2009 International Conference on Future Computer and Communication
An algorithm for visibility graph recognition on simple case
Hossein Morady
Gholamreza Dehghani
Islamic Azad University, Allame Majlesi Branch Isfahan, Islamic republic of Iran
[email protected]
Islamic Azad University, Allame Majlesi Branch Isfahan, Islamic republic of Iran
[email protected]
The Case that mentioned and problem solved on it, is as following. Graph has two maximal cliques that these maximal cliques have only two common nodes (it is clear that other nodes of maximal cliques may have other relations together) and union of nodes of maximal cliques is exactly equal with set of graph nodes. Example of described graph is shown in fig 1.
Abstract— In this paper, Visibility graph recognition problem will be attended. Since, in graph theory, this problem is a famous open problem and because of complexity, no solution has been found for it, so far. Therefore here it is restricted to the special class of graphs and then solved. Because of having related properties with main problem, one special case was chosen. Finally an approach for visibility graph recognition on this special case of graphs has been developed. Proposed approach gets graphs which satisfy this special case and recognizes whether it is a visibility graph. Keywords-Computational geometry, graph, interval graph, Polygon, Visibility graph.
I.
INTRODUCTION
Visibility graphs are fundamental structures in computational geometry. They have applications in areas such as graphics and robotics. To decide whether a given graph is the visibility graph of some simple polygon, is not known to be NP, nor is it known to be NP-hard. It is only known to be PSPACE[2]. From the characterization standpoint, Ghosh in [5] presented first four necessary conditions for internal visibility graphs of a simple polygon but it was shown by Everett [6] that they are not sufficient even for triconnected graphs. Coullard and Lubiw [7] have shown some additional necessary conditions, but also have shown that they are not sufficient. Abello et al. in [8] have strengthened these results by showing that the proposed conditions are not sufficient, even for triconnected graphs. Abello & Kummar have also shown some other necessary conditions [3]. The only complete characterizations obtained to date have been for internal visibility graphs of spiral polygons. Positive results have also been obtained in orthogonal polygons [2]. However, most of these results are fairly specialized and shed little light on the general problem. In this paper an approach for visibility graphs recognition on the simple case of graphs presented. In continue, we discuss about this approach. II.
Figure 1.
In this paper, visibility graph recognition on case which described, has studied and simple approach for it proposed. This approach is based on interval graphs. It is necessary to introduce interval graphs. brief explanation about these graphs presented III.
A graph G = (V, E) with vertices V1,...,Vn is a interval graph, iff there exists open intervals such as T1,...,Tn such that: T1∩Tn ≠ Φ ↔ (V1, Vn) Є E Let G is an arbitrary graph and V0 be the one node of G. set of nodes of G that have relation with V0, named R(V0). R(V0) could decomposed to four classes. These classes are as following. i) Interior V0, In(V0): x Є In(V0) ↔ ((( x,y) Є E(G) ↔ (V0 , y) Є E(G) ) ٨ ( Э y0 , (x0,y0) ¢ E(G) ٨ (V0, y0) Є E(G) ) ٨ (x Є R(V0) ) ) Interior of V0 is subset of R(V0) such that (x Є In(V0)) iff each node was related with x then is related with V0 and
PROBLEM DEFINITION
978-0-7695-3591-3/09 $25.00 © 2009 IEEE DOI 10.1109/ICFCC.2009.130
INTERVAL GRAPHS
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there is the node that is related with V0 and is not related with x. ii) Equal V0, Eq(V0): x Є Eq(V0) ↔ ((( x,y) Є E(G) ↔ (V0 , y) Є E(G) ) ٨ (x Є R(V0) ) ) Equal of V0 is subset of R(V0) such that (x Є Eq(V0)) iff each node was related with x then is related with V0 and reverse. iii) Dominant V0, Do(V0): x Є Do (V0) ↔ (((V0,y) Є E(G) ↔ (y , x) Є E(G) ) ٨ ( Э y0 , (x0,y0) Є E(G) ٨ (V0, y0) ¢ E(G) ) ٨ (x Є R(V0) ) ) Scilicet x Є Do (V0) ↔ V0 Є In(X0 ) iv) ∆(V0): Δ( V0) = R(V0) – (Eq(V0) U Do(V0) U In(V0) ) Clearly, these four classes are distinct and each graph could decompose to these classes IV.
V1
x
V2
y Figure 2.
Connect all nodes of V1 to x and y and continue to cross other area. Distance among crossing points is named viewpoint of that node. V1
x
V2
AN IMPORTANT DEFINITION
Elementary dominant: node A is Elementary dominant of node B if satisfy these conditions: 1. B Є In(A) 2. Node A has left element that, is not related with B or this left element has common interior with A that this common interior is not related with B. 3. Node A has right element that, is not related with B or this right element has common interior with A that this common interior is not related with B. V.
y Figure 3.
