EGGN 512 Computer Vision Sample Problem Set - SOLUTIONS The exam will be closed book, but handwritten notes are allowed. The problems below are representative of exam problems (although there may be more problems than would appear on the actual exam). Some of the problems below are drawn from previous exams. For reference, here are some equations (these will be provided on the exam): A B The rotation matrix for XYZ fixed angles is: RXYZ X , Y , Z RZ ( Z ) RY (Y ) RX ( X ) 0 cz sz 0 cy 0 sy 1 0 sz cz 0 0 1 0 0 cx sx 0 0 1 sy 0 cy 0 sx cx where cx cos( X ), sy sin(Y ), etc The matrix for a rotation about the axis k by an angle is k x k x v c k x k y v k z s k x k z v k y s Rk k x k y v k z s k y k y v c k y k z v k x s k k v k s k k v k s k z k z v c y y z x x z where c cos , s sin , v 1 cos T kˆ k , k , k x y z 1 EGGN 512 Computer Vision 1. Describe what kind of image the following Matlab code will generate, and draw a sketch. I=zeros(128,128); for i=1:128 for j=1:128 if (i-64)*(i-64) + (j-64)*(j-64) < 40*40 I(i,j)=j; else I(i,j)=i; end end end Solution: The background increases from dark at the top of the 128x128 image to light at the bottom of the image. In the middle is a circle of radius 40, whose intensity increases from left to right. 2. Write a Matlab program that will generate an image of a checkerboard, consisting of black and white squares in a 8x8 pattern. Each square is 10x10 pixels. Solution: There are many possible ways to do this. Here is one: N = 10; % Size of each square, in pixels I = false(8*N,8*N); for i=1:2*N:8*N I(:, i:i+N-1) = true; end I = xor(I, I'); imshow(I, []); 2 EGGN 512 Computer Vision 3. A digital camera is modeled as a pinhole camera with focal length f. It has 512x512 sensor elements (pixels), where the center pixel (x,y) = (256,256) corresponds to the optical axis. A point P has 3-D coordinates (1m, 2m, 8m) in camera coordinates, and projects to pixel (x,y) = (356,456). Find the pixel projection of point Q, if Q has 3-D coordinates (-3m, -1m, 16m) in camera coordinates. Solution: We can solve for focal length since we know from point P, (356-256) = f (1 m/ 8 m). So f = 800 pixels. For Q: (x-x0) = f X/Z = (800) (-3)/(16) = -150, or x = -150+256 = 106 (y-y0) = f Y/Z = (800) (-1)/(16) = -50, or y = -50+256 = 206 4. Correlate the horizontal Sobel filter with the image below. You may assume that the image is padded with zeros beyond the visible boundaries of the image. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 Solution: The horizontal Sobel filter is -1 0 +1 -2 0 +2 -1 0 +1 Correlation is defined as g ( x, y) m /2 n /2 h( s, t ) f ( x s, y t ) h f . We will get s m /2 t n /2 0 0 0 0 0 0 0 0 3 EGGN 512 Computer Vision 0 0 0 0 0 0 0 0 0 1 1 0 0 -1 -1 0 0 3 3 0 0 -3 -3 0 0 4 4 0 0 -4 -4 0 0 4 4 0 0 -4 -4 0 0 4 4 0 0 -4 -4 0 0 3 3 0 0 -3 -3 0 5. A square pixel camera looks down at a workbench. Points P1 and P2 have (x,y) image coordinates (100,100) and (100,200), respectively. The locations of points P1 and P2 on the X,Y plane of the workbench are (500mm, 300mm) and (1000mm, 300mm), respectively. What is the X,Y location of point P3 on the workbench if its image coordinates are (250,400)? We will treat this as a rotation, followed by a scaling, followed by a translation. The rotation angle is w – i = 0 – 90 = -90. Next determine the scale factor: The distance between P1 and P2 is 100 pixels in the image, and 500 mm on the workbench. So the scale s = 500 mm/100 pixels = 5 mm/pixel. x w 1 0 x0 s 0 0 cos y w 0 1 y 0 0 s 0 sin 1 0 0 1 0 0 1 0 sin cos 0 0 xi 0 yi 1 1 x w xi s cos yi s sin x0 y w xi s sin yi s cos y 0 Using the