You can still affect the parity while solving the 3rd center by throwing in an extra slice move. This is how you can approach this:

• Make a 2x1 block
• Make a 2x1 block on another face

So far nothing different than what you would normally do. Now you can go two ways. There is the regular way to complete the center:

• Rotate the second 2x1 block so that it's in a different inner slice as the first 2x1 block
• Align the 2 2x1 blocks by means of a slice move

Another way to complete the center is to do it like this:

• Rotate the second 2x1 block so that it's in the same inner slice as the first 2x1 block
• Move the inner slice that does not contain the two half centers by one quarter turn
• Rotate the second 2x1 block so that it's in a different inner slice as the first 2x1 block (i.e. do a half turn of that face).
• Align the 2 2x1 blocks by means of a slice move

Doing it like this, you will end up with the opposite parity compared to the standard way. As you can see, it's 2 moves longer. So on average you could fix the parity using 1 extra move.

If you determine the parity during pre-inspection and you count the number of inner slice moves you perform as you solve the centers, you can fix the parity by solving the 3rd center in one of the two ways described above. Deciding which way to go can be anticipated while constructing the separate two 2x1 blocks.

 

[Kenneth]

Rw2 U2 F2 l F2 U2 Rw2 will solve parity after the centres are done, and it does only changes in the middle layers, not right and left layers, if you only mind the centres you can change l to Lw.

 

[Stefan]

Kenneth, you're swapping two centers.

Lars, thanks for the explanation, though for third center it's easier to see. Sneaky though that Arnaud omitted the "on average" for the one move. Ts ts ts.

 

[AvGalen]

Kenneth, you're swapping two centers.

Lars, thanks for the explanation, though for third center it's easier to see. Sneaky though that Arnaud omitted the "on average" for the one move. Ts ts ts.

That wasn't me being sneaky. It was Lars being incomplete when he told me about this in the car when we were playing around with this idea.

We knew about Chris counting cycles, but that takes to long for a speedcubing method. Then I thought counting "misoriented semi-pairs" would work, but it turned out that was harder than expected for a speedsolve.

Lars was the one that told me it would be best to do fix parity after the third center and that it could be done in 1 move. I guess I remembered badly.

The idea is still interesting for me, but I need to play Stepmania, learn PLL's, develop a better way for big-cube centers (all-rows) and switch to "non-flipping-over-the-clock"-solving first.

It's a good thing there are hardly any competitions left this year :rolleyes:

 

[JLarsen]

I solve the bottom layer just using the way I learned with the petrus method. Permute corners, rotate corners, permute edges. And your assumption was correct, I see that I have a parity problem in the correct bad edges step.

The explanation of the bottom layer you're talking about sounds like the last layer method that Lars uses. It's called CMLL. Perhaps you meant last layer?

 

[Kenneth]

Kenneth, you're swapping two centers.

Ah, darn, but it was a good alg

 

[deco122392]

hmmm ok so what you need are one of 2 algs fom what i can guess but im sure ohters probably have better ones then what i can give you. so here is on that flips the up front edge pairand swaps it with the up back edge pair.r2 B2 r' U2 r' U2 B2 r' B2 r B2 r' B2 r2 B2. ok scence im having a memory laps i cant seem to remember the single edge flip i cant give it to you but the algs are everywere.
and just for kicks ill give you this alg to play with i use it sometimes. switches the front up and back up edge: r2 U2 r2 u2 r2 u2 U2 or the version i use whichi think is faster :U2 r2 U2 r2 U2 r2 u2 hope that helps a bit

 

[fredqbr]

We knew about Chris counting cycles, but that takes to long for a speedcubing method. Then I thought counting "misoriented semi-pairs" would work, but it turned out that was harder than expected for a speedsolve.

Arnaud,
how do you count "misoriented semi-pairs" ?
It doesn't work to know the parity.
You have to count the 24-cycle, or the 12-cycle combined with orientation.

 

[wongxiao]

Okay, I'm a little confused here (being a noob to 4-cubes of course).

How exactly do you determine parity this early in the game..? I usually don't know until I'm treating it like a 3-Cube, during the Fredrich OLL/PLL steps.