This all started for me with Sicherman dice. They are only a few
decades old (first discovered, as far as I've been able to tell, in the
late '70s).
Sicherman (among others) considered 2d6 and wondered if there were any
other way to label their faces with integers that preserves the
familiar bilinear statistics resulting from rolling them. If we look
at the possible rolls, considering the sum of the numbers showing,
die A 1 2 3 4 5 6
die B +---+---+---+---+---+---+
1 | 2 | 3 | 4 | 5 | 6 | 7 |
+---+---+---+---+---+---+
2 | 3 | 4 | 5 | 6 | 7 | 8 |
+---+---+---+---+---+---+
3 | 4 | 5 | 6 | 7 | 8 | 9 |
+---+---+---+---+---+---+
4 | 5 | 6 | 7 | 8 | 9 |10 |
+---+---+---+---+---+---+
5 | 6 | 7 | 8 | 9 |10 |11 |
+---+---+---+---+---+---+
6 | 7 | 8 | 9 |10 |11 |12 |
+---+---+---+---+---+---+
we get a frequency histogram
2 *
3 **
4 ***
5 ****
6 *****
7 ******
8 *****
9 ****
10 ***
11 **
12 *
Sicherman wondered if there were any other way to label the dice faces
with other integers that results in the same frequency histogram.
It turns out there are an infinite number of ways of doing this, but if
we eliminate changes of the "add 1 to each number on die A and subtract
1 from each number on die B" sort by requiring that each die's minimum
number is 1, the answer is unique (up to swapping the dice and
reordering each die's faces, of course): die A has 1 2 2 3 3 4 and die
B has 1 3 4 5 6 8. This can easily be verified by simple exhaustive
search; counting to 11^12=3138428376721 is impractical, but it is easy
to cut the search space down in various ways (as a rudimentary example,
fix one face of each die at 1 and eliminate 1 as a possible value for
all other faces, and it shrinks to 10^10), and it doesn't take much of
that before exhaustive search is within reach of even a modest desktop.
As a (past) role-playing gamer, I am very aware that six is not the
only possible number of sides for a die. I got curious about the
analogous questions for other die sizes. My program was easily adapted
to other die sizes, including sizes that are difficult to realize
physically (such as d7), and I rapidly found there is a lot of
structure hiding beneath this apparently simple question, and the
answer above is intimately related to the ways the number 6 can be
factored. (It turns out to be closely related to factoring
polynomials; stick with me and we'll get there.)
In particular, consider a d6 as a cross product of a d2 and a d3 (with
suitable labelings). There are two ways to do this:
1---2---3 1---3---5
| | | | | |
4---5---6 2---4---6
There's no need to draw out the whole matrix, though; if we note the
size and increment of the arithmetic progressions in question
1x3 2x3
1---2---3 1---3---5
3x2 | | | 1x2 | | |
4---5---6 2---4---6
(where NxM means a progression of M elements with difference N), the
description of the progressions captures the labeling.
To avoid having to constantly subtract and add 1, below, I'm going to
switch to 0-based numbering, so that a d6 has numbers 0 through 5 and
the histogram of two of them has numbers 0 through 10. This is
strictly a convenience change; by adding and subtracting 1 at suitable
times everything works just as well without it, just slightly more
annoyingly. Sicherman's alternative labeling becomes 0 1 1 2 2 3 on
one die and 0 2 3 4 5 7 on the other.
Here's that diagram renumbered that way:
1x3 2x3
0---1---2 0---2---4
3x2 | | | 1x2 | | |
3---4---5 1---3---5
Now, let's make a small change:
1x3 2x3
0---1---2 0---2---4
1x2 | | | 3x2 | | |
1---2---3 3---5---7
All I've done here is swap the "3x2" with the "1x2", then recompute the
numbers based on the new progressions - and Sicherman's alternative
labeling (in its 0-origin form) has appeared as if out of nowhere.
