Solving a Nesting Problem using the Collision Free Region

Transcription

1 Solving a Nesting Problem using the Collision Free Region Master Thesis August 31 th, 2016 Irene Reijntjes Supervisor: dr. ir. C.A.J. Hurkens

2 Abstract This report investigates an irregular shape packing problem, or alternatively a nesting problem. An irregular object is given, which could contain defects. Per object, a set of items is given that consists of at most two different item shapes in different orientations. The goal is to place as many items as possible inside the object, where the items are not allowed to overlap and the defects need to be avoided. This problem is approached by defining a region for every item that corresponds to feasible placements. A few different optimization techniques are applied and compared. These include a bottom-left heuristic and a combination with a branch and bound method. i

3 Acknowledgments The final months of my master, I worked enthusiastically on this project. I would like to thank my supervisor Cor Hurkens for introducing me to Jos van Esch and the project he was working on. Jos introduced me to multiple people and companies that were all working together on this. This gave me great insights in the world of business and how people from different backgrounds can all work together on one project. Furthermore, I would like to thank CQM, and in particular Geert Teeuwen, Judith van Rijswick and Jan van Doremalen for welcoming me with open arms and supporting me during this project. Without your help, I would have never been able to obtain the results presented in this report. I found the brainstorm sessions we had, also with Cor Hurkens, very useful and eye-opening. During the project I was not always confident on whether I would manage to find good results. This increased the satisfaction when I finally did pull it off. ii

4 Contents 1 Introduction Cutting and Packing Problems Nesting Problem Previous Work Data Management Items Convex hull Object Boundary Polygons Defects Inner-Fit Polygon Minkowski difference Algorithm Finding the extreme vertices Finding the intersection points Implementation No-Fit Polygon Minkowski sum Algorithm Collision Free Region Initial CFR CFR algorithm Optimization Algorithm Preprocessing Greedy heuristics Greedy Bottom-Left Heuristic (GBL) Counting Bottom-Left Heuristic (CBL) Branch and Bound Heuristics Bottom-Left and Top-Right (BLTR) Counting Branch and Bound Algorithm (CBNB) Results and Conclusion Results iii

5 7.2 Conclusion Discussion and Recommendations Discussion Recommendations and future work iv

6 Chapter 1 Introduction In this chapter we will give a short introduction to the irregular shape packing problem. Section 1.1 will provide an introduction to the variety of cutting and packing problems. A systematic organization of these problems will be presented. Then, in section 1.2, an introduction will be given to the specific type of problem that is considered in this report. Finally, in section 1.3 previous applied methods for nesting problems are described. In chapter 2, some concepts required for later computations are explained. In chapter 3 it is explained what the inner-fit polygon is and how this concept can be used in the nesting problem. The no-fit polygon is described in chapter 4. Its use in the nesting problem is also explained. Chapter 5 combines the inner-fit polygon and no-fit polygon into the collision free region, which directly corresponds to legal item placements. A few algorithms that will lead to feasible placements using the CFR are given in chapter 6. The results of these algorithms applied to the same input set are given in chapter 7. Finally, chapter 8 discusses the results of the proposed methods and provides recommendations that could improve the algorithms and lead to better solution. 1.1 Cutting and Packing Problems C&P (cutting and packing) problems is a collective name for various problems, that essentially have the same structure but appear under different names in the literature. A few examples of C&P problems include cutting stock, trim loss, bin packing, vehicle loading, nesting and scheduling problems. All these problems have two properties in common. ˆ The input data consists of two groups whose elements define geometric bodies of fixed shapes: the stock, which will be referred to as object from now on; the list of small items, which will be referred to as items. 1

7 ˆ A cutting or packing problem realizes patterns that are geometric combinations of items assigned to objects. This means that every C&P problem consists of two sets of input data: objects and items. While the objectives may be different, the items should always be placed inside the objects without overlapping one another. There is a great variety of C&P problems, each of which with a slightly different objective function. To be able to categorize these problems, a typology has been developed. This is a systematic organization of problems into homogeneous categories on the basis of a given characterizing criteria. It was first described by Dyckhoff [1], and was later improved by Wäscher [2]. The dimension and objective of a C&P problem will be the first property by which the categorization takes place. We distinguish between 1-, 2-, 3- and n-dimensional problems, where n > 3. A 1-dimensional problem would be for example the cutting of bars of certain length (objects), in smaller pieces of given length (items). Cutting out certain patterns (items) from a piece of textile or leather (object) is a 2-dimensional problem. An example of a 3-dimensional problem is vehicle loading, where for instance boxes (items) need to be transported in containers (objects). For the objective function, there is a subdivision into two situations: output maximization and input minimization. In the first situation, a set of small items has to be assigned to a given set of large objects. However, the set of objects is not sufficient for the assignment of all items. Therefore, a subset of items needs to be assigned to the object while maximizing the value. This value is often a measurement for the unused object. An example of a problem with this kind of objective is the knapsack problem. In this situation the object is a knapsack of certain weight capacity, and the items are boxes with a certain value and weight. The goal is to pack as many items as possible without exceeding the capacity of the knapsack and maximize the value of the items packed. In the second situation, the set of objects is sufficient, so all small items can be placed. The goal is now to assign these to a subset of the objects and maximize the value. The bin packing problem is an example of such a problem. We are given items of fixed dimension and objects, the bins, of fixed size. The goal is to find the smallest number of bins necessary so that all items are assigned to a bin. The division into basic categories continues with the next criterion which is the assortment of the items. There are three different cases distinguished: identical items, weakly heterogeneous assortment and strongly heterogeneous assortment. The first category applies when all items are of the same shape and size. The second applies when the items can be grouped into relatively few classes, for which the items are identical with respect to shape and size. Items with identical shape and size but different orientation are considered different kinds of items. Finally, the third category applies when only few items are of the same shape and size. This includes the situation where every item is allowed in any possible orientation. Furthermore, we make a distinction between regular and irregular items. Regular items include rectangles, boxes, circles and balls. 2

