just now, we have analysed
corresponding to
the regular tetrahedron
how to colour the vertex
we will be thinking
just now, we have used ABCD this kind of
symbols to represent its vertices
but at the end, during the calculation
are they still useful
in Burnside's Lemma and
Polya's Theorem
what we are considering？
actually, we are considering the permutation structure
maybe, we do not even need
to write how does each permutation look like
we still take this kind of tetrahedron as an example
we do have ABCD these 4 vertices
but when we are calculating, we need to
know for this kind of operations
how many cycles are there
we just need to write out the
pattern of the cycles
for example, when we are considering the vertex and the face centre
performing positive/negative 120 degree rotation
we will realize a vertex is fixed
the other 3 vertices in one cycle
therefore, we just need to write it as
one 1 order cycle
one 3 order cycle
That is enough
Besides the pattern of the permutations
there is still a question like this
how many this kind of permutations are there?
we just need to find how many angles are there
how many vertices
there are total of 4 different vertices
there are 2 different angles
therefore, there are total of 8 permutation
which are one 1 order cycle with
one 3 order cycle
besides, we look at middle points of two oppositing edges
while performing exchange, it is nothing but just
the exchange between 2 vertices
there are two 2-order cycle
how many corresponding edge pairs are there
in total, there are 3
next, fixed permutation is very simple
total of 4 vertices
each vertex is 1 order cycle
therefore, there are four 1 order cycle
at this time, we just need to simply
list all the permutation pattern
do not need to understand the actual details
we just directly adopt the corresponding
Polya's Theorem
write in the corresponding several order cycle
corresponding several colour power series into it
it managed to get the answer directly .
next, let us look at another question
for 3 different types of beads
for example,
I have 3 kinds of beads with different colours
I want to string a 4 beads necklace
how many types of solutions are there
It looks very similar with
the circular permutation
but, there are differences
which is we have only
3 different colours of beads
but, if we need to make a 4 beads necklace
1 colour must be repeated
therefore, this a repeatable circle permutation
at the same time, this arrangement can be rotated
it is actually corresponding to
perform rotation within a space
therefore, we may need to consider
its permutation group
it should have a problem
if you asked do I need to consider flipping?
sometimes, it is a very realistic problem
what does it mean by considering flipping
if after flipping, it could be the
same as the previous one in some cases
then, we may need to consider this movement
if after it was turned over and it would be changed for all solutions
then, we do not need to consider this flipping
therefore, everyone should think clearly
in reality, do we actually need to consider about it
it contains a practical significance
besides, with or without flipping
does it hold the same answer ?
we can solve it separately
for example, if we consider after the flipping
the bead are split into half
rounded beads
after it was turned over, it then became flat
it simply means
that its looking was changed
therefore, after flipping, it is treated as different
we consider only rotation without flipping
for this kind of 4 points on a flat surface
how many different rotations are there
means how many different rotation angles around the centre point
for example, positive/negative 90 degree
each vertex may
become the next vertex
one 4 order cycle,
how many angles are there
there are total of 2 different angles
positive/negative 90 degree
other than those
there is still
another 180 degree rotation
performing two-by-two exchange
therefore, there are two 2 order cycle
besides, there is still one fixed permutation
directly apply Polya's Theorem
corresponding to 3 colours colouring,
therefore it is
3 to the power of 1, added up with 3 to the power of 1
because, there are 2 types of different
positive and negative 90 degree
then, added up with 3 square
then, added up with 3 to the power of 4
then, divided by all the number of permutation - 4
the answer is exactly 24
if we considered
this bead is round shaped
after we turned it over
it is still a necklace
then, we need to consider flipping as well
at this time if we are considering flipping
similarly, we are going to write
rotation of positive and negative 90 degree around the centre axis
rotation of 180 degree around the centre axis
besides, we need to find its symmetry axis to perform the flipping
at this time, we will realize
there are 2 types of symmetry axes
one type of the symmetry axes is over their
corresponding centre of the necklace
which is when the symmetry axis does not
go through the bead
we will realize actually it is
performing two-by-two flipping exchange
therefore, two 2 order cycles
how many this kind of symmetry axes?
it is nothing more that this angle and that angle
there are total of 2
we will also realize
if we split the bead over its centre point
it can still produce another symmetry axis
whereas for this symmetry axis, its
permutation is different
because the bead over the axis is fixed
perform exchange between 2 beads
therefore, its permutation can be written as
2 1 order cycles
then, one 2 order cycle
how many different axes are there
similarly, there are 2 different choices
then, there is still fixed one
at this point, we can look at
actually, in total it contains
how many different permutations
in total, there are 8 types of different permutations
whereas, corresponding to
applying 3 colours colouring into Polya's Theorem
the answer is 21
it can be seen that with
without considering flipping
the answers are different