coordinates of P1: 500 mm = (100 pix)(5 mm/pix) 0 – (100 pix)(5 mm/pix) (-1) + x0 = 500 mm + x0 300 mm = (100 pix)(5 mm/pix) (-1) + (100 pix)(5 mm/pix) (0) + y0 = -500 mm + y0 or x0 = 0, y0 = 800 mm For P3: xw = (250 pix)(5 mm/pix) 0 – (400 pix)(5 mm/pix) (-1) + 0 = 2000 mm 4 EGGN 512 Computer Vision yw = (250 pix)(5 mm/pix) (-1) + (400 pix)(5 mm/pix) 0 + 800 mm = -450 mm 6. A sphere of radius = 1 meter is projected onto an image. The camera has a focal length of 10 mm, and the each pixel corresponds to 0.01 mm on the image plane. The sphere projects to a circle of radius = 20 pixels on the image plane. How far away is the sphere from the camera (ie., what is its Z coordinate)? Solution: The sphere has a projected radius of 20 pixels or 0.2 mm. By similar triangles, 0.2 mm/10 mm = 1 meter / Z, or Z = 50 meters 7. A camera is mounted on the end effector of a robot arm as shown below. The end effector coordinate system {E} has its Z axis pointing up, its Y axis pointing to the right in the figure, and the X axis pointing out of the page. The origin of the camera {C} is located at a position of (X,Y,Z) = (0, 10 cm, 5 cm) with respect to {E}. The X axis of the camera is aligned with the X axis of the end effector. The Z axis of {C}is tilted down at an angle of 45 degrees with respect to the Y axis of {E}. Write the 4x4 transformation matrix relating the pose of the camera with respect to the end effector, ECT . {C} y z z {E} y Solution: The columns of the rotation matrix CE R are the unit vectors of {C} with respect to {E}. The X axis of the camera is [ 1, 0, 0 ]T with respect to {E}. The Y axis of the camera is [ 0, -0.707, -0.707 ]T with respect to {E}. The Z axis of the camera is [ 0, 0.707, -0.707 ]T with respect to {E}. So 0 0 0 1 0 0.707 0.707 10 E CT 0 0.707 0.707 5 0 0 1 0 5 EGGN 512 Computer Vision It is also possible to do this one by plugging into the formula for the rotation matrix for XYZ angles … we have a single rotation of -135 degrees about the x axis. 8. Image A is modified by the affine transform below. Sketch image “B”. xB 1 0.1 10 xA 20 y A yB 0.2 1 1 0 0 1 1 Image A (the x,y coordinates of the corners of the square are given): 10,10 10,50 50,10 50,50 The corners of the square map as follows: xB = xA – 0.1 yA + 10 yB = 0.2 xA + yA + 20 So (10,10) -> (10-1+10, 2+10+20) = (19, 32) (10,50) -> (10-5+10, 2+50+20) = (15, 72) (50,10) -> (50-1+10, 10+10+20) = (59, 40) (50,50) -> (50-5+10, 10+50+20) = (55, 80) Image B 19,32 15,72 59,40 55,80 6 EGGN 512 Computer Vision 9. The perspective projection matrix for a camera is given below. This matrix models both the intrinsic parameters of the camera and the extrinsic parameters (i.e., its pose in the world). 0 100 500 1500 M 500 100 0 0 0 1 0 10 A point is located at (X=1, Y=0, Z=1) in world coordinates. What image point does it project to? Solution: p = MP = [2000; 500; 10]. We divide by the third element to get the x,y image coordinates: x = 2000/10 = 200 y = 500/10 = 50 10. A pinhole camera (with focal length = 1) observes a flat wall. The camera is oriented with its optical axis 45 degrees from the normal to the wall. The distance to the wall (along the optical axis) is 10 meters. A top-down view is shown below. d=10 m =45° The camera points directly at the origin of the wall plane. The X axis of the wall points to the right along the wall, and the Y axis points down. The mapping of points from the wall plane to the image plane can be described by a projective transform, or homography, such that x1 a11 a12 a13 X Wall x2 a21 a22 a23 YWall , ximg x1 / x3 , yimg x2 / x3 x a 3 31 a32 a33 1 Which of the following matrices is the correct projective transform? 