Image Processing Reference
In-Depth Information
where L
k
(x) is the straight-line segment joining f(
x
k
)tof(
x
k
þ
1
) and is given by
f
ð
x
k
þ
1
Þ
f [xi
ðÞ
x
k
þ
1
x
k
L
k
(
x
) ¼
ð
x
x
k
Þ
f
x
ðÞ
(
6
:
64
)
for
first and the last points for interpolation,
the number of grid points has to be greater than or equal to three, N
x
k
x
x
k
þ
1
. Since we always need the
3. Now assume
that N
¼
x
2
between x
1
and x
M
such that the total MSE given by Equation 6.63 is minimized. Let the solution be x
j
,
where 1
3, which means that we need to
find one grid point
<
j
<
N. The MMSE is given by
X
X
j
M
1
M
1
M
2
2
E*
¼
j
fx
ðÞ
L
1
x
ðÞ
j
þ
j
fx
ðÞ
L
2
x
ðÞ
j
(
6
:
65
)
i
¼
1
i
¼
j
þ
1
where L
1
(x) and L
2
(x) are
fx
j
fx
ðÞ
x
j
x
1
L
1
(
x
) ¼
ð
x
x
1
Þ
fx
ðÞ
(
6
:
66
)
fx
ðÞ
fx
j
x
M
x
j
þ
fx
j
L
2
(
x
) ¼
x
x
j
(
6
:
67
)
E* and the index j are found by the following optimization problem:
j
¼ arg min
k
[
E
(
k
)]
X
X
k
M
1
M
1
M
2
2
¼ arg min
k
j
fx
ðÞ
L
1
x
ðÞ
j
þ
j
fx
ðÞ
L
2
x
ðÞ
j
(
6
:
68
)
i
¼
1
i
¼
k
þ
1
E*
¼
E
(
j
)
(
6
:
69
)
This means that if we start from x
1
and wish to locate one grid point between x
1
and x
M
to approximate f(x), that point will be f(x),
j
and the corresponding MMSE will be E*. We
de
ne a new index, j
1
¼
j, and error, E
1
¼
E*, and assign these two numbers to x
1
and
write it as {j
1
, E
1
}. We repeat this process for x
2
through x
N
2
and form the array of
numbers as shown in column 1 of Table 6.12. We call this the 1-D single-stage grid
allocation algorithm. In this array, Ei
i
is the MSE between the original function and the
linearly interpolated function using grid points
over the closed interval of
[x
i
x
N
]. These indices and their corresponding MSEs are required for the two-stage
optimal search. In the two-stage search, we try to locate two optimal grid points
between x
i
and x
N
such that the total MSE is minimized. Using dynamic programming,
we need to minimize
b
x
i
x
j
x
N
c
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