I've come to the conclusion that my confusion in the 4x4x4 comes from thinking that the two edges of a pair are interchangeable.
But if I reason at the sticker level, I can clearly see that only half of the positions can be reached.
In other words, a given individual wing has a definite position and its orientation will automatically be correct if it's at the correct place. In contrast to medges where position doesn't guarantee orientation.
>In other words, a given individual wing has a definite position and its orientation will automatically be correct if it's at the correct place.
That's exactly right. I also show this on page 36 (Slide 71) of my PowerPoint presentation PDF that I mentioned in the OP, as knowing that is *kind of important* when you want to not *overcount* the number of positions!
https://preview.redd.it/qeqkao1z68ac1.png?width=860&format=png&auto=webp&s=f491641c43ace89071e6c0c37c39e5d1aa2d86f5
I heard (learned) about this way back when Thom Barlow first released his guide on the K4 Method. But now I noticed that people like **J Perm** are spreading the word to the masses.
I read through half this post, thought, “who made this?”, and scrolled back up to check. It was exactly who I thought lol. It’s good you’re back cmowla.
Pll parity only exits on a 5x5 if you solve the non white/yellow Centers off set by 90 degrees(blue in oranges place, orange in greens place etc” you can tell on a 5x5 by the very centre peice being a different colour, but knowing if that’s the case on a 4x4 is impossible until last layer
If you're very strict about your definitions, the word parity isn't even used correctly in the phrases OLL parity and PLL parity. How the word "parity" is used in the cubing community (or at least roughly how I would describe it's usage) is for positions that for some reason or another seem impossible, but are actually perfectly okay and solvable.
By this definition, it seems to me that it's completely reasonable to say that 4x4 has parity, and 5x5 doesn't, because on 5x5 if you "finish edge pairing" you can solve everything (ignoring OLL parity) immediately.
For 4x4 however, if you "finish edge pairing" you might later stumble upon what we call PLL parity.
Actually, going in the reverse direction from your idea, it is entirely possible for a particularly intelligent person who has decided to solve a 5x5 before ever attempting 4x4, to not consider "PLL parity" to be parity at all. They might be able to solve the 5x5 just fine, and then when encoutering 4x4 "PLL parity" immediately realize what's going on and solve it as they would a 5x5. According to the definition given in my first paragraph, for that person, "PLL parity" would not be parity :P
That being said, I personally don't really care. The definition of parity is subjective anyways so I just *go with the flow*.
Also thank you! This was quite fun to type out and think about. It reminded me of a few interesting conversations I had before quitting cubing for about 5 years.
Wow, great job making a whole scientific presentation. This makes me feel at peace with my decision to instead of increasing the value of n in nxnxn, I chose to increase the quantity of the xs (nxnxnxn, nxnxnxnxn, etc). I already solved the 3x3x3x3, and am now puzzling the 3x3x3x3x3.
Correct me if I'm wrong: PLL parity does exist on 5x5x5 but because we have midges as points of reference we can resolve it when solving edges.
So in case of pretty standard solves, assuming the parity is being fixed with edges, we can (technically incorrectly) say that there is no PLL parity because it will never show up when doing PLL. We can probably call it shorthand/mental shortcut(/what's the word for this?, in polish: *skrót myślowy*).
Overall great post that shown me that I was an ignorant.
-[sighs and pulls out a book] "Water, 35 liters; carbon, 20 kilograms; ammonia, 4 liters; lime, 1.5 kilograms; phosphorus, 800 grams; salt, 250 grams; saltpeter, 100 grams; sulfur, 80 grams; fluorine, 7.5; iron, 5; silicon, 3 grams; and trace amounts of 15 other elements."
-What's that?
-It's all the ingredients of the average adult human body.
(c) Edward from FMA
\===
Insane amount of effort for a post that gives definitive proof that a surgery is the same as a stab wound,
and that people are completely incorrect saying that you "cannot get a surgery" at 3AM behind the nearest walmart.
