Image Processing Reference
In-Depth Information
In the following, we use m
o
and c
o
to denote possible original values of m
and c, respectively, so that digitization of l : y = m
o
x + c
o
yields D
o
. That is,
D(l : y = m
o
x + c) = D
o
.
Theorem 3.6. (The I R Algorithm) If the upper and lower bounds of m and
c are defined by the following algorithm: [38]
For k ≥ 0,
c
k+1
l
{y
i
−m
u
.i},0 ≤ i≤ n,
= max
i
c
k+1
u
{y
i
+ 1−m
l
.i},0 ≤ i ≤n,
= min
i
m
k+1
l
−c
u
)/i},1 ≤ i≤ n,and
= max
i
{(y
i
m
k+1
u
{(y
i
+ 1−c
l
)/i}1 ≤i ≤ n
= min
i
and c
l
= y
0
,c
u
= y
0
+ 1,m
l
= −(1/n),m
u
= (n + 1)/n, then, there exist
c
l
,c
u
,m
l
,m
u
such that,
k→∞
m
l
= m
l
, lim
k→∞
c
u
= c
u
, lim
k→∞
m
u
= m
u
, lim
k→∞
c
l
= c
l
........(a)
lim
c
l
< c
o
< c
u
and m
l
< m
o
< m
u
.........(b).
Proof: From D
o
we get, y
i
= ⌊m
o
i + c
o
⌋∀i,0 ≤i ≤ n.
Using the definition floor function and rearranging terms, we can write
(y
i
+ 1−c
o
)/i > m
o
≥ (y
i
−c
o
)/i.
Let c
l
and c
u
be such that c
l
< c
o
< c
u
. So, from the above inequalities
we get,
(y
i
+ 1−c
l
)/i > m
o
≥ (y
i
−c
u
)/i.
Thus, m
k+1
u
= min
i
((y
i
+ 1−c
l
)/i) > m
o
> max
i
((y
i
−c
u
)/i) = m
k+1
.
l
Similarly, we can show that c
k+1
l
< c
o
< c
k+1
holds if we have m
l
< m
o
<
u
m
u
.
The fact that c
l
< c
o
< c
u
and m
l
< m
o
< m
u
hold, ∀k,k ≥ 0 can be
easily proved by induction on k. The inductive basis holds for k = 0 as
c
l
= y
0
≤c
o
< ⌊c
o
+ 1⌋ = y
0
+ 1 = c
u
and
≤ 1 < (n + 1)/n = m
u
.
Now, let us assume that c
l
< c
o
< c
u
and m
l
< m
o
< m
u
hold for some k,
k > 0.
From the above derivation we have
m
l
= −1/n < 0 ≤ m
o
c
l
< c
o
< c
u
implies m
k+1
< m
o
< m
k+
u
and
l
