28/10/05
FMC#100 (sorted by HTM)
Sort by HTM Sort by QTM Sort by STM Sort by SQTM Scramble: L' R' D R (B2 D2 R B F2)*4 U B R' D2 (L' R2 U' F U')*4 F2 L F B
Style: classic - To celebrate FMC#100, we had a special Team event.
Teams of 3 were made, and could work together to solve the scramble.
The winning team was the team who scored the lowest number of moves, averaged over all 4 metrics. |
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THE WINNERS!
The Commutants -
Per Kristen Fredlund
Bob Burton
Sachin Shirwalkar
| F R' B' U R' U R2 U B' D' F U L' D' F L B L' F' L B' U' L' D R2 F
HTM:26
QTM:[28]
STM:{26}
SQTM:{[28]} |
Average number of moves - 27
Umm. This should maybe be recognisable as an edges first solution? It was found like this : F R' B' U R' U R2 U B' D' F L'.R2 F
Then insert L U L' D' L:U' L' D at the dot and L' F L B L' F' L B' at the colon. Had some other starts with less luck for insertions
or that were bad in qtm, giving bad combined metric. So we stick to this one :-) Not the favorite alltime scramble this one :-P
comments: Space is unlimited |
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2nd place!
The Fellowship of the Cube -
Lars Petrus
Gilles Roux
Cyril Castella
| F' R B D R' D F' D U R2 L' U L U F U' L F2 L' F' U' F2 U' F2 U F2 R'
HTM:27
QTM:[32]
STM:{27}
SQTM:{[32]} |
Average number of moves - 29.5
Quote Do R' before scrambling for a better view of the steps :
10 moves pseudo 2x2x3, plus one CE pair then L' U L orients two edges while building another CE pair then U F U' prepares these two pairs to be oriented
then L F2 L' is the miracle orientation of the CE pairs AND you get 2 additionnal CE pairs end : finish a 2x2x1 with F', then F2L
(if you consider Orange as cross color) with U' F2 U' F2 and then it's over...
Lars : I think 27 moves without any insertions is pretty spectacular.
Cyril and Gilles : Congrats Lars !! |
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3rd Place!
Masters of All Challenges -
Mirek Goljan
Alexander Ooms
Chris Hardwick
| F' R D R' D' B D' F' U' B U B2 R' B R L2 B L B' L F R2 U' L' U R2 U' L
HTM:28
QTM:[32]
STM:{28}
SQTM:{[32]} |
Average number of moves - 30
We found this anniversary scramble unusually tough. Here is our straighforward solution.
Any smarter ways of solving led to longer moves :-( . The first 7 moves solve 2x2x3 without one corner.
F' R D R' D' B D' _ F' U' B U B2 R' B R F_ F' L2 B L B' L F U' and solve 3 remaining corners with U R2 U' L' U R2 U' L. |
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4th Place!
SM -
Sebastian Dumitrescu
Mikael Nyberg
Ravi Fernando
| B' L' B R B' L B R' L2 D2 R B F2 U L2 F D' B2 D F' D2 B' R' B' R B R D L2
HTM:29
QTM:[36]
STM:{29}
SQTM:{[36]} |
Average number of moves - 32.5
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5th Place!
Team Neutron -
Charles Tsai
Francois Sechet
Parker Gaitley
| R' B2 F U L' F2 U2 R U' R B' U2 B U' B U' B' U' B U' B2 U' B U' B' U2 B U B' R
HTM:30
QTM:[36]
STM:{30}
SQTM:{[36]} |
Average number of moves - 33
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6th Place!
Team SveKub -
Kenneth Gustavsson
Gunnar Krig
Kare Krig
| F2 R' F B2 U' F2 U2 D' F' R' F2 L' F U R' U' R B' U' B U B' U' B R' U2 F' U' F U R F U R U' R' F' U'
HTM:38
QTM:[44]
STM:{38}
SQTM:{[44]} |
Average number of moves - 41
Using my (Kenneth) "Fish'n'Chips" method.
The F2L was carefully built "Petrus style". Then a "keyhole" for last C/E-pair and LL-edges orientation. The "Fish-step" solves edges
pemutation and one of the corners, here in 12 moves. Then I got a skip in the "Chips" step (that normaly solves the last three corners).
