回到最初，从52步的 Thistlethwaite 开始

aubell

page 2 Getting from G1 into G2. What has been achieved so far (although it doesn't look like it) is the

correct orientation of edge pieces! In the present stage, the same is accomplished for corners, and also

the edge pieces FU, FD, BU, BD are brought into their slice. As is well known, corners do not in general

have a natural orientation, but here, roughly speaking, we shall line them all up the same way. More

precisely, note that each corner piece has either a L-facet or a R-facet: on completion of this stage

each of these facets will lie on either the L or the R face. In fact, the same will be true of the eight

edge pieces with L or R facets, in view of the statement above regarding FU, FD, BU, BD. There are

1082565 cases to consider here, this number being the product of 3^7(total number of corner orientations)

, and 12C4 (total number of arrangements of the set {FU, FD, BU, BD} amongst the twelve edge positions).

Surprisingly, with a certain amount of practice it is possible to get through this stage in at most 17

moves without tables; the same is most certainly not true of the next stage although there are only 29400

cases. However, with a few pages of tables this figure of 17 may be reduced to 13; with a great deal more

computation, it should be possible to reduce it further to 10. To "prove" that 10 moves were sufficient,

one would run through all 10-move sequences on the computer, and check that 1082565 different cases

resulted. This would take no more than a few hours of computer time, in view of certain short cuts

available by considering symmetries. Now to business. The twist of a corner is measured by looking at its

L or R facet and observing how this has been rotated in relation to the adjacent L or R face. Note that

quarter-turns of F and B faces alter the twist of corners, whereas all other moves in the group G1 have

no effect. Now in at most 4 moves in G1 try to obtain a position the edge pieces FU, FD, BU, BD are all

in the UD-slice. If this is not possible, get them all in the U-face in at most 4 moves.

第二页 从G1进入G2 在纠正了棱的方向以后，已经前进了一大步（尽管看起来并非如此）！在这个步骤中，角的方向也

要被翻正，并且FU,FD,BU,BD四个棱也会回到它们应该在的中间层。大家都知道，角并没有所谓自然的朝向，但这里，严

格的讲，我们要让所有的角朝同样的方向。准确的说，注意：每个角块都会有L或R面的贴纸：在完成这一步以后，L或R

面的贴纸只能朝向左或右。事实上，上一步对含LR面贴纸的8个棱也是同样的效果，看FU,FD,BU,BD四个棱。这一步有

1082565种情形。这个数字是3的7次方（角的朝向）乘以12选4（中层四个棱在12个位置中的某处）得出的的。令人吃惊

的是，不用表，仅通过一定数量的尝试，这一步可以通过17步完成。下一个步骤只有29400种情形，自然不要那么多步数

。在使用几页表之后，数字17可以缩减为13；用大量计算以后，也许可以缩减到10.为了“证明”10步是足够的，你可以

在电脑上遍历所有10步长的序列，检查结果是否覆盖这1082565种状态。这可以在几小时之内完成。角的朝向是看L,R面

的贴纸是否在LR面。注意，90度F,B面的选择会翻动棱块，其余G1群操作没有影响。这样，最多用G1中4步操作，使得

FU,FD,BU,BD四个棱回到UD之间的中层。如果这些不可能。用最多4步集中到U面。

（现代的方法是：看角，看UD面的贴纸，朝向上下；UD之间的E层棱先回到E层。“古代”是看LR面的贴纸，M层的棱先回到M层。现在也知道，17步一点也不令人吃惊，因为整个魔方只要20步，如果这个步骤就用17步，其它步骤还怎么混呢？但这种方法有意思，先用公式生成所有的状态，在还原的时候，查找状态，然后把公式逆过来做。）

zxy6350479

内容看完 头晕 很复杂

aubell

page 3 Then refer to the appropriate detailed table. The words listed in these tables need to be

inverted, as mentioned earlier. Getting from G2 into G3. This is the trickiest stage theoretically, and

may be broken down for purposes of clarification into two sub-stages: first get corners into their

natural orbits, and second permute the corners within their orbits so as to obtain one of the 96 corner

permutations in the squares group (G3), while at the same time sorting out the edge pieces into their

correct slices. The table of initial moves on the first page of Stage 3 tables does part of the first of

these substages, and the detailed tables do the rest. After performing the initial moves, the set of

corners out of orbit will be one of three possibilities (modulo symmetries): (i) the empty set; (ii) the

