Als Manuskript gedruckt

Technische Universitat DresdenHerausgeber: Der Rektor

Containment of a pair of rotating objects within a

container of minimal area or perimeter

G. Scheithauer, Y. Stoyan, J. Bennell, T. Romanova, A. Pankratov

MATH-NM-02-2012

May 2012

Contents

1 Introduction 3

2 Related work 4

3 The concept of phi-functions 6

4 Mathematical model of the containment problem 8

5 General solution strategy 10

5.1 The branching tree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

5.2 The basic algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

6 Conclusions and Outlook 15

7 Appendix 15

7.1 Appendix A: Basic phi-functions . . . . . . . . . . . . . . . . . . . . . . . . . . 15

7.2 Appendix B: Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Containment of a pair of rotating objects

within a container of minimal area or perimeter

G. Scheithauer1, Y. Stoyan2, J. Bennell3,

T. Romanova2, A. Pankratov2

1 Technische Universitat Dresden, Germany2 National Academy of Science of Ukraine,

3 Southampton Management School, United Kingdom

Abstract

Cutting and packing problems arise in many fields of applications and theory. Whendealing with irregular objects, an important subproblem is the identification of theoptimal clustering of two objects. Within this paper we consider the case, where twoirregular one-connected objects whose frontier formed by circular and/or line segmentsand which can be free rotated, should be placed such that the enclosing rectangle orcircle has minimal area, or alternatively, has minimal perimeter. We propose a solutionstrategy which is based on the concept of phi-functions and provide some examples.Moreover, for the sake of completeness, we give a comprehensive collection of basicphi-functions.

1 Introduction

Cutting and packing problems, also called placement or allocation problems, are of highinterest in application as well as in theory. It is well known that even the one-dimensionalversion of optimal usage of a given resource, the classical knapsack problem, belongs to theclass of NP-hard optimization problems. For that reason, most of work related to cuttingand packing problems is dealing with heuristic approaches. Nevertheless, the development ofexact solution methods is an important task to broaden the range of optimal solvable cases.

In this respect, the investigation of optimization problems where so-called regular objects(meaning rectangles or parallelepipeds/boxes) have to be packed, is considered in a higherpercentage of the literature. One reason for that can be the fact that stronger optimalitycriterion or bounds are available in contrast to cases where irregular objects (already forcircles) have to be considered.

One of the successful applicable approaches in case of irregular objects, which leads in generalto multi-extremal non-linear optimization problems, is the concept of phi-functions proposedby Stoyan [64] which allows the computation of local optima, and in some cases, even ofglobal optimal solutions.

In this paper, we are going to apply the concept of phi-functions to the counterpart of max-imum yield optimization, namely to the computation of minimal material usage needed toobtain all desired objects. In detail, we will investigate the following two-dimensional prob-lem: Let be given two (irregular one-connected ) objects whose frontier formed by circularand/or straight line segments. Find a containing region (rectangle or circle) of minimal area oralternatively, of minimal perimeter such that the two objects can be placed completely inside

the containing region without overlapping. We assume that the orientation of the objects isarbitrarily, that means, free rotations of the objects are permitted.

The paper is organized as follows. In the next section, we try to give an overview on relatedwork. Then, for completeness, we describe the concept of phi-functions and apply it to theproblem under consideration. We will consider convex and non-convex polygons, circles andobjects whose frontier is a combination of straight-line and circular-arc segments (such objectsare given in Appendix A together with corresponding phi-functions for pairs of it). The generalsolution strategy is given together with some examples, and is followed by an outlook. Due tothe length of the paper, the extension of the approach to the case of more than two objects isleft for a forthcoming paper which will also contain results of numerical experiments of reallife problems.

2 Related work

Note, in the following overview only some of the huge humber of related articles are mentionedin order to find a balance between the wide field of related areas and length of this section.

Solution approaches to irregular nesting problems are analysed by Dowsland and Dows-

land [31]. A tutorial covering the core geometric methodologies currently applied by re-searches in cutting and packing of irregular shapes is presented by Bennell and Oliveira

[15]. Tools of mathematical modeling of arbitrary object packing problems are given by Ben-

nell, Scheithauer, Stoyan and Romanova [14].

Complexity investigations for cutting and packing problems can be found for instance byChazelle, Edelsbrunner and Guibas [20], Blazewicz, Drozdowski, Soniewicki

and Walkowiak [16], Li and Milenkovic [47], and Chlebk and Chlebkov [24].

Among the plurality of two-dimensional cutting and packing problems the so-called strip

packing problem (SPP) is of high interest. In the SPP a given set of objects has to be placedfeasibly within a strip of given (fixed) width W while minimizing the height H needed, i. e.a rectangle of minimal area is looked for. (A packing pattern is feasible if all objects arecontained completely within the rectangle W × H and do not overlap each other.) In thiscase, the objective to minimize the area is equivalent to minimizing the perimeter. This doesnot remain true if the width W is also variable so that the objective becomes non-linear whenthe total area has to be minimized.

For the SPP many variants are considered. In case of rectangular objects (rectangles) two mainproblem classes occur. In the first, the cutting or packing pattern has to possess the so-calledguillotine-property, i. e. all objects can be obtained by a sequence of (orthogonal) guillotinecuts. Very well-known heuristics are Next-Fit, First-Fit and Best-Fit (cf. Coffman et al. [25],Baker and Schwarz [10], Imahori and Yagiura [43]) applied in offline situations as wellas for online cases. If the guillotine-property is not demanded then Bottom up, Left justified-type heuristics are frequently applied (Baker, Coffman and Rivest [8]). In the lattercase, exact methods are proposed based on ILP (integer linear programming) formulations(Padberg [61], Belov and Scheithauer [12]) or a graph-theoretical model (Fekete andSchepers [33]).

The problem of packing circles of various radii is addressed in George, George and Lamar

[34], Stoyan and Yaskov [66]. The packing of identical or non-identical circles within acontaining circle with minimal radius is investigated in Graham [35] and Wang et al. [69],

Zhang and Deng [70], Huang et al. [40], Addis, Locatelli and Schoen [1], respectively.The more general case of circle packing in which the containing region can have differentshapes (circle, triangle, square, etc.) is considered in Bansal and Sviridenko[13].

There exists also work concerning three-dimensional objects, e. g. in Egeblad, Nielsen andOdgaard [32], Bortfeldt and Mack [17]. In Stoyan, Yaskov and Scheithauer [67]a phi-function based approach is proposed for packing solid spheres into a strip.

Minimum perimeter rectangles that enclose congruent non-overlapping circles are computedby Lubachevsky and Graham [48]. Maximizing the number of identical circles packablewithin a square, rectangle or circle is also frequently investigated (cf. e. g. Szabo et al. [68],Graham et al. [35], Cui [27]).

Another broad field in SPP is concerned to the placement of polygonal (convex or non-convex)objects on a (two-dimensional) strip of fixed width and minimal height (or length). Naturalapplications can be found, for instance, in textile or steel industry. Exemplified we refer toHaistermann and Lengauer [39], Oliveira and Ferreira [59], Egeblad, Nielsen andOdgaard [32], Burke et al. [18] where various solution approaches are presented. Meta-heuristics like Simulated Annaeling (Marques, Bispo and Sentieiro [49], Dowsland [30])andGenetic Algorithms (Martens [50]) (and other) find also application to compute good(but in general not optimal) solutions of such packing problems.

Packing problems with irregular objects, whose frontier can be described by a sequence ofline- and arc-segments, are tackled by Burke et al. [19] using the line and arc no-fit polygon.That paper extends the orbital sliding method of calculating no-fit polygons to enable it tohandle arcs and then shows the resultant no-fit polygons being utilised successfully on thetwo-dimensional irregular packing problem.

Several publications are addressed to containment problems. One type of such problems is asfollows: Does a set of given objects (or a single object) fit feasibly within a given containingregion? Sometimes rotation of objects is permitted, sometimes not.

The single polygon-containment problem with translations only has been studied in Baker,

Fortune and Mahaney [9]. Rotation of the polygon is additionally allowed in Martin andStephenson [51], Avnaim and Boissonnat [5], Grinde and Cavalier [37], Milenkovic

[57]. Two- and three-polygon problems are considered in Avnaim and Boissonnat [6],Grinde and Cavalier [38].