This was the best solution any of us came up with but we also had a 39 (STM) and a number of 42-43 moves solutions. |
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SM -
Sebastian Dumitrescu
Mikael Nyberg
Ravi Fernando
| F' R B D' R L2 U2 L2 R' U R2 U2 R2 U R' B' R B R' U' R' U R2 B U' L U L' B' R' U'
HTM:31
QTM:[38]
STM:{31}
SQTM:{[38]} |
Average number of moves - 34.5
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Team Neutron -
Charles Tsai
Francois Sechet
Parker Gaitley
| L2 D2 R B U L2 D2 R2 B2 U' B2 U B' R B' R' B R' B' D' R' D R' D' R2 D B D L2
HTM:29
QTM:[38]
STM:{29}
SQTM:{[38]} |
Average number of moves - 33.5
L2 D2 R B U L2 D2 [pseudo-2x2x3] R2 B2 U' B2 U [get into 2-gen] B' R B' R' B R' B'
(D' R' D R' D' R2 D) B D L2 [fix pseudo-2x2x3] |
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Team Neutron -
Charles Tsai
Francois Sechet
Parker Gaitley
| R' B2 F U L' F2 U2 R2 U' L U R' U' {L' R} B2 U' R B' U B R2 U' L' U R2 U' {L R'} U' R
HTM:31
QTM:[38]
STM:{29}
SQTM:{[36]} |
Average number of moves - 33.5
Same start as our other solution but with a different ending. R' B2 F U L' F2 U2 R U' R then
B2 U' (R (B' U B U') R') U' (and finally R) leaves 5 corners to solve. R' B2 F U L' F2 U2 R U'
(U R U' L U R' U' L') R B2 U' R B' U B U' (U R2 U' L' U R2 U' L) R' U' R is the final result. Got 2 slice moves out of it. |
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The Fellowship of the Cube -
Lars Petrus
Gilles Roux
Cyril Castella
| L F L D2 B U' F2 U2 R' F' r U' r2 R F2 U {M2} B2 U' {M'} U B2 U L2
HTM:26
QTM:[36]
STM:{24}
SQTM:{[33]} |
Average number of moves - 29.75
Really close to Lars' solution, since average of the four metrics is still under 30 ! Following Gilles' style :
first part consists in building a 1x2x3 and an opposite 1x2x2, then second part completes the latter and solves the corners,
and finally the remaining edges are solved in the last part. Nice ergonomy, would say Zbigniew ;-) Gilles and Cyril spent hours on trying to solve the edges with insertions ...
but apparently it's hard ! |
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The Fellowship of the Cube -
Lars Petrus
Gilles Roux
Cyril Castella
| {F' B} D2 F2 L2 B D L' F L2 B U' B' {L R'} F R2 F' R' F' R F R F' R F' R F2 R2
HTM:29
QTM:[36]
STM:{27}
SQTM:{[34]} |
Average number of moves - 31.5
Genetic solver solution
I thought this could be interesting to test my automatic solver on this scramble ... well, this time it was quite efficient ! 10 moves 2x2x3, 5 more moves for getting into 2-Gen group and
15 for solving it (btw it's interesting that there is no CE pair connected until the 8th move in this phase ... I doubt any human would be able to think so !).
Note that I generated this solution just before submitting it, so we did not use it at all for optimizing our other solutions ! This is a really interesting application of Charles Tsai's method
(though I suppose he would not have used the same alg for the second part ...) |
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The Fellowship of the Cube -
Lars Petrus
Gilles Roux
Cyril Castella
| R' F' L F2 R' B' R F' R' B R2 F' {L' R} D R' D' B' D2 L U L' D2 {L' R} B' {L R'} U B' U B' {U D'}
HTM:34
QTM:[38]
STM:{30}
SQTM:{[34]} |
Average number of moves - 34
"Classical insertion" solution
That one is absolutely not impressive, but you said we could post any solution we wanted :) Add D' before scambling, then solve the 2x2x3 and orient
the edges with F' R D R' D' B' D2 L U L' D2 {L' R} B' {L R'} U B' U B' U, this leaves 4 corners in 21 HTM (+1). Insert (R'F'LF:RF'L'F) at the very beginning and(FR'B'RF'R'BR)
at the colon: fairly few half turns (only 4), but way too long solution to win. |
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The Fellowship of the Cube -
Lars Petrus
Gilles Roux
Cyril Castella
| F R2 U' D' F' R2 F2 L' U' L U L' F L' F2 U' F2 U' L' U B' U B U2 L' B' U' B U L2 F U F' U' L'
HTM:35
QTM:[42]
STM:{35}
SQTM:{[42]} |
Average number of moves - 38.5
Our incredible Ultimate blindfolded-like team solve
Rules were the as follows : each of us can add up to 4 moves to the current partial solution.
