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

[CuBeOrDiE]

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.

 

[Sir E Brum]

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...

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

http://www.colemak.com/

 

[CuBeOrDiE]

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...

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

http://www.colemak.com/

lol

 

[MichaelP.]

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...

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

http://www.colemak.com/

Touché

 

[Logan]

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...

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

http://www.colemak.com/

Touché

Souché

 

[qqwref]

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...

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

http://www.colemak.com/

Touché

Souché

Silxdń

 

[Boxcarcrzy12]

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

 

[Ranzha]

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?

 

[cmhardw]

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?

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

 

[Ranzha]

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?

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

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

 

[CubesOfTheWorld]

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.

 

[cmhardw]

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.

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

 

[imaghost]

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.

 

[Ranzha]

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.

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

On a 2x2x2 cube:
oh boy.

You want a real number as the answer?

 

[Marco Aurelio]

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

That's right, isn't it?

 

[InfernoTowel]

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.

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

On a 2x2x2 cube:
oh boy.

You want a real number as the answer?

The formula, when applied to the 2x2x2 cube, gives 1.

 

[JeffDelucia]

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.

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

On a 2x2x2 cube:
oh boy.

You want a real number as the answer?

The formula, when applied to the 2x2x2 cube, gives 1.

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

 

[MichaelErskine]

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

...that you can see

 

[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.

 

[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