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



## CuBeOrDiE (Apr 28, 2010)

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


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## ZamHalen (Apr 28, 2010)

Dude...you're about to be flamed.(Fair warning)


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## 4Chan (Apr 28, 2010)

Why?

It seems like a legitimate question.


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## CuBeOrDiE (Apr 28, 2010)

Why? Because I'm curious.


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## ZamHalen (Apr 28, 2010)

Because I don't think he asked clearly enough for some of the more brutal posters.(I don't understand the question.)
(And for some reason people don't treat him kindly on *this* forum.Even though he gave me some pretty good advice while trying to progress last summer.)


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## 4Chan (Apr 28, 2010)

ZamHalen said:


> Because I don't think he asked clearly enough for some of the more brutal posters.(I don't understand the question.)



>I don't understand the question.


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## Logan (Apr 28, 2010)

ZamHalen said:


> Because I don't think he asked clearly enough for some of the more brutal posters.(I don't understand the question.)



I understand it perfectly. He wants to know the "number of solved positions for a 7x7".


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## CuBeOrDiE (Apr 28, 2010)

What do you mean? You mean I worded it badly? I mean that, say for instance you took two of the corner centers on the blue side of a 7x7 and swapped them. The cube is still solved, but the pieces have been swapped, so this is two "solved" positions. I way "solved" because the cube is solved, but the two pieces are in different spots. So the question is, how many of these "solved" positions are there in total?


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## ZamHalen (Apr 28, 2010)

yeah i got it.I had just looked over it too quickly.


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## CuBeOrDiE (Apr 28, 2010)

K. So what do you think the answer is?


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## DaijoCube (Apr 28, 2010)

I suck at this kind of maths.


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## ZamHalen (Apr 28, 2010)

I don't know I'm good at math but I really hate it.


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## miniGOINGS (Apr 28, 2010)

Hmm. Interesting question! I have no idea how I might even go about calculating this... Congrats, you stumped me! I need to start learning about bigcube theory.


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## cmhardw (Apr 28, 2010)

CuBeOrDiE said:


> 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



I did a calculation of this for the 20x20x20 cube here.

For the 7x7x7 there would be:
(4!)^36 / 2^6 = (4!)^30 * 2^12 * 3^6
That's approximately 7 * 10^47 "solved" states where you could move around centers of the same color and still appear solved.

And for the n x n x n cube it would be:
(4!)^[6*floor((n-2)^2 / 4)] / 2^floor[(n-2)^2 / 4]

or

2^[17*floor((n-2)^2 / 4)] * 3^[6*floor((n-2)^2 / 4)]

"solved" states.

Here floor(x) means the greatest integer less than or equal to x.

Chris


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## koreancuber (Apr 28, 2010)

Lol, Stefan Pochmann is on his way.
EDIT: OR lucas garron and chris hardwick.


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## CuBeOrDiE (Apr 28, 2010)

Personally, I got unreasonably huge numbers. Here's my logic, explained with a 5x5 because its easier to explain: 

Consider 1 side

On this side, there are 4 outer corner centers, which can be placed 4! ways, 4 inner corner centers, which can also be placed 4! ways, as well as inner and outer center "edge" pieces, if you know what I mean. They both can, too, be calculeted with 4!. 

So 1 side has 4! x 4! x 4! x 4! combinations. Mulsiply that by 6 sides and you got it! 

PS- I don't think edges can be interchanged. If they were, the stickers would be swapped. Unless you have an even layered cube... hmm...


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## MichaelP. (Apr 28, 2010)

CuBeOrDiE said:


> Personally, I got unreasonably huge numbers. Here's my logic, explained with a 5x5 because its easier to explain:
> 
> Consider 1 side
> 
> ...


S and T aren't even close on the keyboard!


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## CuBeOrDiE (Apr 28, 2010)

whoops. typo fail. i wonder how that happened?


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## CuBeOrDiE (Apr 28, 2010)

cmhardw said:


> CuBeOrDiE said:
> 
> 
> > 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?
> ...



Hmm. Interesting. I got 80263249920. So I guess that wasn't insanely unreasonable...


