# How do you inspect ZZ solve with Y-axis color neutrality?

##### Member
I can track and inspect EO edges pretty fast like around 8-9 seconds but I am learning to be y-axis color neutral, so how do I inspect bad edges for blue front and red front??? do you cubes inspect the blue front first and then the red front or do you inspect both axis simultaneously.......

please tell me how to do it in depth

#### Escher

##### Babby
When deciding on an axis a good start is to quickly scan U/D faces for their respective colours, then check the E ring for U/D colours. This gives you a really quick but somewhat inaccurate metric for working out which EO is probably going to be easiest, since if you have say, 7 oriented stickers you can guarantee a 2 or 4 edge EO case by checking where the remaining U/D edge is in the E ring and rotating if needed to orient it.

Next you want to check for misoriented equator edges in the E ring since a rotation won't fix these.

Finally you want to look at your E ring edges in the U or D layers, if you have more than 2 that are bad you'll probably want to rotate.

So example inspection;

L2 R2 B2 R2 D R2 U2 F2 D F2 U L F' R' F D F' D' U F'

Immediately we see only 1 U/D edge in the U layer, meaning there must be 3 in the E ring. We then check the bottom layer and discover there are 4 U/D edges. [U/D layer check done]

Next we check the E ring - on the starting orientation we have 3 good edges and 1 stuck edge. This is somewhat unfortunate, because if we had to rotate to solve U we would be sacrificing 3 E edges.

Then we check E ring edges in the U/D layers. We know D has all 4 U/D pieces so we just have to check U, of which 2 are already oriented, and 1 misoriented. Therefore we have a 4 edge case. Following rotation rules for orientation we know that if we applied a y to the cube, we would have 8 misoriented edges.

If you know the rules for how a y move effects U/D respecting orientation then you can work it out relatively quickly. Think of the orientations as a table, and applying a y move as a toggle that effects certain types of positions only. As a quick rule of thumb, if a piece is not in it's own (U/D, or E) ring, a y move will toggle it's orientation.

##### Member
When deciding on an axis a good start is to quickly scan U/D faces for their respective colours, then check the E ring for U/D colours. This gives you a really quick but somewhat inaccurate metric for working out which EO is probably going to be easiest, since if you have say, 7 oriented stickers you can guarantee a 2 or 4 edge EO case by checking where the remaining U/D edge is in the E ring and rotating if needed to orient it.

Next you want to check for misoriented equator edges in the E ring since a rotation won't fix these.

Finally you want to look at your E ring edges in the U or D layers, if you have more than 2 that are bad you'll probably want to rotate.

So example inspection;

L2 R2 B2 R2 D R2 U2 F2 D F2 U L F' R' F D F' D' U F'

Immediately we see only 1 U/D edge in the U layer, meaning there must be 3 in the E ring. We then check the bottom layer and discover there are 4 U/D edges. [U/D layer check done]

Next we check the E ring - on the starting orientation we have 3 good edges and 1 stuck edge. This is somewhat unfortunate, because if we had to rotate to solve U we would be sacrificing 3 E edges.

Then we check E ring edges in the U/D layers. We know D has all 4 U/D pieces so we just have to check U, of which 2 are already oriented, and 1 misoriented. Therefore we have a 4 edge case. Following rotation rules for orientation we know that if we applied a y to the cube, we would have 8 misoriented edges.

If you know the rules for how a y move effects U/D respecting orientation then you can work it out relatively quickly. Think of the orientations as a table, and applying a y move as a toggle that effects certain types of positions only. As a quick rule of thumb, if a piece is not in it's own (U/D, or E) ring, a y move will toggle it's orientation.
thank you soooooooooo much........Can you give me the rotation rules you said about

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#### Escher

##### Babby
Sorry, wrote the post in a hurry.

Basically;

U/D Layer
U/D piece oriented -> y -> oriented
U/D piece misoriented -> y -> misoriented
E layer piece oriented -> y -> misoriented
E layer piece misoriented -> y -> oriented

E Layer
U/D piece oriented -> y -> misoriented
U/D piece misoriented -> y -> oriented
E layer piece oriented -> y -> oriented
E layer piece misoriented -> y -> misoriented

You see the logic?

