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[Help Thread] Blindfolded discussion and help

mark49152

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I have been experimenting with solving parity (M2/OP edges first) by memoing corners first and if parity exists, exchanging UB and UL during edge memo, so I shoot them to opposite places in anticipation of an odd number of corner swaps fixing that. This avoids a parity swap step. It also means there is always an even number of edge swaps, whereas if I had not swapped those targets, there would obviously be an odd number. So there is one more or one less edge swap.

My question is this: Will there always be one more edge swap, always one less, or 50/50 depending on cycle breaks? Statistically, if we count a dedicated parity fix as equal to any other swap, is there a swap count advantage to fixing parity by exchanging targets?

(My guess is that it's 50/50, so in half of parity solves I'd be replacing parity fix with an extra swap, and in the other half I'd be saving two steps, the fix and a swap.)
 

YY

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I have been experimenting with solving parity (M2/OP edges first) by memoing corners first and if parity exists, exchanging UB and UL during edge memo, so I shoot them to opposite places in anticipation of an odd number of corner swaps fixing that. This avoids a parity swap step. It also means there is always an even number of edge swaps, whereas if I had not swapped those targets, there would obviously be an odd number. So there is one more or one less edge swap.

My question is this: Will there always be one more edge swap, always one less, or 50/50 depending on cycle breaks? Statistically, if we count a dedicated parity fix as equal to any other swap, is there a swap count advantage to fixing parity by exchanging targets?

(My guess is that it's 50/50, so in half of parity solves I'd be replacing parity fix with an extra swap, and in the other half I'd be saving two steps, the fix and a swap.)
This question may be more complicated than you think. There are not only one more or one less edge swap patterns, but also three more or less edge swaps patterns and some more or less edge flips patterns.
I try to solve it by dividing into some cases, but this discussion is a bit long. If there are any mistakes or you know smarter solutions, please tell us. If you want to see the conclusion, please see the bottom.

In the following, I will write two exchanging slots as [A], , and sticker that must be in these slots as A, B. Another sticker of the pieces are a, b. I will use N to mean that this is not the sticker of the piece A, B. Another sticker of this piece is n. N',N1,N2,... is the same.
(For example, if white is U and green is F, [A], is UL, UB(or UB,UL) and A, B is white, a, b is blue, orange(or orange,blue))

Pattern 1-1: B in [A], A in
In this case, we can reduce 3 letters(ABA or BAB) by exchanging two slots.

Pattern 1-2: b in [A], A in (B in [A], a in case is the same discussion)
In this case, we can reduce 3 letters, but 1 edge flip is added.

Pattern 1-3: b in [A], a in
In this case, we can reduce 3 letters, but 2 edge flips are added.


Pattern 2-1: A in [A], B in
3 letters are added.

Pattern 2-2: a in [A], B in (A in [A], b in )
3 letters are added and 1 edge flip is removed.

Pattern 2-3: a in [A], b in
3 letters are added and 2 edge flips are removed.

Now, the probability of 1-1 is equal to that of 2-1. That of 1-2 and 2-2, 1-3 and 2-3 are also the same.


Pattern 3-1: N in [A], A in (B in [A], N in )
B is in [N']. Before exchanging, the order of letters is N', B, A, N. After exchanging, the order of letters is N', A, N and the correct piece is in .So, we can reduce 1 letter.

Pattern 3-2: N in [A], a in (b in [A], N in )
Almost the same discussion as 3-1(N'Ban→N'AN), except for plus 1 edge flip in .


Pattern 4-1: A in [A], N in (N in [A], B in )
B is in [N']. Before exchanging, the order is N', B, N. After exchanging, the order is N', A, B, N. So, 1 letter is added.

Pattern 4-2: a in [A], N in (N in [A], b in )
Almost the same discussion as 4-1(N'BN→N'Abn), except for minus 1 edge flip in [A].

Now, the probability of 3-1 is equal to that of 4-1. That of 3-2 and 4-2 are also the same.


Pattern 5: N1 in [A], N2 in
A is in [N3], B is in [N4].
5-1: A and B is in the same loop
The order of letters is N3, A, N1, ... , N4(or n4), B(or b), N2(or n2), ... , N3(or n3).
After exchanging, the order of letters becomes N3, B, N2, ... , N3(or n3) and N4, A, N1, ... , N4(or n4).
So, one loop increases and 1 letter is added.

5-2: A and B is in the other loop
The order is N3, A, N1, ..., N3 and N4, B, N2, ... , N4.
After exchanging, the order becomes N3, B, N2, ... , N4, A, N1, ... , N3.(omit the small letter)
So, one loop decreases and 1 letter is reduced.

