An Upper Bound for the Number of Different
Positions of the Fully-Colored Magic120Cell
The following is an upper bound for the number of permutations
of the fully-colored Magic120Cell. To establish it is the exact
answer (which I am virtually certain of), a number of algorithms
must be found, which I will describe in my explanation.
Here is the upper bound:
(600!/2)*(1200!/2)*(720!/2)*((2^720)/2)*((6^1200)/2)*((12^600)/3)
The exact value was computed using the free computer algebra system
Yacas; here it is in its full glory:
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00000000000000000000000000000000000000000000000000000000000000000000000
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000000000000000000000000000000000
What a number!
Here is an outline of how I derived this upper bound:
First of all, we can count that Magic120Cell has 120
1-colored immobile center pieces, 720 2-colored pieces
(120*12/2), 1200 3-colored pieces (120*30/3), and 600
4-colored pieces (120*20/4); each piece being uniquely
colored. There are three types of rotations to consider
when examining how Magic120Cell can be scrambled
(or solved!)
When a dodecahedral face rotates 72 degrees about a line
connecting the midpoints of two opposite pentagonal faces,
there are two 5-cycles of the 2-colored pieces, six 5-cycles
of the 3-colored pieces, and four 5-cycles of the 4-colored
pieces. Since 5-cycles are even permutations, all permutations
of each type of piece, including facelets, will be even when
this rotation is used.
A dodecahedral face can also rotate 180 degrees about
a line connecting the midpoints of two opposite edges. When
this happens, there are six 2-cycles of the 2-colored pieces,
14 2-cycles of the 3-colored pieces, and ten 2-cycles of the
4-colored pieces. 2-cycles are odd permutations, but
note that each type of piece performs an even number of
2-cycles, making the total for each type of piece even.
Hence, once again, all permutations of each type of piece,
including facelets, will be even when this rotation is used.
Finally, a dodecahedral face can be rotated 120 degrees about
two opposite verticies. For this rotation, there are four
3-cycles of 2-colored pieces, ten 3-cycles of 3-colored pieces,
and six 3-cycles of 4-colored pieces. Since 3-cycles are
even permutations, all permutations of each type of piece,
including facelets, will be even for this rotation. Thus
we see that no matter what type of rotation is used, an even
permutation of pieces and facelets will occur for each
type of piece.
Therefore, the 720 2-coloreds can be permuted
720!/2 ways, dividing by 2 because of the even parity.
Similarly, the 3-coloreds and 4-coloreds can be permuted
1200!/2 and 600!/2 ways, respectively. Multiplying
these three terms together, we obtain an upper bound
for the number of ways the pieces can be permuted without
regards to orientation:
(600!/2)*(1200!/2)*(720!/2)
To show this number is exact, we will have to find 3 algorithms:
one that performs a 3-cycle of any three 2-coloreds without
affecting any other pieces, a 3-cycle of any three 3-coloreds
without affecting any other pieces, and a 3-cycle of any
three 4-coloreds without affecting any other pieces. These
three algorithms, when combined with each other and conjugates
(setup moves), can produce any possible permutation of the
pieces.
Now for the orientations. Since the facelets also undergo
even permutations, all orientations will be limited to even
permutations of facelets.
Therefore 719 2-coloreds can be oriented in any of 2 ways each,
but the last will be determined by the others because of the
even parity. This results in
(2^720)/2
or 2^719 ways of orienting the 2-coloreds. To show
this number is exact, we must find an algorithm that
flips two 2-colored pieces without affecting the others.
For the 3-coloreds and 4-coloreds, I followed closely
the methods of Keane and Kamack in their paper,
"The Rubik Tesseract", modifying their arguments as
neccesary to apply them to Magic120Cell.
Any 3-colored piece can be oriented in 6 different ways.
(not three, because in four dimensions we can reflect
3-colored pieces as well as twist them!) Notice that a
twist (a 3-cycle of the facelets on that piece) is an even
permutation, while a reflection (a 2-cycle of two of the
facelets on that piece) is an odd permutation. Since the
total parity of all of the 3-coloreds must be even,
the first 1199 3-coloreds can be oriented in 6 ways each,
while the last can be oriented in only 3. (If the first
1199 3-coloreds total to an even permutation, the last
3-colored must be one of the 3 even twists, while if
they total to an odd permutation, the last 3-colored
must be one of the 3 odd reflections) This gives a total
of
(6^1200)/2
or (6^1199)*3 ways of orienting the 3-colored pieces.
To show this number is exact, we must find an algorithm
that twists one 3-colored piece without affecting the others,
and an algorithm that reflects two 3-colored pieces without
affecting the others.
Finally, the toughest part. The orientation of the 4-coloreds
required me to generalize the group theory based solution for
the corners of the 3^4 cube by Keane and Kamack to
Magic120Cell.
In their paper, Keane and Kamack first describe that there
are 24 permutations of the facelets of a 4-colored piece,
comprising the S4 group (The symmetric group on four letters.)
They describe orientations using cycle notation of the four faces,
labeled a, b, c, and d.
The 24 different orientations can be broken down into four
crosses, (ab)(cd), (ac)(bd), (ad)(bc), and I, the identity; eight
3-cycles, called twists; six 2-cycles, and six 4-cycles. It
turns out that for n-colored pieces in n-dimensional space,
only even permutations of the facelets can occur, because
the odd permutations are n-dimensional mirror images
(This rule also applies to the 3-dimensional Rubik's Cube,
and to Magic120Cell.) Thus, 2-cycles and 4-cycles cannot
occur in 4-space because they are odd permutations.
