Image Processing Reference
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4. As an exception to symmetric and isotropic (Section 2.2.1.1) neighbor-
hood sets, many anisotropic yet symmetric neighborhoods in 2-D also
define metrics [62, 141].
N 1
=
{(±1, 0), (0, ±1)}
d N 1 (x)
=
|x 1 | + x 2 |
N 2 A
=
{(±1, 0), ±(1, 1)}
d N 2 A (x)
=
max(|2x 1 − x 2 |,|x 2 |)
N 2 B
=
{(±1, 0), ±(1, −1)}
d N 2 B (x)
=
max(|2x 1 + x 2 |, |x 2 |)
N 2 C
=
{(0, ±1), ±(1, 1)}
d N 2 C (x)
=
max(|2x 2 + x 1 |, |x 1 |)
N 2 D
=
{(0, ±1), ±(1, −1)}
d N 2 D (x)
=
max(|2x 2 − x 1 |,|x 1 |)
N 3 A
=
{(±1, 0), (0, ±1), ±(1, 1)}
d N 3 A (x)
=
max(|x 1 |, |x 2 |, |x 1 − x 2 |)
N 3 B
=
{(±1, 0), (0, ±1), ±(1, −1)}
d N 3 B (x)
=
max(|x 1 |, |x 2 |, |x 1 + x 2 |)
N 4 A
=
{(±1, 0), (±1, ±1)}
d N 4 A (x)
=
max(2⌈(|x 1 | − |x 2 |)/2⌉, 0) + |x 2 |
N 4 B
=
{(0, ±1), (±1, ±1)}
d N 4 B (x)
=
max(2⌈(|x 2 | − |x 1 |)/2⌉, 0) + |x 1 |
N 5
=
{(±1, 0), (0, ±1), (±1, ±1)}
d N 5 (x)
=
max(|x 1 |, |x 2 |)
(a) Prove that the above d
s are metrics and give the lengths of the
shortest paths for the corresponding neighborhood sets.
(b) Study the disks of these distances.
(c) Two metrics d 1 and d 2 over Z 2 are isomorphic if there exists a bijec-
tion λ : Z 2 → Z 2 such that ∀x,y ∈ Z 2 , d 1 (λ(x),λ(y)) = d 2 (x,y)
and d 2
−1 (y)) = d 1 (x,y). Prove that the following sets
of metrics are isomorphic to each other: {d N 2 A ,d N 2 B ,d N 2 C ,d N 2 D
−1 (x),λ
},
{d N 3 A ,d N 3 B
} and {d N 4 A ,d N 4 B
}.
5. The t−cost-m−Neighbor (or tCmN) [167] neighborhood set N(t,m,n),
is defined ∀n,m,t ∈ P, 1 ≤ m ≤ n, 1 ≤ t ≤ n as {w :
w ∈ {0,±1} n ,h n (w) ≤ m} with an associated cost function δ(w) =
min(t,h n (w)) where h n (·) is the component sum function (Definition
2.5). It induces tCmN paths π(u,v;t,m : n) and the tCmN norm de-
fined as ∀x ∈ Z n ,
d(x; t,m : n)
max i=0 S i (x),
=
P
P
i
j=1 f j (x) + [minimum(1, (t − i)/(m − i))
n
j=i+1 f j (x)],
S i (x)
=
0 ≤ i ≤ n,
where
(t − i)/(m − i), 0 ≤ (t − i)/(m − i) < 1,
1,
minimum(1, (t − i)/(m − i))
=
(t − i) ≥ (m − i).
and f i (·) is the component function (Definition 2.4).
Note that, for i = m, minimum(1, (t − i)/0) = 1 and 0 ≤ minimum(1, (t − i)/(m −
i)) ≤ 1.
Prove the following for the tCmN neighborhood [167] ∀n,1 ≤ m ≤ n,
∀t,1 ≤ t ≤n:
(a) d(t,m : n) is a metric over Z n .
(b) ∀x ∈ Z n , d(x;t,m : n) = |π
(x;t,m : n)|.
(c) d(t,m : n) is a generalization of m-Neighbor [60] and t-cost [58]
distances. That is, d m = d(1,m : n) and D t = d(t,n : n).
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