Determining Differences between Two Sets of Polygons

Transcription

1 Determining Differences between Two Sets of Polygons MATEJ GOMBOŠI, BORUT ŽALIK Institute for Computer Science Faculty of Electrical Engineering and Computer Science, University of Maribor Smetanova 7, SI-000 Maribor SLOVENIA Abstract: - In this article, a new algorithm for determining differences between two sets of polygons is considered. It has been proven that it is enough to test only the polygon edges to perform the comparison. The algorithm works in two steps. First, the polygon edges are inserted into data structure and eliminated by a simple criterion. In the second part, the algorithm analyses the remaining edges and prepares a report about the detected differences. To speed-up the geometric search, the algorithm uses a combination of a uniform space subdivision and binary searching trees. It has been shown that the algorithm is numerically stable, and works well also with the limited set of non-simple polygons occurring in practice. The algorithm's time complexity is O(n log n), and space complexity O(n), where n is the common number of edges in both sets of polygons. That has been confirmed by experiments using data sets from a land cadastre. Key-Words Geographical information systems (GIS), polygons, polygon comparison, computational geometry, algorithm. Introduction In this paper we discuss an algorithm for the problem that appears frequently in the land cadastre. As the land cadastre is changing during the time (parcels are split, merged or their textual attributes are updated), the land owners, the authorities supervising the cadastral information, and the companies developing GIS solutions, are many times interested in changes occurring during the time. Figure shows a typical example of cadastral data. Although the left and the right set of polygons seem completely identical, there are 993 geometrical differences. each polygon of the first set with all polygons in the second set, resulting in O(n ) time complexity (n means the number of polygons). Better solution would be to determine the Boolean intersection upon two sets of polygons using sweep-line technique [5],[]. However, polygons in land cadastre coincide along the edges, which represents the boundary case for Boolean operations. In this paper, an alternative approach to determining the differences between two sets of polygons is introduced. As will be shown, the algorithm is fast and easy to implement. Background Definition : Let us have n points p 0, p,..., p n- in the plane and let us connect them by n line segments l 0 = p 0 p, l = p p,..., l n- = p n- p 0. They describe a simple polygon if [7]: point p i is a junction of exactly two line segments, non-neighbouring edges do not intersect neither touch. Figure : Two sets of different polygons. By the knowledge of authors, this problem has not been considered adequately up to now. The simplest solution would be to use the brute force algorithm comparing Points p i, where line segments meet, are usually considered as vertices (v i ), and the line segments as edges (e i ). Any simple polygon splits the plane into two domains, a bounded interior and the unbounded exterior, if it does

2 not contain holes. Otherwise, the plane is divided into finite number of domains, among which only one is unbounded. A closed sequence of edges bordering the bounded domain and the unbounded one is known as a loop L [6]. The holes are also described by the closed sequence of edges named rings (R i, 0 i < r, where r is the number of rings in the polygon). The rings can be nested. Polygons violating these conditions are considered as non-simple polygons. The majority of algorithms in computer graphics and computational geometry require simple polygons at the input. However, in some applications (e.g., GIS), the set of simple polygons is not sufficient to describe all possible cases. Therefore, a set of so-called legal polygons has been introduced by Žalik []. For the purpose of this application, the definition is modified a bit. Definition : The polygon is legal, if it consists of one or finite number of simple polygons, or if at polygon vertices more than two edges meet. In figure a and b the legal polygons are shown. Figure a shows an example of a polygon consisting of three loops. This situation can be found in land cadastre when a parcel is split in more pieces by a street, for example. In figure b at some polygon's vertices more than two edges meet. The polygon in figure c is non-simple and non-legal, because its edges have a common point, which is not a polygon vertex. v 8 v v 0 v 6 v 3 v 7 v 5 v 9 v 7 v 8 v 6 v v 4 v 0 v v 0 a) b) c) Figure : Legal (a, b) and non-legal polygon (c). Let us have two sets of polygons Φ = {P 0, P,, P ϕ } and Θ = {Q 0, Q,, Q θ }, and let us denote set of all edges existing in Φ and Θ as E Φ and E Θ. Let us define: Definition 3: Edge e iφ = v i, v i, from E Φ and edge e jθ = v j, v j, from E Θ are twins if { v i,, v i, }= { v j,, v j, }. Definition 4: Two sets of polygons are equivalent if every edge from set E Φ has its twin in E Θ and vice versa. Let us suppose edges e iφ from E Φ and e iθ from E Θ are twins. If they are not, they differ in at least one position v 5 v 4 v 3 v v 3 v v of their vertices. In this case E Φ E Θ and therefore, Φ and Θ are different. Using Definition 4, it is easy to see that the problem of determining whether two sets of polygons are equivalent can be transformed to the search of twin edges regardless of any other polygon information. The polygons can even be non-legal and non-simple for that conclusion. In the application mentioned in the introduction, we know that the set of polygons Θ has been derived from the set Φ. We are not interested only in whether two sets are the same or not, but we would like to know, which polygons (parcels) have been changed. 3 The algorithm The algorithm for determining the differences between two sets of polygons works in three steps: initialization of the geometric searching structure, elimination of twin edges, and analysis of the remaining edges. 3. Initialization of the geometric searching structure One of the most basic tasks of the computational geometry is determination, whether two geometric entities are the same. Best results are achieved when some computer memory is dedicated for the adequate geometric searching data structure. There are two main ideas, how this should be done: Uniform plane subdivision splits the area of interest into equally sized cells [8]. It has been shown that the answer about presence of a geometric entity could be returned in a constant time O(), if the number of cells is large enough []. However, in practice this cannot be always assured, especially, if the number of the elements is huge, and if they are distributed highly non-uniformly. Despite that, the uniform plane subdivision remains among the most favourite speed-up techniques applied at many geometric problems[],[3],[4]). When highly non-uniform distribution of the data is expected, hierarchical plane subdivisions become attractive. There are different ideas about organization of the data, as for example quad trees, K-D trees, or BSP trees (see for a large collection of possibilities in [9],[0]. All the operations and the queries require logarithmic time and their implementation is a bit more complicated, too. A combination of both approaches is used in proposed algorithm. The area where the polygons are located is split into equally sized cells at first. Edges in each cell are stored in a binary searching tree, which is organized

