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 Post subject: Dioctipoid "math" (counting permutations)Posted: Fri Jan 13, 2012 5:08 pm

Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
EDIT: Andreas Nortmann has pointed out that I was not careful and counted indistinguishable pieces in different orbits as though they were in the same orbit. I have corrected the numbers below by calculating just one orbit and then squaring that for the second orbit.

The Dioctipoid author has a page up which attempts to calculate the number of permutations of the puzzle, alluding to the total permutations being somewhat correlated with the difficulty of the puzzle: http://www.dioctipoid.com/Math.html. The math looks pretty broken to me but before I contact the author I'd hate to be making mistakes too so I'd like to run my calculations by the community first.

First, the Dioctipoid is a Rex Cube / FTO and has octahedral symmetry. It has:
• 12 Dino edges
• 6 Skewb centers / FTO corners
• 24 Rex cube points / FTO center triangles

You can see the correspondence between the Rex Cube and Diotipoid in this VIDEO.

Each of these pieces have unusual restrictions that most twisty puzzles don't tend to have:
• The orientation of the Dino edges is fixed and dependent on their position
• The Skewb centers have 4-fold symmetry but only 180-degree twists are reachable
• The Rex cube points are in two orbits
• The orientation of the last Skewb center is determined by the orientation of the other 5
• Due to these properties there are only 12 essentially different orientations to the overall puzzle

And there are parity restrictions on each piece:
• The Dino edge are restricted to even permutations
• The Skewb centers are restricted to even permutations
• The permutation of each orbit of the Rex cube points is even

Given the above, a super-stickered Dioctipoid / FTO / Rex cube should have:

((12! / 2) * (6! / 2) * ((12! / 2)^2) * (2^6 / 2)) / 12
= 13188400838457446891520000000
or
= 1.32 * 10^28

Now, the Dioctipoid 1.0 doesn't have any coloring on the Dino edges at all so they might as well not even be on the puzzle. Also there are 6 groups of 4 identically colored Rex cube points (2 in each orbit).

That should mean:
• The overall orientation of the puzzle still isn't 24 because the parity restriction on the Skewb centers still narrows it to 12
• The per-orbit parity restrictions on the Rex cube points don't apply the same way due to the duplicate colors
• The Skewb centers do not have any visible orientation

Which I count as:
((6! / 2) * ((12! / (2!^6))^2)) / 12
= 1680487300800000
or
= 1.68 * 10^15

The Dioctipoid 2.0 is the same as the 1.0 except that the Dino edges are colored in two groups of 6 and do not have visible orientation.

I count that as:
((12! / (2!^6)) * (6! / 2) * ((12! / (2!^6))^2)) / 12
= 12577439154107520000000
or
= 1.26 * 10^22

These numbers are many orders of magnitude off from the numbers reported by the Dioctipoid author. With the help from Taus and Andreas I feel pretty good about them though.

_________________
Prior to using my real name I posted under the account named bmenrigh.

Last edited by Brandon Enright on Sun Jan 15, 2012 12:48 pm, edited 1 time in total.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Fri Jan 13, 2012 7:34 pm

Joined: Sat Jan 01, 2011 12:54 pm
Location: Copenhagen, Denmark
I tried confirming your numbers in GAP, but I can't get any of it to make sense. It's possible that I made a mistake somewhere, although I'm fairly certain that I've got the definitions correct. Note that my definition is not the true super Rex cube, as I do not distinguish orientation of the centres. Thus, the size is off by 2^5. Also, I do not have a fixed orientation of the cube, hence the size is also 24 times too big.

Here's the code if you want to experiment yourself:

Code:
#             U----------+
#             |    01    |
#             |          |
#             |  02  03  |
#             |          |
#             |04  05  06|
#             |          |
#             |  07  08  |
#             |          |
#             |    09    |
#  L----------F----------R----------B----------+
#  |    10    |    19    |    28    |    37    |
#  |          |          |          |          |
#  |  11  12  |  20  21  |  29  30  |  38  39  |
#  |          |          |          |          |
#  |13  14  15|22  23  24|31  32  33|40  41  42|
#  |          |          |          |          |
#  |  16  17  |  25  26  |  34  35  |  43  44  |
#  |          |          |          |          |
#  |    18    |    27    |    36    |    45    |
#  +----------D----------+----------+----------+
#             |    46    |
#             |          |
#             |  47  48  |
#             |          |
#             |49  50  51|
#             |          |
#             |  52  53  |
#             |          |
#             |    54    |
#             +----------+

rexgroup := Group((03, 20, 34)
(05, 23, 32)
(06, 19, 31)
(07, 26, 30)
(08, 21, 29)
(09, 24, 28),    # Corner twist.

(19, 22, 27, 24)
(20, 25, 26, 21)
(23)

(01, 13, 54, 33)
(02, 16, 53, 30)
(03, 11, 52, 35)
(04, 18, 51, 28)
(05, 14, 50, 32)
(06, 10, 49, 36)
(07, 17, 48, 29)
(08, 12, 47, 34)
(09, 15, 46, 31)

(37, 42, 45, 40)
(38, 39, 44, 43)
(41),            # Rotation.

(01, 04, 09, 06)
(02, 07, 08, 03)
(05)

(10, 19, 28, 37)
(11, 20, 29, 38)
(12, 21, 30, 39)
(13, 22, 31, 40)
(14, 23, 32, 41)
(15, 24, 33, 42)
(16, 25, 34, 43)
(17, 26, 35, 44)
(18, 27, 36, 45)

(46, 51, 54, 49)
(47, 48, 53, 52)
(50));            # Rotation.

rexfaces := [[01..09],[10..18],[19..27],[28..36],[37..45],[46..54]];

dioctopoid1faces := [[01, 04, 06, 09,
10, 13, 15, 18,
19, 22, 24, 27,
28, 31, 33, 36,
37, 40, 42, 45,
46, 49, 51, 54],
[02, 03, 05, 07, 08],
[11, 12, 14, 16, 17],
[20, 21, 23, 25, 26],
[29, 30, 32, 34, 35],
[38, 39, 41, 43, 44],
[47, 48, 50, 52, 53]];

dioctopoid2faces := [Set([02, 03, 05, 07, 08,   37, 01, 09, 19]),
Set([11, 12, 14, 16, 17,   04, 10, 18, 49]),
Set([20, 21, 23, 25, 26,   15, 22, 24, 31]),
Set([29, 30, 32, 34, 35,   06, 28, 36, 51]),
Set([38, 39, 41, 43, 44,   33, 40, 42, 13]),
Set([47, 48, 50, 52, 53,   27, 46, 54, 45])];

rexidsol := Stabilizer(rexgroup,rexfaces,OnTuplesSets);
dioctopoid1idsol := Stabilizer(rexgroup,dioctopoid1faces,OnTuplesSets);
dioctopoid2idsol := Stabilizer(rexgroup,dioctopoid2faces,OnTuplesSets);

rexsize := Size(rexgroup);

Here are the corresponding sizes we get:
Code:
gap> rexsize;
9891300628843085168640000000
gap> rexsize/Size(rexidsol);
9659473270354575360000000
gap> rexsize/Size(dioctopoid1idsol);
40331695219200000
gap> rexsize/Size(dioctopoid2idsol);
301858539698580480000000

I'll try to look at it again tomorrow.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Fri Jan 13, 2012 8:00 pm

Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
Taus wrote:
[...]
Here are the corresponding sizes we get:
Code:
gap> rexsize;
9891300628843085168640000000
gap> rexsize/Size(rexidsol);
9659473270354575360000000
gap> rexsize/Size(dioctopoid1idsol);
40331695219200000
gap> rexsize/Size(dioctopoid2idsol);
301858539698580480000000

I'll try to look at it again tomorrow.
Thanks a lot, defining that group looks like a lot of work!

Your rexsize number is the same as mine:

? 13188400838457446891520000000 / 9891300628843085168640000000
%1 = 4/3

Since you're not counting the 2^5 center orientations but you are counting the 24 orientations that are the same:

? 32 / 24
%1 = 4/3

Our other numbers are very different. (EDIT: This has been explained, see below.)

I don't have any GAP experience but I'm downloading and compiling it now so I'll see if I can sort out the differences.

_________________
Prior to using my real name I posted under the account named bmenrigh.

Last edited by Brandon Enright on Sun Jan 15, 2012 12:49 pm, edited 1 time in total.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Sat Jan 14, 2012 11:42 pm

Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
EDIT: With help from Andreas I have corrected the calculations below to treat the orbits more independently.

Since I went through all of the work to think about these puzzles so thoroughly I figure I should capitalize on on that and also count the permutations for the FTO and Rex cube.

Following from my original post, the FTO has corners (Skewb centers) with visible orientation, Dino edges with visible orientation, and duplicate colored Rex cube points (8 groups of 3) which allow for any permutation in each orbit.

It should have:

((12! / 2) * (6! / 2) * ((12! / (3!^4))^2) * (2^6 / 2)) / 12
= 31408133379194880000000
or
= 3.14 * 10^22

The Rex Cube is the same as the FTO except that the centers do not have visible orientation and there are 6 groups of 4 identical points so it should have:

((12! / 2) * (6! / 2) * ((12! / (2!^6))^2)) / 12
= 402478052931440640000000
or
= 4.02 * 10^23

EDIT: The discrepancies below have been explained, see the post by Andreas below.
The Rex cube number also doesn't match the GAP results

I just realized that all of my numbers differ from the GAP results by a constant factor:

Rex Cube:
? 9659473270354575360000000 / 8626501477440000000
%1 = 1119744

Dioctipoid 1.0:
? 40331695219200000 / 36018675000
%2 = 1119744

Dioctipoid 2.0:
? 301858539698580480000000 / 269578171170000000
%3 = 1119744

1119744 is 2^9 * 3^7 but I can't think of why GAP would be counting those factors. At least if my numbers are wrong and GAP is right, I'm consistently wrong.