In Milenkovic [54], Milenkovic and Daniels [56], Milenkovic [55] the translationalcontainment of multiple polygons within another polygon is investigated, whereas the moregeneral case, allowing also rotations, is considered in Milenkovic [58]. In Milenkovic andSacks [53] the authors offer an implementation of Minkowski sum for objects bounded by linesegments and circular arcs. Grinde and Cavalier [37] use a mathematical programmingformulation/model to solve the following containment problem: Given a convex polygon Qand a (not necessary convex) polygon P , can P be translated and/or rotated such that Pfits within Q? This method is used in an algorithm to place small (polygonal) items into(polygonal) holes obtained after placing larger items.

The minimal-area convex enclosure problem consists in finding the relative position of twosimple polygons to each other such that the area of their convex enclosure is minimized. It isinvestigated in Grinde and Cavalier [36]. The technique used searches along the envelope(or no-fit polygon). Dori and Ben-Bassat [29] consider the problem of circumscribing aconvex polygon by a polygon of fewer sides with minimal area addition.

The best approximation of a convex figure C by a pair of homothetic rectangles (i. e. theyare similar and parallel), r and R with r ⊆ C ⊆ R, is studied by Schwarzkopf, Fuchs,

Rote and Welzl [63]. Approximations (inscribing and enclosing) for planar convex setsusing axially symmetric polygons are studied by Ahn et al. [3]. Finding the largest areaaxis-parallel rectangle in a polygon is considered by Daniels, Milenkovic and Roth [28].

The computation of the minimum bounding box for a given object (and other problems) isconsidered by Martin and Stephenson [51].

The problem of enclosing a set of objects by two minimum area rectangles is addressed byBecker et al. [11]. The enclosure of n points in the plane by two non-convex shapes,rectilinear convex hull and L-shape, having minimal area, is investigated by Bae et al. [7].

Covering (partition) a rectilinear polygon by (into) a minimal number of axis-parallel rectan-gles are considered by Anil Kumar and Ramesh [4] and Cheng and Lin [21]. A minimum-area rectangle enclosing the projection of a higher-dimensional set (a polytope) is constructedby Kuno [46].

The decision problem, given a set of points in the plane and a set of rectangles (which can berotated), is there a placement of the rectangles such that all points are covered, is studied byHuang and Wang [41]. A similar problem is addressed by Ahn et al. [2].

Hche and Liebling [42] investigates the computation of minimum area simple (not-necessaryconvex) pentagons whose vertexes are elements of a given point set. Melissen and Schuur

[52] computes the minimal radius needed when a rectangle is covered by six or seven circles.An optimal algorithm for finding minimal area triangles enclosing a given convex polygon isproposed by O’Rourke, Aggarwal, Maddila and Baldwin [60].

Paper Kallrath [44] considers a placement of circles and polygons within rectangular areaof minimal area.

The smallest square in which all rectangles of size 1/n×1/(n+1), n = 1, 2, . . . , can be packed,and the rectangle with smallest area in which all squares of size 1/n×1/n, n = 3, 4, 5, . . . canbe packed are presented by Jennings [45].

Heuristic approaches to large-scale periodic packing of irregular shapes on a rectangular sheetare also of interest (cf. Costa, Gomes and Oliveira [26].

Most of the approaches dealing with the interaction of two (or only few) objects (containmentor covering) are used in placement algorithms for larger instances as local decision rules. In thispaper, we use a uniform description, the concept of phi-functions, to compute (approximately)a minimum area or minimum perimeter rectangle, or a circle of minimum area, containingtwo objects which can be translated and/or rotated. (The proposed approach can easily beextended to a higher number of objects.) As objects we consider rectangles, polygons, circlesand objects whose frontier is formed by straight-line and circular-arc segments.

3 The concept of phi-functions

Two-dimensional cutting and packing problems with irregular shaped objects occur infootwear, textile, metal cutting, etc. In order to model such optimization problems the in-teraction of two objects with respect to non-overlapping, intersection or contact has to be

described in an appropriate manner. It has been proved that the usage of so-called phi-functions for pairs of geometric objects is very helpful in this respect, cf. Chernov, Stoyan

and Romanova [23], Stoyan and Chugay [62], Stoyan and Yaskov [65].

In any case, we assume that any object T considered here, is a model of a real two-dimensionalobject, i. e. T is phi-object. Canonically closed point set T ⊂ R2 (T = cl∗(T ) = cl(int(T ))having the same homotopic type as its interior is called phi-object. In particular, any phi-object is called a phi-polygon if its frontier is shaped by means of straight lines, rays orline segments Bennell, Scheithauer, Stoyan and Romanova [14]. To handle arbitrarygeometric objects we present them by finite unions of so-called basic objects. In Chernov,

Stoyan and Romanova [23] is proved that each phi-object bounded by line segments and

circular arcs may be decomposed into basic phi-objects: T =n⋃

i=1

Ti, with Ti ∈ ℑ, where

ℑ is the set of basic objects. For details we refer the reader to Chernov, Stoyan andRomanova [23].

The translation of object T ⊂ IR2 by vector u = (ux, uy) ∈ IR2 and rotation of it (with respectto its reference point) by angle θ ∈ [0, 2π) is defined by

T (u, θ) = {t : t = u+D(θ)t ∀ t ∈ T (0, 0)}

where T (0, 0) denotes the non-translated and non-rotated object T , and where D(θ) is givenby

D(θ) =

(cos θ sin θ− sin θ cos θ

)and D−1(θ) =

(cos θ − sin θsin θ cos θ

).

Note, D(θ) defines clock-wise rotation. For analytical descriptions of relations between apair of objects T1(u

1, θ1) and T2(u2, θ2) we employ phi-function technique. Phi-functions

allow us to distinguish the following three cases: T1(u1, θ1) and T2(u

2, θ2) are intersecting sothat T1(u

1, θ1) and T2(u2, θ2) have common interior points; T1(u

1, θ1) and T2(u2, θ2) do not

intersect, i. e. T1(u1, θ1) and T2(u

2, θ2) do not have common points; T1(u1, θ1) and T2(u

2, θ2)are in contact, i. e. T1(u

1, θ1) and T2(u2, θ2) have only common frontier points.

Definition Any everywhere defined continuous function Φ : IR6 → IR is called phi-functionof T1 and T2 if it possesses the following characteristic properties:

Φ(u1, θ1, u2, θ2)

> 0, if T1(u1, θ1) ∩ T2(u

2, θ2) = ∅

= 0, ifintT1(u

1, θ1) ∩ intT2(u2, θ2) = ∅

frT1(u1, θ1) ∩ frT2(u

2, θ2) 6= ∅

< 0, if intT1(u1, θ1) ∩ intT2(u

2, θ2) 6= ∅.

Definition and basic features of phi-functions one may find in Bennell, Scheithauer,

Stoyan and Romanova [14], Chernov, Stoyan and Romanova [23].

Now let

A =

nA⋃

i=1

Ai, B =

nB⋃

j=1

Bj , and Ai, Bj ∈ ℑ. (1)

A phi-function to characterise the non-overlapping of the pair of objects A and B boundedby line segments and circular arcs has the form

ΦAB = min{Φij, i = 1, 2, ..., nA, j = 1, 2, ..., nB},

where Φij denotes a basic phi-function for the pair of objects Ai and Bj with and Ai, Bj ∈ ℑ.The complete class of basic phi-functions is given in Chernov, Stoyan, Romanova andPankratov [22]. We provide the phi-functions in Appendix A.

4 Mathematical model of the containment problem

In the problem under consideration, we are looking for a axis-parallel rectangle

R = R(a, b) = {(x, y) : 0 ≤ x ≤ a, 0 ≤ y ≤ b}

of size a× b in fixed position, or alternatively, for a containing circle

C = C(r) = {(x, y) : x2 + y2 ≤ r}

with origin as center point. In the following, let C0(p) denote the desired containing region,i. e. C0(p) represents either the rectangle R(a, b), or the circle C(r), and p denotes the vector ofmetrical characteristics belonging to the containing region (either p = (a, b) or p = (r)). Theapproach proposed below can be applied also for other types of containing regions. Moreover,the number of objects to be placed is not restricted to two.

Let Fk = Fk(p), k = 1, 2, 3 denote objective functions:

F1(a, b) = ab, F2(a, b) = a+ b, F3(r) = r.

The objective is either to minimize the area, or alternatively, the perimeter, under the con-dition that two objects, named A and B, can be placed completely within C0(p) withoutoverlap. With other words, we are also looking for translation vectors uA and uB and anglesθA and θB such that

A(uA, θA) ⊂ C0(p), B(uB, θB) ⊂ C0(p), A(uA, θA) ∩ intB(uB, θB) = ∅.