Basic assumption in this approach is each viewpoint considered as an interval. After secondary graph construction, these intervals correspond to nodes of interval graph. It is clear that if secondary graph be an interval graph then we can show, arrangement for nodes of V1 and V2 such that nodes of V2–{x,y}which are related with nodes of V1, lie exactly on viewpoint of corresponding node. Node position changing on boundary of first area makes position changing of start point and end point of viewpoint in next area. So interval graph haven’t elementary dominant because if in an interval graph node A be elementary dominant of B, then interval of B must be subinterval of interval of A and interval of B couldn’t exited from interval of A. Proof ii) If secondary graph of G is interval graph which haven’t elementary dominant, we must show, arrangement for nodes of V1 and V2 such that nodes of V1 and V2 which are related with together, see together. Let resulted interval graph have M component. Also there are some nodes of V1 and V2 that are not relate with node of against area. Set of this node in first area and second area named N1 and N2, respectively. Also number of this node in first area and second area assumed K1 and K2, respectively. M + K1 + K2, pieces distinctly assume on the boundary of first area. Obtain viewpoints of these pieces and named corresponding pieces. Now set perimeter of two area and distance among common nodes such that could consider M + K1 + K2, pieces (assume distance among common nodes is little).
APPROACH
Let G has two maximal cliques’ m1 and m2 that these maximal cliques have common nodes x and y and union of nodes of maximal cliques is exactly equal with set of graph nodes. Set of nodes of maximal cliques’ m1 and m2 shown by V1 and V2 respectively. The secondary graph of G is constructed. The secondary graph construction of G is as following, nodes of secondary graph are V1– {x,y}. Two nodes of this graph are related with together if those were related with common node of V1– {x,y. Two nodes of this graph are not related with together if there is not common node of V1– {x,y} that was related with those nodes. Theorem 1: Graph G that satisfies mentioned condition, is a visibility graph iff secondary graph of G is an interval graphs which haven’t elementary dominant. Proof: to prove this theorem, two following problems must be proven: i) If G is a visibility graph, then secondary graph of G, is interval graph which haven’t elementary dominant. ii) If secondary graph of G, is interval graph which haven’t elementary dominant, then G is a visibility graph. Proof i) general view of case is shown by figure 2.
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of graphs and finally solving this problem for general graphs.
Start from an arbitrary component and select any piece arbitrarily. The nodes of this piece, arrange on first area in accordance with arrangement resulted from corresponding interval graph of component. The nodes that should see together lie on corresponding piece in areas. Repeat this for other components to all components arranged. All K1+ M nodes of N1 lie in first area and All K2 nodes of N2 lie in corresponding viewpoints of second area. So obtain target arrangement. In accordance with mentioned approach above, to recognize visibility graph for graphs which satisfy described conditions, first maximal cliques and their common nodes determined then secondary graph constructed. Finally determine whether secondary graph is interval graph. If secondary graph isn’t interval graph then graph isn’t visibility graph. Otherwise if interval graph haven’t elementary dominant then graph is visibility graph. Otherwise graph isn’t visibility graph.
REFERENCES [1]
[2]
[3]
[4] [5]
CONCLUSION
In this paper, recognition visibility graph problem considered and restricted this problem to simple case and solved. But this is couldn’t be goal. This work could be start point for solving visibility graph recognition problem in generalized case. Future works in this area are solving recognition visibility graph problem, for other special class
[6] [7]
[8]
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Chiuyuan Chen, Department of Applied Mathematics, National Chiao Tung University, A necessary condition for a graph to be the visibility graph of a simple polygon, Theoretical Computer Science 276 (2002) 417–424. Chiuyuan Chen, Kaiping Wu, Department of Applied Mathematics, National Chiao Tung University, Disproving a conjecture on planar visibility graphs Theoretical Computer Science 255 (2001) 659–665. James Abello and Krishna Kumar, Visibility Graphs and Oriented Matroids , Information Visualization Research, Shannon Laboratories, AT&T Labs-Research,180 Park Avenue, Florham Park, NJ 07932, USA,
[email protected]. Kurt Mehlhorn , Certifying Graph Recognition Algorithms , technical Report ,. Cambridge University Press, 1999. S.K. Ghosh. On Recognizing and Characterizing Visibility Graphs of Simple Polygons, pp. 132–139.LNCS, 318. Springer-Verlag, Berlin, 1988. H. Everett. Visibility Graph Recognition. Ph.D. Dissertation, Department of Computer Science, University of Toronto, 1990. C. Coullard and A. Lubiw. Distance Visibility Graphs. In Proceedings of the ACM Symposium on Computational Geometry, pp. 290–302, 1991. J. Abello, L. Hua and C. Pisupati. On Visibility Graphs of Simple Polygons. Congressus Numerantium, 90 (1992), 119–128.
