# A bunch of questions I need answers...

#### LexCubing

##### Member
First you are done with EOLine and the DR corners are in their respective places not caring about their orientation.

You are going the make the BL square but the edge/s needed are stuck in the "right block". How many unique cases if I were to solve the square while preserving the DR corners?

Next question. I'm going solve the the FL pair while solving CP. How many unique cases are there including when the edge is stuck?

The L block is done. I'm going solve the BR square. Hoe many unique cases are there that solves it while preserving the DFR corner?

Note everytime I say preserve the DR corners, I just want to preserve their permutation.

Last question. The F2L corner is inserted ( its orientation isn't necessarily solved ) CP and EO is solved. How many unique cases?

#### xyzzy

##### Member
I'm not going to spoonfeed the answers here, but here's some tips. Carefully enumerate every possible case for the edges, along with every possible case for the corners, then multiply those numbers and you get the numbers you want. (Note: does not work if there are parity constraints, but none of the questions you're asking for involve that. If there is a parity constraint, you have to enumerate the even-parity edge cases with the even-parity corner cases, plus the odd-parity edge cases with the odd-parity corner cases.)

So, for your first question, you care about two edges: BL and DL. (Assuming scramble orientation, these are blue-orange and yellow-orange. Adjust accordingly if you use a different orientation for EO.) How many cases are there if only the blue-orange edge is stuck? How many cases are there if only the yellow-orange edge is stuck? How many cases are there if both edges are stuck? Add up these numbers and there's your answer.

Next question. I'm going solve the the FL pair while solving CP. How many unique cases are there including when the edge is stuck?
This is moderately tricky and is also the only one I'm going to answer directly. When you say "unique", I'm going to assume you mean unique up to AUFs. Either the yellow-orange-green corner is in the correct position (case #1) or it's not (case #2). In either case, the matching edge can be in any of three positions, stuck in the right block.

In case #1, there are three possible CPs on the last layer (no-swap, adjacent, diagonal) up to AUF. This contributes 9 subcases

In case #2, without loss of generality, AUF the corner to a fixed location (e.g. UFR). This leaves 4! = 24 possible permutations of the four corners we haven't accounted for (DRF, ULB, UBR, URF), Since we don't care about AUF "premoves" (if you don't know NISS, think of it as AUF applied after you have solved CP), this cuts the number of cases down by a factor of 4 to 6. This contributes 18 subcases.

Now, the yellow-orange-green corner can also start off in an arbitrary orientation, so there are (9 + 18) × 3 = 81 cases in total.

• AlphaSheep

#### LexCubing

##### Member
I'm not going to spoonfeed the answers here, but here's some tips. Carefully enumerate every possible case for the edges, along with every possible case for the corners, then multiply those numbers and you get the numbers you want. (Note: does not work if there are parity constraints, but none of the questions you're asking for involve that. If there is a parity constraint, you have to enumerate the even-parity edge cases with the even-parity corner cases, plus the odd-parity edge cases with the odd-parity corner cases.)

So, for your first question, you care about two edges: BL and DL. (Assuming scramble orientation, these are blue-orange and yellow-orange. Adjust accordingly if you use a different orientation for EO.) How many cases are there if only the blue-orange edge is stuck? How many cases are there if only the yellow-orange edge is stuck? How many cases are there if both edges are stuck? Add up these numbers and there's your answer.

This is moderately tricky and is also the only one I'm going to answer directly. When you say "unique", I'm going to assume you mean unique up to AUFs. Either the yellow-orange-green corner is in the correct position (case #1) or it's not (case #2). In either case, the matching edge can be in any of three positions, stuck in the right block.

In case #1, there are three possible CPs on the last layer (no-swap, adjacent, diagonal) up to AUF. This contributes 9 subcases

In case #2, without loss of generality, AUF the corner to a fixed location (e.g. UFR). This leaves 4! = 24 possible permutations of the four corners we haven't accounted for (DRF, ULB, UBR, URF), Since we don't care about AUF "premoves" (if you don't know NISS, think of it as AUF applied after you have solved CP), this cuts the number of cases down by a factor of 4 to 6. This contributes 18 subcases.

Now, the yellow-orange-green corner can also start off in an arbitrary orientation, so there are (9 + 18) × 3 = 81 cases in total.
Thanks.

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