• Welcome to the Speedsolving.com, home of the web's largest puzzle community!
    You are currently viewing our forum as a guest which gives you limited access to join discussions and access our other features.

    Registration is fast, simple and absolutely free so please, join our community of 35,000+ people from around the world today!

    If you are already a member, simply login to hide this message and begin participating in the community!

Probability Thread

cmhardw

Premium Member
Joined
Apr 5, 2006
Messages
4,106
Likes
134
Location
Atlanta, Georgia
WCA
2003HARD01
Okay, I've given the parity issue more thought, and have determined that I was right to doubt my first impression. Parity excludes some of the scenarios I was thinking were possible. If we assume that the observer is clever enough to calculate the corner parity of a scrambled cube in his head, then the result increases to 6/7, or about 85.714%.
Do you mean permutation parity? The corners can have either even or odd permutation in a legal state. Perhaps I don't follow?

The third and final scenario is that the two pieces share exactly two colors, i.e. they belong adjacent to each other. Let us call the two colors they have in common color A and color B. As before, there are three parity cases. In one case, an orientation exists that reveals stickers A and B, but not one that reveals stickers B and A. In another parity case, an orientation exists that reveals stickers B and A, but not one that reveals stickers A and B. (This bit is what tripped me up the first time 'round.) It seems like you would be able to confuse them, but if you have perfect understanding of parity, it should be possible to determine which corner is which just by seeing A and B or B and A along with all the other stickers that are in view.
I don't follow this. If the two corners in DBL and DBR are ones that belong in UFL and UFR when solved, then I could see a F sticker color at LDB and a U sticker color at RDB in two possible ways, both of which are allowed since I can have even or odd permutation parity in the corners.

Am I not following your argument? Am I making a mistake in my logic somewhere?
 
Joined
Apr 3, 2014
Messages
72
Likes
0
Do you mean permutation parity? The corners can have either even or odd permutation in a legal state. Perhaps I don't follow?
I'm talking about orientation parity. Perhaps I am using the wrong term, since true parity should only be either odd or even, whereas corner orientation is tristate, but I don't know of a better term to use, so I'm sticking with that one for now. If you turn just one corner clockwise 120 degrees, you reach an illegal cube state. If you then turn any other corner counter-clockwise 120 degrees, the cube state is made legal again. This is what I'm talking about.

I don't follow this. If the two corners in DBL and DBR are ones that belong in UFL and UFR when solved, then I could see a F sticker color at LDB and a U sticker color at RDB in two possible ways, both of which are allowed since I can have even or odd permutation parity in the corners.

Am I not following your argument? Am I making a mistake in my logic somewhere?
Okay, let me give some examples. I have here a solved 2x2 with yellow on top, blue in front, and red on the right. Grab your 2x2 and you can follow along at home. For the purposes of this discussion, X will mean the left side sticker of the DBL corner, and Y will mean the right side sticker of the DBR corner.

As I look at the solved cube, X is orange and Y is red, and since I know the DB corners contain only one red sticker and one orange sticker, I know the full state of the cube without looking at the D or B faces.

Now apply this swap algorithm: U F U' F D R D2 R F D2 F. It swaps the DB corners while preserving which stickers are exposed at X and Y.
The result is that now X is red and Y is orange. This is still a uniquely identifiable state.

Now apply this rotation algorithm: R F' D R' D R D2 F2 R' D'. It rotates the DBR corner clockwise 120 degrees and the DBL corner counterclockwise 120 degrees. (Between these two algorithms, we can bring about any of the six legal states of those two corners without affecting the rest of the cube.)

As we look at the cube, X and Y are both green. If we were to execute the above swap algorithm again, they would still both be green, and so this state is not uniquely identifiable.

Now execute the rotation algorithm again. X and Y are now both white, and the same argument applies. Because we can use the swap algorithm to reach a new state that looks the same, this position is also confusable. So far, we have operated entirely inside one orientation parity state of the DB corner pair.

Now we will execute a new algorithm which I will call the parity algorithm: R U2 R' U R2 U' R2 U R' U2. This one changes the parity state of the DB corners by twisting the UFR corner. This change in parity is readily observable without viewing the B or D faces.

If you've followed along properly, you should see that X is now white and Y is now green. White exists on both of the DB corners, and green exists on both of the DB corners. But I assert that this state is still uniquely identifiable without viewing the D or B faces. To demonstrate why, we will cycle through all six possible DB corner pair states for the current orientation parity state:

State

X

Y

Notes

#1

White

Green

This is the starting state. Now perform the rotation algorithm.

