1
1
P ROOF
OF
2
L EMMA 3
2.1
For convenience, let
π∈ΠS,t
− 0
h−
I (S, t) , 0max βI (π ),
π ∈ΠS,t
βI (π) , βI (π)+ − βI (π)− .
−
Therefore, ϕI (S, t) = h+
I (S, t) − hI (S, t) and dI (S, t) =
minπ∈ΠS,t βI (π).
We first restate Lemma 5 in our paper as follows.
Lemma 1. Given an N -level image F and a seed set S,
the distance transform ΘSϕ (F) obeys the linear-threshold
superposition:
=
N
−1
X
ΘSϕ (BkF ).
(1)
k=0
Proof: To avoid unnecessary clutter in notation, we use
Θ to denote ΘSϕ , and Bk to denote BkF .
Recall that for each pixel t,
−
Θ(Bk )(t) = h+
Bk (S, t) − hBk (S, t).
When Bk (t) = 1, we have h+
Bk (S, t) = 1. In this case,
Θ(Bk )(t) = 0 iff there is a path π joining a seed s ∈ S
and t, such that βB−k (π) = 1. This means that π is in the
−
white set of Bk , and it follows that βF
(π) > k. Thus,
when Bk (t) = 1, Θ(Bk )(t) = 0 iff there exists a path
−
π joining a seed s and t, where βF
(π) > k. Namely,
when Bk (t) = 1,
Θ(Bk )(t) = 0 ⇐⇒ k < h−
F (S, t).
Since k < h−
F (S, t) indicates Bk (t) = 1, it follows that
Bk (t) = 1 ∧ Θ(Bk )(t) = 0 ⇐⇒ k < h−
F (S, t).
Similarly, we have
Bk (t) = 0 ∧ Θ(Bk )(t) = 0 ⇐⇒ k ≥ h+
F (S, t).
It follows that
+
Θ(Bk )(t) = 0 ⇐⇒ k < h−
F (S, t) ∨ k ≥ hF (S, t).
Note that Θ(Bk )(t) is either 0 or 1. Therefore,
+
Θ(Bk )(t) = 1 ⇐⇒ h−
F (S, t) ≤ k < hF (S, t).
Then for each t,
N
−1
X
k=0
Θ(Bk )(t) =
N
−1
X
+
δ h−
F (S, t) ≤ k < hF (S, t)
k=0
−
= h+
F (S, t) − hF (S, t)
= Θ(F).
OF
T HEOREM 5
Preliminaries
We first introduce several concepts and results that
are necessary for our proof of Theorem 5 in our paper.
In particular, we will prove a generalized version of
Alexander’s lemma.
+
h+
I (S, t) , min βI (π),
ΘSϕ (F)
P ROOF
2.1.1 Alexander’s Lemma
The following is a version of Alexander’s lemma [1].
Lemma 2. Let s and t be two points in the topological
space D = [0, 1]2 , and P and Q be two disjoint closed sets
in D. If there exist a path from s to t in D \ P and a path
from s to t in D \ Q, then there exists a path from s to t
in D \ (P ∪ Q).
Proof: See Alexander’s lemma for simply connected
spaces in [2] and Theorem 8.1, p. 100 in [1].
An intuitive interpretation is shown in Fig. 1(a). As
P and Q are disjoint closed sets, they cannot “touch”
each other’s boundaries. Then if neither P nor Q can
block s from t, P ∪ Q cannot, either,
The following result is a generalization of Lemma
2. An interpretation is shown in Fig. 1(b).
Theorem 3. Consider the topological space D = [0, 1]2 .
Let S ⊂ D be a connected set, t ∈ D a point, and P and
Q two disjoint closed sets in D. If there exist a path from
S to t in D \ P and a path from S to t in D \ Q, then
there exists a path from S to t in D \ (P ∪ Q).
Proof: We will first modify P and Q while keeping the
preceding assumptions about P and Q unchanged.