_________________
Prior to using my real name I posted under the account named bmenrigh.

Last edited by Brandon Enright on Sun Jan 15, 2012 12:56 pm, edited 2 times in total.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Sun Jan 15, 2012 9:16 am

Joined: Mon Aug 02, 2004 7:03 am
Location: Koblenz, Germany
bmenrigh wrote:
The Rex Cube is the same as the FTO except that the centers do not have visible orientation and there are 6 groups of 4 identical points so it should have:

((12! / 2) * (6! / 2) * ((12! * 12!) / (4!^6))) / 12
= 8626501477440000000
or
= 8.63 * 10^18

The Rex cube number also doesn't match the GAP results
I can explain this:
On the FTO identically colored X-Faces (aka points) are in the same orbit.
On the Rex Cube identically colored X-Faces are in different orbits.
Correct is:

((12! / 2) * (6! / 2) * ((12! * 12!) / (2!^12))) / 12
= 402478052931440640000000
Compared with this Taus number is to high by a factor of 24576 = 24 * 1024.
24 steems from the global orientation Taus allows.
1024=2^10 because Taus assumes the X-Faces are distinguishable.

On a side note:
In this thread I see Taus method to define a puzzle group the second time. I must say it is very clever and I am tempted to adapt it for me. Its advantage is one needs to define only three generators no matter how many axis are there.
Its disadvantage is that it introduces global even in puzzles like the 3x3x3 where there could be pieces fixed in space. To overcome this I used Taus definition and GAP to create more traditional generators:
Code:
gens:=GeneratorsOfGroup(rexgroup);
newgens:=ConjugacyClass(Group([gens[2],gens[3]],gens[1]));
AsList(newgens);
Code:
[ (3,20,34)(5,23,32)(6,19,31)(7,26,30)(8,21,29)(9,24,28),
(12,52,26)(14,50,23)(15,49,27)(16,48,20)(17,47,25)(18,46,22),
(30,48,44)(32,50,41)(33,51,45)(34,52,38)(35,53,43)(36,54,40),
(1,42,10)(2,39,11)(3,44,12)(4,37,13)(5,41,14)(7,38,16),
(21,47,35)(23,50,32)(24,46,36)(25,53,29)(26,48,34)(27,51,31),
(2,17,21)(4,15,19)(5,14,23)(7,12,20)(8,11,25)(9,10,22),
(11,43,47)(13,45,49)(14,41,50)(16,44,52)(17,39,53)(18,42,54),
(1,28,40)(2,29,43)(3,30,38)(5,32,41)(6,33,37)(8,35,39) ]

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Sun Jan 15, 2012 12:40 pm

Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
Andreas Nortmann wrote:
bmenrigh wrote:
The Rex Cube is the same as the FTO except that the centers do not have visible orientation and there are 6 groups of 4 identical points so it should have:

((12! / 2) * (6! / 2) * ((12! * 12!) / (4!^6))) / 12
= 8626501477440000000
or
= 8.63 * 10^18
I can explain this:
On the FTO identically colored X-Faces (aka points) are in the same orbit.
On the Rex Cube identically colored X-Faces are in different orbits.
Correct is:

((12! / 2) * (6! / 2) * ((12! * 12!) / (2!^12))) / 12
= 402478052931440640000000
Good catch, that was quite dumb of me. In the beginning I couldn't decide if I should do it per-orbit or take the short route like I did. I reasoned that since 4^6 is the same as 2^12 it didn't matter. Of course the factorial totally changes that. That also explain where those extra 7 factors of 3 came from in 1119744. 6 of them from my mistake, and one of them from 24.

I will go back and correct my numbers.

Andreas Nortmann wrote:
Compared with this Taus number is to high by a factor of 24576 = 24 * 1024.
24 steems from the global orientation Taus allows.
1024=2^10 because Taus assumes the X-Faces are distinguishable.
I don't understand this factor of 1024. I thought doing Stabilizer(...) with rexfaces := [[01..09],[10..18],[19..27],[28..36],[37..45],[46..54]] was like saying groups of pieces were indistinguishable. What would have to be done to make the X-faces indistinguishable?

Andreas Nortmann wrote:
On a side note:
In this thread I see Taus method to define a puzzle group the second time. I must say it is very clever and I am tempted to adapt it for me. Its advantage is one needs to define only three generators no matter how many axis are there.
Its disadvantage is that it introduces global even in puzzles like the 3x3x3 where there could be pieces fixed in space. To overcome this I used Taus definition and GAP to create more traditional generators:
Code:
gens:=GeneratorsOfGroup(rexgroup);
newgens:=ConjugacyClass(Group([gens[2],gens[3]],gens[1]));
AsList(newgens);
Code:
[ (3,20,34)(5,23,32)(6,19,31)(7,26,30)(8,21,29)(9,24,28),
(12,52,26)(14,50,23)(15,49,27)(16,48,20)(17,47,25)(18,46,22),
(30,48,44)(32,50,41)(33,51,45)(34,52,38)(35,53,43)(36,54,40),
(1,42,10)(2,39,11)(3,44,12)(4,37,13)(5,41,14)(7,38,16),
(21,47,35)(23,50,32)(24,46,36)(25,53,29)(26,48,34)(27,51,31),
(2,17,21)(4,15,19)(5,14,23)(7,12,20)(8,11,25)(9,10,22),
(11,43,47)(13,45,49)(14,41,50)(16,44,52)(17,39,53)(18,42,54),
(1,28,40)(2,29,43)(3,30,38)(5,32,41)(6,33,37)(8,35,39) ]
My solving program takes a very similar approach. To define a puzzle I define which pieces move in a twist and two rotations of the whole puzzle. In this way the program finds all of the orientations as well as all of the pieces and their properties.

One subtle limitation though is that I currently have to define the rotation of the whole puzzle along the same axis as one of the twists. For both the edge turning cube and edge turning dodecahedron only two rotations isn't sufficient to reach all of the orientations of the puzzle (GuiltyBystander ran into this before I did). If you define rotations about a face or corner you won't run into this issue.

_________________
Prior to using my real name I posted under the account named bmenrigh.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Mon Jan 16, 2012 11:17 am

Joined: Mon Aug 02, 2004 7:03 am
Location: Koblenz, Germany
bmenrigh wrote:
Andreas Nortmann wrote:
Compared with this Taus number is to high by a factor of 24576 = 24 * 1024.
24 steems from the global orientation Taus allows.
1024=2^10 because Taus assumes the X-Faces are distinguishable.
I don't understand this factor of 1024. I thought doing Stabilizer(...) with rexfaces := [[01..09],[10..18],[19..27],[28..36],[37..45],[46..54]] was like saying groups of pieces were indistinguishable. What would have to be done to make the X-faces indistinguishable?

That was a quick but devastating mistake. I wanted to write "because Taus assumes the X-Faces are INdistinguishable."

Stabilizer(...) creates a new subgroup were the mentioned pieces stay within there defined Sets, Tuples, Points, etc. It is like saying saying "Create the subgroup with distinguishable permutations of the Super Rex Cube where the Dioctipoid's permutations all look the same." Therefore Taus calculates the Dioctipoid's size by dividing the Super Rex Cube's size by the reduced size.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Sat Jan 21, 2012 12:34 pm

Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
While twisting one of my Dioctipoids this morning it occurred to me that because neither has any 2-color pieces, you can solve the puzzles into different color schemes. I'd expect that most solvers would be perfectly happy to solve the puzzles into any color scheme, not just the original one.

This is somewhat outside of the scope of the "standard" way to count permutations so I don't think it should be included in the "official" numbers.

Also, I'm not quite sure how to count the number of states that are reduced by this. It isn't just 6! / 2 because some permutations of the colors are the same as re-orienting the puzzle and those were already counted.

Since only even permutations of the colors are possible, I suspect only 3-cycles and 5-cycles of the colors are allowed. I think this would be done as (6 choose 3) * 2 + (6 choose 5) * 4 which is 64. If this is correct the Dioctipoid numbers could be reduced by a factor of 64 by counting alternate color schemes.

Edit: as I think about this more, I think even my 64 is an over-count. Take a 3-cycle of three colors around the vertex of a cube. If you cycle those you can "undo" the cycle by re-orienting the puzzle around the vertex, thereby causing three other colors to look cycled instead. My 64 answer counts this twice.

_________________
Prior to using my real name I posted under the account named bmenrigh.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Thu Feb 28, 2013 4:57 pm

Joined: Sat Aug 13, 2011 5:19 pm
Hi All,

I've been checking my math and will be the first to admit that i'm not perfect at sums. For instance, I think my 6! in my analysis of Rubik's Cube is wrong and should be deleted, advice sought.

A sincere thanks due for buying one, really appreciated - have you solved it?

I'm still stuck on fluke-ing out three stars on a good day,a repeatable solution algorithm still escapes me.

So, on the subject of sums, there are 24 triangle bits so there should be a 24!/x term in there somewhere, I still don't know if the x is 2, 3, or 4.