Gilles, your slice moves count as one move - one can propose moves that undo one or more moves - when one has added some moves to the partial solution,
any other team member can continue (otherwise it might be a bit long ...) - one can add premoves (or setup moves) before the scramble. This counts as one move. -
I claim we can use FMC companion to avoid too much scrambling ...
(FR2U'D') [1, Cyril] (F'R2F2L') [2, Gilles] (U'LUL') [3, Lars] (FL'F2U')[4, C] (F2U'L'U) [5, G] (B'UBU2) [6, L] (L'B'U'B )[7, C] (UL2FU) [8, G] (F'U'L') [9, G] (35,42,35,42)
Quotes: [1] Just tried to form as many pairs as possible. After 4 moves we get three pairs, allowing both Gilles' 1x2x3 and Lars' 2x2x3 to be created [2] I finished the 2x2x2
initiated by Cyril and thought about extended it (building other pairs). [4] I decided not to decide which 2x2x3 we should continue with and finished the RWG 2x2x1
[5] Difficult choice. When I considered finishing the first two layers while orienting edges, it didn't look very good for a sub-40, and
I couldn't find any easy last-layer sequence ahead. That's why I decided to finish the first two layers while solving last-layer corners.
I thought that this way, at least sub-40 STM would be easy, even if I wasn't sure of this idea because the other team members may prefer HTM and thinking with outer-slices.
[7] The tricky part ... I spent a lot of time for these for moves ! F2L and corners solved after 24 moves, and 4 edges remaining ...
The good thing is that I do not know even the half of Fridrich OLL algs So I had to think about it with the ones I know and basically tried all of them ...
trying to get lucky ... And I was ! The "T-pattern" OLL L'B'U'BUL leads to another "T-pattern" ... and just had to hope that Gilles would see that for the next moves
[8] I couldn't understand Cyril's moves. No slice, maybe some kind of optimized sequence to solve this last-layer case?
I tried to rebuild the layers in different ways, and after 5 minutes I saw the light. [9] Just the end of the LL as explained in [7]
We haven't had contacts with Lars since a looong time, so Gilles and I decided to finish the solve with the three last moves, after we had been waiting for Lars for more than one month ;-)
Needless to say we are absolutely amazed by such a solve ! |
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The Fellowship of the Cube -
Lars Petrus
Gilles Roux
Cyril Castella
| x z {M'} U2 y' x' U' R' U R U' R' U x' R2 U2 r U2 r U' r' U r' U' {M'} U2 r' U' R x' U' R U R' x U2 R' U2 R U' {M'} U {M} U' {M'} U {M} U' {M'} U2 {M}
HTM:52
QTM:[60]
STM:{44}
SQTM:{[52]} |
Average number of moves - 52
Speed solving solution
Gilles found that one in 14.54 seconds ... Every cube rotations are included to follow the steps one by one. |
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The Fellowship of the Cube -
Lars Petrus
Gilles Roux
Cyril Castella
| x z {M'} U2 y' x' U' R' U R U' R' U x' R2 U2 r U2 r U' r' U r' U' {M'} U2 r' U' R x' U' R U R' x U2 R' U2 R U' {M'} U {M} U' {M'} U {M} U' {M'} U2 {M}
HTM:52
QTM:[60]
STM:{44}
SQTM:{[52]} |
Nearly good solutions
... Here are some cases we had to abandon,
since they leave some edges or some corners ... and we did not find good endings ....
*F' R D R' D' B' D2 L U L' D2 {L' R} B' {L R'} U B' U2 [as U B' U U' B U] F U' B' U F' D' leaves only three twisted
corners in (25,28,23,26) and we had no good alg to deal with it !
*(LFLD2BU'F2)(U2R'F'rU'r'F2)(M2U2L2) solves everything but 4 edges in (18, 26, 17, 24)
*L F L (FU2D2B2U2D2F) D2 B U' F2 U2 R' F' L F' L' F2 R2 L2 D2 L2 leaves two flipped edges in (25,38,22,32) (actually the same as above with an insertion) |
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