COMP 4500 Sohar University Computer Vision and Image Processing Lab Report

Editing part:

Please include the application of all images I shared with you.

Please include screenshots from Matlab like the answer sample I am attaching

Introduction to MATLAB Digital Image Processing
Steps and Discussion:
-First step : Write row vector v= 24
66 4 2
9
Figure 1: Creating Row Vector ( v)
As shown in figure 1=> The syntax is as follows v= [24 66 4 2 9]. A row vector can
be thought of as a 1*N-dimensional array. The output is a Row Vector
V=
24 66 4 2 9
1
-Second step: Display the fourth element
Figure 2: Displaying fourth element of Row Vector (v)
As shown in figure 2 =>We use the following syntax to display the fourth element of
Row Vector v [ v (4) ] . A single index value can be used to access the items of a Row
Vector. The first element of a vector is v (1), the second element is v (2), and so on. As a
result, we have:
Ans=
2
2
– Third step: Change the row vector to column vector
Figure 3: Converting Row Vector v to Column Vector (w)
As shown in figure 3 => The basic syntax for converting Row Vector v to Column
Vector w is w=v.’ To transform a row vector to a column vector, use the transpose
operator (.’). As a result, we have:
W=
24
66
4
2
9
3
– Fourth step: Display the first three element of v
Figure 4: Displaying first three elements of Row vector (v)
As shown in figure 4 => to make the first three components visible We use the following
syntax for Row Vector v [ v (1:3) ]. The first three elements of Row Vector v. are 24, 66,
and 4. As a result, we have:
ans =
24
66
4
4
– Fifth step: Display the index will count started at the last element decreased by 2 and
stop when it is reached the first element
Figure 5 : Displaying last element of Row Vector v which is decreased by 2 until it reaches first element
As shown in figure 5 => We use the basic syntax to display the last element Row Vector
v and then decrease by two elements until we reach the first element: v(end: -2:1) The
index count begins at the final element, decreases by two, and ends at the first element.
As a result, we have:
ans=
9 4 24
5
– Sixth
step: Write 3×3 matrix A= 12 52
30 7
9
2
44
6
11
Figure 6: Creating 3*3 Matrix (A)
As shown in figure 6 => A= [12 52 44; 30 7 6; 9 2 11] is the syntax for creating the
needed Matrix. Matrixes can be represented as a series of row vectors separated by
semicolons and enclosed by squared brackets. The syntax produces the following result:
A=
12
52
44
30
7
6
9
2
11
6
– Seventh step: Write submatrix T2 which contain rows 1&2 and columns 1,2,3 of A
Figure 7: Creating Sub-Matrix T2 contain Rows 1, 2 and Columns 1, 2, 3 of
Matrix (A)
As shown in figure 7 => By giving a vector of values for both the row and column
indices, the T2 Sub-Matrix can be retrieved. The needed Sub-Matrix is built using the
syntax indicated in figure 7 => T2=A ([1 2], [1 2 3]) . T2 is the Sub-Matrix containing
rows 1 and 2 as well as columns 1, 2 and 3. The syntax produces the following result:
T2=
12
52
44
30
7
6
7
– Eighth step: Write submatrix T3 which contain the entire 3rd column only of A
Figure 8 : Creating Sub-Matrix T3 contain entire 3rd column of Matrix (A)
As shown in figure 8 => A colon in the column index position selects all of a Matrix’s
columns. The needed Sub-Matrix is produced using the syntax: T3= A(: ,3), as illustrated
in figure (8). The syntax produces the following result:
T3=
44
6
11
8
– Ninth step: Write submatrix T4 which contain the second row only of A
Figure 9 : Creating Sub-Matrix T4 containing entire second row of Matrix (A)
As shown in figure 9=> When a colon appears in the Row index position, it picks all of
the rows in a Matrix. The needed Sub-Matrix is formed using the syntax: T4= A(2,:) , as
seen in Figure 9. The syntax produces the following result:
T4=
30 7
6
9
– Step tenth: Create a copy D of matrix A
Figure 10 : Create a copy Matrix D of Matrix (A)
As shown in figure 10 => The syntax for making a copy of a matrix is D=A; or when I
want to show all of matrix D, I write D =A In addition, I will write in logical order and
set the third column to one as follows D = logical ([0 0 1; 0 0 1; 0 0 1]) the result,
D=
0
0 1
0
0 1
0
0 1
10
– Eleventh step: Assign zero to all elements in the 3rd column of D
Figure 11 : Assign zero to all elements in 3rd Column of Matrix (D)
As shown in figure 11=> A colon in the column index position selects all of a Matrix’s
columns. Figure 11 shows how to use the syntax D (:,3) = 0. The Matrix is represented
by the letter D. The value 0 will be assigned to all entries in D’s third column using this
approach. The syntax produces the following result:
11
D=
12
52
0
30
7
0
9
2
0
But in logical do A(D) then, A(D)= [0 0 0].
12
– Twelfth step: Using Hilbert matrix (5)
Figure 12 : Using Hilbert 5×5 Matrix