The latter condition is fulfilled if and only if ΦA,B(uA, θA, u

B, θB) ≥ 0. The remaining twoconditions can also be modelled using phi-functions if instead of C0(p) the closure of itscomplement, C∗

0(p) = cl(IR2 \ C(p)), is used. Let ΦA(u

A, θA, a, b) denote the phi-function forthe pair of objects A and C∗

0. Analogously, let ΦB(u

B, θB, a, b) denote the phi-function for thepair of objects B and C∗

0. Consequently, we formulate the following problem:

Compute translation vectors uA, uB ∈ IR2, rotation angles θA, θb ∈ [0, 2π), and sizeparameters p such that Fk(p) (k ∈ {1, 2, 3}) attains its minimum and

ΦA,B(uA, θA, u

B, θB) ≥ 0, ΦA(uA, θA, p) ≥ 0, ΦB(u

B, θB, p) ≥ 0.

Because of a symmetry argument, in case of a rectangular containing region, we may restrictθA to be in [0, π). In case of a circular containing region θA = 0 is possible. Let W ⊂ IR8

denote the corresponding set of feasible solutions (the solution space). Then, we can definethe following optimisation problem:

Fk(u∗) = min{Fk(u) : u ∈ W}, (k ∈ {1, 2, 3}), (2)

where u = (uA, θA, uB, θB, p).

It is obvious, W 6= ∅, and since Fk is bounded from below, the problem is always solvable. Butthe problem is difficult to solve for several reasons. The objective F1 is non-linear (quadratic)as well as the phi-functions which are composed by min- and max-combinations of affine-linear and non-linear functions including sin- and cos-terms (cf. Chernov, Stoyan andRomanova [23]). The set W of feasible solutions is non-convex, leading to many localextrema. Each object, we consider here, can be represented in the form

T = {(x, y, θ) : minj=1,...,n

fj(x, y, θ) ≤ 0}

with fj(x, y, θ) = maxk=1,...,njfjk(x, y, θ), j = 1, . . . , n, where fjk are radical free affine-linear

and non-linear functions including sin- and cos-terms. It follows from definition of T as a unionof basic objects Tj , j = 1, . . . , n, where Tj = {(x, y, θ) : fj(x, y, θ) ≤ 0}.

Example 1

We consider two arbitrary shaped phi-objects, A and B, whose frontier is formed by linesegments and circular arcs. Each object is given by an ordered collection of frontier elements:l1, l2, ..., ln, where li ∈ {−1, 0, 1}; li = 1 corresponds to a convex arc, li = 0 correspondsto a line segment, and li = −1 corresponds to a concave arc. Each arc is given by a tuplegi = (x1, y1, xc, yc, x2, y2, r), where (x1, y1) and (x2, y2) are coordinates of the end points,(xc, yc) is the center point of the circular arc and r is the radius of it in the eigen coordinatesystem of object T . Each straight-line segment is given by a tuple gi = (x1, y1, x2, y2), where(x1, y1) and (x2, y2) are the end points of it in the eigen coordinate system of object T . Theinput data of two objects A and B (Fig. 1) are given the following table.

code (x1, y1) (xc, yc) (x2, y2) rObject A

1 (-3.557281,-0.198854) (-3.557281, 20.801146) (9.517820, 4.368199) 211 (9.517820, 4.368199) (4.083638, 8.06959) (7.7911394, 13.49961) 6.575-1 (7.7911394, 13.49961) (-3.557281, 11.351146) (-3.557281, -0.198854) 11.55

Object B

code (x1, y1) (xc, yc) (x2, y2) r0 (7.166523, 7.707352) - (-0.217565, 14.419766) -0 (-0.217565, 14.419766) - (-4.247752, 7.699269) -0 (-4.247752, 7.699269) - (-6.714476, -3.445282) --1 (-6.714476, -3.445282) (7.176598, -6.519918) (7.166523, 7.707352) 14.227274

In our example (Fig. 1), according to formula (1)

A = V⋃

D, B = H⋃

K and nA = nB = 2.

Assuming n′ = nA · nB we have non-overlapping constraint ΦAB ≥ 0, where

ΦAB = min{Φk : k = 1, 2, ..., n′ = 4} = min{ΦV H , ΦV K , ΦDH , ΦDK},

where ΦV H , ΦV K , ΦDH , ΦDK are phi-functions for basic objects V and H , V and K, D andH , D and K.

Using phi-functions the containment constraint (A ⊂ C0, B ⊂ C0) has the form:

min{ΦC∗

0A, ΦC∗

0B} ≥ 0,

C

VA

A

A

B

B

B

K

H

D

R

Figure 1: a) Objects A and B, b) Decomposition of A and A into basic objects, c) minimalenclosing rectangle, d) minimal enclosing circle

ΦC∗

0A = min{ΦC∗

0Ai

: i = 1, 2, ..., nA = 2}, ΦC∗

0B = min{ΦC∗

0Bj

: j = 1, 2, ..., nB = 2}.

Assuming n′′ = nA + nB we have

min{ΦC∗

0Tk

: k = 1, 2, ..., n′′ = 4, Tk ∈ ℑ} = min{ΦC∗

0V , ΦC∗

0D, ΦC∗

0H , ΦC∗

0K},

where ΦC∗

0V , ΦC∗

0D, ΦC∗

0H , ΦC∗

0K are basic phi-functions for objects C∗ and V , C∗ and D, C∗

and H , C∗ and K.

5 General solution strategy

5.1 The branching tree

Since the considered problem has a non-convex disconnected feasible region W we try topresent W as a union of subsets Ws, s = 1, 2, ..., η. Each Ws is determined by a system ofinequalities with infinitely differentiable functions extracted from our phi-functions. A phi-function of two objects A and B has in general the form

ΦA,B(uA, θA, u

B, θB) = minj

maxksjk(u

A, θA, uB, θB)

where the functions fjk can also contain min- and max-terms. Since minj fjk(x, y) ≥ 0 isequivalent to fjk(x, y) ≥ 0 for all j, and maxk fjk(x, y) ≥ 0 means at least one of the inequal-ities, say fj,k0(x, y) ≥ 0 has to be fulfilled, each of these terms can be considered as a systemof (in general non-linear) inequalities (all max-terms are dissolved). This can be done by abranching scheme. It means that for each inequality ΦAB ≥ 0 we may construct a tree, calledphi-tree. To each terminal node of the phi-tree there corresponds a system of inequalities withinfinitely differentiable functions.

Now we form a finite number of subproblems

Fks(u∗s) = min{Fk(u) : u ∈ Ws} (k ∈ {1, 2, 3}), s = 1, . . . , η, (3)

where u = (uA, θA, uB, θB, p).

Clearly, the desired solution u∗ can be obtained by inspecting and exactly solving all of thesesubproblems, i. e.

u∗ = min{u∗s, s = 1, . . . , η}.

resulting subproblems are attacked by standard techniques (interior point method, feasibledirection method) for local optimization.

The construction of all subproblems by means of a branching tree is illustrated by the followingtwo examples.

Example 2

Let Di = {(x, y) : gi(x, y) ≤ 0, hi(x, y) ≤ 0} where gi(x, y) = (x − xi)2 + (y − yi)

2 − r2iand hi(x, y) = aix + biy + ci such that int(Di) 6= ∅ and hi(xi, yi) > 0, i = 1, 2, be circularsegments with radii ri, and let C0 be a circular container of radius r. For simplicity, we assumer < ri, i = 1, 2, so that the containment of Di within C0 is guaranteed if the two intersectionpoints (xij , yij), j = 1, 2 of gi and hi belong to C0. The objective is to minimize r so thatint(D1) ∩D2 = ∅ and Di ⊂ C0, i ∈ {1, 2}. Then the mathematical model of the problem canbe written in the form

F (u) = r → min s. t. u ∈ W (4)

where u = (r, x1, y1, θ1, x2, y2, θ2) ∈ IR7 and

W = {u ∈ R7 : ΦD1D2(u) ≥ 0, ΦC∗Di

(u) ≥ 0, r ≤ ri, i = 1, 2}.

The number of terminate nodes of the branching tree in this example equals η = 19 as we willsee. In general, the number of leaves may be very large and therefore, a drawback of the basicapproach. Further investigations are required to reduce it drastically. Moreover, the resultingsubsets of W are not disjunctive in general.

