一式真能解万方?

tm__xk

我之前请你弄出高阶那些东东,是为了验证单用若干层转90度的方法是否足以消除扰动..
比方说是否可以只扰动B1和E1..你的探讨证明了这是不可能的..
相信可推到所有高阶,甚至算上内部的块..


然后,现在我突然发现其实没必要弄这些....囧
因为可到达的状态本来就是若干这样的操作叠加的..而只考虑扰动的话,不同方向的同一层效果是一样的....-_-||||

tm__xk

已经知道,任给一打乱状态,总能通过选择若干层转90度来消除所有扰动.
此楼将给出一个选择这些层的方法.

沿用<[原创]一式解万方>中特征值的定义.
不考虑中层转,不考虑(i,m,m).

如果我没看错,(p,q,r)构成至多2个簇,且构成1个簇当且仅当{p,q,r,m}中有重复元素(其中m=(n+1)/2)(包括内部块).
故当第l层转动90度时,(i,j,k)不改变状态当且仅当以下两个条件至少有一个满足:
I.l<>i,j,k
II.{i,j,k}={s,l,l}且s<>l

故l层转90度不改变(i,j,k)状态当且仅当以下三个条件至少有一个成立:
I.i,j,k,l两两互异
II.{i,j,k}={s,s,t}且s<>t,l<>t
III.{i,j,k}={s,s,s}且l<>s

对一个打乱状态,记S为小于m的若干个正整数的集合,使得:对S所有的元素i,将某个第i层转90度,即可消除所有扰动.
对任意的1<=l<m,l属于S当且仅当(1,1,l)有扰动.


说到这我才意识到这个结论是多么的显然........= =||||
不管了,照发....毕竟这也算是理论上的证明....(自我安慰中....囧)

tm__xk

DEVIL'S ALGORITHM FOR RUBIK'S 3x3 CUBE
***************************************************************

First, define the following algorithms:

(COC1) = F D2 F' R' D2 R U R' D2 R F D2 F' U'
(COC2) = F (COC1) F'
(COC3) = L F (COC1) F' L'
(COC4) = F' (COC1) F
(COC5) = F2 (COC1) F2
(COC6) = L' F2 (COC1) F2 L
(COC7) = D' F' (COC1) F D

(EOC1) = L R' F L R' D L R' B2 L' R D L' R F L' R U2
(EOC2) = L F (EOC1) F' L'
(EOC3) = R' F' (EOC1) F R
(EOC4) = F (EOC1) F'
(EOC5) = F' (EOC1) F
(EOC6) = U L U' (EOC1) U L' U'
(EOC7) = U' R' U (EOC1) U' R U
(EOC8) = F2 (EOC1) F2
(EOC9) = D (EOC8) D'
(EOC10) = D' (EOC8) D
(EOC11) = D2 (EOC8) D2

(EPC2) = U F2 U F2 D' L2 B2 U' B2 D L2 F2 D2 B2 D2 F2
(EPC3) = U' D L2 U D' B2
(EPC4) = U F2 U F2 D' L2 B2 D' R2 U R2 F2 D2 B2 U2 B2
(EPC5) = B L' B U2 D2 F' R F' U2 D2
(EPC6) = L D' F2 U' B2 U' R U2 R' F2 R2 F2 D' R2 D F2 D
(EPC7) = B2 U R B2 D2 B2 R B2 U' D' L2 D' B2
(EPC8) = B2 L F' U' B2 U F L' R2 B2 U' D2 L2 D' B2 U
(EPC9) = B2 R B2 D2 B2 R B2 U' L2 U D' F2 U' F2 B2
(EPC10) = D2 R' F U B2 U' F' R U2 B2 D' F2 R2 U F2 U F2
(EPC11) = B2 R2 U' L' D2 B2 U2 R' U2 R2 F2 L2 R2 U R2
(EPC12) = D2 L' B2 U2 F2 R U L2 U' F2 R2 U' R2 U2 F2 D L2 D'

(CPC2) = U B2 U' L2 D L2 B2 D B2 D' B2
(CPC3) = B R' B L2 B' R B L2 B2
(CPC4) = U F2 B2 U' F2 B2 U' F2 L2 R2 B2 R2
(CPC5) = R2 U' F2 D R2 U F2 D' R2 D F2 R2 D' F2
(CPC6) = U' F2 D F2 U L2 F2 D L2 D F2 L2 D2 F2
(CPC7) = U2 D' L2 U' F2 L2 U' B2 L2 R2 U' L2 B2
(CPC8) = F2 D' R2 U2 L2 B2 U F2 L2 U2 F2 L2 U F2 R2

-------------------------------------------------

Then "build" the Devil's algorithm by using the above definitions like this. Note how each new algorithm uses the previous one in some way.