As shown in figure 12=> A linear indexing expression is one that indexes a matrix or
higher-dimensional array with a single subscript. We’ll use a 5×5 Hilbert matrix to
demonstrate the notion, as illustrated in figure 12 . The needed Hilbert is built using the
syntax: H=hilb (5)
13
H=
1.00000 0.50000 0.33333 0.25000 0.20000
0.50000 0.33333 0.25000 0.20000 0.16667
0.33333 0.25000 0.20000 0.16667 0.14286
0.25000 0.20000 0.16667 0.14286 0.12500
0.20000 0.16667 0.14286 0.12500 0.11111
– Thirteenth step: Display the element #17
Figure 13 : Display the element number 17
As shown in figure 13=> Using the syntax H=(17) , display the element number 17 The
syntax produces the following result:
ans = 0.20000
14
– Fourteenth step: Here the pixel values at these row and column coordinates:
Row= [2 1 5];
Col= [1 3 4]; to extract the desired pixel value
Figure 14 : The pixel values at given Row and Column
As shown in figure 14 => As demonstrated in Figure 14, required pixel values are
retrieved using the Syntax: Row r= [2 1 5] and Column c= [1 3 4].
r= [2 1 5];
c= [1 3 4];
15
H (r,c)
ans=
0.50000 0.25000
0.20000
1.00000 0.33333
0.25000
0.20000
0.12500
0.14286
– Fifteenth step: Write the linear indexing equation
Figure 15 : The linear indexing equation
As shown in figure 15 => As we learned the following is the formula for the linear
indexing equation:
M = size(H, 1);
linear_indices = M*(c − 1) + r
linear_indices =
2
11
20
16
H(linear_indices)
= M*(c − 1) + r = 2
linear indices of H is 0.50000
= M*(c − 1) + r = 11 linear indices of H is 0.33333
= M*(c − 1) + r = 20 linear indices of H is 0.12500
ans=
Conclusion
0.50000 :
0.33333
0.12500
We noticed that in this experiment is how to use and how to deal with the main concepts
of processing and understanded it include the indexing matrices, indexing vetoers and
logical indexing on Octave. In short, this lap we used in work effectively without any
problems or facing in doing and implemented this lap.
17
H(linear_indices)
= M*(c-1) +r= 2
linear indices of H is 0.50000
= M*(c-1) +r= 11 linear indices of H is 0.33333
=
= M*(c-1) +r= 20 linear indices of H is 0.12500
ans=
0.50000
0.33333
0.12500
We noticed that in this experiment is how to use and how to deal with the main concepts
of processing and understanded it include the indexing matrices, indexing vetoers and
logical indexing on Octave. In short, this lap we used in work effectively without any
problems or facing in doing and implemented this lap
TTIVI SP: WINU TIIU IMIOAI MUVAINIG vquanon
Octave
File Edit Debug Window Help News
Current Directory: C:\Users\HP
File Browser
х
Command Window
C:/Users/Hp
>> H- hilb(5)
Name
.config
3D Objects
> 2 Contacts
> Desktop
> Documents
>Downloads
Workspace
1.00000
0.50000
0.33333
0.25000
0.20000
0.50000
0.33333
0.25000
0.20000
0.16667
0.33333
0.25000
0.20000
0.16667
0.14286
0.25000
0.20000
0.16667
0.14286
0.12500
0.20000
0.16667
0.14286
0.12500
0.lllll
>> H(17)
ans = 0.20000
>> r[2 i 5);
>> c=(1 3 4);
>> H(r, c)
ans =
Filter D
Name
Class
Dimension
Value
Attribute
double
5x 5
(1, 0.50000, 0.33…
H
M
double
1×1
5
0.50000
1.00000
0.20000
0.25000
0.33333
0.14286
182932
0.20000
0.25000
0.12500
ans
double
1×3
c
double
1×3
[0.50000, 0.3333…
(1,3,4]
[2, 11,20]
[2, 1,5]
linear_indices
double
1×3
>> M =size (H, 1);
>> linear_indices – M* (C-1) +r
linear_indices =
r
double
1×3
2 11
20
Command History
Filter D
>> H(linear_indices)
ans-
H= hilb(5)
H(17)
0.50000
0.33333
0.12500
Figure 15 : The linear indexing equation
As shown in figure 15 => As we learned the following is the formula for the linear
indexing equation:
M = size(H, 1);
=
linear_indices = M*(c − 1) +r
linear indices =
=
2.
11
20
16
– Fourteenth step: Here the pixel values at these row and column coordinates:
Row= [2 1 5];
Col= [1 3 4]; to extract the desired pixel value
Octave
File Edit Debug Window Help News
Current Directory: C:\Users\Hp
File Browser
Command Window
C:/Users/Hp
>> H=hilb(5)
H =
A
Name
.config
> 3D Objects
> 2 Contacts
Desktop
Documents
>Downloads
Workspace
Filter o
1.00000
0.50000
0.33333
0.25000
0.20000
0.50000
0.33333
0.25000
0.20000
0.16667
0.33333
0.25000
0.20000
0.16667
0.14286
0.25000
0.20000
0.16667
0.14286
0.12500
0.20000
0.16667
0.14286
0.12500
0.lllll
5 x
>> r=[2 1 5);
>> c=(1 3 4);
>> H(r,c)
ans =
Name
Class
Dimension
Value
Attribute
А
double
3×3
0.50000
1.00000
0.20000
0.25000
0.33333
0.14286
0.20000
0.25000
0.12500
D
double
3×3
H
double
5×5
>>
[12, 52, 44; 30, 7…
[12, 52,0; 30, 7, …
[1, 0.50000, 0.33…
[0.50000, 0.2500…
(1,3,4)
[2.1,5)
ans
double
3×3
C
double
1×3
r
double
1×3
Figure 14 : The pixel values at given Row and Column
As shown in figure 14 => As demonstrated in Figure 14, required pixel values are
retrieved using the Syntax: Row r= [2 1 5] and Column c=
1
[1 3 4].
r=
= [2 1 5];
c= [1 3 4];
15