Due to the dissection of W = ∪ηs=1

Ws, problem (4) can be reduced to the following optimisa-tion problem:

F (u∗) = min{F (u∗1), F (u∗

2), ..., F (u∗η)} (5)

where

F (u∗s) = min{F (u) : u ∈ Ws}, s = 1, 2, ..., η. (6)

In order to construct subregions Ws the dissection of W is obtained by ensuring the validityof all restrictions. Note, in order to ensure Di(u) ⊂ C0 for i = 1, 2 the following constraintsbelong to each system of inequalities defining one of the sets Ws:

r2 − (xij + xi)2 − (yij + yi)

2 ≥ 0, j = 1, 2, i = 1, 2. (7)

Let us now consider restriction ΦD1D2(u) ≥ 0. Since the two circular segments can be consid-

ered as intersection of two primary objects, a triangle and a circle, we have

D1 = T1⋂

C1 and D2 = T2⋂

C2

where T1, T2 are appropriate triangles and C1, C2 corresponding circles. Hence,

ΦD1D2= max{ΦD1T2

, ΦD1C2}, (8)

and moreover,

ΦD1T2= max{ΦT1T2

,ΦC1T2}, ΦD1C2

= max{ΦC2T1,ΦC1C2

}. (9)

Consequently, we have to consider phi-functions of primary objects ΦT1T2, ΦC1T2

, ΦC2T1and

ΦC1C2.

Let (x′i, y′i), i = 1, 2, 3, be the vertices of T1, and (x′′i , y

′′i ), i = 1, 2, 3, those of T2 (with x

′i = x1j ,

x′′i = x2i, i = 1, 2). Furthermore, we suppose a representation of the two triangles as follows:

T1 = {(x, y) : ϕi = α′ix+ β ′

iy + γ′i ≤ 0, i = 1, 2, 3},

T2 = {(x, y) : ψj = α′′jx+ β ′′

j y + γ′′j ≤ 0, i = 1, 2, 3}.

Phi-function for two convex polygon K1 and K2 is defined as follows, e.g. Chernov, Stoyan

and Romanova [23].

Let (x′i, y′i), i = 1, 2, ..., m1, be vertices of K1, and (x′′j , y

′′j ), j = 1, 2, ..., m2, be vertices of K2,

and ϕi = α′ix+ β ′

iy + γ′i = 0 and ψj = α′′jx+ β ′′

j y + γ′′j = 0 are side equations of K1 and K2,subject to ϕi ≤ 0 and ψj ≤ 0 for all points belonging to K1 and K2 respectively, then

ΦK1K2= max{ max

1≤i≤3

min1≤j≤3

ϕij , max1≤j≤3

min1≤i≤3

ψji }, (10)

where ϕij = α′ix

′′j + β ′

iy′′j + γ′i, ψji = α′′

jx′i + β ′′

j y′i + γ′′j for all i and j.

We define ΦT1T2= ΦK1K2

for m1 = m2 = 3.

For the two circles Ci of radii ri and center points (xCi, yCi

), i = 1, 2, we have

ΦC1C2= (xC1

− xC2)2 + (yC1

− yC2)2 − (r1 + r2)

2. (11)

Phi-functions ΦC1T2and ΦC2T1

are defined as follows:

ΦCT = max{χi, min{ωi, ψi}, i = 1, 2, 3}, (12)

where

χi = αix+ βiy + γi − r with α2

i + β2

i = 1,ωi = (xC − xi)

2 + (yC − yi)2 − r2,

ψi = (βi−1 − βi)(xC − xi)− (αi−1 − αi)(yC − yi) + r(αi−1βi − αiβi−1).

We use notations χ′i, ω

′i, ψ

′i for ΦC1T2

and χ′′i , ω

′′i , ψ

′′i for ΦC2T1

.

Since each phi-function is a composition of minima and maxima of smooth functionsand taking into account relations (8)–(12), the branching tree (also called phi-tree) forΦD1D2

(x1, y1, θ1, x2, y2, θ2) ≥ 0 has the form (υ11 is the root node, and υi−1,kij is the j-th

node of i-th level of the tree which is an offspring of the k-th node of i− 1-th level of the treefor constraint ΦD1D2

≥ 0):

Level 1: υ11 ↔ ΦD1D2≥ 0

Level 2: υ1121

↔ ΦD1T2≥ 0, υ11

22↔ ΦD1C2

≥ 0

Level 3: υ2131

↔ ΦT1T2≥ 0, υ21

32↔ ΦC1T2

≥ 0, υ2233

↔ ΦT1C2≥ 0, υ22

34↔ ΦC1C2

≥ 0

Level 4: (19 nodes)υ3141

↔ {ϕ1j ≥ 0, j = 1, 2, 3}, υ3142

↔ {ϕ2j ≥ 0, j = 1, 2, 3}, υ3143

↔ {ϕ3j ≥ 0, j = 1, 2, 3},υ3144

↔ {ψ1i ≥ 0, i = 1, 2, 3}, υ3145

↔ {ψ2i ≥ 0, i = 1, 2, 3}, υ3146

↔ {ψ3i ≥ 0, i = 1, 2, 3},υ3247

↔ χ′1≥ 0, υ32

48↔ χ′

2≥ 0, υ32

49↔ χ′

3≥ 0, υ32

4,10 ↔ {ω′1≥ 0, ψ′

1≥ 0},

υ324,11 ↔ {ω′

2≥ 0, ψ′

2≥ 0}, υ32

4,12 ↔ {ω′3≥ 0, ψ′

3≥ 0}, υ33

4,13 ↔ χ′′1≥ 0, υ33

4,14 ↔ χ′′2≥ 0,

υ334,15 ↔ χ′′

3≥ 0, υ33

4,16 ↔ {ω′′1≥ 0, ψ′′

1≥ 0}, υ33

4,17 ↔ {ω′′2≥ 0, ψ′′

2≥ 0},

υ334,18 ↔ {ω′′

3≥ 0, ψ′′

3≥ 0}, υ34

4,19 ↔ ω ≥ 0.

Thus the number of terminal nodes η for ΦD1D2≥ 0 is 19. To each terminal node υk there

corresponds a system of inequalities Λk(u) ≥ 0, k = 1, ..., 19.

Example 3 is added into the paper just to demonstrate the approach for computing a globalsolution.

Example 3

Given two non-rotatable triangles, the task is to find an enclosing rectangle of minimal perime-ter. Let a, b be the variable length and width of enveloping rectangle, respectively, then theobjective function is F (u) = a+ b, and the vector of variables is u = (a, b, x1, y1, x2, y2).

Polygon (triangle) K1 has corner points (2,−1), (0, 2) and (−2, 0), and triangle K2 has cornerpoints (0, 0), (3, 2) and (0, 2) (cf. Fig. 2).

Fig. 2 shows the six cases to be considered in order to get a global minima (all subproblemsare linear). The solution is F ∗ = min{F (u∗i ), i = 1, . . . , 6} = F (u∗

4) = 7.666668 with global

minimum at point u∗4= (4.000000, 3.666668, 2.000000, 1.000000, −0.000000, 1.666668).

5.2 The basic algorithm

The overall solution algorithm contains the following steps:

1. According to the branching scheme select one of the subproblems (3).

2. Find a starting point for subproblem (3).

3. Search for a local minima for subproblem (3).

4. Search for the global minimum for subproblem (3) or an approximation of it:

5. If some termination criterion is met then stop, otherwise select a new subproblem oftype (3) and continue with 2.

u1

u2

K 1

K 2

u1*

u2*

u3*

u4*

u5*

u6*

(a)

(b)

Figure 2: a) Polygons K1 and K2, b) Placement of K1 and K2 corresponding to point u∗i of alocal minimum, i = 1, . . . , 6

At this time, we consider the following realisations of subproblem (3):

1. Both objects A and B are (one-connected) polygons. Here we consider four problemclasses: without or with rotations and linear or quadratic objective function.In the first problem class (P1) (without rotations, perimeter minimization) subproblem(3) becomes a linear problem. Class (P2) (without rotations, area minimization) hasquadratic objective but linear constraints. In classes (P3) and (P4) (with rotations,perimeter or area minimization) non-linear constraints with sin- and cos-terms occur.

