In 3OP, you orient the pieces (in this case edges) and the permute them separately.Originally Posted by cusx
In TuRBo, you orient and permute edges at the same time.
TuRBo is like 3OP edges on steroids i guess?
In 3OP, you orient the pieces (in this case edges) and the permute them separately.
In TuRBo, you orient and permute edges at the same time.
TuRBo is like 3OP edges on steroids i guess?
TuRBo!
In 3op you Orient and THEN permute!
in TuRBo you orient and permute at the same time.
and nice bump btw!
I got the gist of everything up to part 4. Then it got really confusing.
I think I made it more confusing by trying to simplify part 4. Is there anything you specifically you find confusing, or just part 4 in general?
The even swaps and how you end with a 2-cycle rather than a 3 cycle. You tried explaining it to me once and I still couldn't get it. I don't blame you its just a little over my head.
It's not over your head, it just needs to be explained to you in the right way.
I'll try again, explaining in a slightly different way from how I first tried.
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*I'm going to stop using the term '2-cycle', and replace it with parity alg. I probably should've done this in the first place.
*1 swap = swapping 2 pieces with each other.
-A 3-cycle has 2 swaps (ie. for UF>UR>UL, the swaps are: UF-UR and UL-UF)
-A parity alg (such as y-perm) also has 2 swaps (but instead of swapping one type of piece, there is 1 swap of corners and 1 swap of edges).
The total number of swaps needed to solve a cube will ALWAYS be even, this means a 3BLD solve can be deduced to 2 situations:
#1: odd edges and odd corner swaps. (parity)
#2: even edges and even corner swaps.
Imagine yourself in situation #2. Since 3-cycles are even (2 swaps), you can solve the cube with nothing but 3-cycles.
Now imagine yourself in situation #1. No matter how many 3-cycles you do, you won't be able to solve it, because 2 swaps won't ever add to give an odd number of swaps. Therefore you must apply a parity alg to perform 1 edge swap and 1 corner swap.
This provides a detailed explanation of why there will always be an even number of swaps.
Oh I get it now. Thanks Zane, your awesomeIt's not over your head, it just needs to be explained to you in the right way.
I'll try again, explaining in a slightly different way from how I first tried.
---------------------------------------------------------------------------
*I'm going to stop using the term '2-cycle', and replace it with parity alg. I probably should've done this in the first place.
*1 swap = swapping 2 pieces with each other.
-A 3-cycle has 2 swaps (ie. for UF>UR>UL, the swaps are: UF-UR and UL-UF)
-A parity alg (such as y-perm) also has 2 swaps (but instead of swapping one type of piece, there is 1 swap of corners and 1 swap of edges).
The total number of swaps needed to solve a cube will ALWAYS be even, this means a 3BLD solve can be deduced to 2 situations:
#1: odd edges and odd corner swaps. (parity)
#2: even edges and even corner swaps.
Imagine yourself in situation #2. Since 3-cycles are even (2 swaps), you can solve the cube with nothing but 3-cycles.
Now imagine yourself in situation #1. No matter how many 3-cycles you do, you won't be able to solve it, because 2 swaps won't ever add to give an odd number of swaps. Therefore you must apply a parity alg to perform 1 edge swap and 1 corner swap.
This provides a detailed explanation of why there will always be an even number of swaps.
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