# My F2L study - Results



## Stefan (Jan 4, 2008)

I'm not finished with the analysis, but you can see a first version here:
http://stefan-pochmann.info/misc/F2Lstudy/

masterofthebass, CorwinShiu, cmhardw, badmephisto, Erik, isaacthecuber - You're not yet included because at least one of your solutions don't work. Please check them and if you can, fix them and PM me when you're done.

Lucas Garron's solves are longer because he actually orients the last layer during the end of F2L. His start is Fridrich-like, though, so his solves fit my purpose of only having very similar solves/solvers in this first test.

You can see from the CSL columns that people went different paths very early, and at latest shortly after the cross no to solves remain on the same path.

The two big questions I had in mind when I planned this:

*1) How meaningful are the terms "lucky scramble" or "easy scramble"?
2) How important is it that we get the same scrambles in competition rounds?*

My analysis is supposed to offer some data to help answer these questions. Something I didn't have in mind at first is to compare the length/fingerfriendliness of the solves. I decided to do this when I saw that Harris had so very short solutions.

More to come...


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## AvGalen (Jan 5, 2008)

Very strange that nobody actually solved the F2L the same way. Especially if you take into account that Guus and I have found an idenctial Fewest Move solution once.

I didn't expect everyone to find the same cross, but I was expecting at least some solves to be the same.


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## Stefan (Jan 5, 2008)

I could imagine a lot of people think so. At least all those who start a thread like "Easiest scramble ever you must try this". Unless every step is really very obvious and somewhat forces you to do specific moves, I think it's unlikely that two people come up with the same solution.



AvGalen said:


> Guus and I have found an idenctial Fewest Move solution once.


That's one incidence among how many possible? Many possible pairs of cubers times many FMC attempts. So while it is surprising, you've probably also had a lot of non-identical solves that you just didn't count. With more attempts, maybe we'll also come across two identical F2L speedsolves.


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## Stefan (Jan 5, 2008)

I just noticed that for all three scrambles, Jason Baum was involved in the pair of solves with the longest common start. So for now I reward him the nonserious title of "most common cuber".


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## watermelon (Jan 5, 2008)

Stefan, if possible, could you change my name from "watermelon" to "Mitchell Stern" on your F2L study page?

As for the page itself, the results are interesting, but not surprising. Every cuber I have met has his/her own style (not necessarily just fingertricks, but what that cuber looks for during a solve, what that cuber sees first, etc.). After looking at the page, I notice that my solves are among the least ugly, but also some of the longest. I guess I should work on optimizing my F2L .

Edit: After looking through some of the other solves, it seems that quite a few of the solutions which were shorter than mine simply got easier pairs and cases. Aside from a few simple F2L tricks and slightly shorter crosses, no one seemed to be doing anything out of the ordinary...


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## Johannes91 (Jan 5, 2008)

```
CSL    Normalized solve
1      F2 L2 B D B D2 ...
5      F2 L2 B D B U2 D2 ...
```
Should that CSL be counted as 6?


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## Kenneth (Jan 5, 2008)

StefanPochmann said:


> *1) How meaningful are the terms "lucky scramble" or "easy scramble"?
> 2) How important is it that we get the same scrambles in competition rounds?*
> 
> ..............
> ...



You must in the cases of "easy scrambles" also consider the fact that those often has got obvious situations makeing the cubers choose the same path - like a four turn double x-cross and such. That does not happen equally often if you take any random scramble.


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## Pedro (Jan 5, 2008)

damn...I have some of the longer solutions 

maybe I should work more on my cross...


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## Kenneth (Jan 5, 2008)

Average number of turns for F2L is 33.83

I did not count Lucas' solves because of his orientation style.

Most people says Fridrich F2L is 32-33 turns on average. Here we are close to 34 and this is the elite (small sample doe), the average cuber is probably around 38 or something.


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## Lt-UnReaL (Jan 5, 2008)

I'm not sure why Lucas added his OLL and PLL to the solves, and also not sure why you put added them into your F2L study.


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## Stefan (Jan 6, 2008)

Updated version with more solves:
http://stefan-pochmann.info/misc/F2Lstudy/



Johannes91 said:


> ```
> CSL    Normalized solve
> 1      F2 L2 B D B D2 ...
> 5      F2 L2 B D B U2 D2 ...
> ...


Yes. Actually there are more problems and since it's not that bad and I don't see a way to fix it properly, I'll leave it as it is (also see new explanation on the new page).



Kenneth said:


> You must in the cases of "easy scrambles" also consider the fact that those often has got obvious situations makeing the cubers choose the same path - like a four turn double x-cross and such. That does not happen equally often if you take any random scramble.


Good point. Maybe I'll include an "easy scramble" in a future version.


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## badmephisto (Jan 6, 2008)

nice thx for updated version. You collected some nice data there.
the CSL problem is pretty interesting, you should indeed treat
D U L and 
U D L as CSL = 3.
I don't really see a quick way of solving it though, and as you point out there could be some problems. Maybe some kind of a finite state machine for the implementation would do the trick? I don't like the argument of it "not happening too often" though


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## Dyste (Jan 6, 2008)

Hm.., what if fewest move F2L became part of the weekly competition?