We can even do things a bit more radical:
1x2 2x3
0---1 0---2---4
1x2 | | 2x3 | | |
1---2 2---4---6
| | |
4---6---8
We now have a d4 and a d9 instead of 2d6, but, upon working out the
frequencies,
0 2 2 4 4 4 6 6 8
0 0 2 2 4 4 4 6 6 8
1 1 3 3 5 5 5 7 7 9
1 1 3 3 5 5 5 7 7 9
2 2 4 4 6 6 6 8 8 10
we see the same histogram of sums reappear.
We can also swap things around different ways, eg 1x2/3x2 on the d4 and
1x3/2x3 on the d9
1x2 1x3
0---1 0---1---2
3x2 | | 2x3 | | |
3---4 2---3---4
| | |
4---5---6
which leads to dice 0 1 3 4 and 0 1 2 2 3 4 4 5 6, or 3x2/3x2 and
1x3/1x3
3x2 1x3
0---3 0---1---2
3x2 | | 1x3 | | |
3---6 1---2---3
| | |
2---3---4
0 3 3 6 0 1 1 2 2 2 3 3 4
or even go with a d3 and a d12 (because 12 has three factors instead of
two, the diagram grows a third dimension):
1x3 1x3
0---1---2 0---1---2
3x2 |\ |\ |\
3-\-4-\-5 \
\ \ \ \ \ \
3x2 \ 3---4---5
\| \| \|
6---7---8
0 1 2 0 1 2 3 3 4 4 5 5 6 7 8
There clearly is some underlying structure going on here. I spent a
good deal of time investigating this from this point of view, computing
(via exhaustive search of various sorts) alternative labelings for d12,
d30, and eventually even d210 (210 being interesting because it is the
smallest number with four distinct prime factors). The details are not
important here, because once I started looking at it this way I found a
point of view that links it directly to a pre-existing field of
combinatorics.
Consider the generating series for a traditional d6
2 3 4 5 6
[1] x + x + x + x + x + x
The generating series for the sum of 2d6 is of course the square of
this, or
2 3 4 5 6 7 8 9 10 11 12
[2] x + 2 x + 3 x + 4 x + 5 x + 6 x + 5 x + 4 x + 3 x + 2 x + x
This can, naturally, be factored back into two copies of the original
generating polynomial. But it can also be factored as
2 3 4 3 4 5 6 8
[3] (x + 2 x + 2 x + x ) (x + x + x + x + x + x )
which corresponds to Sicherman dice.
Simplifying by dividing everything by x (which corresponds to
subtracting 1 from all the dice labels), we can write the generating
polynomial for a 0-orgin d6 as
2 3 4 5
[4] 1 + x + x + x + x + x
This can be factored in two ways (among others)
2 4
[5] (1 + x) (1 + x + x )
2 3
[6] (1 + x + x ) (1 + x )
which correspond to the two ways of diagramming shown above. This of
course allows us to factor [2] (or, rather, its 0-origin form) as
2 3 2 4
[7] (1 + x) (1 + x + x ) (1 + x ) (1 + x + x )
and choosing to multiply out the left two terms and the right two terms
gives us
2 3 2 3 4 5 7
[8] (1 + 2 x + 2 x + x ) (1 + x + x + x + x + x )
which corresponds exactly to the 0-origin Sicherman labeling.
Thus, finding Sicherman-style labelings is isomorphic to finding
factorizations of generating polynomials.
As the Wikipedia page for "Sicherman dice" points out, the theory of
cyclotomic polynomials is of use here; one factorization cited on that
page amounts to writing a 0-origin d6 as the product of three factors:
2 2
[9] (1 + x) (1 + x + x ) (1 - x + x )
The last of these factors does not correspond to a die; viewed as a
die, it is a d1 with one face labeled 0, one face labeled 2, and minus
one face labeled 1. However, nothing says we have to use an
irreducible factorization like that rather than the other
factorizations (into non-irreducible polynomials) like [7].
It's not hard to see that this business of swapping axes around, as
outlined above when generating a d4 and a d9 with the statistics of
2d6, is equivalent to writing generating polynomials as products of
polynomials with coefficients all 1 or 0 and exponents forming
arithmetic progressions. Incidentally, these can always be written as
fractions; for example,
6
4 2 x - 1
[10] x + x + 1 = ------
2
x - 1
The generating polynomial for an MxN axis is (X^(M*N)-1)/(X^M-1).