8 1.2 Nesting Problem The problem dealt with in this report is 2 dimensional. We are given a set of objects and a set of items. There is a certain demand for every type of item that needs to be met, and it is not required that every object is used. This implies that the objective of the problem is to minimize the input. Namely, there are more objects than necessary for the required items. The next categorization step is the assortment of items. The items are not identical so they form a heterogeneous set. Per object, no more than two types of items may be placed, with a number of allowed orientations. The items form a weakly heterogeneous set. The basic problem type corresponding to these characteristics is the cutting stock problem (CSP). The CSP deals with cutting standardsized objects into pieces of specified size, while minimizing the material wasted. The problem type can be further refined by analyzing the characteristics of the objects, these are not standard-sized. The objects are irregular and can be non-homogeneous. Regular objects are for instance rectangles and circles. An object can be non-homogeneous, because it could contain defects. This can occur in natural products such as leather. Due to these features, the problem is categorized as an irregular cutting stock problem, which is often referred to as the nesting problem. This report concerns a special case of the nesting problem where the objects are available one by one. For every object there is only a subset of items available that are allowed to be placed. This subset contains at most two different types of items. By a different type, we mean an item with a different shape. Items that have the same shape but a different orientation are considered as the same type of item. Per item, approximately 20 different orientations are allowed. In the remainder of this report, we will therefore talk about one object being the input together with a set of items that is allowed to be placed in this specific object. The CSP has been proved to be NP-hard in the strong sense by Garey and Johnson [3]. As a result, solution methodologies utilize heuristics. 1.3 Previous Work A defining characteristic of nesting problems is the requirement to develop powerful geometric tools to handle the wide variety and complexity of shapes that need to be packed. The question that needs to be answered is: Given a position on the object of two items, do they overlap, touch or are they separated?. This question is easily answered for us humans, but turns out to be the first struggle when implementing heuristic algorithms. Over the past years, researchers have proposed a variety of methods, each with their own advantages and disadvantages. Not only the speed is important, but also the complexity of the implementation. The most common approaches are the raster method, direct trigonometry, the no-fit polygon and the phi-function. [4] ˆ Raster Method The raster method approach divides the object into discrete areas. This reduces the geometric information to a grid represented by a matrix. Initially all raster boxes are empty, which corresponds to the zero matrix. 3

9 The placement of an item inside the object sets value 1 to the raster boxes that are covered by the item. Even if the item only covers a small part of the raster box, it is set to be occupied. This leads to inaccuracies, especially for irregular items. To increase accuracy, the size of the raster boxes can be decreased, but this leads to higher computation times. ˆ Direct Trigonometry This method deals with irregular shapes in a better way, since it does not approximate the items, but uses the items themselves. The immediate drawback is that the computational effort is very high. Every time the position of an item is changed, it needs to be tested for overlap with all items that have already been placed again. ˆ No-fit Polygon The no-fit polygon is a region representing the possible translation vectors that can be applied to a polygon such that the polygon does not overlap other (already placed) polygons. It is chosen to use this concept for the overlap detection in this report. Its precise definition and implementation are described in chapter 4. ˆ Phi-functions The phi-function is a mathematical expression that represents the mutual positions of two objects. When the phi-function is normalized, its value is the Euclidean distance between the two items. This property helps us move items in such a way that they touch one another. This happens precisely when their normalized phi-function is equal to 0. The lack of an algorithmic process for generating the phi-functions for arbitrary shapes, is the likely the reason that it has not been broadly applied. The no-fit polygon has become an increasingly popular option since it is more efficient than direct trigonometry, while maintaining the accuracy of this former method since the true items are used. For this reason, it was chosen to use the no-fit polygon in this report over the other available options. 4

