Recognizing 4x4x4 Corner Parity

[Bigcc32]

I am relatively new to the 4x4x4 cube, and was taking a closer look at the parity cases. I don't yet have all of the case algorithms memorized (but I can still solve the cube completely without them). I have completely memorized only 2 (simply because they are really the only ones that are needed to solve it), but I am obviously not taking the quickest or most efficient route by any means because some times I get parity cases that can be dealt with using a different algorithm, but I use the one I know thus causing myself to have to resolve the last layer. I learned to recognize any the edge parities early (really simple) and now I can save myself the trouble of finishing and resolving because of a case I don't have an algorithm for. I was wondering if there was actually a surefire way to identify if your corner pieces were out of place before trying to place them in their correct spots (after solving the last edges). This seems to be really inefficient and I know I could just learn the other corner algs, but I tend to like fairly intuitive methods as well and figured if I could get by this way, I would.

Does this make sense? Is there actually a way to identify your corner parities before orienting the edges?

 

[blade740]

Walk the plank.

Edit: erm, I mean, PLL parity + T perm.

 

[Ellis]

What do you mean by corner parity? When you solve the whole cube and you have just 2 corners that need to be permuted? Thats an edge PLL parity thats set up in such a way that all the edges look correct. I don't think there are special algs for these cases, you'd just do PLL parity+perm (usually T or N perm, or if you don't like N like me you can make that case a V perm). Thats normally how it goes, so I wouldn't consider doing PLL Parity+T perm messing up the last layer.

If what you're really asking is if there's a surefire way tell if you have a double parity right before you begin to fix your OLL parity in order to avoid doing two algorithms then... well I'm sure there is a way but the time it takes to recognize such a case might make it not worth it. Half of the time that you try and recognize if you do actually have double parity, you won't have it.

I don't think people typically learn more than two parity algorithms for the 4x4.

 

[Spyyder]

im too lazy to memorize the corner parity so i just do an A Perm and it sets it up with another pll and an edge parity.

 

[Stefan]

II learned to recognize any the edge parities early (really simple)

How early? And how?

What do you mean by corner parity? When you solve the whole cube and you have just 2 corners that need to be permuted? Thats an edge PLL parity ...

No it's not.

 

[hr.mohr]

Dan Harris has some stuff about solving OLL and parity simultaneously using setup moves.

 

[mpohl100]

What do you mean by corner parity? When you solve the whole cube and you have just 2 corners that need to be permuted? Thats an edge PLL parity ...

No it's not. [/QUOTE]

i also thought that there is only edge parity cause to permute 2 corners you actually switch two edges and then you do a normal 3x3 PLL alg.
I'd like to know your reason why there exists a corner PLL parity.

 

[Tomarse]

There is no corner parity, when you get that case, do a T perm, and you'll soon realise its not corner parity ;p

 

[Stefan]

there is only edge parity cause to permute 2 corners you actually switch two edges and then you do a normal 3x3 PLL alg.

Just because that's the only way you know how to solve it, doesn't mean that's what it is.

For a normal 3x3x3 (and if you take the centers as reference), edge parity and corner parity always match, giving an overall even parity. But do any quarter turn to a solved 3x3x3 and both corner and edge parity are odd. For the pseudo 3x3x3 on a 4x4x4, this is not the case, parity of edges and corners are independent. And if everything is solved except two corners are swapped, then that's (odd) corner parity and not edges parity.

People should really stop using "parity" as kind of a magic synonym for "trouble" and learn what it really is. I recommend Jaap's articles, particularly this one:
http://www.geocities.com/jaapsch/puzzles/theory.htm

 

[masterofthebass]

Actually, 4x4 PLL 'parity' isn't even a parity... It's just 2 2cycles of edges. I don't really think that this can be approached the same was as parity comes up on the 3x3. If your corner parity is odd on 4x4, then your overall edge parity is the same, except for when you have this case os 2 2cycles.

 

[mpohl100]

there is only edge parity cause to permute 2 corners you actually switch two edges and then you do a normal 3x3 PLL alg.

Just because that's the only way you know how to solve it, doesn't mean that's what it is.

For a normal 3x3x3 (and if you take the centers as reference), edge parity and corner parity always match, giving an overall even parity. But do any quarter turn to a solved 3x3x3 and both corner and edge parity are odd. For the pseudo 3x3x3 on a 4x4x4, this is not the case, parity of edges and corners are independent. And if everything is solved except two corners are swapped, then that's (odd) corner parity and not edges parity.

People should really stop using "parity" as kind of a magic synonym for "trouble" and learn what it really is. I recommend Jaap's articles, particularly this one:
http://www.geocities.com/jaapsch/puzzles/theory.htm

thank you Stefan for clearing that up.
I simply didn't know that till ten minutes ago.

 

[Stefan]

Actually, 4x4 PLL 'parity' isn't even a parity

Not for the 4x4x4, I agree. Though for the "pseudo 3x3x3".

It's just 2 2cycles of edges.

For example, yes. Or two 2-cycles of edges plus a corner 3-cycle. Or any of the many other cases. But yes, this parity can be resolved by two edge swaps.

And maybe we should say "sedges" in addition to "dedges" to make discussions clearer...

 

[Bigcc32]

There is no corner parity, when you get that case, do a T perm, and you'll soon realise its not corner parity ;p

That answered the question, and I realized that in most cases the Tperm results in the same amount of resolve time (or more) as the algorithm I was originally using.