i would look for edges that are incorrectly flipped but in the right spot
So you mean: Let X= set of of already-permuted edges. This implies you ust remember 12-|X|. This seems like a good approach, but it's a bit counter-intuitive if you want to remember the already-permuted edges last rather than first. That way, which exact pieces you need to orient doesn't have to stick in your head when you are memorizing edges/corners.

what memo do you use? if you use letters just think about all the letters not used.
I use 1-syll letter pairs. ^That is kind of a slow approach for me since I then have to spend the time decoding my phrases when continuing to code (memorize). When I memorize, I just try to list off all the letter pairs, throw them into a 1-syllable word one at a time, and completely forget about what pieces I actually memorized until I go into the execution. (it is an audio loop after all)

after awhile you will be able to keep track of pieces that are already solved and pieces that arent
I figured this might be it. That I'll get some "intuitive" tracking of what pieces I haven't already coded into my 1-syllable phrases yet. I was just curious if there were more explicit methods. Thanks for the advice.
 
I figured this might be it. That I'll get some "intuitive" tracking of what pieces I haven't already coded into my 1-syllable phrases yet. I was just curious if there were more explicit methods. Thanks for the advice.

that was one of my problems when i started to do BLD i always got corners right and edges were always off because i thought the cycle was done but i had 2 edges un-orientated or i had 2 edges or 3 edges wrong but my buffer solved youll pick up on it after a couple of solves
keep practicing and dont let DNFs end it
ive had alot of DNFs before my first success and my first was about 20 minutes long because memo. I spent alot of time making sure i got it right. After time memo will speed up
 
Sorry, here's another question: Considering audio loops, is there a more efficient approach of chunking than 2 stickers per syllable?
I've been trying to read as much as possible on this (I went through almost every page on this thread and Memo Method) and the two most efficient methods I've read so far are Chris' preset 440 1-syllables (my current method), which is 2 stickers per 1-syllable, and consonants+vowels. Consonants as pieces and vowels as orientation still only leads to 2 stickers per 1-syllable.

I've heard a little bit of speculation that using consonants+vowels could in fact lead to more than 2 stickers per 1-syllable, but I couldn't find any efficient approach in setting this up. I could certainly see cases where this is possible, but I can't find a way to extrapolate this to work for all such combinations. Any insight from advanced BLDers or theorists?
 
I've heard a little bit of speculation that using consonants+vowels could in fact lead to more than 2 stickers per 1-syllable, but I couldn't find any efficient approach in setting this up. I could certainly see cases where this is possible, but I can't find a way to extrapolate this to work for all such combinations. Any insight from advanced BLDers or theorists?

Audio is already so fast and simple that I think a more complex method might not be worth it just to condense it a little more. Furthermore, using one word per pair works very nicely with 3-cycle methods, as each word represents one commutator. After a while you could even recognize the algorithm directly from the word instead of thinking about the pieces.

Having said this, I'm still very interested to hear any proposals.
 
After that involved question that went right over my head there....

When I start a new cycle, do I have to do the first number in that cycle again at the end of the cycle?

Also, white U blue F orange L, corectly oriented blues and greens are F U B D in L and R orbits, and U and D in M orbit?

Also, any links to memory sport sites?

There will probably be a lot of questions over the coming days so stand by.
 
After that involved question that went right over my head there....

When I start a new cycle, do I have to do the first number in that cycle again at the end of the cycle?

Also, white U blue F orange L, corectly oriented blues and greens are F U B D in L and R orbits, and U and D in M orbit?

Also, any links to memory sport sites?

There will probably be a lot of questions over the coming days so stand by.
For the first cycle you don't have to do the first number in the cycle again, because that's simply the buffer piece being solved. For cycles after that you do have do the first number again.
Shouldn't orange be on R if you have white and blue like that? Anyway, that is correct. So you're learning 3OP? Did you look into Old Pochmann? It combines orientation and permutation, and I wouldn't say it's any harder than 3OP to learn.
I don't know of any memory sport sites that provide much info I'm afraid.
 
For the first cycle you don't have to do the first number in the cycle again, because that's simply the buffer piece being solved. For cycles after that you do have do the first number again.
Shouldn't orange be on R if you have white and blue like that? Anyway, that is correct. So you're learning 3OP? Did you look into Old Pochmann? It combines orientation and permutation, and I wouldn't say it's any harder than 3OP to learn.
I don't know of any memory sport sites that provide much info I'm afraid.

