yimo.guo@ee.oulu.fi22.09.2011

Digital Image Processing Exercise 1

Exercises:

- Questions : one week before class

- Solutions : the day we have class

- Slides along with Matlab code (if have) : after class

1 gigabyte = ( 8 × 10243 ) bits

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

2. (Exam 4.12.2004) Perform connected component analysis of the follow-ing binary image. Use the two-scan labeling algorithm and representresults after each scan by using letters (a,b,c,. . . ) as labels.

(a) Assume 4-connectivity.

From left to right, top to bottom. Examine each pixel P and its neighbor pixels: left (x1) and up (x2).

First scan:

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-neighborsof pixel q have values from V .

ii. The shortest 8-path from p to q, considering 8 neighborhood of one pixel.

3

022

21

1 2 1 1

2

1 (q)

(p) 1 0 1 2

V = {0, 1}

p = (3; 0); (3; 1); (2; 2); (1; 2); (0; 3) = q

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. m-adjacency (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

022

21

1 2 1 1

2

1 (q)

(p) 1 0 1 2

V = {0, 1}

1 2 1

0 1

Intersection set of N4(p) and N4(q) is {1, 2}

Thus, the length of this path is 5.

(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) = qThe 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) = qThe 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) = qThe length of this path is 6.

Notice that these paths are not unique.

It is easily verifiedthat another path of the same length exists between p and q.

Equalization of an image histogram is the cumulative density function.

(a) Perform histogram equalization given the following histogram.(r=Gray level, n=number of occurrences)

First, calculate the probability pk for each gray level:

pk = nk/sum(nk)

4.

Second, compute the discrete cumulative density function sk.

Finally, round to the nearest discrete value available:

The equalized histogram is:

x/7

(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 targetdistribution.(inverse transform z = G-1(s), where G(z) is a mapping that equalizes the target distribution)

Compute this mapping:

First, cumulate the probability.

Next, apply the inverse transform z = G-1(s), by finding the closest sk for each sk

’ computed in part (a).

part (a)

Thus, the histogram resulting from the transform is:

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:

(Matlab code)

P(x,y)

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