12.6.05
all ways of colouring a 3x3 square with two colours, excluding rotations and reflections
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I'm sure there's a general formula for these things.
3 x 3 matrices with entries in the integers modulo 2? (e.g green=1,red=0). Then you can add and multiply them together and, moreover, it's a ring. It will be a finite ring, and you can classify these things.
The difficulty, as far as I can tell, is in getting your structure to recognise when two things are 'the same' i.e. a rotation or reflection of each other - but one can factor out by the group of symmetries (D8) to get another ring. What you want is the 'orbits' of each group element and a way to tell which are the same.
Hey ho. Will think about it ...
3 x 3 matrices with entries in the integers modulo 2? (e.g green=1,red=0). Then you can add and multiply them together and, moreover, it's a ring. It will be a finite ring, and you can classify these things.
The difficulty, as far as I can tell, is in getting your structure to recognise when two things are 'the same' i.e. a rotation or reflection of each other - but one can factor out by the group of symmetries (D8) to get another ring. What you want is the 'orbits' of each group element and a way to tell which are the same.
Hey ho. Will think about it ...
# posted by 
: 2:30 pm
: 2:30 pm
working out how many there are supposed to be is easy.
in fact you can work out a expression for the general nxn square, i think.
an easy way to list them all is what i haven't worked out yet, which is why i did this lot by hand to get a feel for it.
for a general rectangle as well.
although i'm sure you can find it in a book of algorithms.
without symmetries it's just 2^2n, you can write them in order.
perhaps you can write D8 as 8 matrices? i wouldn't know
in fact you can work out a expression for the general nxn square, i think.
an easy way to list them all is what i haven't worked out yet, which is why i did this lot by hand to get a feel for it.
for a general rectangle as well.
although i'm sure you can find it in a book of algorithms.
without symmetries it's just 2^2n, you can write them in order.
perhaps you can write D8 as 8 matrices? i wouldn't know
In the 2D plane you can write D8 as 8 matrices. Each rotation/reflection is a linear map on the 2D plane, and so you just have to say where the basis elements (your (1,0,0..)'s etc) go.
In your case - Since multiplication by matrices is distributive, you only have to specify where the addative generators (i.e. the elements of the second collums in your picture) go under each element of the group.
Incidently, that you may find a matrix (multiplication by which gives you the rotation of any given matrix) is a consequence of the fact that the determinant of all the generator matrices is zero, and hence there is an invertible matrix going from one to the other.
All right, so I'm not quite sure if the same matrix will rotate all the generators in the right way. I'll think about it.
P.S. - Have linked here in my new webpage. And removed pron link.
In your case - Since multiplication by matrices is distributive, you only have to specify where the addative generators (i.e. the elements of the second collums in your picture) go under each element of the group.
Incidently, that you may find a matrix (multiplication by which gives you the rotation of any given matrix) is a consequence of the fact that the determinant of all the generator matrices is zero, and hence there is an invertible matrix going from one to the other.
All right, so I'm not quite sure if the same matrix will rotate all the generators in the right way. I'll think about it.
P.S. - Have linked here in my new webpage. And removed pron link.
# posted by : 4:43 pm
: 4:43 pm
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# posted by : 9:37 pm
: 9:37 pm << Home