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## Johannes91 (Jan 6, 2008)

badmephisto said:


> the CSL problem is pretty interesting, you should indeed treat
> D U L and
> U D L as CSL = 3.


There can't be such bugs in the normalized solves, because when two adjacent moves are U and D (or F and B, or L and R), they always come in the same order. I agree with Stefan that it's fine to keep the count as it is, because there can't be any big errors.

@Stefan:
You could add a "cursor: pointer;" rule to make ".fakelink"s look more like real links.
Why _x_7 instead of _x_3? (just curious)


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## Lucas Garron (Jan 6, 2008)

I don't like the speedsolving server; it has a downtime of, like, 1/3 for me. Thus, my message never got posted, and I have to retype dstuff from 2 days ago:

1. Before you hand out the title of "ugliest solve": For #3, I solved cross on the face that starts out as U (you said to scramble with cross color on D, but I didn't interpret that to use the D color for cross...). Should I redo?

2. Can you count "smoothness" in the solves? R'U and RU' are smooth, RU and R'U' are not. The only thing is, half turns complicate this... Could you count the ratio of adjacent smooth quarter turns to non-smooth?


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## badmephisto (Jan 6, 2008)

Johannes91 said:


> badmephisto said:
> 
> 
> > the CSL problem is pretty interesting, you should indeed treat
> ...



ok, if they come in the same order that changes things


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## Kenneth (Jan 6, 2008)

Lucas Garron said:


> ..... half turns complicate this...



Not if the notation U2 and U2' are used. If I found it is better to do U2' than U2 in an alg I sometimes also notate it like that.

When it comes to metrics I have thougth of a model where RU' is = 1.5 and RU is = 2.5 because of the smoothness.


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## Stefan (Jan 6, 2008)

Johannes91 said:


> I agree with Stefan that it's fine to keep the count as it is, because there can't be any big errors.


I could improve it, maybe I will if I have more time. Idea is to sort the solves not alphabetically (B<D<F<L<R<U) but axiswise (e.g. U<D<L<R<F<B). Then the most similar solves should really be next to each other, and then I could treat your exception with special code.



Johannes91 said:


> Why _x_7 instead of _x_3? (just curious)


Because I sometimes actually do U3, but never U7. So for showing the raw original solves, if someone actually does and writes U3, I don't want to turn it into U'. Ideally Perl would have a general encode function and JavaScript a matching decode function...


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## Stefan (Jan 6, 2008)

Lucas Garron said:


> 1. Before you hand out the title of "ugliest solve": For #3, I solved cross on the face that starts out as U (you said to scramble with cross color on D, but I didn't interpret that to use the D color for cross...). Should I redo?


I'm going to show a second table for each scramble, showing the raw original solves instead of the normalized ones, so that a better "ugliness" can be computed for that. The normalized solves were really only intended for the CSL values, and I added the "ugly" column as an afterthought. I'll mention that it's not always meaningful.



Lucas Garron said:


> 2. Can you count "smoothness" in the solves? R'U and RU' are smooth, RU and R'U' are not. The only thing is, half turns complicate this... Could you count the ratio of adjacent smooth quarter turns to non-smooth?


Good suggestion, thanks. I'll do that.

If you guys have more suggestions for interesting statistics, let me know.


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## Kenneth (Jan 6, 2008)

Average number of turns for the whole sample would be nice (maybe with and also without Lucas' OLL solves). I can calculate it manually but why should I when you are a master programmer? I like automation =)


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## Tim Reynolds (Jan 6, 2008)

Lucas Garron said:


> 2. Can you count "smoothness" in the solves? R'U and RU' are smooth, RU and R'U' are not. The only thing is, half turns complicate this... Could you count the ratio of adjacent smooth quarter turns to non-smooth?



One note for that is that R U2 R' is either RU UR' or RU' U'R', both of which are not smooth in one place. I guess that doesn't work for R U2 R, but you could just assume that that's R U2' R and smooth. So if it's A X2 B, if A and B are the same way, it's smooth, otherwise, it's not smooth once.


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## philkt731 (Jan 6, 2008)

and also, if someone did R' F R F', which could be just as fast as U R U' R', the ugliness would go up a lot even though it is actually quite nice


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## Dene (Jan 7, 2008)

Ok, what about getting people to do the same scramble, but from a different side, so they have to use a different cross, and see if there are similarities between solves of the same scramble despite using a different cross?


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## Swordsman Kirby (Jan 7, 2008)

philkt731 said:


> and also, if someone did R' F R F', which could be just as fast as U R U' R', the ugliness would go up a lot even though it is actually quite nice



Yes, in certain cases, some are faster for me. For example: R U R' U R U' R' is slower for me than R U' R2' F R F'.


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## pjk (Jan 8, 2008)

Stefan,
I replied to your PM, not sure if you got it or just haven't had time. I edited my third solution in my initial post, so it should be correct now.

Thanks


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## Stefan (Jan 9, 2008)

I just didn't have the time. Might continue tomorrow.


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