Number of "Solved" Positions for a 7x7!?

[CubesOfTheWorld]

If there are more than 1 solved states for a 7x7, which I don't think there are, then someone should make a video on it.

 

[hawkmp4]

If there are more than 1 solved states for a 7x7, which I don't think there are, then someone should make a video on it.

You've clearly ignored the entire thread. If you'd read it carefully, you'd understand what we're discussing.

 

[Ranzha]

1,432,626,401,430,044,535,008,733,970,965,769,702, 373,707,479,678,713,856,000,000 is the answer I got awhile back. Is this or isn't this right? Refer to the post above to verify.

floor((7-2)^2/4) = 6...
And Python says 2^(17*6)*3^(6*6) = 761070218101112490867290611780470776618217897984

Oh, I see what I did wrong.
I went too far.

1432626401430044535008733970965769702373707479678713856000000 / 761070218101112490867290611780470776618217897984 = 1882384000000.

1882384000000 = 2^10 x 5^6 x 7^6.
Yeah.

But what does that mean, I wonder...?

 

[hawkmp4]

1,432,626,401,430,044,535,008,733,970,965,769,702, 373,707,479,678,713,856,000,000 is the answer I got awhile back. Is this or isn't this right? Refer to the post above to verify.

floor((7-2)^2/4) = 6...
And Python says 2^(17*6)*3^(6*6) = 761070218101112490867290611780470776618217897984

Oh, I see what I did wrong.
I went too far.

1432626401430044535008733970965769702373707479678713856000000 / 761070218101112490867290611780470776618217897984 = 1882384000000.

1882384000000 = 2^10 x 5^6 x 7^6.
Yeah.

But what does that mean, I wonder...?

What does what mean?

 

[Ranzha]

1,432,626,401,430,044,535,008,733,970,965,769,702, 373,707,479,678,713,856,000,000 is the answer I got awhile back. Is this or isn't this right? Refer to the post above to verify.

floor((7-2)^2/4) = 6...
And Python says 2^(17*6)*3^(6*6) = 761070218101112490867290611780470776618217897984

Oh, I see what I did wrong.
I went too far.

1432626401430044535008733970965769702373707479678713856000000 / 761070218101112490867290611780470776618217897984 = 1882384000000.

1882384000000 = 2^10 x 5^6 x 7^6.
Yeah.

But what does that mean, I wonder...?

What does what mean?

What does the quotient 1882384000000 mean?

 

[hawkmp4]

Well, I'm not sure how you got your number (1432626401430044535008733970965769702373707479678713856000000), especially considering there are factors of 5 and 7 in there, so the meaning of the quotient depends on the meaning of the aforementioned number.
I'm guessing that your number was simply an arithmetic error, and so the quotient is simply the factor by which your number erred.

 

[wheezyg]

I am considering this question too for the 5^3 and the 7^3. This fact that you (cubeordie) pointed out in your question for n^3 where n>3 seems to make these cubes not mathematical groups b/c the identity would not be unique. (That is by defining the identity as a solved state.) At least that is my hunch and I am exploring that myself now. I will post what o figure out as far as how many of these "solved states" there are. It is a very good and interesting question.

 

[MichaelErskine]

Isn't there a general formula for a nxnxn picture cube?

 

[uberCuber]

Isn't there a general formula for a nxnxn picture cube?

there's a general formula for how many total positions a picture cube has but i'm confused as to how that would help find the number of 'solved' positions on a non-picture cube :confused:

 

[MichaelErskine]

Isn't there a general formula for a nxnxn picture cube?

there's a general formula for how many total positions a picture cube has but i'm confused as to how that would help find the number of 'solved' positions on a non-picture cube :confused:

Here's a clue: a - b

EDIT: no forget that! The answer might be devilishly close if I could just interpret http://en.wikipedia.org/wiki/V-Cube_7#Permutations properly: perhaps 25!^6

 

[uberCuber]

Isn't there a general formula for a nxnxn picture cube?

there's a general formula for how many total positions a picture cube has but i'm confused as to how that would help find the number of 'solved' positions on a non-picture cube :confused:

Here's a clue: a - b

wow i am dumb :fp..i will try to find the formula again..i cant remember where i saw it last...

EDIT: it was on Chris's page: http://www.speedcubing.com/chris/cubecombos.html

 

[myboringlife92]

i know that these pieces rotate in 3's just as most pieces can, but they can be rotated in 3's across faces...
for the 7x7 idk...
but for the 5x5 i believe there are 9130086859014144 different formations that will be solved but have switched center pieces.
tell me if you want to see where that calculation came from because i actually thought it out.

 

[hawkmp4]

i know that these pieces rotate in 3's just as most pieces can, but they can be rotated in 3's across faces...
for the 7x7 idk...
but for the 5x5 i believe there are 9130086859014144 different formations that will be solved but have switched center pieces.
tell me if you want to see where that calculation came from because i actually thought it out.

Yeah, numbers without justification aren't worth much. Does that confirm the finding above? (subtracting the two results of formulas on Chris' page?)

 

[cmhardw]

...but for the 5x5 i believe there are 9130086859014144 different formations that will be solved but have switched center pieces.
tell me if you want to see where that calculation came from because i actually thought it out.

I confirm that same number as well for the number of "solved" states to a 5x5x5. I for one am interested in your thought process as to how you got there, if you wouldn't mind explaining. I'm always interested in methods to calculate things like this.

Yeah, numbers without justification aren't worth much. Does that confirm the finding above? (subtracting the two results of formulas on Chris' page?)

I think this subtraction method would not be as simple to use in practice as it might sound.