3 according to the smallest x coordinate of both vertices. In this way, we reduce the drawback of the uniform space subdivision when non-uniform distribution of the input data is expected and the depth of an individual searching tree is reduced. Figure 3 schematically shows the organization of the data structure for the geometric search. Figure 3: Geometric data storage. The most interesting question when using the uniform space subdivision is a suitable selection of the number and the size of the cells. One could guess: more cells, better results. However, that is only partially true. Larger number of cells requires longer initialization time and consumes more memory. Therefore, different simple heuristics have been proposed to construct the most suitable uniform plane subdivision [3]. We calculate the number of cells (equation ()) by taking into consideration an expected number of edges inside a set (n), and the shape of the bounding box (ratio) surrounding the polygons. NoOfCells = ratio n n NoOfCells = x y ratio () As we do not want to count all the edges at the initial stage of the algorithm, their number has been estimated as shown in equation (). The constant 8.5 has been obtained by considering over 5 million parcels of the national land cadastre. n = 8. 5NoOfPolygo ns () taking the polygons one after another. The edges of the polygon are inserted into described data structures. Every edge is equipped with the following data: Coordinates of both edge vertices. Information about the polygons to which the edge belongs: o polygon indices, o flags indicating to which polygon set the edge belongs, o if the edge is part of a ring, the following additional data are stored: index of the parent polygon, consecutive number of the hole inside parent polygon. Edge counter (EC) identifies how many times the edge has been met. Its role will be explained later. In a land cadastre, the holes in polygons are usually not empty, because other parcels cover them. Figure 4 shows a common example. The loop of the polygon P, covers all edges of the ring R 0,0 in polygon P 0. The parent index of the ring edges would in this case be 0. P 0 P R Figure 4: Polygon information. After processing the first set of polygons, all edges are waiting in our data structure. Those edges being met only once are located on the border of the set of polygons, while the inner edges are visited twice. Edge counter EC stores these values (figure 5 shows an example). 0,0 Variable ratio stores the ratio of the sides of the bounding box (equation (3)). x max x min ratio = (3) y max y min The size of the cells is now simply determined as shown in equation (4): CellSize x xmax xmin = NoOfCells x CellSize y ymax ymin = (4) NoOfCells 3. Elimination of twin edges At first, the algorithm uses the data in the first file, y Figure 5: Interior and outer border edges. After that, the polygons from the second set are being used. The algorithm solves concurrently two main tasks:. It examines what changes occur in the geometrical part of parcel descriptions. 3

