2 Light and Shading 1

Contents

• Light and Shading 1
• Color 6
• Linear Filters 11
• Local Image Features 14
• Texture 18
• Segmentation by Clustering 20

10 Grouping and Model Fitting 23 11 Tracking 28 12 Registration 31 13 Smooth Surfaces and Their Outlines 33 15 Learning to Classify 35 16 Classifying Images 37 i© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All Rights Reserved. This material is protected under all copyright law as they currently exist.No portion of this material may be reproduced, in any form or by any means, without written permission from the publisher.No SM for chapter 1, 7, 8 and 14 (Computer Vision A Modern Approach 2e David Forsyth Jean Ponce) (Solution Manual all Chapter) 1 / 4

C H A P T E R 2

Light and Shading

PROBLEMS

2.1.We see a diffuse sphere centered at the origin, with radius oneand albedoρ, in an orthographic camera, looking down thez-axis. This sphere is illuminated by a distant point light source whose source direction is (0,0,1). There is no other illumination. Show that the shading field in the camerais ρ p 1−x 2 −y 2

Solution:

The surface is (x; y; p 1−x 2 −y 2

• We get two tangent vectors by partial differentiation;

they are (1,0, p) and (0,1, q) wherep=−x= p 1−x 2 −y 2 andq=−y= p 1−x 2 −y 2 .You can verify that the unit normal is ~ N= (−p;−q;1) p

• +p

2 +q 2 and the shading must beρ Γ

(0,0,1)∙

~ N ∙ , which yields 1 p

• +p

2 +q 2 = p 1−x 2 −y 2 2.2.What shapes can the shadow of a sphere take if it is cast on a plane and the source is a point source?

Solution:

These are conic sections with one important exception - you only get one half of the hyperbola.

2.3.We have a square area source and a square occluder, both parallel to a plane.The source is the same size as the occluder, and they are vertically above one another with their centers aligned.(a)What is the shape of the umbra?

Solution:

Square. This is rather a special case. You can construct the umbra by constructing all points on the plane that can see no part of the source. This is asquare directly below the occluder.(b)What is the shape of the outside boundary of the penumbra?1© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All Rights Reserved. This material is protected under all copyright law as they currently exist.No portion of this material may be reproduced, in any form or by any means, without written permission from the publisher. 2 / 4

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Solution:

This is quite a nasty question. Easiest way to construct the answer is to think about an arbitrary point on the occluder; construct a cone over the source, whose vertex is this point. Now intersect that cone with the plane — the resultingsquare region on the plane consists of all points such that this point on the occluder blocks some point on the source.Finally, take the union of all such regions (i.e. over all points on the occluder); that’s the penumbra. The envelope of the boundaries of these regions is the boundary of the penumbra. It’s a square.

2.4.We have a square area source and a square occluder, both parallel to a plane.The edge length of the source is now twice that of the occluder, and they are vertically above one another with their centers aligned.(a)What is the shape of the umbra?

Solution:

Depending on the distances between area source, occluder and plane, either there isn’t one, or it’s square.(b)What is the shape of the outside boundary of the penumbra?

Solution:

same as in previous exercise; This is quite a nasty question.Easiest way to construct the answer is to think about an arbitrary point on the occluder; construct a cone over the source, whose vertex is this point. Now intersect that cone with the plane — the resulting square region on the plane consists of all points such that this point on the occluder blocks some point on the source. Finally, take the union of all such regions (i.e. over all points on the occluder); that’s the penumbra. The envelope of the boundaries of these regions is the boundary of the penumbra. It’s a square.

2.5.We have a square area source and a square occluder, both parallel to a plane.The edge length of the source is now half that of the occluder,and they are vertically above one another with their centers aligned.(a)What is the shape of the umbra?

Solution:

Square (b)What is the shape of the outside boundary of the penumbra?Solution:This is quite a nasty question. Easiest way to construct the answer is to think about an arbitrary point on the occluder; construct a cone over the source, whose vertex is this point. Now intersect that cone with the plane — the resultingsquare region on the plane consists of all points such that this point on the occluder blocks some point on the source.Finally, take the union of all such regions (i.e. over all points on the occluder); that’s the penumbra. The envelope of the boundaries of these regions is the boundary of the penumbra. It’s a square.

2.6.A small sphere casts a shadow on a larger sphere. Describe thepossible shadow boundaries that occur.© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All Rights Reserved. This material is protected under all copyright law as they currently exist.  No portion of this material may be reproduced, in any form or by any means, without written permission from the publisher. 3 / 4

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Solution:

The question is slightly poorly posed,mea culpa, because it doesn’t say what the source is. Assume it’s a point source. There are two interesting cases: the point source is on the line connecting the two sphere centers; or it’s off. If it’s on, then the shadow must be a circle. If it’s off, it’s the intersection of a sphere and a right circular cone.

2.7.Explain why it is difficult to use shadow boundaries to infer shape, particularly if the shadow is cast onto a curved surface.

Solution:

As the previous exercise suggested, quite simple geometries lead to very complex shadows; it becomes hard to tell whether the shadow is complex becauseit was a simple shadow cone intersected with a complex surface, or because it was a complex shadow cone intersected with a simple surface.

2.8.As in Figure 2.18, a small patch views an infinite plane at unitdistance. The patch is sufficiently small that it reflects a trivial quantityof light onto the plane. The plane has radiosityB(x; y) = 1 + sinax. The patch and the plane are parallel to one another. We move the patch around parallel to the plane, and consider its radiosity at various points.(a)Show that if one translates the patch, its radiosity varies periodically with its position inx.

Solution:

If you translate the patch by (2=a) units, it’s in the same position with respect to the radiosity pattern as it was; and the plane is infinite.(b)Fix the patch’s center at (0,0); determine aclosed formexpression for the radiosity of the patch at this point as a function ofa. You’ll need a table of integrals for this (if you don’t, you’re entitled to feel very pleased with yourself).

Solution:

This is worked out in detail in (Haddon and Forsyth 1997) 2.9.If one looks across a large bay in the daytime, it is often hardto distinguish the mountains on the opposite side; near sunset, they are clearly visible. This phenomenon has to do with scattering of light by air—a large volume of air is actually a source. Explain what is happening. We have modeled air as a vacuum and asserted that no energy is lost along a straight line in a vacuum.Use your explanation to give an estimate of the kind of scalesover which that model is acceptable.

Solution:

During the day, the sun illuminates the air, so the air between the viewer and the far side of the bay glows (from the light scattered into the viewing direction). This reduces the contrast of the stuff on the other side of the bay — the viewer sees a bright light superimposed on so can’t distinguish the edges of objects, etc. In the evening, the air does not glow, and so you can see the far side of the bay. Generally, you can think of air as a vacuum on scales up to kilometers; if you’re very lucky and the air is very clear, perhaps 100 km.

2.10.Read the bookColour and Light in Nature, by Lynch and Livingstone, pub-© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All Rights Reserved. This material is protected under all copyright law as they currently exist.  No portion of this material may be reproduced, in any form or by any means, without written permission from the publisher.

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