On the subject of orders of magnitude, 42! is significantly higher than 26!, the scale of legal permutations between the two will be related and eventually proved.

10^22 is maybe on the low side when compared against the the MIT Rubik analysis figure of 10^19.

Best regards,

Gary (Dioctipoid Author)

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Thu Feb 28, 2013 5:58 pm

Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
Hi Gary, welcome to the forum! Normally bumping an old topic is poor form and against the rules but I think we can make an exception for the puzzle designer

Designersaurus wrote:
A sincere thanks due for buying one, really appreciated - have you solved it?
I've physically solved both 1.0 and 2.0 more than once each.

Designersaurus wrote:
So, on the subject of sums, there are 24 triangle bits so there should be a 24!/x term in there somewhere, I still don't know if the x is 2, 3, or 4.
At first blush it sure seems like this would be the case but it isn't. Even though there are 24 triangles, they come in two orbits that never mix. This was just discussed in another thread recently: Piece orbits on the Master Skewb. There are two orbits, each with 12 pieces. Also, the permutation of each orbit is always even.

This means instead of 24! / x it's actually (12! / 2) * (12! / 2). Edit: these numbers assume pieces are distinguishable. Since there are 6 pairs of identical pieces in each orbit, the actual number for the Dioctipoid triangles is (12! / (2! ^ 6)) * (12! / (2! ^ 6)) and you don't have to worry about the parity of the orbits.

It is the orbits of the pieces and the twist restrictions on the diamonds and center squares that makes counting the group size a bit trickier.

_________________
Prior to using my real name I posted under the account named bmenrigh.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Tue Mar 05, 2013 5:47 pm

Joined: Sat Aug 13, 2011 5:19 pm
Hi All,

Retraction for me here and also to be posted up on my own site.

Bmenrigh is indeed right, some work and serious thinking with a prototype marked up with a pen shows that the triangles behave in a counter-intuitive manner. A triangle touching "a pole" or rotation axis will not migrate to the opposing pole. Even if you spend a couple of hours with it and ask nicely. I really thought the puzzle was fully polymorphic, so these mistakes are mine.

Also, and again by observation, the squares orientate "North/South" and refuse to assume an "East/West" Orientation x2? The diamonds also appear in a fixed orientation pattern so i think the orientation is also x1.

So guessing the parities for squares, triangles and diamonds as 2, 3 and 3 respectively and arguing by analogy to the MIT paper i get :-

(6! . 2^6) . 12! . 12! . 12!
2 3 3 3

= 9.38 x 10^28

So, hardly mathematical due to the guess work, but submitted here for discussion or review.

Mind if I ask that we keep this forum up and running without worrying about bumping? I'm interested to see what this number actually is.

Regards and respect,

Gary

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Tue Mar 05, 2013 11:20 pm

Joined: Thu Dec 21, 2006 5:32 pm
Location: Bay Area, CA
Designersaurus wrote:
Mind if I ask that we keep this forum up and running without worrying about bumping? I'm interested to see what this number actually is.
Part of the definition of a bump is bringing up an old thread without adding significant value. The classic case is a "Cool mod!" post on a puzzle introduced years back.

In the situation you present, someone posting later on with good arguments for a better number would be adding value and therefore most likely justifying the bump. Note that "with good arguments for a better number" is the key issue here. Someone making an uneducated guess wouldn't be adding much and therefore it would be considered a bump.

Dave

_________________

LitwinPuzzles.com has info on my puzzles.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Tue Mar 05, 2013 11:21 pm

Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
Designersaurus wrote:
Retraction for me here and also to be posted up on my own site.
I'm not sure you have to issue a retraction. A correction will do

Designersaurus wrote:
Bmenrigh is indeed right, some work and serious thinking with a prototype marked up with a pen shows that the triangles behave in a counter-intuitive manner. A triangle touching "a pole" or rotation axis will not migrate to the opposing pole. Even if you spend a couple of hours with it and ask nicely. I really thought the puzzle was fully polymorphic, so these mistakes are mine.

As you know, your Dioctipoid puzzles have octahedral symmetry which I mentioned before can be thought of two fused tetrahedrons and the pieces don't mix.

Carl Hoff (user wwwmwww) has a very good demo / discussion about viewing octahedral puzzles in this way here:
http://twistypuzzles.com/forum/viewtopic.php?p=230926#p230926
I'd recommend reading Carl's initial post.

The specific image that is very useful for you is:

Red lines are the cuts created by one of the tetrahedrons and blue lines are from the other.
wwwmwww wrote:
So you have a Master Skewb. But in the first form above you can notice that no matter how you scramble the puzzle Blue parts NEVER mix with Red parts and the Blue cuts never mix with the Red cuts. Its really two Tetrahedral puzzles in one and as I don't have a Master Skewb I didn't realize till now that you couldn't flip the edge pieces. This picture shows you why.

Designersaurus wrote:
Also, and again by observation, the squares orientate "North/South" and refuse to assume an "East/West" Orientation x2? The diamonds also appear in a fixed orientation pattern so i think the orientation is also x1.
Yep. This is also apparent from the image above. You can't twist those center squares by 90 degrees because that would make the red cuts overlap with the blue cuts and vice-versa. The diamonds on your Dioctipoids are the edges in that image above and they can't be twisted in place for the same reason.

Designersaurus wrote:
So guessing the parities for squares, triangles and diamonds as 2, 3 and 3 respectively and arguing by analogy to the MIT paper i get :-

(6! . 2^6) . 12! . 12! . 12!
2 3 3 3

= 9.38 x 10^28

So, hardly mathematical due to the guess work, but submitted here for discussion or review.

Alright it looks like you're trying to calculate the "Super Dioctipoid" where every piece is unique and orientation is visible.

Let's break down your calculation into piece types:

Squares: (6! * 2^6) / 2
Diamonds (edges): 12! / 3
Triangles: (12! * 12!) / 3

I don't understand where you get the 3 for the Diamonds or 3 for Triangles from.

Here is how I think the correct calculation goes:

Squares:
There are 6 of them however it is not possible to swap two squares relative to the other pieces because their permutation parity is always even. This reduces their contribution by a factor of 2. Also, you can not twist a single square independently of the others. If the twist of 5 squares is already define then the 6th square is forced into a single orientation. This further reduces the squares contribution by 2.

Therefore squares should be ((6! / 2) * ((2^6) / 2))

Diamonds:
As you've already noticed, diamonds are forced into an orientation based on the position they are in so you don't have to worry about twisting them. It isn't possible to swap two diamonds so their permutation parity is always even. This reduces their contribution by a factor of 2.

Therefore diamonds should be (12! / 2)

Triangles:
As we've discussed before, triangles are in two orbits or 12 pieces each. Each orbit's permutation parity is always even which reduces the contribution of each orbit by a factor of 2 (4 total).

Therefor triangles should be ((12! / 2) * (12! / 2))

Puzzle orientation:
One thing your calculation left out is that you can take the Dioctipoid in any state including the solved state and reorient it in space 24 ways. These re-orientations don't change anything about the puzzles state but the result of reorientation is that the puzzles looks like it is in a different state. Because we've allowed all of the pieces to move we haven't accounted for these duplicate states caused by reorientation.

At first you'd think you need to divide the count by 24 however because the diamond pieces only have one orientation, you can tell the difference between 12 of the 24 orientations. Put another way, you can fix one of the edges in space and not let it move. If you do this it will prevent you from over counting the states.

Therefore the whole calculation should be reduced by a factor of 12 for duplicate states that collapse via reorientation of the puzzle.

Putting all that together I get:
(((6! / 2) * ((2^6) / 2)) * (12! / 2) * ((12! / 2) * (12! / 2))) / 12
= 13188400838457446891520000000
= 1.31 * 10^28

Designersaurus wrote:
Mind if I ask that we keep this forum up and running without worrying about bumping? I'm interested to see what this number actually is.

As long as you keep posting and I feel like I can provide a meaningful response you can count on it

I come to this forum for the theoretical and mathy stuff and a discussion of solving. If I had my way there'd be a lot more of that. I welcome as many posts to this thread as it takes for us to tease out all of the details of counting the states of these puzzles!

_________________
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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Wed Mar 06, 2013 8:37 am

Joined: Thu Sep 17, 2009 6:07 am
Location: Germany, Bavaria
I found this an interesting Bump!
I had missed this topic and Gary's puzzle completely and have ordered both versions of the Dioctipoid today.
Is there any reason that the two versions are harder to solve than the FTO?
Probably, the colour scheme makes it a bit more confusing?

As these puzzles and the normal FTO have exchangable pieces, what do we get, if we count the visually distinguishable permutations only? (FTO / Disoctipoid V1 / V2) (No Superstickers assumed)

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Wed Mar 06, 2013 6:40 pm

Joined: Sat Aug 13, 2011 5:19 pm
Hi, taking time out after a rough day at work so this may not be strictly to the post but thanks to all for making allowances.

Credit where due, my industrial partners Piers & Paul had a significant & truly innovative input that makes the Dioctipoid what it is i.e. it's not just mine. My design was immature until they fixed it (unless I was going to N.C. machine the thing from solid metal at 0.03mm tols and i did consider it).

Vielen dank, Konrad. Really hope you enjoy! And yes, the idea behind the posts is to find the baseline number for a unique dioctipoid and offer this up as a comparison against the 1,0 and 2,0 as a basis for comparison.