10 Chapter 2 Data Management In this chapter the available data and the representation is described. Both the object and all items are required as input, which are all represented by polygons. In turn, a polygon is described by a list of vertices in counterclockwise order. In section 2.1, the further specification of the items will be described. It will be explained in section how the convex hull of a polygon can be obtained efficiently. This algorithm applies to the object as well. Then, in section 2.2, the specifications of the object polygon are given. Finally, in section 2.2.1, it will be explained what boundary polygons are, and how they can be obtained from a polygon and its convex hull. 2.1 Items The item polygons, in contrast to the object polygon, can be freely moved in the plane. The location of an item is entirely defined by the location of its reference point. The reference point of a polygon is defined to be the lower left corner of the axis-aligned minimum bounding box around the polygon. This is equivalent to (x r, y r ), where x r is the smallest x-coordinate occurring in the polygon and y r is the smallest y-coordinate occurring in the polygon. An example of a polygon with corresponding reference point is given in Figure 2.1. The item polygons are provided as polygons with reference point (0, 0). The polygon can then easily be placed with its reference point on any position (x, y) by simply applying the translation (x, y) to all vertices. In general, if the reference point would be (x r, y r ), and the item needs to be placed with its reference point on (x p, y p ), then the translation (x p x r, y p y r ) needs to applied. It is clear that this translation vector is easily obtained when (x r, y r ) = (0, 0), which is why this situation is preferred. When referring to Figure 2.1: An example of a polygon with reference point. 5

11 the placement of an item on a certain position inside the object, we will always mean the placement of the items reference point at this certain position. Another advantage of defining the item polygons with the origin as reference will become clear in section Convex hull For various algorithms that will be explained later on in chapters 3 and 4, the convex hull of the object and item polygons is required. The convex hull of a polygon is the smallest convex set that contains the polygon. A convex set is a set with the property that all points that are on a line segment between two points that are contained in the set, are also contained in the set. A fast algorithm for such a computation has been developed by Melkman, and described in [5]. The input for this algorithm needs to be a simple polygon. This is a polygon with no self-intersecting edges. The object and item polygons are simple and so this algorithm can be applied. The algorithm computes the convex hull of a polygon with n vertices in O(n) time. It starts with three vertices v 0, v 1 and v 2 that form a counterclockwise triangle. In other words, the edge from v 1 to v 2 is a left turn with respect to the edge from v 0 to v 1. These vertices are stored in a deque D, which consists of all vertices that are in the convex hull so far. The first and the last element of a deque, the bottom and top, are the same. In this case D[bot]=D[top]=v 2, D[bot+2]=D[top 1]= v 1 and D[bot+1]=D[top 2]= v 0. This situation is sketched in Figure 2.2. Now the next vertex in the polygon, v 3, is considered. It needs to be checked whether it is contained in the current polygon that is described by the vertices in the deque. This can be done by finding its location with respect to the bottom two vertices of the deque and the top two Figure 2.2: Adding next vertex v 3 to the current convex hull described by the vertices in the deque. vertices of the deque. If both the path D[bot], D[bot+1], v 3 and the path D[top 1], D[top], v 3 make a left turn, then v 3 is already contained in the convex hull described by the deque and we can continue with v 4. To determine whether the edges make a left or right turn, we use the function given in equation (2.1). turn = (x 1 x 0 )(y 2 y 0 ) (x 2 x 0 )(y 1 y 0 ), (2.1) where v 0 = (x 0, y 0 ), v 1 = (x 1, y 1 ) and v 2 = (x 2, y 2 ). There are three possible options for the result: { > 0 if the segment v 1, v 2, v 3 makes a left turn; turn 0 if the segment v 1, v 2, v 3 makes a right turn or is straigt. This is an efficient method for finding the turn of two line segments with respect to one another, without having to compute the exact angle they make. 6

12 If v 3 does not make a left turn with both the top and the bottom of the deque, the vertex needs to be added to the deque on both ends, after possible removal of vertices from the top and bottom. The bottom of the deque is considered first. If D[bot], D[bot+1], v 3 is a right turn, D[bot] needs to be erased and the new D[bot] is D[bot+1]. Now, it needs to be checked again whether the edges make a left turn. This is repeated until the bottom three vertices make a left turn. In the example shown in Figure 2.2, the path v 2, v 0, v 3 makes a right turn, implying that the bottom vertex is removed. Now D[bot]=v 0 and D[bot+1]=v 1. This implies that D[bot], D[bot+1], v 3 is equal to v 0, v 1, v 3. The corresponding edges make a left turn indeed, and hence D[bot]=v 3, D[bot+1]=v 0 and D[bot+2]=v 1. Now it needs to be checked whether the top vertices form a left turn. If D[top 1], D[top], v 3 is a left turn, we are done and we put D[top]=v 3. If it is a right turn, D[top] is erased and the new D[top]=D[top 1]. This is repeated until the three current vertices make a left turn. In the situation shown in Figure 2.2, the top vertices form a left turn so we can simply add v 3 as D[top]. This yields the following final deque from bottom to top: v 3, v 0, v 1, v 2, v 3. This process is repeated until every vertex of the input polygon is visited exactly once. It is checked whether it needs to be added and if it does, it will be added twice, while other vertices in the deque could be removed once. Therefore, the running of this algorithm to compute the convex hull of a simple polygon is O(n), where n is the number of vertices of the input polygon. 2.2 Object The object polygon has a certain fixed location in the plane, so it cannot be moved or translated. This polygon is therefore not required to have its reference point on the origin. The coordinates of the vertices directly correspond to the location in the plane. The convex hull of the object is useful, which can be computed using the algorithm presented in section Boundary Polygons Working with the convex hull instead of the true polygon has the disadvantage that it could be the case that is not accurate enough for computations. This happens especially when the difference between the actual object polygon and its convex hull is large. It could result in the incorrect placement of items. If an item is contained in the convex hull of the object, this does not imply that the item is also contained in the object polygon itself. We can work around this issue, by storing the boundary polygons. How these boundary polygons can be used in the nesting algorithm will become clear in chapter 5. Consider a polygon A and its convex hull CH(A). Then the set of boundary polygons will be represented by A = A \ CH(A). In Figure 2.3 an example of an irregular polygon is shown. Its convex hull is bounded by the blue line and the set of boundary polygons consists of all polygons filled with red. 7

