The One-Answer Question Thread [closed]

[guysensei1]

Are there +2's for BLD? If you forget to undo one setup move like U or D ?

+2 exists for every event except FMC.

 

[Future Cuber]

+2 exists for every event except FMC.

What happens at FMC when you finish a solve with D

 

[guysensei1]

What happens at FMC when you finish a solve with D

DNF

 

[uberCuber]

+2 exists for every event except FMC.

and clock (in terms of one-move-away, of course there are still other possible penalties)

 

[guysensei1]

Is there an alg subset that solves LL edges while disregarding LL corners?

 

[yoinneroid]

Is there an alg subset that solves LL edges while disregarding LL corners?

http://www.speedsolving.com/wiki/index.php/LLEF

 

[10461394944000]

Oooohhh yea....My bad....... thnx for pointing that out

if you write E in an fmc solution you will get a dnf

Consider two non-zero numbers x and y such that

x = y.
Then x2 = xy.
Subtract the same thing from both sides:
x2 - y2 = xy - y2.
Dividing by (x-y), obtain
x + y = y.
Since x = y, we see that
2 y = y.
Thus 2 = 1, since we started with y nonzero.
Subtracting 1 from both sides,
1 = 0.

That was jus 4 fun please dont debate on that

congralutations you can divide by 0

 

[guysensei1]

it has twice the number of cases of CLL and the algorithms that solves them are mostly long (the worst LL case of them all is in this group, it needs 16 turns optimally (HTM))

What is 'the worst LL case of them all'?

 

[Future Cuber]

congralutations you can divide by 0

Did you read the last line.....??????

 

[brian724080]

Consider two non-zero numbers x and y such that

x = y.
Then x2 = xy.
Subtract the same thing from both sides:
x2 - y2 = xy - y2.
Dividing by (x-y), obtain
x + y = y.
Since x = y, we see that
2 y = y.
Thus 2 = 1, since we started with y nonzero.
Subtracting 1 from both sides,
1 = 0.

That was jus 4 fun please dont debate on that

I think you mean x^2 and y^2 for x2 and y2

Also, you can't divide by x - y because x = y, which means x - y = 0

 

[Future Cuber]

I think you mean x^2 and y^2 for x2 and y2

Also, you can't divide by x - y because x = y, which means x - y = 0

Chill.... i was jus kidding.....

 

[brian724080]

Chill.... i was jus kidding.....

Yea, I saw the last bit of the post. Couldn't resist...

 

[PJKCuber]

Guys, since everybody is telling me to get off the forums and practice for a while, can anybody just check the if following things I want to learn are right
Goals Before September
2x2 Sub 6 Average of 100 (LBL). Learn how to make effecient layers. One look 1st layer. Then learn CLL after being sub 5.
3x3 Sub 20 Average of 100. Learn full PLL,Hike up metronome to 90bps for Look ahead and learn a few F2L algs.
4x4 Sub 1:30 Average of 100. Nothing much but look ahead like crazy during 3-2-3. Become CN using Yau.
3BLD: Success
3x3OH: Sub 1 Average of 100: Learn Full OH PLL.
And most : Do not get on the forums!!!!!

 

[TDM]

Also, you can't divide by x - y because x = y, which means x - y = 0

Yeah, that problem's easy to see. This one annoyed me for a while because of something I didn't know and couldn't be bothered to look up.

 

[EvelienV]

Do you guys know of any algorithms where they use numbers to solve a 3x3x3 cube? It's not a double rotation but something like this (3,2,2) F L U

 

[Hypocrism]

Guys, since everybody is telling me to get off the forums and practice for a while, can anybody just check the if following things I want to learn are right
Goals Before September
2x2 Sub 6 Average of 100 (LBL). Learn how to make effecient layers. One look 1st layer. Then learn CLL after being sub 5.
3x3 Sub 20 Average of 100. Learn full PLL,Hike up metronome to 90bps for Look ahead and learn a few F2L algs.
4x4 Sub 1:30 Average of 100. Nothing much but look ahead like crazy during 3-2-3. Become CN using Yau.
3BLD: Success
3x3OH: Sub 1 Average of 100: Learn Full OH PLL.
And most : Do not get on the forums!!!!!

There's no order, just do stuff faster.

 

[goodatthis]

Consider two non-zero numbers x and y such that

x = y.
Then x2 = xy.
Subtract the same thing from both sides:
x2 - y2 = xy - y2.
Dividing by (x-y), obtain
x + y = y.
Since x = y, we see that
2 y = y.
Thus 2 = 1, since we started with y nonzero.
Subtracting 1 from both sides,
1 = 0.

That was jus 4 fun please dont debate on that

Here's why your "proof" doesn't work: let's say x=2.

2=2 (correct)
2^2=2*2 (correct)
(2^2)-(2^2)=(2*2)-(2^2) (correct)
Here's where you're problem is, x-y=0, and you can't divide by zero. Making silly variables to hide the fact that this is impossible isn't a "cool math trick."

If you wanted to create a proof on what 0^0 equals, that would be neat.

Also, I read the last line.

 

[Future Cuber]

Here's why your "proof" doesn't work: let's say x=2.

2=2 (correct)
2^2=2*2 (correct)
(2^2)-(2^2)=(2*2)-(2^2) (correct)
Here's where you're problem is, x-y=0, and you can't divide by zero. Making silly variables to hide the fact that this is impossible isn't a "cool math trick."

If you wanted to create a proof on what 0^0 equals, that would be neat.

Also, I read the last line.

1.Its not my proof .... google it ..you'll find it
2.i read your last line ...so i am assuming that this post is fun as well
..................

 

[slinky773]

Does anyone know an algorithm for 2x2 Ortega PBLL where the top layer is solved and the bottom layer is a Y-Perm? I'm really trying to get rid of cube rotations for 2x2, so I'm trying to find algorithms for permuting only the bottom layer without a z2 or anything. I found the algorithm for bottom T-Perm (R' U R' U' R' F R2 U' R' U' R U R' F' R2) that Kennan posted here, but I can't find the Y-Perm.

 

[TDM]

Does anyone know an algorithm for 2x2 Ortega PBLL where the top layer is solved and the bottom layer is a Y-Perm? I'm really trying to get rid of cube rotations for 2x2, so I'm trying to find algorithms for permuting only the bottom layer without a z2 or anything. I found the algorithm for bottom T-Perm (R' U R' U' R' F R2 U' R' U' R U R' F' R2) that Kennan posted here, but I can't find the Y-Perm.

Y perm on top:
R U' R' U' F2 U' R U R' D R2
Y perm on bottom can be done by cancelling the end with R2 F2 R2, which is just like inserting an F2 before the end:
R U' R' U' F2 U' R U R' D F2 R2

also that adj swap on D in your post is genius (and yet so obvious), thanks kennan
shame it doesn't work with my alg... R F R F' R U2 R' U R U2 R... meh