I think the unusual restrictions are what makes these puzzles intrinsically difficult. A question : the triangles are in pairs, a star may be in an incorrect relationship to allow for parity elsewhere?

Mit freundlichen Grüßen,

Gary

(Not a german speaker but i do understand ein bißchen und zwei Bier bitte ).

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Wed Mar 06, 2013 7:08 pm

Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
Designersaurus wrote:
I think the unusual restrictions are what makes these puzzles intrinsically difficult.
Octahedral puzzles are a bit more complicated to analyze however these complications don't really translate into solving difficulty.

Designersaurus wrote:
A question : the triangles are in pairs, a star may be in an incorrect relationship to allow for parity elsewhere?
I wouldn't say the triangles are in pairs. Usually when pieces are said to "come in" or "be in" pairs that means that two pieces that are physically separated from each other act as a single piece and move in tandem. There are many examples of this across the twisty puzzle landscape.

In the case of this puzzle, it would be better to say the triangles "come in groups" or "are in two independent orbits".

I'm not sure what you mean by "a star" but perhaps you mean the 4-triangle + center square arrangement for the color scheme? As far as parities are concerned, the color scheme has no effect for this puzzle. All moves perform an even permutation for all pieces.

With duplicate triangles in an orbit you can make a 3-cycle look like a single pair of pieces are swapped but almost every twisty puzzler knows how to use duplicate pieces in a 3-cycle to resolve this pseudo-parity.

_________________
Prior to using my real name I posted under the account named bmenrigh.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Thu Mar 14, 2013 6:04 pm

Joined: Thu Sep 17, 2009 6:07 am
Location: Germany, Bavaria
My calculation for a physical Dioctipoid V1.0 (see my photo below)
bmenrigh wrote:
...
Here is how I think the correct calculation goes:

Squares:
There are 6 of them however it is not possible to swap two squares relative to the other pieces because their permutation parity is always even. This reduces their contribution by a factor of 2. Also, you can not twist a single square independently of the others. If the twist of 5 squares is already define then the 6th square is forced into a single orientation. This further reduces the squares contribution by 2.

Therefore squares should be ((6! / 2) * ((2^6) / 2))

Diamonds:
As you've already noticed, diamonds are forced into an orientation based on the position they are in so you don't have to worry about twisting them. It isn't possible to swap two diamonds so their permutation parity is always even. This reduces their contribution by a factor of 2.

Therefore diamonds should be (12! / 2)

Triangles:
As we've discussed before, triangles are in two orbits or 12 pieces each. Each orbit's permutation parity is always even which reduces the contribution of each orbit by a factor of 2 (4 total).

Therefore triangles should be ((12! / 2) * (12! / 2))

Puzzle orientation:
One thing your calculation left out is that you can take the Dioctipoid in any state including the solved state and reorient it in space 24 ways. These re-orientations don't change anything about the puzzles state but the result of reorientation is that the puzzles looks like it is in a different state. Because we've allowed all of the pieces to move we haven't accounted for these duplicate states caused by reorientation.

At first you'd think you need to divide the count by 24 however because the diamond pieces only have one orientation, you can tell the difference between 12 of the 24 orientations. Put another way, you can fix one of the edges in space and not let it move. If you do this it will prevent you from over counting the states.

Therefore the whole calculation should be reduced by a factor of 12 for duplicate states that collapse via reorientation of the puzzle.

Putting all that together I get:
(((6! / 2) * ((2^6) / 2)) * (12! / 2) * ((12! / 2) * (12! / 2))) / 12
= 13188400838457446891520000000
= 1.31 * 10^28
...
Brandon's (=bmenrigh's) calculation is for a "super stickered" Dioctipoid where each piece is distinguishable from each other.
On a real V1.0 the number is much lower.
Here is a photo of mine:

Here is how I think the V1.0 specific calculation goes:

Squares: (equivalent to the corners of a Face Turning Octahedron)
Because the squares have no visible orientation we get a factor of
((6! / 2) =360

Diamonds:
The 12 Diaomonds are identical. Therfore, they contribute a factor of 1

Triangles:
As Brandon (bmenrigh) discussed above, triangles are in two orbits or 12 pieces each. Each orbit contains 6 groups of two pieces which look identical.
Therefore all triangles should contribute this factor ((12! / 2 ^6) * (12! / 2^6))= 56,016,243,360,000
(Visually we can get a "parity" when solving it. This is fixed by a 3-cycle involving two identically looking triangles.)

Puzzle orientation:
Please look at Brandon's explanation below, why we are overcounting so far by a factor of 12 (Originally I was not certain, but had assumed a reduction factor of 6).
Putting all that together I get:
((6! / 2) * 1* (12! / 2^6) * (12! / 2^6)) / [color=#0000FF]12 = 1,680,487,300,800,000 = 1.681 * 10^15
This is the same figure as in the corrected post of Brandon on top of this topic from January 2012.

BTW, because the stars (=squares & surrounding triangles) have no given solved state, We can identify (6!/2)/12 = 30 solved states for the V1.0.

EDIT: I had to change the Puzzle Orientation factor to 12.
Brandon explains it clearly in the following post.

EDIT2: Another correction in the number of solved states. I had 30 before corrected it to 60 and after a later post by Brandon I'm back to 30

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Thu Mar 14, 2013 7:23 pm

Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
[...]Because the 6 squares of the V1.0 Dioctipoid have no orientation, the diamonds are identical (all white) and the two triangle orbits contain 6 groups of two identically looking pieces, we have to reduce the calculation by a factor of 6 (instead of 12 in Brandon's calculation above).
Therefore the whole calculation should be reduced by a factor of 6 for duplicate states that collapse via reorientation of the puzzle (turning the whole puzzle around in space).

When you divide the whole calculation by some factor (24 or 12 or 6 or whatever) it's because you over-counted. Sometimes you over-count due to parity, sometimes it's due to orientation, and sometimes it's other reasons.

An octahedron (and cube) have 24 orientations so generally you'd end up over-counting by a factor of 24. For a puzzle like the Rubik's cube this doesn't happen because you always just hold the 6 centers fixed. For a puzzle like the 2x2x2 the typical calculation assumes all 8 corners can move and all but one can twist which results in (8! * 3^7) which is an over-count by a factor of 24.

There are a few ways to handle this sort of case. The approach I take is to just divide by 24 which is:
? (8! * 3^7) / 24
% = 3674160

Another approach would be to hold a single corner fixed in place. Then only 7 corners can move and only 6 of them can twist:
? (7! * 3^6)
% = 3674160

This avoids the over-count so there is no need to divide by the duplicate states due to orientation.

So lets try something similar for the Dioctipoid 1.0. We'll hold a single square fixed in place.

That way the 5 free squares can still move. That calculation would be:
? (5! / 2) * (12! / (2^6)) * (12! / (2^6))
% = 3360974601600000

I think this is the general logic you've followed to suggest a factor of 6 rather than 12.

There is a trick though. It's still an over-count. If you hold a center in place, it's still possible to solve the puzzle into the rotated 180-degree color-scheme. You'd still have to divide by 2 for this over-count.

You'd want to hold an edge fixed in space rather than a square which seems tricky because on the Dioctipoid 1.0 the edges are all indistinguishable.

Here is another argument why it couldn't be 6. If you accept that the fully super-stickered version has 12 orientations that have to be removed, then removing pieces / making them identical could never decrease the number of duplicate orientations.

_________________
Prior to using my real name I posted under the account named bmenrigh.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Fri Mar 15, 2013 6:29 am

Joined: Thu Sep 17, 2009 6:07 am
Location: Germany, Bavaria
Thanks Brandon!
I have corrected my last post above.

I'm still waiting for version 2.0.
These two puzzles arrived after ordering V1.0 and V2.0 from Moulding Innovations:

As you can see, the package labelled V2.0 contains something like V1.0 (just the scripture "Dioctipoid" is missing)
Gary has promised to fix this.

@Gary: Your website still contains calculations where I can not follow (e.g. a Rubik's Cube with centre orientation can be in 88,580,102,706,155,200,000,000 or 8,858*10^22 states. A normal Rubik's Cube has 4.3 * 10^19 states. A "super stickered" Rubik's Cube has 2,048 as much. You wrote: 63,777,673,948,431,800,000,000,000 = 6.4 x 10^25)

You should consider your physical puzzles and consider as well the difference between V1.0 and V2.0.

You mentioned above that you have not solved it yet. I do not know, if you are interested in solving twisty puzzles so much, but a good idea would be to look at the Face Turning Octahedron (FTO) and compare it with your puzzles.
Actually, a Dioctipoid V1.0 is quite easy to solve, if you know essential parts of the FTO method.
If you consider any configuration of the squares as "solved", you have to solve the triangles only.
This is done by a single move sequence.
Can you run the Gelatinbrain program on your computer?
The FTO is Face Turning Octahedron 4.1.2
The following 8 moves will do a clean 3-cycle of FTO centres (equivalent to Dioctipoid triangles):
[UFL&2, UBR, UFL', UBR', UFL'&2, UBR, UFL, UBR']
You can cut and paste it into Gelatinbrain 4.1.2.
The Gelatinbrain octahedron notation is equivalent to the notation of a vertex turning hexahedron and not so easy to understand at the first glance.

