What Constitutes Lucky in BLD?

[Leo]

Well I've been wondering this for awhile. What constitutes a "lucky" case in blindfold cubing? I mean other than the obvious whole step skip, such as for three cycle, no edge orientation needed. But what else?

 

[joey]

"For blindfold solving a lucky case is defined by:
1) more than 5 corners are correctly oriented, or
2) more than 8 edges are correctly oriented, or
3) more than 3 corners are correctly positioned, or
4) more than 4 edges are correctly positioned."

Taken from speedcubing.com
But this doesn't all make sense for non orient methods.
So I just use:
"1) more than 3 corners are correctly positioned, or
2) more than 4 edges are correctly positioned."

But then does a piece which is in the correct place but wrong orientation count here? I don't think so, because I have to remember to flip it, which is not a primary part of my method.

I think we should try set some non-method specific rules, which could be difficult.

 

[Stefan]

I don't understand the separation of corners and edges. You can have 7 pieces completely solved and not call it lucky? That's more than a third of the puzzle! I noticed Rowe's "non-lucky" unofficial record had six pieces solved, and I think that's quite lucky.

Edit: Um, actually I don't understand Rowe's notation. Looks like an 8-cycle of edges and two 3-cyles of corners which is impossible.

 

[joey]

I wonder if it would be better to give two definitions, depending on wether orientation is used or not?

For someone using COCPEOEP, a skip of CO and EO, would definitely be lucky, but for me who doesn't orient, is that lucky? It *would* be an easy solve, but since they are not parts to my method, is it lucky?

If we said soemthing like, more than 4 pieces solved is lucky. What if a piece was correctly permuted, but incorrectly oriented? Does that piece come under "solved" pieces?

 

[Stefan]

A general rule could be something like "Luck is when you have 20% fewer steps than average" and then that could be translated to each method.

 

[Mike Hughey]

That would be a nice rule (although sometimes it could be tough to calculate). I had a 4x4x4 BLD yesterday that I decided was lucky because there were only 3 corners permuted (and 3 misoriented). But if I counted every 3-cycle that I do, I would imagine the total on that solve was really pretty average, since the centers weren't all that great.

But "lucky" is awfully subjective anyway.

 

[joey]

But "lucky" is awfully subjective anyway.

Well, I guess thats what we are trying to do. Make "lucky" non-subjective, by defining a simple set of rules.

"Easy" on the other hand, will always be subjective. For example, yesterday me and alex yu were racing BLD. We came to a scramble, of which I got a low 1-min, but he had a realtively high DNF. It was because I found the scramble ridiculously easy, but because he has a different memo method and excution for him the scramble was very hard. I don't think its possible to really create equally fair scrambles.

 

[alexc]

But "lucky" is awfully subjective anyway.

Well, I guess thats what we are trying to do. Make "lucky" non-subjective, by defining a simple set of rules.

"Easy" on the other hand, will always be subjective. For example, yesterday me and alex yu were racing BLD. We came to a scramble, of which I got a low 1-min, but he had a realtively high DNF. It was because I found the scramble ridiculously easy, but because he has a different memo method and excution for him the scramble was very hard. I don't think its possible to really create equally fair scrambles.

Yes, I agree with you. A major part of whether a solve is easy or not is by your memo and execution method. Memo because a solve with many two cycles is generally easier to do visually than with images. (at least I find that) Execution obviously because of different buffers, whether you orient first, etc.

 

[hait2]

to answer thread title

this:
http://upload.wikimedia.org/wikipedia/commons/6/61/Rubiks_cube_solved.jpg