Digital Image Processing
Exercise 1
yimo.guo@ee.oulu.fi
22.09.2011
Exercises:
- Questions : one week before class
- Solutions : the day we have class
- Slides along with Matlab code (if have) : after class
1. One of the several HDTV formats is 1080p24, which means video stream of full
frames of 1920×1080 pixels at frame rate 24 fps. If each pixel has 24 bits of
intensity resolution (8 bits each for red, green and blue channels), how many
gigabytes are needed for 2 hours of HDTV video without compression?
HDTV (High-definition television)
formats is 1080p24.
Each pixel has 24 bits of intensity
resolution.
Frame rate 24 fps.
Video size is nb × nf bits
1 gigabyte = ( 8 × 10243 ) bits
2. (Exam 4.12.2004) Perform connected component analysis of the following binary image. Use the two-scan labeling algorithm and represent
results after each scan by using letters (a,b,c,. . . ) as labels.
(a) Assume 4-connectivity.
First scan:
From left to right, top to bottom.
Examine each pixel P and its neighbor pixels:
left (x1) and up (x2).
Second scan:
Image is scanned and pixels are given final
labels according to the equivalences found
during the first scan.
(b) Assume 8-connectivity.
This picture is scanned in the similar
way as with 4-connectivity, but now
we examine four neighbors of P
(also the diagonal neighbors).
We notice that a is equal to b.
They are given final label A in the
second scan.
(Matlab code)
3. (Gonzalez & Woods 2007, Ex. 2.15) Consider the image segment shown.
(a) Let V = {0, 1} and compute
the lengths of the shortest
4-, 8-, and m-path between
p and q. If a particular path
does not exist between
these points, explain why.
i. There is no 4-path between p and q,
as none of the 4-neighbors
of pixel q have values from V .
ii. The shortest 8-path from p to q,
considering 8 neighborhood of
one pixel.
3
1
2
1 (q)
V = {0, 1}
2
2
0
2
1
2
1
1
p = (3; 0); (3; 1); (2; 2); (1; 2); (0; 3) = q
(p) 1
0
1
2
The length is N - 1 where N is the
number of pixels on the path.
The length of the shortest 8-path is 4.
iii. The shortest m-path from p to q. madjacency (Page 67) :
Two pixels p and q with values from
V are m-adjacent if:
(i) q is in N4(p), or
(ii) q is in ND(p) and the intersection set
of N4(p) N4(q) has no pixels whose
values are from V.
3
1
2
1 (q)
2
2
0
2
1
2
1
1
(p) 1
0
1
2
Thus, the length of this path is 5.
V = {0, 1}
1
2
1
0
1
Intersection set of N4(p) and N4(q) is
{1, 2}
(b) Let V = {1, 2} and compute
the lengths of the shortest
4-, 8-, and m-path between
p and q. If a particular path
does not exist between
these points, explain why.
i. One possibility for 4-path:
p = (3; 0); (2; 0); (2; 1); (2; 2); (2; 3); (1; 3); (0; 3) = q
The length of this path is 6.
ii. One possibility for the shortest 8-path:
p = (3; 0); (2; 1); (1; 1); (0; 2); (0; 3) = q
The length of the shortest path is 4.
iii. One possibility for the shortest m-path:
p = (3; 0); (2; 0); (2; 1); (1; 1); (0; 1); (0; 2); (0; 3) = q
The length of this path is 6.
Notice that these paths
are not unique.
It is easily verified
that another path of the
same length exists
between p and q.
4.
(a) Perform histogram equalization given the
following histogram.
(r=Gray level, n=number of occurrences)
Equalization of an image
histogram is the cumulative
density function.
First, calculate the probability
pk for each gray level:
pk = nk/sum(nk)
Second, compute the discrete
cumulative density function sk.
Finally, round to the nearest
discrete value available:
x/7
The equalized histogram is:
(b) Perform histogram specification of the
previous histogram using the specified
histogram shown in the following table.
(r=Gray level, p=probability of occurrences)
Transform the histogram into a
given distribution.
First, equalize the histogram.
(in part (a) )
Second, change the equalized
histogram into the given target
distribution.
(inverse transform z = G-1(s),
where G(z) is a mapping that
equalizes the target
distribution)
Compute this mapping:
First, cumulate the probability.
part (a)
Thus, the histogram resulting from the transform is:
Next, apply the inverse
transform z = G-1(s), by
finding the closest sk for
each sk’ computed in part
(a).
5. An image is corrupted by additive uncorrelated, zero-average noise
yielding
How is the signal-to-noise ratio affected if you average these K images?
Signal-to-noise power ratio:
For single image:
zero-average noise:
The average image:
Its signal-to-noise power ratio:
Finally, the signal-to-noise ratio becomes:
P(x,y)
(Matlab code)
6. (Exam 2.12.2005) Explain different methods for handling border
pixels with mask operations.
(a) operator modification
operator is modified for exceptions where some of the necessary
neighbors are missing
 often complex
 seldom used

(b) adding zeroes
easy to perform
 often used

(c) reflecting
usually better than adding zeroes
 often used

(d) image is considered to be cyclic
seldom used
 one should have some reason for assuming the image to be periodic

(e) only the pixels that have all the necessary neighbors are processed
the only ‘right’ way
 processed image is smaller than original