13 These boundary polygons can be obtained from A and CH(A). Let w 1,..., w n be the vertices of polygon CH(A). Then the vertices of polygon A are w 1, v 1,1,..., v 1,k1, w 2, v 2,1,..., v 2,k2, w 3,..., w n, v n,1,..., v n,kn. Here k i is the number of vertices of polygon A between vertex w i and w i+1. It could be the case that there is no boundary polygon between two successive vertices of CH(A). In that case k i = 0, which corresponds to the situation where two successive vertices of A are both in CH(A). The boundary polygon A i, between w i and w i+1 is given by the vertices w i, v i,1,..., v i,ki, w i+1. The set of boundary polygons for A is A = {A 1,..., A n }. Here A n is the boundary polygon between w n and w Defects In certain applications, for instance leather cutting, the object may contain defects, or lower quality regions. These need to Figure 2.3: Example of an irregular polygon with its convex hull and set of boundary polygons. be avoided when placing the items inside the object. How this can be dealt with in the algorithm, will be explained in chapter 4. The defects are given by either polygons or lines, depending on whether the defect is a hole or a crack. Lower quality regions can be defined by a polygon as well. 8

14 Chapter 3 Inner-Fit Polygon In this chapter, the use of the IFP (Inner-Fit Polygon) will be explained. The IFP is a polygon describing the possible translation vectors for the placement of items inside an object. Since the items are located in such a way that their reference point is the origin, these translation vectors directly indicate the placement of the item. Whenever an item is placed with its reference point inside or on the boundary of the IFP, it is located inside the object or touching its boundary. This is illustrated in Figure 3.1, where the items are placed with their reference point on the boundary of the IFP, resulting in the items touching the boundary of the object. Note that instead if the actual object, its convex hull is used. (a) (b) (c) (d) Figure 3.1: Examples of the IFP: the region in blue is the inner-fit polygon for the items with the convex hull of the objects. The bottom-left corner of the grey box corresponds to the reference point. 9

15 This is an important concept for a nesting problem. If the IFP can be determined, it is not necessary to check for intersection between an item and the object. Since these geometric checks can be both difficult and time consuming, the use of the IFP is is preferred. In section 3.1, the relation between the Minkowski difference and the IFP is described. In section 3.2 the algorithm to compute the IFP will be described. 3.1 Minkowski difference The IFP of two polygons can be calculated using the Minkowski difference. Definition 3.1 (Minkowski Difference). A B = b B A b, (3.1) where A b = {a + b a A} represents the set A translated by b B. We are interested in the placement of an item B, into an object A, where both A and B are polygons. We claim that the IFP of B into A is given by A ( B). To show that this claim holds, the following theorem needs to be proved. Theorem 3.2. x A ( B) B x A Proof. ( ): Let x A ( B). Then, by definition x {a + b a A}. b B This indicates that x {a + b a A}, b B implying that b B, â A such that x = â + b. Rewriting this expression yields b + x = â which is equivalent to b + x = â, for b B. Note that b + x B x and â A. Since for all elements of polygon B an such an element in polygon A needs to exist, we indeed showed that B x A. ( ): Let B x A. This is equivalent to {b + x b B A}. So every element of the form b + x with b B is an element of A. This indicates that x A b for b B, which is equivalent to x A + b for b B. For all b B, a A such that x = a + b. So x {a + b a A}, b B which is equivalent to x A ( B). In Figure 3.2 an example of the Minkowski difference of two polygons is depicted. Every edge of polygon A is translated by every vertex of polygon B. This results in n new (overlapping) polygons, where n is the number of vertices of B. The intersection of these polygons is the Minkowski difference A B. 10

