Image Processing Reference
In-Depth Information
(B) b
l
< b
o
< b
u
and a
l
< a
o
< a
u
.
Proof: First, we establish Part (A).
By induction on k, we can easily prove that
(i)a
k−1
u
6 a
k−2
u
implies b
l
> b
k−1
,k > 2, and
l
(ii)b
k−1
l
> b
k−2
l
implies a
u
6 a
k−
u
,k > 2.
Applying (i) and (ii) alternately we get,
a
u
> a
u
> a
u
> ... > a
u
> a
k+1
> ..... > a
o
, and
u
b
l
6 b
l
6 b
l
6 ... 6 b
l
6 b
k+1
6 ..... < b
o
.
l
such that a
u
= a
k
′
, or there
exists an infinite sequence of a
u
satisfying a
u
< a
k−
u
,k > 1. It implies that
a
u
,k > 0 is a monotone decreasing sequence. Again, it is found that a
u
> a
o
.
Hence, lim
k→∞
a
u
= a
u
. Numerically, we can estimate a
u
by selecting a small
quantity ǫ > 0 and terminating with |a
k−1
u
Hence either there exists a k = k
′
u
,k > k
′
−a
u
| < ǫ.
Other cases can be derived likewise. Part (B) follows from Part (A).
€
From the last theorem it is clear that there exist p and q, 0 6 p 6 x
0
and
0 6 q 6 y
0
such that (p,y
p
) ∈D
Y
,(x
q
,q) ∈ D
o
X
, and
(1−q
2
/b
l
) and b
l
= y
p
/
(1−p
2
/a
u
).
a
u
= (x
q
+ 1)/
(5.1)
So E
ul
= E(a
u
,b
l
) passes through at least two grid points, namely (p,y
p
)
and (x
q+1
,q). Intuitively, the iteration stops by hitting the extreme grid points
(p,y
p
) and (x
q
+ 1,q). This can be compared with the case of a circle where
r
u
=
((x
q
+ 1)
2
+ q
2
) gives an extreme of the radius value. Similar results
also hold for a
l
,b
u
and E
lu
Let us consider an example now.
Example 5.4. Take a
o
= 12.5, b
o
= 10.5, and ǫ = 10
−6
. D
o
X
=
(12,12,12,11,11,10,10,9,8,6,3) and D
Y
= (10,10,10,10,9,9,9,8,8,7,6,4,2).
The convergence can be easily established. We get a
l
= 12.220202, a
u
=
12.533591(q = 3 and x
q
= 11), b
l
= 10.392305 (p = 8 and y
p
= 8),
b
u
= 10.583005.
We state various properties of these bounds leading to the reconstruction
algorithm.
Lemma 5.3. For any D
o
, q < y
p
and p < x
q
+ 1, where p,q,x
q
,y
p
are as
defined in Eq. 5.1 [41].
Proof: Since (p,y
p
) ∈ D
Y
, we have b
o
(1 − p
2
/a
o
) > y
p
. Replacing p by
(1−y
p
/b
l
) (from b
l
) and rearranging terms we get,
a
u