In my notation it would be r L R' L' r' L R L' (we call this a [1,3] commutator but you do not need to understand this term )

This picture shows the result on a solved FTO (the right diagram shows the backview and is included to show that nothing else has changed):

My notation (the part only, we need here):
Imagine an octahedron sitting on a table with one vertex pointing at you, the face to your left is named L, the one to your right is R.
The black arrow wants to show the turn of layer r, which is the inner layer below face R.
L is a clockwise turn by 120° of face L. L' is counter clockwise.

The purple arrows show how the 3 centres are moving arround.
At first glance you would think it is a swap of two edges, but actually two centres of the same colour are involved.

Maybe, it is not so easy to transpone this sequence to a Dioctipoid?
On a solved Dioctipoid three differently coloured triangles would be cycled.

This picture shows the result of the sequence on a solved Dictipoid V1.0:

To your left you see the start situation, to your right the result. The arrows show how the triangles have moved arround.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Fri Mar 15, 2013 3:01 pm

Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
BTW, because the stars (=squares & surrounding triangles) have no given solved state, I think we can identify 5!/2 = 60 solved states for the V1.0.

I've given this some more thought.

There are 6 colors and their permutation is always even so that's 6! / 2 = 360 possible permutations. Since reorienting each permutation gives you 11 others (12 orientations) there are actually (6! / 2) / 12 = 30 distinct solved states.

A related question is how many distinct color-independent positions are there? That is, if you took any scramble and did a color-3-cycle of say, red, green, and blue, the scramble would look the same modulo the color change. I think the answer is that you can take this calculation:

? ((6! / 2) * (12! / (2^6)) * (12! / (2^6))) / 12
% = 1680487300800000

Any legal position in the 1680487300800000 can be turned into (6! / 2) other positions by cycling the colors so that just removes the (6! / 2) term resulting in:

? ((12! / (2^6)) * (12! / (2^6))) / 12
% = 4668020280000

_________________
Prior to using my real name I posted under the account named bmenrigh.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Fri Mar 15, 2013 4:21 pm

Joined: Thu Sep 17, 2009 6:07 am
Location: Germany, Bavaria
Thanks again Brandon!

I have solved it now several times to a specific colour scheme or to randomly chosen ones.
The puzzle turns not badly, but it could be better after lubrication.
Has somebody an advice, how I can safely disassemble it?
In a fully assembled state lubrication will not be very successful.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Fri Mar 15, 2013 4:35 pm

Joined: Sun Mar 15, 2009 12:00 am
Location: Jarrow, England
The outer pieces just clip into the inner floating machanism - see here for pictures. You just gently lever up the parts with a screwdriver and they will click apart.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Sat Mar 16, 2013 5:07 am

Joined: Thu Sep 17, 2009 6:07 am
Location: Germany, Bavaria
Thank you Gus!

I have done now a Search for "Dioctipoid" (what I should have done in the first place )
I'm really surprised how many posts I have found and I have no idea, how I have managed it to miss all of them, back in 2011.

When Gary (Designersaurus) bumped this topic, I heard the first time about these puzzles.
I assumed by mistake that this topic was the only one about the Dioctipoid, because there was no link to other forum threads. As Rox has said several times: The search button is your friend
There is even a member with the name Dioctipoid (actually he signed once with "Paul").

Back to the puzzle:
Gus, I plugged off the outer parts without difficulties, actually using the device I had ordered following your advice for taking off the tiles of the Cubetwist Bandaged 3x3x3.
After lubrication the puzzle turns really smooth.

Unfortunately, I'm still waiting for a resolution of the mistake that I received two puzzles V1.0.
The V1.0 is not hard to solve (if you have solved the FTO before, you may find it "exceptionally easy" like Brandon.)
The V2.0 is certainly the more interesting one.
Still, I view the Dioctipoid as a puzzle in its own right and as visually quite different from the Rex or FTO.
It is not completely trivial getting used to the spherical shape.

I can really recommend it, especially at a price of GBP 9.99.

EDIT: I just read feedback from Moulding Innovations. V2.0 is in the mail. I had overlooked the mail in my Spam folder for several days.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Sun Mar 17, 2013 8:30 am

Joined: Sat Aug 13, 2011 5:19 pm
Hello there,

Again, thanks for buying them and advertising the fact - that's a nice thing to say!

RE Math, I'm still working on it when I find the time. Bmenrigh has posted some excellent points with respect to orientation over-count. So when I find the time I intend to post a more coherent article (up for a form of peer review).

What I am thinking seriously about is the parity problem. How to prove that the parity is 2 and not 3? While my engineering math is up to the task of design the theoretical stuff (or proof) is outside my competence, but its fun to think about.

So proofs invited : Are we certain that the parity of (say) the triangle positions are 2 and not 3? The turns are 1/3, not 1/4.

Best regards to all,

Gary

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Sun Mar 17, 2013 1:27 pm

Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
Almost everything I know about counting permutations, parity, twist, and other tricky things I learned from Jaap.

His site is the best, most coherent place for twisty puzzle knowledge out there:
http://www.jaapsch.net/puzzles/

If you want the position calculations for a bunch of different puzzles see:
http://www.jaapsch.net/puzzles/puzstats.htm

http://www.jaapsch.net/puzzles/theory.htm

Speaking for myself, I understood the general effect of a permutation parity: you can't swap two pieces all by themselves when a permutation has a parity restriction. What I didn't understand is the exact details of why or how you could identify such a restriction. The 24 triangles in two different orbits is complicated enough. How can you figure out what the restrictions are when you have so many pieces moving? Fortunately it turns out to be rather easy in most cases.

Ignore permutations for a moment and imagine a simple game where I give you a bunch of numbers in a list and a starting number. You can pick any number from the list and add or subtract it from your starting number to get a new number.

For example:

I give you {2, 4, 8, 16}
You start at 0

There are many, many ways to solve this. You could do 16 + 4 + 2 or 16 + 8 - 2 - 2 or 2 + 2 + 2 ....

You can string together these in an infinite number of ways to get to your goal.

What if your goal is 5? You shouldn't need to do an infinite number of calculations to figure out that's impossible.

Checking to see if 5 could possibly be reachable is super easy. To get to an odd number you either need to start with an odd number or have an odd number available to you in your list. You started at 0 and that's even. Your list only had even numbers. 5 (and 7 and 9 and ...) aren't reachable.

This is called a parity.

Now, this addition / subtraction game is identical to permutations. Instead of looking at adding and subtracting numbers you look at adding and subtracting swaps.

Take 3 different colored pieces:

O O O

If I permute them, how many swaps does it take to reach this?:
O O O

One swap. Very easy to see.

How many swaps does it take to reach this 3-cycle?:
O O O

It's harder to see but it takes 2 swaps. In general, the minimum number of swaps to reach a 3-cycle is always 2 swaps. In fact, a 3-cycle can never take an odd number of swaps. You might want to arrange a bunch of different coins in a row and count the swaps needed to do 3-cycles, 4-cycles, 5-cycles, etc.

What you'll learn is that an N-cycle takes N-1 swaps. A 3-cycle needs 2 swaps, a 4-cycle needs 3, and so on. That makes a 3-cycle an "even permutation" because it takes an even number of swaps. 4-cycles are odd. 5-cycles are even, etc. Yes, an N-cycle is even when N is odd and odd when N is even. This is because the even and odd terms apply to the number of swaps in the permutation, not the order of the cycle.

So to figure out what sort of restrictions there are for permutations on a twisty puzzle like your Dioctipoid, you have to make a list of the operations available to you.

For simple puzzles like the Dioctipoid there is only one turn type. By symmetry, looking at one of the 8 turns is just as good as looking at all 8 of them.

When you do a turn on the Dioctipoid, there are three triangle tips pointing at the axis you're turning that cycle and there are another 6 around the equator of the turn that cycle. The three tips pointing at the axis 3-cycle. It's a bit more complicated but the 6 tips around the equator of a turn actually do two 3-cycles, one cycle for each orbit. The path they take in their cycles looks like a Star of David:
Attachment:

star-of-david.gif [ 3.54 KiB | Viewed 3191 times ]

So one turn of the Dioctipoid does one 3-cycle in one orbit and two 3-cycles in another orbit. 3-cycles are even permutations. Since this is the only way to turn the puzzle, it's impossible to perform an odd permutation of the triangles. Therefor you can never have just two triangles swapped.

As you know, when you count the number of permutations of N-pieces it's just N! or N * (N-1) * (N-1) ... * 2 * 1

When the permutation is restricted to only even permutations you need to eliminate the 2 term because it isn't possible to swap two pieces.

Your suggestion of eliminating the 3 term just doesn't make sense. I've never seen a group where you needed to cancel a factor of 3 in the permutation. I can't imagine what strange structure a group would have to need that and I'm not even sure something like that would be physically possible.

Proving that you don't need to eliminate a factor of 4 is easy. Find any pure 3-cycle for an orbit. This uses 2 swaps therefor a factor of 4 is too much (this also proves 6, 8, 10, .. etc.).

I'd encourage you to practice counting the number of permutations of various puzzles. I'd start with the Megaminx and then move to the Kilominx / Impossiball. Then try the Rubik's cube. If you can get that one try calculating the Curvy Copter (without jumbling).

_________________
Prior to using my real name I posted under the account named bmenrigh.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Sun Mar 17, 2013 2:15 pm

Joined: Thu Sep 17, 2009 6:07 am
Location: Germany, Bavaria
Hi Gary,

(When I wanted to submit this, I saw that Brandon posted in the meantime. So, there is a possible overlap of our two responses I'll submit my post as it stands now, anyway.)

