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Random Cubing Discussion

cmhardw

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I feel this warrants a double post. If people don't agree then either I or another mod can merge the two.

5x5x5 expected number of semi-pairs:
When solving 5x5x5, assuming you have no preference between a solved tredge vs. two semi-pairs, then you are expected to have 1 semi-paired edge at the start of edge pairing when solving using the reduction method. This also assumes that you use no techniques to either create or preserve semi-pairs during the building of centers. Proof.
 

Mike Hughey

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Cool. Would it be stretching too much to ask what are the odds that you'll have at least 1 semi-paired edge?

For AvG solving (like Arnaud and I use), a solved tredge is in a way disadvantageous vs. two semi-pairs, because then we don't have an "easy" place to start. Generally I do like Erik Akkersdijk says he does - I just go ahead and pick a random edge and solve it if I don't find a semi-pair immediately, so I don't really find it all that disadvantageous most of the time. The problem is if I pick a random edge and it is immediately a "problem edge", but that happens rather rarely.

For freeslice solving, I'd think a solved tredge would probably be advantageous vs. two semi-pairs. Interesting how the two methods are kind of opposite on this - I never really thought about that before.
 

cmhardw

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Cool. Would it be stretching too much to ask what are the odds that you'll have at least 1 semi-paired edge?

Nope :)

The probability of having at least one semi-pair is:
\( 1-\frac{1}{24!}\sum_{i=0}^{24} \left( \begin{array}{ccc} 24 \\ i \end{array}\right)(24-i)!(-1)^i\approx0.632121 \)

There is an approximately \( \frac{100(e-1)}{e}\% \) chance that you will have at least one semi-pair ;)

For AvG solving (like Arnaud and I use), a solved tredge is in a way disadvantageous vs. two semi-pairs, because then we don't have an "easy" place to start. Generally I do like Erik Akkersdijk says he does - I just go ahead and pick a random edge and solve it if I don't find a semi-pair immediately, so I don't really find it all that disadvantageous most of the time. The problem is if I pick a random edge and it is immediately a "problem edge", but that happens rather rarely.

For freeslice solving, I'd think a solved tredge would probably be advantageous vs. two semi-pairs. Interesting how the two methods are kind of opposite on this - I never really thought about that before.

If you or any 5x5x5 solvers have input on how the two cases compare, that is how a solved tredge compares to two semi-pairs, I can try to calculate a form of expected value for number of (semi-pairs + tredges). I would need to know something like "a tredge is 1.5 semi-pair to me" or "a tredge is 0.8 of a semi-pair to me". I still can't guarantee that I know exactly how to calculate that kind of a weighted average, but I have some ideas.
 

qqwref

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In terms of moves, a tredge is worth two semi-pairs to me, but in terms of time it might be a little less since it can confuse me a bit (I think I have 5 edges left but there are only 4). Not sure how to calculate the impact of that though.
 
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At odds of 1 in 373,248 to 1 I got 3 (Nb) perms in a row.

OK, maybe not THAT unlikely as I handscrambled so I may influence the PLL, but that is insane.

And I've only ever had one LL skip...
 

uberCuber

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OK, maybe not THAT unlikely as I handscrambled so I may influence the PLL, but that is insane.

Unless your F2L solutions were absurdly short, I wouldn't say the handscrambling would affect the very last step of the solve, since there would have been numerous different ways to solve your scrambles, most of which would not have resulted in the Nb.
 
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I did DNF one solve when I aw the N perm because it annoyed me xD, then started scrambling from the Nb position, but I doubt that would have much of an effect
 

cmhardw

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Tonight I tried doing 5x5x5 double turns only, and it was really hard! It's a lot like 3x3x3 double turns only, but with the element of trying to use reduction too. You really have to use order of a permutation to figure out ways to cycle small numbers of t-centers or to fix centers. It probably took me 20 minutes to figure it out, but now I'm really hooked and want to try this again!
 

Mollerz

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Tonight I tried doing 5x5x5 double turns only, and it was really hard! It's a lot like 3x3x3 double turns only, but with the element of trying to use reduction too. You really have to use order of a permutation to figure out ways to cycle small numbers of t-centers or to fix centers. It probably took me 20 minutes to figure it out, but now I'm really hooked and want to try this again!
That sounds like something we should do at a UK comp. brb
 

Cubetastic

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I just made a set of algorithms for what I call Partial Winter Variation

Its exactly like winter variation but instead of orienting the corners and edges with the algorithms while inserting the last f2l pair, mine only orients the edges so you only need to know 2-look oll.

It has a total of 14 algorithms (counting mirrors)

Is it worth implementing?
 

JonnyWhoopes

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I just made a set of algorithms for what I call Partial Winter Variation

Its exactly like winter variation but instead of orienting the corners and edges with the algorithms while inserting the last f2l pair, mine only orients the edges so you only need to know 2-look oll.

It has a total of 14 algorithms (counting mirrors)

Is it worth implementing?

First off, I think you misunderstand what Winter Variation actually is.

Second off, you aren't the first to come up with the idea. Others have tried it and have found it to have minimal value. Look up both ZBF2L and VH.
 

cubernya

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I just made a set of algorithms for what I call Partial Winter Variation

Its exactly like winter variation but instead of orienting the corners and edges with the algorithms while inserting the last f2l pair, mine only orients the edges so you only need to know 2-look oll.

It has a total of 14 algorithms (counting mirrors)

Is it worth implementing?

VHLS
 

AvGalen

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Cool. Would it be stretching too much to ask what are the odds that you'll have at least 1 semi-paired edge?

For AvG solving (like Arnaud and I use), a solved tredge is in a way disadvantageous vs. two semi-pairs, because then we don't have an "easy" place to start. Generally I do like Erik Akkersdijk says he does - I just go ahead and pick a random edge and solve it if I don't find a semi-pair immediately, so I don't really find it all that disadvantageous most of the time. The problem is if I pick a random edge and it is immediately a "problem edge", but that happens rather rarely.

For freeslice solving, I'd think a solved tredge would probably be advantageous vs. two semi-pairs. Interesting how the two methods are kind of opposite on this - I never really thought about that before.

I missed these posts, but just wanted to let you and Chris know that I read and appreciated it.
Technically a tredge is simply 2 semi-pairs, but for "bigcubes method" it is a big advantage to have a tredge while for AvG-pairing I would much prefer to have 2 semi-pairs instead of 1 tredge. It also seems to me that the way I do semi-pairs I benefit a lot from "short cycles" where I get "a free semi-pair" every time. I don't think there is an easy way of putting a general weight on such cases that can be put into an easy formula
 

Lchu613

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If someone does a good job of swapping two opposite colors on your solved cube, then it is impossible for you to tell which colors they swapped. Cool!

My friend trolled me, he made a tower of cubes and told me my cube (which was in the middle) had the orange and red stickered backwards. Little did I know, the cube was upside-down.
 
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