# BLD CO



## drkmagicard (Jul 4, 2007)

For Corner Ori.
What algorithm would I do if I have 2 clockwise and 1 counter clockwise?
thanks


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## Karthik (Jul 4, 2007)

I dont think that is possible.You have to have more mis-oriented corners.Only 2 clockwise and 1 counter clockwise is not possible I guess.


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## AvGalen (Jul 4, 2007)

correct!

Use this trick:
Clockwise is +1
Counterclockwise is -1

All clockwise and counterclockwise totals should be divisible by 3 (the number of positions of the corners). This means the following are allowed on a cube (top and bottow, so 8 at most): -6, -3, 0, 3, 6

If you have your orientation on the top (+2 -1 = +1) there will be a -1 (or 2 +1's) at the bottom or the cube is unsolvable.


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## drkmagicard (Jul 4, 2007)

well i always have trouble when I have an odd number of corners not oriented.
so for example right now I have 5 peices that need to be flipped.
clockwise: 3
counter clockwise: 2

Ive only memorized (12) and (13) from macky's site
what should I do that will solve this... I can learn the others


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## Karthik (Jul 4, 2007)

Taken from PJK's guide:
(12): U'AUA'
(13): U'AU2A'U'
(123 ccw): U'AUAUAU'
(123 cw): U'A'UA'UA'U'
where A=(R'D'RD)x2 and A'=(D'R'DR)x2
For more clarification,see PJK's guide.


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## drkmagicard (Jul 4, 2007)

So I would do a 
(123 cw): U'A'UA'UA'U'
for the three clockwise
then a (12) for the last 2

also what about the case were i have 3cw and 1 ccw?


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## hait2 (Jul 5, 2007)

i think you misunderstand

don't think of co as individual pieces, they are pairs or triplets. you can only have a cw-ccw pair, or a ccw triplet or a cw triplet. then of course you can have some combination thereof, but always comes down to cw-ccw pairs or cw*3 or ccw*3
edit: i suggest you figure out for yourself why that is true. it will help you a lot in your understanding of the cube in general. of course it's not necessary to solve bld though

knowing this, having 3ccw and 2cw is impossible, because you have a ccw triplet, and then.. not a cw triplet, nor a cw-ccw pair. so you're in trouble! 
of course, you can also look at it as having 2 cw-ccw pairs and 1 extra ccw, which once again, spells trouble 

3cw and 2ccw is impossible im afraid, similarly to 3cw and 1ccw, for the reasons above


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## drkmagicard (Jul 5, 2007)

ok yeah i see what you mean
because I only memorized how to flip 2 corners, so I only saw solving CO as finding 2 pairs
thanks a lot


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## Karthik (Jul 5, 2007)

> _Originally posted by drkmagicard_@Jul 5 2007, 02:32 AM
> * because I only memorized how to flip 2 corners, so I only saw solving CO as finding 2 pairs
> *


 Even if you see them as pairs you can do it.
Eg:If you 3 ccw pieces at 1,2,3 then you can do 1,2(1-ccw and 2-cw) and 2,3(2-cw and 3-ccw)
Isnt it obvious??


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## AvGalen (Jul 5, 2007)

I strongly disagree with this:
"don't think of co as individual pieces, they are pairs or triplets."

You SHOULD think of them as individual pieces that need to be in a group that "cancels out (-3, 0 or 3 as I described before).

If you would have 4 corners that are (in order) cw, cw, ccw, ccw this a group that cancels out (+1 +1 -1 -1 = 0) and I would solve it as A' U A' U A U A U. I wouldn't solve it as 2 groups of pairs (A' U2 A U2 and U A' U2 A U)


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## Karthik (Jul 5, 2007)

> _Originally posted by AvGalen_@Jul 5 2007, 05:54 AM
> * I would solve it as A' U A' U A U A U. I wouldn't solve it as 2 groups of pairs (A' U2 A U2 and U A' U2 A U) *


 Did you make up that alg??Or is it a standard one??
BTW I dint recommend that CO be seen in pairs.I just said that it was also possible


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## hait2 (Jul 6, 2007)

> _Originally posted by AvGalen_@Jul 5 2007, 05:54 AM
> * I strongly disagree with this:
> "don't think of co as individual pieces, they are pairs or triplets."
> 
> ...


yes i agree with you disagreeing with what i said. i tend to oversimplify things to get the point across. eventually they'd figure out on their own the 'true' limitations and why they are there (it's probably more fun to do this and it makes you feel great when you figure out "oh hey, i can do THIS".. i mean the bld epiphany thread is a prime example of this)

i also solve them in whatever way as long as it cancels out =)

edit: on a completely unrelated note, when do you think i can replace beginner cuber with intermediate cuber in my sig? i'm thinking sub30 single and/or sub4 bld, but others may think differently. is it possible to draw a line between beginner/intermediate/advanced?


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## AvGalen (Jul 6, 2007)

karthikputhraya: I "made up" that alg, but let me put it this way:

If you do A, you turn a corner ccw and if you do A', you turn a corner cw. If the corners are grouped in pairs (-3, 0, 3 as I mentioned before) in the U-layer, then all you have to do is:

1) Do U untill an unsolved corner is at the "working position"
2) Do A or A', depending on the cw/ccw direction the misoriented corner needs to be turned
3) Repeat from step 1 untill all misoriented corners are oriented.

Examples
A = clockwise Alg = (FDF'D')*2
A' = counterclockwise Alg = (DFD'F')*2
Working Position = Up-Front-Right

Example1a:
Scramble = R U2 R2 U' R2 U' R2 U2 R
Analysis = +1-1-1+1
Solution: A U A' U A' U A U
Example1b:
Scramble = R U2 R2 U' R2 U' R2 U2 R U
Analysis = -1 -1 +1 +1
Solution: A' U A' U A U A U 
Example2:
Scramble = R U R' U R U' R' U R U2 R'
Analysis = +1 -1 +1 -1
Solution: A U A' U A U A' U

hait2: I am glad we agree on this! What about the following characteristics....

+60 = Beginner
30-60 = Intermediate
20-30 = Advanced
15-20 = Expert
Sub 15 = World Class

I am talking about averages, not single solves and definatively not lucky single solves! All of the above times are just my definition for the 3x3x3.


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## hait2 (Jul 7, 2007)

> _Originally posted by AvGalen_@Jul 6 2007, 11:21 AM
> * hait2: I am glad we agree on this! What about the following characteristics....
> 
> +60 = Beginner
> ...


 sounds good to me! must work to become advanced.. all that's left is practice now, i probably won't learn any more algs
looking forward to steady (hopefully) improvement in your competitions AvG ^_^


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## hdskull (Jul 31, 2007)

yay! i'm advanced! haha 

okay arnaud, so basically what u'r saying is all 1 are solved with A and all -1 are solved with A' ?


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