Square-1: CSP algs swapped when CS is flipped?

[Hazel]

Today I started learning CSP. I've found that with some cases, when the CS is flipped upside down, which solving algorithm is odd and which one is even changes, while some do not change. For example, the CS case in this image with the numbering scheme indicated (from Hashtag Cuber's website). The top solution is the odd parity solution for CSP usually (and the bottom solution even), but if I do (6,6), now the top solution is for even parity, and the bottom for odd. With other cases—for example, 8/Star—the mirror does not equal switched algorithms. How do you remember which cases have this switched alg phenomenon and which don't? Is there an easy way to know without having to just memorize for each case?

 

[IsThatA4x4]

Today I started learning CSP. I've found that with some cases, when the CS is flipped upside down, which solving algorithm is odd and which one is even changes, while some do not change. For example, the CS case in this image with the numbering scheme indicated (from Hashtag Cuber's website). The top solution is the odd parity solution for CSP usually (and the bottom solution even), but if I do (6,6), now the top solution is for even parity, and the bottom for odd. With other cases—for example, 8/Star—the mirror does not equal switched algorithms. How do you remember which cases have this switched alg phenomenon and which don't? Is there an easy way to know without having to just memorize for each case?

View attachment 25690

Yes, when you have a 6 edges/2 edges case (and only in those cases) flipping the cube inverts the parity count. I normally deal with this by just flipping the cube so that the 6 edges are always on top when tracing, so I always get the parity count that matches with the solution I do.

 

[NebraskanSpeedsolver]

Yes, when you have a 6 edges/2 edges case (and only in those cases) flipping the cube inverts the parity count. I normally deal with this by just flipping the cube so that the 6 edges are always on top when tracing, so I always get the parity count that matches with the solution I do.

Well this is odd...
it says that you quoted MY post, but both a). I don't do squan and b). I haven't replied to this thread (up until now)
weird.

 

[Hazel]

Yes, when you have a 6 edges/2 edges case (and only in those cases) flipping the cube inverts the parity count. I normally deal with this by just flipping the cube so that the 6 edges are always on top when tracing, so I always get the parity count that matches with the solution I do.

if the equator isn't flipped, that'd lead to some pretty awkward solutions wouldn't it?
Also, that rule doesn't apply to Right 4-2/Paired Edges or Left 4-2/Paired Edges, those are the same mirrored or not

 

[IsThatA4x4]

Well this is odd...
it says that you quoted MY post, but both a). I don't do squan and b). I haven't replied to this thread (up until now)
weird.

Sorry about that, I think I had one of your posts from another thread quoted accidentally and when I realised I deleted it straight away but it still gave you an alert I guess.

 

[IsThatA4x4]

if the equator isn't flipped, that'd lead to some pretty awkward solutions wouldn't it?
Also, that rule doesn't apply to Right 4-2/Paired Edges or Left 4-2/Paired Edges, those are the same mirrored or not

If the equator isn't flipped, I still put 6 on top to trace, then flip the cube over to do the same solution. For left/right 4-2/pair, it does invert parity count depending on whether you have 6 or 2 on top to trace, I'll give an example scramble here so I can explain better maybe:
(3,2)/(-3,-3)/(-2,-5)/(-3,0)/(2,-1)/(-5,-2)/(5,-1)/(3,-2)/(-3,-3)/(-2,-1)/(-2,0)/(0,2) Right 4-2/pair
In this case I trace from these positions:

For this case, even would be to go to scallop/scallop, and odd would be to go to shield/shield.
Pretending the equator wasn't flipped here, I would flip the cube so that 4-2 is on top and trace even, so I can flip the cube back over and go to scallop/scallop.
However, if I had instead not flipped the cube over and traced with pair on top, I would have traced odd, gone to shield/shield, and got parity.