When working with diagrams, I found that the ways of writing a single
dN corresponded to the ways of building N up by multiplying its factors
together. For example, a d12 can be diagrammed three different ways:
4x3 2x3 1x3
1---5---9 1---3---5 1---2---3
1x2 |\ |\ |\ 1x2 |\ |\ |\ 3x2 |\ |\ |\
2-\-6-\10 \ 2-\-4-\-6 \ 4-\-5-\-6 \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
2x2 \ 3---7--11 6x2 \ 7---9--11 6x2 \ 7---8---9
\| \| \| \| \| \| \| \| \|
4---8--12 8--10--12 10--11--12
which correspond to the three orders of multiplying 2, 2, and 3
together to yield 12: 1->2->4->12 (2,2,3); 1->2->6->12 (2,3,2); and
1->3->6->12 (3,2,2). In terms of generating polynomials, this makes
sense; the first of the above three, for example, corresponds to
writing
12 2 4 12
x - 1 x - 1 x - 1 x - 1
[11] ------- = ------ * ------ * -------
2 4
x - 1 x - 1 x - 1 x - 1
When dealing with diagrams, I noticed something. Working with d6s, I
wondered why I couldn't make anything work with axes like 2x2 or 3x3
that don't appear in any of the ways of drawing the original dice.
(Considering a d2 bearing 0 2 and a putative d18, it's easy to see
that, quite aside from all other considerations, there's no way to fill
in the other numbers on the d18 that works - the odd sums never work
out.) From the generating polynomial point of view, this means that
x^2 + 1 either doesn't divide the 0-origin analog of [2] - [2] divided
by x^2 - or the quotient involves terms with negative coefficients,
which make no sense as die labelings. (In this case, as it happens,
it's the former; division leaves a remainder of 2x.)
I formed two conjectures based on my work with diagrams:
Conjecture A: any dice that "work" can be obtained by factoring the
original dice and recombining them (as outlined above).
Conjecture B: an axis labeled MxN is possible only when M*N divides the
original die size (6, here).
Conjecture A clearly implies conjecture B.
In terms of generating polynomials, these amount to saying that the
generating polynomial for the joint roll of the standard dice admits
only certain factorizations - those based on (x^(M*N)-1)/(x^M-1)
factors, as sketched above. I see no obvious reason this should be so.
What little exhaustive search I've been able to do indicates that these
conjectures are true, but exhaustive search becomes impractical long
before numbers large enough to really hold interest here are reached.
I did exhaustive search on 3d6 and found that the only alternative
labelings consisted of a Sicherman labeling plus a standard d6. This
fits with conjecture A; there is no way to factor 3d6 and reassemble
the factors to form three six-siders that doesn't involve at least one
standard d6.
I did exhaustive search on some of the smaller prime sizes, such as d5
and d7, and found that they have no Sicherman-style alternative
labelings. This also follows from cyclotomic polynomial theory, since
(x^N-1)/(x-1) - the generating function for a dN - is irreducible
whenever N is prime, so the product of two such cannot be factored any
other way at all, never mind any other way that actually corresponds to
dice.
I'm going to go back to drawing diagrams, since polynomial factoring
theory does not, as far as I know, have anything corresponding to the
constraints we are imposing. These constraints basically mean
nonnegative coefficients, not necessarily in the factors, but in the
products resulting after multiplying them back into generating
polynomials for (putative) dice.
The next composite size is 8. The basic diagram is
1x2
0---1
2x2 |\ |\
2-\-3 \
\ \ \ \
4x2 \ 4---5
\| \|
6---7
which of course can be drawn any of various ways. This means our prime
dice, to coin a term, are two each of 1x2, 2x2, and 4x2; omitting the
"x2", which is always the same when working with the d8, and writing
only the other numbers, we can get various ways of reassembling these
prime dice into two d8s, or a d4 and a d16, or even a d2 and a d32.
