ZBLL Superman’s clairvoyance (2 Opposite-color edges hunt)

[Tito_H]

Mr. OreKehStrah describes the following Con in his document in regard to the Tran-Style V2, one of the ZBLL recognition method.

“For any given angle, you will only see 2 edges so you will need to peek to the left to see a 3rd edge to know the position of the other 2 edges.”

In my previous article “ZBLL Superman’s clairvoyance (the blind-side 2 EP colors guess)”, I have learnt the way to guess the colors of 2 blind-side edges.

This time I want to present my idea to tackle the Con, applying the similar logics used in its article.

<Mission impossible>
To guess each position of the 2 opposite-color edges and its color, without AUF or peeking to the left.

<Purpose>
To mitigate the following Con of the Tran-Style V2.
“For any given angle, you will only see 2 edges so you will need to peek to the left to see a 3rd edge to know the position of the other 2 edges.”

<Condition>
1. T set is used as an example.
2. Cube position: Yellow Center up.
3. The COLL orientation is shown as the pictures of the source: https://speedcubedb.com/a/3x3/ZBLLT.

<Term>
1. Color order: The order of 2 colors. There are 3 orders, which are named as follows.

• “O(Opposite)”: Same concept used in the ZBLL BH recognition method.
• “AS(Adjacent Same)”: Adjacent color set with the Same order as the cube side center colors.
• “AR(Adjacent Reverse)”: Adjacent color set with the Reverse order to the cube side center colors.

2. Color group: A group containing the 2 opposite colors. There are 2 groups, which are named as follows.

• “Dark”: whichever color of Blue or Green.
• “Bright”: whichever color of Red or Orange.

<Process>
1. Recognize the COLL swap type.

2. Recognize the Color group of imaginary permuted F-face, swapping and twisting corners as necessary to permute them.

3. Find out the opposite color set of edges by the following cases/steps.

1) COLL of swap type (5 types, 60 cases):
A) Imaginary permuted F-face and UF edge are both “Dark”, or both “Bright”:

If Color order of UF & UR is “AS”, then UF & UL, UR & UB are opposite color set.

If Color order of UF & UR is “AR”, then UF & UB, UR & UL are opposite color set.

If Color order of UF & UR is “O”, then Color order of UF & UL is “AS”.

B) Imaginary permuted F-face and UF edge are “Dark” and “Bright”, or “Bright” and “Dark”:

If Color order of UF & UR is “AS”, then UF & UB, UR & UL are opposite color set.

If Color order of UF & UR is “AR”, then UF & UL, UR & UB are opposite color set.

If Color order of UF & UR is “O”, then Color order of UR & UB is “AS”.

2) COLL of NO swap type (1 type, 12 cases):

There are at least 2 ways.
One way is to invert the above lines A) and B), and proceed with the remaining steps.

-> Change this;
A) F-face and UF are both “Dark”, or both “Bright”:
-> to this;
A) F-face and UF are “Dark” and “Bright”, or “Bright” and “Dark”:

and,
-> Change this;
B) F-face and UF are “Dark” and “Bright”, or “Bright” and “Dark”:
-> to this;
B) F-face and UF are both “Dark”, or both “Bright”:

Another way is to change the step 2, and proceed with the remaining steps as they are.

-> Change this;
Recog the Color group of imaginary permuted F-face, ….
-> to this;
Recog the Color group of imaginary permuted F-face, …, and if it is “Dark”, consider F-face is “Bight”, and vice versa.

4. As the colors of UF & UR are visible, now it is possible to guess where the opposite-color edge and what its color is.

Mission completed!

I will send some examples later.

 

[Tito_H]

I chose 6 cases as examples. The numbering after each picture corresponds to that described in <Process> of my above article.

T37

1. Front swap.
2. Permuted F-face is Bright(Red).
3. F-face and UF edge are both Bright. Color order of UF & UR is “AS”.
4. Consequently, UF & UL, UR & UB are opposite color set.

