The One-Answer Question Thread [closed]

[miniGOINGS]

While doing ZZ or Petrus, what is the chance of a corner orientation skip after F2L if no seperate techniques are applied?

 

[cmhardw]

While doing ZZ or Petrus, what is the chance of a corner orientation skip after F2L if no seperate techniques are applied?

How many possible orientations are there for a given corner? How many corners are there on the cube? Also, how do the orientations of pieces relate to each other? For example, can all 4 LL corners be twisted once clockwise away from "correctly oriented"? Lastly, how many of those possible states do you consider to be the solved state?

Chris

 

[miniGOINGS]

While doing ZZ or Petrus, what is the chance of a corner orientation skip after F2L if no seperate techniques are applied?

How many possible orientations are there for a given corner? How many corners are there on the cube? Also, how do the orientations of pieces relate to each other? For example, can all 4 LL corners be twisted once clockwise away from "correctly oriented"? Lastly, how many of those possible states do you consider to be the solved state?

Chris

Haha, ok. 3 differnet orientations. 4 corners. The sum of all of the orientations adds up to a multiple of 3 if counter clockwise equals 2 and clockwise equals 1. The solved state has all corners oriented correctly, with the U layer colour up.

Thanks for the quick reply, Chris.

 

[KubeKid73]

I have a cube with CubeSmith tiles on it with the white V-Cube color scheme and I want to put the V logo on the center of the black face. I tried using white out but it looks horrible. How can I make a good V on it? And I don't want to cut the tile.

Bump my question. Also:
I want to make a magnetic dice cube. http://www.instructables.com/id/Magnetic-Rubik_s-Dice-Cube/ And it says I need a drill press (the kind of drill that's mounted and you just push the drill bit down into what you want to drill.) but I don't have one, so do you think I would be able to just use a regular power drill, or would it be too hard and not accurate enough?

And:
Are either of these QJ brand? http://www.dealextreme.com/details.dx/sku.16332 http://www.dealextreme.com/details.dx/sku.21302
If not, do I need a PayPal account to order from DP. The way they say to set up the email it says to give your PayPal email address, but I don't have a PayPal account.

Last one:
QJ or Meffert's Megaminx? I will choose which one to get by the first response.

 

[miniGOINGS]

Ok, but actually, what is the chance of a CO skip after Petrus or ZZ F2L?

 

[cmhardw]

Ok, but actually, what is the chance of a CO skip after Petrus or ZZ F2L?

Do Petrus or ZZ F2L influence corner orientation on the last layer at all? This of course assumes, like you said, that no additional special techniques are applied other than the goals of the methods themselves.

Chris

 

[waffle=ijm]

Ok, but actually, what is the chance of a CO skip after Petrus or ZZ F2L?

Does Petrus or ZZ F2L influence corner orientation on the last layer at all? This of course assumes, like you said, that no additional special techniques are applied other than the goals of the methods themselves.

Chris

when you use EJF2L, then yes

 

[miniGOINGS]

Ok, but actually, what is the chance of a CO skip after Petrus or ZZ F2L?

Does Petrus or ZZ F2L influence corner orientation on the last layer at all?

Chris

No, none. Let's put it this way. The LL stickers somehow become invisible during F2L untill F2L is solved. Then you have a _ chance of the corners being correctly oriented.

 

[cmhardw]

No, none. Let's put it this way. The LL stickers somehow become invisible during F2L untill F2L is solved. Then you have a _ chance of the corners being correctly oriented.

I think you misunderstand what I'm trying to do. I completely refuse to answer your question, because it is so fundamentally related to the laws of the cube that you should already know how to figure this out. If you don't know how to figure this out, please read this.

Chris

 

[iSpinz]

I have a cube with CubeSmith tiles on it with the white V-Cube color scheme and I want to put the V logo on the center of the black face. I tried using white out but it looks horrible. How can I make a good V on it? And I don't want to cut the tile.

Bump my question. Also:
I want to make a magnetic dice cube. http://www.instructables.com/id/Magnetic-Rubik_s-Dice-Cube/ And it says I need a drill press (the kind of drill that's mounted and you just push the drill bit down into what you want to drill.) but I don't have one, so do you think I would be able to just use a regular power drill, or would it be too hard and not accurate enough?

And:
Are either of these QJ brand? http://www.dealextreme.com/details.dx/sku.16332 http://www.dealextreme.com/details.dx/sku.21302
If not, do I need a PayPal account to order from DP. The way they say to set up the email it says to give your PayPal email address, but I don't have a PayPal account.

Last one:
QJ or Meffert's Megaminx? I will choose which one to get by the first response.

The second one is QJ. Get it. It's awesome. Period.

 

[miniGOINGS]

No, none. Let's put it this way. The LL stickers somehow become invisible during F2L untill F2L is solved. Then you have a _ chance of the corners being correctly oriented.

I think you misunderstand what I'm trying to do. I completely refuse to answer your question, because it is so fundamentally related to the laws of the cube that you should already know how to figure this out. If you don't know how to figure this out, please read this.