Maybe a different source helps.
Jaap calculates on his page the number of possible permutation of an FTO (Face Turning Octahedron).
My pictures here want to shed light on the relationship between the FTO and the Dioctipoid:

I find the FTO a bit easier to explain what happens if you do a single turn.
I have labelled the 8 faces with my own notation.
I hope you have any kind of octahedron around to look at and you should better understand what I'm talking about.
The lower part shows the FTO re-stickered like a Dioctopoid V1.0.
The FTO corners (pieces with four stickers) correspond to Dioctipoid Squares.
The FTO edges (pieces with two stickers) correspond to Dioctipoid Diamonds.
The FTO centres (pieces with one sticker) correspond to Dioctipoid Triangles.

Here is a picture showing the 8 "faces" of a Dioctipoid V1.0.

If you fix the orientation of a solved FTO in space and just turn faces, you permute 3 corners, 3 edges, 9 centres by a single turn.
Note that the centres on U, F, DR, DL are in one orbit (in one group) and the centres on R,L,B,U are in the second orbit.
Centres of one orbit can never reach the other orbit.

Let's make a U turn as an example:
corners 3-cycle: URFL -> UL(DL)B -> UB(DR)R
edges 3-cycle: UR-> UL -> UB
centres on U face 3-cycle (I use the name of the adjacent corner for a centre): URFL -> UL(DL)B -> UB(DR)R
centres on R,L,B are cycled in two 3-cycles.

We recollect that a 3-cycle is always an even permutation.
Even in this context means the following:
If we start with a given order of elements (like the solved state of the FTO) and permute the elements somehow the resulting order can require either an even number or an odd number of swaps of two elements (transpositions) to go back to the initial order.

Lets take {1,2,3} as an example where we have a start order 1 2 3. (1 is in place 1, 2 in place 2, 3 in place 3)
A 3-cycle 2 3 1 requires an even number of swaps to recreate the initial order:
1. 2 <-> 3: 3 2 1
2. 1 <-> 3: 1 2 3
Any 3- cycle is an even permutation of the initial order.

Our U turn above are just 5 different even permutations, but this means that the overall permutations of all three piece types are even.

BTW, "parity" of a permutation is either even or odd (parity = 2 or 3 doesn't make sense).
If we start with a given state (the FTO solved state in my picture) and permute all pieces in 3-cycles only, we will always need an even number of swaps to recreate the initial state, we have always even parity of all permuted piece types.
If we now calculate the permutations of 6 elements (the FTO corners) as 6!, we count all possible permutations those with even parity and those with odd parity.
Because we know that legal FTO turns produce even permutations only, we know that we have over counted by a factor of two.

Here is the Wikipedia link to Permutation.
Quote:
All permutations are then classified as even or odd, according to the parity of the transpositions in any such expression.
In twisty puzzle language the term "parity" is sometimes used in a sloppy way expressing something like "strange situation", but I doubt that you ever read about "parity = 3".

In the case of the Dioctopoid we can see situations that seem to require a single transposition of two triangles. This comes from the fact that in each orbit (group of triangles) we have six groups of two indentically looking triangles.
So, what is required indeed is a 3-cycle involving two identical triangles.

I hope this helps.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Mon Mar 18, 2013 6:30 am

Joined: Thu Sep 17, 2009 6:07 am
Location: Germany, Bavaria
....
In the case of the Dioctopoid we can see situations that seem to require a single transposition of two triangles. This comes from the fact that in each orbit (group of triangles) we have six groups of two indentically looking triangles.
So, what is required indeed is a 3-cycle involving two identical triangles.....
I want to elaborate this last remark.
This picture sequence explains most of it (trust me, the Dioctipoid is solved unless the last two Triangles ):

The first picture shows the situation where seemingly a single swap is needed (indicated by the double arrow). That would mean odd parity!
A single swap is impossible on a puzzle that can do even permutations only, as Brandon and I have explained in some length above.

The second picture shows the 3-cycle we have to do instead.

Note that I have shown the sequence r L R' L' r' L R L' before. Here we need an anti clockwise cycle and use just the inverse
L R' L' r L R L' r'.
I hope that after all of this the calculation quoted from above makes sense:
Quote:
Here is how I think the V1.0 specific calculation goes:

Squares: (equivalent to the corners of a Face Turning Octahedron)
Because the squares have no visible orientation we get a factor of
((6! / 2) =360

Diamonds:
The 12 Diaomonds are identical. Therfore, they contribute a factor of 1

Triangles:
As Brandon (bmenrigh) discussed above, triangles are in two orbits or 12 pieces each. Each orbit contains 6 groups of two pieces which look identical.
Therefore all triangles should contribute this factor ((12! / 2 ^6) * (12! / 2^6))= 56,016,243,360,000
(Visually we can get an odd "parity" when solving it. This is fixed by a 3-cycle involving two identically looking triangles.)

Puzzle orientation:
Please look at Brandon's explanation above (March 15th), why we are overcounting so far by a factor of 12

Putting all that together I get:
((6! / 2) * 1* (12! / 2^6) * (12! / 2^6)) / 12 = 1,680,487,300,800,000 = 1.681 * 10^15
This is the same figure as in the corrected post of Brandon on top of this topic from January 2012.

BTW, because the stars (=squares & surrounding triangles) have no given solved state, we can identify (6!/2)/12 = 30 solved states for the V1.0.
The most tricky part may still be the contribution of the Triangles.
If you accept the existence of the two orbits, we look at just one orbit first.
If we had 12 different triangles in one orbit we would get 12!.
But we do have six groups of two identical triangles.
Here I googled one of many explanations why the 12! has to be divided by (2!*2!*2!*2!*2!*2!) = 2^6.
So one orbit contributes (12!/2^6).
So we get for the two orbits: (12! / 2^6) * (12! / 2^6))

EDIT: Today, I have received the V2.0
Thanks Gary, thanks Paul!

Now, we should not have a problem to adapt the calculation above to the V2.0.

Only the term for the Diamonds will change from 1 to (12!/2^6)= 7,484,400
The logic for this is the very same as for one orbit of Triangles!

The overall calculation is now
((6! / 2) * (12!/2^6) * (12! / 2^6) * (12! / 2^6)) / 12 = 12,577,439,154,107,500,000,000 =1.26 * 10^22.

(same figure as in Brandon's original post)

To solve the Diamonds the following 3-cycle comes in handy:
Gelatinbrain: [UFL&2, URF, UFL'&2, URF&2, UFL&2, URF', UFL'&2, URF'&2] or for those familiar with the commutator/ conjugate notation: [[UFL&2:URF],URF&2]
In my notation: r U r' u r U' r' u'

The reult shown on a solved V2.0 as in the picture above:

The arrows show which 3 Diamonds have moved after executing the move sequence (and nothing else!).

I tought, we shouldn't start a seperate topic for the Dioctipoid in the Solving forum, because the twisty puzzlers familiar with the FTO will have no problem to search for FTO solutions. It requires a bit time getting used to the different shape and the different colour scheme, otherwise any known FTO strategy can be applied.

I included here the two 3-cycles for Triangles and Diamonds for the convenience of the Dioctipoid father Gary in this math oriented thread.

EDIT: I included a picture showing the result of the Diamonds 3-cycle above.

_________________

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Tue Mar 26, 2013 7:23 pm

Joined: Sat Aug 13, 2011 5:19 pm
Hi All,

I've had another go at the math on this and plan on submitting this thread to a suitable peer. However, the number I get for a unique solution dioctipoid (which I do intend to make available) is :-

1,055 X 10^29, method as described on http://www.dioctipoid.com/Math.html

Very close to the 10^28 figure so on order-of-magnitude i think we are close there. We are in disagreement on the parity figure but consider, as the MIT paper applies a parity of 0,5 to the whole device not to each individual term, so have I.

I'll freely admit that I could still be way wrong but I am still very interested in knowing the number, so criticism of the math truly invited.

Respect and regards,

Gary

PS: Who's solved it, what version and on what date? Please write to me and let me know, I'll respect privacy ie i would like to publish Avatars/nicknames on a hero-wall type of page.

info@designersaurusrexltd,com

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Wed Mar 27, 2013 12:42 am

Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
I find your math page very hard to read. Here is what I see:
Attachment:

dioctipoid_math_page.png [ 20.04 KiB | Viewed 2902 times ]

I highly recommend using MathJax. Using the TeX input:
Code:
<p>$$6! \cdot 2^6 \cdot 11! \cdot 12! \cdot 12! \over 2 \cdot 2$$</p>

Produces:
Attachment:

dioctipoid_mathjax.png [ 1.39 KiB | Viewed 2902 times ]

In the case of your calculation, you hold an edge fixed so that results in the 11! term instead of a 12! term.

For the two 2 terms in the denominator, you say:

"Only 1/2 of the permutations have the rotations of the central squares in possible
orientations."

Which is basically saying you can't twist a single square in isolation.

And:

"But in the Dioctipoid, the permutations of the pieces are dependent on the positions of
related pieces and only 1/2 of these have the correct segment arrangement parity; we can
not manipulate a single piece in isolation for instance."

This statement is true but it's not the whole story.

You reference the MIT paper for using a factor of 2 in this case.