#2

Red

White

Now perform the rotation algorithm again.

#3

Green

Orange

Now perform the swap algorithm.

#4

Orange

Green

Now perform the rotation algorithm again.

#5

White

Red

Now perform the rotation algorithm one last time.

#6

Green

White

This is the sixth and final state.



Notice that each of these six states has a distinct set of values for X and Y. This means that if you see X is white and Y is green, you know that the DB corners are in state #1, whereas if you see that X is green and Y is white, you know that they are in state #6. There is no ambiguity here because of parity.

In order to reach the other White/Green states, you would have to change the orientation of one of the other six corners, which would be visible without viewing the D or B faces, and so the observer, understanding parity perfectly, would be able to figure out what you had done and would still not be fooled. This is why the answer to question A is 7/6 (~85.714%) and not 16/21 (~76.19%).

Does this sufficiently answer your question?
 
Joined
Dec 8, 2013
Messages
1,017
Likes
39
YouTube
IRNjuggle28
But you could have a single dedge flip on the DB edge? There are also different ways you can move the centers.

I think it would be interesting to see these stats for both regular cubes and super cubes.
What you said only supports the conclusion that you can determine a 4x4 permutation from UFRL 0% of the time. Yes. On each solve, there is the possibility of the DB edge either being flipped, or not. Ergo, there will never be a 4x4 permutation that can be identified without looking at the DB edge. Ergo, there are no 4x4 permutations that can be identified from seeing the U, F, L, and R faces.
 
Joined
Apr 3, 2014
Messages
72
Likes
0
Excellent analysis! Thanks for writing this out, that was very clear!
You are most welcome. Perhaps the reason that this one was clearer is that I wrote it while fully awake, whereas my previous attempt was written while falling asleep at the keyboard.
 
Joined
Jan 11, 2014
Messages
836
Likes
11
Location
NY
WCA
2014CAVA01
YouTube
Goodatthis123
So I'm basically asking for trouble here, and I'm not even sure how one would calculate it and/or if someone even would, but here goes.

So there's something like 10^996 combinations on a Petaminx. (Not sure on the base but I know the exponent)
Now I've heard rumors and seen photos of an Examinx, a 13 layered minx. (Assuming you consider megaminx as 3 layered)
Now let's say you had super cube stickers on an Examinx. How many combinations would this puzzle have?
I'm not completely sure if minxes can have super cube centers, but I'm just going to assume they can.
 

TDM

Member
Staff member
Joined
Mar 7, 2013
Messages
7,009
Likes
318
Location
Oxfordshire, UK
WCA
2013MEND03
YouTube
TDM028
So I'm basically asking for trouble here, and I'm not even sure how one would calculate it and/or if someone even would, but here goes.

So there's something like 10^996 combinations on a Petaminx. (Not sure on the base but I know the exponent)
Now I've heard rumors and seen photos of an Examinx, a 13 layered minx. (Assuming you consider megaminx as 3 layered)
Now let's say you had super cube stickers on an Examinx. How many combinations would this puzzle have?
I'm not completely sure if minxes can have super cube centers, but I'm just going to assume they can.
Permutations...
So each type of 'centre piece' (i.e. anything not on the outer layer) has 60 positions it could go in to, so for each one there are 60! possible permutations. There are 30 types, so for centres alone we have 60!^30, which is a huge number. About 4.03*10^2457. This is assuming this is a supercube, as you said in your post.
There are 30 edges in the middle, so 30! permutations.
60 permutations for each of the five other types of edge, so 60!^5.
20 corners, 20! permutations.
And divide by 2 (is it 2?) for parity.
Orientations...
Centres have no orientation.
Middle edges: 30 edges, so 2^29.
Other edges have no orientation.
Corners: 3^19.

In total...
(60!^30*30!*60!^5*20!*2^29*3^19)/2 = 3.23*10^2935
so almost 3000 digits, unless I made a mistake somewhere
 

cmhardw

Premium Member
Joined
Apr 5, 2006
Messages
4,106
Likes
134
Location
Atlanta, Georgia
WCA
2003HARD01
erm... I barely know anything about just Megaminxes, so I don't really know about this. Is it I don't need to divide by two because it's a supercube (or superdodecahedron), so I would have included parity there?
Odd permutation parity, for any piece orbit, is impossible on a megaminx (and it's relatives).
 

cmhardw

Premium Member
Joined
Apr 5, 2006
Messages
4,106
Likes
134
Location
Atlanta, Georgia
WCA
2003HARD01
So does that mean you don't have to divide by two?
If odd permutation parity is impossible, then you have to divide by 2 for every piece orbit on the puzzle. This is the case because you were calculating the cases for a super-minx, where the concept of permutation parity makes sense for all piece types.
 

