# A Few Questions About basic 3BLD



## TomH (Apr 9, 2013)

So i have just in the last few days began to pick up blind solving. im using OP for corners and edges and i pretty much get the idea. My memory is horrible, as im also a mixed martial artist and get punched in the head every day, so remembering things isnt exactly my specialization xD anyway, my questions include these:

i memo corners and then edges, solving edges first. Where can i find an in depth tutorial on parity for solving this way? like how to tell if you have it pre solve etc? it seems every time that i think there is parity, there is not, and when i think there is there isnt. my best solve so far is two edges swapped and two corners swapped, and i think i just dont understand the whole parity idea.

secondly, what is an easier way to distinguish letter pairs in words when memorizing for people who use it? i use audio for edges but many times when i use letter pairs for corners, il mix up my words. Say im using GirlFriend as my word for g and f, il accidently think its GirlfRiend, as in g and r. and so on and so on. is it just my horrid memory, or is there a way to make this easier?

sorry it was a lot of typing for two questions, i like to get my questions as clear as possible!

Thanks, Tom.


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## Noahaha (Apr 9, 2013)

Ok! Fortunately for you, all your problems are really easy to fix:

1. If you memorize in letter pairs, you have parity when you have an unpaired letter for corners and edges a.k.a. an odd number of targets for both. You can't have odd for one and even for the other. 

2. Parity happens when you have an odd number of corner targets and an odd number of edge targets. It's special because after doing your odd number of edge targets, you haven't just solved the edges, you've disrupted other things as well, unlike solving in even number where everything is back to normal.

3. If you're using M2 for edges, solve parity simply by doing D' L2 D M2 D' L2 D after edges and before corners. If you're using Old Pochmann, just do an R-perm in the same place: R U R' F' R U2 R' U2 R' F R U R U2 R' U'

4. There are 3 good ways to fix your letter pair problem:

You can standardize your letter pairs, as in make every one be first letter/last letter or first letter of first word/first letter of second word in special cases.

You can make a list. Very time consuming, but very effective.

You can just try to use the same thing for the same letter pair every time, and slowly eliminate your overlaps. As soon as you mess up girlfriend for example, make a decision about what letters it denotes and stick with it. This is what I did.


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## sneaklyfox (Apr 9, 2013)

If you are mistaking "GirlFriend" as G and R then perhaps it's because you're better with audio and there's a lot of R sound in "girlfriend". Do what works for you. You can try audio for corners too and see if it's better. I'm not a BLD expert though so I'll let someone else answer the parity question.

Edit: Ok, Noah already answered the parity question.


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## JF1zl3 (Apr 9, 2013)

I'm no pro at BLD yet (just did my first successful solve last night) but the parity for solving edges first in M2 method is this:
(D' L2 D) M2 (D' L2 D)
It can be performed anytime after solving your edges including DURING corner the corner solve if you would want to for some reason. I recommend either during the transition from Edges to Corners, or at the very end of the solve, after corners and edges.
A way to recognize parity is this:
If you have and odd number of targets, then you have parity. If you have an even number of targets, then you do NOT have parity.
An easy way to tell if you have an odd or even number of targets is if at the end of your memo, you have an extra letter that isn't a part of a letter pair. And, one of the rules of the cube is that if you have Odd Corners, then you also have Odd Edges, no matter what.
So here is an example for Detecting parity just from your Edge memo:

AB XF TG LE SO C
That memo would have parity, so after executing those edge targets you would do the Parity fix I posted above. And the corner memo would have an odd number of targets as well.

XJ HT YU SD LO
This would have no parity since all letters are paired, so you would just continue to do the corner solve. And the corner memo would have an even number of targets as well.

As for your second question, I would just recommend using smaller words so they can't mislead you as easily. But I also don't really understand how you derive the R from Girl Friend. The F is the first letter of the second word, not the R, so I am a bit confused.

And I learned most of this information from Noahaha in his BLD tutorials on cubing world just so you know. That tutorial will definitely help you if you haven't seen it yet.

Edit: Got beat to the punch lol.


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## TomH (Apr 9, 2013)

Thanks for all the feedback! most likely just need more practice for the remembering which letter is which thing, i mean ive only ben even attempting to blind solve for about 2 or 3 days hahah. and thanks for the parity fix noah! i thought it was an r perm i just couldnt remember and i was always kind of scared to try it mid solve in fear of totally messing it up c:


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## Riley (Apr 9, 2013)

Sighted solves should help a lot. 

