Graphics Reference
In-Depth Information
In this case,
(
)
(
[
]
)
=
(
[
()()
◊◊◊
()
]
)
∂
o
f
vv
◊◊◊
v
∂
f
v
f
v
f
v
q
#
q
01
q
q
0
1
q
q
i
Â
=
()()
◊◊◊
()
◊◊◊
()
[
]
1
f
v
f
v
f
v
0
i
q
i
=
0
q
i
Â
[
ˆ
]
=
()
1
f
vvv
◊◊◊
◊◊◊
#
0
i
q
q
-
1
i
=
0
q
Ê
Á
ˆ
˜
i
Â
[
ˆ
]
=
f
-
()
1
vvv
◊◊◊
◊◊◊
#
0
i
q
q
-
1
i
=
0
=
(
)
[
]
f
o
∂
(
vv
◊◊◊
v
)
.
#
q
01
q
q
-
1
Case 2.
The vertices f(
v
1
), f(
v
2
),..., and f(
v
q
) are all distinct, but f(
v
0
) = f(
v
1
).
The assumption f(
v
0
) = f(
v
1
) implies that
(
)
(
[
]
)
=
()
=
∂
o
f
vv
◊◊◊
v
∂
00
qq
#
01
q
q
and
q
Ê
Á
ˆ
˜
(
)
Â
i
(
[
]
)
=
()
()
[
ˆ
]
f
o
∂
vv
◊◊◊
v
f
1
v
◊◊◊
v
◊◊◊
v
#
q
01
q
#
0
i
q
q
-
1
q
-
1
i
=
0
q
Â
i
[
]
=
()()
ˆ
1
f
vvv
◊◊◊
◊◊◊
#
0
i
q
q
-
1
i
=
0
0
1
=
() ()()
◊◊◊
[
(()
]
+
()()()
◊◊◊
()
[
]
1
f
vv
f
f
v
1
f
vv
f
f
v
1
2
q
0
2
q
2
q
+
()
◊
+◊◊◊+
()
◊
10
1 0
[
()( )
◊◊◊
()
]
-
[
()()
◊◊◊
()
]
=
f
vv
f
f
v
f
v v
f
f
v
1
2
q
0
2
q
=
0
.
Case 3.
The vertices f(
v
0
), f(
v
1
),..., f(
v
i
),..., f(
v
q
) are all distinct, but f(
v
i
) = f(
v
j
)
for some i < j.
This case follows easily from Case 2 because [
v
0
v
1
···
v
q
] =±[
v
i
v
j
v
0
v
1
···
ˆ
i
···
ˆ
j
···
v
q
].
Case 4.
There exist distinct indices i, j, and k, such that f(
v
i
) = f(
v
j
) = f(
v
k
).
In this case,
(
)
(
[
]
)
=
()
=
∂
o
f
vv
◊◊◊
v
∂
00
qq
#
01
q
q
and
