Difference between OLL and PLL parity on 4x4

[cmhardw]

What a great explanation! As someone commented in the probability thread, you have a professor's knack for distilling and demystifying complicated puzzle theory into simple understandable chunks. Thanks for taking the time. I have a couple questions about this:

Glad to help I remember when people had to explain it to me, or help me through it. I'm glad to be able to pass that on.

Conceptually, I get that the corners and dedges, independently, will have odd or even parity. For OLL parity, I know that the parity state is determined by whether the total number (scramble + solve) of inner-layer turns (QTM) is odd or even. (1) Is that the same measure you mean when you say "the permutation parity of edges"?

The total number of inner layer turns (QTM) of the (scramble + solution to centers) is a measure of whether you will have OLL parity or not. In this case we are referring to the permutation parity of 24 wing edges. In my previous explanation I was referring to the permutation parity of 12 edge groups (dedges).

For the remainder of this post I am referring to the permutation parity of 24 wing edges.

A good way to remember how the parities are created, or not, is:

Scramble:
This has either an even or odd number of inner layer quarter turns. This means that the scrambled 4x4x4 cube has the 24 wings edges in some permutation, and that permutation will have either even or odd permutation parity.

Solution to centers (OLL parity):
Your solution to pairing up all the centers will contain either an even or an odd number of inner layer quarter turns. If your solution to centers has an even number of inner-layer quarter turns, then it is classified as an "even permutation" on wings. If your solution to centers has an odd number of inner-layer quarter turns, then it is classified as an "odd permutation" on wings.

- If your original scramble has even permutation parity of wings and you perform an even permutation on wings while solving centers, then the wings will still have even permutation parity (no OLL parity).
- If your original scramble has even permutation parity of wings, and you perform an odd permutation on wings while solving centers, then the wings will have odd permutation parity (OLL parity).
- If your original scramble has odd permutation parity of wings, and you perform an even permutation on wings while solving centers, then the wings will have odd permutation parity (OLL parity).
- If your original scramble has odd permutation parity of wings, and you perform an odd permutation on wings while solving centers, then the wings will have even permutation parity (no OLL parity).

Basically, the OLL parity is created (or not) the moment you execute the last slice turn to solve all the centers. The OLL parity state, either having it or not, remains unchanged when solving edges, because you are using conjugate maneuvers of (inner-layer turn) -> (Outer slice turn(s)) -> (inner layer turn) which performs an even number of inner layer turns at each iteration of pairing up edge groups. The effect of this is to leave the permutation parity of the 24 wings unchanged during the paring of edges.

You either have or don't have OLL parity at the moment you solve centers, and this state remains unchanged throughout the solution to pairing up edges unless you, the solver, change it by using an "OLL parity alg".

Clarification on OLL parity:
OLL Parity has absolutely nothing to do with the centers themselves. It only has to do with whether you solved the centers using an even, or an odd number of inner-slice quarter turns. The actual positions, or permutation, of the center pieces has absolutely no effect on whether you have OLL parity or not.

PLL Parity:
See my post above.

(2) What is the measure of the parity state of the corners -- the total number of outer-layer turns (QTM)?

Yes, counting from the moment you do the first turn to start scrambling the cube, the permutation parity of the corners is determined by whether you have done a even or an odd number of outer-slice quarter turns.

 

[advincubing]

Wow -- super helpful! Again.

I've almost got. This part is crystal clear:

OLL parity: If inner layer turns (QTM of scramble + solution to centers) is odd, there will be odd OLL parity (colloquially, "parity" or "parity error"). The subsequent inner layer turns related to pairing the edge pieces into dedges will not affect the parity state vis-a-vis OLL parity -- if odd before edge pairing, it will stay odd; if even before edge pairing it will stay even.

As is this part:

PLL parity: PLL parity cases are the result of there being a difference (one being odd and the other being even) between (a) the corner parity state ("determined by whether you have done an even or an odd number of outer-slice quarter turns") and (b) *something*.

I had thought that (b) was the OLL parity state. But I feel like you were suggesting something else, when you wrote this:

The total number of inner layer turns (QTM) of the (scramble + solution to centers) is a measure of whether you will have OLL parity or not. In this case we are referring to the permutation parity of 24 wing edges. In my previous explanation I was referring to the permutation parity of 12 edge groups (dedges).

For the remainder of this post I am referring to the permutation parity of 24 wing edges.

