# Being sure you are done with your edges (Old Pochmann)



## rsquaredcuber (Apr 1, 2014)

How can you be completely sure that you are done with all of your edges in an Old Pochmann blindfold solve, and if you still have edges, how can you quickly identify which ones are unsolved?


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## cmhardw (Apr 1, 2014)

Do you mean during memorization?

If so, you sort of develop a feel for it over time. A memorization will sometimes feel "too short" when you still have a small cycle of edges left to memorize. If you want to be thorough you can place your fingers on the edges as you rehearse your memory to mark which ones you've memorized. There are 12 edges and you have 10 fingers, so try to pick a slice (say the L slice) to mentally track edges on rather than put a finger on any edge on that slice. If you have a finger on every edge in the M and R slice, and you mentally marked every edge on the L slice then you got all of them during your memorization. If not, then the edges you're not touching or mentally "marking" are the ones left to memorize.


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## DeeDubb (Apr 1, 2014)

cmhardw said:


> Do you mean during memorization?
> 
> If so, you sort of develop a feel for it over time. A memorization will sometimes feel "too short" when you still have a small cycle of edges left to memorize. If you want to be thorough you can place your fingers on the edges as you rehearse your memory to mark which ones you've memorized. There are 12 edges and you have 10 fingers, so try to pick a slice (say the L slice) to mentally track edges on rather than put a finger on any edge on that slice. If you have a finger on every edge in the M and R slice, and you mentally marked every edge on the L slice then you got all of them during your memorization. If not, then the edges you're not touching or mentally "marking" are the ones left to memorize.



Is it possible to just do the math? Like, if there's no cycle breaks, skips, or flipped edges, it's 11 moves, right? A cycle break would be +1, a flipped edge+1, and a skip -1, so wouldn't you just be able to deduct exactly how many moves there should be?


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## cmhardw (Apr 1, 2014)

DeeDubb said:


> Is it possible to just do the math? Like, if there's no cycle breaks, skips, or flipped edges, it's 11 moves, right? A cycle break would be +1, a flipped edge+1, and a skip -1, so wouldn't you just be able to deduct exactly how many moves there should be?



This is certainly possible, but it is usually done during the review phase of memorizing. The problem with this is that most top BLD solvers recommend to review little to not at all, so this counting phase would ideally be skipped on 3x3 BLD to save time. On 4x4 BLD or 5x5 BLD, for those like me who are not world class, it is good to do a count like this to make sure you have memorized all 24 wing edges. I doubt that the 4x4 BLD and 5x5 BLD people at the very top are doing this counting phase, but who knows? Perhaps they are, they just do it very efficiently.


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## DeeDubb (Apr 1, 2014)

cmhardw said:


> This is certainly possible, but it is usually done during the review phase of memorizing. The problem with this is that most top BLD solvers recommend to review little to not at all, so this counting phase would ideally be skipped on 3x3 BLD to save time. On 4x4 BLD or 5x5 BLD, for those like me who are not world class, it is good to do a count like this to make sure you have memorized all 24 wing edges. I doubt that the 4x4 BLD and 5x5 BLD people at the very top are doing this counting phase, but who knows? Perhaps they are, they just do it very efficiently.



Thanks. Good to know. I'm still so terrible at it that I have plenty of time to second guess myself, haha.


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## rsquaredcuber (Apr 3, 2014)

DeeDubb said:


> Is it possible to just do the math? Like, if there's no cycle breaks, skips, or flipped edges, it's 11 moves, right? A cycle break would be +1, a flipped edge+1, and a skip -1, so wouldn't you just be able to deduct exactly how many moves there should be?



Yes, but could you mathematically deduct which EXACT edge(s) you had left? If so, this is exactly what I am looking for.


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## rsquaredcuber (Apr 3, 2014)

cmhardw said:


> This is certainly possible, but it is usually done during the review phase of memorizing. The problem with this is that most top BLD solvers recommend to review little to not at all, so this counting phase would ideally be skipped on 3x3 BLD to save time. On 4x4 BLD or 5x5 BLD, for those like me who are not world class, it is good to do a count like this to make sure you have memorized all 24 wing edges. I doubt that the 4x4 BLD and 5x5 BLD people at the very top are doing this counting phase, but who knows? Perhaps they are, they just do it very efficiently.



Chris, I greatly appreciate this idea and it is helpful, yet I still crave a way to efficiently figure what edges are left mathematically and identify them. Thanks, though!


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## TheOneOnTheLeft (Apr 3, 2014)

(edge targets memorised) + (number of solved pieces) - (number of cycle breaks) = 11


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## DeeDubb (Apr 3, 2014)

TheOneOnTheLeft said:


> (edge targets memorised) + (number of solved pieces) - (number of cycle breaks) = 11



Wouldn't flipped edges add a +1?


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## mDiPalma (Apr 3, 2014)

DeeDubb said:


> Wouldn't flipped edges add a +1?



There are better ways to solve flipped edges in BLD than shooting to the same piece twice.


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## TheOneOnTheLeft (Apr 3, 2014)

DeeDubb said:


> Wouldn't flipped edges add a +1?



This is assuming you solve flipped edges as a separate cycle. So if UL is flipped, you memorise UL then LU. This adds two edge targets and one cycle break. 2 - 1 = 1 more piece solved. If you solve flipped edges with a more direct method, then you could add " + (number of flipped edges)" to the left side of the formula.


