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2
K
where
1
. Thus, for a symmetry of order K, there exists a
set of K transformations
R
β
k
K
β
k
=
k
,
k
=
0
,...,
K
−
1
for which Eq. 1 holds. The use of homogeneous
coordinates in Eq. 2 allows us to handle non-centered symmetries.
An example of rotational symmetry is given in Fig. 1a. Equation 2 implies that
det
(
T
C
K
)=
1
(3)
where the invariant set
S
I
, is the center of rotational symmetry.
Given a rotation operator
T
C
K
, the center of rotation (that is also the center of rota-
tional symmetry)
X
c
, is invariant under
T
C
K
, and can be computed as the eigenvector
of
T
C
K
corresponding to the eigenvalue
λ
=
1
T
C
K
X
c
=
X
c
.
(4)
2.2
Reflectional Symmetry
n
is reflectionally symmetric with
Definition 2 (Reflectional symmetry).
AsetS
∈
R
respect to the vector (reflection axis)
cos
α
0
,
sin
α
0
with a reflectional transform
T
D
K
,
if
∀
x
i
∈
S
, ∃
x
j
∈
S, s.t.
x
j
=
T
D
K
x
i
.
(5)
2
,T
D
K
is given by
where for x
i
∈
R
⎛
⎞
⎛
⎞
cos 2
α
0
sin 2
α
0
0
x
y
1
⎝
⎠
⎝
⎠
.
T
D
K
(
x
,
y
)=
sin 2
α
0
−
cos 2
α
0
0
(6)
0
0
1
A set S has reflectional symmetry of order K, if there are K angles
α
k
that satisfy
Eq. 5.
An example of reflectional symmetry is given in Fig. 1b. where
α
0
is the angle of
the reflection axis, and Eq. 6 implies that
det
(
T
D
K
)=
−
1
.
(7)
Similar to the rotational symmetry case, the points on the symmetry axis form an
invariant set
X
R
that corresponds to the eigenspace of
T
D
K
X
R
=
X
R
.
(8)
Conversely to Eq. 4, the eigenspace corresponding to
λ
=
1 is of rank 2, in accor-
dance to
S
I
being a line.
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