ottozing
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Before consuming this, the reader is assumed to already be at least somewhat familiar with the system outlined in this video (or this thread if you prefer written guides). The method covered here is very similar, but the major improvement comes from no longer needing to do a BLD style parity trace to determine odd or even parity (the slowest step by far, especially if you're not an amazing BLD solver).
In place of that, we mostly reapply the same system of determining the difference between 4a, 4b, and 4c, but on varying sets of pieces. Credit to Jackson Warley for the 2c system & special thanks to Elliot Wiecek for verifying whether or not the stuff that was being cooked in the Discord group chat actually worked or not. Everything else here was either labbed by me or is just "common knowledge".
Quick terminology rundown since I know some old guides swap a/b for 4a/4b:
0c - Corners are oriented on all 3 axes (8c is the same thing but a slice move away)
2c - 2 corners are misoriented on 2 axes (6c is the same thing but a slice move away)
4a - 4 misoriented corners which can all be set up to U (or D) using only half turns
4b - 4 misoriented corners which cannot be set up to U (or D) using only half turns
4c - On one side of the scramble (in NISS terms), the corners are in 4a but on the other side they're in 4b
3/1 2/2 4/0 1/1 2/0 - Distribution of oriented/misoriented corners across U/D (we will be assuming the DR is solvable with U D R2 L2 F2 B2 in all cases)
Cases:
0c0/0c3/0c4
For these it's useful to know what HTR corners look like visually. The main thing to pay attention to is what kinds of patterns are/aren't possible in terms of opposite sides (we'll use U/D for this too). You can either have a face, "stacked" bars (R2 or R2 U2, but not R2 U), or slashes (F2 U2 R2). If all 6 sides are one of these 3 "real" patterns, it's HTR. If all 6 sides don't have a real pattern (either a 3/4 face or unstacked bars), it's 0c4. If it's a mix of the two, it's 0c3.
Another way to think about this which is more in line with how other cases get recognized is doing a 4a/4b comparison trace of the F, R, and if needed U faces. All 3 being 4a means it's 0c0, 1/3 being 4b would be 0c3, and 3/3 being 4b would be 0c4.
2c3/2c4/2c5
There are two possible methods for this that I'm a fan of (Jackson's one & an application of an old idea called "mental swap"). The mental swap one is literally just mentally swapping the 2 corners in your head and visualizing the resulting state. If it looks like HTR (or 0c0), it's 2c5, if it looks like 0c3, it's 2c4, if it looks like 0c4, it's 2c3.
Jackson's method goes like this:
1 - Bring the 2 misoriented corners next to each other, let's say UFR UFL (but something like UFR DFR is also possible)
2 - Trace both of the two layers that don't contain the 2c. Here this would be the D face followed by the B face
3 - 4a/4a = 2c5, 4a/4b = 2c4, 4b/4b = it might be 2c3 or 2c4 so we need one more step
4 - If you got 4b/4b, trace the DF and UB corners, 4b again = 2c3, 4a = 2c4 (another way to think of this step is doing either D2 or B2 & retracing D if you did B2, or retracing B if you did D2).
Note that if you instead have the 2c in say UFR DFR, you still trace the same way but would do L and B, followed by FL and BR corners if necessary. I recommend being comfortable with both so you can always trace without needing a setup move & can instead just invert the last move of your DR solution
4a1/4a2
1 - Bring the 4 corners such that they're 2/2 across U/D and adjacent to each other rather than diagonal (involves inverting the last move of your DR solution in the worst case scenario)
2 - Trace 2 oriented corners in U and 2 misoriented corners in D, 4b = 4a1, 4a = 4a2
4b2/4b5
1 - Bring the 4 corners such that they're 2/2 across U/D. Note that you can avoid doing any setup moves by rotating. For example, if you do U R2 U R2 on a solved cube, y z' sets them up to be around the UF and DL edges or UFR+UFL and DFL+DBL. When we get to step 3, we'll also be assuming the corners are in these 4 locations
2 - Trace 2 oriented corners in U and 2 misoriented corners in D, 4a = 4b2, 4b = it might be 4b2 or 4b5 so we need one more step
3 - Looking at the 4 bad corners, two of them will be stacked (here they're UFL and DFL). We want to trace the stacked corners along with the closest oriented corner on U/D to these stacked corners (in this case it would be tracing UFL+UBL and DFL+DFR). 4b = 4b5, 4a = 4b2
Side note; if you do a setup move to form the initial 2/2 split instead of rotating, this version of tracing also tells you whether or not you need to do the fake-HTR U2 fix from the side you're looking at the case on. It's possible that the rotating way could also do this/already does, but I can't confirm that fact.
4c3/4c4 (from the b side)
1 - Bring the 4 corners such that they're 2/2 across U/D (same as 4b2/4b5 including the rotation trick)
2 - Trace 2 oriented corners in U and 2 misoriented corners in D, 4a = 4c3, 4b = 4c4
4c3/4c4 (from the a side)
1 - Bring the 4 corners such that they're 2/2 across U/D and adjacent to each other (same as 4a1/4a2)
2 - Trace 2 oriented corners in U and 2 misoriented corners in D, then trace all 4 corners in U
3 - 4b, 4a = 4c3 and ready to qt into 4b2 (from this position), 4b, 4b = 4c4, 4a, 4b = 4c3 but need half turn before a qt brings the case into 4b2 (from this position)
In place of that, we mostly reapply the same system of determining the difference between 4a, 4b, and 4c, but on varying sets of pieces. Credit to Jackson Warley for the 2c system & special thanks to Elliot Wiecek for verifying whether or not the stuff that was being cooked in the Discord group chat actually worked or not. Everything else here was either labbed by me or is just "common knowledge".