4 . It checks the associated non-geometric information. Because of that, we have to get the identifier of the corresponding polygon being located in the first set. Let us suppose we are processing polygon Q i from the second set Θ. The edges of Q i are inserted one by one into data structure. There are two possibilities:. The edge does not exist in the data structure. In this case, a new edge equipped by all the necessary information is inserted.. The edge meets its twin. The information found in the corresponding record is extended with the information carried by the considered edge, and after that, the edge counter is increased. Let us clarify the actions using a few simple examples. In figure 6, two sets Φ and Θ of identical polygons exists. The set Φ has already been processed, and let us suppose, we start with polygon Q 0 from set Θ. The algorithm easily discovers that all the edges of Q 0 meet their twins in data structure. The ECs of those edges are increased and have the value 3 now. As mentioned, we also have to access the covered polygon in set Φ to check the textual information. Each vertex stores the pointers to at most two polygons from set Φ. In our case, we would obtain the following list for pairs of pointers to polygons P 0 P 4, P P 4, P P 4, and P3P4. Actually, the algorithm remembers only the first obtained pair and compares it with the arriving pairs. As soon as two different pairs are obtained, the polygon pointer pointing to the same polygon represents the solution. If the pair represents the border edge of the polygon set, it points exactly to the searched polygon. In our case, the searched polygon is P 4. It is easy to compare the associated textual information of both polygons (Q 0 and P 4 ) and report differences. P P0 P4 Q0 Q P3 Figure 6: Two sets of identical polygons. A special case when determining the covered polygon appears at legal polygons (see section ). Figure 7 shows such an example. It consists of three polygons. Polygon P 0 is not simple, because at vertex v more than two edges meet. Polygon P 0 contains only one ring (R 0,0 ), which coincides with the loop of polygon P. The polygon P fills the empty space caused by the legal Q Q3 P Q4 loop of polygon P 0. The second set Θ contains exactly the same polygons as the first set. In this case, problems occur at determination of the identifier of the polygon P. Namely, all pairs of pointers to polygons are the same (P 0 and P ). If that happens, the bounding boxes of both polygons are compared and the wrong polygon is easily eliminated. Similar situation appears with the polygons covering the holes (see polygon P and ring R 0,0 in figure 7). All the edges belonging to the ring are common to polygon P 0 and P. However, this time the edges of polygon P 0 carry the information that they also belong to the hole and therefore, P is easily selected. P0 v P Figure 7: Determination of the covered polygon. In the case that not all polygons are the same, edge counter EC takes the values presented in table. EC Actions needed edge was not covered and has to be analysed if edge is the border edge of the data set (it has only one polygon pointer), it should be removed, otherwise, it has to be analysed 3 edge has to be analysed 4 edge should be removed Table : The meaning of edge counter. Figure 8 shows some characteristic possibilities. The input sets are shown in figures 8a in 8b, while figure 8c shows the values of EC. Let us analyze the remaining edges. All edges of polygon Q 0 have the value EC =. This means, that Q 0 has been added and that it has no common edges with the rest of polygons in Θ. Similarly, we also know that such polygon has been erased if it existed in Φ. Very similar case appears, when some edges of the same polygon have EC = 3 and some (edges of polygons Q 3 and Q 5 ). In this case, the polygon has been added (erased), but this time it is connected with the other polygons from the set. However, these situations are a generalisation of the problem and should not normally occur in land cadastre. The next two examples are characteristic for land cadastre operations. All the remaining edges have EC =. The remaining edge, which has border polygons P 4 R0,0 P 4

5 and P 5 indicates merging of these two polygons. This edge has pointers to two different polygons from the first set. Similarly, the remaining edges between Q 3 and Q 4 denote splitting. The edges have two pointers to two polygons in the second set. P 5 P4 P P0 P P3 Q0 Q7 Q Q6 Q Q4 a) b) c) Figure 8: Differences analysis: the input (a, b) and output (c). 4 Time and space complexity estimation Let us suppose there are ϕ polygons in the first set having n edges, and θ polygons in the second set with m edges. We can suppose m n. At first, cells of the uniform subdivision are allocated. This is done according to equation (), and therefore, the initialization is done in linear time O(n). For every edge, the insertion into data structure is done in a logarithmic time, because binary tree is used for storing the edges. The average number of edges in a cell is only c, where c tends to in the case of uniform distribution of data. However, in the worst case, all edges except one are located in one cell, and then c = n. Therefore, the total time for inserting all edges is T(n) = O(n log n). Adequately, the processing of the second polygon set is estimated with O(n log n), too. At the end, there are k n edges in the data structure. Each is accessible in the logarithmic time. By the stored information, every edge is processed in a constant time, and therefore this is also finished in O(n log n). However, if both sets are completely the same, this step is not needed at all, because k = 0. Therefore, we can conclude that the time complexity of the presented algorithm is O(n log n). 3 Q3 Q5 In the worst case, we have to store every edge from both sets. For that we need O(n) memory locations. The analysis of the edges does not require additional space. Therefore, the algorithm solves the given task with linear space complexity O(n). In the continuation we give the actual run times of the algorithm while solving our testing data sets. The polygon sets present the real data from a GIS database. Here, three pairs of sets of polygons are tested and table introduces the basic information about them. Table 3 gives used number of cells (NoOfCells x, NoOfCells y ) of the uniform plane subdivision, and spent CPU time for individual steps of the algorithm. Each data set has been measured three times with minor deviation in spent CPU time (up to 4%). It is interesting to see, how efficient the proposed heuristic introduced in subsection 3. is. Table 4 gives the total spent CPU time and used amount of memory for the third pair of polygon sets regarding the number of cells of the uniform plane subdivision. As seen, the shortest CPU time is achieved when number of cells obtained by the heuristic is applied (graph in figure 9a gives graphical representation). As mentioned before, larger number of cells does not mean faster execution time of the algorithm as too much time is spent on memory allocation and initialisation of the huge number of cells. Clearly, memory requirement is almost linearly dependent on number of cells as shown in figure 9b. Table 5 and figure 0 show the influence of varying the parameter for average number of polygon edges. The results confirm that the estimation is adequate. Parameter Pair Pair Pair 3 Num. of polygons in each set Num. of edges in each set Num. of changed polygons (geometry) Num. of changed polygons (attributes) Num. of changed edges Table : Information about the input sets. Parameter Pair Pair Pair 3 (NumOfCells x, NumOfCells y ) adding edges (. set) adding edges (. set) ESRI files access total.3..5 Table 3: CPU time (s) for testing pair of polygons. Let us also consider the memory complexity. As already mentioned, for the uniform grid, O(n) cells are needed. 5

