Cơ sở dữ liệu hình ảnh - Chương 6 pdf

6. FINDING BASIC SHAPES 6.1 Combining Edges Bits of edges, even when they have been joined up in some way by using, for example, crack edge relaxation, are not very useful in themself unless they are used to enhance a previous image. From identification point of view it is more useful to determine structure of lines, equations, lengths, thickness There are a variety of edge-combining methods in literature. These include edge following and Hough transforms. 6.2 Hough Transform This technique allows to discover shapes from image edges. It assumes that a primitive edge detection has already been performed on an image. It attempts to combine edges into lines, where a sequence of edge pixels in a line indicates that a real edge exists. As well as detecting straight lines, versions of the Hough transform can be used to detect regular or non-regular shapes, though, as will be seen, the most generalized Hough transform, which will detect a two dimensional specific shape of any size or orientation, requires a lot of processing power in order to be able to do its work in a reasonably finite time. 6.2.1 Basic principle of the straight-line Hough transform After primitive edge detection and then thresholding to keep only pixels with a strong edge gradient, the scree n may look like Figure 6.1. Figure 6.1 Screen after primitive edge detection and thresholding (only significant edge pixel shown). A straight line connecting a sequence of pixels can be expressed in the form: y = mx + c If we can evaluate values for m and c such that the line passes through a number of the pixels that are set, then we have a usable representation of a straight line. The Hough transform takes the above image and converts into a new image (what is termed) in a new space. In fact, it transforms each significant edge pixel in (x,y) space into a straight line in this new space. Original data Line to be found 1 2 3 4 Figure 6.2 Original data. Clearly, many lines go through a single point (x, y), e.g. a horizontal line can be draw through the point, a vertical line, and all the lines at different angles between these. However, each line will have a slope (m) and intercept (c) such that the above equation holds true. A little manipulation of the above equation gives: c = (−x)m + y y x Gives Transposed 3 1 3 = m . 1 + c c = −1m + 3 2 2 2 = m . 2 + c c = −2m + 3 3 4 3 = m . 4 + c c = −3m + 3 0 4 0 = m . 4 + c c = −4m + 3 3 Three line coincide here 0 3 c = − 1m+3 c = − 2m+2 c = − 4m+3 c = − 4m c m Figure 6.3. Accumulator array in (m,c) space. Maximum in the accumulator array is 3 at (−1,4), suggesting that a line y = −1x + 4 goes through three of the original data points. We know the value of x and y (the position where the pixel may be on an edge), but in this form. the equation now represents a straight line in (m,c) space, i.e. with a horizontal m-axis and a vertical c-axis, each (x,y) edge pixel corresponds to a straight line on this new (m,c) graph. We need space to be available to hold this set of lines in an array (called the accumulator array). Then for every (x,y) point, each element that lies on the corresponding line in the (m,c) accumulator array can be incremented. So that after the first point in the (x, y) space has been processed, there will be a line of 1 st in the (m,c) array. This plotting in the (m, c) array is done using an enhanced form of Bresenham’s algorithm, which will plot a wide, straight line (so that at the ends crossing lines are not missed). At the end of processing all the (x,y) pixels, the highest value in the (m,c) accumulator array indicates that a large number of lines cross in that array at some points (m’,c’). The value in this element corresponds to the same number of pixels being in the straight line in the (x,y) space and the position of this element gives the equation of the line in the (x,y) space, and the position of this element gives the equation of the line in (x,y) space: y = m’x + c’ 6.2.2 