For GF, I use GirlFriend, and for GR, I use GiRl. 

But yea, after some practice, you'll slowly have a "set" list embedded into your memory and that problem soon won't be an issue.


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## chardison1980 (Apr 9, 2013)

I used to have the same problem, I don't remember on here who told me but try this, when you go to start solving take one of your feet start flat on the ground, when you do your first alg lift your foot up, (first target piece) when you go your alg for your second target piece put your foot down keep doing that for each piece if you get to the end of your pieces and your foot is up you've got parity if its down your good to go, 

Try it sighted a few times and you'll see what I'm talking about.


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## applemobile (Apr 9, 2013)

chardison1980 said:


> I used to have the same problem, I don't remember on here who told me but try this, when you go to start solving take one of your feet start flat on the ground, when you do your first alg lift your foot up, (first target piece) when you go your alg for your second target piece put your foot down keep doing that for each piece if you get to the end of your pieces and your foot is up you've got parity if its down your good to go,
> 
> Try it sighted a few times and you'll see what I'm talking about.




It's a lot easier to just memo pairs, or memo even numbers of things in a roman rooms method. You will know of you have pairity long before you even start solving.


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## A Leman (Apr 9, 2013)

I distingish my pairs by the letters of the first sylable so GF could be golf or goofy and GR could be Garados, Gerard, Garra, a gear etc. An exception to this method is that I think of some words as acronyms and treat them like Acronyms ie: the EU(European Union) and DQ (Dairy Queen). I would have thought of girlfriend as GL which could also be Gollum or a goal post. I would honestly also of think someone like a girlfriend by first name instead of "girlfriend".
I managed to make this system work on my list for for every letter pair (I had to stretch it a bit on some pairs, but I remember them).


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## applemobile (Apr 9, 2013)

GF should only ever be remembered as a gorilla cocking up one leg, tilting its head and passing wind. GF = Gorilla fart.


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## Username (Apr 9, 2013)

applemobile said:


> GF should only ever be remembered as a gorilla cocking up one leg, tilting its head and passing wind. GF = Gorilla fart.



Thatnk you for a new image for GF


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## TomH (Apr 9, 2013)

i also kind of have problems with really weird pairs, like il get X B or something and just sit there like ".....what."


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## Noahaha (Apr 9, 2013)

TomH said:


> i also kind of have problems with really weird pairs, like il get X B or something and just sit there like ".....what."



I talk about some ways to deal with that in part 4 of my BLD tutorial.


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## A Leman (Apr 9, 2013)

TomH said:


> i also kind of have problems with really weird pairs, like il get X B or something and just sit there like ".....what."



Xbox?


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## Coolster01 (Apr 9, 2013)

TomH said:


> i also kind of have problems with really weird pairs, like il get X B or something and just sit there like ".....what."



XB27 is the username of a guy on speedsolving (a.k.a. The inventor of the x cube 4).


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## Ollie (Apr 9, 2013)

List of letter pairs.


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## applemobile (Apr 9, 2013)

TomH said:


> i also kind of have problems with really weird pairs, like il get X B or something and just sit there like ".....what."




Think outside the box.........Cross Bow.


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## kunparekh18 (Apr 10, 2013)

applemobile said:


> Think outside the box.........Cross Bow.



Excellent!! 

Sent from my A75 using Tapatalk 2


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## TomH (Apr 11, 2013)

GOT MY FIRST SUCESS  hahah i kind of switched my memo and started using journey for edges. time was something like 13 minuets i believe? im pumped! ive just found that when i get a weird pair or something il sit there and think about the letters and attribute them to a random object and il somehow remember. dont know why but it works. ima get a sub 10 tomorrow, it will happen!!!!!


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## JF1zl3 (Apr 14, 2013)

TomH said:


> GOT MY FIRST SUCESS  hahah i kind of switched my memo and started using journey for edges. time was something like 13 minuets i believe? im pumped! ive just found that when i get a weird pair or something il sit there and think about the letters and attribute them to a random object and il somehow remember. dont know why but it works. ima get a sub 10 tomorrow, it will happen!!!!!



Awesome job!!!!
Do you attribute both letters if the pair to one object, or does each letter get its own object?