Can you elaborate on the distinction? What is (b) if not the OLL parity state, as locked in after forming the centers?

Thanks so much for all that you've contributed to this topic. I've learned so much over the past couple days, and am now really intrigued by puzzle theory.

 

[cmhardw]

But, on a 5x5, if the centre stickers (the fixed centres, not the 3x3 centres) are removed, wouldn't it be possible to have PLL parity?

In a way, yes. You could have the permutation parity of the centralmost edges be different from the permutation parity of the corners (what everyone is calling "void cube parity"). Then, you could solve the 24 wing edges "correctly" around the centralmost edges. This would then appear to be that two tredges are swapped on the cube. At least I think this is what you mean with "PLL parity" on a 5x5x5. Does that sound right?

Then you can have a void cube parity. (PLL parity can't, because of the middle pieces of the edges.)

I guess you sort of could have PLL parity, see my above description

I hadn't though about it that way. PLL parity is just 2 3-cycles of edges anyway though. So it can occur as void cube parity or the even/odd thing. Oh wait, void cube parity has to do with an even/odd number of slice moves.

PLL parity: PLL parity cases are the result of there being a difference (one being odd and the other being even) between (a) the corner parity state ("determined by whether you have done an even or an odd number of outer-slice quarter turns") and (b) *something*.

I had thought that (b) was the OLL parity state. But I feel like you were suggesting something else, when you wrote this:

Can you elaborate on the distinction? What is (b) if not the OLL parity state, as locked in after forming the centers?

(b) is the permutation parity of the 12 paired up dedges the moment you complete the pairing of all edges on the cube. PLL parity (or the lack of PLL parity) is "locked in" so to speak the moment you complete the edge pairing step. This assume that you use the solved center groups to define the cube's "solved" state.

The permutation parity of the 12 paired up dedges is an artificial construct of sorts. You as the solver are creating this in your mind so to speak. This is the goal of the reduction method in fact, to reduce the 24 wing edges to a state that can be treated as if it were 12 dedge groups that can be solved using only outer-slice turns (assuming OLL or PLL parity are not present).

Also, read elrog's comment. Elrog is describing how PLL parity is not really a parity when viewing the cube as having 24 wing edges. If you view the cube as having 24 wing edges, then PLL parity is just two three cycles, which is classified as an even permutation, hence no odd permutation parity is present here. However, if you change your viewpoint to be such that the 4x4x4 cube is composed of 12 dedges, then you indeed have odd permutation parity for the dedges when you have PLL parity.

PLL parity is only a "parity" error if you think of the 4x4x4 as composed of 12 dedges. Since everyone who solves with standard reduction does view the cube as composed of 12 dedges, then this is why PLL parity is called a "parity error".

Thanks so much for all that you've contributed to this topic. I've learned so much over the past couple days, and am now really intrigued by puzzle theory.

Sure thing, glad to help!

 

[advincubing]

(b) is the permutation parity of the 12 paired up dedges the moment you complete the pairing of all edges on the cube. PLL parity (or the lack of PLL parity) is "locked in" so to speak the moment you complete the edge pairing step.

Meaning: The number (odd vs. even) of inner slice moves (QTM) of the scramble + solution through reduction to 12 dedges?

If that's not right, then I have some sort of crazy mental block for getting my head around the dedge parity!?! If you recorded on paper every move it took you to get from a solved cube through a scramble through reduction to 12 dedges, what would you count to see if the dedges have odd or even parity?

 

[cmhardw]

Meaning: The number (odd vs. even) of inner slice moves (QTM) of the scramble + solution through reduction to 12 dedges?

The permutation parity of the 12 paired up dedges is related to the number of outer-slice quarter turns it would take to solve all the dedges correctly in relation to the paired up center groups.

If that's not right, then I have some sort of crazy mental block for getting my head around the dedge parity!?! If you recorded on paper every move it took you to get from a solved cube through a scramble through reduction to 12 dedges, what would you count to see if the dedges have odd or even parity?

The dedge parity (PLL parity) is related only to outer-slice quarter turns. If you consider inner slice quarter turns then PLL parity is not possible. Considering only inner-slice quarter turns is like considering that there are 24 wing edges. 24 wing edges can be in either even or odd permutation parity. If you have all the dedges paired up, but you are examiningg the number of inner slice quarter turns, then you are checking for OLL parity. 24 wing edges in odd permutation parity is OLL parity. 24 wing edges in even permutation parity is "not OLL parity."