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## cmhardw (Apr 28, 2010)

I fixed some typos in my formula, try it again.

Chris


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## CuBeOrDiE (Apr 28, 2010)

cmhardw said:


> I fixed some typos in my formula, try it again.
> 
> Chris



I got what you got now! But previously, I meant I got that number myself before posting the thread. Sorry for being unclear.


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## Sir E Brum (Apr 28, 2010)

MichaelP. said:


> CuBeOrDiE said:
> 
> 
> > Personally, I got unreasonably huge numbers. Here's my logic, explained with a 5x5 because its easier to explain:
> ...



http://www.colemak.com/


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## CuBeOrDiE (Apr 28, 2010)

Sir E Brum said:


> MichaelP. said:
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> 
> > CuBeOrDiE said:
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 lol


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## MichaelP. (Apr 28, 2010)

Sir E Brum said:


> MichaelP. said:
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> 
> > CuBeOrDiE said:
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Touché


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## Logan (Apr 28, 2010)

MichaelP. said:


> Sir E Brum said:
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> > MichaelP. said:
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Souché


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## qqwref (Apr 28, 2010)

Logan said:


> MichaelP. said:
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> 
> > Sir E Brum said:
> ...


Silxdń


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## Boxcarcrzy12 (Apr 28, 2010)

Ok there are 1 for the standard, 2 for the dazzler, and idk how many for the illusion.


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## Ranzha (Apr 28, 2010)

Boxcarcrzy12 said:


> Ok there are 1 for the standard, 2 for the dazzler, and idk how many for the illusion.



Wrong, sir.
The centre pieces on a 7x7x7 can be swapped without you knowing it.
This is why when someone has a signed cube, it's sometimes difficult to get that signature back solved after the puzzle is scrambled.

Here's my explanation:

Attempt #1: failed.


Spoiler



I will assume that by no means a solved case is different from a similar solved case if two core-centre pieces (the ones that determine the face) are twisted.
Theorem: No two individual pieces can be swapped.
Because of this, the very LAST two centre pieces counted must be predetermined. This would mean you would divide by two, no?

So! As it has been done before, let's start out with a 5x5x5 cube.



Spoiler



*Visualisation on a 5x5x5*
On a 5x5x5, the eight "centre pieces" (centre-corners and centre-edges) around the actual centre piece (the one that determines the colour of the face) could be swapped in any way at this point.

*Centre-corners*
Since the four centre-corners can be in any positions at this point, if one were to take them out and position them in the four slots, one by one:
The first slot can have any of the four corner-centres (4)
The second slot can have any of the remaining three corner-centres (3)
The third slot can have any of the remaining two corner-centres (2)
The last slot can have but the one remaining corner-centre (1).

This means that there are *4•3•2•1* (which can be notated *4!* Or 4 factorial, which means 4•3•2•1) possible outcomes for the corner-centres of this side. 4•3•2•1 = *24*.

*Centre-edges*
Since there are four centre-edges per face which can go in any slot, using the method I used for the centre-corners, one can deduce the possible outcomes for this is *4!*, or *24*.

*Wrapping up the first side*
Since there are 24 outcomes for centre-corners, and within those there are 24 outcomes for centre-edges, one can deduce that, for one side, there would be 24² possible outcomes for this one face. That totals *576 outcomes for one face*.

*Covering all six sides*
Because there are six sides of the cube, one would think that the total number of cases would be 576•6. However, since the last two pieces inserted are predetermined, as “No two individual pieces can be swapped”, one must divide by two. Thus:
*576•6÷2 = 1,728.*

Am I doin’ it right?



Since there are 1,728 solved positions on a 5x5x5 cube, we can deduce that there will be 1,728 positions for the inner centres on the 7x7x7. However, to be correct, one would need to multiply by 2 to get the actual number of positions. If, however, one were to solve the inner centres last, there would be but 1,728. I’ll stick with this number to gain free range in the outer centre calculation.



Spoiler



*Visualisation on a 7x7x7*
Since we’ve already covered the inner centre (the eight pieces surrounding each core-attached centre piece) as 1,728 outcomes, whose division by two allows free range in the outer centre, I shall move on to that outer centre.