16 The Minkowski difference represents the feasible set of translation vectors which can be applied to B such that it is contained in polygon A. When the applied translation vector is on one of the edges of the Minkowski difference, then the two polygons touch one another. Due to the fact that the item polygons are located such that their reference point is the origin, the Minkowski difference polygon is placed inside the object polygon, so that it is directly translates into possible placements for polygon B. Figure 3.2: The Minkowski difference of polygons A and B: the intersection of translated line segments of A by vertices of B. As proven in Theorem 3.2, the IFP is related to the Minkowski difference in the sense that IFP AB = A ( B). Hence, the IFP of A and B can be computed using the Minkowski difference. Notice that the input needs to be provided in the correct way. Namely, B is required as input and not B. The first step, before being able to implement the Minkowski difference algorithm, is to determine B from B. Every vertex v i = (x i, y i ) of B is replaced by v i = ( x i, y i ), which results in the mirrored polygon. In order to be consistent, this polygon is translated in such a way that the reference point is the origin. To do so, from the original polygon, the largest x-coordinate, x max, and the largest y-coordinate, y max, are selected. Every vertex of the mirrored polygon is translated by (x max, y max ) to obtain the new polygon of which the origin is its reference point. We will describe an algorithm to compute the Minkowski difference for convex polygons. In fact, it is irrelevant whether polygon B is convex. Namely A ( B) = A ( CH(B)). This is due to the fact that a vertex in a cavity of polygon B cannot be an extreme vertex for any edge of A. From here on, polygon CH(B) will be referred to as B. How to deal with the non-convexity of polygon A will be explained in chapter 5. The algorithm presented in section 3.2 follows the approach provided in [6]. 3.2 Algorithm Firstly, the global structure of the algorithm will be described. Afterwards, the steps will be explained in detail. Input for the algorithm consists of two polygons A and B, where both polygons are convex. These are represented by a list of their vertices, which have to be in counterclockwise ordering. The starting vertex is the vertex with the largest x-coordinate. If there are multiple vertices with that same largest x-coordinate, then the vertex of these with the largest y coordinate is the starting point. Let polygon A consist of the vertices a 0,..., a m 1 with edge a i from a i mod m to a (i+1) mod m, and polygon B consist of the vertices b 0,..., b n 1 with edge bj from b j mod n to b (j+1) mod n. For the remainder of this report mod n will be omitted. 11

17 The algorithm consist of two main steps: 1. For every edge of A, find the corresponding extreme vertex of polygon B. Move all edges a i of A by their corresponding vertex b j of B. New line segments c i = a i b j are obtained; 2. Find the intersection points of the line segments in order to construct the resulting polygon. The steps in the above enumeration will be explained in detail in the sections and respectively Finding the extreme vertices The first step in the algorithm is to find the vertex from B that is the extreme vertex in the outer normal direction n i of a i. This vertex will correspond to the vertex of polygon B that will touch the edge a i, if it is moved as far as possible in the direction n i. This vertex can be found by looking at the direction of the edges of polygon A and B with respect to one another. The correct edge is the first edge of B that is not making a right turn relative to the current edge of A. The relative direction of edge a i = (a i, a i+1 ) and edge b i = (b i, b i+1 ) is computed by det( a i, b j ) = a i b i a i+1 b i+1 = a ib i+1 a i+1 b i. (3.2) There are three possible options for the direction: < 0 if b j is to the right with resprect to a i det( a i, b j ) > 0 if b j is to the left with resprect to a i = 0 if b j is parallel to a i Vertex b j is the extreme point for edge a i if and only if b j is the first edge of polygon B not making a right turn compared with a i. Equivalently, if it is the first edge of polygon B for which det( a i, b j ) 0. The starting point for both polygons is the point with the largest x-coordinate. For the first edge of these polygons, there are two options. Either edge b 0 is the correct edge, or it is not. If it is the correct edge, vertex b 0 is the extreme vertex of a 0. A new edge c 0 is defined by c 0 = a 0 b 0 = (a 0 b 0, a 1 b 0 ) and we continue with a 1. If it was not the correct edge, we continue with b 1. We keep doing this, until the correct edge is found. If an edge of B is visited and is not the matching edge for the current edge a i, it cannot be the matching edge of any other edge from A that has not been visited yet. This follows from the convexity of the polygons and the fact that the vertices are traversed in counterclockwise ordering, starting with the rightmost points. If an edge from B is skipped, we move to the next edge. This edge is always a left turn compared to the previous visited edge, implying that one of the next edges has to provide the correct extreme vertex. Once an edge from polygon B is skipped, it does not have to be revisited, since it is impossible for it to be an extreme vertex for any other edge of polygon A. This process is repeated until there is a matching vertex from polygon B for 12

18 every edge from polygon A. The algorithm loops over all edges of polygon A and all edges of polygon B once. This gives rise to a computational complexity of O(m + n), where m is the number of vertices of polygon A and n is the number of vertices of polygon B Finding the intersection points We have now obtained m oriented line segments c 0,..., c m 1 that can be used to obtain the IFP. The IFP is the set of points that is left from, or on all the line segments c i. Not all line segments will be in the IFP, as can be seen in Figure 3.3. The correct IFP can be determined by finding intersection points between the line segments. The method described below follows the implementation presented in chapter 7.2 from [7]. Consider two line segments v = (a, b) and w = (c, d). Define det(a, b) = x a y b y a x b. Let det1 = det(b a, c d), then det1 = 0 if and only if the two line segments are parallel. If det1 is non-zero, there can be an intersection point. In order to find the intersection point if there is one, two other determinants need to be computed. Define t = det(c a, c d) and u = det(b a, c a) and t u det1 let t = det1 and u =. There is an intersection point between the two line segments if and only if 0 t, u 1. If this condition holds, the intersection point is give by p = a (1 t ) + b t. The pseudocode for this algorithm is given in algorithm 1. The output consists of a Boolean variable together with a point. If there is an intersection point, the function returns the value true, together with the intersection point. If there is no intersection point, the function returns false and a dummy point. 3.3 Implementation Figure 3.3: An example of the translated line segments from which the IFP is obtained. Although the algorithm consists of two different steps, it is possible to cover both these steps in the same iteration loop. This is done by storing the matching vertex of polygon B together with the corresponding edge of polygon A. For every vertex of polygon B we store a list of translated edges of polygon A that match with this vertex. The first list that is found, is added to a list that could eventually describe the IFP. As soon as a second list is found, the intersection point between these lists is determined if there is one. Assume that there is a list of vertices a i,..., a j 1 corresponding to vertex b i, and a list of vertices a j,..., a k corresponding to vertex b j. Here, b j is not necessarily the direct successor of b i, since not all vertices of polygon B are extreme vertices. Every time a path of line segments for a vector from B is found, we will determine whether there is an intersection point with a previous path of line segment. Let the list that has already been processed be B1 and the current list to be processed be B2. We start looking for an intersection point between the last edge b 1 of the first 13