(COT1) = (COC1)3
(COT2) = [(COC2)(COT1)]3
(COT3) = [(COC3)(COT2)]3
(COT4) = [(COC4)(COT3)]3
(COT5) = [(COC5)(COT4)]3
(COT6) = [(COC6)(COT5)]3
(CO) = (COT7) = [(COC7)(COT6)]3 (48,600 moves)

(EOT1) = [(EOC1)(CO)]2
(EOT2) = [(EOC2)(EOT1)]2
(EOT3) = [(EOC3)(EOT2)]2
(EOT4) = [(EOC4)(EOT3)]2
(EOT5) = [(EOC5)(EOT4)]2
(EOT6) = [(EOC6)(EOT5)]2
(EOT7) = [(EOC7)(EOT6)]2
(EOT8) = [(EOC8)(EOT7)]2
(EOT9) = [(EOC9)(EOT8)]2
(EOT1O) = [(EOC10)(EOT9)]2
(O) = (EO) = (EOT11) = [(EOC11)(EOT1O)]2 (199,228,148 moves)

(EPT2) = [(EPC2)(O)]2
(EPT3) = [(EPC3)(EPT2)]3
(EPT4) = [(EPC4)(EPT3)]4
(EPT5) = [(EPC5)(EPT4)]5
(EPT6) = [(EPC6)(EPT5)]6
(EPT7) = [(EPC7)(EPT6)]7
(EPT8) = [(EPC8)(EPT7)]8
(EPT9) = [(EPC9)(EPT8)]9
(EPT10) = [(EPC10)(EPT9)]10
(EPT11) = [(EPC11)(EPT10)]11
(EP) = (EPT12) = [(EPC12)(EPT11)]12 (95,430,612,313,219,476 moves)

(CPT2) = [(CPC2)(EP)]2
(CPT3) = [(CPC3)(CPT2)]3
(CPT4) = [(CPC4)(CPT3)]4
(CPT5) = [(CPC5)(CPT4)]5
(CPT6) = [(CPC6)(CPT5)]6
(CPT7) = [(CPC7)(CPT6)]7
(D) = (CP) = (CPT8) = [(CPC8)(CPT7)]8 (3,847,762,288,469,010,006,992 moves)

(D) is Devil's Algorithm. If you apply it to the cube, it will be solved at some point before you have done the algorithm once. As you can see, it is terribly long, nearly a thousand times more moves than there are possible positions.




呃........再给个链接:http://folk.ntnu.no/eivindfo/docs/devalg.txt

tm__xk

我只能说....确实可以称之为"一式"........囧rz

tm__xk

LZ可没说非得按邱志红的想法来定义"一式",只是问大家的想法....
某种意义上讲,这也是"一个式子"....算是"一式"也未尝不可吧........囧

tm__xk

我觉得你可能有些误解了..

31L一大堆都只是为了叙述..可以把前面一大堆都代进最后一个式子,然后前面的就都没用了..只需要最后一个式子..
当然直接把最后一个式子直接写出来也是可以的,就是太长长长长了而已....

我换个说法..假设总共可能有N种状态(当然,N=8!*12!*3^8*2^12/12..):S1,S2,...,SN
对每个i,从Si当然可以变到S(i+1).
把从S1到S2,从S2到S3,从S3到S4,...,从S(N-1)到SN的转换公式依次写下组成"一个"很长很长很长很长的式子....
这个式子就足以解了:从头做到尾,必有某个时刻是还原了的....(如果你够运气,能活到那个时候....)

tm__xk

不可以.
前者每一步都根据目前状态从"若干"公式中作选择;后者只有一个公式,不必理会初态,从头开始照做就可以了,只是需要选择停止的时刻.

tm__xk

其实我觉得两者没多大可比性....
我觉得你只是为了拉出"共同性",硬是给两者都套上了"取舍"这个词....

"对一批“子公式”的取舍"
如果我定义"可以解决"="有限步之内出现还原态",而不是"有限步之内终态为还原态"..那这点可以忽略..-_-||||

tm__xk

能否再详细一些?我对"提取部分片段"不是太理解..

tm__xk

LR'U'RUL'U'R'UR=(R'L)U'RUL'U'R'URR'L(L'R).

[ 本帖最后由 tm__xk 于 2010-2-22 19:00 编辑 ]