*Outer centre-corners*
The outer centre-corners follow the same principle of the 5x5x5 centre-corners. This means there are *24* possible outcomes per side.

*Outer middle centre-edges*
These are NOT the wing centre-edges.
These pieces follow the same principle of the 5x5x5 centre-corners. This means there are *24* possible outcomes per side.

*Wing centre-edges*
Now, there are EIGHT of these per side. This means there are 8•7•6•5•4•3•2•1 or *8!* outcomes per side. This equates to *40,320*.

*Wrapping up the outer centre*
Since there are 24 outcomes for outer centre-corners, and within those there are 24 outcomes for outer middle centre-edges, and within those there are 40,320 outcomes for the wing centre-edges, one can deduce that, for one side, there would be 24•24•40,320 possible outcomes for the outer centre on one face. That totals *23,224,320 outcomes for the outer centre on one face*.

*Wrapping up the outer centres on all six sides*
Since there are 23,224,320 outcomes for the outer centres on one face, and there are six faces, there would be 23,224,320•6 possible outcomes for the outer centres all over the cube. That totals *139,345,920 outcomes for the outer centres on all faces*.

*Covering all centre pieces on all six sides*
So! We’ve calculated that there are 139,345,920 possible outcomes for the outer centres on a 7x7x7, and 1,728 outcomes for the inner centres on a 7x7x7, where the theorem “No two individual pieces can be swapped” has already been accounted for. Multiply these two together, and you get the result!
Thus:
139,345,920•1,728 = *240,789,749,760*.

Did I do it right?






EDIT: 4-28-2010: 18:54 GMT-7

Attempt #2: on the right track moar?


Spoiler



I will assume that by no means a solved case is different from a similar solved case if two core-centre pieces (the ones that determine the face) are twisted.
Theorem: No two individual pieces can be swapped.
Because of this, the very LAST two centre pieces counted must be predetermined. This would mean you would divide by two, no?

So! As it has been done before, let's start out with a 5x5x5 cube.



Spoiler



*Visualisation on a 5x5x5*
On a 5x5x5, the eight "centre pieces" (centre-corners and centre-edges) around the actual centre piece (the one that determines the colour of the face) could be swapped in any way at this point.

*X-centres*
Since the four X-centres can be in any positions at this point, if one were to take them out and position them in the four slots, one by one:
The first slot can have any of the four X-centres (4)
The second slot can have any of the remaining three X-centres (3)
The third slot can have any of the remaining two X-centres (2)
The last slot can have but the one remaining X-centres (1).

This means that there are *4•3•2•1* (which can be notated *4!* Or 4 factorial, which means 4•3•2•1) possible outcomes for the X-centres of this side. 4•3•2•1 = *24*.

*T-centres*
Since there are four T-centres per face which can go in any slot, using the method I used for the X-centres, one can deduce the possible outcomes for this is *4!*, or *24*.

*Wrapping up the first side*
Since there are 24 outcomes for X-centres, and within those there are 24 outcomes for T-centres, one can deduce that, for one side, there would be 24² possible outcomes for this one face. That totals *576 outcomes for one face*.

*Covering all six sides*
Because there are six sides of the cube, one would think that the total number of cases would be 576•6. However, each combination per side totals 576, and on two sides would be 576 on one side times 576 on the other sides. As “No two individual pieces can be swapped”, one must divide by two multiplied by itself. Thus:
*576^6÷2^2 = 9,130,086,859,014,144.*

Am I doin’ it right?



Since there are 9,130,086,859,014,144 solved positions on a 5x5x5 cube, we can deduce that there will be 9,130,086,859,014,144 positions for the inner centres on the 7x7x7. However, to be correct, one would need to multiply by 2^2 (which is 4) to get the actual number of positions. If, however, one were to solve the inner centres last, there would be but 9,130,086,859,014,144. I’ll stick with this number to gain free range in the outer centre calculation.