*paraphrasing* Use the supercube combinations formula, and approach it by doing a - b

Let's define a and b more precisely:

a = Total number of possible states to the n x n x n supercube
b = Total number of positions where any of the following are true (or some combination of these conditions are true)
b1) Not all corners are solved
b1a) Not all central edges are solved (if n is odd)
b2) Not all wings are solved
b3) Not all centers are confined to their same colored face

Calculating b directly would not necessarily be a simple calculation in my opinion. You would have to break up this calculation into 7 (15 if n is odd) sub-cases to consider all possible combinations of the above conditions, and then count the number of supercube combinations for each sub-case.

I personally think it would be easier to count the number of "solved" states to a n x n x n cube directly. It was defined earlier in the thread, but remember that the definition of these "solved" states more precisely is this: The number of states on the n x n x n supercube where all corners, wings, and central edges if n is odd are solved. In addition every center is confined to the face of the same color, but centers are not necessarily in their solved locations.

One last comment, the b we have defined above here would be calculated easiest by directly calculating the "solved" states to the n x n x n supercube, and then subtracting that number from the total number of supercube combinations. That is my opinion at least.

Chris

 

[cmhardw]

Actually, I just thought of a huge oversight to these calculations that many of us have been making, except for perhaps Ranzha and the others who were getting much larger numbers than most of us.

If we're going to consider the fact that the moveable centers can be in different positions noticeably on the n x n x n cube, then we also have to consider the fact that the central most centers on an odd cube can rotate noticeably for these "solved" states. I did not take this into account in my formula earlier in the thread, and this would mean that any calculation with my formula for an odd cube is incorrect. Calculations for even cubes with my old formula were correct

I would like to submit a new formula that takes this fact into consideration for the odd cubes. This new formula counts the following:

1) If n is even, it counts how many ways the centers of a given color can be scrambled on that same face such that the n x n x n cube still appears to be solved, but considering the cube as an n x n x n supercube the centers may not be in their correctly solved locations on that face.

2) If n is odd, it counts the number of positions the moveable centers can have in the exact same way that it does for the n is even case. It also factors in that the centralmost centers can rotate in place. Considering the cube as an n x n x n cube it will still appear to be solved, but considering it as an n x n x n supercube it is not necessarily solved.

Spoiler

\( 2^{17\lfloor\frac{(n-2)^2}{4}\rfloor + 11(n \ mod \ 2)} \times 3^{6\lfloor\frac{(n-2)^2}{4}\rfloor} \)

Notice that with n=2 there is 1 "solved" state, but with n=3 there are 2048 "solved" states.

Using this corrected formula the 5x5x5 has the following number of "solved" states:

Spoiler

\( 2^{45} \times 3^{12} = 18\ 698417\ 887260\ 966912 \)

Chris

 

[myboringlife92]

for the 5x5x5 my calculations were like this:
i figured on each face there are 4 center-corners and 4 center-edges. I know that these rotate in 3's just as most pieces do, but they can rotate in3's across faces also.

so to start with i just considered one face's center-corners...
4 possibilities for "piece 1"
3 possiblities for "piece 2"
2 possibilities for "piece 3"
1 possibility for the final piece.

so for center-corners on the 5x5x5 for one face there are 4x3x2 (or 24) potential combinations. the same is true for faces 1-5.. so we have so far 24^5 possibilities counted.

For the 6th face, there are 4 possibilities for the first piece, 3 for the second, but since there can never be a 1-1 switch, there must be one configuration for the last two pieces, leaving the possibility multiplier for the final face at 4x3 (or 12)

So, for the center-corners there are 24^5 x 12 (or 95551488) possibilities.

Finally, we square that number because the center-edges work just the same as the center corners, leaving us with 9130086859014144 possible configurations.

 

[myboringlife92]

oh yes i just read this after posting my last post with the calculation. It would be 2048x as much. and this assumes we are being color-specific. For example we could have two switched on red and two switched on orange on one cube and have two switched on blue and two switched on green on another cube. This would be counted as two separate permutations on our calculations, but would count as only one if we were to look at the colors indifferently.

Also i have seen some of your videos on the 5x5. I was wondering if you know of any sites showing methods that dont do centers first, then group edges, and then 3x3x3 solve? I really hate that method. Just boring to me.
I actually developed my own method for solving the 5x5 that i think is pretty interesting.

 

[mrCage]

Hint(s)!

Think orbitals. One can permute centers in same orbital on same face and still retain the solved state. However not all such permutations are possible. A swap in an orbital on a face is possible only if you do a swap in same orbital on another face also. Now the "problem" is to find the number of orbitals for the nxnxn cube. Not that hard!!

Per

PS! (7x7x7 calculation, 6 center orbitals)

Single orbital, first face 4!=24
Single orbital, all faces 24˄6/2=95551488

All orbitals, all faces 95551488˄6=7.6107.....e+47

Extreme centers orientations 4˄6/2=2048

Grand total 7.6107.....e+47*2048=1.558671....e+51

A better tool would give me all digits!!

PPS!! I found a better tool (http://world.std.com/~reinhold/BigNumCalc.html)
1558671806671078381296211172926404150514110255071232 is the accurate answer.

 

[mrCage]

...but for the 5x5 i believe there are 9130086859014144 different formations that will be solved but have switched center pieces.
tell me if you want to see where that calculation came from because i actually thought it out.

I confirm that same number as well for the number of "solved" states to a 5x5x5. I for one am interested in your thought process as to how you got there, if you wouldn't mind explaining. I'm always interested in methods to calculate things like this.

I get the same number for 5x5x5 cube disregarding extreme centers rotations!! Follow my 7x7x7 calculations with 2 orbitals instead of 6...

Per

 

[bootmii]

A hard number

As we all know, the centers of the 7x7 have one sticker. So I think that it's logical to conclude some may be swapped and the cube still remain solved. Is this correct? And if so, how many total "solved" positions are there?

thnx

720. 6 factorial is 720.