TDM

Member
Staff member
Joined
Mar 7, 2013
Messages
7,009
Likes
318
Location
Oxfordshire, UK
WCA
2013MEND03
YouTube
TDM028
If odd permutation parity is impossible, then you have to divide by 2 for every piece orbit on the puzzle. This is the case because you were calculating the cases for a super-minx, where the concept of permutation parity makes sense for all piece types.
Right, so when you say every orbit, does that mean you divide by 2 six times (i.e. dividing by 64)? I'm really confused...
 
Joined
May 7, 2006
Messages
7,287
Likes
40
WCA
2003POCH01
YouTube
StefanPochmann
If odd permutation parity is impossible, then you have to divide by 2 for every piece orbit on the puzzle.
Do you mean when they're independent, or is the independence already included in the term "piece orbit"? Because I think the 3x3x3 has two piece orbits (corners and edges) and we divide by 2 only once.
 
Last edited:
Joined
Jun 29, 2008
Messages
1,802
Likes
0
WCA
2008COUR01
Now I've heard rumors and seen photos of an Examinx, a 13 layered minx. (Assuming you consider megaminx as 3 layered)
11-layered, not 13. We already have:
- megaminx: 3 layers
- gigaminx: 5 layers
- teraminx: 7 layers
-petaminx: 9 layers.
So examinx should have 11.
 
Last edited:

cmhardw

Premium Member
Joined
Apr 5, 2006
Messages
4,106
Likes
134
Location
Atlanta, Georgia
WCA
2003HARD01
Right, so when you say every orbit, does that mean you divide by 2 six times (i.e. dividing by 64)? I'm really confused...
If an examinx is an 11 layered minx, then you divide by 2^36 for the superexaminx.
There are 30 center orbits, 1 corner orbit, 1 middle edge orbit, 4 wing edge orbits.

Do you mean when they're independent, or is the independence already included in the term "piece orbit"? Because I think the 3x3x3 has two piece orbits (corners and edges) and we divide by 2 only once.
Independence of the orbits is not necessarily true on the examinx, but that is irrelevant here.

Look at a quarter turn on an examinx. No matter what face type is turned you are either performing a 5 cycle of pieces in an orbit or two 5-cycles of pieces in an orbit (for x-centers). An odd parity permutation is impossible. You must divide the overcount calculation for the examinx by 2^36 (divide by one factor of 2 for each of the 36 orbits on the puzzle).
 
Last edited:

TDM

Member
Staff member
Joined
Mar 7, 2013
Messages
7,009
Likes
318
Location
Oxfordshire, UK
WCA
2013MEND03
YouTube
TDM028
I get it now... I didn't know what 'orbit' meant. I also looked at the first clear image on google images for 'examinx', which has 13 layers, not 11, so my calculations were wrong anyway. I think I understand what you're saying now, thanks for correcting me.
 
Joined
Sep 17, 2009
Messages
891
Likes
36
Location
New Orleans, LA
YouTube
4EverTrying
qqwref helped me with this in the past, and I recall that the number of positions on the minx of order n (for all n>2) is:





Abbreviating the above product as (a)(b)(c)(d),

Numerator of (a)
The numerator represents the number of permutations and orientations of 20 corners.

First Factor in the Denominator of (a)
We divide by (20*3) for even minx's, as we can fix a corner to be solved on any even minx and thus divide the total number of positions by how much of the product any single corner contributes. We may also interpret dividing by 60 for the even minx to be that we instead fix one non-fixed piece or one wing edge, as each of them also contribute a factor of 60 into the total number of positions.

Second Factor in the Denominator of (a)
We divide by (3) by the minx law of corner orientations.

Third Factor in the Denominator of (a)
We divide by (2) since only even permutation corner permutations are allowed.
Numerator of (b)
The numerator represents the number of permutations and orientations of 30 middle edges.

First Factor in the Denominator of (b)
We divide by (2) by the minx law of middle edge orientations.

