Let {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,2 0,21,22,23,24} = the solved state.

By the scramble, we have {15,21,13,4,22,18,12,2,1,14,24,9,19,16,3,6,11,23,2 0,8,5,10,17,7} (23-cycle)

That is,

(As you can see, I made a similar “translation key” as was done for wing edges).

*Sketch*
By

**Lemma 1**, the following representation represents

*all* 23-cycles.

{23,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1 9,20,21,22,24}

Unfortunately, the 23-cycle is one of the very worst cases for big cube parts. Only one piece is solved (4 in our actual scramble) (or 24 in the representation above) and therefore we cannot have the luxury promised by

**Corollary (1)** to be able to solve it in one commutator. We could solve it in one commutator using an odd number 2-cycle solution if the corners, middle edges, and + center pieces all had at least 2 pieces solved. The + center pieces do (as we will see next), but the middle edges and corners do not.

In the document, I actually solved the

*sketch* above to prove that all even permutation cycle classes for 24 objects could be solved in 2 commutators or less (considering permutations only—not taking orientations into account).

Just copying and pasting the solution directly from page 22 of the document,

Let's break it up into 5 sections by marking 8 numbers/pieces.

{

**23**,1,2,3,

**4**,

**5**,6,7,8,

**9**,

**10**,11,12,13,

**14**,

**15**,16,17,18,

**19**,20,21,22,24}

Iteration I: (4↔23)(5↔9)(10↔14)(15↔19)

** 4 2-cycle**
{4,1,2,3,

**23**,9,6,7,8,

**5**,14,11,12,13,

**10**,19,16,17,18,

**15**,20,21,22,24}

Iteration II: (20↔23)(5↔15)(22↔24)(7↔9)

** 4 2-cycle**
{4,1,2,3,

**20**,7,6,9,8,

**15**,14,11,12,13,

**10**,19,16,17,18,

**5**,23,21,24,22}

Iteration III: (1↔4)(2↔3)(11↔14)(12↔13)(16↔19)(17↔18)(21↔23)(22↔2 4)

** 8 2-cycle**
{1,4,3,2,

**20**,7,6,9,8,

**15**,11,14,13,12,

**10**,16,19,18,17,

**5**,21,23,22,24}

Iteration IV: (2↔4)(12↔14)(17↔19)(22↔23)(5↔20)(10↔15)(6↔7)(8↔ 9)

** 8 2-cycle**
= solved state.

**Note**: If an even permutation cycle class does not have at least 2 pieces solved and the number of required 2-cycles for one commutator is odd, the solution to the 23-cycle above is yet another solving technique which you can use (splitting up the large cycles into increments of 4…).

*Back to our scramble*.

If we translate the solution to the

*sketch* to our actual scramble, we find that one solution to our 23-cycle is the following:

Iteration I: (7↔15)(24↔18)(6↔22)(5↔20)

** 4 2-cycle**
→ {7,21,13,4,6,24,12,2,1,14,18,9,19,16,3,22,11,23,5, 8,20,10,17,15}

Iteration II: (19↔15)(24↔5)(3↔4)(17↔18)

** 4 2-cycle**
→ {7,21,13,3,6,5,12,2,1,14,17,9,15,16,4,22,11,23,24, 8,20,10,18,19}

Iteration III: (1↔7)(9↔12)(16↔22)(14↔10)(21↔20)(2↔8)(13↔15)(3↔ 4)

** 8 2-cycle**
→ {1,20,15,4,6,5,9,8,7,10,17,12,13,22,3,16,11,23,24, 2,21,14,18,19}

Iteration IV: (9↔7)(14↔22)(2↔20)(3↔15)(24↔19)(6↔5)(11↔17)(23↔ 18)

** 8 2-cycle**
→ {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,2 0,21,22,23,24}

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