Solution #1

gilles · Message par **gilles** » lun. août 22, 2005 9:24 am

Here's my first proposal, following my way of cubing.

(LFLD2BU'F2) is the most promising beginning I found (in the train between Tours and Paris). It gives you a 1x2x3 and a 1x2x2. From here, you may have different ideas than mine.

Then, you can solve the last F2L pair with (r'U'RU'R2FRF').
It's not the shortest sequence, but then you can solve the last corners with (FR'F'RU2RU2R') and make some moves disappear.

(r'U'RU'R2FRF'FR'F'RU2RU2R') -> (r'U'(R)U'(R)'U2(R)U2(R)').

Note that this last sequence is flexible. (R) means you can use R or r, it does not have any impact on U corners or L/R-1x2x3 blocks.

Looks good till now. But I couldn't find a happy ending with the 6 last edges.

My best is :
(LFLD2BU'F2) (r'U'RU'r'U2rU2R') (UM2B2U'M'UB2UL2)
HTM : 27
QTM : 36
SHTM : 25
SQTM : 33

I'll try to see if there's a way to insert an early edge 3-cycle or something that leads to a shorter solution. 27 is not good enough! :-)

lars · Message par **lars** » mar. sept. 06, 2005 9:38 am

So... what does r mean in this notation?

Message par **cyril** » mar. sept. 06, 2005 12:40 pm

r is a double layer move : you turn the R layer and the M (central, vertical slice) together

gilles · Message par **gilles** » mar. sept. 06, 2005 1:28 pm

The solution, as it is written, works in Cube Companion, if you need to figure out how it looks like.

Message par **cyril** » dim. oct. 09, 2005 12:29 am

Hi,

I took Gille's first moves and translated them into classic notation :

(LFLD2BU'F2) (r'U'RU'r'U2rU2R') ->L F L D2 B U' F2 L' B' R B' L' D2 L B2 R' B2

That leaves 5 edges ... The best insertion I found was L F L D2 .B .(LR'D2L'RF2)U' F2 L' B' R B' L' D2 L B2 R' B2, leaving 3 edges in (23,30,21,28 ) ... + layers adjustments ... probably not interesting to get a solution better than Lars', but I'll try anyway ...

edit : it's actually 4 edges ... sorry

Message par **cyril** » lun. oct. 10, 2005 1:29 pm

Okay ...
let's start again ...

A good insertion after gille's start is

L F L (D2B'RL'D2R'L[B'D2)D2 .B] .U' F2 L' B' R B' L' D2 L B2 R' B2, which leaves 3 edges after (22,28, 20, 26) I did not find a good way to finish ... perhaps you Gilles know some more edges algs ? Good luck !

gilles · Message par **gilles** » lun. oct. 10, 2005 6:23 pm

Thanks for trying to improve the solutions Cyril, that's team spirit. I looked at the other suggestions, but couldn't find anything.

Tonight, I'll look at this FMC for the last time, then we'll have to think about sending solutions.

For now, we have:
- An elegant QTM-wise solution,
- Another solution that's slightly better STM-wise.
- An incredible team solve.
- A speed solve.

Message par **cyril** » lun. oct. 10, 2005 8:25 pm

I don't think we are going to find much better than yours and Lars ... I've spent hours on this scamble and yet I don't find anything better

Then thing is, Dan hasn't posted #99 results yet, so I guess we have one week more for posting the solutions ... I mean, we are not going to improve the solutions, but perhaps I will at last have time to try to speedsolve the cube your way and time it (a solve is a solve ... no ?) that could be fun if you time yourself on your speedsolving solution and send a time below 10 secs ...

Good luck for this evening

and do not hesitate to send partial solutions ...

PS : et profites-en pour finir le team solve, il manque toujours trois mouvements avant d'établir le record du monde pour ces 30 prochaines années 8)

gilles · Message par **gilles** » mar. oct. 11, 2005 11:10 am

I found another sequence that solves the last F2L pair and the corners, try this:

(LFLD2BU'F2)(U2R'F'rU'R'F2)
or
(LFLD2BU'F2)(U2R'F'rU'r'F2)

(LFLD2BU'F2)(U2R'F'rU'r'F2)(M2U2L2) solves everything but 4 edges in (18, 26, 17, 24). Those last 4 edges end up in a very sad configuration

gilles · Message par **gilles** » mar. oct. 11, 2005 12:11 pm

cyril a écrit :Okay ...
let's start again ...

A good insertion after gille's start is

L F L (D2B'RL'D2R'L[B'D2)D2 .B] .U' F2 L' B' R B' L' D2 L B2 R' B2, which leaves 3 edges after (22,28, 20, 26) I did not find a good way to finish ... perhaps you Gilles know some more edges algs ? Good luck !

U LU'M'UL'U'R2 leads to (31,38,28,35)

With a good insertion of that kind of commutators, I guess you can lower the move count a bit, but sub-27 HTM looks difficult.

Message par **cyril** » mar. oct. 11, 2005 1:29 pm

... I'm trying to solve the 4 edges ...
... I thought there would be a miracle, since after the first three moves they are in a really good-looking configuration (cross pattern on two layers that one can solve with FU2D2B2U2D2F)
... well ... indeed I thought that it was the case

and got a "nearly 25 moves" HTM solution

Do you know any 1-move alg to flip two edges

?

L F L (FU2D2B2U2D2F) D2 B U' F2 U2 R' F' L F' L' F2 R2 L2 D2 L2

gilles · Message par **gilles** » mar. oct. 11, 2005 2:23 pm

cyril a écrit :
Do you know any 1-move alg to flip two edges ?

Yes, but with a rather different metric, and it requires a screwdriver.

gilles · Message par **gilles** » mer. oct. 12, 2005 11:47 am

gilles a écrit : (LFLD2BU'F2)(U2R'F'rU'R'F2)
or
(LFLD2BU'F2)(U2R'F'rU'r'F2)

OR (LFLD2BU'F2)(U2R'F'rU'M*R'F2)!

This way I found a rather short ending (solving the last L-edge, then the R-edge and... the others

):

(LFLD2BU'F2)(U2R'F'rU'r2RF2)(UM2B2U'M'UB2UL2)
or (LFLD2BU'F2)(U2R'F'rU'Mr'F2)(UM2B2U'M'UB2UL2) (lowers SQTM score)

(26,36,24,32)
It does not beat Lars' 32 QTM.

gilles · Message par **gilles** » jeu. oct. 13, 2005 5:34 pm

gilles a écrit : (LFLD2BU'F2)(U2R'F'rU'r2RF2)(UM2B2U'M'UB2UL2)

By the way, did you notice that the first and third sequence are the same that in my first solution?

(LFLD2BU'F2)(r'U'RU'r'U2rU2R')(UM2B2U'M'UB2UL2)

r'U'RU'r'U2rU2R'=U2R'F'rU'r2RF2