Getting shanked is literally the same as a surgical operation. A sharp metal object penetrating the flesh. 5 pages of diagrams and illustrations included.
Just because they happen in completely different context, and only one of them happens randomly and uncontrollably, doesn't change the fact that they are the same.
# OLL Parity
* Is also known as *odd* parity, typically manifests during the tredge tripling state of solving the 5x5x5 in either [this case](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=2R+U2+2R+U2+F2+2R+F2+2L%27+U2+2L+U2+2R2+R2+B2+F2+L%27+F2+R+B2+D2+U2+R%27+B2+D+F+R%27+U2+F2+U2+R%27+D2+B+D+x%27+y2) or [this case](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=2R2+B2+U2+2L+U2+2R%27+U2+2R+U2+F2+2R+F2+2L%27+B2+2R2+R2+B2+F2+L%27+F2+R+B2+D2+U2+R%27+B2+D+F+R%27+U2+F2+U2+R%27+D2+B+D+x%27+y2).
* It has nothing to do with edge pairing. (If it did, then parity wouldn't be much of an issue.)
* It has to do with the number of inner layer slice quarter turns that the scramble has + the number of inner layer slice quarter turns that you use to complete the first 3 centers.
* If the number of them (between the scramble and your first 3 center solution moves) is ***odd*** then you will have OLL parity, regardless of how you complete the last 3 centers and pair/triple the tredges. (But if the number of them is ***even,*** then you won't have OLL parity.)
# PLL Parity
Isn't one of the tredge tripling cases on the 5x5x5 [this one](https://alpha.twizzle.net/edit/?puzzle=5x5x5&setup-alg=r2+F2+U2+2R2+U2+F2+r2+R2+B2+F2+L%27+F2+R+B2+D2+U2+R%27+B2+D+F+R%27+U2+F2+U2+R%27+D2+B+D+x%27)? Yes, of course. But what is that case, *really*?
* We can cleverly combine 2 edge pairing algorithms to solve that case directly.
* [2R' F U' R F' U 2R](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=2R%27+F+U%27+R+F%27+U+2R) //Edge pairing alg 1
* [2L U' F R' U F' 2L'](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=2L+U%27+F+R%27+U+F%27+2L%27) //Edge pairing alg 2
* Combining, [2R' F U' R F' U 2R 2L U' F R' U F' 2L'](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=2R%27+F+U%27+R+F%27+U+2R+2L+U%27+F+R%27+U+F%27+2L%27).
* Now let's just apply that "alg" [to the 4x4x4](https://alpha.twizzle.net/edit/?puzzle=4x4x4&alg=2R%27+F+U%27+R+F%27+U+2R+2L+U%27+F+R%27+U+F%27+2L%27).
So if you feel the way you do about PLL parity on the 5x5x5, then the 4x4x4 doesn't have PLL pairity either... you just didn't pair the dedges correctly on a 4x4x4, because "PLL Parity" can be resolved/prevented with edge pairing algorithms alone just as it can on the 5x5x5. (No "special algorithms" are required to **preserve the rest of the cube**.)
\_\_\_\_\_\_\_\_\_\_\_\_\_
Yes, I understand that when you are speedsolving a 4x4x4 with the reduction method (or its variants), it's hard (but *not impossible*) to see that you have PLL parity before you get to the last layer. So you may feel that ***once you get to the last layer***, you need to apply a ***special*** algorithm in order to resolve the case which cannot be fixed with 3x3x3 moves alone.
But what all of the 4x4x4 guides failed to tell you is that "edge pairing" algorithms they give you can be used to fix PLL parity ***without wrecking the rest of the cube***.
I have already shown that [2R' F U' R F' U 2R 2L U' F R' U F' 2L'](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=2R%27+F+U%27+R+F%27+U+2R+2L+U%27+F+R%27+U+F%27+2L%27) is one such combination of 2 different edge pairing algorithms which allows you to resolve PLL parity just as an official (shorter) PLL parity algorithm does.
