1 look last layer thing

[zzcuberman]

Can you use 1ll on the d and p shapes just from info from the top?

 

[Imsoosm]

I'm pretty sure you have to have at least 5 unoriented pieces, because if you have 5 unoriented pieces, the three remaining pieces would either be 2 edges and 1 corner / 2 corners and 1 edge, and there are no PLLs that solve 2 of edges/corners and 1 of corners/edges.

In this case D and P would probably be no, because they only have 4 unoriented pieces.

So the dot cases, square shapes, knight moves, L shapes, line shapes, and OLLs 7, 8, 9, 10, 11, 12, 29, 30 can all be recognized just from the top.

 

[zzcuberman]

Ok

I'm pretty sure you have to have at least 5 unoriented pieces, because if you have 5 unoriented pieces, the three remaining pieces would either be 2 edges and 1 corner / 2 corners and 1 edge, and there are no PLLs that solve 2 of edges/corners and 1 of corners/edges.

In this case D and P would probably be no, because they only have 4 unoriented pieces.

So the dot cases, square shapes, knight moves, L shapes, line shapes, and OLLs 7, 8, 9, 10, 11, 12, 29, 30 can all be recognized only from the top.

Awesome thanks

 

[xyzzy]

So the dot cases, square shapes, knight moves, L shapes, line shapes, and OLLs 7, 8, 9, 10, 11, 12, 29, 30 can all be recognized just from the top.

If you have (at least) two oriented edges, it's always impossible to deduce PLL from only the top face.

Case (i) zero or two twisted corners: You can swap the two oriented edges and the two oriented corners.
Case (ii) four twisted corners: By pigeonhole principle, two of the corners will have the same up-facing sticker; swap those two and the two oriented edges.
Case (iii) three twisted corners: For every S CLL case, there's a corresponding AS CLL case with the same U colours and the opposite permutation parity, and vice versa. Change the corners accordingly and swap the two oriented edges. (Example: S vs AS.)

---

So that leaves only the dot OLLs. For the sake of brevity, I'll use the same numbering as Roux's CMLL numbers (but with our usual letters for OCLL names).

S/AS: Always possible to deduce.

O: Always impossible to deduce (do an H perm).

H/pi: The cases with only two distinct U colours on the corners (H1, H6, pi3, pi5) can always be converted to an identical looking case with an E perm or H perm. The rest are always possible to deduce: two corners will have the same U sticker, so the identities of the remaining two corners are fixed; since we know permutation parity, we can also identify the two corners with the same U sticker.

T/U/L: The cases with two identical U colours on the twisted corners (L2, L3, U2, U6, T1, T4) are always impossible to deduce (do an E perm or H perm). The cases with two opposite U colours on the twisted corners (L4, L5, U1, U4, T2, T6) are always impossible to deduce (wlog the colours are red and orange; swap the two red corners and the two orange corners and twist accordingly). The remaining cases are always possible to deduce: the identities of the two twisted corners are uniquely determined and permutation parity lets you identify the two oriented corners.

tl;dr summary: If you can see all four edges and two adjacent colours on the twisted corners, then you can deduce the 1LLL. Otherwise, you can't.

 

[zzcuberman]

If you have (at least) two oriented edges, it's always impossible to deduce PLL from only the top face.

Case (i) zero or two twisted corners: You can swap the two oriented edges and the two oriented corners.
Case (ii) four twisted corners: By pigeonhole principle, two of the corners will have the same up-facing sticker; swap those two and the two oriented edges.
Case (iii) three twisted corners: For every S CLL case, there's a corresponding AS CLL case with the same U colours and the opposite permutation parity, and vice versa. Change the corners accordingly and swap the two oriented edges. (Example: S vs AS.)

---

So that leaves only the dot OLLs. For the sake of brevity, I'll use the same numbering as Roux's CMLL numbers (but with our usual letters for OCLL names).

S/AS: Always possible to deduce.

O: Always impossible to deduce (do an H perm).

H/pi: The cases with only two distinct U colours on the corners (H1, H6, pi3, pi5) can always be converted to an identical looking case with an E perm or H perm. The rest are always possible to deduce: two corners will have the same U sticker, so the identities of the remaining two corners are fixed; since we know permutation parity, we can also identify the two corners with the same U sticker.

T/U/L: The cases with two identical U colours on the twisted corners (L2, L3, U2, U6, T1, T4) are always impossible to deduce (do an E perm or H perm). The cases with two opposite U colours on the twisted corners (L4, L5, U1, U4, T2, T6) are always impossible to deduce (wlog the colours are red and orange; swap the two red corners and the two orange corners and twist accordingly). The remaining cases are always possible to deduce: the identities of the two twisted corners are uniquely determined and permutation parity lets you identify the two oriented corners.

tl;dr summary: If you can see all four edges and two adjacent colours on the twisted corners, then you can deduce the 1LLL. Otherwise, you can't.

I'll just do 1ll with reg zbll recognition lol