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Mathematical Optimization in Graphics and Vision - Luiz Velho - Impa

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2.7. THE PROBLEM OF POSING PROBLEMS 29<br />

L<strong>in</strong>ear programs<br />

A very important situation occurs when both the objective <strong>and</strong> the constra<strong>in</strong>t functions<br />

are l<strong>in</strong>ear <strong>and</strong>, moreover, the constra<strong>in</strong>ts are def<strong>in</strong>ed either by equalities or<br />

<strong>in</strong>equalities, that is, the constra<strong>in</strong>ts are given by l<strong>in</strong>ear equations <strong>and</strong> l<strong>in</strong>ear <strong>in</strong>equalities.<br />

As we have remarked before, the solution set is a solid polyhedron<br />

of the euclidean space. <strong>Optimization</strong> problems of this nature are called l<strong>in</strong>ear<br />

programs <strong>and</strong> their study constitute a sub-area of optimization. In fact, several<br />

practical problems can be posed as l<strong>in</strong>ear programs, <strong>and</strong> there exists a huge number<br />

of techniques that solve them, exploit<strong>in</strong>g its characteristics: the l<strong>in</strong>earity of<br />

the objective function <strong>and</strong> the piecewise l<strong>in</strong>ear structure of the solution set (polyhedra).<br />

2.7 The problem of pos<strong>in</strong>g problems<br />

Most of the problems <strong>in</strong> general, <strong>and</strong> <strong>in</strong> particular the problems from computer<br />

graphics, can be posed us<strong>in</strong>g the concept of operator between two spaces. 1 These<br />

problems can be classified <strong>in</strong> two categories: direct <strong>and</strong> <strong>in</strong>verse problems.<br />

Direct problem. Given two spaces O 1 <strong>and</strong> O 2 of graphical objects, an operator<br />

T : O 1 → O 2 , <strong>and</strong> a graphical object x ∈ O 1 . Problem: compute the object<br />

y = T (x) ∈ O 2 . That is, we are given the operator T <strong>and</strong> a po<strong>in</strong>t x <strong>in</strong> its doma<strong>in</strong>,<br />

<strong>and</strong> we must compute the image y = T (x)<br />

Because T is an operator, it is not multi-valued, therefore the direct problem<br />

always has a unique solution.<br />

Inverse problems.<br />

There are two types of <strong>in</strong>verse problems:<br />

Inverse problem of first k<strong>in</strong>d. Given two spaces O 1 e O 2 of graphic objects,<br />

an operator T : O 1 → O 2 , <strong>and</strong> an element y ∈ O 2 , determ<strong>in</strong>e <strong>and</strong> object x ∈ O 1<br />

tal que T (x) = y. That is, we are given an operator T <strong>and</strong> a po<strong>in</strong>t y, belong<strong>in</strong>g to<br />

1 In some books the use of the term operator implies that it is l<strong>in</strong>ear. Here we use the term to<br />

mean a cont<strong>in</strong>uous function.

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