2. Objects A and B are composed objects.In similarity to above, four problem classes (P5) – (P8) are considered due to not allowingor allowing rotations and the two objective functions. In any case, non-linear (quadraticor with sin- and cos-terms) constraints (inequalities) are present.

In case, the branching tree is completely considered and all subproblems (3) are solved tooptimality, we obtain a global minimum for problem classes (P1) and (P2).

6 Conclusions and Outlook

In the paper a basic approach is presented to handle placement problems with irregular shape.In particular, we consider objects whose frontiers formed by straight-line and/or circular-arcsegments. We investigated the problem of enclosing two such objects by a rectangle or circleof minimal area or perimeter by means of phi-functions.

Since the number of subproblems to be solved increases rapidly when the objects become morecomplicated, pruning rules to reduce drastically that number are needed and will be part offuture research.

7 Appendix

7.1 Appendix A: Basic phi-functions

Let A ⊂ IR2 and B ⊂ IR2 be phi-objects, bounded by circular arcs and line segments. Weallow free rotations and translations of objects A and B in IR2. Vector u = (xt, yt, θ) definesthe arrangement of A in IR2. Heregy θ is the rotation parameter, and (xt, yt) is the translationvector of A. Each point (x0, y0) ∈ A in its eigen coordinate system is transformed into point(x, y) according to

x = x0 · cos θ + y0 · sin θ + xt, y = −x0 · sin θ + y0 · cos θ + yt,

and each straight line L0 = {(x, y) ∈ IR2 |α0x+ β0y+ γ0 = 0,√α2

0+ β2

0= 1.} is transformed

into straight line L = {(x, y) ∈ IR2 |αx+ βy + γ = 0, where

α = α0 · cos θ + β0 · sin θ, β = −α0 · sin θ + β0 · cos θ, γ = γ0 − α · xt − β · yt.

Let us introduce a set of basic phi-objects:C is a circle, C∗ = IR2\int(C) (cf. Fig. 3a) its complement, K is a convex phi-polygon (cf.

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(a) (b)

(c) (d) (e)

(f) (g)

Figure 3: Basic objects

Fig. 3b); K∗ = IR2\int(K) (cf. Fig. 3c), D = K ∩C is a circular segment (disk segment) (cf.Fig. 3d), H = C∗ ∩K is a ”hat” (cf. Fig. 3), V = G∩W or V = H ∩W is a ”horn” (cf. Fig.3), and a set of auxiliary objects:G = D ∩ C∗ (cf. Fig. 3), W = K ′ ∪D or W = K ′′ ∩ C (cf. Fig. 3).

Remark 1. We suppose that two sides of K are tangents to the end points of the arc of objectsD, H , W (cf. Fig. 3).

Remark 2. Each angle arc size of V is less then π.

Now we give a complete class of basic radical-free phi-functions:{ΦCC , ΦCK , ΦCD, ΦCH , ΦCV , ΦCC∗ , ΦCK∗ , ΦKK , ΦKD, ΦKH , ΦKV , ΦKC∗, ΦKK∗, ΦDD, ΦDH ,ΦDV , ΦKV , ΦKC∗ , ΦKK∗, ΦHH , ΦHV , ΦHC∗ , ΦHK∗ , ΦV V , ΦV C∗ , ΦV K∗}.

• Objects C∗ and C, rC∗ ≥ rC

ΦC∗C = −(xC∗ − xC)2 − (yC∗ − yC)

2 + (rC∗ − rC)2. (13)

• Objects C∗ and K

ΦC∗K = min{ωi, i = 1, 2, ..., m} with ωi = (xC − xi)2 + (yC − yi)

2 − r2. (14)

• Objects D and E ∈ {C,K,D}

ΦDE = max{ΦCE ,ΦKE}. (15)

• Objects K∗ and D

ΦK∗D = mini=1,...,m

max{ΦPiCD,ΦPiKD

}. (16)

• Objects C∗ and DLet D = K ∩ CD, where CD is a circle of radius rD with center (xD, yD).

ΦC∗D = min{ψ0,max{ΦC∗CD, χ1,−χ2}}, (17)

where

ψ0 = mini=1,2

(r2C − (xC − xi)2 − (yC − yi)

2),

χi = (xD − xC)(yi − yD)− (yD − yC)(xi − xD), i = 1, 2.

If rD ≥ rC then we use ΦC∗D = ψ0.

• Objects W and E ∈ {C,K,D, P,W} W = K ∩ C (cf. Fig. 3)

ΦWE = max{ΦKE,ΦCE}. (18)

• Objects W and E ∈ {C∗, K∗} W = K ∪D (cf. Fig. 3)

ΦWE = min{ΦKE,ΦDE}. (19)

G

C

K1

2K

G

C

t1

t2

tt

t

t

11

12

21

22 L

L

1

2

G

C

K 1

2K

t1

t2

K

G K1

2K

t1

t2

CG

P P

PC

D

T

1p

2p

D

D

(a) (b)

(c)

Figure 4: a) Objects G and C, b) Objects G and K c) Objects G and D for rD < rC

• Objects G and C

ΦGC = max{ΦC∗

GC ,ΦPGC ,min{ω1, ψ1, ω2, ψ2}}, (20)

where ωi = (x− xi)2 + (y − yi)

2 − (r)2, i = 1, 2, and ψi = 0 are equations of straightlines Li, passing through points ti1 and ti2, ψi(OG) > 0, i = 1, 2, OG is center point ofCG (Fig. 4a)).

• Objects G and K

ΦKG = min{ΦK1K ,ΦK2K , ψ}, (21)

where ψ = mini=1,...,m

max{r2C − (xC − xi)2 − (yC − yi)

2, αxi + βyi + γ}, rC is radius of

C = IR2\int(C∗) (Fig. 4 b)).

• Objects G and DIf rD < rC then phi-function of D and H is defined in the form (Fig. 4 c))

ΦDG = max{ΦPGTD,ΦC∗

GD,ΦGCD

, ϕ1, ϕ2}, (22)

here [ϕ1 = min{f p1CG , f t1PD ,−f t2PD ,ΦDK2}, ϕ2 = min{f p2CG , f t2PD ,−f t1PD ,ΦDK1

}, i. e.ϕi = min{f piCG , f tiPD ,−f t3−iPD ,ΦDK3−i

} where f piCG is a function of belonging pointpi to CG, + i = 1, 2, i. e. f piCG = r2G − (xG − xi)

2 − (yG − yi)2, and where (x1, y1) and

(x2, y2) are coordinates of endpoints p1 and p2 of D; (xG, yG) are coordinates of centerpoint of CG;

f tiPD is a function of belonging point ti toPD where PD ={αDx + βDy + γD ≤ 0},LD = {(x, y) ∈ IR2 : αDx+ βDy + γD = 0}, pi ∈ LD, TD ⊂ PD, i = 1, 2.

If rD ≥ rC , then functions ϕ1, ϕ2 in (22) take the form

φ1 = min{f p1CG , f t1PD ,−f t2PD ,ΦDK2, χ1},

φ2 = min{f p2CG , f t2PD ,−f t1PD ,ΦDK1,−χ2},

χi = (xD − xC)(yi − yD)− (yD − yC)(xi − xD), i = 1, 2.

• Objects W and G W = K ∪D (cf. Fig. 3)

ΦWG = min{ΦKG,ΦDG}. (23)

• Objects H and E ∈ {C,K,D, P,W} H = K ∩G

ΦHE = max{ΦGE,ΦKE}. (24)

• Objects H and E ∈ {C∗, K∗}

ΦHE = ΦKE. (25)

• Objects H and HLetH ′ = G′

⋂T ′ be a hat andH ′′ = G′′

⋂T ′′ another hat. Equivalently, H ′ = (C ′)∗

⋂T ′

and H ′′ = (C ′′)∗⋂T ′′. For the hat H ′ we use notation G′, C ′, T ′, etc., as defined above,

and for the hat H ′′ the respective notation G′′, C ′′, T ′′, etc.