 

[advincubing]

The permutation parity of the 12 paired up dedges is related to the number of outer-slice quarter turns it would take to solve all the dedges correctly in relation to the paired up center groups.

That makes perfect sense. Without thinking about it too deeply, I'm pretty confident that is NOT as simple as counting the outer slice turns from scramble through the solve to reduction, right? (Counting the outer-slice turns would give you the corner parity, right?)

 

[cmhardw]

That makes perfect sense. Without thinking about it too deeply, I'm pretty confident that is NOT as simple as counting the outer slice turns from scramble through the solve to reduction, right? (Counting the outer-slice turns would give you the corner parity, right?)

Counting the number of outer-slice or inner-slice turns starting with the scramble through your solution will certainly help you determine the permutation parity of a piece orbit. However this can be very impractical. Permutation parity can always be calculated from any "scrambled" state as long as you know what the "solved state" looks like.

Even more practically speaking at the end of the reduction steps you can look at the cycles for the corners and the cycles for the dedge groups the same way you would memorize a blindfolded solve. Calculating the parities of the corner permutation and the dedge permutation at this point will tell you at the end of the reduction steps whether you have PLL parity. Using this same technique(s) you can calculate the permutation parity of the 24 wing edges and know at the end of the reduction steps whether you have OLL parity.

A while ago people were trying 4x4x4 fewest moves on the forum here, mostly for fun but also a little competitively. Using the above techniques of calculating permutation parity you can do a reduction solve that will, guaranteed, have neither OLL nor PLL parity. All you have to do is calculate the permutation parity of the 24 wings after executing the scramble. You can also count the number of inner-slice quarter turns in the scramble to get this result. Then, you control the number of inner-slice quarter turns you do to solve centers such as to create even permutation parity in the 24 wing edges once the centers are solved. Then, near the end of pairing the dedge groups up you calculate the permutation parity of the corners. As you build the last dedges, make sure that in doing so you give them the same permutation parity as what the permutation parity of the corners will be at the end of edge pairing. Voila, both OLL and PLL parity can be avoided 100% of the time if you want to.

 

[advincubing]

I'm sure I'm driving everyone crazy here, but trying to put this all together....

I had asked:

PLL parity cases are the result of there being a difference (one being odd and the other being even) between (a) the corner parity state ("determined by whether you have done an even or an odd number of outer-slice quarter turns") and (b) *something*.

And you responded:

(b) is the permutation parity of the 12 paired up dedges the moment you complete the pairing of all edges on the cube. PLL parity (or the lack of PLL parity) is "locked in" so to speak the moment you complete the edge pairing step. This assume that you use the solved center groups to define the cube's "solved" state. ... PLL parity is only a "parity" error if you think of the 4x4x4 as composed of 12 dedges. Since everyone who solves with standard reduction does view the cube as composed of 12 dedges, then this is why PLL parity is called a "parity error".

And separately you added:

The permutation parity of the 12 paired up dedges is related to the number of outer-slice quarter turns it would take to solve all the dedges correctly in relation to the paired up center groups.

Putting this all back to (a) vs. (b) isn't working for me. Is that even a helpful construct? If (a) is a measure of outer-slice turns, I don't see how (b) can also be the same measure. Unless (a) is backwards looking (how many have been done to get to reduction?) and (b) is forward looking (how many would it take to get the dedges into their proper position vis-a-vis the center pairs?)....

By, the way, I found superantoniovivaldi's parity video really helpful, especially at around 19 mins, where he explains "breaking the rules of reduction." The cups are a great tool.

 

[cmhardw]

PLL parity cases are the result of there being a difference (one being odd and the other being even) between (a) the corner parity state ("determined by whether you have done an even or an odd number of outer-slice quarter turns") and (b) *something*.

(a) is the permutation parity of the corners
(b) is the permutation parity of the 12 dedge groups once you complete the edge pairing steps.

The permutation parity of the corners can be calculated by knowing whether it would take an even or an odd number of outer-slice quarter turns to solve all the corners in relation to the paired up center groups. The permutation parity of the dedges can be calculated by knowing whether it would take an even or an odd number of outer-slice quarter turns to solve all the dedges in relation to the paired up center groups.

There are 4 possible cases here, two of which are "PLL parity" and two of which are "not PLL parity".