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## TomH (Apr 14, 2013)

both letters in the pair to a single object


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## sk8erman41 (Jul 3, 2013)

I'm having the same problem with distinguishing parity. My confusion comes from cycles. Its kinda hard to explain what I mean in words so forgive me if my writing is confusing but I'll try to spell it out as clearly as possible. When I get a piece that belongs in my buffer (I use classic pochmann and speffz) and I shoot to a different edge or corner, do I skip the letter for the buffer or put it in my memo as another letter? Also, I sort of understand...


Zane_C said:


> Formula to reassure yourself that all pieces have been cycled: *t = u + c*
> If *t = u + c*, you have cycled all the pieces.
> 
> Where:
> ...



But, I guess I am confused because wouldn't the number of cycle breaks change depending on what edge you choose to shoot the buffer piece to? In other words, couldn't you create more cycle breaks for the same solve depending on what cubie you choose to shoot the buffer to? You know you are back to the end of the cycle when you shoot a correct piece to the arbitrary cubie that you shot the buffer to correct? I am having trouble thinking that I am done with corners/edges because I finish a cycle and my buffer is in the correct spot but other pieces are still messed up. I hope this makes some sense.


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## etshy (Jul 3, 2013)

sk8erman41 said:


> I'm having the same problem with distinguishing parity. My confusion comes from cycles. Its kinda hard to explain what I mean in words so forgive me if my writing is confusing but I'll try to spell it out as clearly as possible. When I get a piece that belongs in my buffer (I use classic pochmann and speffz) and I shoot to a different edge or corner, do I skip the letter for the buffer or put it in my memo as another letter? Also, I sort of understand...
> 
> 
> But, I guess I am confused because wouldn't the number of cycle breaks change depending on what edge you choose to shoot the buffer piece to? In other words, couldn't you create more cycle breaks for the same solve depending on what cubie you choose to shoot the buffer to? You know you are back to the end of the cycle when you shoot a correct piece to the arbitrary cubie that you shot the buffer to correct? I am having trouble thinking that I am done with corners/edges because I finish a cycle and my buffer is in the correct spot but other pieces are still messed up. I hope this makes some sense.



For the 1st question , yes you skip it from your memo , you never shoot to your buffer , because it's physically impossible 

For the second one , You will always have at least u targets ( by zane's equation ) and each time you break into a new cycle , you just add 1 to this u , so basically if you broke into a new cycle and then you returned to the same piece again , but the number of targets that you already shoot to is less than u+1 , then there is more pieces to solve , so you will have to break into a new cycle and do the same thing again 

I hope this helps  good luck


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## sk8erman41 (Jul 3, 2013)

I think I am getting it, same as everything... practice practice practice. Thanks


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## sk8erman41 (Jul 8, 2013)

Alright, still confused about T = U + C...
Using the following scramble using classic Pochmann and speffzs I get the following for edges...

U B R2 D' U2 L' U2 R2 B R' U2 L' R2 D' U2 R' F' L' R' D' U B' U' B' L' B2 D U' F' U

O C K D R S N J F A (G/X is flipped edge)

wouldn't T = 11 (twelve edges - the one that is just misoriented)
U = 10 (twelve edges - 1 buffer - 1 flipped edge)
C = 0 (no cycles needed)

therefore, T=11=10 (doesnt work)? Where am I going wrong?


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## Renslay (Jul 8, 2013)

sk8erman41 said:


> Alright, still confused about T = U + C...
> Using the following scramble using classic Pochmann and speffzs I get the following for edges...
> 
> U B R2 D' U2 L' U2 R2 B R' U2 L' R2 D' U2 R' F' L' R' D' U B' U' B' L' B2 D U' F' U
> ...



You have 10 targets (10 edges) and 1 misoriented edge, which means 11 edges do deal with. And you don't have to deal with the buffer. So I don't see the problem. 

T is 10. You have to start from 11 edges, not 12 (don't count the buffer). -> _eleven_ edges minus the one that is just misoriented

T = U + C
10 = 10 + 0


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## etshy (Jul 8, 2013)

you shouldn't count the buffer, because once all other edges are permuted and oriented correctly , the buffer will always be solved correctly , so basically T=11(all edges except buffer)-1(flipped edge)=10


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## tseitsei (Jul 8, 2013)

sk8erman41 said:


> Alright, still confused about T = U + C...
> Using the following scramble using classic Pochmann and speffzs I get the following for edges...
> 
> U B R2 D' U2 L' U2 R2 B R' U2 L' R2 D' U2 R' F' L' R' D' U B' U' B' L' B2 D U' F' U
> ...