This is, however, the first size for which three dice do anything new;
3d8 can be factored and reassembled as (to name just one possibility)
1x2/1x2/1x2 + 2x2/2x2/2x2 + 4x2/4x2/4x2, or
0 1 1 1 2 2 2 3 - 0 2 2 2 4 4 4 6 - 0 4 4 4 8 8 8 12. (There are of
course a lot of other ways of assembling three d8s out of those
factors, many of which don't involve any standard d8s.) I'll be
confining myself to considering only two base dice for most of the rest
of this, though.
The d9 has nothing new to offer. We can do 1x3/1x3 and 3x3/3x3, we can
do d3/d27, and that's about it here.
The d10 is mildly interesting, because its factors are unequal, but it
doesn't really have anything to say we didn't get from the d6.
The d12 begins to get interesting. 12 factors as 2x2x3, leading to
diagrams for the base dice as shown above.
If we restrict ourselves to plans that generate two 12-siders, we can
diagram the resulting (Sichermanoid) dice as
Ax3 Dx3
0---x---x 0---x---x
Bx2 |\ |\ |\ Ex2 |\ |\ |\
x-\-x-\-x \ x-\-x-\-x \
\ \ \ \ \ \ \ \ \ \ \ \
Cx2 \ x---x---x Fx2 \ x---x---x
\| \| \| \| \| \|
x---x---x x---x---x
I then did exhaustive search on A,B,C,D,E,F values taken from divisors
of 12 (1,2,3,4,6). To cut down on repetitions, I required that if ABC
is different from DEF, that ABC be lower; I also required that B<=C and
E<=F (since swapping E and F, or B and C, does not change anything
important).
Twenty-three sets produce values that work, ie, that give the same
frequency table as two traditional d12s; when looked at from the point
of view of the labels on the dice, there are only eight distinct
labelings - most of them correspond to more than one set of A-F values.
Specifically:
die 1 die 2 A B C D E F
0 1 1 2 2 2 3 3 3 4 4 5 0 3 4 6 7 8 9 10 11 13 14 17 1 1 2 4 3 6
0 1 1 2 2 3 3 4 4 5 5 6 0 2 4 6 6 8 8 10 10 12 14 16 1 1 3 2 6 6
0 1 1 2 2 3 3 4 4 5 5 6 0 2 4 6 6 8 8 10 10 12 14 16 1 1 3 4 2 6
0 1 1 2 2 3 3 4 4 5 5 6 0 2 4 6 6 8 8 10 10 12 14 16 2 1 1 2 6 6
0 1 1 2 2 3 3 4 4 5 5 6 0 2 4 6 6 8 8 10 10 12 14 16 2 1 1 4 2 6
0 1 1 2 2 3 6 7 7 8 8 9 0 2 3 4 5 6 7 8 9 10 11 13 1 1 6 2 3 6
0 1 1 2 2 3 6 7 7 8 8 9 0 2 3 4 5 6 7 8 9 10 11 13 1 1 6 4 2 3
0 1 1 2 4 5 5 6 8 9 9 10 0 2 2 4 4 6 6 8 8 10 10 12 4 1 1 2 2 6
0 1 1 2 4 5 5 6 8 9 9 10 0 2 2 4 4 6 6 8 8 10 10 12 4 1 1 4 2 2
0 1 2 2 3 3 4 4 5 5 6 7 0 1 4 5 6 7 8 9 10 11 14 15 1 2 3 4 1 6
0 1 2 2 3 3 4 4 5 5 6 7 0 1 4 5 6 7 8 9 10 11 14 15 2 1 2 4 1 6
0 1 2 2 3 4 6 7 8 8 9 10 0 1 3 4 4 5 7 8 8 9 11 12 1 2 6 4 1 3
0 1 2 3 3 4 4 5 5 6 7 8 0 1 2 6 6 7 7 8 8 12 13 14 1 3 3 1 6 6
0 1 2 3 3 4 4 5 5 6 7 8 0 1 2 6 6 7 7 8 8 12 13 14 2 1 3 1 6 6
0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 11 1 3 6 1 3 6
0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 11 1 3 6 2 1 6
0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 11 1 3 6 4 1 2
0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 11 2 1 6 1 3 6
0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 11 2 1 6 2 1 6
0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 11 2 1 6 4 1 2
0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 11 4 1 2 1 3 6
0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 11 4 1 2 2 1 6
0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 11 4 1 2 4 1 2
Looking at these, the first thing to note is that all these A/B/C/D/E/F
sets satisfy the conjecture mentioned above. The next thing of
interest is that each of them has exactly two 1s in it; this makes
sense, in that another 1 would lead to too many 1s on the dice and thus
too many 1s in the frequency table.