Step 1. Let P denote D \ P and CPt ⊂ P denote the
connected component of P that contains t. Then we
change P and Q to P1 and Q1 respectively, s.t.
P1 = P ∪ P \ CPt ,
Q1 = Q \ P \ CPt .
Step 2. Following the same notations, we make
similar changes to P1 and Q1 :
t
Q2 = Q1 ∪ Q1 \ CQ
,
1
t
P2 = P1 \ Q1 \ CQ
.
1
Fig. 2 gives an example to show how the above
steps work. In this example, each step fills the holes
inside one of the sets. Generally, as we shall prove in
the following, these two steps make the complement
of P2 ∪ Q2 a path connected set that contains t.
We now show that the assumptions for P and Q are
inherited by P2 and Q2 . For this purpose, it suffices
to show that these assumptions hold for P1 and Q1 ,
because step 2 is a symmetric operation of step 1.
First, we show P1 and Q1 are two disjoint closed set
in D. It is straightforward to see that P1 and Q1 are
disjoint. D = [0, 1]2 is locally connected, and P = D\P
is open in D. Each connected component of an open
2
P
Then we have
P
S
t
Q
s
D \ (P2 ∪ Q2 ) = P2 ∪ Q2
(a)
= P 2 ∩ Q2
t
Q
t ∩ Ct
= P1 ∩ C Q
Q
1
1
t
t
= P 1 ∪ C Q1 ∩ CQ
1
(b)
= P 1 ∩ Q2
Fig. 1: An illustration of Lemma 2 and Theorem 3. P and Q are
two disjoint closed sets in [0, 1]2 . In (a), there is a path from s to
t which does not meet P , and a path from s to t which does not
meet Q (see the two dash lines). Then there exist a path (the solid
line) from s to t which does not meet P ∪ Q. In (b), a generalized
scenario is shown, where the seed set S is not necessarily singleton,
but a connected set. If S and t can be connected by paths that do
not meet P and Q respectively, then S and t can be connected by
a path that does not meet P ∪ Q.
P
P1
P2
t
Q
t
Q1
(a)
t
Q2
(b)
(c)
Fig. 2: An example to show the effects of the two modification
steps. In (a), the set shown in red, P , and the set shown in green,
Q, are two disjoint closed set in [0, 1]2 . After step 1, the hole inside
P is filled (see P1 in (b)). Similarly, after step 2, the hole inside
Q1 is filled (see Q2 in (3)). Generally, these two steps make the
complement of P2 ∪ Q2 a path connected set that contains t.
set in a locally connected space is open in that space
[3] (Theorem 25.3 at p. 161). It follows that P \ CPt is
an open subset in D, because P \ CPt is the union of
all the components of P except CPt . Thus, Q1 is closed
in D. Furthermore, it is easy to see that P1 = D \ CPt ,
so P1 is closed in D.
Second, we show there exist a path from S to t in
D \ P1 and a path from S to t in D \ Q1 . As discussed
above, D \ P1 = CPt . Recall that CPt is the connected
component of D \ P that contains t. Therefore, the
path from S to t in D \ P must be included in D \ P1 .
Moreover, the path from S to t in D \ Q is included
in D \ Q1 for D \ Q ⊂ D \ Q1 .
Similarly, P2 and Q2 have all the properties discussed above. Now we show that D \(P2 ∪Q2 ) is path
connected. It is easy to verify the following facts:
P ∪ Q ⊂ P2 ∪ Q2 ,
(2)
CPt ,
(3)
t
P2 = P1 ∩ C Q
,
1
(4)
t
CQ
.