Hence, each 4-colored piece can only be oriented in 12 ways.
The even permutations are all possible, and form the alternating
group A4. Keane and Kamack continue by observing that the
crosses are a normal subgroup of the alternating group they
call N. So,
N = [I, (ab)(cd), (ac)(bd), (ad)(bc)]
The cosets of N consist of the twists, which they call S and Z:
S = [(abc), (adb), (acd), (bdc)]
Z = [(acb), (abd), (adc), (bcd)]
They then note that the sets N, S, and Z form the quotient-group
of A4 by N, in which N acts as the identity:
A4/N = [N, S, Z]
We can then see that the group multiplication table is:
N S Z
N N S Z
S S Z N
Z Z N S
From this table, we can see that this quotient-group is isomorphic
to the group of residue classes, mod 3. This means that we can assign
the number 0 to N, the number 1 to S, and the number -1 to Z,
and adding these numbers mod 3 is the same as taking the product
of elements of these three subgroups.
Notice that this entire argument applies equally well to
Magic120Cell as to the tesseract. Now the only thing left
to show is that the sum of the orientations of the 4-coloreds
(counting 0 for an orientation in N, etc.) mod 3 is always the
same, whether to the tesseract or Magic120Cell.
The orientations can be defined by assigning, to each 4-colored
piece, a letter to each facelet and each position of each
facelet. Then each orientation can be described by a 4-letter
string (e.g. ABCD) relative to the position it is occupying.
When pieces or cubies rotate in a cycle, their facelets undergo
n seperate cycles if they have n facelets. The important thing
is that there are always n disjoint cycles. In the tesseract,
every corner rotation boils down to four 4-cycles of facelets
for each cycle of four cubies. In Magic120Cell, it is the same
except it is four 5-cycles. If we can show that in
cycles of any length, the sum of the orientations of the pieces
does not change, we will have proved this for both the tesseract
and Magic120Cell.
Consider four 2-cycles:
ABCD 1
ABCD 2
Each row represents a 4-colored piece. The actual 4-cycles
are vertical in direction. For example:
ABCD 1
CDAB 2
This means that facelet A on piece 1 goes where facelet C
on piece 2 was, etc. In this example, piece 1 performed an
N-twist. Now notice that since we are dealing with 4-cycles,
the facelets of piece 2 must return to the original positions
of the facelets of piece 1. Therefore, piece 2 also performed
an N-twist. It can be checked that if piece 1 performs a Z-twist,
piece 2 performs an S-twist, and if piece 1 performs an S-twist,
piece 2 performs a Z-twist. Therefore, the sum of the values
does not change, and equals zero. (N=0, Z=-1, S=1)
Now we can do a proof by induction to show that four cycles
of any length has this property. Assume that 4 k-cycles always
sum to zero:
ABCD 1
.
.
.
ABCD k (not neccesarily this orientation)
We can think of this as a single k-cycle of the pieces
(1 2 ... k). Now consider adding a piece, to produce the
cycle (1 2 ... k (k+1)). The piece k now replaces piece k+1,
(and hence may change the type of twist it performs) and piece
k+1 replaces piece 1.
Notice that we can write (1 2 ... k (k+1)) as a product of the
two cycles (1 2 ... k)(1 (k+1)). Now by induction, the elements
of the first cycle sum to zero. But notice that the second cycle
also sums to zero by induction. Thus, the cycle (1 2 ... (k+1))
sums to zero as well.
Therefore, no matter what the length of the cycles, the
sum of the values of the orientations equals zero. This
means it is true for both the tesseract and Magic120Cell.
So now we know the sums of the orientations, mod 3, always
equal zero in Magic120Cell. Since an N-twist is 0, we can have
an isolated N-twist without affecting any other pieces.
The value S - Z must therefore be congruent to 0, mod 3.
This means that the first 599 4-colored pieces can each be
in any of 12 orientations. If the value of orientations
up to that point is 0, the remaining value must be an N-twist.
If it is 1, the remaining value must be a Z-twist, and if
it is -1, the remaining value must be an S-twist. In each case
there are four possible orientations left for the last
4-colored piece. Therefore, the upper bound for the orientations
of the 4-colored pieces is
(12^600)/3
or (12^599)*4. To prove this is exact, we must find algorithms
that show that any N-twist can be performed without affecting
the rest of the pieces, and algorithms showing that any Z-twist
can be performed along with any S-twist without affecting any
other pieces. Keane and Kamack were able to find these 20
algorithms, all generated from a single pair of twists. This
technique can probably also be applied to Magic120Cell.
When we multiply these figures with the ones for permutations,
we arrive at the final answer.
I believe that before long, all of the algorithms neccesary to
make this figure exact will be found. I wish to thank Roice
Nelson with the deepest gratitude for creating Magic120Cell,
which without, I would have never had the inspiration to
perform this calculation.
-David Smith
Update: Roice has discovered 3 algorithms that perform 3-cycles
on the 2-colored, 3-colored, and 4-colored pieces. They are
actually less restrictive than the ones that I said would be
required. The algorithm that cycles the 2-coloreds scrambles
the 3-coloreds and 4-coloreds, the one that cycles the
3-coloreds scrambles the 4-coloreds and preserves the 2-coloreds,
and the one that cycles the 4-coloreds preserve the rest of
the pieces. Therefore, as long as they are applied in that
order, any possible permutation of pieces can be achieved.
I am sure that before long the same will also be true for
orientations!