Spoiler



*Visualisation on a 7x7x7*
Since we’ve already covered the inner centre (the eight pieces surrounding each core-attached centre piece) as 9,130,086,859,014,144 outcomes, whose division by two allows free range in the outer centre, I shall move on to that outer centre.

*Outer X-centres*
The outer centre-corners follow the same principle of the 5x5x5 X-centres. This means there are *24* possible outcomes per side.

*Outer T-centres*
These pieces follow the same principle of the 5x5x5 X-centres. This means there are *24* possible outcomes per side.

*Oblique centres*
Now, there are EIGHT of these per side. This means there are 8•7•6•5•4•3•2•1 or *8!* outcomes per side. This equates to *40,320*.

*Wrapping up the outer centre*
Since there are 24 outcomes for outer X-centres, and within those there are 24 outcomes for outer T-centres, and within those there are 40,320 outcomes for the oblique centres, one can deduce that, for one side, there would be 24•24•40,320 possible outcomes for the outer centre on one face. That totals *23,224,320 outcomes for the outer centre on one face*.

*Wrapping up one face*
Since there are 23,224,320 possible outcomes for one outer centre on a face of a 7x7x7 and 576 possible outcomes for one inner centre of on a face of a 7x7x7, you’d multiply these go get one total face.
Thus:
23,224,320 × 576 = *13,377,208,320*. Yes, the first four digits in that are 1337.

*Wrapping up the outer centres on all six sides*
Since there are 23,224,320 outcomes for the outer centres on one face, and there are six faces, there would be 23,224,320^6 possible outcomes for the outer centres all over the cube. That totals * 156,912,680,410,659,076,957,309,986,198,597,402,624,000,000 outcomes for the outer centres on all faces*.

*Covering all centre pieces on all six sides*
So! We’ve calculated that there are 13,377,208,320 possible outcomes one centre on a 7x7x7, and accounting the theorem “No two individual pieces can be swapped”, you’d multiply to the sixth power, and divide by 2^2 as we did in the 5x5x5 example.
Thus:
13,377,208,320^6 ÷ 2^2 = *1,432,626,401,430,044,535,008,733,970,965,769,702,373,707,479,678,713,856,000,000*.

Did I do it right?


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## cmhardw (Apr 28, 2010)

Ranzha V. Emodrach said:


> Spoiler
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> 
> 
> ...



You're on the right track, but the number is *significantly* larger than your number.



Spoiler



I agree with you that each face has 576 different ways the centers could be arranged. Now let's look at the U and F faces. Each face has 576 ways to be arranged, but *for each and every arrangement of the U face* the F face has 576 arrangements. So it's 576*576=576^2 not 576*2 to count the total number of arrangements considering *both* faces. For 6 faces it's 576^6 not 576*6 to count the total number of arrangements.

Your second step is also on the right track but not quite. You do need to divide by 2 to fix, or determine, the last two centers in each center orbit. However you need to do this *for each and every orbit of centers* on the cube. So on the 5x5x5 cube you would divide by 2^2 or 4 not by 2.

For the 5x5x5 there would be:
576^6 / 2^2 = 9130 086859 014144 "solved" states.



Let me explain my formula from the second page, and hopefully that will help:


Spoiler



For the nxnxn cube there are:
(4!)^[6*floor((n-2)^2 / 4)] / 2^[floor((n-2)^2 / 4)]
"solved" states allowing the centers to be rearranged.

Remember that floor(x) means the greatest integer less than or equal to x.

So here is what the (4!)^[6*floor((n-2) / 4)] term is doing. Let's look at the nxnxn cube. On every nxn face of pieces there are (n-2)^2 center pieces. So for example on the 4x4x4 cube their are (4-2)^2 = 4 center pieces. On the 5x5x5 cube their are (5-2)^2 = 9 center pieces. This counts the central most center which does not move, but only spins.

