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Multiples of 3 for Speedsolving steps:

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LanceTheBlueKnight
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Hey guys, I was up all night last night wondering about this thing, I wonder if anyone can shed some light on this. It seems that for the majority of all steps there are a total number of cases which is a multiple of 3. For example:

Basic CFOP:
21 PLL cases, 21 is a multiple of 3.
57 OLL cases, 57 is a multiple of 3.
42 F2L cases, 42 is a multiple of 3.

More advanced methods:
306 ZBF2L cases, 306 is a multiple of 3.
1212 LL cases, 1212 is a multiple of 3.

For MGLS:
21 ELS cases, 21 is a multiple of 3.
104 CLS cases, which admittedly is not a multiple of 3.

For COLL:
40 cases, again not a multiple of 3.

But the curious thing, I think, is the OLL, PLL, and Full LL. Looking at these three steps independantly I thought it was very strange that they should all be multiples of 3. Why do you suppose that the last layer behaves this way?

Of course, you also have to take into account the 'solved' case for each of these, which makes them not multiples of 3... AAAARGH!!!

Am I on to something, or am I way off? What do you think?
 
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abr71310
#2
Don't they all involve corners, which technically has 3 possible "ways" it can be rotated??

Maybe I'm just too tired to thing, but I thought it had something to do with combinations / permutations of specific things (especially the math formulae in which we got that many combinations for, anyhow).
 

AvGalen

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arnaudvg
#4
there are a total number of cases
Do you consider the solved position as a case? Because that would change all of your assumptions, wouldn't it?

You should write down where the number 21 for PLL's (example) originates from. As you will soon find out it comes from 72 which is divisible by 3 AND by 4 (it includes the solved state) for a good reason

57 OLL originates from 216 (divisible by 3 and 4 again)

I hope this will stimulate your research
 
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fanwuq
#5
COLL is 42, you forgot the Y perm and J perm cases and solved case?

Wait, let me check now.

COLL-
U 6, T 6, L 6, pi 6, H 4, Sune 6, antisune 6, J, Y, Solved.

That is 42 excluding the solved case.

I guess it is an interesting question because some OLL and PLL cancel out (T perm and U2 T perm U2 are the same case by you definition, but FURU'R'F' and FRUR'U'F' and F'U'L'ULF are different cases.)
 
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cuBerBruce
#6
1212 LL cases includes the solved case and counts mirrors and inverses as the same case, while 21 PLLs does not count the solved case and counts inverses and mirrors (if distinct) as separate cases. If you count PLLs the same way as you counted LL cases, you get 14 (not divisible by 3).

When reducing cases by using symmetry, one tends to get "oddball" numbers whose prime factors don't seem to have any apparent relationship to the total number of arrangements. For example, OLL has 216 arrangements total (including solved case), but gets reduced by rotational symmetry to 58 cases (when you include the solved case) which is 2*29.
 
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Sa967St

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Sa967St
#9
104 CLS cases, which admittedly is not a multiple of 3.
"-" cases (DRF corner at FUR): 27
"+" cases (DRF corner at RFR): 27
"O" cases (DRF corner at URF): 27
"I" cases (DRF corner at FRD): 8
"Im" cases (DRF corner at RDF): 8
"C" cases (DRF corner at RDF): 7

The number of cases where the DRF corner is on the U layer are all multiples of 3.
If you include the solved C case, the total of all CLS cases would be a multiple of 3 (105)
 
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