1) PLL parity does not exist when the corners would require an even number of outer-slice quarter turns to solve, and the dedges would require an even number of outer-slice quarter turns to solve.
2) PLL parity does not exist when the corners would require an odd number of outer-slice quarter turns to solve, and the dedges would require an odd number of outer-slice quarter turns to solve.
3) PLL parity exists when the corners would require an even number of outer-slice quarter turns to solve, and the dedges would require an odd number of outer-slice quarter turns to solve.
4) PLL parity exists when the corners would require an odd number of outer-slice quarter turns to solve, and the dedges would require an even number of outer-slice quarter turns to solve.

Putting this all back to (a) vs. (b) isn't working for me. Is that even a helpful construct? If (a) is a measure of outer-slice turns, I don't see how (b) can also be the same measure. Unless (a) is backwards looking (how many have been done to get to reduction?) and (b) is forward looking (how many would it take to get the dedges into their proper position vis-a-vis the center pairs?)....

Counting the number of outer-slice quarter turns from the end of the reduction steps through to the solved state is a measure of the permutation parity for both the corners and the dedges. PLL parity happens when that one measure reports different results for the corners than for the dedges. PLL parity does not happen when that one measure reports the same results for both corners and dedges.

By, the way, I found superantoniovivaldi's parity video really helpful, especially at around 19 mins, where he explains "breaking the rules of reduction." The cups are a great tool.

I haven't seen this video yet, and it is very late for me tonight. I will check this out tomorrow though!

 

[advincubing]

The permutation parity of the corners can be calculated by knowing whether it would take an even or an odd number of outer-slice quarter turns to solve all the corners in relation to the paired up center groups. The permutation parity of the dedges can be calculated by knowing whether it would take an even or an odd number of outer-slice quarter turns to solve all the dedges in relation to the paired up center groups.

Got it. This makes sense to me. Thanks for your patience and clear explanations.

I would have thought that *how* you manipulate the corners and dedges into a solved position would influence PLL parity -- i.e., which F2L steps and PLL perms you use. But you wrote this earlier:

PLL parity (or the lack of PLL parity) is "locked in" so to speak the moment you complete the edge pairing step. This assume that you use the solved center groups to define the cube's "solved" state.

Is it true, then, that no combination of F2L and (3x3) PLLs that would place the corners and reduced edges into a solved state will affect the parity state that is locked in when you're done with edge pairing -- i.e, that any combination of F2L and "normal" (non-parity) PLLs will have an even parity as between the corners and dedges? And that is why you need a PLL parity algorithm that will change the parity state of either the corners or dedges.....

 

[cmhardw]

Thanks for your patience and clear explanations.

No problem. Parity on 4x4 is a very involved subject, and it takes time to learn it.

Is it true, then, that no combination of F2L and (3x3) PLLs that would place the corners and reduced edges into a solved state will affect the parity state that is locked in when you're done with edge pairing -- i.e, that any combination of F2L and "normal" (non-parity) PLLs will have an even parity as between the corners and dedges? And that is why you need a PLL parity algorithm that will change the parity state of either the corners or dedges.....

It is true that no combination of F2L and (3x3) PLLs will affect the PLL parity state. Whether you have PLL parity or not is decided by the end of edge pairing and will remain unless you execute a PLL parity algorithm.

To understand why this is, we have to look at what a quarter turn on a 3x3x3 does when viewed as a permutation on corners and edges. When you turn a quarter turn of a side on 3x3x3 you are executing a 4-cycle on corners, as well as a 4-cycle on edges. This means that a quarter turn on 3x3x3 is an odd permutation on corners and an odd permutation on edges. This means that every quarter turn you do on a 3x3x3 will change the permutation parity of both corners and edges.

Now let's examine a reduced 4x4x4. This, by all appearances, has the look of a 3x3x3 cube. However, your reduced 4x4x4 has 4 possible states:

There are 4 possible cases here, two of which are "PLL parity" and two of which are "not PLL parity".

1) PLL parity does not exist when the corners would require an even number of outer-slice quarter turns to solve, and the dedges would require an even number of outer-slice quarter turns to solve.
2) PLL parity does not exist when the corners would require an odd number of outer-slice quarter turns to solve, and the dedges would require an odd number of outer-slice quarter turns to solve.
3) PLL parity exists when the corners would require an even number of outer-slice quarter turns to solve, and the dedges would require an odd number of outer-slice quarter turns to solve.
4) PLL parity exists when the corners would require an odd number of outer-slice quarter turns to solve, and the dedges would require an even number of outer-slice quarter turns to solve.