ΦH′H′′

= max{ΦT ′H′′

,ΦG′T ′′

, ω, τ}, (26)

where

ω = min{α′2x′′1+ β ′

2y′′1+ γ′

2, α′′

2x′1+ β ′′

2y′1+ γ′′

2,

(rC′)2 − (xC′ − x′′3)2 − (yC′ − y′′

3)2,

(rC′′)2 − (xC′′ − x′3)2 − (yC′′ − y′

3)2},

τ = min{α′1x′′3+ β ′

1y′′3+ γ′

1, α′′

1x′3+ β ′′

1y′3+ γ′′

1,

(rC′)2 − (xC′ − x′′1)2 − (yC′ − y′′

1)2,

(rC′′)2 − (xC′′ − x′1)2 − (yC′′ − y′

1)2},

here (x′i, y′i) are the coordinates of the vertices and α′

ix+ β ′iy + γ′i = 0, i = 1, 2, are the

equations of lines containing the two straight sides of H ′, respectively; (xC′ , yC′) andrC′ are the coordinates of the center and the radius of the arc bounding H ′. Similarnotation apply to the hat H ′′.

• Objects V and E ∈ {C,K,D,W} V = G ∩W

ΦV E = max{ΦGE ,ΦWE}. (27)

• Objects V and E ∈ {C∗, K∗} V = G ∩W , W = D ∪K

ΦV E = min{ΦDE , fpE}, (28)

where f pE is a function of belonging point p to E∗ = IR2\int(E).

• Objects V and HLet V = HV ∩WV , then

ΦHV = max{ΦHHV,ΦHWV

}. (29)

• Objects V and VLet Vi = Hi ∩Wi, i=1,2, then

ΦV1V2= max{ΦH1H2

, ΦW2G1, ΦW1G2

, ΦW1W2}. (30)

Formulas (11)-(30) complete the class of basic phi-functions.

7.2 Appendix B: Example 1

(a) Optimal covering of objects A and B by a circle

Point of the global minimum:u∗ = (r∗, x∗

1, y∗

1, θ∗

1, x∗

2, y∗

2, θ∗

2) = (9.623201,−1.100537,−8.179520, 0,−0.310665, 5.370833,−2.808011)

F (u∗) = r∗ = 9.623201

The system on which the best solution has been reached:

−(7.7046 ∗ sin θ2 + 7.1753 ∗ cos θ2 + x2)2 − (−7.1753 ∗ sin θ2 + 7.7046 ∗ cos θ2 + y2)

2 + r2 ≥ 0−(14.4198 ∗ sin θ2 − 0.2176 ∗ cos θ2 + x2)

2 − (0.2176 ∗ sin θ2 + 14.4198 ∗ cos θ2 + y2)2 + r2 ≥ 0

−(7.6999 ∗ sin θ2 − 4.2457 ∗ cos θ2 + x2)2 − (4.2457 ∗ sin θ2 + 7.6999 ∗ cos θ2 + y2)

2 + r2 ≥ 0−(7.7074 ∗ sin θ2 + 7.1665 ∗ cos θ2 + x2)

2 − (−7.1665 ∗ sin θ2 + 7.7074 ∗ cos θ2 + y2)2 + r2 ≥ 0

−(7.6993 ∗ sin θ2 − 4.2478 ∗ cos θ2 + x2)2 − (4.2478 ∗ sin θ2 + 7.6993 ∗ cos θ2 + y2)

2 + r2 ≥ 0−(−3.4453 ∗ sin θ2 − 6.7145 ∗ cos θ2 + x2)

2 − (6.7145 ∗ sin θ2 − 3.4453 ∗ cos θ2 + y2)2 + r2 ≥ 0

−(x1 − 3.5573)2 − (y1 − 0.1989)2 + r2 ≥ 0−(x1 + 9.5178)2 − (y1 + 4.3682)2 + r2 ≥ 016.4329 ∗ x1 + 13.0751 ∗ y1 + 213.5205 ≥ 0−(x1 + 7.7916)2 − (y1 + 13.4971)2 + r2 ≥ 0−(x1 + 9.5239)2 − (y1 + 4.3741)2 + r2 ≥ 0−(x1 + 7.7911)2 − (y1 + 13.5097)2 + r2 ≥ 0−(x1 + 4.0898)2 − (y1 + 8.0755)2 − 13.1500 ∗ r + r2 + 43.2306 ≥ 0−(7.7046 ∗ sin θ2 +7.1753 ∗ cos θ2 + x2)

2 +2 ∗ (7.7046 ∗ sin θ2 +7.1753 ∗ cos θ2+ x2) ∗ (x1 − 3.5573)−(x1 − 3.5573)2 − (−7.1753 ∗ sin θ2 + 7.7046 ∗ cos θ2 + y2)

2 + 2 ∗ (−7.1753 ∗ sin θ2 + 7.7046 ∗ cos θ2 +y2) ∗ (y1 + 11.3511) − (y1 + 11.3511)2 + 133.4025 ≥ 0−(14.4198∗ sin θ2− 0.2176∗ cos θ2+x2)

2+2∗ (14.4198∗ sin θ2− 0.2176∗ cos θ2+x2) ∗ (x1− 3.5573)−(x1 − 3.5573)2 − (0.2176 ∗ sin θ2+14.4198 ∗ cos θ2 + y2)

2 +2 ∗ (0.2176 ∗ sin θ2 +14.4198 ∗ cos θ2+ y2) ∗(y1 + 11.3511) − (y1 + 11.3511)2 + 133.4025 ≥ 0−(7.6999 ∗ sin θ2 − 4.2457 ∗ cos θ2 + x2)

2 +2 ∗ (7.6999 ∗ sin θ2 − 4.2457 ∗ cos θ2+ x2) ∗ (x1 − 3.5573)−(x1 − 3.5573)2 − (4.2457 ∗ sin θ2 + 7.6999 ∗ cos θ2 + y2)

2 + 2 ∗ (4.2457 ∗ sin θ2 + 7.6999 ∗ cos θ2 + y2) ∗(y1 + 11.3511) − (y1 + 11.3511)2 + 133.4025 ≥ 0(− sin θ2 + 0.0004 ∗ cos θ2) ∗ (x1 + 9.5239) + (−0.0004 ∗ sin θ2 − cos θ2) ∗ (y1 + 4.3741) − (− sin θ2 +0.0004 ∗ cos θ2) ∗ x2 − (−0.0004 ∗ sin θ2 − cos θ2) ∗ y2 + 7.7016 ≥ 0(− sin θ2 + 0.0004 ∗ cos θ2) ∗ (x1 + 13.2253) + (−0.0004 ∗ sin θ2 − cos θ2) ∗ (y1 + 9.8083) − (− sin θ2 +0.0004 ∗ cos θ2) ∗ x2 − (−0.0004 ∗ sin θ2 − cos θ2) ∗ y2 + 7.7016 ≥ 0(− sin θ2 + 0.0004 ∗ cos θ2) ∗ (x1 + 7.7911) + (−0.0004 ∗ sin θ2 − cos θ2) ∗ (y1 + 13.5097) − (− sin θ2 +0.0004 ∗ cos θ2) ∗ x2 − (−0.0004 ∗ sin θ2 − cos θ2) ∗ y2 + 7.7016 ≥ 0−63858901.3280 ∗ sin θ2 − 76954796.6093 ∗ cos θ2 ≥ 088969116.2709 ∗ sin θ2 − 45656284.8902 ∗ cos θ2 ≥ 0(sin θ2 − 0.0007 ∗ cos θ2) ∗ (x1 − 3.5573) + (0.0007 ∗ sin θ2 + cos θ2) ∗ (y1 − 0.1989)− (sin θ2 − 0.0007 ∗cos θ2) ∗ x2 − (0.0007 ∗ sin θ2 + cos θ2) ∗ y2 − 7.7023 ≥ 0(0.2161 ∗ sin θ2 − 0.9764 ∗ cos θ2) ∗ (x1 +7.7916) + (0.9764 ∗ sin θ2 +0.2161 ∗ cos θ2) ∗ (y1 +13.4971)−(0.2161 ∗ sin θ2 − 0.9764 ∗ cos θ2) ∗ x2 − (0.9764 ∗ sin θ2 + 0.2161 ∗ cos θ2) ∗ y2 − 5.8113 ≥ 0−(7.6993 ∗ sin θ2 − 4.2478 ∗ cos θ2 + x2)

2 +2 ∗ (7.6993 ∗ sin θ2 − 4.2478 ∗ cos θ2+ x2) ∗ (x1 − 3.5573)−(x1 − 3.5573)2 − (4.2478 ∗ sin θ2 + 7.6993 ∗ cos θ2 + y2)