1
(5)
P1 =
Q2 =
Therefore, for each x ∈ D\(P2 ∪Q2 ), we have x ∈ P 1 =
CPt (see Eqn. 3). Note that every open connected set
in a locally path connected space is path connected
[3] (Exercise 4 at p. 162). Therefore, P 1 is also path
connected, and x and t are path connected in P 1 . It
follows that there exists a path joining x and t in P 2 ,
because P 1 ⊂ P 2 . Similarly, x ∈ D \ (P2 ∪ Q2 ) ⇒ x ∈
t
Q2 = CQ
(see Eqn. 5), so there exists a path joining
1
x and t in Q2 . Recall that P2 and Q2 are two disjoint
closed sets in D. According to Lemma 2, there exists a
path joining x and t in D\(P2 ∪Q2 ). Thus, D\(P2 ∪Q2 )
is path connected.
Since S is connected, there exists a point s ∈ S
contained in D \ (P2 ∪ Q2 ). To see this, recall that
there exist a path in P 2 joining S and t, and a path in
Q2 joining S and t, so S \ P2 6= ∅ and S \ Q2 6= ∅.
If S ⊂ P2 ∪ Q2 , then S can be divided into two
disjoint nonempty closed sets in S (S ∩P2 and S ∩Q2 ),
contradicting the assumption that S is connected.
D \ (P2 ∪ Q2 ) is path connected, so there is a path
from s to t in D\(P2 ∪Q2 ). This path must be included
in D \ (P ∪ Q), because (P ∪ Q) ⊂ (P2 ∪ Q2 ). Thus,
there is a path from S to t in D \ (P ∪ Q).
2.1.2
Hyxel and Supercover
The following concepts of hyxel and supercover are
for k-D digital images. A k-D digital image I can be
represented by a pair (V, I), where V ⊂ Zk is a set of
regular grid points, I : V → R is a function that maps
each grid point x ∈ V to a real value I(x). Moreover,
we assume k-D digital images are rectangular, i.e.
Qk
V = i=1 {1, · · · , ni }. In particular, we are interested
in the case when k = 2.
Next, we introduce hyxel [4], [5], a useful concept
in digital topology.
Definition 1. A hyxel Hx for a grid point x in a k-D
image I = (V, I) is a unit k-D cube centered at t, i.e.
1
1
1
1
Hx = x1 − , x1 +
× · · · × xk − , xk +
,
2
2
2
2
where x = (x1 , · · · , xk ) ∈ V ⊂ Zk . Hyxels are the
generalization of pixels in the 2-D images.
Since hyxels and grid points in Zk have a one-toone correspondence, the connectivity of a set of hyxels
can be induced by the adjacency relationship between
grid points.
3
Definition 2. A sequence of hyxels is a nadj -path if the
corresponding sequence of grid points is a path in
terms of n-adjacency. A set of hyxels is nadj -connected
if any pair of hyxels in this set are connected by a
nadj -path contained in this set.
Definition 3. The supercover S(M ) of a point set M
in Rk is the set of all hyxels that meet M .
The following is a result given in [4].
Lemma 4. Let M ⊂ Rk be a connected set. Then S(M )
is 2kadj -connected.
Proof: See [4], [5]. Note that 2kadj -connectivity is equivalent to the concept of (k − 1)-connectivity defined in
[4].
2.2
Proof of Theorem 5
We first restate Theorem 5 in our paper as follows.
Definition 4 (Definition 2 in our paper). Given a
grayscale image I, εI = maxi,j |I(i)−I(j)| is the maximum local difference, where pixel i and j are arbitrary
8-adjacent neighbours.
Theorem 5 (Theorem 5 in our paper). Given a 2-D
digital image I with 4-connected paths, if the seed set S is
connected, then ΘSϕ (I) approximates the MBD transform
ΘSd (I) with errors no more than 2εI , i.e. for each pixel t,
0 ≤ dI (S, t) − ϕI (S, t) ≤ 2εI .
e t) = h+ (S,
e t) − h− (S,
e t)
ϕIe (S,
e
e
I
I
π∈ΠS,t
e
e t) = inf β e (π).
dIe (S,
I
π∈ΠS,t
e
π 0 ∈ΠS,t
e
e
e 0
e 0
h−
e (S, t) − < min I(π (x)) ≤ max I(π (x)) <
I
x
x
e
h+
e (S, t) + .
