Problem 2.1
The diameter,x, of the retinal image corresponding to the dot is obtained from similar triangles, as shown in Fig. P2.1. That is,
(d=2)
0:2 = (x=2) 0:014
which givesx= 0:07d. From the discussion in Section 2.1.1, and taking some liberties of interpretation, we can think of the fovea as a square sensor array having on the order of 337,000 elements, which translates into an array of size580£580elements. Assuming equal spacing between elements, this gives 580 elements and 579 spaces on a line 1.5 mm long. The size of each element and each space is thens = [(1:5mm)=1;159] = 1:3£10¡6m. If the size (on the fovea) of the imaged dot is less than the size of a single resolution element, we assume that the dot will be invisible to the eye. In other words, the eye will not detect a dot if its diameter,d, is such that0:07(d)<1:3£10¡6m, or d <18:6£10¡6m.
Figure P2.1
8 Chapter 2 Problem Solutions
Problem 2.3
¸=c=v= 2:998£108(m/s)=60(1/s)= 4:99£106m= 5000Km.
Problem 2.6
One possible solution is to equip a monochrome camera with a mechanical device that sequentially places a red, a green, and a blue pass®lter in front of the lens. The strongest camera response determines the color. If all three responses are approximately equal, the object is white. A faster system would utilize three different cameras, each equipped with an individual®lter. The analysis would be then based on polling the response of each camera. This system would be a little more expensive, but it would be faster and more reliable. Note that both solutions assume that the®eld of view of the camera(s) is such that it is completely®lled by a uniform color [i.e., the camera(s) is(are) focused on a part of the vehicle where only its color is seen. Otherwise further analysis would be required to isolate the region of uniform color, which is all that is of interest in solving this problem].
Problem 2.9
(a) The total amount of data (including the start and stop bit) in an 8-bit,1024£1024 image, is(1024)2£[8 + 2]bits. The total time required to transmit this image over a At 56K baud link is(1024)2£[8 + 2]=56000 = 187:25sec or about 3.1 min. (b) At 750K this time goes down to about 14 sec.
Problem 2.11
Letpandqbe as shown in Fig. P2.11. Then, (a)S1andS2are not 4-connected because qis not in the setN4(p)u(b)S1andS2are 8-connected becauseqis in the setN8(p)u (c)S1andS2arem-connected because (i)qis inND(p), and (ii) the setN4(p)\N4(q) is empty.
Problem 2.12 9
Figure P2.11
Problem 2.12
The solution to this problem consists of de®ning all possible neighborhood shapes to go from a diagonal segment to a corresponding 4-connected segment, as shown in Fig.
P2.12. The algorithm then simply looks for the appropriate match every time a diagonal segment is encountered in the boundary.
Figure P2.12
Problem 2.15
(a) WhenV =f0;1g, 4-path does not exist betweenpandqbecause it is impossible to
10 Chapter 2 Problem Solutions
get fromptoqby traveling along points that are both 4-adjacent and also have values fromV. Figure P2.15(a) shows this conditionuit is not possible to get toq. The shortest 8-path is shown in Fig. P2.15(b)uits length is 4. In this case the length of shortestm- and 8-paths is the same. Both of these shortest paths are unique in this case. (b) One possibility for the shortest 4-path whenV =f1;2gis shown in Fig. P2.15(c)uits length is 6. It is easily veri®ed that another 4-path of the same length exists betweenpandq.
One possibility for the shortest 8-path (it is not unique) is shown in Fig. P2.15(d)u its length is 4. The length of a shortestm-path similarly is 4.
Figure P2.15
Problem 2.16
(a) A shortest 4-path between a pointpwith coordinates(x; y)and a pointqwith coor- dinates(s; t)is shown in Fig. P2.16, where the assumption is that all points along the path are fromV. The length of the segments of the path arejx¡sjandjy¡tj, respec- tively. The total path length isjx¡sj+jy¡tj, which we recognize as the de®nition of theD4distance, as given in Eq. (2.5-16). (Recall that this distance is independent of any paths that may exist between the points.) TheD4distance obviously is equal to the length of the shortest 4-path when the length of the path isjx¡sj+jy¡tj. This oc- curs whenever we can get fromptoqby following a path whose elements (1) are from V;and (2) are arranged in such a way that we can traverse the path fromptoqby mak- ing turns in at most two directions (e.g., right and up). (b) The path may of may not be unique, depending onV and the values of the points along the way.
Problem 2.18 11
Figure P2.16
Problem 2.18
With reference to Eq. (2.6-1), letHdenote the neighborhood sum operator, letS1 and S2denote two different small subimage areas of the same size, and letS1+S2denote the corresponding pixel-by-pixel sum of the elements inS1andS2, as explained in Section 2.5.4. Note that the size of the neighborhood (i.e., number of pixels) is not changed by this pixel-by-pixel sum. The operator Hcomputes the sum of pixel values is a given neighborhood. Then,H(aS1+bS2)means: (1) multiplying the pixels in each of the subimage areas by the constants shown, (2) adding the pixel-by-pixel values fromS1and S2(which produces a single subimage area), and (3) computing the sum of the values of all the pixels in that single subimage area. Letap1andbp2denote two arbitrary (but corresponding) pixels fromaS1+bS2. Then we can write
H(aS1+bS2) = X
p12S1andp22S2
ap1+bp2
= X
p12S1
ap1+ X
p22S2
bp2
= a X
p12S1
p1+b X
p22S2
p2
= aH(S1) +bH(S2)
which, according to Eq. (2.6-1), indicates thatHis a linear operator.