Quick terminology rundown since I know some old guides swap a/b for 4a/4b:
0c - Corners are oriented on all 3 axes (8c is the same thing but a slice move away)
2c - 2 corners are misoriented on 2 axes (6c is the same thing but a slice move away)
4a - 4 misoriented corners which can all be set up to U (or D) using only half turns
4b - 4 misoriented corners which cannot be set up to U (or D) using only half turns
4c - On one side of the scramble (in NISS terms), the corners are in 4a but on the other side they're in 4b
3/1 2/2 4/0 1/1 2/0 - Distribution of oriented/misoriented corners across U/D (we will be assuming the DR is solvable with U D R2 L2 F2 B2 in all cases)
Cases:
0c0/0c3/0c4
For these it's useful to know what HTR corners look like visually. The main thing to pay attention to is what kinds of patterns are/aren't possible in terms of opposite sides (we'll use U/D for this too). You can either have a face, "stacked" bars (R2 or R2 U2, but not R2 U), or slashes (F2 U2 R2). If all 6 sides are one of these 3 "real" patterns, it's HTR. If all 6 sides don't have a real pattern (either a 3/4 face or unstacked bars), it's 0c4. If it's a mix of the two, it's 0c3.
Another way to think about this which is more in line with how other cases get recognized is doing a 4a/4b comparison trace of the F, R, and if needed U faces. All 3 being 4a means it's 0c0, 1/3 being 4b would be 0c3, and 3/3 being 4b would be 0c4.
2c3/2c4/2c5
There are two possible methods for this that I'm a fan of (Jackson's one & an application of an old idea called "mental swap"). The mental swap one is literally just mentally swapping the 2 corners in your head and visualizing the resulting state. If it looks like HTR (or 0c0), it's 2c5, if it looks like 0c3, it's 2c4, if it looks like 0c4, it's 2c3.
Jackson's method goes like this:
1 - Bring the 2 misoriented corners next to each other, let's say UFR UFL (but something like UFR DFR is also possible)
2 - Trace both of the two layers that don't contain the 2c. Here this would be the D face followed by the B face
3 - 4a/4a = 2c5, 4a/4b = 2c4, 4b/4b = it might be 2c3 or 2c4 so we need one more step
4 - If you got 4b/4b, trace the DF and UB corners, 4b again = 2c3, 4a = 2c4 (another way to think of this step is doing either D2 or B2 & retracing D if you did B2, or retracing B if you did D2).
Note that if you instead have the 2c in say UFR DFR, you still trace the same way but would do L and B, followed by FL and BR corners if necessary. I recommend being comfortable with both so you can always trace without needing a setup move & can instead just invert the last move of your DR solution
4a1/4a2
1 - Bring the 4 corners such that they're 2/2 across U/D and adjacent to each other rather than diagonal (involves inverting the last move of your DR solution in the worst case scenario)
2 - Trace 2 oriented corners in U and 2 misoriented corners in D, 4b = 4a1, 4a = 4a2
4b2/4b5
1 - Bring the 4 corners such that they're 2/2 across U/D. Note that you can avoid doing any setup moves by rotating. For example, if you do U R2 U R2 on a solved cube, y z' sets them up to be around the UF and DL edges or UFR+UFL and DFL+DBL. When we get to step 3, we'll also be assuming the corners are in these 4 locations
2 - Trace 2 oriented corners in U and 2 misoriented corners in D, 4a = 4b2, 4b = it might be 4b2 or 4b5 so we need one more step
3 - Looking at the 4 bad corners, two of them will be stacked (here they're UFL and DFL). We want to trace the stacked corners along with the closest oriented corner on U/D to these stacked corners (in this case it would be tracing UFL+UBL and DFL+DFR). 4b = 4b5, 4a = 4b2
Side note; if you do a setup move to form the initial 2/2 split instead of rotating, this version of tracing also tells you whether or not you need to do the fake-HTR U2 fix from the side you're looking at the case on. It's possible that the rotating way could also do this/already does, but I can't confirm that fact.
4c3/4c4 (from the b side)
1 - Bring the 4 corners such that they're 2/2 across U/D (same as 4b2/4b5 including the rotation trick)
2 - Trace 2 oriented corners in U and 2 misoriented corners in D, 4a = 4c3, 4b = 4c4
4c3/4c4 (from the a side)
1 - Bring the 4 corners such that they're 2/2 across U/D and adjacent to each other (same as 4a1/4a2)
2 - Trace 2 oriented corners in U and 2 misoriented corners in D, then trace all 4 corners in U
3 - 4b, 4a = 4c3 and ready to qt into 4b2 (from this position), 4b, 4b = 4c4, 4a, 4b = 4c3 but need half turn before a qt brings the case into 4b2 (from this position)
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