2 + 2 ∗ (4.2478 ∗ sin θ2 + 7.6993 ∗ cos θ2 + y2) ∗(y1 + 11.3511) − (y1 + 11.3511)2 + 133.4025 ≥ 0(0.2161 ∗ sin θ2 − 0.9764 ∗ cos θ2) ∗ (x1 + 9.5239) + (0.9764 ∗ sin θ2 + 0.2161 ∗ cos θ2) ∗ (y1 + 4.3741) −(0.2161 ∗ sin θ2 − 0.9764 ∗ cos θ2) ∗ x2 − (0.9764 ∗ sin θ2 + 0.2161 ∗ cos θ2) ∗ y2 − 5.8113 ≥ 0(0.2161 ∗ sin θ2 − 0.9764 ∗ cos θ2) ∗ (x1 +13.2253) + (0.9764 ∗ sin θ2 +0.2161 ∗ cos θ2) ∗ (y1 +9.8083)−(0.2161 ∗ sin θ2 − 0.9764 ∗ cos θ2) ∗ x2 − (0.9764 ∗ sin θ2 + 0.2161 ∗ cos θ2) ∗ y2 − 5.8113 ≥ 0(0.2161 ∗ sin θ2 − 0.9764 ∗ cos θ2) ∗ (x1 +7.7911) + (0.9764 ∗ sin θ2 +0.2161 ∗ cos θ2) ∗ (y1 +13.5097)−(0.2161 ∗ sin θ2 − 0.9764 ∗ cos θ2) ∗ x2 − (0.9764 ∗ sin θ2 + 0.2161 ∗ cos θ2) ∗ y2 − 5.8113 ≥ 0

b) Optimal clustering objects A and B into a rectangle R

Point of the global minimum: u∗ = (a∗, b∗, x∗1, y∗

1, θ

,1x∗2, y∗

2, θ∗

2) =

((19.566388, 10.054259, 15.403537, 2.613369, 4.020905, 7.311720, 2.451621, 7.313793)

F (u∗) = 196.725538

The system on which the best solution has been reached:

7.7046 ∗ sin θ2 + 7.1753 ∗ cos θ2 + x2 ≥ 014.4198 ∗ sin θ2 − 0.2176 ∗ cos θ2 + x2 ≥ 07.6999 ∗ sin θ2 − 4.2457 ∗ cos θ2 + x2 ≥ 0−7.1753 ∗ sin θ2 + 7.7046 ∗ cos θ2 + y2 ≥ 00.2176 ∗ sin θ2 + 14.4198 ∗ cos θ2 + y2 ≥ 04.2457 ∗ sin θ2 + 7.6999 ∗ cos θ2 + y2 ≥ 07.7074 ∗ sin θ2 + 7.1665 ∗ cos θ2 + x2 ≥ 07.6993 ∗ sin θ2 − 4.2478 ∗ cos θ2 + x2 ≥ 0−3.4453 ∗ sin θ2 − 6.7145 ∗ cos θ2 + x2 ≥ 0−7.1665 ∗ sin θ2 + 7.7074 ∗ cos θ2 + y2 ≥ 04.2478 ∗ sin θ2 + 7.6993 ∗ cos θ2 + y2 ≥ 06.7145 ∗ sin θ2 − 3.4453 ∗ cos θ2 + y2 ≥ 0−7.7046 ∗ sin θ2 − 7.1753 ∗ cos θ2 − x2 + a ≥ 0−14.4198 ∗ sin θ2 + 0.2176 ∗ cos θ2 − x2 + a ≥ 0−7.6999 ∗ sin θ2 + 4.2457 ∗ cos θ2 − x2 + a ≥ 07.1753 ∗ sin θ2 − 7.7046 ∗ cos θ2 − y2 + b ≥ 0−0.2176 ∗ sin θ2 − 14.4198 ∗ cos θ2 − y2 + b ≥ 0−4.2457 ∗ sin θ2 − 7.6999 ∗ cos θ2 − y2 + b ≥ 0−7.7074 ∗ sin θ2 − 7.1665 ∗ cos θ2 − x2 + a ≥ 0−7.6993 ∗ sin θ2 + 4.2478 ∗ cos θ2 − x2 + a ≥ 03.4453 ∗ sin θ2 + 6.7145 ∗ cos θ2 − x2 + a ≥ 07.1665 ∗ sin θ2 − 7.7074 ∗ cos θ2 − y2 + b ≥ 0−4.2478 ∗ sin θ2 − 7.6993 ∗ cos θ2 − y2 + b ≥ 0−6.7145 ∗ sin θ2 + 3.4453 ∗ cos θ2 − y2 + b ≥ 0−0.1989 ∗ sin θ1 − 3.5573 ∗ cos θ1 + x1 ≥ 0−0.1989 ∗ sin θ1 + 3.7779 ∗ cos θ1 + x1 ≥ 04.3682 ∗ sin θ1 + 9.5178 ∗ cos θ1 + x1 ≥ 03.5573 ∗ sin θ1 − 0.1989 ∗ cos θ1 + y1 ≥ 0−3.7779 ∗ sin θ1 − 0.1989 ∗ cos θ1 + y1 ≥ 0−9.5178 ∗ sin θ1 + 4.3682 ∗ cos θ1 + y1 ≥ 013.4971 ∗ sin θ1 + 7.7916 ∗ cos θ1 + x1 ≥ 0−7.7916 ∗ sin θ1 + 13.4971 ∗ cos θ1 + y1 ≥ 08.0755 ∗ sin θ1 + 4.0898 ∗ cos θ1 + x1 − 6.5750 ≥ 0−9.5239 ∗ sin θ1 + 4.3741 ∗ cos θ1 + y1 ≥ 0−13.2253 ∗ sin θ1 + 9.8083 ∗ cos θ1 + y1 ≥ 0−7.7911 ∗ sin θ1 + 13.5097 ∗ cos θ1 + y1 ≥ 00.1989 ∗ sin θ1 + 3.5573 ∗ cos θ1 − x1 + a ≥ 00.1989 ∗ sin θ1 − 3.7779 ∗ cos θ1 − x1 + a ≥ 0−4.3682 ∗ sin θ1 − 9.5178 ∗ cos θ1 − x1 + a ≥ 0−3.5573 ∗ sin θ1 − 20.8011 ∗ cos θ1 − y1 + b− 21 ≥ 0−13.4971 ∗ sin θ1 − 7.7916 ∗ cos θ1 − x1 + a ≥ 07.7916 ∗ sin θ1 − 13.4971 ∗ cos θ1 − y1 + b ≥ 0−4.3741 ∗ sin θ1 − 9.5239 ∗ cos θ1 − x1 + a ≥ 0−9.8083 ∗ sin θ1 − 13.2253 ∗ cos θ1 − x1 + a ≥ 0−13.5097 ∗ sin θ1 − 7.7911 ∗ cos θ1 − x1 + a ≥ 0

9.5239 ∗ sin θ1 − 4.3741 ∗ cos θ1 − y1 + b ≥ 013.2253 ∗ sin θ1 − 9.8083 ∗ cos θ1 − y1 + b ≥ 07.7911 ∗ sin θ1 − 13.5097 ∗ cos θ1 − y1 + b ≥ 0