Problems There are serious problems in using (m,c) space. For each pixel, m may properly vary from minus infinity to infinity (i.e. straight line upwards). Clearly this is unsatisfactory: no accumulator array can be set up with enough elements. There are alternatives, such as using two accumulator array, with m ranging from −1≤ m ≤ +1 in one and −1≤ 1/m ≤ +1 in the second. It is safer, though requiring more calculation, to use angles, transforming to polar coordinates (r, θ ), where xcos θ + ysin θ = r. y=a 1 x+b 1 y=a 2 x+b 2 y=a 3 x+b 3 y=a 4 x+b 4 y=a 5 x+b 5 Point(x,y) Figure 6.4 Family of lines (Cartesian coordinates) through the point (x,y). r (x,y) One of many possible lines through (x,y), e.g. y=ax+b Shotest distance from origin to line defines the line in term of r and θ θ x y (x,y) x y (y-x tan θ )sin θ y-x tan θ x/cos θ xtan θ θθθ θ θ θ θ θ θ θθ θ sincossin cos sin1 cos sin sin cos sin)tan( cos 2 2 yxyx xy x xy x r +=+         − = −+= −+= Figure 6.5 Relationship between Cartesian straight line and polar defined line. Technique 6.1. Real straight-edge discovery using the Hough transform. USE. This technique is used to find out and connect substantial straight edges already found using and edge detector. OPERATION. For each edge pixel value I(x,y), vary θ from 0 o to 360 o and calculate r = xcos θ + ysin θ . Given an accumulator array size (N+M,360), increment those elements in the array that lie in box (b x b) with center (r, θ ). Clearly if the box is (1x1), only one element of the array is incremented; if the box is 3 x 3, nine elements are incremented. This gives a "thick" line in the new space so that intersections are not missed. Finally, look for the highest values in the accumulator arrays (r, θ) and thus identify the pair (r, θ) that are most likely to indicate a line in (x,y) space. This method can be enhanced in a number of ways: 1. Instead of just incrementing the cells in the accumulator array, the gradient of the edges, prior to thresholding, could be added to the cell, thus plotting a measure of the likelihood of this being an edge. 2. Gradient direction can be taken into account. If this suggest s that the direction of the real edge lies between two angles θ 1 and θ 2 , then only the elements in the (r, θ) array that lies in θ 1 < θ < θ 2 that are plotted. 3. The incrementing box does not need to be uniform. It is known that the best estimate of (r, θ) is at the center of the box, so this element is incremented by a large figure than the elements around that center element. Note that the line length is not given, so that the lines go to infinity as it stands. Three approaches may be considered: 1. Pass 3 x 3 median filter over the image original and subtracting the value of the center pixel in the window from the result. This tends to find some corners of images, thus enabling line endings to be estimated. 2. Set up four further accumulator array. This first pair can hold the most north-east position on the line and the second pair the most south-west position, these positions being updated as and when a pixel contributes to the corresponding accumulating element in the main array. 3. Again with four further accumulator array, let the main accumulator array be increased by w for some pixel (x,y). Increase this first pair by wx and wy and the second by (wx) 2 and (wy) 2 . At the end of the operation a good estimate of the line is: mean of lines ± 2σ where σ is the standard deviation, i.e. ( ) 2 2         − ± = ∑ ∑ ∑ ∑ ∑ ∑ w wx w wx w wx estimate line of End for the x range and the similar expression for the y range. This makes some big assumption regarding the distribution of edge pixels, e.g. it assumes that the distribution is not skewed to one end of the line, and also many not always be appropriate. The Hough technique is good for finding straight lines. It is even better for finding circles. Again the algorithm requires significant edge pixels to be identified so some edge detector must be passed over the original