The next thing of interest appears upon factoring the A-F values and
looking at the merged set of factors (which we can do by multiplying
them together and factoring the result):
A B C D E F product factors
1 1 2 4 3 6 144 2 2 2 2 3 3
1 1 3 2 6 6 216 2 2 2 3 3 3
1 1 3 4 2 6 144 2 2 2 2 3 3
1 1 6 2 3 6 216 2 2 2 3 3 3
1 1 6 4 2 3 144 2 2 2 2 3 3
1 2 3 4 1 6 144 2 2 2 2 3 3
1 2 6 4 1 3 144 2 2 2 2 3 3
1 3 3 1 6 6 324 2 2 3 3 3 3
1 3 6 1 3 6 324 2 2 3 3 3 3
1 3 6 2 1 6 216 2 2 2 3 3 3
1 3 6 4 1 2 144 2 2 2 2 3 3
2 1 1 2 6 6 144 2 2 2 2 3 3
2 1 1 4 2 6 96 2 2 2 2 2 3
2 1 2 4 1 6 96 2 2 2 2 2 3
2 1 3 1 6 6 216 2 2 2 3 3 3
2 1 6 1 3 6 216 2 2 2 3 3 3
2 1 6 2 1 6 144 2 2 2 2 3 3
2 1 6 4 1 2 96 2 2 2 2 2 3
4 1 1 2 2 6 96 2 2 2 2 2 3
4 1 1 4 2 2 64 2 2 2 2 2 2
4 1 2 1 3 6 144 2 2 2 2 3 3
4 1 2 2 1 6 96 2 2 2 2 2 3
4 1 2 4 1 2 64 2 2 2 2 2 2
The first thing that appears is that the factor list has the same
number of factors in every case. Conjecture A would explain this, but
I see no other obvious explanation. That they aren't all the same is
interesting; it may be related to conjecture A and the asymmetry of the
various diagrams that yield standard d12s.
d13, d14, and d15 have nothing new to offer.
For the d16, using a 2x2x2x2 diagram, exhaustive search similar to that
for 12 finds 10 die pairs:
die 1 die 2 die 1 die 2
0 1 1 2 2 2 3 3 3 3 4 4 4 5 5 6 0 4 4 8 8 8 12 12 12 12 16 16 16 20 20 24 1 1 2 2 4 4 8 8
0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 0 2 4 6 8 8 10 10 12 12 14 14 16 18 20 22 1 1 2 4 2 4 8 8
0 1 1 2 2 3 3 4 8 9 9 10 10 11 11 12 0 2 4 4 6 6 8 8 10 10 12 12 14 14 16 18 1 1 2 8 2 4 4 8
0 1 1 2 4 4 5 5 5 5 6 6 8 9 9 10 0 2 2 4 8 8 10 10 10 10 12 12 16 18 18 20 1 1 4 4 2 2 8 8
0 1 1 2 4 5 5 6 8 9 9 10 12 13 13 14 0 2 2 4 4 6 6 8 8 10 10 12 12 14 14 16 1 1 4 8 2 2 4 8
0 1 1 2 8 8 9 9 9 9 10 10 16 17 17 18 0 2 2 4 4 4 6 6 6 6 8 8 8 10 10 12 1 1 8 8 2 2 4 4
0 1 2 2 3 3 4 4 5 5 6 6 7 7 8 9 0 1 4 5 8 8 9 9 12 12 13 13 16 17 20 21 1 2 2 4 1 4 8 8
0 1 2 2 3 3 4 5 8 9 10 10 11 11 12 13 0 1 4 4 5 5 8 8 9 9 12 12 13 13 16 17 1 2 2 8 1 4 4 8
0 1 2 3 4 4 5 5 6 6 7 7 8 9 10 11 0 1 2 3 8 8 9 9 10 10 11 11 16 17 18 19 1 2 4 4 1 2 8 8
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 2 4 8 1 2 4 8
Another curious thing appears: the diagram step values always have
exactly two each of 1, 2, 4, and 8, and each die labeling corresponds
to exactly one set of numbers.