Now we must look for how many center orbits there are. Well that's where the floor(x) function comes in handy. If we take the (n-2)^2 center pieces and divide by 4, we will find how many center orbits are on an *even* cube. So for the 4x4x4 cube there are (4-2)^2 / 4 = 1 center orbit, namely the x-centers or corner-centers. For the 6x6x6 cube their are (6-2)^2 / 4 = 4 center orbits. There are both center corner orbits (inner and outer) as well as the two oblique orbits, for a total of 4. Now, this expression does not work for odd cubes, because you end up with a strange answer. For the 5x5x5 cube there are *not* (5-2)^2 / 4 = 2.25 center orbits. However there *are*:
floor[(5-2)^2 / 4] = floor(2.25) = 2 center orbits to the 5x5x5.

Ok so floor((n-2)^2 / 4) finds how many center orbits there are on the nxnxn cube. By center orbit I mean different orbits of *movable* centers, so on an odd nxnxn cube we are not counting the central most center. Since each orbit on a face can be arranged into 4! positions then on any given face there are:
(4!)^floor((n-2)^2 / 4) ways to arrange the centers are on that face.

However, there are 6 faces, so for all six faces there are:
[(4!)^floor((n-2)^2 / 4)]^6 = (4!)^[6*floor((n-2)^2 / 4)]

So the first term of my formula, the (4!)^[6*floor((n-2)^2 / 4)] term, is finding how many ways there are arrange the centers on all 6 faces if there were no restrictions. However, as mentioned in the other spoiler, we must divide by 2 for each and every center orbit on the cube. Remember that on the nxnxn cube there are floor((n-2)^2 / 4) center orbits. So we have to divide by:
2^[floor((n-2)^2 / 4)]

This gives the final formula of:
(4!)^[6*floor((n-2)^2 / 4)] / 2^[floor((n-2)^2 / 4)]

The second version of the formula is just doing some algebra on the one above by rewriting 4! as 2^3*3 and then combining like terms, this gives you:

2^[17*floor((n-2)^2 / 4)] * 3^[6*floor((n-2)^2 / 4)]



Hope this helps,
Chris


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## Ranzha (Apr 29, 2010)

cmhardw said:


> Ranzha V. Emodrach said:
> 
> 
> > Spoiler
> ...



Thanks! I didn't take that into account.
Wow, now I have the right forumla, methinks.


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## CubesOfTheWorld (Apr 29, 2010)

CuBeOrDiE said:


> 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



1. Final answer.

The corners and edges have to be correct too, which is why there is 1 solved state; the centers also do not change in relation to each other. This was a very logical question though, but you were just thinking too hard.


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## cmhardw (Apr 29, 2010)

CubesOfTheWorld said:


> CuBeOrDiE said:
> 
> 
> > 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?
> ...



I interpreted the question the following way:

On an n x n x n supercube how many different possible combinations are there with all corners and edges completely solved, and all centers on their correct face, but not necessarily in their solved locations?

I thought the wording clearly denoted that this is how we were to interpret the question. Yes there is only one solved state to a standard n x n x n cube, but here we are considering the partially "solved" states of the n x n x n supercube.

That is at least how I thought of the question.
Chris


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## imaghost (Apr 29, 2010)

the centers do change
you can have a rotated center piece on a normal 3x3, why can't you have a rotated center piece on a 7x7? It can, and does happen.


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## Ranzha (Apr 29, 2010)

cmhardw said:


> CubesOfTheWorld said:
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> 
> > CuBeOrDiE said:
> ...



On a 2x2x2 cube:
oh boy.

You want a real number as the answer?


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## Marco Aurelio (May 21, 2010)

Wow! So here is the answer: 240789749760 solved positions for the 7x7

That's right, isn't it?


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## InfernoTowel (Jun 27, 2010)

Ranzha V. Emodrach said:


> cmhardw said:
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> 
> > CubesOfTheWorld said:
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The formula, when applied to the 2x2x2 cube, gives 1.


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## JeffDelucia (Jun 27, 2010)

InfernoTowel said:


> Ranzha V. Emodrach said:
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> > cmhardw said:
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Of course because there are no interchangeable pieces on the 2x2.


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## MichaelErskine (Jun 27, 2010)

JeffDelucia said:


> Of course because there are no interchangeable pieces on the 2x2.



...that you can see


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## Ranzha (Jul 2, 2010)

*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.