image before it is transformed using the Hough technique. Technique 6.2. Real circle discovery using the Hough transform. USE. Finding circles from an edge-detected image. OPERATION. If the object is to search for circles of a known radius R, say, then the following identity can be used: ( ) ( ) 2 22 Rbyax =−+− where (a,b) is the centre of the circle. Again in (x,y) space all pixels or, an edge are identified (by thresholding) or every pixel with I(x,y) > 0 is processed. A circle of elements is incremented in the (a,b) accumulator array centre (0<a< M–1, 0<b<N-1), radius R for each edge pixel to be processed. Bresenham's circle drawing algorithm can be used to increment the circle elements quickly. Finally, the highest values ill the (a,b) array, indicate coincident edges in (a, b) space corresponding to a number of pixels on the edge of the same circle in space. Circle to be found Figure 6.6. Original data in (x,y) domain. Again it is possible to reduce the amount of work by using the gradient direction to indicate the likely arc within which the circle centre is expected to lie. Figure 6.7 illustrates this technique. It is possible to look for the following types of circles: different radii plot in (a,b,R) space different radii, same vertical centres plot in (b,R) space different radii, same horizontal centres plot in (a,R) space Four cicles coincide here only Figure 6.7 Illustration of Hough circle transform (looking for circles radius 1/√2). Corresponding accumulator circles in (a,b) domain. If the circle radius is known to be one of three values, say, then (a,b,R) space can be three planes of (a,b) arrays. The following points are important: 1. As the number of unknown parameters increases, the amount of processing increases exponentially. 2. The Hough technique above can be used to discover any edge that can be expressed as a simple identity. 3. The generalized Hough transform can also be used to discover shapes that can not be represented by simple mathematical identities. This is described below. Technique 6.3. The generalized Hough transform. USE. Find a known shape  in its most general form-of any size or orientation in an image. In practice it is best to go for a known size and orientation. OPERATION. Some preparation is needed prior to the analysis of the image. Given the object boundary, and assuming that the object in the image is of the same size and orientation (otherwise a number of accumulator arrays have to beset up for different sizes and orientations), a ‘centre’ (x,y) is chosen somewhere within the boundary of the object. The boundary is then traversed and after every step d alone the boundary the angle of the boundary tangent with respect to horizontal is noted, and the x difference and y difference of the boundary position from the centre point are also noted. For every pixel I(x, y) in the edge-detected image, the gradient direction is found. The accumulator array (same size as the image) is then incremented by 1 for each such element. Finally, the highest-valued elements in the accumulator array point to the possible centres of the object in the image. 6.3 Bresenham’s Algorithm Bresenham’s line algorithm is an efficient method for scan-converting straight lines in that it uses only integer addition, subtraction, and multiplication by 2. As a very well known fact, the computer can perform the operations of integer addition and subtraction very rapidly. The computer is also time-efficient when performing integer multiplication and division by powers of 2. The algorithm described in the following is a modified version of the Bresenham algorithm. It is commonly referred to as the midpoint line algorithm. U D x k x k + 1 d 1 d 2 y y k y k +1 Figure 6.8 Midpoint algorithm  The equation of a straight line in 2-dimensional space can be written in an implicit form as F(x, y) = ax + by + c = 0 From the slope-intercept form y dy dx x B= + we can bring it to the implicit form as dy x dx y Bdx⋅ − ⋅ + = 0 So a = dy, b = −dx, c = Bdx  Suppose that point (x i , y i ) has been plotted. We move x i to x i + 1. The problem is to select between two pixels, U(x i + 1, y i + 1) and D(x i + 1, y i ). For this purpose, we consider the middle pixel M(x i + 1, y i + 1 2 ). We have d = F(M) = a(x i + 1) + b( y i + 1 2 ) + c If d > 0 , choose U d < 0 , choose D d = 0 , choose either U or D, so choose U. - When D is chosen, M is incremented one step in the x direction. So d new = F(x i +2, y i + 1 2 ) = a(x i + 2) + b(y i + 1 2 ) + c while d old = F(x i + 1, y i + 1 2 ) = a (x i + 1) + b (y i + 1 2 ) + c So the increment in d (denoted d D ) is d D = d new − d old = a = dy - When U (x i + 1, y i + 1) is chosen, M is incremented one step in both directions: d new = F (x i +2, y i + 3 2 ) = a (x i + 2) + b( y i + 3 2 ) + c = d old + a + b So the increment in d (denoted d U ) is d U = a + b = dy − dx In summary, at each step, the algorithm chooses between two pixels based on the sign of d. It updates d by adding d D or d U to the old value.  First, we have the point (x 1 , y 1 ). So M (x 1 +1, y 1 + 1 2 ) and F(M) = a(x 1 + 1) + b (y 1 + 1 2 ) + c = F(x 1 , y 1 ) + a + b/2 Since F (x 1 , y 1 ) = 0, we have d = d 1 = dy − dx/2 In order to avoid a division by 2, we use 2d 1 instead. Afterward, 2d is used. So, with d used in place of 2d, we have First set d 1 = 2dy − dx If d i ≥ 0 then x i+1 = x i + 1, y i+1 = y i + 1 and d i+1 = d i + 2 (dy − dx) If d i < 0 then x i+1 = x i + 1, y i+1 = y i d i+1 = d i + 2dy The algorithm can be summarized as follows: Midpoint Line Algorithm [Scan-convert the line between (x 1 , y 1 ) and (x 2 , y 2 )] dx = x 2 − x 1 ; dy = y 2 − y 1 ; d = 2*dy − dx; /* initial value of d */ dD = 2*dy; /* increment used to move D */ dU = 2*(dy − dx); /* increment used to move U */ x = x 1 ; y = y 1 ; Plot Point (x, y); /* the first pixel */ While (x < x 1 ) if d <0 then d = d + dD; / * choose D */ x = x + 1; else d = d + dU; /* choose U */ x = x + 1; y = y + 1; endif Plot Point (x, y); /* the selected pixel closest to the line */ EndWhile Remark The described algorithm works only for those lines with slope between 0 and 1. It is generalized to lines with arbitrary slope by considering the symmetry between the various octants and quadrants of the xy-plane. Example. Scan-convert the line between (5, 8) and (9, 11). Since for the points, x < y, consequently the algorithm can apply. Here dy = 11 − 8 = 3, dx = 9 − 5 = 4 First d 1 = 2dy − dx = 6 − 4 = 2 > 0 So the new point is (6, 9) and d 2 = d 1 + 2 (dy − dx) = 2 + 2(−1) = 0 ⇒ the chosen pixel is (7, 10) and d 3 = d 2 + 2 (dy − dx) = 0 +2(−1) = −2 < 0 the chosen pixel is (8, 10), then d 4 = d 3 + 2dy = −1 +6 = 5 > 0 The chosen pixel is (9, 11). 6.3.2 Circle incrementation A circle is a symmetrical figure. Any circle-generating algorithm can take advantage of the circle’s symmetry to plot eight points for each value that the algorithm calculates. Eight-way symmetry is used by reflecting each calculated point around each 45° axis. For example, if point 1 in Figure 6.9 were calculated with a circle algorithm, seven more points could be found by reflection. The reflection is accomplished by reversing the x, y coordinates as in point 2, reversing the x, y coordinates and reflecting about the y axis as in point 3, reflecting about the y ② ③ ① ⑧ ⑦ ⑥ ⑤ ④ (2, 8) (y, x) (-2, 8) (-y, x) y (8, 2) (x, y) x (x, -y) (8, -2) (y, -x) (2, -8) (-y, -x) (-2, -8) (-x, -y) (-8, -2) (-8, 2) (-x, y) 9 9 [...]... cannot deal with occlusion or part views of objects, nor can it deal with three-dimensional objects in different orientations 6. 6 Exercises 6. 6.1 Using standard graph paper, perform a straight line Hough transform on the binary pixels array shown in the following figure transforming into (m,c) space Figure 6. 8 Binary array 6. 6.2 A library object has the following ordered interest point classification... or Lb = s 2 2 and 2 If the inverse of the square L matrix is calculated, then the values for a to f can be evaluated and the error determinated This is calculate as L-1L a = L-1 s1 and L-1L b = L-1 s2 Resulting in a = L-1s1 and b = L-1s2 6. 