2004-10-22日的较少步数还原问题

cube_master

2. L+ F2 L- U+ R- F+ R+

cube_master

3. U+ F+ U+ F2 L+ F+ L-

cube_master

4. R+ B2 L- B- L+ B- R-

cube_master

5. R2 U+ F+ B- R2 F- B+ U+ R2 U+

还猪哥哥

在cube_master的基础上加以改进,41步 U2 F B2 D3 R D2 B2 F3 D B3 L2 B3 L F2 L3 U R3 F R U F R3 F R F2 B L2 F3 L3 F L3 B3 B2 U L3 R B2 L R3 U B2 U2

合并最后两步,36步 U2 F B2 D3 R D2 B2 F3 D B3 L2 B3 L F2 L3 U R3 F R U F R3 F R F2 B L3 D3 L3 D2 R F2 R3 D3 L2 B3

大烟头

还猪哥哥，合并最后两步有决窍吗？这可不能靠最少步软件的，不然算你无效喔[em05].如果是记公式，那可要记很多啊！是吗？

cube_master

唉，还是老外的水平高几个级数！

大烟头

看看这期的最佳答案：http://www.necrophagous.co.uk/cubestation/fmcresults/fmcresults221004.php

28步：[1] U' R2 U B U2 L D L R2 B2 [2] U2 R' B' R' B [3] R U2 R' U R2 U' R2 U' R' U R' U' R - HTM = 28 moves QTM = [36 moves] STM = {28 moves}

用老猫的贴助手编了一下贴出来看看：

1, 　2*2*3（10步） U' R2 U B U2 L D L R2 B2

2。　

U2 R' B' R' B

3。

R U2 R' U R2 U' R2 U' R' U R' U' R

cube_master

唉，还是老外的水平高几个级数！

cube_master

22/10/04 FMC #60			Scrambling algorithm: B L U R (F2 D' U2 R')*5 F L U (D2 F' R L)*3 F L U B
Pos.	Name	Country	Solution	Quote
=1 [=1] {=1}	Charles Tsai 603	USA	[1] U' R2 U B U2 L D L R2 B2 [2] U2 R' B' R' B [3] R U2 R' U R2 U' R2 U' R' U R' U' R - HTM = 28 moves QTM = [36 moves] STM = {28 moves}	Saved by a shorter-than-usual step 3.
=1 [=1] {=1}	Mirek Goljan 709	USA/Czech	B' R' D R' D L F2 U' F2 R2 F R F2 U F U2 R U F2 D B' R B' R' U2 B' U R2 - HTM = 28 moves QTM = [36 moves] STM = {28 moves}	After not an unusual start B' R' D R' D L F2 U'. D B2 U' 2X2X3 is done. Continue U B R B' R' U2 B' U R2 and 2 edges yet to be flipped and 3 corners twisted. This can be done in 9 turns like R2 F R F2 U F U2 R U after the first U' turn at the dot sign if we help with F2
3 [=3] {3}	Sebastian Dumitrescu 601	Romania	D' L' F' {U2 D2} F2 B L2 F2 R B F U' L' U' {L R'} F R L F' L' U2 F' U' F2 U' F2 U2 F U' - HTM = 31 moves QTM = [40 moves] STM = {29 moves}	This position was remarcable in that you couldn't build a corner-edge pair in one move, you needed two moves, so I tried to find another start than forming pairs. I exploited the two corners WOB and WOG, making a nice WO 2x2x3 in ten moves D' L' F' U2 D2 F2 B L2.R B , with the F2 move inserted at the dot orienting two edges. Then F U'. L F' L' is step 2. The step 3 moves U2 F' U' F2 U' F2 U2 F U' are a Rune which solves the corners and two edges, the last 5 edges being permuted L' U' L R' F R in the middle of step 2, at the dot as usually. I like this solution, I was very satisfied when I solved those 5 edges, even if moves didn't cancel out. Next week I'll do sub-30!
4 [=3] {4}	Zbigniew Zborowski 582	Poland	[1] U' B' U' L B' R U R' U' D2 F' U' F U' R' U' R [/1] D2 [2] F U F' L' B' U' B U' [= U L [/2] [3] L' U2] L U2 L F' L B L B' L F [/3] D - HTM = 37 moves QTM = [40 moves] STM = {37 moves}	just do not look at my (bad) solution zz
5 [5] {5}	Thomas Bischof 394 provisional	USA	L D' R2 D2 L' U' R U R' U2 L U' L U2 L' U' L U L' R2 U2 R U R' U {R2 L2} U2 F' L2 F U2 L U' L U L F' L' F U' - HTM = 41 moves QTM = [52 moves] STM = {40 moves}	I seem to be having a good streak with FMC so far: both times, I have had a 41-ish move solution on the second try. Here's this one: L D' R2 D2 : Cross L' U' R U R' : F2L pair one U2 L U' L U2 L' U' L U L' : F2L pair two R2 U2 R U R' U R2 : F2L pair three L2 U2 F' L2 F U2 : F2L pair four L U' L U L F' L' F U' : OLL with PLL skip

[此贴子已经被作者于2004-10-30 23:29:47编辑过]