I
(7)
(8)
e t) = ϕ e (S,
e t), we need to show
In order to prove dIe (S,
I
that for any > 0, there exists a path π ∈ ΠS,t
e such
that
e
e
e
h−
I(π(x))
≤ max I(π(x))
<
e (S, t) − < min
I
x
x
(9)
e t) + .
h+
(
S,
e
I
Equivalently, we need to show there exists a path
e
π ∈ ΠS,t
e in D \ (P ∪ Q), where P = {x ∈ D : I(x) ≤
− e
+ e
e
hIe (S, t) − } and Q = {x ∈ D : I(x) ≥ hIe (S, t) + }. It
is easy to see that P and Q are two disjoint closed set
in D, and there exist paths from Se to t in D \ P and
e
D \ Q respectively (see the definition of h+
e (S, t) and
I
− e
hIe (S, t) in Eqn. 7). According to Theorem 3, Eqn. 9 is
(10)
Then the supercover voxelization of π 0 is applied. According to Lemma 4, the resultant pixel set includes a
e
4adj -path π that joins S and t. Because |I(x)−I(y)|
≤ εI
for any y ∈ D that is covered by hyxel (pixel) Hx , we
have
e
h−
e (S, t) − εI − < min I(π(x)) ≤ max I(π(x)) <
I
x
x
e
h+
e (S, t) + εI + .
I
(11)
Similarly, we have
−
+
e
h−
e (S, t) − εI − < hI (S, t) ≤ hI (S, t) <
I
e
h+
e (S, t) + εI + .
I
(6)
Proof: First we need to show that in the continuous
e where D ⊂ [0, 1]2 , Ie : D → R is
setting Ie = (D, I),
e t)
continuous, and the seed set Se is connected, ϕIe (S,
e
and dIe (S, t) are equivalent (compare Theorem 1 in
[6]).
As discussed in [6], in the continuous setting, the
e t) and d e (S,
e t) becomes
definition of ϕIe (S,
I
= inf βIe+ (π) − sup βIe− (π 0 ),
e t) in the continuous
e t) = d e (S,
proved, and thus ϕIe (S,
I
setting.
Returning to the proof of Theorem 5, we do the
same trick as in [6] to translate the digital version
of the problem to the continuous setting. We can
e by bilinearly
get a continuous image Ie = {D, I}
interpolating the digital image I = {V, I}. Note that
the seed set Se in Ie is the union of the hyxels (pixels)
of the 4adj -connected seeds in I. Thus, Se is connected
in D. According to Eqn. 9, for any > 0, there exists
a path π 0 ∈ ΠS,t
e such that
(12)
As → 0, Eqn. 11 indicates
e t) + 2εI ,
dI (S, t) ≤ ϕIe (S,
(13)
and Eqn. 12 indicates
e t) + 2εI .
ϕI (S, t) ≤ ϕIe (S,
(14)
Each discrete 4adj -path on the digital image I has
e which is constructed
a continuous counterpart in I,
by linking the line segments between consecutive
pixels on the discrete path. It is easy to see that
the values on this continuous path are the linear
interpolations of the values on the discrete path, for
Ie is obtained by bilinearly interpolating I. Thus,
e t) ≤ dI (S, t) and ϕ e (S,
e t) ≤ ϕI (S, t). Moreover,
dIe (S,
I
e
e
recall that ϕIe (S, t) = dIe (S, t), and ϕI (S, t) is a lower
bound of dI (S, t) [6]. Then we have
e t) ≤ ϕI (S, t) ≤ dI (S, t) ≤ ϕ e (S,
e t) + 2εI . (15)
ϕIe (S,
I
It immediately follows that
0 ≤ dI (S, t) − ϕI (S, t) ≤ 2εI .
(16)
Remark. The above result can be easily generalized to
k-D images. Its proof is analogous to the ones in [6],
[7]. The basic idea is to reduce the problem in the high
dimension to the 2-D case via homotopy [6].
4
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