Now, I should consider the probability of 5-1 and 5-2.
I set the begin of loop to A. Next letter is N1.
In [N1], there may be N3(n3) or N4(n4) or N(n).(N is not equal to N1,2,3,4)
1. If N3(n3), the loop is finished because A(a) comes next, and this case is 5-2 because B(b) doesn't appear yet.
2. If N4(n4), next B(b) comes and it is before A(a). Loop still continues, but we can decide that this case is 5-1.
3. If N(n), next letter is same discussion as N1 because N3(n3) and N4(n4) don't appear yet.

And, the probability of N3(n3) or N4(n4) comes are the same in all the case of [N] before N3 and N4 come. Therefore, by repetition of this discussion, we can understand that the probability of 5-1 and 5-2 are the same.


Considering all of above cases, simple average of all cases is no plus or minus letter and edge flip.

In conclusion, we can save only a parity fix in average by exchanging targets. However, edge flip problem is a bit complicated. If there are already one flip, plus one flip is a small problem in the length of execution. I think this calculation is troublesome, so I didn't try to solve it.

Furthermore, this conclusion is in the case of using M2/OP method. If we use 3 cycle methods, I think the conclusion is almost the same as above, but in small part varies with parity methods we use.
 

mark49152

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This question may be more complicated than you think.
That's a great reply, thank you very much. I'll digest it in detail later, but my first thought is that I would avoid trying to exchange those edges if it were obvious that it would create more work, like if they formed a cycle alone, or if one of them were already solved. Not sure of that would make a significant difference, but I think you answered the question, which is that on average there's no saving beyond the parity fix.
 

Damien Porter

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I'm having trouble with consistency. I've done 50 attempts, and 7 of them were successes. Is it normal for someone this new to have such a low success rate?
It depends how fast you went from not knowing how to blind solve and doing it. Concider taking a step back and just practicing edges and corners separatly. Increase your accuracy on them separately, work on the most problematic then start doing full attempts again.

Having said that 7/50 isn't that bad. Blind solving isn't an easy thing to learn.
 

mark49152

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I'm having trouble with consistency. I've done 50 attempts, and 7 of them were successes. Is it normal for someone this new to have such a low success rate?
As Damien said, practise edges and corners separately. Also practise solving sighted so you can exactly how the solve is working. What method do you use?

I am a fairly new BLDer. At first I would recommend focusing on success not time - so go slow, master your letter scheme, develop a memo method, and master your chosen execution method. You should get to the point where if your memory is correct, your solve is correct, with no failures due to execution screw ups.

Then when you start to think about getting faster, I think it is good to have failures. If you don't have failures, you aren't pushing yourself hard enough. I try to stay just over 50% - so I push my speed as much as I can but if my success rate drops below 50% I'll slow down a bit.
 

josh42732

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Quick question out of curiosity: Assuming you use 1 letter per piece memorized, what is the breakdown of the average number of letters memorized needed to compete the corners? Thanks

-Doug
For corners, due to my experience of BLD solving, I'd say around 8.

And to piggyback on what he said, what is the average for 4BLD?
 

tseitsei

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I dont have any real numbers just an approximation from my own countless solves but I would say for corners it's less than 8 if you memo twists visually (high 7) and more than 8 if you memo twist with letters (low 8).

For 4bld I would say something like 16 for centers 24 for wings and 8 for corners approximately...
 

dovshmuel

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I know that this may have been answered several times already, but how do you deal with twisted corners with OP? Or having to shoot to BU and BU shoots to FD using m2?
 

h2f

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Twisted corners you can deal with algs like: (R U R' U') (R U R' U') L' (U R U' R') (U R U' R') L or with OP - shoot your piece to buffer, next shoot your piece to proper spot on cube.

In 3bld for shooting to BU you do special alg. Theres no need to shoot to FD - its your buffer from other angle. In big blds BU - special alg same like in 3bld, FD - do the setup l2 and next you do alg for BU.
 
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hedgehoghead

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This is aimed at beginners learning TURBO. A list of basic corner/edge algs for TURBO BLD. The algorithms are aimed at learners who learn algs visually. Rather than looking at daunting lists, I've tried to present the info as concisely as possible. You can print them out as index cards to carry around as you learn.

www.facebook.com/turbobld
 

Nikhil Soares

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I wanted to learn 5bld. I've already learnt 4bld. The only problem in 5bld is that I don't know the parities! I don't know the alg not how to use them. The method I use is U2,r2,and OP corners.Please help me. Wanted to learn it before my nationals'Indian Nationals 2016'
 
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