Kirjava
Colourful
I've always executed this as [U'RU2r'U,L2]
I've always executed this as [U'RU2r'U,L2]
antoineccantin
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Random super easy recog 1LLL case:
F D R' U' R D' R2 U R' U' R2 U F'
So F J-perm F'. Pretty fast (I'm sure there's something better though).
F D R' U' R D' R2 U R' U' R2 U F'
So F J-perm F'. Pretty fast (I'm sure there's something better though).
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Julian
Veteran
Yup, JJ and 4te taught me that a while ago.
googlebleh
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Schmidt
Veteran
Random conspiracy therory: The perfect SS 4x4x4 has already been made. They just make minor adjustments for each new version, so we have to buy them all, instead of releasing the ultimate cube!
Ranzha
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Julian
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maxcube
Regular
Gotta love almost setting a new PB solve
... that +2 ;~;
... that +2 ;~;
Ninja Storm
Veteran
This is why you get your hands on a v3 and never buy another .-.
cmhardw
Premium Member
Expected number of solved edge and tredge groups (4x4 and 5x5)
4x4x4 speedsolving:
When solving 4x4x4 using standard reduction (not Yau), at the beginning of the edge pairing step you are expected to have \( \frac{12}{23} \) of an edge group already paired up. Basically this means that about half of your solves you will have no edge groups already paired up, and about half your solves you will have one edge group already paired up (expected value). Proof.
5x5x5 speedsolving:
When solving 5x5x5 using standard reduction, at the beginning of the edge pairing step you are expected to have \( \frac{1}{46} \) of a tredge group already paired up. Basically this means that about 1 in every 46 solves there will be one tredge group already paired up, the rest of your solves will have no tredge groups already paired up (expected value). Proof.
4x4x4 speedsolving:
When solving 4x4x4 using standard reduction (not Yau), at the beginning of the edge pairing step you are expected to have \( \frac{12}{23} \) of an edge group already paired up. Basically this means that about half of your solves you will have no edge groups already paired up, and about half your solves you will have one edge group already paired up (expected value). Proof.
5x5x5 speedsolving:
When solving 5x5x5 using standard reduction, at the beginning of the edge pairing step you are expected to have \( \frac{1}{46} \) of a tredge group already paired up. Basically this means that about 1 in every 46 solves there will be one tredge group already paired up, the rest of your solves will have no tredge groups already paired up (expected value). Proof.
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rj
Veteran
I dual between CFOP and Roux, depending on the scramble. Is that normal?
aceofspades98
Regular
That is method neutrality. Not very common, but it has been used before.
kcl
Legend
I used to when I was close in speed between roux and CFOP. Now I'm twice as fast with CFOP so roux is just for fun.
rj
Veteran
OK, thanks!
AvGalen
Premium Member
How about semi-pairs on 5x5x5? (meaning 1 center edge has 1 outer edge attached)
In my experience this happens about half of the solves on 5x5x5 but because I save semi-pairs during general center-solving and final centers I only have to solve roughly 20 out of 24 semi-pairs
cmhardw
Premium Member
At the moment I'm not exactly sure how I would calculate that. I'm interested to look into this more, and I hope my math skills are up to the challenge
Thanks Arnaud! This will be a fun math problem to puzzle over for a few days! If I figure this out I'll certainly post about it.
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Thanks for asking this - I was about to ask the same thing.
And Chris, thanks for doing those calculations - I've always wondered what the numbers were for that.
cmhardw
Premium Member
Thinking about this more, I will count semi-pairs as strictly semi-pairs only. Basically I will count the expected value of having a midge and exactly one wing solved next to it. I will not count tredges in this calculation at all. I was unsure before how to calculate this as I wasn't sure how to handle fully solved tredges. The easy answer is I won't, and I should (hopefully) be able to calculate this tonight after work
You're welcome Mike, and I too am curious about these numbers!--edit--
I just realized that the 5x5x5 wing parity case where 10 tredges are solved and 2 tredges are each a semi pair with one wing swapped with the other semi-pair will count as a "2 semi-pair" case. There's nothing really I can do about this, unless you guys want me to somehow treat solved tredges as 2 semi-pairs. I don't really know how to handle that calculation yet (treating solved tredges as 2 semi-pairs). I will calculate the expected number of true semi-pairs later tonight, even though this will count the wing parity case as in the same class of cases where wings and midges are totally scrambled, with only two semi-pair edges created. I'm still curious of that result, it will at least work as a good estimate for semi-pairs even though it doesn't factor in that solved tredges are "better" than scrambled tredges.
tl;dr
I feel like we should count solved tredges somehow as "better" than scrambled tredges. I will try to look into counting solved tredges as 2 semi-pairs even though I currently don't know how to do that yet (maybe I do, but not sure).
--edit-- #2
Question: Is a solved tredge "better", "worse" or "same as" two semi-pairs in the strict sense of the term "semi-pair"?
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cmhardw
Premium Member
If a solved tredge is treated exactly the same as two semi-pairs in the strict sense of that term, then I know how to do this calculation, and can have it done by tonight.
I am still curious to know how 5x5x5 solvers view a solved tredge vs. two semi-pairs. If 5x5x5 solvers view them as not the same, then the calculation I run tonight will function as a great estimate, but it will not really model how solvers use the reduction method.
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