Image Processing Reference
In-Depth Information
r,
r(α
2
+α
3
)
p
,
rα
3
p
of H(B;r), by x(r) where x =
, the difference between x(r)
and x(r) is bounded by the following theorem from [154].
Theorem 2.27. For any r > 0 and for any sorted B, we have
1. r−x
1
(r) = 0;
−x
2
(r) ≤ α
1
−
α
p
p
4
r(α
2
+α
3
)
p
p even
2. 0 ≤
≤
€
(p
2
−1)
4p
−x
3
(r) ≤α
3
−
α
p
rα
3
p
p odd
Perimeter, Area, Volume and Shape Features
Next we present a few theorems (from [128]) related to the computation of
the geometric features of the digital discs (and spheres) when the N-Sequence
is intrinsically sorted.
Theorem 2.28. The perimeter P and area A of a digital disk of radius r for
an octagonal distance in 2-D are given by
S({1
α
1
,2
α
2
};r)
= P(β;r)
=
4rP(β) and
H({1
α
1
,2
α
2
};r)
= A(β;r)
= r
2
F(β)
where P(β) = (2−
√
√
α
1
+α
2
=
α
p
€
Theorem 2.29. The volume V and the surface area A of the polyhedron of
radius r for a 3-D octagonal metric are given by:
2, F(β) = 2 + 4β−2β
2
and β =
α
2
2)β +
S({1
α
1
,2
α
2
,3
α
3
};r)
4r
2
G(β,γ) and
= A(β;r)
=
H({1
α
1
,2
α
2
,3
α
3
};r)
4
3
r
3
T(β,γ)
= V (β;r)
=
where
T(β,γ) = {1 + 3β + 3γ + 6βγ + 3β
2
− 6γ
2
+ 3βγ
2
− 6β
2
γ − 2β
3
+ γ
3
},
G(β,γ) = {β
2
(3 − 2
√
√
√
√
√
√
√
√
3) +γ
2
(
3 − 3) +βγ(2
3 − 6
2 + 6) +β(2
3) +γ(6
2 − 4
3) +
3}
α
1
+α
2
+α
3
=
α
2
+α
3
α
2
+α
3
α
1
+α
2
+α
3
=
α
p
.
α
3
β =
and γ =
€
p
It may be noted that for m = 0, n = 0, an octagonal face in the digital
sphere degenerates to a point, a rectangular one degenerates to a straight line
segment, and a hexagon degenerates to a triangle. Similarly, for m = 0, n = 0,
octagon degenerates to a square, a rectangle degenerates to a straight line,
and a hexagon degenerates to a triangle, and for m = 1, n = 1, the octagon
degenerates to a square, a rectangle degenerates to a straight line, and the
hexagon degenerates to a point. But in all such cases, the expression given in
Theorem 2.29 holds.
Next we compute the shape feature based on the above measurements. In
2-D, the shape feature is defined as:
16(β(2−
√
√
(perimeter)
2
(
area
)
(4rP(β))
2
r
2
F(β)
2)
2
2)+
ψ
2
(β)
=
=
=
2+4β−2β
2