Second Factor in the Denominator of (b)
We divide by (2) since only even permutation middle edge permutations are allowed.
This factor tells us how many positions of the wing edges there are.

There are 30(2) = 60 wing edges in every wing orbit (that is, there are 2 wing edges of each wing edge orbit in each composite edge), and only half of the permutations are allowed (odd permutations cannot exist on a minx).

The minx^n has the same number of wing edge orbits that the nxnxn cube has.
This factor tells us how many positions of the non-fixed centers there are.

There are 5(12) = 60 non-fixed center pieces in every non-fixed center piece orbit (that is, there are 5 non-fixed center pieces in each of the 12 faces per orbit).

Since every face has 5 of the same color centers, there are 5! equivalent positions regarding each of the 12 color sets of non-fixed centers, and this is true for every orbit of non-fixed centers. Thus we divide by 5! raised to the number of non-fixed center orbits.

We do not also divide by 2 because, for example, we can picture 2 discolored centers as either being a 3-cycle (even permutation) or a 2-cycle (odd permutation), since this is not the super minx.

Lastly, the minx^n has the same number of non-fixed center orbits as the nxnxn cube.

As I did here, the number of positions formula of the minx^n can be written as a product of powers of consecutive prime numbers. This time, a product of all primes from 1 to 59 instead of just 1 to 23.





[HR][/HR]For the superminx, I get that we simply multiply the number of positions formula for the minx^n by


(this is the number of "solved positions" of the minx^n). For example minx^11 has this number/ 10^501 solved positions.

Let the above product be represented by (e)(f).
This is a factor pertaining to the number of positions the non-fixed centers contribute.

Numerator of (e)
Non-fixed centers are distinguishable on the superminx^n. Therefore we do not divide by (5!)^12 raised to the number of non-fixed center orbits anymore. We include this to simply counteract doing this in the number of positions formula for the regular 12 colored minx^n.

The Denominator of (e)
We divide by 2 because only even permutations of non-fixed centers are allowed on the superminx^n.
This is the fixed center of the odd minx^n factor.

There are 12 fixed centers on the odd minx^n, and each can be rotated in 5 ways.

We do not divide by 2 here because if we do an outer face quarter turn to a face on a solved odd layer superminx^n, we can solve back all pieces using commutators (except for the fixed center) without rotating the face itself.
We multiply (a)(b)(c)(d)(e)(f) to get the following formula for the superminx^n (for n>2):

Again, we can write this formula as a product of powers of consecutive primes. To make things simple, we can just convert the superminx factor to a product of the following consecutive primes, and we can manually merge the exponents of the common prime numbers both products have to have a formula for the superminx^n as a product of powers of consecutive prime numbers from 1 to 59.


So I'm basically asking for trouble here, and I'm not even sure how one would calculate it and/or if someone even would, but here goes.

So there's something like 10^996 combinations on a Petaminx. (Not sure on the base but I know the exponent)
Now I've heard rumors and seen photos of an Examinx, a 13 layered minx. (Assuming you consider megaminx as 3 layered)
Now let's say you had super cube stickers on an Examinx. How many combinations would this puzzle have?
I'm not completely sure if minxes can have super cube centers, but I'm just going to assume they can.
10^996 is correct for minx^9.

For minx^11, using the formula above for the regular 12-color minx^n, I get 10^1533. Or, this number of positions.

For the superminx^11, I get 10^2035 or this number of positions.

For the superexaminx (an 11 layered minx) I get:

(60! / 2)^34 *(20! / 2) (3^19)(30! / 2)(2^29)
Chris, I get:
(60!/2)^24*(20!/2) (3^19) (30!/2) (2^29) (5^12)

(Don't we count fixed center rotations?)
 
Last edited:

cmhardw

Premium Member
Joined
Apr 5, 2006
Messages
4,106
Likes
134
Location
Atlanta, Georgia
WCA
2003HARD01
Chris, I get:
(60!/2)^24*(20!/2) (3^19) (30!/2) (2^29) (5^12)

(Don't we count fixed center rotations?)
I see what I did wrong. I forgot about centralmost center rotations, but I also thought that the largest outer pentagon of centers had the same number of layers as he minx itself (obviously incorrect). I can see now that the pentagon rings from outside to inside have 8, 6, 4, 2 orbits or 20 center orbits plus the 4 wing orbits for an exponent of 24 and not 34.
 
Top