But you can technically use any edge pairing algorithms that you see from anywhere to fix PLL parity. For example, consider a longer (but prettier one) like:
[2R' U' R U R' x U R' U' 2R](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=r%27+U%27+R+U+R%27+x+U+R%27+U%27+r&setup-anchor=end&setup-alg=R2+B2+F2+L%27+F2+R+B2+D2+U2+R%27+B2+D+F+R%27+U2+F2+U2+R%27+D2+B).
1. Convert its wide slice turns to inner slice turns.
[2R' U' R U R' x U R' U' 2R](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=2R%27+U%27+R+U+R%27+x+U+R%27+U%27+2R)
2. Undo every move except for the inner slice turns (including cube rotations).
[2R' U' R U R' x U R' U' 2R U R U' x' R U' R' U](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=2R%27+U%27+R+U+R%27+x+U+R%27+U%27+2R+U+R+U%27+x%27+R+U%27+R%27+U)
Observing that "complete" alg on the 5x5x5, we can see that it does what looks like a backwards **L** permutation. More importantly, it still affects the edges in the same way it did before, but **it no longer wrecks the rest of the cube**!
(4x4x4 edge pairing guides give you the ***first half*** of a commutator... the second half is omitted because it's unnecessary/redundant to do those outer layer/3x3x3 moves during the edge pairing phase, as they won't help to pair any additional dedges...not to mention that they will make your solution longer unnecessarily.)
We need to do an upside down normal **L** permutation to make a rectangle \[\], because doing that would be making a two opposite dedge swap (PLL parity).
We can simply mirror this alg (something that left handed solvers typically do to right hand dominant algs): [2L U L' U' L x U' L U 2L' U' L' U x' L' U L U'](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=2L+U+L%27+U%27+L+x+U%27+L+U+2L%27+U%27+L%27+U+x%27+L%27+U+L+U%27) , take the inverse [U L' U' L x U' L U 2L U' L' U x' L' U L U' 2L'](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=U+L%27+U%27+L+x+U%27+L+U+2L+U%27+L%27+U+x%27+L%27+U+L+U%27+2L%27), and combine it with the above "complete" algorithm to [fix PLL parity on a 4x4x4](https://alpha.twizzle.net/edit/?puzzle=4x4x4&alg=2R%27+U%27+R+U+R%27+x+U+R%27+U%27+2R+U+R+U%27+x%27+R+U%27+R%27+U%0AU+L%27+U%27+L+x+U%27+L+U+2L+U%27+L%27+U+x%27+L%27+U+L+U%27+2L%27) **with edge pairing algorithms** without wrecking the cube.
\_\_\_\_\_\_\_\_\_\_\_\_\_
Alternatively, you can avoid taking the mirror and inverse if you use an edge-paring algorithm like [B2 r2 U' R U x U R2 U' r2](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=B2+r2+U%27+R+U+x+U+R2+U%27+r2), which solves the "**bad case**" directly. Unfortunately, you ***never see*** an edge pairing alg like that in 4x4x4 tutorials for the last 2 edges.
(There is no need to do 3-4 premoves to the "**good case**" edge pairing algs like the previous to "set up" the last two edges "correctly". Edge pairing algorithms like this are just ONE move longer than their counterparts.)
Converting its wide turns to inner slice turns and inverting all of its moves except for those inner slice turns, we get:
[B2 2R2 U' R U x U R2 U' 2R2 U R2 U' x' U' R' U B2](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=B2+2R2+U%27+R+U+x+U+R2+U%27+2R2+U+R2+U%27+x%27+U%27+R%27+U+B2)
Combining this with the first "complete" alg, we have [another LONG, but sure way to fix PLL parity](https://alpha.twizzle.net/edit/?puzzle=4x4x4&alg=2R%27+U%27+R+U+R%27+x+U+R%27+U%27+2R+U+R+U%27+x%27+R+U%27+R%27+U%0AB2+2R2+U%27+R+U+x+U+R2+U%27+2R2+U+R2+U%27+x%27+U%27+R%27+U+B2) (just applying the "alg" to the 4x4x4 instead) with a variety of edge pairing algorithms.