6 Size of UPS CPU time Memory = = = = = = = = Table 4: CPU time (s) and memory requirements (Mb). Estimated number of polygon edges CPU time ( 97) (56 484) (956 84).5 0 (4 969).8 5 ( ) (49 939) 30. Table 5: CPU time for estimated number of edges. CPU Time [s] Memory [Mb] 34,00 3,00 30,00 8,00 6,00 4,00,00 0,00 0,0 0,5,0,5,0,5 3,0 3,5 4,0 4,5 5,0 5,5 6,0 Number of cells (in Millions) ,0 0,5,0,5,0,5 3,0 3,5 4,0 4,5 5,0 5,5 6,0 a) Number of cells (in Millions) Figure 9: CPU time (a) and memory requirement (b). CPU Time [s] b) Estimated number of edges Figure 0: Estimated number of polygon edges.. 5 Conclusion The paper introduces a new algorithm for comparing two sets of polygons. An efficient algorithm has been obtained by reducing the comparison of two polygon sets to their edges. An efficient data structure has been proposed consisting of the uniform plane subdivision at the first level, and the binary searching trees at the second. For the uniform plane subdivision, adequate heuristics have been introduced. The heuristics have been confirmed by experiments, too. The theoretical estimation for the worst case time complexity of the geometric part of the algorithm is O(n log n) and space complexity O(n), where n is the number of polygon edges. The algorithm copes well also with the limited set of non-simple polygons (named legal polygons), which appear occasionally in practice. Their existence does not slow down the algorithm neither spoils its stability. The algorithm has been used successfully in practice. References: [] Bentley, J., Weide, B., Andrew, C., Optimal Expected-Time Algorithms for the Closest Point Problems, ACM Transaction on Mathematical Software, Vol. 6, No. 4, 980, pp [] Brobst, S., Grant, S., Thompson, F., Partitioning Very Large Database Tables with Oracle8. Oracle Magazine, Vol., No., 999, pp [3] Fang, T., Piegl, L., Delaunay Triangulation Using a Uniform Grid. IEEE Computer Graphics & Applications, Vol. 3, No. 3, 993, pp [4] Fujimoto, A., Tanaka, T., Iwata, K., ARTS: Accelerated Ray-Tracing System. IEEE Computer Graphics & Applications, Vol. 6, No. 4, 986, pp [5] [6] Mortenson, M., Geometric Modelling, John Wiley, 985. [7] O Rourke, J., Computational Geometry in C, Cambridge University Press, 993. [8] Preparata, F., Shamos, M., Computational Geometry: An Introduction, Springer, 985. [9] Samet, H., The Design and Analysis of Spatial Data Structures, Addison Wesley, 989. [0] Samet, H., Applications of Spatial Data Structures, Addison Wesley, 990. [] Vatti, B., A generic solution to polygon clipping. Communications of the ACM, Vol. 35, No. 7, 99, pp [] Žalik, B., Merging a set of polygons. Computers & Graphics, Vol. 5, No., 00, pp The algorithm has been written in C++, and tested on a PC with Intel Celeron 300 MHz processor, and Gbytes RAM under Windows 000 operating system. 6