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## hawkmp4 (Jul 2, 2010)

Ranzha V. Emodrach said:


> *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


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## CubesOfTheWorld (Jul 2, 2010)

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.


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## hawkmp4 (Jul 2, 2010)

CubesOfTheWorld said:


> 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.


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## Ranzha (Jul 3, 2010)

hawkmp4 said:


> Ranzha V. Emodrach said:
> 
> 
> > *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.
> ...



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...?


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## hawkmp4 (Jul 3, 2010)

Ranzha V. Emodrach said:


> hawkmp4 said:
> 
> 
> > Ranzha V. Emodrach said:
> ...


What does what mean?


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## Ranzha (Jul 3, 2010)

hawkmp4 said:


> Ranzha V. Emodrach said:
> 
> 
> > hawkmp4 said:
> ...



What does the quotient 1882384000000 mean?


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## hawkmp4 (Jul 3, 2010)

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.


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## wheezyg (Jul 21, 2010)

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.


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## MichaelErskine (Jul 21, 2010)

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


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## uberCuber (Jul 21, 2010)

MichaelErskine said:


> 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:


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## MichaelErskine (Jul 21, 2010)

uberCuber said:


> MichaelErskine said:
> 
> 
> > Isn't there a general formula for a nxnxn picture cube?
> ...



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


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## uberCuber (Jul 21, 2010)

MichaelErskine said:


> uberCuber said:
> 
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> > MichaelErskine said:
> ...



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


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## myboringlife92 (Jul 22, 2010)

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.


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## hawkmp4 (Jul 22, 2010)

myboringlife92 said:


> 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?)


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## cmhardw (Jul 22, 2010)

myboringlife92 said:


> ...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.



hawkmp4 said:


> 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


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## cmhardw (Jul 22, 2010)

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


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## myboringlife92 (Jul 23, 2010)

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.


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## myboringlife92 (Jul 23, 2010)

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.


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## mrCage (Jul 23, 2010)

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.


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## mrCage (Jul 24, 2010)

cmhardw said:


> myboringlife92 said:
> 
> 
> > ...but for the 5x5 i believe there are 9130086859014144 different formations that will be solved but have switched center pieces.
> ...


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

Per


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## bootmii (Aug 3, 2010)

*A hard number*



CuBeOrDiE said:


> 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.


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## hawkmp4 (Aug 3, 2010)

bootmii said:


> CuBeOrDiE said:
> 
> 
> > 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?
> ...


What's the justification for that number?


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## uberCuber (Aug 3, 2010)

bootmii said:


> CuBeOrDiE said:
> 
> 
> > 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?
> ...



did you read anything else in this thread at all??

oh thats right, his name is "boot me"....


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## Weston (Aug 3, 2010)

hawkmp4 said:


> bootmii said:
> 
> 
> > CuBeOrDiE said:
> ...



Lol He took the number of fixed centers on a cube, 6, and just factorial'd that. Which makes no sense in so many ways.
He didnt understand the question, and even if he had, it would still be wrong.


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## hawkmp4 (Aug 3, 2010)

I know it's BS- I just wanted to see what he came up with for a reason.


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## MichaelErskine (Aug 4, 2010)

We need _*more*_ math geeks


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## hawkmp4 (Aug 4, 2010)

MichaelErskine said:


> We need _*more*_ math geeks


Have you seen the xkcd math forum? 

...It's great. I love it.


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## hawkmp4 (Sep 17, 2010)

I felt this was the appropriate place for this question. I'm interested in finding the number of unique positions of a Gigaminx. Edges and corners are simple. But I'm afraid I don't understand the calculations in this thread regarding centres. As I understand it, there are two different kinds of Gigaminx centre pieces, the trapezoidal ones and the rhomb...rhomboid (?) ones. 
The Gigaminx only allows even permutations of each type of piece. So, clearly, 3-cycles within one centre are allowable while still being solved. But 2 2-cycles, with them both on the same face or on different faces, are allowable as well. 2 4-cycles, a 5-cycle, etc.