Sorry, your right about orange.
I'm using purely t perms and j perms where appropriate (yes- I fully capable of 2H j perms! :p) that's pochmann?

I'm having real problems assigning 24 instead of 8 for corners, but orient and permute is easy (ish). I assign 1-8 then orient after permute. Is this wrong?
 
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Audio is already so fast and simple that I think a more complex method might not be worth it just to condense it a little more. Furthermore, using one word per pair works very nicely with 3-cycle methods, as each word represents one commutator. After a while you could even recognize the algorithm directly from the word instead of thinking about the pieces.

Having said this, I'm still very interested to hear any proposals.
That's certainly true, but if condensing it would not complicate the decoding procedure, then there would be no reason to not condense it. It's like doing CFOP with 4LLL as opposed to 2LLL, since 4LLL is "fast enough". I agree that the pair system works great as it is, and it may be very complicated if a more efficient chunking method puts 3 stickers per syllable. However, there are certainly cases where chunking consonant + vowel + vowel + consonant would lead to 4 stickers per syllable, but the problem si that there's not enough consonants and vowels to assign one of each per sticker. Anyone figure out how to chunk 3-4 stickers per syllable?
 
It's also a lot of possible combinations to condense it more:

3 stickers per syllable would be 7920 words for edges, and 5670 for corners. (Not counting set of words to represent pairs, as number of targets will not always be a multiple of 3.)

4 stickers per syllable would be 126720 words for edges, and 68040 for corners. (Not counting set of words to represent triplets or pairs, as number of targets will not always be a multiple of 4.)

(Assuming letters on every sticker but the fixed buffer piece.)
 
Using such a short audio loop, you don't care so much about the sounds being words, but that each sound is different. For instance, consider an edge memory of "borg sigh cade", where each letter represents a sticker. As long as each consonant and vowel is audibly different, it doesn't matter what diction the actual phrases make. This is much easier than "bow rag sin ghost can dead". Of course, the problem here would be how to condense your stickers to be "borg sigh cade".

More generally, if you could super-condense the method to represent 4 stickers per syllable, you'd only need 5 phrases to memorize the entire cube, which is easily sub-10 memo time. I think I have a method of doing 3 stickers per syllable, but I haven't found a way to condense it even further. And as previously said, 3 stickers per syllable could get messy.
 
If I remember correctly, those who use audio loop with sentences simply combine multiple letters into a recognizable word or phrase whenever this is convenient.

STRG could be "STRonG"
DRML could be "DReMeL"
etc.

You could prepare a very large collection of easy words or phrases for certain 4-letter combinations. The key being that you would not have to prepare every possible 4-letter combination. There would be a certain amount of ad libbing involved in this method. If I understand correctly, people are already using this approach for sentence, audio loop methods.
 
I think I have a method of doing 3 stickers per syllable, but I haven't found a way to condense it even further. And as previously said, 3 stickers per syllable could get messy.

Still worth looking into. Care to share some details?

If I understand correctly, people are already using this approach for sentence, audio loop methods.

Yea, a lot of people make 4-letter words for sentences when it's convenient or obvious, but I don't think it's as common for pure audio loop (completely incoherent strings of random sounds/words).
 
Sorry, your right about orange.
I'm using purely t perms and j perms where appropriate (yes- I fully capable of 2H j perms! :p) that's pochmann?

I'm having real problems assigning 24 instead of 8 for corners, but orient and permute is easy (ish). I assign 1-8 then orient after permute. Is this wrong?
Orient then permute sounds like a very weird plan. With Pochmann you would usually orient at the same time as permute. So with edges, you look at which sticker is in UR, and set up the appropriate sticker to UL before T perm. So if UR belonged in DF, you'd do D' L2 to setup, but if it belonged in FD, you'd do M' to setup to J perm. This way, you do orientation at the same time. The only piece that should be oriented at the end are the stationary flipped ones (ones that are already in the correct place, but flipped/twisted)
 
How are the multi-blind WCA rankings calculated? Is it solve time per cube solved? Is there some sort of penalty for unsolved ones in the rankings? Does it count better to solve more cubes, even if your rate will obviously stay the same?

I'm sorry for teh n00b qu3stion(s) pl0x.
 
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