19 Algorithm 1 Find the intersection point between two line segments if it exists 1: function findintersectionpoint( v, w) Input: Two line segments v = (a, b) and w = (c, d). Output: A Boolean variable indicating whether there is an intersection point. If there is one, it is returned, if there is no, a dummy point is returned 2: det1 det(b a, c d) 3: if det1 is nonzero then 4: t det(c a,c d) det1 5: u det(b a,c a) det1 6: if 0 t, u 1 then 7: result a (1 t) + b t 8: return (true, result) 9: else 10: return (f alse, a) 11: end if 12: else 13: return (f alse, a) 14: end if 15: end function list B1 and the first edge b 2 of the last list B2. These two line segments will be tested for intersection using the function 1. If there is no intersection point between these segments, one of the two needs to be removed. Which of the two line segments, will be decided by investigating the direction of the second edge with respect to the first. Let b 1 = (v 1, v 2 ) and b 2 = (w 1, w 2 ). Let e 1 = (v 1, w 1 ) and e 2 = (v 1, w 2 ). Now, we compute d 1 = det( b 1, e 1 ) and d 2 = det( b 1, e 2 ) using the function presented in equation (3.2). We make a distinction between six different outcome scenarios: 1. d 1 = 0 2. d 2 = 0 3. d 1 d 2 < 0 i.e. the signs are different (a) d 1 < 0, d 2 > 0 (b) d 1 > 0, d 2 < 0 4. d 1 d 2 > 0 i.e. the signs are the same (a) d 1 > 0, d 2 > 0 (b) d 1 < 0, d 2 < 0 For case 1, a possible situation is depicted in Figure 3.4a. The edges of interest are s 5 from B1 and t 1 from B2. The starting point of t 1, w 1, is on the line extending s 5. By convexity, the endpoint of t 1, w 2, is left of this line. Since no intersection point was found between the line segments s 5 and t 1, w 1 is located either after v 6 or before v 5. If the former is the case, there will be no intersection between the paths. Therefore, it does not matter which of the two segments will be erased. If the latter is the case, edge t 1 could be the edge providing for the intersection. Imagine for example that w 2 would not be present. Then the 14

20 intersection point will be caused by the line segment from w 1 to w 3 and the segment from v 3 to v 4. This indicates that s 5 should be removed, since this segment cannot cause an intersection point between the two paths. (a) (b) Figure 3.4: Visualization of situations that could occur for respectively case 1 and case 2. For case 2, a possible situation is depicted in Figure 3.4b. The edges of interest are s 5 from B1 and t 1 from B2. The endpoint of t 1, w 2, is on the line extending s 5. It follows by convexity that the starting point of t 1, w 1, is right of this line. Since no intersection point was found between the line segments s 5 and t 1, w 2 is located either after v 6 or before v 5. If the former is the case, there can be no intersection between the two paths, so it is irrelevant which edge is removed. If the latter is the case, edge t 1 cannot be causing an intersection point since all other edges of path B1 are left of the extended line from s 5. Therefore, edge t 1 is erased in this situation. If case 3 applies, there is another subdivision into two options. Option 3a, implying that the starting point of b 2 is to the right of b 1 while the endpoint of b 2 is to the left. In this situation b 1 has to be removed. The other option is case 3b. This corresponds to the starting point of b 2 being left of b 1 while the endpoint is right. This cannot occur in the setting we have, so whenever the signs of d 1 and d 2 are different, we can erase b 1. Lastly, we have case 4, which applies if d 1 and d 2 have the same sign. It is again split up into two options. Either case 4a applies, indicating that b 2 is completely to the left of b 1. Or case 4b applies, meaning that b 2 is completely to the right of b 1. In the problem setting we have here, the former case can not occur. Hence, whenever d 1 and d 2 have the same sign, b 2 is to the right of b 1 leading to the decision to erase b2. Summarizing, there are two options: ˆ b 1 is erased, this occurs for case 1 and for case 3; ˆ b 2 is erased, this occurs for case 2 and for case 4. The pseudocode for the algorithm that decides which of the line segments needs to be erased is presented in algorithm 2. 15