Chris

...wow. Just realized how simply this is. It's like so easy. I'm so stupid, how did I not get that.

 

[cmhardw]

...wow. Just realized how simply this is. It's like so easy. I'm so stupid, how did I not get that.

Quoting from the site link I sent you:

Seven [of the corners] can be oriented independently, and the orientation of the eighth depends on the preceding seven, giving 3^7 (2,187) possibilities.

That counts how many orientations for all corners. Now, how would you apply this to only the last layer corners?

Chris

 

[miniGOINGS]

...wow. Just realized how simply this is. It's like so easy. I'm so stupid, how did I not get that.

Quoting from the site link I sent you:

Seven [of the corners] can be oriented independently, and the orientation of the eighth depends on the preceding seven, giving 3^7 (2,187) possibilities.

That counts how many orientations for all corners. Now, how would you apply this to only the last layer corners?

Chris

If 4 of the corners are solved, 3 can be oriented any way, but the last one is dependant on the others. The first one has 3 orientations, the second has 3, the third has 3 and the last only has 1 because it is independant. Three cubed is 27. Only one of the cases is solved, this gives 1/27 or ~3.7% the all of the corners are oriented correctly.

 

[cmhardw]

If 4 of the corners are solved, 3 can be oriented any way, but the last one is dependant on the others. The first one has 3 orientations, the second has 3, the third has 3 and the last only has 1 because it is independant. Three cubed is 27. Only one of the cases is solved, this gives 1/27 or ~3.7% the all of the corners are oriented correctly.

And you answered your own question You can use that same idea for any related question about odds of skipping certain steps when looking at different kinds of methods.

Chris

 

[miniGOINGS]

If 4 of the corners are solved, 3 can be oriented any way, but the last one is dependant on the others. The first one has 3 orientations, the second has 3, the third has 3 and the last only has 1 because it is independant. Three cubed is 27. Only one of the cases is solved, this gives 1/27 or ~3.7% the all of the corners are oriented correctly.

And you answered your own question You can use that same idea for any related question about odds of skipping certain steps when looking at different kinds of methods.

Chris

Wow, I thought you were being a complete jerk, but now I realize that you wanted to help me understand and be able to answer questions I may have in the future. Thank you.

 

[cmhardw]

Wow, I thought you were being a complete jerk, but now I realize that you wanted to help me understand and be able to answer questions I may have in the future. Thank you.

No, I was just trying to show you that you already knew how to do this, even if you didn't know that you knew.

So to test you, the probability of an OLL skip after finishing Fridrich F2L (assuming no special techniques are used other than solving a standard cross and four F2L pairs) is 1/216. Using the same ideas as from your corners question, prove this is true.

If you can do this problem, you can handle many different kinds of method questions you might have.

Chris

 

[miniGOINGS]

Wow, I thought you were being a complete jerk, but now I realize that you wanted to help me understand and be able to answer questions I may have in the future. Thank you.

No, I was just trying to show you that you already knew how to do this, even if you didn't know that you knew.

So to test you, the probability of an OLL skip after finishing Fridrich F2L (assuming no special techniques are used other than solving a standard cross and four F2L pairs) are 1/216. Using the same ideas as from your corners question, prove this is true.

If you can do this problem, you can handle many different kinds of method questions you might have.

Chris

We know that the chance of a corner orientation skip is 1/27. Three of the edges can be oriented in any way, but the fourth is fixed, depending on the other three. If each edge has 2 different orientations, 2 cubed equals 8. Eight edge orientation cases times 27 corner orientations equals 216.

 

[cmhardw]

We know that the chance of a corner orientation skip is 1/27. Three of the edges can be oriented in any way, but the fourth is fixed, depending on the other three. If each edge has 2 different orientations, 2 cubed equals 8. Eight edge orientation cases times 27 corner orientations equals 216.

Yep

 

[miniGOINGS]

We know that the chance of a corner orientation skip is 1/27. Three of the edges can be oriented in any way, but the fourth is fixed, depending on the other three. If each edge has 2 different orientations, 2 cubed equals 8. Eight edge orientation cases times 27 corner orientations equals 216.

Yep

You just gained a whole lot of respect from me. The reason for my question, I was wondering if I should learn PLL so that if I ever got a corner orientation skip, I would have a 1LLL solve with ZZ. Thanks.

 

[cmhardw]

You just gained a whole lot of respect from me. The reason for my question, I was wondering if I should learn PLL so that if I ever got a corner orientation skip, I would have a 1LLL solve with ZZ. Thanks.

If you think it's worth it, 1/27 chance is not too terrible. Plus full PLL is only 21 algs, which is not too bad considering the chance of being able to use one of those cases.

To give you an idea, if you do an average of 5 in a competition you have a
1-(26/27)^5 = 17% chance approximately of getting at least one corner orientation skip in your average. That's if you use no special techniques. If you work on partial corner orientation control, your chances obviously get better.

Chris