# Keeping track of the number of cycles



## toastman (Feb 2, 2011)

So I've started BLD cubing using letter pairs. Surprised to find that (with all the techiques I've read on the forum) recalling ~20 letters is quite do-able. However, I'm finding it tough keeping track of the number of edge "cycles" I have when memorizing.

The only reason you need to know this number is for the "#items = (#wrong cubies) + (#cycles) - 2" formula, to ensure you haven't forgotten a piece. I find remembering #cycles tough. How do you guys do it?

Do you keep an extra "cycle count" number in your head?

An idea I had was to use the letter "Z" to denote "End of Cycle". In which case #items = #wrong cubies. The upside is you don't need to track an extra number in your memorization phase. The downside is you have to memo more letters, and (I think) you lose the benefit of "If you finish edges in the middle of a letter-pair you have to do parity".

I've also heard of someone who starts the memo of each cycle using a famous person. #famous people = #cycles.

Any other methods? Or do you just brute-force it.


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## aronpm (Feb 2, 2011)

Why do you need to know the number of cycles? It's not important.


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## RyanReese09 (Feb 2, 2011)

Say I have this scramble (Scramble WG hold it YO)
Ah first scramble I tried had no new cycles needed for edges, let me try another..
U B' L B R' L' B R U2 F' U2 D' B F U' R2 B F R L' F2 R2 B2 D2 F
Ok
Depending on your lettering scheme this will vary but for my purposes just bare with me
Holding it YO I start with the UR piece. I'd memo edges.
This is my memo I got
EQ BV JM OT WG KL
I use images to memo, so this is what I have memorized (I won't say what actions they are doing together)
An Elephant and a Queen
Batman and a Virgin
JigglyPuff and a Magikarp
Olive and the Triforce
Werewolf and a Gorilla
Kangaroo and a Lapras


No new cycles are needed until the very end where it is K L. Now in a real solve I'd just do a pure 2flip for those edges but some people would break into a new cycle. Notice in my memo I don't even recognize the fact that I have a new cycle. I count the images in my head, and if it adds up to me having all the edges accounted for, then good, I'm done. It might also help to have a finger stuck to the pieces that get solved while your tracking memo. Like, starting memo, I'd put a finger on the YO edge spot because my first edge needs to go there, then I'd place a finger on the LD spot to know that that one is done, etc etc. I don't think people actually use that formula in solves, bit stupid


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## toastman (Feb 2, 2011)

The question I'm trying to answer is "Have I memod enough letters or am I missing some cubies?"

I'm having a hard time coming up with an example, but think of it this way. Say you had a horrible scramble and you had to break into a new cycle 3 times. I.e, your buffer piece kept on being getting solved. In addition, (If UR is your buffer), DF and DB (Letters U and W) need to be swapped.

So, say your memo (for argument's sake, this is an extreme case) was:

A C D B (New Cycle) J K L J (New Cycle) R S T R (New Cycle), so in your head you have
AC DB JK LJ RS TR 
And you need to ask yourself "That's 6 pairs/12 letters, is that enough or have I forgotten something?"

You somehow need to know that you've used 3 cycles, so that when you apply
#items = (#wrong cubies) + (#cycles) - 2 you go "There are 2 items missing", and you figure out DF and DB need their own cycle, so you need to add UW to your memo.

If you forget at any point "I'm in my nth cycle", you're in danger of missing the last 2 pieces and DNF. In addition, if you know "There are 2 pieces missing, let's find them", you don't waste time trying to figure out "Do I have all the pieces covered"?

Reason I ask is I wasted like 2 minutes yesterday on a BLD trying to figure out "Do I have all the pieces I need covered" ~

This sort of problem I'd imagine is far more common on the 4x4 (and the megaminx, if you're some sort of freak).


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## Cubenovice (Feb 2, 2011)

You can just "code" your cycles into regular letter pair images.
suppose after DR AL your buffer is back in its spot..
Now you have to store it ito another location, just memo the letter of the sticker position you are placing it, suppose E.
The next target will be the 2nd letter of the image pair. DR AL EQ etc etc

This way you end up with a memo that consists of letter pairs only, no specific info on cycles, just a long string of targets.
Parity is indicated by the last letters: if you end on a letter PAIR there is no parity, if you end on a single letter you have parity.
In case of a single letter you can indeed use a list of specific words to indicate a single letter. For instance the "famous people"

Some of this is currently being discussed in the "memory methods" thread


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## Innocence (Feb 2, 2011)

toastman said:


> The question I'm trying to answer is "Have I memod enough letters or am I missing some cubies?"
> 
> I'm having a hard time coming up with an example, but think of it this way. Say you had a horrible scramble and you had to break into a new cycle 3 times. I.e, your buffer piece kept on being getting solved. In addition, (If UR is your buffer), DF and DB (Letters U and W) need to be swapped.
> 
> ...