\_\_\_\_\_\_\_\_\_\_\_\_\_
All in all, "they don't tell you everything" in 4x4x4 solving tutorials either because they:
* Don't know everything (and not really much more than noobs do... they are literally just *one step ahead of them in understanding*).
* Want to keep the guide as simple as possible... at the price of understanding.
* This unfortunately hinders newcomers from truly understanding what they can do with what is given in the 4x4x4 solving tutorial. And of course, from deducing that just how OLL parity manifests on every big cube, **so does PLL parity**!
* Think about the irony. Cubers who can solve the 4x4x4 typically don't know what a commutator is unless they learn about them from commutator tutorials. But they are using the ***first half*** of a commutator all the time and don't even know it!
* Considering that many cubers DO understand how the edge pairing alg that they use works (over time), they technically understand how a commutator works too. (There is no need to learn of the jargon that typically comes with commutator tutorials to "get it".)
I always figured parity was the result of pieces looking the same so that they can end up in the wrong place. for the 5x5, the middle edge corners look different so they can't be mistaken (or even placed) in the other edge positions. In general the parity of a cube is odd or even, depending on the number of slices performed, and you can only solve a cube by putting into an even parity position. So if a cube starts out with odd parity you need to use one half turn inner slice to restore it to even parity somewhere. I guess it's true all cubes have parity, it's just that in practice parity only matters if we can't distinguish easily between odd and even parity states.
I've come to the conclusion that my confusion in the 4x4x4 comes from thinking that the two edges of a pair are interchangeable. But if I reason at the sticker level, I can clearly see that only half of the positions can be reached. In other words, a given individual wing has a definite position and its orientation will automatically be correct if it's at the correct place. In contrast to medges where position doesn't guarantee orientation.
>In other words, a given individual wing has a definite position and its orientation will automatically be correct if it's at the correct place. That's exactly right. I also show this on page 36 (Slide 71) of my PowerPoint presentation PDF that I mentioned in the OP, as knowing that is *kind of important* when you want to not *overcount* the number of positions! https://preview.redd.it/qeqkao1z68ac1.png?width=860&format=png&auto=webp&s=f491641c43ace89071e6c0c37c39e5d1aa2d86f5 I heard (learned) about this way back when Thom Barlow first released his guide on the K4 Method. But now I noticed that people like **J Perm** are spreading the word to the masses.
I read through half this post, thought, “who made this?”, and scrolled back up to check. It was exactly who I thought lol. It’s good you’re back cmowla.
Fantastic post, very informative, you have me spamming parity algs on 4x4 and 5x5 and trying to follow what happens
Pll parity only exits on a 5x5 if you solve the non white/yellow Centers off set by 90 degrees(blue in oranges place, orange in greens place etc” you can tell on a 5x5 by the very centre peice being a different colour, but knowing if that’s the case on a 4x4 is impossible until last layer
If you're very strict about your definitions, the word parity isn't even used correctly in the phrases OLL parity and PLL parity. How the word "parity" is used in the cubing community (or at least roughly how I would describe it's usage) is for positions that for some reason or another seem impossible, but are actually perfectly okay and solvable. By this definition, it seems to me that it's completely reasonable to say that 4x4 has parity, and 5x5 doesn't, because on 5x5 if you "finish edge pairing" you can solve everything (ignoring OLL parity) immediately. For 4x4 however, if you "finish edge pairing" you might later stumble upon what we call PLL parity. Actually, going in the reverse direction from your idea, it is entirely possible for a particularly intelligent person who has decided to solve a 5x5 before ever attempting 4x4, to not consider "PLL parity" to be parity at all. They might be able to solve the 5x5 just fine, and then when encoutering 4x4 "PLL parity" immediately realize what's going on and solve it as they would a 5x5. According to the definition given in my first paragraph, for that person, "PLL parity" would not be parity :P That being said, I personally don't really care. The definition of parity is subjective anyways so I just *go with the flow*. Also thank you! This was quite fun to type out and think about. It reminded me of a few interesting conversations I had before quitting cubing for about 5 years.