 

[Hazel]

If the equator isn't flipped, I still put 6 on top to trace, then flip the cube over to do the same solution. For left/right 4-2/pair, it does invert parity count depending on whether you have 6 or 2 on top to trace, I'll give an example scramble here so I can explain better maybe:
(3,2)/(-3,-3)/(-2,-5)/(-3,0)/(2,-1)/(-5,-2)/(5,-1)/(3,-2)/(-3,-3)/(-2,-1)/(-2,0)/(0,2) Right 4-2/pair
In this case I trace from these positions:
View attachment 25713
For this case, even would be to go to scallop/scallop, and odd would be to go to shield/shield.
Pretending the equator wasn't flipped here, I would flip the cube so that 4-2 is on top and trace even, so I can flip the cube back over and go to scallop/scallop.
However, if I had instead not flipped the cube over and traced with pair on top, I would have traced odd, gone to shield/shield, and got parity.

I always trace from the same side - for example for this case, I always trace starting with the 6-edge side, so whether I flip the cube or not won't affect the parity I trace.

Here is an example scramble:
(1,-3)/(-4,-1)/(-5,-2)/(-1,-4)/(1,-2)/(5,-4)/(1,-2)/(0,-4)/(-3,0)/(5,-2)/(4,0)/(-4,0)

This scramble is the mirror of the "original" cubeshape. I trace odd parity. If I do the odd parity algorithm on this scramble, I still get parity—I have to do the even parity algorithm instead, from the bottom. If I do a z2 at the beginning, or (6,6) and solve the cube from there, I have to do the odd parity algorithm as intended. If I'm dealing with the CS's mirror, it's switched. For some other cases though, for example the one you gave an image of, that switching is NOT the case. If I get odd parity, with the mirror or not, I do the odd parity algorithm. Both my scramble and the cubeshape you showed have 6 edges on one side and 2 on the other, yet they don't follow the same rule.

 

[IsThatA4x4]

I always trace from the same side - for example for this case, I always trace starting with the 6-edge side, so whether I flip the cube or not won't affect the parity I trace.

Here is an example scramble:
(1,-3)/(-4,-1)/(-5,-2)/(-1,-4)/(1,-2)/(5,-4)/(1,-2)/(0,-4)/(-3,0)/(5,-2)/(4,0)/(-4,0)

This scramble is the mirror of the "original" cubeshape. I trace odd parity. If I do the odd parity algorithm on this scramble, I still get parity—I have to do the even parity algorithm instead, from the bottom. If I do a z2 at the beginning, or (6,6) and solve the cube from there, I have to do the odd parity algorithm as intended. If I'm dealing with the CS's mirror, it's switched. For some other cases though, for example the one you gave an image of, that switching is NOT the case. If I get odd parity, with the mirror or not, I do the odd parity algorithm. Both my scramble and the cubeshape you showed have 6 edges on one side and 2 on the other, yet they don't follow the same rule.

Edit: sorry for the obscene amount of text (as well as initially misunderstanding your problem), I find it quite difficult to explain this

I think I see the issue now, it's not with the tracing (because you always trace with 6 on top) so it's not to do with 6 edges/2 edges, but just how cases mirror in general.

You will find that, for example, when your even solution for one mirror is holding something in the back; for the other mirror your even solution would be holding it in the front instead (because if you did an x2 to preserve how your right slice affects cs, the back and front switch). The reason this doesn't happen for cases like 4-2/pair, is because the odd and even solutions differ by entirely changing the path you take through cs (i.e. go to scallops instead of shields) so the position you start in isn't so important. This does however happen for 4-1-1/pair, because both solutions involve going to good pawns in 1 slice, joining the same blocks, just with different starting positions.

Spoiler: example, you can see the single edge of the muffin doing this

So, in general, when you notice that both even and odd algs go through the same cs path, and/or when it is especially important to start in a particular position (e.g. holding the L in the back to go to scallop/kite with 4-2/L), this rule will apply with back/front switching - effectively you have to do this on a case-by-case basis (as a sidenote, the doc I used to learn full csp from gave written explanations, rather than algs, which made it clearer to see when this applies, so you might need to do some experimentation to see if back/front matters).