Everything between 16 and 30 is either prime, has only two prime
factors, or has at least two prime factors which are equal. None of
these give us anything new, possibly excepting 24 (2*2*2*3).
But 30 is the smallest number with three distinct prime factors, and it
has some interest for us. Using a diagram like the one used above for
d12, but larger
A or D
0---x---x---x---x
B |\ |\ |\ |\ |\
or x-\-x-\-x-\-x-\-x \
E |\ \|\ \|\ \|\ \|\ \
x-\-x-\-x-\-x-\-x \ \
\ \ \ \ \ \ \ \ \ \ \
C \ \ x---x---x---x---x
or F \ \|\ \|\ \|\ \|\ \|
\ x---x---x---x---x
\| \| \| \| \|
x---x---x---x---x
Stock d30s can be generated by six A-B-C triples: 1-5-15, 1-10-5,
3-1-15, 6-1-3, 2-10-1, and 6-2-1. Interestingly, looking at the A
values, I see 1, 1, 2, 3, 6, 6; the B values, 1, 1, 2, 5, 10, 10; the C
values, 1, 1, 3, 5, 15, 15: in each case, 1, 1, X, Y, X*Y, X*Y, where
X*Y is 30 divided by the size of the diagram in the corresponding
dimension. I wonder if that is significant.
I did a similar exhaustive search, finding 13 die pairs, corresponding
to 69 distinct parameter sets (with de-duping rules similar, though not
quite identical since all factors are different, to the ones used for
the d12). Here are the die pairs and the parameter sets A-F that
generate them:
0 1 1 2 2 2 3 3 3 3 4 4 4 4 4 5 5 5 5 5 6 6 6 6 7 7 7 8 8 9 - 0 5 6 10 11 12 15 16 17 18 20 21 22 23 24 25 26 27 28 29 31 32 33 34 37 38 39 43 44 49
1 1 3 6 5 15
1 1 3 6 10 5
1 2 1 6 5 15
1 2 1 6 10 5
0 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 7 7 7 8 8 8 9 9 9 10 10 11 - 0 3 6 9 10 12 13 15 16 18 19 20 21 22 23 24 25 26 27 28 29 31 32 34 35 37 38 41 44 47
1 1 5 3 10 15
1 1 5 6 10 3
2 1 1 3 10 15
2 1 1 6 10 3
0 1 1 2 2 2 3 3 3 4 4 4 5 5 6 15 16 16 17 17 17 18 18 18 19 19 19 20 20 21 - 0 3 5 6 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 31 32 34 37
1 1 15 3 5 15
1 1 15 3 10 5
1 1 15 6 5 3
0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 - 0 2 4 6 8 10 12 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 31 33 35 37 39 41 43
1 5 1 2 10 15
1 5 1 6 2 15
3 1 1 2 10 15
3 1 1 6 2 15
0 1 1 2 2 3 3 4 4 5 10 11 11 12 12 13 13 14 14 15 20 21 21 22 22 23 23 24 24 25 - 0 2 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 31 33
1 10 1 2 5 15
1 10 1 2 10 5
1 10 1 6 2 5
0 1 1 2 2 3 6 7 7 8 8 9 12 13 13 14 14 15 18 19 19 20 20 21 24 25 25 26 26 27 - 0 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 31
6 1 1 2 10 3
6 1 1 3 2 15
6 1 1 6 2 3
0 1 2 2 3 3 4 4 4 5 5 5 6 6 6 7 7 7 8 8 8 9 9 9 10 10 11 11 12 13 - 0 1 6 7 10 11 12 13 16 17 18 19 20 21 22 23 24 25 26 27 28 29 32 33 34 35 38 39 44 45
1 2 5 6 10 1
2 1 3 6 10 1
2 2 1 6 10 1
0 1 2 2 3 3 4 4 4 5 5 6 6 7 8 15 16 17 17 18 18 19 19 19 20 20 21 21 22 23 - 0 1 5 6 6 7 10 11 11 12 12 13 16 17 17 18 18 19 22 23 23 24 24 25 28 29 29 30 34 35