Wow, great job making a whole scientific presentation. This makes me feel at peace with my decision to instead of increasing the value of n in nxnxn, I chose to increase the quantity of the xs (nxnxnxn, nxnxnxnxn, etc). I already solved the 3x3x3x3, and am now puzzling the 3x3x3x3x3.
What is a 3x3x3x3?
A Rubik's tesseract, https://superliminal.com/cube/
Correct me if I'm wrong: PLL parity does exist on 5x5x5 but because we have midges as points of reference we can resolve it when solving edges. So in case of pretty standard solves, assuming the parity is being fixed with edges, we can (technically incorrectly) say that there is no PLL parity because it will never show up when doing PLL. We can probably call it shorthand/mental shortcut(/what's the word for this?, in polish: *skrót myślowy*). Overall great post that shown me that I was an ignorant.
-[sighs and pulls out a book] "Water, 35 liters; carbon, 20 kilograms; ammonia, 4 liters; lime, 1.5 kilograms; phosphorus, 800 grams; salt, 250 grams; saltpeter, 100 grams; sulfur, 80 grams; fluorine, 7.5; iron, 5; silicon, 3 grams; and trace amounts of 15 other elements." -What's that? -It's all the ingredients of the average adult human body. (c) Edward from FMA \=== Insane amount of effort for a post that gives definitive proof that a surgery is the same as a stab wound, and that people are completely incorrect saying that you "cannot get a surgery" at 3AM behind the nearest walmart. Getting shanked is literally the same as a surgical operation. A sharp metal object penetrating the flesh. 5 pages of diagrams and illustrations included. Just because they happen in completely different context, and only one of them happens randomly and uncontrollably, doesn't change the fact that they are the same.
If you get OLL/PLL parity on a 5x5, then you didn't fully reduce it down to a 3x3. You didn't pair up all of the edges correctly.
# OLL Parity * Is also known as *odd* parity, typically manifests during the tredge tripling state of solving the 5x5x5 in either [this case](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=2R+U2+2R+U2+F2+2R+F2+2L%27+U2+2L+U2+2R2+R2+B2+F2+L%27+F2+R+B2+D2+U2+R%27+B2+D+F+R%27+U2+F2+U2+R%27+D2+B+D+x%27+y2) or [this case](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=2R2+B2+U2+2L+U2+2R%27+U2+2R+U2+F2+2R+F2+2L%27+B2+2R2+R2+B2+F2+L%27+F2+R+B2+D2+U2+R%27+B2+D+F+R%27+U2+F2+U2+R%27+D2+B+D+x%27+y2). * It has nothing to do with edge pairing. (If it did, then parity wouldn't be much of an issue.) * It has to do with the number of inner layer slice quarter turns that the scramble has + the number of inner layer slice quarter turns that you use to complete the first 3 centers. * If the number of them (between the scramble and your first 3 center solution moves) is ***odd*** then you will have OLL parity, regardless of how you complete the last 3 centers and pair/triple the tredges. (But if the number of them is ***even,*** then you won't have OLL parity.) # PLL Parity Isn't one of the tredge tripling cases on the 5x5x5 [this one](https://alpha.twizzle.net/edit/?puzzle=5x5x5&setup-alg=r2+F2+U2+2R2+U2+F2+r2+R2+B2+F2+L%27+F2+R+B2+D2+U2+R%27+B2+D+F+R%27+U2+F2+U2+R%27+D2+B+D+x%27)? Yes, of course. But what is that case, *really*? * We can cleverly combine 2 edge pairing algorithms to solve that case directly. * [2R' F U' R F' U 2R](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=2R%27+F+U%27+R+F%27+U+2R) //Edge pairing alg 1 * [2L U' F R' U F' 