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 Post subject: Number of positions of a 17x17
PostPosted: Mon Jan 13, 2014 4:29 pm 
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I'd like to figure out how many reachable and discernible positions are possible for Oskar's Over the Top 17x17x17 cube.

I remember years ago seeing Richard Carr and Chris Hardwick's formulas for finding the positions of an NxNxN cube: http://www.speedcubing.com/chris/cubecombos.html
http://www.ws.binghamton.edu/fridrich/Richard/Cubes.pdf

Using the formula from Chris I've gotten this: http://bit.ly/1a2Tdhy
However I don't trust the answer because when I try it for a 3x3x3 I get this: http://bit.ly/L3BOtR

Ultimately I'd like to know the number of positions for both Oskar's Over the Top 17x17x17, and for a generic 17x17x17. (Both ignoring edges and corners, and including them.)

Any help would be appreciated! :)

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 Post subject: Re: Number of positions of a 17x17
PostPosted: Mon Jan 13, 2014 4:47 pm 
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In another thread I calculated the positions of a super 17x17x17, which I did with my own approach of handling each piece type one at a time. So I think I'll show it here as well.
For Oskar's 17x17x17, there are 56 groups of centers(ignoring the central center pieces), each of which contain 24 centers. 4 in each set have the same color, and there are 6 colors for them as well. This means there are:
((24!^56)/(4!^336))=4.371*10^868 possible combinations.
For a normal 17x17x17, we must account for the edges and corners.
Corner orientations: (3^7)
Corner permutations: (8!)
Middle edge permutations: (12!/2)
Middle edge orientations: (2^11)
There are 7 groups of wing edges, each containing 24 edges, which gives (24!^7) for their permutations, which are not restricted to even parity.
Multiplying these all together gives:
(8!)*(3^7)*(12!/2)*(2^11)*((24!^56)/(4!^336))*(24!^7)=6.691*10^1054 possible combinations.
(EDIT: fixed an error after Brandon pointed it out)
Are you making a "Why is the Rubik's cube so hard" episode 3? If so, maybe you could use Planck volumes to make a comparison. :lol:

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Last edited by benpuzzles on Mon Jan 13, 2014 5:36 pm, edited 2 times in total.

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 Post subject: Re: Number of positions of a 17x17
PostPosted: Mon Jan 13, 2014 5:19 pm 
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Joined: Thu Dec 31, 2009 8:54 pm
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I'm a bit busy right now but I'll make a generic NxNxN function later.

In the mean time, my calculation is:

8 corners: 8! and 3^3 twist, modulo parity and twist restriction
12 middle edges: 12! and 2^12 twist modulo parity with corners and twist restriction
7 edge wing types: (24!)^7
56 center piece types: (24!)^42 modulo 4 duplicate colors on each face for each type

? ((8! * 12!) / 2) * (3^8 / 3) * (2^12 / 2) * ((24!)^7) * ((24!) / ((4!) ^ 6))^56 =
669092608710520096261408314575991967111408122691540707290601365
294496257802119618956938205705136041636028689428016336273634131
487726647385709719884121474908504692670910698985371460377688900
699349198842497638186290806683678986850334593701338440753224464
740484033975924212665646410310537811828359510439026667039347182
757336297730724281196033862808102327432941067250179060157266025
054048093556007135154007603434085100547748064670636958246371249
119454463174658330555208369758612382449403973332343369712706870
923838041336318861143098538193323362829868347779481784646568888
023722509270749811402466088245770360947102010990952406412565132
175988024238740278224215845876500391255162029122054815404278641
999475767222218668661025073508769221156288818802031152122167665
036654264459567862643991333029626496008847360000000000000000000
000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000


That's 6.69 * 10^1054


I need to double-check the calculation and make it generic for any size.

Ben: your 12 denominator term in ((24!/12)^56) is definitely wrong.

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 Post subject: Re: Number of positions of a 17x17
PostPosted: Mon Jan 13, 2014 5:24 pm 
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In Chris Hardwick's formula, the brackets that look like ⌊ and ⌋ represent the floor function, meaning round down to the next integer. The other brackets ⌈ and ⌉ in Richard Carr's formula represent the ceiling function, meaning round up to the next integer. See http://en.wikipedia.org/wiki/Floor_and_ceiling_functions for more information.

So for a 3x3x3 you want put in n = 3, and get: http://bit.ly/1cXrhJs, which is the right answer.

And for a 17x17x17, put n = 17, and get: http://bit.ly/1lWHV0S. The other formula gives exactly the same thing: http://bit.ly/1eEb11I.


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 Post subject: Re: Number of positions of a 17x17
PostPosted: Mon Jan 13, 2014 5:26 pm 
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Joined: Mon Jun 07, 2010 11:24 am
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Thanks guys for the help!

Ben: You guessed it! I'm doing brainstorming on the topic but we'll see if I ever get to making a video on it. :)

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 Post subject: Re: Number of positions of a 17x17
PostPosted: Mon Jan 13, 2014 6:03 pm 
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All right, now how about a 17-layer minx? :lol:


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 Post subject: Re: Number of positions of a 17x17
PostPosted: Mon Jan 13, 2014 6:17 pm 
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Jared wrote:
All right, now how about a 17-layer minx? :lol:

viewtopic.php?f=1&t=24058&p=285637&hilit=oskarminx#p285637

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