Maximum Number of Moves for Oka (Pyraminx) 1st Step

riffz

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So I have 2 related questions:

What is the maximum number of moves to solve the first 2 edges if you choose ANY 2 to solve?

What is the maximum number of moves to solve the first 2 edges if you choose the best 2 to start with?
 
so essentially CN vs. non-CN?

Oops, no. I just reread the question.

It's like, roll a die/flip a coin to choose which edges to solve.

Basically, what is the optimal number of moves to solve the worst 2 edges. For example, in Fridrich the worst cross(es) is (are) 8 moves.
 
Sorry, I'm still a bit confused, what is the difference between the two questions? One is the average move count for the block of Oka method with CN, the second is without CN?
 
Sorry, I'm still a bit confused, what is the difference between the two questions? One is the average move count for the block of Oka method with CN, the second is without CN?

No. He is simply asking how many moves it would take on average for the first block of oka. For example, cride5 posted an average of how many moves he takes to do EOline. OP is asking for the same thing except for the first block of oka. It really isn't that difficult to understand lol.

EDIT: Ok I missed the second question :p. Ok here's what he's asking. The first one is how many moves would it take in order to solve the block even if it wasn't the most "lucky." I believe the second question means what would the move count if you picked the easiest one to solve.

At OP: Honestly I don't believe that you can calculate this. I'm sure you can calculate the average number of moves for the block in general. However "best" is subjective and you need to be more specific.
 
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Sorry, I'm still a bit confused, what is the difference between the two questions? One is the average move count for the block of Oka method with CN, the second is without CN?

Technically, he never says optimal, but we'll assume that's what he means...

1st. Maximum number of optimal moves to solve any 2 edges.

2nd. Maximum number of optimal moves to solve the best 2 edges.

Example:

All combinations of 2 edges can be solved from any state in at most _ number of moves.

From any scramble, 2 edges can be solved in at most _ number of moves.
 
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Instead of just the maximum, I'd be interested in seeing the entire distributions for fixed orientation, worst orientation, and best orientation (CN). Shouldn't be too hard to calculate given the number of total Pyraminx positions (under a million).
 
For the first question, an easier way of putting it is;

The maximum number of optimal moves needed to solve 2 edges (for example, red-green and red-yellow).
 
For the first question, an easier way of putting it is;

The maximum number of optimal moves needed to solve 2 edges (for example, red-green and red-yellow).

Correct.

And to clarify, the second question was meant to ask what is the maximum possible number of moves required to solve the 2 edges that will require the least moves optimally. This question is much harder to answer than the second one (I believe).

I'm not concerned as much with average move-count because I want to make sure I'm not using more moves than necessary to solve any situation.
 
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