# What's the most optimal rotation to get to the least # of bad edges?



## LexCubing (Mar 21, 2017)

The definition of a good edge changes when you rotate. Some edges remain unaffected when rotating and some are.

For n bad edges what's the most optimal rotation to get the least # of bad edges?


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## obelisk477 (Mar 21, 2017)

LexCubing said:


> The definition of a good edge changes when you rotate. Some edges remain unaffected when rotating and some are.
> 
> For n bad edges what's the most optimal rotation to get the least # of bad edges?



Are you referring to F2L or ZZ? The answer may be simplified if it's one or the other.


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## xyzzy (Mar 21, 2017)

n bad edges _where_? And what are the colours of the bad edges?


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## shadowslice e (Mar 21, 2017)

The cube is symmetrical so each orientation has an equal chance of getting a certain number of bag edges.


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## mDiPalma (Mar 21, 2017)

edges that are in their desired slice will not change orientations if the centers are shifted in a way that emulates a rotation of that slice


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## Teoidus (Mar 21, 2017)

To answer question directly: no rotation at all will give you "optimal" number of bad edges, as expected value is 6 for every orientation and any fixed rotation, on average, won't change anything. (Sometimes rotating will give you a better case than not rotating, sometimes it'll give you a worse case... these will balance out as sample size increases)

More interestingly, you can look at how advantageous it is to use 1 fixed front vs 2 fronts (e.g. y neutral) vs 3 (full neutral) assuming that you always pick the front with fewest number of bad edges. iirc you get severely diminishing returns and 3 almost provides no additional benefit


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## AlphaSheep (Mar 22, 2017)

The way I understand the question is this:
Given that there are n bad edges on the first orientation you inspected and given no further information, then is there a rule that lets you determine whether one of the remaining two orientations is going to be better without having to actually inspect them?

The answer is no. If you have n bad edges, and are choosing between an x vs y rotation, then the probability distribution for the number of bad edges on the other two orientations is identical.

There are however rules you can use if you take into account the position and type of the bad edges, which @mDiPalma explained quite succinctly. Basically, a y rotation will only change the orientation of edges that belong in the E slice if they are in the U/D layers, and edges that are in the E slice that belong in the U/D layers. I used to use this when I was still bothering with y neutrality to save having to inspect both sides. You can apply the same idea to x rotations and the M slice.


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