0 10 1 0 0 0.707 0 0 .707 0 1 1 1 0 (c) 0 .707 0 (d) 1 0 0 (a) 0 1 0 (b) 0 0 0 1 0.707 0 10 0 0 0 10 0 1 7 EGGN 512 Computer Vision Solution: First transform the wall points to the camera’s coordinate system, C W CR H W 0 Now, C W C Cam Cam P Wall H Wall P , where t Worg . 1 R is the rotation matrix that describes the orientation of frame {W} with respect to frame {C} (see slide 6 of Lecture 3). The two frames differ by a rotation about the Y axis. Wz Wy Wx Cz Cy Cx The rotation is -45 degrees (see slide 9 of Lecture 4): Start with the {W} frame aligned with the {C} frame. Point your right thumb in the direction of Wy. Then rotate the {W} frame until you get to the desired orientation. You will have to rotate -45 degrees; i.e., Y = -45°. Wz Cz The arrow shows the direction of rotation for a positive angle Wx +Y Cx The resulting rotation matrix is cos Y 0 sin Y 0.707 0 0.707 C 0 1 0 0 1 0 W R sin 0 cos 0.707 0 0.707 Y Y This makes sense because the columns of the rotation matrix are the unit vectors of the {W} frame, expressed in the representation of the {C} frame. So the first column is the X axis of 8 EGGN 512 Computer Vision {W}. As you can see in the figure, it points in the +X and +Z direction in the {C} frame. The third column is the Z axis of {W}, and it indeed points in the –X and +Z direction in the {C} frame. The full transformation matrix is a rotation about the Y axis by 45 degrees, plus a translation of 10. 0.707 0 0.707 0 0 1 0 0 Cam Wall H 0.707 0 0.707 10 0 0 1 0 So XCam = (0.707) XWall – (0.707) ZWall YCam = YWall ZCam = (0.707) XWall + (0.707) ZWall + 10 But the wall is a plane, so all its Z values are zero: XCam = (0.707) XWall YCam = YWall ZCam = (0.707) XWall + 10 Under image perspective projection, ximg = f XCam / ZCam = (0.707) XWall / ( (0.707) XWall + 10 ) yimg = f YCam / ZCam = YWall / ( (0.707) XWall + 10 ) We can accomplish this with the matrix x1 0.707 0 0 X Wall 1 0 YWall , x2 0 x 0.707 0 10 1 3 ximg x1 / x3 , yimg x2 / x3 11. Write MATLAB code to shrink the size (ie., total number of pixels) of an image I by a factor of four by subsampling. Assume that the dimensions of I are 512x512. (Note for this problem, you are not allowed to use the MATLAB function “imresize”). Solution: for i=1:512 for j=1:512 I2(i/2, j/2) = I(i,j); end end 9 EGGN 512 Computer Vision or for i=1:256 for j=1:256 I2(i, j) = I(2*i,2*j); end end or I2 = I(1:2:end, :); I3 = I2(:, 1:2:end); 12. Image A is mapped to image B using an affine transformation. A square in image A is mapped to parallelogram in image B. In image A, the square has dimension LxL, with its top left corner of the square is at location (L,L). In image B, the parallelogram has the dimensions shown. Its top left corner is also at location (L,L). Give the affine transformation that maps A to B. A B L L L L L/2 Solution: We want to find the elements of the matrix aij such that xB a11 a12 t x x A yB a21 a22 t y y A 1 0 0 1 1 The corners of the square in image A are (L,L) (2L,L) (L,2L) (2L,2L) The corresponding points in the image B are: (L,L) (2L,L) 10 EGGN 512 Computer Vision (3L/2,2L) (5L/2,2L) Obviously yB = yA, so a21 = 0, a22 = 1, ty = 0. To find the rest of the values, write the equations for xB = a11 xA + a12 yA + tx Writing down this equation at each tiepoint yields L = a11 (L) + a12 (L) + tx 2L = a11 (2L) + a12 (L) + tx 3L/2 = a11 (L) + a12 (2L) + tx Solving for these yields a11 = 1, a12 = 1/2, tx = -L/2 The affine transformation is xB 1 1/ 2 L / 2 x A 0 y A yB 0 1 1 0 0 1 1 11

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