Image Processing Reference
In-Depth Information
f
a
(x
i
,i) 6 a
o
= f
a
(f
x
(i,a
o
),i) < f
a
(x
i
−1,i).
Similarly, as f
a
is increasing in y,
f
a
(i,y
i
) 6 a
o
= f
a
(i,f
y
(i,a
o
)) < f
a
(i,y
i
+ 1).
Thus, the two bounds on a
o
are obtained as follow:
a
l
= max(max
i
(f
a
(x
i
,i)),max
i
(f
a
(i,y
i
)))
a
u
= min(min
i
(f
a
(x
i
−1,i)),min
i
(f
a
(i,y
i
+ 1))).
For any value of a such that a
l
6 a < a
u
,
f
a
(x
i
−1,i) > a
u
> a > a
l
> f
a
(x
i
,i).
That is, f
a
(x
i
−1,i) > a > f
a
(x
i
,i).
Hence, x
i
= f
x
(i,f
a
(x
i
,i)) > f
x
(i,a) > f
x
(i,f
a
(x
i
−1,i)) = x
i
−1.
Rearranging, f
x
(i,a) 6 x
i
< f
x
(i,a) + 1.
Consequently, x
i
= ⌈f
x
(i,a)⌉.
Similarly ∀a,a
l
6 a < a
u
,y
i
= ⌊f
y
(i,a)⌋.
In fact, if a < a
l
, it is easy to see that D(f(x,y,a)) will miss some
point (x
i
,i) or (i,y
i
) of D
o
. Also, if a > a
u
, then some point is included
in D(f(x,y,a)) that is not in D
o
. Thus, we have the following theorem.
Theorem 5.13. a
l
6 a < a
u
if and only if D(f(x,y,a)) = D
o
. In other
words, Domain of D
o
= [a
l
,a
u
) [40].
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Example 5.7. Let f : y
2
− 4ax = 0. Here, the monotonicity matrix is the
same as given in Table 5.3. The different functions are listed below.
x = f
x
(y,a) = y
2
/4a
y = f
y
(x,a) =
(4ax)
a = f
a
(x,y) = y
2
/4x
It has been shown in a previous section that
(i
2
/4x
i
),max
i
(y
i
/4i)) and
a
l
=
max(max
i
(i
2
/4(x
i
((y
i
+ 1)
2
/4i)).)
a
u
=
min(min
i
−1)),min
i
These formulae are the same as the ones given in this section.
