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−
α
o
⎛
⎞
Rot
−
100
010
001
2
N
⎝
⎠
·
T
D
k
(
x
0
,
y
0
,
α
o
)=
T
(
x
0
,
y
0
)
−
k
Rot
2
+
α
o
T
N
k
(
−
x
0
,−
y
0
)
,
k
∈
Z
(18)
where
T
(
Δ
x
,
Δ
y
)
is a translation by
(
Δ
x
,
Δ
y
)
,
Rot
(
α
)
is a rotation by
α
about the
origin of axes and
Re f
is a reflection about the
y
axis.
Equation 18 corresponds to shifting the center of symmetry to the origin of axes,
aligning the reflection axis with the Z axis, reflecting the set
S
about the axis
Z
and
inverting the rotation and translation.
Given two reflectional transforms, we get
(
y
)
T
D
k
1
(
x
0
,
y
0
,
α
o
)
T
D
k
2
(
x
0
,
y
0
,
α
o
)=
T
(
x
0
,
y
0
)
Rot
−
k
1
−
α
o
Re f
(
y
)
Rot
2
N
k
1
+
α
o
T
(
−
x
0
,−
y
0
)
·
2
N
T
(
x
0
,
y
0
)
Rot
−
k
2
−
α
o
Re f
(
y
)
Rot
2
N
k
2
+
α
o
T
(
−
x
0
,−
y
0
)=
2
N
Rot
2
N
k
1
+
α
o
Rot
k
2
−
α
o
Rot
−
k
1
−
α
o
Re f
2
N
2
N
T
(
x
0
,
y
0
y
)
(
y
)
−
·
(
2
N
(
k
1
−
k
2
)
)
Rot
Re f
(
y
)
Rot
2
N
k
2
+
α
o
T
(
−
x
0
,−
y
0
)=
N
k
1
−
α
o
Re f
(
y
)
Rot
2
N
(
k
1
−
k
2
)
Re f
(
y
)
T
(
x
0
,
y
0
)
Rot
−
Rot
2
N
k
2
+
α
o
T
(
−
x
0
,−
y
0
)=
2
(
−
(
k
1
−
k
2
)
)
2
N
Rot
T
(
x
0
,
y
0
)
Rot
k
1
−
α
o
Rot
Rot
2
k
2
+
α
o
2
N
2
N
(
k
1
−
k
2
)
N
−
−
T
(
−
x
0
,−
y
0
)=
(
2
N
2
k
1
)
)
Rot
(
k
2
−
T
(
x
0
,
y
0
)
Rot
2
N
2
(
k
2
−
k
1
)
T
(
−
x
0
,−
y
0
)=
Rot
x
0
,
y
0
,
(
k
2
−
k
1
)
4
N
(19)
Rot
x
0
,
k
1
)
is a rotation by
4
N
4
N
y
0
,
(
k
2
−
(
k
2
−
k
1
)
, about the center of symme-
try, and as
k
1
,
k
2
∈
Z
, it corresponds to
C
K
,and
Rot
x
0
4
N
(
T
C
K
=
,
y
0
,
k
2
−
k
1
)
=
T
D
1
·
T
D
2
(20)
Acknowledgements.
We would like to thank authors of [PLC
+
08], for making their test
dataset of symmetric images public. We would also like to thank Shira Chertok for proof-
reading and editing the text.
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