How many of those 43 quintillion+ cases are 'scrambles'?

[IamWEB]

This idea suddenly came to my mind, but I have no idea how to figure it out. If anyone else that can would like to, be my guest.

How many of those total positions the 3x3x3 can be in have no F2L pairs done (for any color, and then how many just for 1 cross?), no more than 2 cross pieces correctly done (for any color, and then how many for 1 cross?), and things like that?

I'm curious about how this turns out. Even if I don't understand, seeing how finding the answer is achieved would be nice.

Anyone else care to know this?

 

[cmhardw]

If you aren't using computers, you could use inclusion-exclusion for finding a count of no pairs done. It would get tricky when looking at how many possible ways can four or more pairs be made, but it must surely be possible still with only pencil and paper. I'm actually interested to try this and see if I can come up with the same answer as those who will inevitably solve this problem much more quickly than I can

--edit--
This might be harder than I thought, because as you start to build 3x1x1 blocks, do you consider this a single pair, or two pairs built at once? What happens when a 3x1x1 shares a corner with a different 3x1x1 block?

Hmm..... This is an interesting problem. I'll have to give this a think.

Chris

 

[Ranzha]

I'm going to figure this out....
ARGH....

 

[IamWEB]

If you aren't using computers, you could use inclusion-exclusion for finding a count of no pairs done. It would get tricky when looking at how many possible ways can four or more pairs be made, but it must surely be possible still with only pencil and paper. I'm actually interested to try this and see if I can come up with the same answer as those who will inevitably solve this problem much more quickly than I can

--edit--
This might be harder than I thought, because as you start to build 3x1x1 blocks, do you consider this a single pair, or two pairs built at once? What happens when a 3x1x1 shares a corner with a different 3x1x1 block?

Hmm..... This is an interesting problem. I'll have to give this a think.

Chris

I would just exclude all blocks, regardless of how many pairs they count as. 1 pair at all and it's excluded.

 

[cmhardw]

--edit--
This might be harder than I thought, because as you start to build 3x1x1 blocks, do you consider this a single pair, or two pairs built at once? What happens when a 3x1x1 shares a corner with a different 3x1x1 block?

Hmm..... This is an interesting problem. I'll have to give this a think.

Chris

I would just exclude all blocks, regardless of how many pairs they count as. 1 pair at all and it's excluded.

Actually it's not really that weird to think this way. A solved cube is essentially 6 F2L's, each consisting of 4 corner/edge pairs. So a solved cube has 24 corner/edge pairs built. A 3x1x1 is always considered as a corner/edge pair built on two opposite corner colors with the same edge, so it's just two corner/edge pairs.

Yes inclusion/exclusion should work just fine as a method to figure out the answer to how many scrambled states have no pairs built. Actually applying it will be very tedious (but also interesting!), but not necessarily extremely difficult.

Chris

 

[daniel0731ex]

just exclude plls, ll, f2l, and cross and you are left with the scrambles

 

[rubiknewbie]

Quite close to 43 quintillion I guess.

 

[miniGOINGS]

Quite close to 43 quintillion I guess.

Ahem. Although that may be right, I think you should try to figure it out, just so you can prove yourself wrong.

 

[cmhardw]

Ok I had a chance to look at it, and although I still think inclusion/exclusion is a viable way to count the number of ways that no corner/edge pair is built, I am getting confused of the setup for the count.

Here are my thoughts so far:
I think it will be easiest to consider corners, and how many of them match with their edges. For example, I would start with the total number of cube states on the cube: 43,252,003,274,489,856,000

Then you will subtract the number of ways that 1 corner can match with exactly one of it's edges (permuting/orienting all other pieces randomly).

From here it gets confusing for me. I am thinking that you adjust the previous overcount by adding back the number of ways that 2 distinct corners can match exactly one of their respective edges. The problem is, where in the count do we consider cases where 1 corner matches exactly two of it's edges? How do we adjust the inclusion exclusion case counting structure to consider such cases? Also, the problem grows more complex when you think about having 4 corners match exactly one of their respective edges, or maybe 2 corners than each match exactly two of their respective edges.

This is an extremely interesting problem! Is there any way this can be done elegantly with pencil and paper, or is it likely that a computer analysis is required here?

Chris

 

[Tim Major]

Quite close to 43 quintillion I guess.

Ahem. Although that may be right, I think you should try to figure it out, just so you can prove yourself wrong.