Unfortunately the Rubik's cube has a more complicated parity restriction. On the Rubik's cube you CAN change the parity of the edges and the corners. A single quarter-turn is a 4-cycle of edges and a 4-cycle of corners. The trick is that their permutation parity is connected. You can't change the permutation parity of the corners without also changing it for the edges, and vice-versa. The MIT paper doesn't call this out as explicitly as it should. They simply count the total number of "cubies" exchanged in a turn. That is, they're counting edges and corners. This works for the Rubik's Cube but is a dangerous shortcut that can't easily be generalized without being explicit about the parity each piece type.

So between both the edges and the corners on a Rubik's cube, there is only a single factor of 2.

The Dioctipoid is a different story. There is no connection between the pieces. Each piece type has a parity restriction independent of the other piece types.

So instead of your single factor of 2 for all parities, you need:

• 2 for the permutation of the squares
• 2 for the permutation of the edges (holding one still isn't enough)
• 2 for the one orbit of the triangles
• 2 for the other orbit of the triangles

All this is in addition to the 2 already there for the twist of the corners. It's three additional factors of 2 beyond what you have right now. If you do that you'll match my calculation.

_________________
Prior to using my real name I posted under the account named bmenrigh.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Wed Mar 27, 2013 6:07 am

Joined: Sat Aug 13, 2011 5:19 pm
Hi again,

Bear in mind that i recognize that I am not a trained mathematician and this is really uncharted territory for me. The number I'm throwing up is a hypothesis (guess) for discussion and clearly caveat-ed as such. I'm also aware that i'm mangling the terminology.

Reading the above posts, the crux of the discrepancy is my reliance on the MIT paper and applying a global parity, bmenrigh applies a parity separately to the subgroups or types. Agreed?

I think bmenrigh is probably more on the money here then I am and may have got the number exactly right in his first post (respect, i appreciate the education here) so let's keep on going. I'll be hitting the books again to try to understand the group theory part of the MIT paper. I published my page way too early, culpa mea.

Also, I need to solve my own puzzle to relate this back to the math and so illustrate which way the sum goes. Bear with me, I was fairly slow solving the Rubik's Cube during my school-days as well so this may take a while.

Regards,

Gary

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Wed Mar 27, 2013 1:53 pm

Joined: Thu Sep 17, 2009 6:07 am
Location: Germany, Bavaria
Designersaurus wrote:
..Reading the above posts, the crux of the discrepancy is my reliance on the MIT paper and applying a global parity, bmenrigh applies a parity separately to the subgroups or types. Agreed?
...
Gary
I do not know the MIT paper you are referring to, but it is crucial to understand two things:

1. Each twisty puzzle has one or more piece types, like corners , edges and centres of the 3x3x3.
A piece of type A can never be permuted with a piece of type B. So, if you can calculate the number of permutations of type A and multiply it with the number of permutations of type B that can be achieved independently (i.e. for each permutation of type A, you can produce all calculated permutations of type B), you'll get the total number of the permutations of both piece types.

2. For most twisty puzzles you cannot get all possible permutations, you could create via physical rearrangement of the puzzle pieces. So, if you are counting first the number of all physical permutations, you have to find out by which factor you have over-counted, if you are interested in the permutations that can be reached by legal turns only. If you can prove e.g. that you can reach such permutations, only, that need an even number of swaps of two elements (the math term for this is "transpositions") of a given piece type, than you have over-counted by a factor of 2. You have to divide the original count by 2.

In the case of the Dioctipoid you can convince yourself that each single turn produces several 3- cycles, but only 3-cycles and a 3-cycle is always an even permutation (requiring an even number of transpositions to go back to the state before).
So, for each piece type you have to divide the number of overall (reachable by physical rearrangements) permutations by 2.

I can confirm that the calculations done by Brandon on March 6th are correct (and they match the figures in its original post).

We should look at each step of the calculations and find out where you agree or not(I'll add a few comments in <blue>):
bmenrigh wrote:
...Alright it looks like you're trying to calculate the "Super Dioctipoid" where every piece is unique and orientation is visible.
....
Here is how I think the correct calculation goes:

Squares:
There are 6 of them however it is not possible to swap two squares relative to the other pieces because their permutation parity is always even. This reduces their contribution by a factor of 2. Also, you can not twist a single square independently of the others. If the twist of 5 squares is already defined then the 6th square is forced into a single orientation. This further reduces the squares contribution by 2.

Therefore squares should be ((6! / 2) <divided by 2, because even permutations are reachable only by twisting>* ((2^6) / 2))<the 2^6 would be reached if each square can be in two orientations; due to the fact that a single square can never be reoriented on its own, we have to divide by 2>

Diamonds:
As you've already noticed, diamonds are forced into an orientation based on the position they are in so you don't have to worry about twisting them. It isn't possible to swap two diamonds so their permutation parity is always even. This reduces their contribution by a factor of 2.

Therefore diamonds should be (12! / 2)<divided by 2, because even permutations are reachable only by twisting>

Triangles:
As we've discussed before, triangles are in two orbits or 12 pieces each. Each orbit's permutation parity is always even which reduces the contribution of each orbit by a factor of 2 (4 total).

Therefore triangles should be ((12! / 2)<divided by 2, because even permutations are reachable only by twisting> * (12! / 2))<divided by 2, because even permutations are reachable only by twisting>

Puzzle orientation:
One thing your calculation left out is that you can take the Dioctipoid in any state including the solved state and reorient it in space 24 ways. These re-orientations don't change anything about the puzzles state but the result of reorientation is that the puzzles looks like it is in a different state. Because we've allowed all of the pieces to move we haven't accounted for these duplicate states caused by reorientation.

At first you'd think you need to divide the count by 24 however because the diamond pieces only have one orientation, you can tell the difference between 12 of the 24 orientations. Put another way, you can fix one of the edges in space and not let it move. If you do this it will prevent you from over counting the states.

Therefore the whole calculation should be reduced by a factor of 12 for duplicate states that collapse via reorientation of the puzzle.<Do you accept the division by 12?>

Putting all that together I get:
(((6! / 2) * ((2^6) / 2)) * (12! / 2) * ((12! / 2) * (12! / 2))) / 12
= 13188400838457446891520000000
= 1.31 * 10^28

....
Let us find out on which part of this calculation we get a discrepancy.

I contibute to this thread only, to convince you that Brandon is perfectly right. An additional voice may help you finding out where you do not follow so far.

_________________

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Wed Mar 27, 2013 3:45 pm

Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
Designersaurus wrote:
Bear in mind that i recognize that I am not a trained mathematician and this is really uncharted territory for me. The number I'm throwing up is a hypothesis (guess) for discussion and clearly caveat-ed as such. I'm also aware that i'm mangling the terminology.
Few few people on this forum have a degree in mathematics. I have a degree in computer science which, depending on the focus, is either a branch of math or a branch of engineering. From a formal training point of view most of us on the forum are not qualified to say anything important about math.

The great thing about math is that most mathematical statements are unambiguously either true of false. There is an answer to the question "How many distinct states does the super-Dioctipoid have?". The answer is an unambiguous, positive, unique integer.

The way mathematicians solve these debates is with a proof of correctness. My mathematical tool chest is too poor to produce a high quality proof.

Another option would be to use a computer to prove the calculation is right. GAP could do that for us. I don't like computer-aided proofs though because they often provide answers without insight. They're certainly great for checking answers though.

Designersaurus wrote:
Reading the above posts, the crux of the discrepancy is my reliance on the MIT paper and applying a global parity, bmenrigh applies a parity separately to the subgroups or types. Agreed?
It seems that way. I can be more explicit about the difference and probably include images of the difference if that'll help.

Designersaurus wrote:
I think bmenrigh is probably more on the money here then I am and may have got the number exactly right in his first post (respect, i appreciate the education here)
To be fair I made a few mistakes the first time I made the post that I corrected with the help of Taus and Andreas.

Designersaurus wrote:
so let's keep on going. I'll be hitting the books again to try to understand the group theory part of the MIT paper. I published my page way too early, culpa mea.
I found the MIT paper somewhat more confusing than it should have been. I'd suggest using Jaap's writings and calculations as a source.

Also, I performed a calculation of the Super-3x3x3 like you've done on your math page. Your calculation is correct but the numerical expansion of it is wrong.

You have 88,580,102,706,155,200,000,000 but the correct number is 88,580,102,706,155,225,088,000. I checked your formula and it seems you must have used a "double" floating point number. Your calculation appears to have overflowed the 53 bit mantissa of a double which caused the roundoff error.

The issue is obvious when you look at the binary expansion of your number:
10010110000011110111111011110100011011100000000000000000000000000000000000000

The correct number is:
10010110000011110111111011110100011011011111111111110100000010011000000000000

Notice you've lost all of the low precision bits? Floating point is treacherous.

_________________
Prior to using my real name I posted under the account named bmenrigh.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Wed Mar 27, 2013 5:29 pm

Joined: Thu Sep 17, 2009 6:07 am
Location: Germany, Bavaria
bmenrigh wrote:
...
Designersaurus wrote:
Reading the above posts, the crux of the discrepancy is my reliance on the MIT paper and applying a global parity, bmenrigh applies a parity separately to the subgroups or types. Agreed?
It seems that way. I can be more explicit about the difference and probably include images of the difference if that'll help.
.....
Aah, I understand now that this is indeed the crucial point of the misunderstanding.
It's all there in Brandon's post above:
bmenrigh wrote:
Unfortunately the Rubik's cube has a more complicated parity restriction. On the Rubik's cube you CAN change the parity of the edges and the corners. A single quarter-turn is a 4-cycle of edges and a 4-cycle of corners. The trick is that their permutation parity is connected. You can't change the permutation parity of the corners without also changing it for the edges, and vice-versa. The MIT paper doesn't call this out as explicitly as it should. They simply count the total number of "cubies" exchanged in a turn. That is, they're counting edges and corners. This works for the Rubik's Cube but is a dangerous shortcut that can't easily be generalized without being explicit about the parity each piece type.