 
To actually answer your question - Just look at the pieces and think about whether you've solved them or not. That's all I do, checking the number of already solved pieces and stuff, then just estimating when I should be done from the length of the 'phrase', and then do a quick scan for forgotten/flipped pieces.


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## whauk (Feb 2, 2011)

when i think that i am done i just take a quick look at all the edges again and ask myself whether i memod them. but i know where your problem is. i once DNFd an official blind solve because i just didnt see a flipped edge in place.
so well my solution is just to look at every spot again and if you are not sure whether a certain piece is part of your memo just look where it has to go and see if it is a new 2 cycle or sth...
its actually not slow. i dont loose more than 10 seconds for this


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## aronpm (Feb 2, 2011)

whauk said:


> its actually not slow. i dont loose more than 10 seconds for this


 
10 seconds is a lot of time!


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## Marcell (Feb 2, 2011)

toastman said:


> This sort of problem I'd imagine is far more common on the 4x4 (and the megaminx, if you're some sort of freak).



Right. After some time you won't need this kind of test for a 3x3, as it doesn't have that much pieces and you can naturally keep track of them.
On 4x4 wings I count the times I have to break into a new cycle on my fingers.


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## whauk (Feb 2, 2011)

aronpm said:


> 10 seconds is a lot of time!


 
not if your average is 5 minutes


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## cisco (Feb 2, 2011)

You should realize there is a single flipped piece, because the last shoot is to the same piece you opened the last cycle with, but the wrong sticker. Sorry, my English sucks with more-than-5-words sentences


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## ssb150388 (Feb 2, 2011)

To start with, I use 3OPfp) and use numbers(another :fp) to memo the edges and visual EO. And I keep counting as I go along.. so when I notice end of a cycle, I just quickly recall how many numbers I have memo-ed. That + 1 is the number of edges actually done as one is the buffer which is automatically solved in the end.

This is easy when there are 1 or 2 cycles.. but becomes difficult when there are more cycles or many swaps(quite rare). 

My memo time is around 1-1:15 with this method. 
And avg ~2:45-3.


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## toastman (Feb 2, 2011)

Marcell said:


> Right. After some time you won't need this kind of test for a 3x3, as it doesn't have that much pieces and you can naturally keep track of them.
> On 4x4 wings I count the times I have to break into a new cycle on my fingers.


 
Yes, I was sort of thinking along these lines.


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## cmhardw (Feb 2, 2011)

Honest answer for me is that on 3x3x3 I just have an intuitive sense of "did I miss any areas of the cube or not?" Notice I didn't say "did I miss any pieces," but instead areas of the cube. Sometimes, after breaking lots of cycles and having a confusing scramble to memo, I just have an intuitive feeling that I never really cycled to the bottom, back, left of the cube. This usually is all I need to find the last missing 2-cycle. As to more rigorous methods, I have two.

If you break new cycles:
Daniel Beyer uses this method, and described it to me. Basically you need to memorize what an ideal scramble should be like in your memo. Ideal for 3x3x3 edges means a pure 12-cycle with no cycle breaks. For me this is 6 images, the last one being a single letter image, a famous person. My memo would look like this:
(PAIR PAIR PAIR) (PAIR PAIR SINGLE) 

These parentheses represent the journey locations, this is still all the same cycle. PAIR means letter pair image, and SINGLE means single letter image. Notice that I don't memorize the first buffer as an image. This is assumed to always be the same buffer. Whether you memorize this as an image or not will not affect how you use this method. You are always comparing your cycles to your ideal cycle, so whatever you ideal cycle looks like for _you_, use that for the rest of the method.

Now let's say I had to break cycles. Let's say my cube is scrambled with (ABCD)(EFGH)(IJKL)
My memo would be:
(A BC DE FG HE IJ KL I)
which I would notice is:
(PAIR PAIR PAIR) (PAIR PAIR PAIR) (SINGLE)

Now realistically I would cram that last single image into the same journey location, but I'll use a 3rd one for illustration's sake. Ok, so the difference between my current memo, and the ideal, is 2 letters. My last location is supposed to be (PAIR PAIR SINGLE) and instead I ended with 
(PAIR PAIR PAIR)(SINGLE)

This means I added a piece onto what used to be a SINGLE at the end, and I added a new Single after all of it. I've added two pieces.