Just as an added thing, sometimes this can be useful - in the example you gave, since we know that doing a 6 on bottom would set the 4-1-1 up to be in the "correct" position for these algs, we also know that the starting position effectively has "inverted" solutions, i.e we can do the "even" alg from the starting position and not end up with parity (that would look like 40/-40/1-4/03/) which avoids having to do a time consuming 06 at the start.

 

[Hazel]

Edit: sorry for the obscene amount of text (as well as initially misunderstanding your problem), I find it quite difficult to explain this

I think I see the issue now, it's not with the tracing (because you always trace with 6 on top) so it's not to do with 6 edges/2 edges, but just how cases mirror in general.

You will find that, for example, when your even solution for one mirror is holding something in the back; for the other mirror your even solution would be holding it in the front instead (because if you did an x2 to preserve how your right slice affects cs, the back and front switch). The reason this doesn't happen for cases like 4-2/pair, is because the odd and even solutions differ by entirely changing the path you take through cs (i.e. go to scallops instead of shields) so the position you start in isn't so important. This does however happen for 4-1-1/pair, because both solutions involve going to good pawns in 1 slice, joining the same blocks, just with different starting positions.

Spoiler: example, you can see the single edge of the muffin doing this

View attachment 25723

So, in general, when you notice that both even and odd algs go through the same cs path, and/or when it is especially important to start in a particular position (e.g. holding the L in the back to go to scallop/kite with 4-2/L), this rule will apply with back/front switching - effectively you have to do this on a case-by-case basis (as a sidenote, the doc I used to learn full csp from gave written explanations, rather than algs, which made it clearer to see when this applies, so you might need to do some experimentation to see if back/front matters).

Just as an added thing, sometimes this can be useful - in the example you gave, since we know that doing a 6 on bottom would set the 4-1-1 up to be in the "correct" position for these algs, we also know that the starting position effectively has "inverted" solutions, i.e we can do the "even" alg from the starting position and not end up with parity (that would look like 40/-40/1-4/03/) which avoids having to do a time consuming 06 at the start.

Thank you—this explanation makes sense. So, what do you think the best solution to my issue is? Just memorize which cases have flipped odd/even with mirrors?

 

[IsThatA4x4]

Thank you—this explanation makes sense. So, what do you think the best solution to my issue is? Just memorize which cases have flipped odd/even with mirrors?

I just memorised on a case-by-case basis, where if I see I need to hold something in the back/front specifically for the solutions then I know that this happens. It was annoying at first, but since so many of this kind of thing shows up throughout CSP, I eventually got used to it and it practically became second nature. There is no way to tell without the solution, so I think this is the only way you can do it.

 

[LBr]

I'm bumping this bc I am interested to see how you found the learning process and how I can help possibly if it's not too late. If you find hashtag's website confusing then there is this doc which I used to learn and my sheet which I would recommend when you know all the angles and just need a reference point for Odd and even solutions. It's a good template so you can even just make your own version and swap odd and even around if your tracing positions are different. And to answer your initial question briefly (nvm if it's too late) basically I would just always from the side with more edges on top and just vizualise which part of the shape you slice away. If you can do that, you should be able to execute from any angle just fine

 

[Hazel]

I'm bumping this bc I am interested to see how you found the learning process and how I can help possibly if it's not too late. If you find hashtag's website confusing then there is this doc which I used to learn and my sheet which I would recommend when you know all the angles and just need a reference point for Odd and even solutions. It's a good template so you can even just make your own version and swap odd and even around if your tracing positions are different. And to answer your initial question briefly (nvm if it's too late) basically I would just always from the side with more edges on top and just vizualise which part of the shape you slice away. If you can do that, you should be able to execute from any angle just fine

I appreciate you wanting to help, but I'm not still learning CSP—I figured if I wasn't interested in Square-1 enough to learn regular cubeshape, it would be a waste of time to try and figure out all the nuances of CSP.