Random Blindfold Cubing Discussion

Tried this just now. Didn't get a success.
Do a 2/2 (or more) MBLD, but execute it alternatingly.
So you would do your first letter on cube 1, then the first letter on cube 2, then the second letter on cube 1 and so on.
Just a silly challenge :P
 
Tried this just now. Didn't get a success.
Do a 2/2 (or more) MBLD, but execute it alternatingly.
So you would do your first letter on cube 1, then the first letter on cube 2, then the second letter on cube 1 and so on.
Just a silly challenge [emoji14]
I guess that would be helpful for remembering and recalling different cubes for MBLD. I might try it later but I'm at a church conference right now
 
Anyone who speaks 2 or more than 2 languages: In which language do you memorize for any BLD? Or do you memorize in more than 1 language?

My native language is Spanish, and I speak English quite well. I also know some Italian, German, Esperanto [yes, that], and I know even less of some other languages, such as Danish, Turkish, French, Hebrew, etc. And I sometimes memorize in a mixture of lots of languages.

Does anyone do soomething similar?
 
Anyone who speaks 2 or more than 2 languages: In which language do you memorize for any BLD? Or do you memorize in more than 1 language?

My native language is Spanish, and I speak English quite well. I also know some Italian, German, Esperanto [yes, that], and I know even less of some other languages, such as Danish, Turkish, French, Hebrew, etc. And I sometimes memorize in a mixture of lots of languages.

Does anyone do soomething similar?

I do speak more than two and I use English (native language) but sometimes when I have letter pairs which has a good word in either French or Spanish or even Hindi I use the language that is the more convenient even if I have a sentence in English with a french word. It all depends on my list of words corresponding to a particular letter pair and somehow the language I was currently using (to make the memo more coherent).
 
Anyone who speaks 2 or more than 2 languages: In which language do you memorize for any BLD? Or do you memorize in more than 1 language?

My native language is Spanish, and I speak English quite well. I also know some Italian, German, Esperanto [yes, that], and I know even less of some other languages, such as Danish, Turkish, French, Hebrew, etc. And I sometimes memorize in a mixture of lots of languages.

Does anyone do soomething similar?
I use some Chinese in my memo
Not much but occasionally for hard pairs like XU
 
For 3BLD I use mostly German and Japanese, for MBLD I Memo my first 4 cubes in German, the next 4 in Japanese, the next in Spanish and the next in English.
After that I start by German again.
 
I memorize mostly spanish (my native language) but often mix other languages, mostly english. In multi I may memorize a whole cube or just edges or corners in english if the pairs fit better. I also sometimes memorize an english word in spanish like "Perro" for the letter pair DG (Dog).
 
Thanks to everyone who replied :D

I'll scramble a cube and I'll write my memo:

VL DM JR GX IQ C[A]

VeLa, DaMe JaRa. GX (gax, probably audio for this), IQ ca.

Translation: Sail, give me Jara. GaX, IQ approximately (from the abbreviation ca.)

Vela, dame Jara -> Spanish
GX -> Nothing
IQ ca. -> English

Weird stuff
 
What's the probability distribution for numbers of edge cycles in a random solve, excluding edge flips? What about corners?

I'm very interested in this as well, but it is a difficult question and I am still on vacation for Worlds :)

Edges one cycle:
Case 1: 12 cycle
11!/12!=1/12 probability

Case 2: 11 cycle and one solved piece (flipped or not doesn't matter)
12*10!/12!=1/11

Case 3: 10 cycle and two "solved" edges
(12C2)*9!/12!=1/20

Case 4: 9 cycle and three "solved"
12C3 * 8!/12!=1/54

Case 5: 8 cycle and four "solved"
12C4 * 7!/12!=1/192

Case 6: 7 cycle and five "solved"
1/((5!)*7)=1/840

Case 7: 6 cycle and six "solved"
1/(6!*6)=1/4320

Case 8: 5 cycle and 7 solved
1/(7!*5)=1/25200

Case 9: 4 cycle and 8 solved
1/(8!*4)=161280

Case 10: 3 cycle and 9 solved
1/(9!*3)=1088640

Case 11: 2 cycle and 10 solved
1/(10!*2)=1/7257600

Case 12: all "solved", but maybe many disoriented
1/12!=1/479001600

Probability of an edge one cycle is:
1/12+1/11+1/20+1/54+1/192+1/840+1/4320+1/25200+1/161280+1/1088640+1/7257600+1/479001600=103447/414720

This is approximately a 24.94% chance of an edge one cycle.

I leave you to do the corner one cycles the same way.

That gets the discussion started. I'll think on this some more after I get home. Interesting question.
 
Back
Top