##### Member
Sorry, wrote the post in a hurry.

Basically;

U/D Layer
U/D piece oriented -> y -> oriented
U/D piece misoriented -> y -> misoriented
E layer piece oriented -> y -> misoriented
E layer piece misoriented -> y -> oriented

E Layer
U/D piece oriented -> y -> misoriented
U/D piece misoriented -> y -> oriented
E layer piece oriented -> y -> oriented
E layer piece misoriented -> y -> misoriented

You see the logic?
Actually, I had already worked that out..............but the problem is I cannot track or keep count of "red/orange front bad edges" and "blue/green front bad edges" simultaneously. My actual question is how do you keep count of the bad edges on both sides?

Then we check E ring edges in the U/D layers. We know D has all 4 U/D pieces so we just have to check U, of which 2 are already oriented, and 1 misoriented. Therefore we have a 4 edge case. Following rotation rules for orientation we know that if we applied a y to the cube, we would have 8 misoriented edges.
I didn't understand this part also...........how did you know 4 bad edges and a rotation will make 8 bad edges..........

Sorry.......I know I am a stupid cuber and thanks for helping me

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#### Escher

##### Babby
You don't need to keep count of bad E layer edges on either side; practise just inspecting one colour and then try to work out what would happen to them if you did a y move.

The misoriented edges part - with this scramble, think of the D layer as always having 2 bad edges in either rotation, so we don't have to do any counting or anything. The U layer has 1 edge that's good in either rotation (UR), so that's 3 that count as good for both. We also have one bad edge in UL that will become good if we rotate, so that's 4 edges we know already that will become good. We know from looking at UF/UB they will become bad, and we have those 2 edges on the bottom that won't change regardless, so there is 4 there. Then we take a look at the E ring and see that all the orientations for the U/D layer pieces are conditional on keeping this rotation, and we have a stuck E piece that's bad.

In sum this means we have 3 unconditionally good pieces (UR, DB/DF) and 1 conditionally good (UL). We then have 1 unconditionally bad piece on the E ring (FL), and everything else is conditionally bad. (The 'condition' in this sentence referring to making a y move). The number of conditionally bad pieces + unconditionally bad pieces = how many bad edges you would have if you rotate. In this case that's 5+3.

Perhaps you might find it easier if you learn to look for unconditionals first, then whatever conditionals you have will help you decide your orientation, so say "4 unconditionals must leave 8 conditionals, of which 6 are good right now, therefore I won't rotate".

Does this help?

#### Petro Leum

##### Member
You don't need to keep count of bad E layer edges on either side; practise just inspecting one colour and then try to work out what would happen to them if you did a y move.

The misoriented edges part - with this scramble, think of the D layer as always having 2 bad edges in either rotation, so we don't have to do any counting or anything. The U layer has 1 edge that's good in either rotation (UR), so that's 3 that count as good for both. We also have one bad edge in UL that will become good if we rotate, so that's 4 edges we know already that will become good. We know from looking at UF/UB they will become bad, and we have those 2 edges on the bottom that won't change regardless, so there is 4 there. Then we take a look at the E ring and see that all the orientations for the U/D layer pieces are conditional on keeping this rotation, and we have a stuck E piece that's bad.

In sum this means we have 3 unconditionally good pieces (UR, DB/DF) and 1 conditionally good (UL). We then have 1 unconditionally bad piece on the E ring (FL), and everything else is conditionally bad. (The 'condition' in this sentence referring to making a y move). The number of conditionally bad pieces + unconditionally bad pieces = how many bad edges you would have if you rotate. In this case that's 5+3.

Perhaps you might find it easier if you learn to look for unconditionals first, then whatever conditionals you have will help you decide your orientation, so say "4 unconditionals must leave 8 conditionals, of which 6 are good right now, therefore I won't rotate".