I am guessing it is nowhere near 43 quintillion. By scrambled, do mean, a non-cuber would say it "looks scrambled" like, has no big blocks, or do you mean any cases that aren't solved. What would happen in a competition, if it turned out a scramble solved itself. Or had the first 2 layers solved? What would be the chances of this? I wish there were more threads like this, as they are the type of thread I like, instead of "do you think an avatar says something about the poster", or whatever that thread was. These are interesting, and I wish there was a cubing site I new of, that just discussed Rubik's cube theory, instead of that type of stupid thread.
Btw, if you mean scrambled, as in "not solved" then yes, it would be very close to 43 quintillion.

 

[cmhardw]

I am guessing it is nowhere near 43 quintillion. By scrambled, do mean, a non-cuber would say it "looks scrambled" like, has no big blocks, or do you mean any cases that aren't solved.

I am interpreting it as the following problem: How many possible positions are there where no corner on the cube matches correctly with any of its three adjacent edges?

Chris

 

[DavidWoner]

I am guessing it is nowhere near 43 quintillion. By scrambled, do mean, a non-cuber would say it "looks scrambled" like, has no big blocks, or do you mean any cases that aren't solved.

I am interpreting it as the following problem: How many possible positions are there where no corner on the cube matches correctly with any of its three adjacent edges?

Chris

And also have no corners or edges that are both permuted and oriented.

 

[Cangurino]

And also have no corners or edges that are both permuted and oriented.

It should be possible to figure out this part by itself. For the edge/corner matchups I have no idea yet.

 

[Tim Major]

That sounds like quite a small fraction. No pairs, and no edges/corners in the correct position. I am guessing less than half. Still, I'm not going to spend awhile thinking about it, though I may later when I'm off the computer.

 

[nigtv]

And also have no corners or edges that are both permuted and oriented.

It should be possible to figure out this part by itself.

That's what I'm thinking. I think I'll try to figure this out in my head before thinking of any of that other stuff...:confused:

 

[LNZ]

By definition, any state of the 3x3x3 cube that is not the solved state is called a scrambled state, regardless of how much it is scrambled.

This is important when working out god's algorithm for any puzzle toy.

The solved state is defined by a 3x3x3 that needs no face turns to solve.

To see this, look and solve a 1x1x1 cube. It needs no face rurns to solve and has therfore only 1 possible state, which is the solved state.

 

[IamWEB]

What I said (or meant if there was confusion) was referring to cases that:

Have no corner/edge pairs done and have no more than 2 edges correctly oriented and permuted for one layer (i.e. not an easy cross for that color).

I think the latter part will be harder to do...

 

[AvGalen]

What I said (or meant if there was confusion) was referring to cases that:

Have no corner/edge pairs done and have no more than 2 edges correctly oriented and permuted for one layer (i.e. not an easy cross for that color).

I think the latter part will be harder to do...

Your definition is really weird. I have had many scrambles had have a corner edge pair connected. Sometimes it is even done.

Your second situation of having 2 cross pieces already oriented and permuted (i.e. white-red and white-orange opposite on the white layer) or completely done (i.e. white-red edge and white-orange edge entirely solved) happens quite often and is a great help when you EO-line or cross. Especially when color neutral. But having two edges solved doesn't mean solving cross is always short, it just means an easier inspection so you can probably plan further than just the cross (x-cross?)

 

[Stefan]

An estimation for "How many states have no matched corner-edge pairs":

Edges alone are harmless. So all edge states are possible. Then each corner has 3 bad spots out of 24, where it would match an edge (or 1 or 2 bad spots, but not often). So a probability of 21/24 that the corner is at a good spot. And (21/24)^8 is about 34%. This individual treatment of pieces/pairs doesn't provide an accurate answer, but I think it gets close. So I estimate 34% of all cases to have no matched corner-edge pairs.

Edit: Reversing the roles of corners and edges leads to (22/24)^12, about 35%.

 

[cmhardw]

An estimation for "How many states have no matched corner-edge pairs":

Edges alone are harmless. So all edge states are possible. Then each corner has 3 bad spots out of 24, where it would match an edge (or 1 or 2 bad spots, but not often). So a probability of 21/24 that the corner is at a good spot. And (21/24)^8 is about 34%. This individual treatment of pieces/pairs doesn't provide an accurate answer, but I think it gets close. So I estimate 34% of all cases to have no matched corner-edge pairs.

Edit: Reversing the roles of corners and edges leads to (22/24)^12, about 35%.

Stefan, I use an extremely similar approach to estimating the expected number of solved x-centers and t-centers on 5x5x5 after the scramble for BLD (assuming you don't optimize an orientation and solve center points after). We've found it to be fairly accurate in practice, so I think your estimation is probably fairly accurate.

Any ideas on how to set up an inclusion-exclusion case structure/strategy/approach to figure out the exact count? If this would be too tedious to be practical, is there a way computers can aid in finding this count?

I think this question (# of cases with no corner edge pairs built) is a simple to pose question, but a challenging concept to apply!

Chris