So between both the edges and the corners on a Rubik's cube, there is only a single factor of 2.

The Dioctipoid is a different story. There is no connection between the pieces. Each piece type has a parity restriction independent of the other piece types.
I don't know if this may help to clarify the difference:

A single turn of a Dioctipoid "face" means
- one 3-cycle of Squares
- one 3-cycle of Diamonds
- one 3-cycle of Triangles in orbit I
- two 3-cycles of Triangles in orbit II
We have 4 piece types (the two orbits of the Triangles can be viewed as different piece types) which can reach even permutations only.
Each piece type can be permuted in pure 3-cycles. ("pure" means all other pieces are not permuted at all). I've shown examples of such pure 3-cycles above.
So all piece groups can independently reach all even permutations. That's why it is mathematically correct to calculate the number of even permutations for all piece groups and we have to divide the contributing factors each by 2.

A single turn of a Rubik's Cube face means
- one 4-cycle of corners (which is an odd permutation)
- one 4-cycle of edges (which is an odd permutation as well)
(- the rotation of the centre piece is no permutation; it just shows up in the 3x3x3 Super Cube calculation)
So, odd permutations of corners can be reached, but never independently.
As you may know from solving the 3x3x3 it is never possible that two corners are swapped, if all other cubies are solved - unless somebody has played a bad joke on you.
You can swap two corners (e.g. with this cuboid sequence R2 U R2 D' R2 U R2 U' R2 U D R2 U') only, if you swap two edges, too.
This is why someone can say "pieces of the 3x3x3" (i.e. corners and edges together) have an even permutation.
Even Jaap does it this way:
Jaap's Puzzle Page wrote:
There are 8 corner pieces with 3 orientations each, 12 edge pieces with 2 orientations each, giving a maximum of 8!·12!·38·212 positions. This limit is not reached because:
The total twist of the cubes is fixed (3)
The total number of edge flips is even (2)
The pieces have an even permutation (2)

This leaves 8!·12!·3^7·2^10 = 43,252,003,274,489,856,000 or 4.3·10^19 positions.
Each individual group - corners or pieces - can be in an odd permuatation, if the other one is too.

I really hope that I help clarifying things and do not add confusion.

In this case:
bmenrigh wrote:
Also, I performed a calculation of the Super-3x3x3 like you've done on your math page. Your calculation is correct but the numerical expansion of it is wrong.

You have 88,580,102,706,155,200,000,000 but the correct number is 88,580,102,706,155,225,088,000. I checked your formula and it seems you must have used a "double" floating point number. Your calculation appears to have overflowed the 53 bit mantissa of a double which caused the roundoff error.
.....
I'm pretty much convinced that Excel is the source of the less precise calculation. I used Excel myself and got the very same deviation as Gary

_________________

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Wed Mar 27, 2013 6:15 pm

Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
I'm pretty much convinced that Excel is the source of the less precise calculation. I used Excel myself and got the very same deviation as Gary

Hmm I guess we've already covered Excel before:

http://twistypuzzles.com/forum/viewtopic.php?p=291999#p291999

In my opinion if a program performs math for a user it better be right or warn the user that the calculation has errors.

The correct calculation for the Rubik's cube is:
? (8! * 12! * 3^8 * 2^12) / (2 * 2 * 3)
% = 43252003274489856000

If you enter =(FACT(8) * FACT(12) * 3^8 * 2^12) / (2 * 2 * 3) into Excel you get:
43252003274489900000

If you Google this number you'll see many others have fallen into the same trap. I wonder how many times Excel has provided incorrect results in situations where precision really matters?

_________________
Prior to using my real name I posted under the account named bmenrigh.

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 Post subject: Re: Dioctipoid "math" (counting permutations)Posted: Thu Mar 28, 2013 7:04 am

Joined: Thu Sep 17, 2009 6:07 am
Location: Germany, Bavaria
There is even a Wikipedia article on Excel precision

Gary, maybe an additional source for the discussed calculation helps.
Have you looked at FTO calculation on Jaap's Puzzle Page?
Here is the calculation quoted directly:
Quote:
The number of positions:
If you were to paint the puzzle with two colours, adjacent faces having different colours, then none of the moves will mix the colours. This shows that the vertex pieces have only two orientations, and the edges only one. Furthermore, the centres split in two sets that don't intermingle. This leaves us with 6 vertex pieces with 2 orientations, 12 edge pieces, and two sets of 12 centre pieces. This gives a combined upper bound of 6!·2^6·12!·(12!)^2 arrangements. This limit is not reached because:

Only an even number of vertex pieces are flipped (2)
The vertex permutation is even (2)
The edge permutation is even (2)
The centres come in identical triplets (3!^8)
The orientation of the puzzle does not matter (12)
Note that the last factor is 12 rather than 24, since we fix the orientation of the puzzle by fixing one unique piece (a vertex or edge), which only has 12 possibilities. The total number of positions is therefore 6!·2^3·11!·(12!)^2 / (3!)^8 = 31,408,133,379,194,880,000,000
Can we convince you that the Dioctipoid is a shape modification of the FTO?
We all agree that the "upper bound" (counting all physically possible permutations) is calculated by 6!·2^6·12!·(12!)^2
Now, by which factor have we over counted?

Obviously, we agree about the /12 (The orientation of the puzzle does not matter ) You take 11! instead of 12! fixing one piece in space.

You divide twice by 2 .
I interpret the first /2 stands for the fact Only an even number of vertex pieces are flipped
The second /2 stands for the even number of permutations and you assume that this is global.

If you follow Jaap, the permutation of corners (Squares) and edges (Diamonds) have to be seen as independent.
I have shown a pure 3-cycle permutation of edges (Diamonds) in my post from March 18th.
That should convince you that Jaap is right counting even permutations for the corners and even permutations of edges only.
If you accept this, another /2 is necessary!

A crucial difference between the Jaap FTO calculation and the Dioctipoid Super Sticker version is this: The centres come in identical triplets (3!^8) for the FTO.
Obviously, we have to eliminated the division by (3!)^8 for the Super Sticker FTO.
Is this the whole story?
In my post from March 15th I have shown a pure 3-cycle of triangles.
Because the piece groups "corners" (Squares) and "edges" are not permuted by the move sequence, this should convince you that the permutations of triangles are independent of the state of corners and edges!

We can get even permutations for triangles in one orbit only - independent of the permutations of everything else.
Yet another /2 is necessary.
And the same argumentation holds for the second orbit.
Yet another /2 is necessary.

So, you should be convinced by now that your calculation is missing three times a division by two.
This accounts for the fact that your number is 8 times higher than the correct value calculated by Brandon.

This may not be a proof in a mathematical sense, because I'm referring to observations in the physical world (move sequences doing pure 3-cycles for different groups of pieces). Still, I hope that it is convincing enough.

You can do me a favour if you use on your page the correct definition of the word "parity" in the mathematical sense. Parity is either even or odd and never 2 or 3 or 1/2.
(You can say that only 1/2 of the number of permuations can be reached, because legal turns lead to permutations with even parity, only.)
Cuber language is here very often something between sloppy or plainly incorrect.
Wikipedia wrote:
Every permutation of a finite set can be expressed as the product of transpositions. Moreover, although many such expressions for a given permutation may exist, there can never be among them both expressions with an even number and expressions with an odd number of transpositions. All permutations are then classified as even or odd, according to the parity of the transpositions in any such expression.
Wikipedia wrote:
In mathematics, the parity of an object states whether it is even or odd.
Quote:
If any total ordering of X is fixed, the parity (oddness or evenness) of a permutation of X can be defined as the parity of the number of inversions ...

EDIT: I corrected a few typos above.

EDIT2: Here is a pure 3-cycle for the corners shown on FTO Gelatinbrain 4.1.2:

In Gelatinbrain notation: [UFL, URF, UFL', DBL, UFL, URF', UFL', URF, UFL, URF, UFL', DBL', UFL, URF', UFL', URF']
= [[UFL:[URF:[UFL':DBL]]],URF]
In my notation: R U R' D R U' R' U R U R' D' R U' R' U' = [R U R' D R U' R', U] = [[R:[U:[R':D]]],U] (16 moves)
The result is shown when you perform this sequence on a solved FTO:

Yet another time only three pieces (corners) are permuted, all other pieces are not.
So, I have shown you 3 pure sequences for the four piece groups: corners, edges, triangles in orbit I, triangles in orbit II.

EDIT3: Synopsis of the argumentation above
1. On the Dioctipoid (and FTO) there are even permutations only. Therefore, for each independently permuted group of pieces the overall count of possible configurations has to be divided by 2 (We can never achieve the odd permutations by legal turns).
2. Four piece groups exist on the Dioctipoid: Squares (6), Diamonds (12), Triangles in orbit I (12), Triangles in orbit II (12)
3. Each piece group can be permuted by even permutations independent of the three others This is shown by defining pure 3-cycles for each piece group.
4. Conclusion: Each individual count of overall countable configurations of a piece group has to be divided by 2.

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