1 2 15 6 5 1
0 1 2 2 3 4 4 5 6 6 7 8 8 9 10 15 16 17 17 18 19 19 20 21 21 22 23 23 24 25 - 0 1 3 4 6 7 9 10 10 11 12 13 13 14 16 17 19 20 20 21 22 23 23 24 26 27 29 30 32 33
2 1 15 3 10 1
0 1 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 16 17 - 0 1 2 6 7 8 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 33 34 35 39 40 41
1 5 3 6 1 15
3 1 3 6 1 15
3 2 1 6 1 15
0 1 2 3 3 4 4 5 6 7 10 11 12 13 13 14 14 15 16 17 20 21 22 23 23 24 24 25 26 27 - 0 1 2 5 6 6 7 7 8 11 12 12 13 13 14 17 18 18 19 19 20 23 24 24 25 25 26 29 30 31
1 10 3 6 1 5
0 1 2 3 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 16 17 18 19 - 0 1 2 3 4 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 35 36 37 38 39
1 5 5 1 10 15
2 5 1 1 10 15
3 1 5 1 10 15
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 - 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
1 5 15 1 5 15
1 5 15 1 10 5
1 5 15 2 10 1
1 5 15 3 1 15
1 5 15 6 1 3
1 5 15 6 2 1
1 10 5 1 5 15
1 10 5 1 10 5
1 10 5 2 10 1
1 10 5 3 1 15
1 10 5 6 1 3
1 10 5 6 2 1
2 10 1 1 5 15
2 10 1 1 10 5
2 10 1 2 10 1
2 10 1 3 1 15
2 10 1 6 1 3
2 10 1 6 2 1
3 1 15 1 5 15
3 1 15 1 10 5
3 1 15 2 10 1
3 1 15 3 1 15
3 1 15 6 1 3
3 1 15 6 2 1
6 1 3 1 5 15
6 1 3 1 10 5
6 1 3 2 10 1
6 1 3 3 1 15
6 1 3 6 1 3
6 1 3 6 2 1
6 2 1 1 5 15
6 2 1 1 10 5
6 2 1 2 10 1
6 2 1 3 1 15
6 2 1 6 1 3
6 2 1 6 2 1
(Incidentally, note these also fit conjecture B above; I haven't done
the tedious work to check conjecture A.) Based on the results for the
d12, I took each set A-F, multiplied them, and factored the product.
Using . for 2, = for 3, and @ for 5, to produce more visually obvious
patterns, here they are, with each line annotated with distribution of
the factors (or, equivalently, their exponents in the factoring) and a
graphic histogram of the number of sets that correspond to that line:
. . . . = = 4 2 +
. . . . = @ 4 1 1 +++
. . . . @ @ 4 2 +
. . . = = = 3 3 +++
. . . = = @ 3 2 1 +++++
. . . = @ @ 3 2 1 +++++
. . . @ @ @ 3 3 +++
. . = = = = 4 2 +
. . = = = @ 3 2 1 +++++
. . = = @ @ 2 2 2 +++++++++++++++
. . = @ @ @ 3 2 1 +++++
. . @ @ @ @ 4 2 +
. = = = = @ 4 1 1 +++
. = = = @ @ 3 2 1 +++++
. = = @ @ @ 3 2 1 +++++
. = @ @ @ @ 4 1 1 +++
= = = = @ @ 4 2 +
= = = @ @ @ 3 3 +++
= = @ @ @ @ 4 2 +
Throwing away the actual factors and concentrating on the histograms
and exponent distribution, I see only five different lines:
4 2 +
3 3 +++
4 1 1 +++
3 2 1 +++++
2 2 2 +++++++++++++++
These look suspiciously related to the partitions of 6:
6
5 1
4 2
4 1 1
3 3
3 2 1
3 1 1 1
2 2 2
2 2 1 1
2 1 1 1 1
1 1 1 1 1 1
where partitions with more than three elements don't appear (they
can't, because 30 has only three distinct prime factors).