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## blah (Apr 3, 2014)

There's no need to actively count pieces if you design a memo system that does it for you for free. What do I mean by this? Consider this analogy: if I told you my credit card number is 39549272037311, you'd have to count to realize that that can't be right because it only has 14 digits. Even worse, no matter how slowly you count, there's always a risk of miscounting when your string of digits gets sufficiently long. But if I told you it was 3954 9272 0373 11, you'd immediately know that it's 14 digits by the time you reach the last digit (without even trying to recall any of the previous digits) because "the system" inserts a space after every 4 digits -- the system is counting for you for free.

I suspect this is different from the "feel you develop over time" that Chris mentioned. There's no iffy feel here. You know _precisely_ how many pieces you've gone through with such a system. And if you count the number of flipped edges/new cycles you broke into using some other system (feet, perhaps?) then you know exactly how many pieces you still need.


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## cmhardw (Apr 3, 2014)

blah said:


> I suspect this is different from the "feel you develop over time" that Chris mentioned. There's no iffy feel here. You know _precisely_ how many pieces you've gone through with such a system. And if you count the number of flipped edges/new cycles you broke into using some other system (*feet, perhaps?*) then you know exactly how many pieces you still need.



That's a good idea. I used to use a base 5 counting method with my feet for 5x5x5 wing edges back when I first stared. The digits are:

0 - foot resting flat in comfortable position, toes pointing straight ahead and heel pointing straight back
1 - heel in rest position, toes slid left
2 - heel in rest position, toes slid right
3 - toes in rest position, heel raised and slid left
4 - toes in rest position, heel raised and slid right

Your left foot is the 5s place and you right foot as the 1s place.



rsquaredcuber said:


> Yes, but could you mathematically deduct which EXACT edge(s) you had left? If so, this is exactly what I am looking for.



As for knowing mathematically _which_ edges are left unsolved, think about that for a second. You have 12 edges, and let's say 10 of them are accounted for in cycles. You glance at the cube and see no permuted edges, so the other two remaining edges are in a 2 cycle. Either you must have a list of edges not used from which you are "crossing off" edges as you memorize, or you must have a list of edges you _have_ memorized to which you are adding edges as you memorize.

I doubt many peolpe, even among top BLD solvers, have a robustly defined method to tell exactly _which_ edges are not yet memorized beyond the "feeling" that you have not cycled that much to the *L* face, say. This is too much work, and there is a chance that you have a 12 cycle of edges and have done this extra step of tracking edges for nothing, because it was not necessary to find straggler edges. If the top BLD solvers memorize a 3x3x3 in 8 or 9 seconds, I doubt they are tracking which edges (or even how many edges) remain to be memorized.

If you really want a method to track which edges remain, try using some sort of physical body marking system (feet, hands, fingers, teeth, tongue, etc.) for how many pieces you've cycled to on the *L*, *M*, and *R* slices. I think this is too much trouble and would result in slower times not faster times, but I would loved to be proved wrong by such a system helping identify unused pieces more quickly.


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## rsquaredcuber (Apr 3, 2014)

TheOneOnTheLeft said:


> (edge targets memorised) + (number of solved pieces) - (number of cycle breaks) = 11



and how would you mathematically identify those edges that are left?


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## blah (Apr 4, 2014)

rsquaredcuber said:


> Chris, I greatly appreciate this idea and it is helpful, yet I still crave a way to *efficiently* figure what edges are left *mathematically* and identify them. Thanks, though!





rsquaredcuber said:


> and how would you *mathematically* identify those edges that are left?


Here's a solution that's both "efficient" and "mathematical."

Label the \( n \) pieces by any \( n \) primes you like: \( p_1,\ldots,p_n \).
Let \( M=\prod_{i=1}^np_i=p_1\times p_2\times\cdots\times p_n \) and memorize it.
When you extend your memorization string by the \( n \)th piece, redefine \( M \) to be \( M/p_n \).
When you've completed a cycle and \( M\neq1 \), then there are still pieces that you haven't traversed. (Otherwise, terminate the algorithm.)
To determine the remaining pieces, factor \( M \) uniquely into primes. This is possible because \( \mathbb Z \) is a Unique Factorization Domain. Alternatively, invoke the Fundamental Theorem of Arithmetic if you're more familiar with that.
Choose any prime factor of \( M \) and start your new cycle from there. You may use the Axiom of Choice to choose this prime, but it's theoretically unnecessary.


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## cmhardw (Apr 4, 2014)

blah said:


> Here's a solution that's both "efficient" and "mathematical."
> 
> ...



That does perfectly answer the question. Also, that video is amazing


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## Noahaha (Apr 4, 2014)

Beginner: use the formula:
# of targets = number of pieces - 1 + number of cycle breaks - number of flipped/twisted/solved pieces. If you have too few, place your fingers on the pieces you have targeted to see what is left.

Intermediate: as you go along, keep track of how many targets there should be. Start developing a sense for what hasn't been targeted.

Advanced: Just know.


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## rsquaredcuber (Apr 5, 2014)

blah said:


> Here's a solution that's both "efficient" and "mathematical."
> 
> Label the \( n \) pieces by any \( n \) primes you like: \( p_1,\ldots,p_n \).
> Let \( M=\prod_{i=1}^np_i=p_1\times p_2\times\cdots\times p_n \) and memorize it.
> ...



Wow, you blew me off my feet. Thanks, although it is quite unlikely that I will use it in a BLD solve...it's cool to know that there is actually a mathematical way!


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