# # of Possible EO (without line) cases



## riffz (Apr 8, 2010)

Hi, I've never learned how to do probability calculations other than very simple factorial related ones, so I have a question:

For the first step of the ZZ method, you must orient all the edge pieces and solve the FD and BD cross pieces. I am not interested in the cross pieces for now, just the orientation of edges.

How many possible cases are there for the orientation of edges? The number of disoriented edges must be a multiple of 2, and there are 12 possible positions.

An answer is appreciated, and an answer with an explanation is greatly appreciated.


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## cmhardw (Apr 8, 2010)

This is a case count where you have to throw out symmetrical cases. If you have stickers you can count this out on your cube by placing stickers on the "mis-oriented" edges.

I believe you can use the Lemma that is not Burnside's for counts like this, but to be honest I never put in the work I probably should have to learn how to apply this.

I don't have the time to do the whole count right now, although I find it interesting. But here is how you would apply it.

Number of cases with 0 flipped edges: 1
Number of cases with 2 flipped edges: 5
- The edges can be contained on the same layer in two ways (opposite and adjacent)
- The edges can be contained on adjacent layers but NOT on the same layer (U and E, or R and M) in 2 ways (UF and FR OR UF and FL)
- The edges can be contained on polar opposite layers (R and L, F and B) in 1 way without duplicating a previous case (UF and DF)

etc..

Notice, though, that the number of cases with 10 flipped edges is EXACTLY the same number as of 2 flipped. This is because for 10 flipped edges you simply count how many cases there are for the *2* non-flipped edges.

Again sorry for the short response. I am getting ready for work, and I saw this thread and found it really interesting.

Chris


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## Cride5 (Apr 8, 2010)

Cases with 0 edges = 12!/(0!*12!) = 1
Cases with 2 edges = 12!/(10!*2!) = 66
Cases with 4 edges = 12!/(8!*4!) = 495
Cases with 6 edges = 12!/(6!*6!) = 924
Cases with 8 edges = 12!/(4!*8!) = 495
Cases with 10 edges = 12!/(2!*10!) = 66
Cases with 12 edges = 12!/(12!*0!) = 1

So total is: *2048* including solved. 

NOTE: This does not take account of symmetries though. As Chris points out, there are significantly fewer after removing symmetrical cases. Just as a quick example, if you consider any case reflected over the S slice as the same, then the number is halved.


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## Sir E Brum (Apr 8, 2010)

riffz said:


> For the first step of the ZZ method, you must orient all the edge pieces and solve the FD and BD *cross* pieces. I am not interested in the *cross *pieces for now, just the orientation of edges.



Just a note of clarification. ZZ does not have a cross, just a line.


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## riffz (Apr 8, 2010)

Sir E Brum said:


> riffz said:
> 
> 
> > For the first step of the ZZ method, you must orient all the edge pieces and solve the FD and BD *cross* pieces. I am not interested in the *cross *pieces for now, just the orientation of edges.
> ...



Lol. Yes, I know. I'm was just describing it in terms of Fridrich because most people here would understand what I mean.

Okay, thanks guys.


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## cuBerBruce (Apr 9, 2010)

Since the OP mentioned ZZ, I assume that the ZZ EO scheme is the edge orientation scheme intended. With that scheme, the applicable group of symmetries is that subgroup of the symmetries of the cube which map the moves {U,U',D,D',L,L',R,R',F2,B2} onto the same set. So the applicable symmetry group is <z,x2,v> where v represents left-right reflection.

Instead of "the Lemma that is not Burnside's," you could also just have a computer do a more brute force counting of the equivalence classes.

Basically if you have a group G, and applicable symmetry group M, you can calculate m'gm for all g in G and all m in M, and thus partition the group G into equivalence classes such that for any g in G, if h = m'gm for any m in M, then h must be place in the same equivalence class as g.

Using GAP to do this, I get the number of cases to be 186 (including the all edges oriented case).


Spoiler





```
gap> Fuf := (7,18);
(7,18)
gap> Fur := (5,26);
(5,26)
gap> Fub := (2,34);
(2,34)
gap> Ful := (4,10);
(4,10)
gap> Fdf := (23,42);
(23,42)
gap> Fdr := (31,45);
(31,45)
gap> Fdb := (39,47);
(39,47)
gap> Fdl := (15,44);
(15,44)
gap> Ffl := (13,20);
(13,20)
gap> Ffr := (21,28);
(21,28)
gap> Fbl := (12,37);
(12,37)
gap> Fbr := (29,36);
(29,36)
gap> Geo := Group(Fuf*Fur,Fuf*Fub,Fuf*Ful,Fuf*Fdf,Fuf*Fdr,Fuf*Fdb,
>       Fuf*Fdl,Fuf*Ffl,Fuf*Ffr,Fuf*Fbl,Fuf*Fbr);
<permutation group with 11 generators>
gap> z := (2,29,47,12)(4,26,45,15)(5,31,44,10)(7,28,42,13)
>      (18,21,23,20)(34,36,39,37);
(2,29,47,12)(4,26,45,15)(5,31,44,10)(7,28,42,13)(18,21,23,20)(34,36,39,37)
gap> z2 := z*z;
(2,47)(4,45)(5,44)(7,42)(10,31)(12,29)(13,28)(15,26)(18,23)(20,21)(34,39)(36,
37)
gap> mir := (4,5)(10,26)(12,29)(13,28)(15,31)(20,21)(36,37)(44,45);
(4,5)(10,26)(12,29)(13,28)(15,31)(20,21)(36,37)(44,45)
gap> x := (2,39,42,18)(4,37,44,20)(5,36,45,21)(7,34,47,23)
>      (10,12,15,13)(26,29,31,28);
(2,39,42,18)(4,37,44,20)(5,36,45,21)(7,34,47,23)(10,12,15,13)(26,29,31,28)gap> x2 := x*x;
(2,42)(4,44)(5,45)(7,47)(10,15)(12,13)(18,39)(20,37)(21,36)(23,34)(26,31)(28,
29)
gap> Gsym := Group(z,x2,mir);
<permutation group with 3 generators>
gap>
gap> MConj := function (G, M)
>   local count, g, h, i, j, m, L, L2;
>   count := 0;
>   L := [];
>   L := Set (G);
>   i := 0;
>   for g in G do
>     i := i + 1;
>     if g in L then
>       count := count + 1;
>       L2 := Set ([ g ]);
>       j := 1;
>       for m in M do;
>         h := (m^-1)*g*m;
>         AddSet (L2, h);
>       od;
>       SubtractSet (L, L2);
>     fi;
>   od;
>   return count;
> end;
function( G, M ) ... end
gap>
gap> MConj (Geo, Gsym);
186
```


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