set {1, 8, 2, 7} ; (iii) the set {1, 5} . The reader then has to calculate which coset of form G3αβ the

permutation of corners lies in, where α is one of 1, (14)(68), (24)(58), (12), (14), (24), and β is one

of 1, (18)(27), (15). Since some of these cosets are reflections of others, it was not necessary to

produce tables for all six possible values of α. The task of reflecting positions if necessary is left

to the reader. I apologise for the slightly anomalous numbering of edge positions for this stage; this

was due originally to a typing error in a programs and has stuck ever since! Getting from G3 to home. In

this final stage one uses only 180 turns. The order of G3 is 96x6912 = 663552. The tables for this stage

give words for producing each of the 6912 edge positions with corners fixed. Therefore one must first

restore the corners (in at most 4 moves), and then use the tables to restore the edges. Considerable

practice is needed to use these tables efficiently, but I have found (after considerable practice) that I

can find the desired move in about 2 minutes. One hint is that when faced with three 3-cycles, consider

the configuration of the fixed pieces. Also it pays to get to know the different sorts of 4-cycle.

第三页 现在看如何正确查表。表中列出的词要反转一下，如前面提到的。

从G2进入G3
理论上讲，这一步最富技巧性。可以分解成两个子步骤：先让角进入自然的轨道，然后在轨道内对其进行排列，以期获

得G3中96种角排列中的一种，同时，让棱进入各自所属的中间层。步骤3的第一页的表做第一步的工作，后面的表完成后

面的工作。在执行初始化操作以后，轨道外的角的集合只会出现下面3种情况：（余子对称群）（1）空集；（2）

{1,8,2,7}的集合；（3）{1,5}的集合。读者要计算商集，G3αβ,角排列处于何种状态， α是1,(14)(68),(24)(58),

(12),(14),(24)中的一个， β 是1,(18)(27),(15)中的一个。因为有些余子集同其他是反射的，所以不必要计算α的所

有六种可能。反射的位置留给读者完成。Thistlethwaite为棱的编号怪异而抱歉，因为最初的打印错误导致！

从G3到还原
最后的步骤只使用180度的旋转。G3的秩是96*6912=663552.这一步的表给出的是角固定以后，棱的6912种排列。因此，

必须先还原角（最多4步），然后使用表来还原棱。要用好这些表需要大量的练习，但Thistlethwaite发现，他大约用两

分钟就能找到期望的解法。一点提示是面对3循环，考虑固定的块。还需了解不同类型的4循环。

（翻译到这里我们知道，尽管下面有很多页的纸张，给出了长长的公式列表，但这个表是不完整的，对称变换要还自己

做。程序和数据不是截然分开的，某些程序可以是另一些程序的数据，某些数据也可以是程序。现代计算机存储和计算

的能力都变强了，分成了4个步骤的表，应该可以完整的给出了；也可以不给出表，全都计算？）

[ 本帖最后由 aubell 于 2011-2-7 03:22 编辑 ]

aubell

从G2进入G3是难点。因为这里有些群论的计算。并非简单的罗列每种状态如何用公式。
这里应该也是精华所在。因为，把一个大的集合，划分成若干小的集合，这种方法
才使得20步得到证明。
四步骤解法有了很多的实现，似乎大都没有再对G2进行分类讨论，而是仰仗计算机的能力，
很暴力的进入G3，这样的实现只能说是退步。
分类讨论是一种重要的思想方法。似乎好像那个20步的证明是直接把G0给分类讨论了。

八目阿修罗

非常感谢lz的翻译
收藏ed

aubell

现在，要开始编码了，再实现一遍Thistlethwaite方法，要使用对称，使用群论中的乘法，使用分类讨论。
制作完整的表。
得鱼忘筌，原始的表的数据就可以不用了。

aubell

按照现代约定
G0=<U D L R F B>
G1=<U D L R F2 B2>
G2=<U D L2 R2 F2 B2>
G3=<U2 D2 L2 R2 F2 B2>

下面是第一个步骤，翻棱用表。尽管这一步不需要表，为了完整，还是做了一个表。
文中提到的其它表，也尽量弄齐。只要不是太大的话。

[ 本帖最后由 aubell 于 2011-2-7 21:19 编辑 ]

aubell

在txt格式的文件中，似乎某些公式有误？莫非在识别转换的时候出错了？
原始的扫描图片作为参详，仔细检查一下。
打算恢复这个算法的原貌。

[ 本帖最后由 aubell 于 2011-2-8 13:38 编辑 ]