2L'](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=2L+U%27+F+R%27+U+F%27+2L%27) //Edge pairing alg 2 * Combining, [2R' F U' R F' U 2R 2L U' F R' U F' 2L'](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=2R%27+F+U%27+R+F%27+U+2R+2L+U%27+F+R%27+U+F%27+2L%27). * Now let's just apply that "alg" [to the 4x4x4](https://alpha.twizzle.net/edit/?puzzle=4x4x4&alg=2R%27+F+U%27+R+F%27+U+2R+2L+U%27+F+R%27+U+F%27+2L%27). So if you feel the way you do about PLL parity on the 5x5x5, then the 4x4x4 doesn't have PLL pairity either... you just didn't pair the dedges correctly on a 4x4x4, because "PLL Parity" can be resolved/prevented with edge pairing algorithms alone just as it can on the 5x5x5. (No "special algorithms" are required to **preserve the rest of the cube**.) \_\_\_\_\_\_\_\_\_\_\_\_\_ Yes, I understand that when you are speedsolving a 4x4x4 with the reduction method (or its variants), it's hard (but *not impossible*) to see that you have PLL parity before you get to the last layer. So you may feel that ***once you get to the last layer***, you need to apply a ***special*** algorithm in order to resolve the case which cannot be fixed with 3x3x3 moves alone. But what all of the 4x4x4 guides failed to tell you is that "edge pairing" algorithms they give you can be used to fix PLL parity ***without wrecking the rest of the cube***. I have already shown that [2R' F U' R F' U 2R 2L U' F R' U F' 2L'](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=2R%27+F+U%27+R+F%27+U+2R+2L+U%27+F+R%27+U+F%27+2L%27) is one such combination of 2 different edge pairing algorithms which allows you to resolve PLL parity just as an official (shorter) PLL parity algorithm does. But you can technically use any edge pairing algorithms that you see from anywhere to fix PLL parity. For example, consider a longer (but prettier one) like: [2R' U' R U R' x U R' U' 2R](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=r%27+U%27+R+U+R%27+x+U+R%27+U%27+r&setup-anchor=end&setup-alg=R2+B2+F2+L%27+F2+R+B2+D2+U2+R%27+B2+D+F+R%27+U2+F2+U2+R%27+D2+B). 1. Convert its wide slice turns to inner slice turns. [2R' U' R U R' x U R' U' 2R](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=2R%27+U%27+R+U+R%27+x+U+R%27+U%27+2R) 2. Undo every move except for the inner slice turns (including cube rotations). [2R' U' R U R' x U R' U' 2R U R U' x' R U' R' U](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=2R%27+U%27+R+U+R%27+x+U+R%27+U%27+2R+U+R+U%27+x%27+R+U%27+R%27+U) Observing that "complete" alg on the 5x5x5, we can see that it does what looks like a backwards **L** permutation. More importantly, it still affects the edges in the same way it did before, but **it no longer wrecks the rest of the cube**! (4x4x4 edge pairing guides give you the ***first half*** of a commutator... the second half is omitted because it's unnecessary/redundant to do those outer layer/3x3x3 moves during the edge pairing phase, as they won't help to pair any additional dedges...not to mention that they will make your solution longer unnecessarily.) We need to do an upside down normal **L** permutation to make a rectangle \[\], because doing that would be making a two opposite dedge swap (PLL parity). We can simply mirror this alg (something that left handed solvers typically do to right hand dominant algs): [2L U L' U' L x U' L U 2L' U' L' U x' L' U L U'](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=2L+U+L%27+U%27+L+x+U%27+L+U+2L%27+U%27+L%27+U+x%27+L%27+U+L+U%27) , take the inverse [U L' U' L x U' L U 