nvm not low 19: 19.44


12 23.19 B2 D' L U' D R2 F2 B' U F2 B' R' U L' D2 L F2 R L2 B' U2 F' D2 L2 R

13 16.32 R' D U' B R' D F' R' B' R' L' D2 L R' F2 U R2 U2 D R2 B2 F' R' F2 B'

14 16.32 D2 F2 L R' U R' B D' U2 R L' B2 D2 R2 D' F2 B2 U L' F2 R B F' D R

15 18.83 R U L U F D2 B L B U2 F2 D F2 L B2 D' B' D2 R2 U2 L' B' U F' B2

16 26.77 B F' U R2 L2 B' D2 F B' U' D B2 F2 L' F L' D2 L' D F' D2 R B' L2 B
 
Yesterday evening I did 125 solves (after that I got really tired, so I stopped there).

And I did a PB!

Avg100: 14.66 (PB, improved by 0.09! Yay! And there was a lot of aweful solves!)
Avg50: 14.44 (almost...)
Avg12: 14.03 (should be better)
Avg5: 13.30 (not bad)
Best: 10.28 (yay, another sub 10.5!)
 
I contantly peeking on the down side to determine the Fuu case. I want to get rid of it.
Tracking the BU edge sometimes really easy, sometimes really hard.
Recently, I started to learn Athefre's method (tracking DF/DB pieces). It's a very interesting method, although it might make the LSE a bit longer. How much longer in worst case? 1 move? Is it worth to drill the method to avoid the peeking?
 
I contantly peeking on the down side to determine the Fuu case. I want to get rid of it.
Tracking the BU edge sometimes really easy, sometimes really hard.
Recently, I started to learn Athefre's method (tracking DF/DB pieces). It's a very interesting method, although it might make the LSE a bit longer. How much longer in worst case? 1 move? Is it worth to drill the method to avoid the peeking?

It doesn't have to be longer, just skip the "put the edges opposite to each other step" and solve the rest intuitively/optimally instead.

What I do is just track the position of a DB/DF edge that goes onto the bottom. This is enough information for me to figure out the case.

I'm pretty sure it is only one move longer in the worst case. I can try to list all of the cases that take a move longer if you want.
 
It doesn't have to be longer, just skip the "put the edges opposite to each other step" and solve the rest intuitively/optimally instead.

What I do is just track the position of a DB/DF edge that goes onto the bottom. This is enough information for me to figure out the case.

I'm pretty sure it is only one move longer in the worst case. I can try to list all of the cases that take a move longer if you want.

Thank you. I try to do it intuitively then.
The list is not necessary.
 
Okay, maybe it's me, but I found a better interpretation for the last part of 4c with Athefre's method.

Technically, when you finished 4b and put the DF/DB edges diagonal, the center shows you the direction of the M move. Simply move it to the direction of the matching edge (matching with UB -> M', matching with UF -> M), then U2, and the rest is pretty obvious.

A few examples:

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Pretty normal:

Setup:
D2 L2 R2 U L' R F D U' L D' U F' x'

Solve:
U' M U M' U M U2 // now it's time to predict the DF/DB edges: after M2, the the one in UB goes to DF
M2 U // diagonal DF/DB. Now the center matches with the UB sticker, so next move is M' U2
M' U2 M U2 M2 // finish

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Pretty normal:

Setup:
U R2 U2 R' U B L U' F L' B' R'

Solve:
M' U M U' M' U2 M' U' // now it's time to predict the DF/DB edges: after M2, the the one in UB goes to DF
M2 U // diagonal DF/DB. Center again matches with the UB sticker, so M' U2 again
M' U2 M' U2 M2 U2 // finish

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Pretty normal:

Setup:
D2 L2 F2 R2 U R2 F2 L' R' B' L2 R2 x

Solve:
M2 U' M' U2 // now it's time to predict the DF/DB edges: after M2, the the one in UB goes to DF
M2 U' // diagonal DF/DB. Center matches with the UF sticker, so it's M U2
M U2 M U2 M2 // finish

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Tricky recognition:

Setup:
R2 B2 L2 U R2 F2 R2 U L R' x

Solve:
M U' M' // tricky one: see the sticker on FD? It's a DF/DB sticker going to DB
U2 M U // diagonal DF/DB. Center matches with the UF sticker, so this time it's M U2
M U2 M // finish

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Tricky recognition:

Setup:
B L' F U2 D2 B' R F' D' F2 B2 U R2 L2

Solve:
M' U M' U M U // tricky: see the FU sticker? It's a DF/DB sticker going to DF
M U' // // diagonal DF/DB. Center matches with the UF sticker, so it's M U2
M U2 M U2 M2 U2 // finish

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

An interesting one with misoriented centers:

Setup:
B F U2 L' U2 B2 U2 L' B F' L2

Solve:
M' U M' U2 M U2 // now it's time to predict the DF/DB edges. Let's say our DF/DB edges are currently on UL and UF. Interestingly, it is still working if we chose the other two edges (currently on UB and UR) as our DF/DB edges! It leads to the same solution.
M2 U' // diagonal DF/DB. Center matches with the UF sticker, so it's M U2
M U2 M U2 M // finish

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Again with misoriented centers:

Setup:
B2 L R' U2 B' L2 R2 F2 L R' D' L R' x'

Solve:
M U M U2 // now it's time to predict the DF/DB edges: we can chose between UB+UL, or UR+UF (it's the same)
M2 U // // diagonal DF/DB. Center matches with the UB sticker, so it's M' U2
M' U2 M U2 M // finish
 
You Rouxers are gonna love this scramble D U R2 U' B2 D R2 U' B2 U F2 L' F U B2 U R' F' U' L U2 LOL 2/3 of the green block is already done.