The formula for counting this way is:
# of pieces still unaccounted for = (# of items in excess or short of the ideal memo) - (# of times you broke a cycle) + (# of solved pieces)

For (# of items in excess or short of the ideal memo), use negative numbers if you fall short of the ideal memo, and you're using the formula (which I don't recommend during a real solve).

Here I would see that:
# of pieces still accounted = 2 - 2 + 0 = 0
So we accounted for everything.

Now, in a real solve you don't think of this in terms of the formula. Basically I look at my memo. If I am short of the ideal situation I need to subtract out one image for each of the cycle breaks (you shoot to the first letter twice). After that I need to add any solved pieces. If doing this does not bring me to my ideal memo length, then there are still pieces left to go.

If my memo goes beyond my ideal cycle length, then I need to subtract out one for each cycle break, then add any solved pieces. If doing this does not bring me to my ideal memo length, then I am missing some pieces.

If you do not break cycles

If you don't break cycles, then you have single letter images that start all new cycles, and you have single letter images to end all odd cycles. Let's look again at the same cycle as above.
(A BC D)(E FG H)(I JK L)

I have put the spaces in the show the pair images and the single images in each cycle.

The memo structure of the cycle we have above would be:
(BC D)(E FG H)(I JK L)
(PAIR SINGLE SINGLE)(PAIR SINGLE SINGLE)(PAIR SINGLE)

When you memorize this way often it becomes very easy to "scan" through your memo looking for single letter images. Here I would see that I have 4 single letter images in the first two locations, when I am supposed to have only one (the one at the end of the second location). This puts me "under" by 3 images. I'm supposed to have PAIR images there, and I only have SINGLEs. This means I have a cycle debt of 3 pieces.

Now I look at the overage beyond the ideal cycle length and see that it consists of a paired image and a single image (3 pieces). So I am "over" by 3 pieces beyond the ideal cycle length. An overage of 3 balances an undershoot of 3, so I have accounted for all pieces. You also have to add in solved pieces in this method too. I personally think counting this way is much easier than counting with the cycle break method, for me personally at least. However, I think there are some disadvantages in that memorizing this way has "wildcard" scrambles with lots of small cycles where you need an extra journey location more than you normally do. I have worked in (optional) journey locations in all of my journeys for this, in case I ever need them, and I sometimes do.

Hope this helps.


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## toastman (Feb 3, 2011)

Awesome post Chris, thanks. I learned some other stuff as well.

I think the thing for me to do on the 3x3 is to just check that there are 6 pairs (12 pieces) in my edge memo. If they're aren't, then quick-scan the cube for solved pieces.

I found the following. In a 3x3 solve, you're extremely unlikely to need 3 cycles. When I'm doing my Memo, if I have to break into a new cycle, I sort of go "Damnit!". If I need 2 cycles, I'm more like "OH BLOODY DAMNIT!". If there were 3, I'd be "FFFFFFUUUUU!!!". So, if I'm at the end of my memo, and I'm quite p*ssed off (OH BLOODY DAMNIT!), I know that I've got two god damned cycles and two extra goddamn stupid letters to memo.

Similarly, when there's a solved piece (or two), I'm like "WOOHOO!" and know I need to factor that in.

(Would one consider this "emotional memory"?)

If there's no "Woohoo!" or "DAMN!" that I'm feeling, and 12 letters in my head (6 pairs, which for me is 2x PVO images), pull that blindfold down baby!

If I were doing a 5x5 or a Megaminx, I think I'd need to count on my fingers, or keep a formal running count.

Thanks everyone! Problem solved.


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## Kynit (Feb 3, 2011)

toastman said:


> If I were doing a 5x5 or a Megaminx, I think I'd need to count on my fingers, or keep a formal running count.


 
A 5x5 has two kinds of edges: centrals (which work the same way as you've described) and wings (which are the same, there are just 24 of them). There's hardly any difference. Just count 4 images instead of 2!


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## Zane_C (Feb 3, 2011)

Kynit said:


> A 5x5 has two kinds of edges: centrals (which work the same way as you've described) and wings (which are the same, there are just 24 of them). There's hardly any difference. Just count 4 images instead of 2!


 
Wings aren't the same as central edges, the permutations are distinct. You can't permute a wing and have it wrongly oriented.


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## aronpm (Feb 3, 2011)

Zane_C said:


> Wings aren't the same as central edges, the permutations are distinct. You can't permute a wing and have it wrongly oriented.


 
For the sake of counting here they are the same


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