Image Processing Reference
In-Depth Information
−a
o
/b
o
).
Again, from (x
q
,q) ∈D
o
X
and x
q
+ 1 = a
u
y
p
> (a
u
−a
o
)/(a
u
/b
l
(1−q
2
/b
l
), we get
q
2
< (a
u
−a
o
)/(a
u
/b
l
−a
o
/b
o
) 6 y
p
.
So, q < y
p
. Similarly, p < x
q
+ 1
€
Theorem 5.4. If D = D(E(a,b)) = D
o
, then both the following conditions
hold [41]: (A) a
l
< a < a
u
and (B) b
l
< b < b
u
.
In other words, R
ul
= Rect((a
l
,b
u
),(a
u
,b
l
)), the rectangle formed by
(a
l
,b
u
) and (a
u
,b
l
) as diagonally opposite corners, properly contains the do-
main of D
o
, i.e., (a,b) ∈ R
ul
.
Proof: We first show that b
l
< b and a < a
u
. We consider three cases, each
of which leads to a contradiction.
Case 1: a 6 a
u
and b 6 b
l
. but (a,b) = (a
u
,b
l
).
This case leads to the fact (p,y
p
) ∈ D, a contradiction.
Case 2: a > a
u
and b > b
l
This case leads to the fact (x
q
,q) ∈D, a contradiction.
Case 3: a > a
u
and b < b
l
Let E ∩E
ul
= (x
′
,y
′
) where E
ul
= E(a
u
,b
l
). If x
′
> p, we use Lemma
5.2, to show (p,y
p
) ∈D, a contradiction. If x
′
6 p, then we use Lemma 5.2 to
show that (x
q
,q) ∈ D, a Contradiction.
Combining all cases b > b
l
and a < a
u
, the rest follows similarly.
€
The above theorem establishes that R
ul
is a rectangular bound on the
domain of the possible a,b values such that D(E(a,b)) = D
o
. It is shown [41]
that this bound could be made tight by first providing an algorithm to find
a possible a (or b) for any given b (or a) between the upper and lower limits,
and then by deriving the boundaries of the domain directly.
Now we state some additional properties that we will use later.
Lemma 5.4. [41] Let E
ul
= E(a
u
,b
l
) and p,q,x
q
,y
p
are as defined in Eq.
5.1. If D = D
o
, then E∩E
ul
= (x
′
,y
′
) such that
′
′
(A) p 6 x
< x
q
+ 1 and (B) q < y
6 y
p
where,
′
′
/a
u
)
2
)}, and
p =
max{p
: y
p
′
= b
l
(1−(p
′
′
/b
l
)
2
)−1)}.
q =
max{q
: x
q
′
= a
u
(1−(q
Proof: Consider any p and q. As (p,y
p
) ∈ D
Y
= D
Y
, y
p
6 b(
(1−p
2
/a
2
).
′
(1 −p
2
/a
u
) >
Now if x
< p, we can easily show (Lemma 5.2) that y
p
= b
l
(1 − p
2
/a
2
). Contradiction. So, x
′
b
> p. Other cases follow similarly. Fi-
nally, the maximum is computed to take the possibility of multiple p's and/or
multiple q's into consideration.
€
