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

[Anthony]

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.

Btw, if you mean scrambled, as in "not solved" then yes, it would be very close to 43 quintillion.

You're off by two hundred fifty-two quadrillion, three trillion, two hundred seventy-four billion, four hundred eighty-nine million, eight hundred fifty-five thousand, nine hundred ninety-nine.

 

[hawkmp4]

Yes, you're correct. But when talking about the cube, especially here, I think it's safe to say 43 quintillion is understood to be all the possible states of the cube.

 

[miniGOINGS]

Btw, if you mean scrambled, as in "not solved" then yes, it would be very close to 43 quintillion.

You're off by two hundred fifty-two quadrillion, three trillion, two hundred seventy-four billion, four hundred eighty-nine million, eight hundred fifty-five thousand, nine hundred ninety-nine.

 

[Cangurino]

A quick probabilistic simulation shows that 56% of all states have no piece in the right spot and oriented correctly.

 

[Lucas Garron]

A quick probabilistic simulation shows that 56% of all states have no piece in the right spot and oriented correctly.

Really?

A quick mathematical calculation shows me ≈0.606 for edges, and ≈0.513 for corners, about 0.311 together.

 

[Stefan]

A quick probabilistic simulation shows that 56% of all states have no piece in the right spot and oriented correctly.

A quick mathematical calculation shows me ≈0.606 for edges, and ≈0.513 for corners, about 0.311 together.

And I estimate (23/24)^20 ≈ 42.7%.

 

[meichenl]

The expected value of the number of corner/edge pairs in a random position (counting a 3x1x1 block as two pairs) is exactly one. This is because any of the 24 possible pairs has a chance 1/24 of occurring in a random scramble. Focusing, for definiteness, on the red/blue/white corner and red/blue edge, once the corner is placed, there is a 1/12 chance that the edge will go in the right slot to make a pair, and a 1/2 chance that once there, it will have the correct orientation.

So, if we sample n random cubes, then each corner edge pair will appear in roughly n/24 of them, and since there are 24 possible pairs, the expected number of corner/edge pairs is exactly one.

We can use this calculation to confirm Stefan's estimate earlier in the thread for the number of positions with no corner/edge pairs. If the corner/edge pairings were independent, the odds of no corner edge pairs would be

(23/24)^24 = .360,

which is not far from 1/e = lim n->inf [(n-1)/n]^n = .367

One way to calculate the exact number by hand, which would be rather tedious, is what Chris already suggested.

The number of positions with no corner/edge pairs is the total number of positions minus the number with at least one corner edge pair.

For a given corner edge pair, there are 1802166803103744000 positions that contain it (1/24 of the possible positions). Multiplying by 24, we get the 43 quintillion cases with at least one pair, except that we've double-counted anything that has two pairs, and triple counted anything with three pairs, etc.

So we need to find how many positions have two pairs. We could do this by breaking it down into three cases:

1. the two pairs have two corners and two edges
2. the two pairs have two corners and one edge
3. the two pairs have one corner and two edges

For each of these, it is not too difficult to count the total possibilities. For example, for two corners and one edge, we first note there are twelve such pairings. Each pairing has six free corners and eleven free edges, making
12*6!*11!/2*2^10*3^5 total such positions.

We could continue for the other cases. But in doing this, we triple count positions with three pairs, and 6-fold count positions with four pairs, etc.

So we'd have to count the number of ways to make three pairs, which involves analyzing more topologies, such as three corners three edges, three corners two edges, two corners three edges, etc.

Eventually, we could analyze everything all the way out to 24 pairs, and cascade our way back.

This sounds tedious enough that I'm not interested in doing it, and makes me think there is probably a better, more clever method possible.

 

[Lucas Garron]

A quick probabilistic simulation shows that 56% of all states have no piece in the right spot and oriented correctly.

A quick mathematical calculation shows me ≈0.606 for edges, and ≈0.513 for corners, about 0.311 together.

And I estimate (23/24)^20 ≈ 42.7%.

And my quick simulation agrees with Stefan (and so does my brain; I really wasn't expecting my code to work - I was mostly saying I can throw out an unsubstantiated answer easily, too). Cangurino, are you sure you didn't forget to complement your answer?

(Shoulda used KroeneckerDelta, but meh.)

 

[krazedkat]

first you must exclude the solved states...

 

[Muesli]

first you must exclude the solved states...

You mean all 1?

 

[Tomk]

first you must exclude the solved states...

You mean all 1?

Sure, if you look at it closely it has a high percentage impact on the outcome...

 

[Kirjava]

Odd definition of 'scramble' you guys have here.

 

[Yes We Can!]

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

Wrong. Just take the center switching alg.

 

[IamWEB]

Odd definition of 'scramble' you guys have here.

This isn't based on anyone's perception of the word, but it seemed like a fitting term to use in the title to keep the title a good size.

 

[JLarsen]

Scrambles with no corner edge pairs are quite rare for me. I think as a block builder I would notice them more often than cross solvers. If I were to guess I'd say no CE pair scrambles come around maybe 20-30% of the time. Might I point out that no CE scrambles are some of the hardest to solve (longest solutions), and that scrambles with 3 or more CE pairs I'd most likely still call scrambled?

 

[Cyrus C.]

I stumbled upon this link:

http://eklhad.net/rubik/enumerat.html

I confess, I have not reviewed it to check the accuracy but from a quick glance it appears to be a good start.

I think that is talking about positions, not scrambles.

 

[Stefan]

correct, the article discusses positions. As far as scrambles are concerned, the final number would be dependent on both the length of the scrambles along with the boundaries of turns per movement.

Please actually read this thread, or at least the first post?

 

[chris410]

correct, the article discusses positions. As far as scrambles are concerned, the final number would be dependent on both the length of the scrambles along with the boundaries of turns per movement.

Please actually read this thread, or at least the first post?

Apologies! (posts removed)