T38

1. Front swap.
2. Permuted F-face is Bright(Orange).
3. F-face and UF edge are Bright & Dark. Color order of UF & UR is “O”.
4. Consequently, Color order of UR & UB is “AS”, which means that UB is Red, and the remaining edge UL is Orange.

T39

1. Front swap.
2. Permuted F-face is Bright(Orange).
3. F-face and UF edge are both Bright. Color order of UF & UR is “O”.
4. Consequently, Color order of UF & UL is “AS”, which means that UL is Green, and the remaining edge UB is Blue.

T40

1. Front swap.
2. Permuted F-face is Dark(Green).
3. F-face and UF edge are Dark and Bright. Color order of UF & UR is “AR”.
4. Consequently, UF & UL, UR & UB are opposite color set.

T41

1. Front swap.
2. Permuted F-face is Bright(Red).
3. F-face and UF edge are Bright and Dark. Color order of UF & UR is “AS”.
4. Consequently, UF & UB, UR & UL are opposite color set.

T42

1. Front swap.
2. Permuted F-face is Bright(Orange).
3. F-face and UF edge are Bright and Dark. Color order of UF & UR is “AS”.
4. Consequently, UF & UB, UR & UL are opposite color set.

 

[Tito_H]

This “Edges hunt” method can be applied without any modification at any angle. Let us check if it is true.

T37 is used as an example.

T37 Angle 1

This is the base angle, and already mission completed yesterday.

1. Front swap.
2. Permuted F-face is Bright(Red).
3. F-face and UF edge are both Bright. Color order of UF & UR is “AS”. So, UF & UL, UR & UB are opposite color set.
4. Consequently, UL is Orange, and UB is Blue.

T37 Angle 2

1. Left swap.
2. Permuted F-face is Dark(Green).
3. F-face and UF edge are both Dark. Color order of UF & UR is “O”. So, Color order of UF & UL is “AS”
4. Consequently, UL is Red, and UB is Orange.

T37 Angle 3

1. Back swap.
2. Permuted F-face is Bright(Orange).
3. F-face and UF edge are Bright and Dark. Color order of UF & UR is “AR”. So, UF & UL, UR & UB are opposite color set.
4. Consequently, UL is Green, and UB is Red.

T37 Angle 4

1. Right swap.
2. Permuted F-face is Dark(Blue).
3. F-face and UF edge are Dark and Bright. Color order of UF & UR is “O”. So, Color order of UR & UB is “AS”
4. Consequently, UB is Green, and UL is Blue.

In any angle, we could hunt the edges.
Mission completed!

 

[Storm V K]

This sounds really good! I Can Think of a con, though; if you are dual colourneutral blue-orange would be AR for yellow top, but AS for white top. Still sounds really cool, though.

 

[Tito_H]

This sounds really good! I Can Think of a con, though; if you are dual colourneutral blue-orange would be AR for yellow top, but AS for white top. Still sounds really cool, though.

Dear Storm V K:
Thank you for your comment.
I was gracefully surprised to have received it to my old post.

I think your comment is correct, since I considered Yellow center top. A mirror concept or point symmetry should be considered and all the contents should be re-written without losing its theory, for a dual colour-neutral.
(Corrected on 16-5-2026)

 

[Tito_H]

Dear Storm V K:
Thank you for your comment.
I was gracefully surprised to have received it to my old post.

I think your comment is correct, since I considered Yellow center top. A mirror concept or point symmetry should be considered and all the contents should be re-written without losing its theory, for a dual colour-neutral.

I thought about this matter again today. As a result, I found that the theory I wrote for Mission Impossible can be applied to dual color-neutral (and even full color-neutral) without any rewriting at all. The reason for this is that, as written in <Term> (see below), AS and AR are based on the center color of the cube.

• “AS(Adjacent Same)”: Adjacent color set with the Same order as the cube side center colors.
• “AR(Adjacent Reverse)”: Adjacent color set with the Reverse order to the cube side center colors.

For example, with a Yellow top, Orange-Blue is AS, but even with a White top, Blue-Orange is still AS. This is because with a White top, the direction in which the center colors of the cube are aligned is reversed.

I apologize for my misunderstanding yesterday, even though it was something I had figured out.