Shouldn't the formula be U = T+C where U = unsolved pieces(not counting flipped) T = number of targets and C = number of cycles(not cycle breaks)

So that would make your scramble for edges: U=11(all edges are unsolved except the flipped one so 12-1=11)
T=10 (12-1buffer-1flipped = 10) and they are all in one cycle so C = 1 => U = T+C = 11 = 10+1


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## sk8erman41 (Jul 8, 2013)

ok then, same scramble..
Corners: X cycle P D U B T M V
T = 7 (8 corners - buffer)
U = 7 (8 corners - buffer no twisted corners)
C = 1
T = 7 = 7 +1 doesn't work? I don't get why I am so confused by this lol


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## etshy (Jul 8, 2013)

T = U+C , you can't get T until you know U and C , you can't know T beforehand 

so here U=7 , C=1 ( one cycle break ) ===> you will shoot to 8 targets = T

Edit : to make it easier , T is the number of Y perms that you will execute ( not including the Y perms to flip the flipped corners)


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## Renslay (Jul 8, 2013)

Am I the only one who think this is a bit over-complicated?

Normally you have to do 11 algs for the edges (11 targets / 11 things to remember / etc). So let's say N = 11. If you have solved pieces (flipped or regular), modify N by substracting the number of these pieces (e.g., 2 solved edges -> N = 9 targets should be memorized). You start memorizing the edges. If you memorized less number of edges then N, then you have to break into a new cycle and add one to N (so break into a new cycle and let N = 10 in our example). And so on. If you reach the number N, stop.

(Basically, starting from N = 11, every flipped/solved edge decrease N by 1, and every cycle-break increase N by 1.)

Example:
N = 11.
One solved edge -> N = 10.

Memorizing edges: item1, item2, item3, item4, item5. The cycle is finished, but 5 < 10. So break into a new cycle with item6, and let N = 11. Continue memorizing: item7, item8, item9. Again, cycle is finished, but 9 < 11. Break into a new cycle with item 10 and let N = 12. Continue: item 11, item 12. Cycle is finished again and 12 = 12, so everything is memorized.

Note that using the equality from above (T = U + C):
U = 11-1 = 10 (eleven edges minus one solved)
C = 2 (2 cycle breaks)
T = U + C = 12, so we have to memorize 12 items. And we memorized 12 items indeed!

Corners are the same, starting from N = 7.


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## Noahaha (Jul 8, 2013)

That's how I think about it.


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## Cubenovice (Jul 8, 2013)

Renslay said:


> Am I the only one who think this is a bit over-complicated?
> 
> ...
> very complicated stuff
> ...



I never understood this U C T stuff, not did I ever want to understand it...

What is there more to say then: 
Uneven number of corner targets in your memo--> parity

(same for edges but you only need to "count" one or the other)


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## sk8erman41 (Jul 9, 2013)

Renslay said:


> Am I the only one who think this is a bit over-complicated?
> 
> Normally you have to do 11 algs for the edges (11 targets / 11 things to remember / etc). So let's say N = 11. If you have solved pieces (flipped or regular), modify N by substracting the number of these pieces (e.g., 2 solved edges -> N = 9 targets should be memorized). You start memorizing the edges. If you memorized less number of edges then N, then you have to break into a new cycle and add one to N (so break into a new cycle and let N = 10 in our example). And so on. If you reach the number N, stop.
> 
> ...



Thank you, that was very helpful and I think that I finally get it. Much appreciated! I know that the time was pathetic, but hey, I still got it... My second ever success just now. 10:30 memo 5:40 execution = 16:10


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## TheOneOnTheLeft (Jul 9, 2013)

Cubenovice said:


> What is there more to say then:
> Uneven number of corner targets in your memo--> parity



It's a way to check you haven't missed any unsolved pieces. Especially as you're coming to the end of edges, maybe you have 12 targets memorised, but you've broken into 2 new cycles. You're used to having about 10-12 targets to memo, so it might feel like you've memoed enough targets for a normal solve, but if you do targets (12) - cycle breaks (2) = 10 < 11 (using the system Renslay described), you can know you need to look for one solved or flipped edge.


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## Renslay (Jul 9, 2013)

Cubenovice said:


> > Am I the only one who think this is a bit over-complicated?
> >
> > ...
> > very complicated stuff
> ...



So you think "every flipped/solved edge decreases N by 1, and every cycle-break increases N by 1" is complicated? Because all the "very complicated stuff" is just an explanation and example.


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