Easy SB as well. I found a short F2B that cancels into CMLL

y
U'R'URU'B
r'U'rFRF'
U'R2DR'U2RD'R2URU'R'
M2UMUM'UM'U2M2U2M'

35stm

alg.cubing.net

Edit: Just saw Renslay's post

Okay, maybe it's me, but I found a better interpretation for the last part of 4c with Athefre's method.

Technically, when you finished 4b and put the DF/DB edges diagonal, the center shows you the direction of the M move. Simply move it to the direction of the matching edge (matching with UB -> M', matching with UF -> M), then U2, and the rest is pretty obvious.

It was already possible to recognize these cases intuitively, as you know the positions of both of the DF/DB edges after UL/UR. I guess this makes it a bit easier then?
 
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Ok so from all this interesting LSE discussion, I remembered a proposal I made a time ago. Basically it's a different way/mindset of solving 4b-4c. The idea is that you put UF and UB edges in bottom of M-slice, use M2 to solve them (just like if they're UL and UR) , then go on to solve UL and UR edges. This "avoids" Fuu cases. But I say this tentatively because I don't know how efficient it is compared to the conventional way and most especially I don't know if this makes sense at all or it's some stupid idea I formulated.

There is also the issue of tracking UF and UB edges since you can't really "cheat" by tracking just the block color stickers (what you would do in normal way). This is one downfall.
 
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Ok so from all this interesting LSE discussion, I remembered a proposal I made a time ago. Basically it's a different way/mindset of solving 4b-4c. The idea is that you put UF and UB edges in bottom of M-slice, use M2 to solve them (just like if they're UL and UR) , then go on to solve UL and UR edges. This "avoids" Fuu cases. But I say this tentatively because I don't know how efficient it is compared to the conventional way and most especially I don't know if this makes sense at all or it's some stupid idea I formulated.

Uhm... What? Like this?

Setup:
F2 L' R U2 B D2 L' R F2 D' x2

U' M U2 M' U2 M' U M' // EO
U2 M U2 M' // put UF/UB to the bottom
M2 // solve UF/UB
U M2 U2 M2 U // solve UL/UR edges(?)
E2 M E2 M // dot case(?)

Nope. I like the good old
U' M U2 M' U2 M U2 M2 U' M' U2 M' U2 M
better.

With this, I think you do addigitonal moves by putting the UF/UB to the bottom, and also increase the chance of the dot case, which is worst/slower than Fuu.
 
You don't have to do UL/UR (again, UF/UB here is a sort of pseudo UL/UR), you can just to a U and then do a kind of pseudo 4c.

So,

Setup:
F2 L' R U2 B D2 L' R F2 D' x2

U' M U2 M' U2 M' U M' // EO
U2 M U2 M' // put UF/UB to the bottom
M2 // solve UF/UB
U // UL/UR
M U2 M2 U2 M' U'

I think it can be shorter in some cases. Obviously for your scramble UL/UR is shorter. I sometimes solve UF/UB edges instead of UL/UR edges when they seem easier.
 
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You don't have to do UL/UR (again, UF/UB here is a sort of pseudo UL/UR), you can just to a U and then do a kind of pseudo 4c.

So, I think it can be shorter in some cases. Obviously for your scramble UL/UR is shorter. I sometimes solve UF/UB edges instead of UL/UR edges when they seem easier.

Aaaaw, I see it now.

Well, obviously, this is just like a U/U' premove if you think it as an FMC solve.

Which means with this, sometimes you have longer, sometimes shorter solutions, and on average the two is the same (both in move length and the number of Fuu cases).
Technically, you treat UF/UB as if they were UL/UR edges. No difference.
But the recognition might be harder, and also many times it can be hard to decide during midsolve whenever this, or the regular way leads to a shorter solution.
 
This is exactly what I thought. Though there are some perks, there are problems with recog and move count. However, I think it could possibly be feasible with more work and discussion. Finding ways to influence UL and UR after UF and UB could be interesting. Also, dots case really isn't bad. I can put up some examples later if needed?
 
I was surprised that you guys don't do a combination of UF/UB and UL/UR like I do. I thought everyone did a combination of the two.

I honestly think there is no problem whatsoever with recog. And as for movecount, it's sometimes more and sometimes less, and pretty much the same most of the time. Just like choosing one block over another; you try to guess which one is faster.

I do think that exclusively doing UF/UB would be slower than exclusively doing UL/UR though. I find I do UL/UR most of the time as well.
 
Sometimes it uses more moves. I know a couple special cases where it definitely saves moves and I am able to do all the cases where UFUB are solved (which I agree is beneficial).

I tried not to do UFUB in the beginning because some cases use more moves than ULUR- even more so if you mess up the (harder) recognition. I do have an issue with your "sometimes more, sometimes less" idea. Does this mean you don't know which cases will use more and which ones will use less?
 
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