Counting is not my strong suit. How would I go about finding the number of solved positions for a Gigaminx?
A more complicated question, how many unique positions of the Gigaminx are there? That is, how many indistinguishable positions are there? Some positions would be distinguishable Gigaminx "supercube" but not on a regular Gigaminx.

Thanks for the help.


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## aminayuko (Jun 6, 2011)

Here is the solution:
make the cubes you are thinking of into stephan pochmans (sorry if i spelled it wrong stephan) super cube design.
that way, the centers have a specific placement and you don't have to suffer doing random math (no offense to anyone who takes math seriously)


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## Lucas Garron (Jun 6, 2011)

Huh?


aminayuko said:


> Here is the solution:
> make the cubes you are thinking of into stephan pochmans (sorry if i spelled it wrong stephan) super cube design.


Since you're aware you might be spelling his name wrong, how about *looking at his name*?
Even Google can correct his name: http://www.google.com/search?q=stephan+pochman



aminayuko said:


> that way, the centers have a specific placement and you don't have to suffer doing random math (no offense to anyone who takes math seriously)


Yes. So?

By the way, I think we should stop keeping track of age, and only remember day and month of birth, your birthday has a specific placement and you don't have to suffer doing random math (no offense to anyone who takes age seriously)


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## d4m4s74 (Jun 6, 2011)

we could also simply remove the years from post dates too, so we don't see this is a huge bump only to state this thread is silly

no wait, this post is from september, so we would still be able to see. Should have waited three more months, aminayoko (sorry if I spelled it wrong whatever, your real name is)


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## aminayuko (Jun 6, 2011)

lucas garron, what i am saying is this http://cube4you.com/eastsheen-super-5x5-cube-a5s-p-134.html sticker mod on the 7x7 so each center piece has a specific place


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## Jaysammey777 (Jun 6, 2011)

come to say, this guy could of just looked it up on google


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## JLarsen (Jun 6, 2011)

Lucas Garron said:


> Huh?
> 
> Since you're aware you might be spelling his name wrong, how about *looking at his name*?
> Even Google can correct his name: http://www.google.com/search?q=stephan+pochman


 


d4m4s74 said:


> Should have waited three more months, aminayoko (sorry if I spelled it wrong whatever, your real name is)


 
Lulz.


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## Ranzha (Jun 6, 2011)

aminayuko said:


> lucas garron, what i am saying is this http://cube4you.com/eastsheen-super-5x5-cube-a5s-p-134.html sticker mod on the 7x7 so each center piece has a specific place


 
So you're saying you found out? Pray tell. I'm actually interested in the answer since I spent a bunch of time trying to figure it out.


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## Stefan (Jun 6, 2011)

Jaysammey777 said:


> come to say, this guy *could of* just looked it up on google


 
Argh!

And looked *what *up?


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## mrCage (Jun 8, 2011)

Stefan said:


> Argh!
> 
> And looked *what *up?



This is not a lingo forum, but i agree Stefan. Treat english properly so others can understand it!!

Per


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## Stefan (Jun 8, 2011)

mrCage said:


> This is not a lingo forum, but i agree Stefan. Treat english properly so others can understand it!!


 
Well, I don't usually point out such mistakes unless they're funny or they're made when mocking others about spelling. Potentially this is a case of the latter kind, if Jay meant my name. Which I don't know, that's why I asked him. It's possible he meant the thread starter, but that was over a year ago and how does Jay know what Google found a year ago? Plus I don't think the question is or was that trivial to answer by googling.


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## trex9966 (Jul 6, 2011)

to my calculations -its late and im tiered so i don't know how accurate i am- say that there is 4398046511104 solved states


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## cubersmith (Jul 6, 2011)

mrCage said:


> This is not a lingo forum, but i agree Stefan. Treat english properly so others can understand it!!
> 
> Per


 
But we could understand him. Theres no need to criticise people's grammar, if you understand their point.


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## cuBerBruce (Jul 6, 2011)

trex9966 said:


> to my calculations -its late and im tiered so i don't know how accurate i am- say that there is 4398046511104 solved states


 
You disagree with the answer given in this [post=422374]post[/post]?


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