Group theory: Colorings of a regular hexagon (wtih 3 colors)
qsymmetry
hexagon with 3 colors (red, blue, green) are there,
if two colorings are rendered equivalent
if one can be obtained from the other
by either a rotation or a reflection (in D_6) ?
I know that this is an application of Burnside's counting argument, but I am having difficulty determining
Fix(g) for each g in D_6.
Incidentally, is there a general formula for (or a systematic way of determining) the number of distinct (as above) colorings of a regular n-gon with k colors ?
(As n and k increases, solving problems of this sort
can be rather tedious to do by hand alone).
Thanks in advance.
Timothy Murphy
> How many distinct colorings of the edges of a regular
> hexagon with 3 colors (red, blue, green) are there,
> if two colorings are rendered equivalent
> if one can be obtained from the other
> by either a rotation or a reflection (in D_6) ?
>
> I know that this is an application of Burnside's counting argument, but I
am having difficulty determining
> Fix(g) for each g in D_6.
Wouldn't have thought it would be that difficult.
You only have to go through the 6 conjugacy classes
(2 kinds of reflection, 4 rotation classes).
Take rotation through 2pi/6.
All edges must be the same colour, so Fix = 3.
--
Timothy Murphy
e-mail: gayleard /at/ eircom.net
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College Dublin
Chip Eastham
Do you allow for using fewer than all 3 colors
on the hexagon?
regards, chip
qsymmetry
We allow for 3 or fewer colors, yes.
Chip Eastham
>
> - Show quoted text -
Hi, qsymmetry:
I wrote a quick program yesterday (while my
internet connection was out because a lightning
strike in the neighborhood). For two colors I
get 13 arrangements, and for three I get 92.
I've not checked the programming carefully yet,
but the basic idea is to run through all the
colors^edges possibilities and retain only the
ones which are lexicographically least among
all the allowed symmetric rearrangements.
regards, chip
Timothy Murphy
>> > > How many distinct colorings of the edges of a
>> > regular
>> > > hexagon with 3 colors (red, blue, green) are
>> > there,
>> > > if two colorings are rendered equivalent
>> > > if one can be obtained from the other
>> > > by either a rotation or a reflection (in D_6) ?
>>
>> > > I know that this is an application of Burnside's
>> > counting argument, but I am having difficulty
>> > determining
>> > > Fix(g) for each g in D_6.
...
> I wrote a quick program yesterday (while my
> internet connection was out because a lightning
> strike in the neighborhood). For two colors I
> get 13 arrangements, and for three I get 92.
>
> I've not checked the programming carefully yet,
> but the basic idea is to run through all the
> colors^edges possibilities and retain only the
> ones which are lexicographically least among
> all the allowed symmetric rearrangements.
Is this really necessary?
Let s denote rotation through 2pi/6.
Then the conjugacy classes are [a],[d],[1],[s],[s^2],[s^3],
where a,d are reflections in an altitude and a diagonal.
Then it is easy to see that
Fix(a) = 3^4, Fix(d) = 3^3,
Fix(1) = 3^6, Fix(s) = 3, Fix(s^2) = 3^2, Fix(s^3) = 3^3.
What the OP called Burnside's argument, and I call Polya's Lemma,
the number of colourings is
1/12 (3 Fix(a) + 3 Fix(d) + Fix(1) + 2 Fix(s) + 2 Fix(s^2) + Fix(s^3))
Chip Eastham
Hi, Timothy:
Obviously not necessary, given your succinct
solution, but perhaps useful as a check to
the OP who was trying to apply that lemma.
[I think I might well fail to realize that
there were two reflection conjugacy classes
and screwed up my counting that way.]
The idea for my program had been kicking
around in my head for some time, and this
was a good excuse to spit it out. Here's
a more complicated question of the sort
that got me thinking about it:
http://groups.google.com/group/sci.math/browse_frm/thread/a8723ef98de9901e/0c67bd1a4e031bfd
regards, chip
Bill Taylor
> I wrote a quick program yesterday (while my
> internet connection was out because a lightning
> strike in the neighborhood). For two colors I
> get 13 arrangements, and for three I get 92.
Yes, 92 is correct, though it only takes a minute or two
to calculate using Burnside/Polya.
Incidentally, of those 92, only 22 are distinct if we
also allow permutations between the colours as giving
"the same colouring".
What we might call "colouring patterns".
It takes only a few more minutes to actually list these,
by hand, and then verify the 92 by multiplying the counts
by 1, 3 or 6 as appropriate.
-- Bill Taylor
Q: What is a consultant?
A: A person who does a con job on you then lives like a sultan.
DavidA
You asked for a general formula. Are you aware of Polya counting, eg:
http://en.wikipedia.org/wiki/P%C3%B3lya_enumeration_theorem