Massachusetts Institute of Technology Department of Computer Science and Electrical Engineering 6.801/6.866 Machine Vision Quiz I Handed out: 2004 Oct. 21st Due on: 2003 Oct. 28th Problem 1: Uniform reflecting properties are a prerequisite for the usual shape from shading methods. Consider now a surface covered by a material of spatially varying reflectance. Suppose that the brightness can be treated as the product of a spatially varying ‘reflectance’ or ‘albedo’ ρ(x, y), and a ‘geometric factor’ R(p, q) that depends only on surface orientation. There are two unknowns—z(x, y) and ρ(x, y)—at every position on the surface, so a single image will not provide enough information to recover both (consider, for example, a photographic print of a rounded object where the bright-ness variations could either be from a rounded object of uniform albedo or from a flat object of varying albedo). Now suppose we take two images under different lighting conditions. (a) Combine the two resulting image irradiance equations in such a way as to eliminate ρ(x, y). Suppose the new—now ‘ρ-free’—equation can be written in the form E�(x, y) = R�(p, q). (b) Show that if the underlying surface actually is a Lambertian reflector. then the isophotes in gradient space are straight lines. When will they be parallel straight lines? (c) Show that the isophotes all go through a common point in gradient space in the case that they are not parallel. Where in gradient space would you expect the highest accuracy in recovering surface orientation? That is, where is ‘brightness’ (in the ρ-free equation) most affected by small changes in surface orientation? Relate this back to the original imaging situation. How should the lighting be arranged to obtain high accuracy? Problem 2: An edge detection method starts by finding the brightness gradient (Ex ,Ey ) at each picture cell using some local estimator of the partial deriva-tives of image brightness E(x, y). The magnitude of the brightness gradient is then computed. Next, “non-maximum suppression’’ keeps for further consider-ation only pixels where the magnitude of the gradient is a local maximum in the direction of the local brightness gradient.� 2 6.801/6.866 Quiz #1 The maximum of the gradient need, of course, not fall exactly on a discrete pixel location, when the gradient magnitude is considered as a continuous func-tion of position, Edge locations can be estimated to finer than pixel resolution if this continuous function of image position is first estimated from the discrete samples of the magnitude of the brightness gradient at pixels. We can approximate the magnitude of the gradient locally as a power series M(x, y) = ax 2 + bxy + cy 2 + dx + ey + f where (x, y) is the displacement from the center of a 3 × 3 region of pixels. (a) Show that the parameters a, b, c, d, e, and f can be estimated from M , Mx , My , Mxx , Mxy and Myy evaluated at the origin i.e. at the center pixel. (b) The following six stencils (a.k.a. computational molecules) can be used to es-timate the parameters a, b, c, d, e and f . Indicate which stencil corresponds to which parameter. (1) k1 + 1 +1 +1 − 2 −2 −2 + 1 +1 +1 (2) k2 − 1 +1 + 1 −1 (3) k3 + 1 +1 +1 − 1 −1 −1 (4) k4 − 1 +2 −1 + 2 +5 +2 − 1 +2 −1 (5) k5 − 1 +1 − 1 +1 − 1 +1 (6) k6 + 1 −2 +1 + 1 −2 +1 + 1 −2 +1 (c) Deter mine suitable values for the six constants {ki }, assuming that the pixel ispacing is . (Hint: apply the stencil to the test functions x yj for i = 0, 1, ... and j = 0, 1, ...). (d) Show that the distance from the origin to the maximum in brightness gradi-ent in the direction of the local brightness gradient is ρ =− Mx Ex + My Ey E2 + E2 E2 E2Mxx x + 2Mxy Ex Ey + Myyyx y (e) This result only applies when the denominator of the above expression is negative. Why? (f) How large can ρ reasonably become? Keep in mind that a picture cell where this computation is performed is one that has survived the non-maximum suppression operation.�� � 3 6.801/6.866 Quiz #1 Problem 3a: In a simple model of brightness in a Scanning Electron Microscope we have a reflectance map 2R(p, q) = p 2 + q . At a singular point, Ex = 0 and Ey = 0. Assume the singular point is at (0, 0). Suppose that the surface near the singular point can be approximated by a power series of the form z = z0 + sx x + sy y + ax 2 + 2bxy + cy 2 (a) First, show that sx = 0 and sy = 0 at the singular point. (b) Then show that Exx − 2Exy + Eyy = k � (a − b)2 + (b − c)2� for some k. What is the value of k? (c) If Exx = Exy = Eyy = 8,what are a, b, and c in the series expansion given above for the surface height near the singular point? Is there a unique answer? Verify that b(Exx − Eyy ) = (a − c)Exy Problem 3b: In solving the shape from shading problem, we derived the char-acteristic strip equations from the image irradiance equation E(x, y) = R(p, q). Here we go in the other direction, showing that solutions of the characteristic strip equation satisfy the image irradiance equation. Show that — when using the method of characteristic strip expansion to solve the shape from shading problem — E(x, y) − R(p, q) is constant along a characteristic strip. That is: d � � E(x, y) − R(p, q) = 0 dξ along the strip. Conclude that, if E(x, y) = R(p, q) at the beginning of the strip, then E(x, y) = R(p, q) all along the strip. (If nothing else, this is a sanity check on the characteristic strip equations). Problem 4: The image brightness gradient is large where an edge is thought to occur. Derivatives of brightness can, however, be used for purposes other than finding edges. We saw this when we computed the perimeter of a “binary’’ object using E2 + E2 y dx dy x D where E(x, y) = 1 in the object and E(x, y) = 0 in the background, with a narrow, smooth transition region in between.� � � 4 6.801/6.866 Quiz #1 For this exercise, consider an image with a circularly symmetric bright spot. Consider polar coordinates with x = r cos θ and y = r sin θ. Let E(x, y) = f(r) where r = x2 + y2 . Suppose that f(r) = 1 for r< R − , and f(r) = 0 for r> R + , Then the above integral of the magnitude of the brightness gradient yields 2πR—i.e. the perimeter of the circular disk as  → 0. (a) Show that Ex = f �(r) x r and Ey = f �(r)y r and Exx …