21 Algorithm 2 Determine the location of two non-intersecting line segments with respect to one another to decide which one to erase. 1: function findlocation( v, w) Input: Two line segments v = (a, b) and w = (c, d). Output: A Boolean variable indicating which line segments needs to be removed. 2: arc v 3: arc1 c a 4: arc2 d a 5: dir1 det(arc, arc1) 6: dir2 det(arc, arc2) 7: if det1 = 0 then 8: location f alse 9: else if det2 = 0 then 10: location true 11: else if det1 det2 < 0 then 12: location f alse 13: else 14: location true 15: end if 16: return location 17: end function To further clarify this procedure, two paths are displayed in Figure 3.5, where the goal is to find the intersection point. Here B1 is the directed path from vertex v 1 to vertex v 6, and B2 is the directed path form vertex w 1 to w 5. Notice that these edges are translated edges from the convex polygon A by their corresponding extreme vertex from polygon B. This implies that every edge within the path makes a left turn with respect to its predecessor edge. Since the edges that form B2 are traversed after the edges from B1, the first edge of B2 makes a left turn with respect to the last edge of B1. This fact helps us decide which of the edges to erase if no intersection point is found between the two. In Figure 3.5a, the edges of interest are s 5 from path B1 and t 1 from B2. The extended line from segment s 5 is drawn to visualize the location of t 1 with respect to s 5. It can be seen that both w 1 and w 2 are on the same side (right) of the line. This corresponds to situation 4 leading to the removal of edge t 1. In Figure 3.5b this edge is removed and the new edge of interest for path B2 is t 2. Now, w 2 is on the other side of the line compared to w 3. Now case 3 applies and hence, s 5 will be removed. The new situation is depicted in Figure 3.5c with edges of interest s 4 and t 2. The extended line from s 4 is drawn and it can be seen that both w 2 and w 3 are on the same side of this line. Here, edge t 2 is removed since the situation that applies is case 4. In Figure 3.5d the edge of interest for path B2 is t 3. Its starting point w 3 is to the right of the line extending s 4 and its endpoint w 4 is to the left of this same line. This implies that case 3 applies which leads to the removal of s 4. This removal induces the situation shown in Figure 3.5e. The edges of interest are s 3 and t 3. These line segments give rise to an intersection point which ends the procedure. In the list where B1 is stored, the last element is erased (in this example this corresponds 16

22 to v 4 ). Then the intersection point is added, followed by all vertices of B2 except the first (in this case w 3 ). As soon as the list is added we move back to the procedure explained section This process will be repeated until all edges of polygon A and all vertices of polygon B have been visited. It could occur that there is no intersection point between two paths. The function that is trying to find the intersection between the paths, will terminate whenever one of the paths is empty. Depending on which of the paths became empty, the algorithm should proceed with either one of them. If B1 is empty, this implies that B2 is left of B1, indicating that B1 cannot be part of the IFP. So, B2 is now saved as the current IFP and the process continues by finding the next list of translated line segments. If on the other hand B2 is empty, this list can simply be skipped since it is to the right compared to B1 and therefore cannot be part of the IFP. Due to the convexity we are certain to find an IFP when all paths are processed. When the last edge of polygon A is reached, the IFP needs to be closed. This is done by finding an intersection point between the front and the back of the list of IFP vertices that we have. If there is no intersection point between the first and the last line segment, either one of them is erased, depending on the location. This is repeated until an intersection point is found. Again, due to convexity, there will be an intersection point. Hence, we have constructed the IFP for two convex polygons in an efficient manner. 17

23 Algorithm 3 Computing the IFP for two convex polygons. 1: function MinkowskiDifference(A,B ) Input: Two convex polygons A and B starting with the rightmost vertex Output: The inner-fit polygon of A and B, computed using IFP=A B 2: Initialize vertexa, vertexb, arca and arcb as the first elements 3: while not all arcs of polygon A are traversed do 4: direction det(arca, arcb) 5: if direction 0 then 6: The starting point of arcb is the extreme vertex for arca 7: arcc (arcaf rom vertexb, arcat o vertexb) 8: add arcc to list of vertexb and go to next arca and vertexa 9: else 10: The correct extreme vertex is not found 11: B2 list corresponding to vertexb 12: if this is the first list found then 13: add it to list of possibleif P 14: else 15: while B2 has not been added do 16: B1 possibleif P 17: b1 last arc of B1, b2 first arc of B2 18: (bool, point) findintersectionpoint(b1, b2) 19: while no intersection do 20: boollocation findlocation(b1, b2) 21: erase b1 or b2 depending on boollocation 22: if B1 is empty then 23: possibleif P B2 24: addedb2 true 25: end if 26: if B2 is empty then 27: addedb2 true 28: end if 29: end while 30: there is an intersection point 31: add point and remaining vertices of B2 to possibleif P 32: addedb2 true 33: end while 34: end if 35: end if 36: go to the next arcb and vertexb 37: end while 38: find intersection between first and last line segments to obtain IFP 39: return IFP 40: end function 18

24 (a) Extend line segment s 5 to visualize the location of t 1 with respect to s 5. (b) Edge t 1 is removed, new edge of interest for B2 is t 2. Investigate its location with respect to s 5. (c) Edge s 5 is removed, new edge of interest for B1 is s 4. (d) Investigate the location of t 2 with respect to s 4. (e) Edge t 2 is removed, new edge of interest for B2 is t 3. Investigate its location with respect to s 4. (f) Edge s 4 is removed, intersection point is found between s 3 and t 3. Figure 3.5: Determining the intersection point between line segments by repeatedly erasing an edge if no intersection point is found. 19