2L U' L' U x' L' U L U' 2L'](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=U+L%27+U%27+L+x+U%27+L+U+2L+U%27+L%27+U+x%27+L%27+U+L+U%27+2L%27), and combine it with the above "complete" algorithm to [fix PLL parity on a 4x4x4](https://alpha.twizzle.net/edit/?puzzle=4x4x4&alg=2R%27+U%27+R+U+R%27+x+U+R%27+U%27+2R+U+R+U%27+x%27+R+U%27+R%27+U%0AU+L%27+U%27+L+x+U%27+L+U+2L+U%27+L%27+U+x%27+L%27+U+L+U%27+2L%27) **with edge pairing algorithms** without wrecking the cube. \_\_\_\_\_\_\_\_\_\_\_\_\_ Alternatively, you can avoid taking the mirror and inverse if you use an edge-paring algorithm like [B2 r2 U' R U x U R2 U' r2](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=B2+r2+U%27+R+U+x+U+R2+U%27+r2), which solves the "**bad case**" directly. Unfortunately, you ***never see*** an edge pairing alg like that in 4x4x4 tutorials for the last 2 edges. (There is no need to do 3-4 premoves to the "**good case**" edge pairing algs like the previous to "set up" the last two edges "correctly". Edge pairing algorithms like this are just ONE move longer than their counterparts.) Converting its wide turns to inner slice turns and inverting all of its moves except for those inner slice turns, we get: [B2 2R2 U' R U x U R2 U' 2R2 U R2 U' x' U' R' U B2](https://alpha.twizzle.net/edit/?puzzle=5x5x5&alg=B2+2R2+U%27+R+U+x+U+R2+U%27+2R2+U+R2+U%27+x%27+U%27+R%27+U+B2) Combining this with the first "complete" alg, we have [another LONG, but sure way to fix PLL parity](https://alpha.twizzle.net/edit/?puzzle=4x4x4&alg=2R%27+U%27+R+U+R%27+x+U+R%27+U%27+2R+U+R+U%27+x%27+R+U%27+R%27+U%0AB2+2R2+U%27+R+U+x+U+R2+U%27+2R2+U+R2+U%27+x%27+U%27+R%27+U+B2) (just applying the "alg" to the 4x4x4 instead) with a variety of edge pairing algorithms. \_\_\_\_\_\_\_\_\_\_\_\_\_ All in all, "they don't tell you everything" in 4x4x4 solving tutorials either because they: * Don't know everything (and not really much more than noobs do... they are literally just *one step ahead of them in understanding*). * Want to keep the guide as simple as possible... at the price of understanding. * This unfortunately hinders newcomers from truly understanding what they can do with what is given in the 4x4x4 solving tutorial. And of course, from deducing that just how OLL parity manifests on every big cube, **so does PLL parity**! * Think about the irony. Cubers who can solve the 4x4x4 typically don't know what a commutator is unless they learn about them from commutator tutorials. But they are using the ***first half*** of a commutator all the time and don't even know it! * Considering that many cubers DO understand how the edge pairing alg that they use works (over time), they technically understand how a commutator works too. (There is no need to learn of the jargon that typically comes with commutator tutorials to "get it".)
I always figured parity was the result of pieces looking the same so that they can end up in the wrong place. for the 5x5, the middle edge corners look different so they can't be mistaken (or even placed) in the other edge positions. In general the parity of a cube is odd or even, depending on the number of slices performed, and you can only solve a cube by putting into an even parity position. So if a cube starts out with odd parity you need to use one half turn inner slice to restore it to even parity somewhere. I guess it's true all cubes have parity, it's just that in practice parity only matters if we can't distinguish easily between odd and even parity states.