Question: Why don't the "6" and "5 1" partitions appear?
Question: Why 6 factors? Because the 1s disappear, I don't really
think it's because there are 6 axes (A through F); after deleting the
1s, there are only 4. (The d12 also has 6 factors; they're just
simpler factors.)
Question: Are the histogram counts related to anything? They are not
just the ways of arranging those factors, not even after dividing by a
constant; consider
4 2 6!/(4!2!)=15 + (1)
3 3 6!/(3!3!)=20 +++ (3)
4 1 1 6!/(4!1!1!)=30 +++ (3)
3 2 1 6!/(3!2!1!)=60 +++++ (5)
2 2 2 6!/(2!2!2!)=90 +++++++++++++++ (15)
Also, in the first pair,
0 1 1 2 2 2 3 3 3 3 4 4 4 4 4 5 5 5 5 5 6 6 6 6 7 7 7 8 8 9
0 5 6 10 11 12 15 16 17 18 20 21 22 23 24 25 26 27 28 29 31 32 33 34 37 38 39 43 44 49
the distribution of numbers on the "lower" is provocatively similar to
the sort of frequency distribution we've been looking at all along.
The discrete derivative of the second die
0 5 6 10 11 12 15 16 17 18 20 21 22 23 24 25 26 27 28 29 31 32 33 34 37 38 39 43 44 49
5 1 4 1 1 3 1 1 1 2 1 1 1 1 1 1 1 1 1 2 1 1 1 3 1 1 4 1 5
feels as though it has something significant lurking. I see a pattern
in that collapsing things that add up to 6 does something interesting:
5 1 4 1 1 3 1 1 1 2 1 1 1 1 1 1 1 1 1 2 1 1 1 3 1 1 4 1 5
---- ------- ---------- ------------- ------------- ---------- ------- ----
This is probably related to the cross-product that generated this die
(which can be 6x5/5x3/15x2 or 6x5/10x3/5x2), but the relationship is
obscure to me.
The next number I would expect to hold any interest is 60; it's the
first number with four prime factors and three _distinct_ prime
factors. I don't expect it to hold much interest, though. That
distinction belongs to 210, the smallest number with four distinct
prime factors (2*3*5*7). Using a diagram like
B or F
0---x---x---x---x
C |\ |\ |\ |\ |\
or x-\-x-\-x-\-x-\-x \
G |\ \|\ \|\ \|\ \|\ \ A or E
x-\-x-\-x-\-x-\-x \ \ . . . [7 copies] . . .
\ \ \ \ \ \ \ \ \ \ \
D \ \ x---x---x---x---x
or H \ \|\ \|\ \|\ \|\ \|
\ x---x---x---x---x
\| \| \| \| \|
x---x---x---x---x
I did the same kind of exhaustive search I did for 12 and 30. It took
a significant time in this case; it involved counting to 2^32 (the
eighth power of the number of divisors of 210, that number being 16),
doing a nontrivial amount of work at each step, but fortunately I have
a machine which turned out to be fast enough to complete the job in a
matter of hours. (If I were convinced conjecture B were true I could
have pruned the search substantially; conjecture A would allow so much
pruning it hardly counts as search - that would be more accurately
called `enumeration'.)
This results in 230 dice pairs, derived from 1956 different A-H sets.
All the sets conform to conjecture A, and thus of course conjecture B
as well.
From the generating polynomial perspective, this implies some things
about the factorizations of a particular degree-418 polynomial.
However, it is not clear that there aren't factors involving negative
coefficients (thus not corresponding to dice) that end up resulting in
valid dice by the time they get multiplied back out again.