5 Problems There are some problems with interest point First, coordinates must be paired beforehand That is, there are known library coordinates, each of which must... allows these goals to be met Scan-converting a circle using Bresenham’s algorithm works are follows If the eightway symmetry of a circle is used to generate a circle, points will only have to be generated through a 45° angle And, if points are generated from 90 to 45°, moves will be made only in the +x and -y directions (see Figure 6. 12) -y 45° +x Figure 6. 12 Circle scan-converted with Bresenham’s algorithm... current iteration, the point of evaluation moves from (xi, yi ) to (xi +1, yi -1 ) The first-order differences are d D = 2( xi − yi ) + 5 ′ d D = 2 [ xi + 1 − ( yi − 1)] + 5 = 2( xi − yi ) + 4 + 5 d U = 2 xi + 3 d U ′ = 2( xi + 1) + 3 Thus the second-order differences are ∆U = 2, ∆D = 4 So the revised algorithm using the second-order differences is as follows: (1) (2) (3) h = 1 − r, x = 0 , y = r , ∆U... point of evaluation moves from (xi, yi ) to (xi+1, yi ) The first-order difference has been calculated as dU = 2xi + 3 At point (xi + 1, yi ), this will be d′U = 2( xi + 1) + 3 Thus the second-order difference is ′ ∆U = d U − d U = 2 Similarly, dD at (xi, yi ) is 2(xi − yi )+5 and at (xi +1, yi ) is d ′ = 2(xi +1− yi ) + 5 Thus D the second-order difference is ∆D = d ′ − d D = 2 D  If D is chosen in... square root of the result must be found The second method of defining a circle makes use of trigonometric functions (see Figure 6. 11): y P = ( x, r 2 − x 2 ) y P=(r cos θ, r sin θ) y r θ x x r cos θ r sin θ x Fig 6. 10 Circle defined with a seconddegree polynomial equation Fig 6. 11 Circle defined with trigonometric functions x = r cosθ where θ r x y = = = = y = r sinθ current angle circle radius x coordinate...Figure 6. 9 Eight-way symmetry of a circle axis as in point 4, switching the signs of x and y as in point 5, reversing the x, y coordinates, reflecting about the y axis and reflecting about the x axis as in point 6, reversing the x, y coordinates and reflecting about the y axis as in point 7, and reflecting about... P(xi, yi) M× O D(xi +1, yi - 1) (a) (b) Figure 6. 13 Bresenham’s Circle Algorithm (Midpoint algorithm) Bresenham Midpoint Circle Algorithm h = 1 − r ; /*initialization */ x = 0; y = r; Plot Point (x, y); While y > x if h < 0 then /* Select U */ dU = 2*x + 3; h = h + dU; x = x + 1; else /* Select D */ dD = 2*(x − y) + 5; h = h − dD; x = x + 1; y = y − 1; endif End While (iii) Second-order differences  If... (1,0), (2,4)} Identify, using the above technique, which of the following two sets of interest points represent a transition, rotation, and/or scaling of the above object: {(1,1), (6, 12), (2,5), (12,23)} {(1,3), (1,12), (-1 ,8), (3 ,6) } Check your answer by showing that a final point maps near to its corresponding known point ... scan-converted with Bresenham’s algorithm The best approximation of the true circle will be described by those pixels in the raster that fall the least distance from the true circle Examine Figures 6. 13(a) and 6. 13(b) Notice that if points are generated from 90 and 45°, each new point closest to the true circle can be found by taking either of two actions: (1) move in the x direction one unit or (2) move . y ② ③ ① ⑧ ⑦ ⑥ ⑤ ④ (2, 8) (y, x) (-2 , 8) (-y, x) y (8, 2) (x, y) x (x, -y) (8, -2 ) (y, -x) (2, -8 ) (-y, -x) (-2 , -8 ) (-x, -y) (-8 , -2 ) (-8 , 2) (-x, y) 9 9 Figure 6. 9 Eight-way symmetry of a circle. axis. evaluated and the error determinated. This is calculate as L -1 L a = L -1 s1 and L -1 L b = L -1 s2 Resulting in a = L -1 s1 and b = L -1 s2. 6. 5 Problems There are some problems with interest point generated from 90 to 45°, moves will be made only in the +x and -y directions (see Figure 6. 12). -y 45° +x Figure 6. 12 Circle scan-converted with Bresenham’s algorithm. The best approximation