Does this help?
Dude, this way of thinking about orientations, adding a second layer to it as "condition", is totally genius.
while i like to just inspect both orientations spearately since the number of unoriented edges is not the only reason to rotate, this certainly serves as a great shortcut once you learn it.

thanks for the inspiration!

#### mDiPalma

##### Member
yeah im with petro

just inspect twice. once in standard orientation. once in y rotation.

it's really not that hard.

#### Petro Leum

##### Member
yeah im with petro

just inspect twice. once in standard orientation. once in y rotation.

it's really not that hard.
you dont even have to plan two entire eolines. just check for number of edges and position of the line edges on first orientation, then rotate y. if second scramble has ALOT more unoriented edges or a really hard case or stupid line edges, you just go back to first oreantation. if not, plan the entire EOLine for the second orientation, rotate y', compare.

after some practice, counting the number of unorianted edges and memorizing the position of the line edges shouldnt take you more than 3 seconds per orientation.

#### mDiPalma

##### Member
yeah and you can always rotate x if you're hardcore enough

#### Petro Leum

##### Member
yeah and you can always rotate x if you're hardcore enough
or you could just use Petrus...

nah jk, that would be too hardcore lolz

#### Z0chary

##### Member
So I have been working on a good ZZ Y-axis CN inspection system and here is what i have so far:

Find the "switchable" edges that would switch from oriented to disoriented and vice versa. This helps me decide if its worth switching.

To find all these edges, find the E-layer edge that are in the U/D face, and then find all the U/D edges that are in the E-layer. While finding these count how many are oriented correctly and how many or disoriented. This is only to find the difference of these edges. I do this by starting at 0 and adding 1 for every oriented switchable edge and subtracting 1 for every disoriented edge.
If you get zero, then solve the orientation that you are most comfortable with. If positive, keep the same orientation from inspection. If negative, then do a y-rotation. The number that you get tells you the difference in disoriented edges for each rotation. +2 means that if you do a y-rotation then you will gain 2 disoriented edges. -2 means that if you do a y-rotation will you will lose 2 disoriented edges.
I'll do some examples below.

My normal orientation is Yellow-Top and Blue-Front.

L' U' F2 L' U B' F D2 B2 L' U' B' F2 R2 D2 L U B F U' B' F R U F U L' B2 R2 F2

On the yellow face, there is 1 disoriented switchable edge (-1).
On the white face, there is 1 disoriented switchable edge and 1 oriented switchable edge (-1 -1 +1 = -1)
On the E-layer, there is 2 oriented switchable edges and 1 disoriented switchable edge. (-1 +2 -1 = 0).
With a difference of 0, that means both rotations will have the same number of disoriented edges.
Both rotations have 6 disoriented edges.

F2 R2 B' U' F D' B' F R2 F2 L2 R' F D' B F' L D B' U2 F L' B' D' U2 B' D B D U'

On the yellow face, there are no switchable edges. (0)
On the white face, there is 1 switchable oriented edge and 1 disoriented switchable edges. (0)
On the E-Layer, there are 2 switchable oriented edges (+2).
This means that with a blue-front, I will have 2 less disoriented edges than if I use a red/orange front.
Blue-front has 4 disoriented edges.
Red-front has 6 disoriented edges.

Pros:
-Fast-ish, allowing one to inspect after finding the better orientation.
-Allows one to see if a certain orientation would be really bad. (If i switch, I'll go from 10 bad edges to 6).

Cons:
-Assumes that in every situation having less bad edges is always better. However sometimes I'll miss an easy eoline with 6 bad edges because i switched to a harder eoline with 4 edges. These situations are rare, but can happen.
-Doesn't let one see both eolines directly.

Overall, I have put a lot of thought into this, but am still open to suggestions on how to improve it. Thanks!

#### Isaac VM

##### Member
So I have been working on a good ZZ Y-axis CN inspection system ...
I think this is a good system, just need to get used to it.

#### gyroninja

##### Member
I typically just inspect D and decide what orientation will be better. Most just the D face will be a good indicator.