25 Chapter 4 No-Fit Polygon In this chapter, the NFP (No-Fit Polygon) will be explained. The NFP is a polygon describing the feasible locations for two polygons such that they do not overlap. Whenever an item is placed with its reference point on the boundary or outside the NFP, it can be concluded that the two polygons do not overlap. So the NFP will help us placing the items close to one another without overlap. An example is shown in Figure 4.1. Here, the black polygon is fixed and another polygon is placed without overlapping the black polygon if its reference point is located outside or on the NFP, which is the red polygon. If the concept of the NFP is used, it is not necessary to perform geometric checks to determine overlap. The NFP of two convex polygons can be constructed efficiently. In section 4.1, the relation between the Minkowski sum and the NFP is described. In section 4.2 the algorithm to compute the NFP will be described. (a) (b) (c) (d) Figure 4.1: Examples of the NFP: the item in black is the fixed item and the items in blue are the free items. The no-fit polygon of the two is shown in red. The bottom-left corner of the grey boxes corresponds to the reference point. 20

26 For the NFP of polygon B that moves freely around polygon A, we use the notation NFP AB. If polygon B is placed on the boundary of NFP AB, the polygons A and B are touching. If it is placed in the interior of NFP AB, the polygons overlap. And if the polygon is placed on on the exterior of NFP AB, then there is no overlap between the polygons. 4.1 Minkowski sum Just as the IFP is equivalent to the Minkowski difference, the NFP is equivalent to the Minkowski sum. Definition 4.1 (Minkowski Sum). A B = b B A b, (4.1) where A b = {a + b a A} represents the set A translated by b. Say that we would like to place two items A and B without overlap. Here, A is the fixed polygon which has been placed already and B is the free polygon, which can be moved around. We claim that the NFP of A and B is given by A ( B). To show that this claim holds, the following theorem is proved. Theorem 4.2. x A ( B) A B x Proof. ( ): Let x A ( B). Then, by definition x {a + b a A}. b B This indicates that b B such that x = a b, a A. Thus, a = b + x, indicating that both a A and a B x. This implies that A B x. ( ): Let y A B x, so we have y A and y B x. This means that b B such that y = b + x. From this, it follows that x = y b = y + ( b) = y + b. Since b B we know that b = b B. Hence, y + b A ( B) so x A ( B). In Figure 4.2 an example of the Minkowski sum of two polygons is depicted. Every edge of polygon A is translated by every vertex of polygon B. This results in n new (overlapping) polygons, where n is the number of vertices of B. The union of these polygons if the Minkowski sum A B. The Minkowski sum represents the set of translation vectors which, when applied to the free polygon B, result in polygons A and B overlapping. Figure 4.2: The Minkowski sum of polygons A and B: the union of translated line segments of A by vertices of B. 21

27 Hence, the set of feasible translation vector consists of all translation vector except the ones contained in the Minkowski sum polygon. This polygon can be translated in such a way the the polygon directly represent the set of unfeasible placements positions of polygon B. This happens automatically when the reference point of polygon B is the origin. As proven in Theorem 4.2, the NFP is related to the Minkowski sum in the sense that NFP AB = A ( B). Hence, the NFP of A and B can be computed using the Minkowski sum. Notice that the input needs to be provided in a correct way. Just as for the IFP, the input needed is B instead of B. However, B itself is needed to determine the location of the NFP with respect to the fixed polygon A. The reversion of B to B will take place in the algorithm itself so that both polygons can be used for calculations. Since a lot of the translated vertices will be in the interior of the NFP, it is unnecessary to compute the additions of all vertices. Because it is also very time consuming, there has been a lot of research to find better and faster methods for the NFP construction. In the case that both polygons are convex, which will be assumed for now, there is an easy method running in O(m + n) time, where m and n are the number of vertices for polygon A and polygon B respectively. 4.2 Algorithm We use the same notation as before for the polygons A, B and their vertices and edges. Polygon A is the fixed polygon and is located with its reference point on the origin. Then polygon B, the free polygon moves around polygon A where the path of its reference point describes the NFP. This path can be constructed by simply putting the edges of polygon A and B = B in increasing slope order. The first step is to find the correct location of the NFP with respect to fixed polygon A. This is done by placing polygon B in a position where it touches polygon A. Let vertex p B y min be the vertex of B with smallest y-coordinate (if there are more than one, the vertex among these with the smallest x-coordinate). Furthermore, let vertex p A y max be the vertex of A with largest y-coordinate (if there are more than one, the vertex among these with the largest x-coordinate). By letting these two vertices, p A y max and p B y min touch one another, we have found a feasible placement where polygons A and B are touching and not overlapping. The reference point of polygon B is added to the list of NFP vertices. From here on we can start adding arcs to the NFP. The list of edges for polygon A is ordered so that the polygon is traversed in the counterclockwise direction and the first edge is the one with starting point p A y max. Since NFP AB is computed by A B, we need the edges of polygon B. Polygon B is constructed by mirroring all vertices (and translating the polygon such that the reference point is the origin). For the NFP construction we only need the edges of the polygon and are not interested in its precise location. Therefore, this translation is not necessary. The edges of polygon B are ordered such that the first edge is the one with starting point p B y max. 22