A Few Questions About basic 3BLD

[TomH]

both letters in the pair to a single object

 

[sk8erman41]

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...

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

Where:
t = No. of targets (excluding targets associated with flipping edges or twisting corners)
u = No. of unsolved pieces (excluding buffer and flipped/twisted pieces)
c = No. of cycle breaks (ie. how many times you break into a new cycle)

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.

 

[etshy]

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

 

[sk8erman41]

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

 

[sk8erman41]

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?

 

[Renslay]

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?

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

 

[etshy]

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

 

[tseitsei]

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?

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

 

[sk8erman41]

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

 

[etshy]

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)

 

[Renslay]

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.

 

[Noahaha]

That's how I think about it.

 

[Cubenovice]

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)

 

[sk8erman41]

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.

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

 

[TheOneOnTheLeft]

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.

 

[Renslay]

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...

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.