If PLL parity is created at the end of edge pairing, then your 4x4x4 cube is in either state #3 or state #4 in my list above. Let's say that your cube is in state #3. What I mean by this is that your corners have even permutation parity, and you dedges have odd permutation parity. When you do a quarter turn of an outer-slice on your 4x4x4, you are 4-cycling the corners on that slice, and 4-cycling the dedges on that slice. This performs an odd permutation on corners, and an odd permutation on dedges. This changes the permutation parity of corners to odd and the permutation parity of your dedges to even. Notice that odd permutation parity for corners, and even permutation parity for dedges is state #4 in the list above.

No combination of outer-slice turns can get you out of the cycle of going back and forth and back again in a loop between states #3 and states #4 above. Performing a PLL parity alg will break you out of the state #3 and state #4 loop and put you into a loop between states #1 and states #2. In the state #1 and state #2 loop, any time you execute a quarter turn of an outer-slice this will take you from state #1 to state #2 and vice versa. Performing a PLL parity alg again puts you back into the state #3 and state #4 loop.

By the way, a 3x3x3 cube is a loop between state #1 and state #2.

A void cube has all the above 4 states, and not having "void cube parity" is a loop between states #1 and #2. Having "void cube parity" is a loop between states #3 and #4.

This may help to get some perspective on how this relates to the 4x4x4. As always, let me know if you have any questions.

 

[advincubing]

The above is crystal clear. I especially appreciated you connecting it back to a 3x3 and the Void Cube:

By the way, a 3x3x3 cube is a loop between state #1 and state #2.

A void cube has all the above 4 states, and not having "void cube parity" is a loop between states #1 and #2. Having "void cube parity" is a loop between states #3 and #4.

This may help to get some perspective on how this relates to the 4x4x4. As always, let me know if you have any questions.

Thanks again!

 

[advincubing]

So, I thought about all of this last week and poked around some puzzle theory sites and what not. This really is such a fascinating topic. Turns out I do have a couple questions, still.

Question 1 relates to this:

To understand why this is, we have to look at what a quarter turn on a 3x3x3 does when viewed as a permutation on corners and edges. When you turn a quarter turn of a side on 3x3x3 you are executing a 4-cycle on corners, as well as a 4-cycle on edges. This means that a quarter turn on 3x3x3 is an odd permutation on corners and an odd permutation on edges. This means that every quarter turn you do on a 3x3x3 will change the permutation parity of both corners and edges.

If an outer-slice turn is a 4-cycle turn on each of the corners and edges, and they're all moving together, I get why an outer-slice turn alone cannot "disrupt" the parity state of the corners vis-a-vis the edges. That answered my earlier questions perfectly. But I don't understand why it loops each between odd/even. I would think that it applies an even number (4) of odd changes (1) to each. Because that's even (4x1=4), I thought that corners (in the aggregate) that were odd or even before a turn would still be odd or even (in the aggregate) after a turn. So, why does a quarter turn toggle as opposed to maintain the parity states of the corners and edges?

Question 2: Dovetailing off the above, What establishes the parity state of the corners? I understand that the measure for purposes of PLL parity is whether it would take an odd or even number of outside layer turns to properly place the corners in relation to the formed centers. But what makes the corners odd, when they are? I would have thought that the parity state of the corners would never leave even (starting from solved, through scramble, through partial solve, etc.). Is it that the way the centers are formed creates a parity "gap," so to speak, making otherwise even corners odd vis-a-vis the combined centers?

 

[cuBerBruce]

The parity of a permutation is odd if you have an odd number of even-length cycles. Odd-length cycles have no effect.

A pure 4-cycle is odd parity, since a 4-cycle has even length and you have exactly 1 such cycle. 1 is odd, so you have odd parity.

If you do U D, you now have 2 4-cycles (considering only the corners). 2 is an even number, so you have even parity.

If you do U R2 B2, you now have a 3-cycle and a 4-cycle (and, of course, a trivial 1-cycle) of corners. The 3-cycle and 1-cycle have odd lengths, so you ignore them. There is 1 even-length cycle, a 4-cycle. 1 is odd, so the corners have odd parity.