(− sin θ1) ∗ (7.7046 ∗ sin θ2 + 7.1753 ∗ cos θ2 + x2) + (− cos θ1) ∗ (−7.1753 ∗ sin θ2 + 7.7046 ∗ cos θ2 +y2)− (− sin θ1) ∗ x1 − (− cos θ1) ∗ y1 − 0.1989 ≥ 0(− sin θ1) ∗ (14.4198 ∗ sin θ2 − 0.2176 ∗ cos θ2 + x2) + (− cos θ1) ∗ (0.2176 ∗ sin θ2 + 14.4198 ∗ cos θ2 +y2)− (− sin θ1) ∗ x1 − (− cos θ1) ∗ y1 − 0.1989 ≥ 0(− sin θ1) ∗ (7.6999 ∗ sin θ2− 4.2457 ∗ cos θ2+x2)+ (− cos θ1) ∗ (4.2457 ∗ sin θ2+7.6999 ∗ cos θ2+ y2)−(− sin θ1) ∗ x1 − (− cos θ1) ∗ y1 − 0.1989 ≥ 0(− sin θ2+0.0004∗cos θ2)∗(4.3741∗sin θ1+9.5239∗cos θ1+x1)+(−0.0004∗sin θ2−cos θ2)∗(−9.5239∗sin θ1+4.3741∗cos θ1+y1)−(− sin θ2+0.0004∗cos θ2)∗x2−(−0.0004∗sin θ2−cos θ2)∗y2+7.7016 ≥ 0(− sin θ2+0.0004∗cos θ2)∗(9.8083∗sin θ1+13.2253∗cos θ1+x1)+(−0.0004∗sin θ2−cos θ2)∗(−13.2253∗sin θ1+9.8083∗cos θ1+y1)−(− sin θ2+0.0004∗cos θ2)∗x2−(−0.0004∗sin θ2−cos θ2)∗y2+7.7016 ≥ 0(− sin θ2+0.0004∗cos θ2)∗(13.5097∗sin θ1+7.7911∗cos θ1+x1)+(−0.0004∗sin θ2−cos θ2)∗(−7.7911∗sin θ1+13.5097∗cos θ1+y1)−(− sin θ2+0.0004∗cos θ2)∗x2−(−0.0004∗sin θ2−cos θ2)∗y2+7.7016 ≥ 0100000000∗(−0.9441∗sin θ1+0.3298∗cos θ1)∗(0.6263∗sin θ2+0.7796∗cos θ2)−100000000∗(−0.3298∗sin θ1 − 0.9441 ∗ cos θ1) ∗ (0.7796 ∗ sin θ2 − 0.6263 ∗ cos θ2) ≥ 0−100000000∗(0.1858∗sin θ1+0.9826∗cos θ1)∗(0.6263∗sin θ2+0.7796∗cos θ2)+100000000∗(−0.9826∗sin θ1 + 0.1858 ∗ cos θ1) ∗ (0.7796 ∗ sin θ2 − 0.6263 ∗ cos θ2) ≥ 0(−0.9441∗ sin θ1+0.3298∗ cos θ1)∗ (7.7074∗ sin θ2+7.1665∗ cos θ2+x2)+ (−0.3298∗ sin θ1−0.9441∗cos θ1) ∗ (−7.1665 ∗ sin θ2+7.7074 ∗ cos θ2+ y2)− (−0.9441 ∗ sin θ1+0.3298 ∗ cos θ1) ∗x1− (−0.3298 ∗sin θ1 − 0.9441 ∗ cos θ1) ∗ y1 + 0.9853 ≥ 0(0.1858∗ sin θ1+0.9826∗ cos θ1)∗ (−3.4453∗ sin θ2−6.7145∗ cos θ2+x2)+ (−0.9826∗ sin θ1+0.1858∗cos θ1) ∗ (6.7145 ∗ sin θ2 − 3.4453 ∗ cos θ2 + y2)− (0.1858 ∗ sin θ1 + 0.9826 ∗ cos θ1) ∗ x1 − (−0.9826 ∗sin θ1 + 0.1858 ∗ cos θ1) ∗ y1 − 10.1637 ≥ 0−(4.3682∗ sin θ1+9.5178∗cos θ1+x1)

2+2∗ (4.3682∗ sin θ1+9.5178∗cos θ1+x1)∗ (−6.5199∗ sin θ2+7.1766 ∗ cos θ2 + x2)− (−6.5199 ∗ sin θ2 +7.1766 ∗ cos θ2 + x2)

2 − (−9.5178 ∗ sin θ1 +4.3682 ∗ cos θ1+y1)

2+2∗ (−9.5178∗ sin θ1+4.3682∗ cos θ1+ y1)∗ (−7.1766∗ sin θ2−6.5199∗ cos θ2+ y2)− (−7.1766∗sin θ2− 6.5199 ∗ cos θ2 + y2)

2 ++202.4153 ≥ 0(0.9441 ∗ sin θ1 − 0.3298 ∗ cos θ1) ∗ (−3.4453 ∗ sin θ2 − 6.7145 ∗ cos θ2 + x2) + (0.3298 ∗ sin θ1 +0.9441 ∗cos θ1) ∗ (6.7145 ∗ sin θ2 − 3.4453 ∗ cos θ2 + y2) − (0.9441 ∗ sin θ1 − 0.3298 ∗ cos θ1) ∗ x1 − (0.3298 ∗sin θ1 + 0.9441 ∗ cos θ1) ∗ y1 − 0.9853 ≥ 0−(0.9441 ∗ sin θ1 − 0.3298 ∗ cos θ1) ∗ (7.7074 ∗ sin θ2 +7.1665 ∗ cos θ2 + x2)− (0.3298 ∗ sin θ1 +0.9441 ∗cos θ1) ∗ (−7.1665 ∗ sin θ2 +7.7074 ∗ cos θ2 + y2)− (−(0.9441 ∗ sin θ1 − 0.3298 ∗ cos θ1) ∗x1 − (0.3298 ∗sin θ1 + 0.9441 ∗ cos θ1) ∗ y1 − 0.9853) ≥ 0−(4.3682∗ sin θ1+9.5178∗cos θ1+x1)

2+2∗ (4.3682∗ sin θ1+9.5178∗cos θ1+x1)∗ (−6.5199∗ sin θ2+7.1766 ∗ cos θ2 + x2)− (−6.5199 ∗ sin θ2 +7.1766 ∗ cos θ2 + x2)

2 − (−9.5178 ∗ sin θ1 +4.3682 ∗ cos θ1+y1)

2+2∗ (−9.5178∗ sin θ1+4.3682∗ cos θ1+ y1)∗ (−7.1766∗ sin θ2−6.5199∗ cos θ2+ y2)− (−7.1766∗sin θ2− 6.5199 ∗ cos θ2 + y2)

2 + 202.4153 ≥ 010000000∗(− sin θ1)∗(−0.0007∗sin θ2−cos θ2)−10000000∗(− cos θ1)∗(− sin θ2+0.0007∗cos θ2) ≥ 0(− sin θ1) ∗ (7.7074 ∗ sin θ2 + 7.1665 ∗ cos θ2 + x2) + (− cos θ1) ∗ (−7.1665 ∗ sin θ2 + 7.7074 ∗ cos θ2 +y2) + (sin θ1) ∗ x1 ++(cos θ1) ∗ y1 − 0.1989 ≥ 0−10000000∗ (− sin θ1)∗ (−0.6263∗ sin θ2−0.7796∗cos θ2)+10000000∗ (− cos θ1)∗ (−0.7796∗ sin θ2+0.6263 ∗ cos θ2) ≥ 0−(4.3741∗ sin θ1+9.5239∗cos θ1+x1)

2+2∗ (4.3741∗ sin θ1+9.5239∗cos θ1+x1)∗ (−6.5199∗ sin θ2+7.1766 ∗ cos θ2 + x2)− (−6.5199 ∗ sin θ2 +7.1766 ∗ cos θ2 + x2)

2 − (−9.5239 ∗ sin θ1 +4.3741 ∗ cos θ1+y1)

2+2∗ (−9.5239∗ sin θ1+4.3741∗ cos θ1+ y1)∗ (−7.1766∗ sin θ2−6.5199∗ cos θ2+ y2)− (−7.1766∗sin θ2− 6.5199 ∗ cos θ2 + y2)

2 ++202.4153 ≥ 0−(13.5097 ∗ sin θ1 + 7.7911 ∗ cos θ1 + x1)

2 + 2 ∗ (13.5097 ∗ sin θ1 + 7.7911 ∗ cos θ1 + x1) ∗ (−6.5199 ∗sin θ2+7.1766 ∗ cos θ2+x2)− (−6.5199 ∗ sin θ2 +7.1766 ∗ cos θ2+x2)

2 − (−7.7911 ∗ sin θ1+13.5097 ∗cos θ1+ y1)

2 +2 ∗ (−7.7911 ∗ sin θ1 +13.5097 ∗ cos θ1 + y1) ∗ (−7.1766 ∗ sin θ2− 6.5199 ∗ cos θ2 + y2)−

(−7.1766 ∗ sin θ2− 6.5199 ∗ cos θ2 + y2)2 ++202.4153 ≥ 0

(8.0755∗ sin θ1+4.0898∗ cos θ1+x1)∗ (−5.4342∗ sin θ1−3.7014∗ cos θ1)− (−4.0898∗ sin θ1+8.0755∗cos θ1+y1)∗(−3.7014∗sin θ1+5.4342∗cos θ1)−(−5.4342∗sin θ1−3.7014∗cos θ1)∗(−6.5199∗sin θ2+7.1766 ∗ cos θ2+x2)+ (−3.7014 ∗ sin θ1+5.4342 ∗ cos θ1) ∗ (−7.1766 ∗ sin θ2− 6.5199 ∗ cos θ2+ y2) ≥ 0

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