A Theory of Rectangularly Dualizable Graphs
AA Theory of Rectangularly Dualizable Graphs
Vinod Kumar ∗ , Krishnendra Shekhawat Abstract
A plane graph is called a rectangular graph if each of its edges can be orientedeither horizontally or vertically, each of its interior regions is a four-sided regionand all interior regions can be fitted in a rectangular enclosure. Only planargraphs can be dualized. If the dual of a plane graph is a rectangular graph , thenthe plane graph is a rectangularly dualizable graph .In 1985, Ko´zmi´nski and Kinnen presented a necessary and sufficient condi-tion for the existence of a rectangularly dualizable graph for a separable con-nected plane graph. In this paper, we present a counter example for which theconditions given by them for separable connected plane graphs fail and hence,we derive a necessary and sufficient condition for a plane graph to be a rectan-gularly dualizable graph . Keywords: plane graph, rectangularly dualizable graph, rectangular floorplan,VLSI circuit.
1. Introduction
The theory of rectangularly dualizable graphs plays an important role infloorplanning, particularly at large scale such as VLSI circuit design. It providesus information at early stage to decide whether a given plane graph can berealized by a rectangular floorplan (RFP). There exists a geometric dualityrelationship between plane graphs and rectangular floorplans (RFPs) which canbe described as follows:An RFP is a partition of a rectangle R into n rectangles R , R , . . . , R n suchthat no four of them meet at a point. A graph G is a dual of a plane graph G if the vertices of G correspond to the regions of G and for every pair ofadjacent vertices of G , the corresponding regions in G are adjacent. A planegraph is called a rectangular graph if each of its edges can be oriented eitherhorizontally or vertically, each of its interior regions is a four-sided region andall interior regions can be fitted in a rectangular enclosure. Only planar graphscan be dualized. If dual of a plane graph is a rectangular graph , then the plane ∗ Corresponding Author
Email addresses: [email protected] (Vinod Kumar), [email protected] (Krishnendra Shekhawat) Department of Mathematics, BITS Pilani, Pilani Campus, Rajasthan-333031, India a r X i v : . [ m a t h . C O ] F e b raph is a rectangularly dualizable graphs (RDG). Thus an RFP can be seen asan embedding of the dual of a planar graph and it can formally be describedas a rectangular dual graph of an RDG, i.e., for the dual of a RDG to be anRFP, we need to assign horizontal and vertical orientations to its edges. For abetter clarification, consider a planar graph G shown in Fig. 1a. We form itsextended graph (Fig. 1b.) by inserting cycle of length 4 at the exterior of G andthen connecting the vertices of the cycle to the exterior vertices of G . Then it isdualized in Fig. 1c. After assigning horizontal or vertical orientation to each ofits edges, an embedding as shown in Fig. 1d. is obtained. In fact, it is an RFP.Thus G is rectangularly dualized to an RFP. This transformation is known asthe rectangular dualization method which is well-studied in the literature. v v v v v v v v v v v v v v v v v v v v v v R R R R R R R R R R R R R R R R R R R R R R a b c dv b v l v r v u Figure 1: Rectangular dualization: (a) plane graph, (b) extended plane graph, (c) rectangulardual graph and (d) rectangular floorplan
RDG application
A VLSI system structure is described by a graph where vertices correspondto component modules and edges correspond to required interconnections. For agiven graph structure of a VLSI circuit, floorplanning is concerned with allocat-ing space to component modules and their interconnections [2]. An embeddingmethod given by Heller [2] enforces interconnection by abutment. Modules aredesigned in such a way that their connectors exactly match with their neighbors.Adapting this methodology, interconnections are coped with a clever design.Due to the advancement of VLSI technology, it is extremely large. On theother hand, an RDG can handle atmost 3 n − n is the number of modules. Consequent to this, its graph may not necessarilybe planar and hence in modern VLSI system, component modules and inter-connections can not be treated as independent entities. In such a situation,not all interconnections can be enforced by abutment. Linking the remaininginterconnections with nonadjacent modules utilize additional routing space. Forpracticality of solution, a graph described by a VLSI circuit can be embeddedin such a way that most of the interconnections can be made by abutment andthe remaining interconnections linking with nonadjacent modules use additional2outing space. For example, in Fig. 2d, R and R , R and R are intercon-nected through shaded areas R and R respectively. These routing areas areanticipated by introducing crossover vertices at a common point of intersectionof edges.The use of an RDG in floorplanning of a VLSI system can be illustrated bythe following example. Consider a graph described by a VLSI system as shownin Fig. 2a. Note that input-output connections between VLSI system andoutside world is represented by arrow heads. Although this graph is not planar,it is planarized by adding cross over vertices as shown in Fig. 2b. In order tosatisfy the necessary adjacency requirements, new edges (red edges) have beenadded in Fig. 2c. After these modifications, it is possible to construct an RFPas shown in Fig. 2d where a component rectangle R i is dualized to a vertex v i . v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v R R R R R R R R R R R a bc dv Figure 2: Constructing an RFP corresponding to a plane graph described by a VLSI system.
Previous Study
Investigations in the literature shows that the rectangular dualization theory[1, 4, 5, 6] of planar graphs is not much emphasized. It is known that every planegraph can be dualized, but not rectangularly dualized [1, 4, 5, 6]. Kozminski andKinnen [1] derived a necessary and sufficient condition for a plane triangulatedgraph to be an RDG and implemented it in quadratic time [7]. Later, Bhaskerand Sahni [4] improved this complexity to linear time implementing the rectan-gular dualization theory given by Kozminski and Kinnen [1]. Rinsma [5] showedthrough a counter example that it is not always possible for a vertex-weightedouter planar graph having 4 vertices of degree 2 to be an RDG. Besides thisproperty, there are infinite outer planar graphs that are not rectangularly du-alized. In fact, an outer planar graph having more than four critical shortcuts These vertices are introduced at the intersection of edges if it exists in order to embed agraph as a plane graph.
Gap in the existing work
Motivated by the following points, we here find a necessary and sufficientcondition for a given plane graph to be an RDG.i. Kozminski and Kinnen [1] found the following necessary and sufficient fora separable connected graph to be an RDG.
Theorem 1.1. [1, Theorem 3] Suppose that G is a separable connectedplane graph with each of its interior faces triangular. G is an RDG if andonly if(a) G has no separating triangle,(b) block neighborhood graph (BNG) is a path,(c) each maximal blocks corresponding to the endpoints of the BNGcontains at most 2 critical corner implying paths,(d) no other maximal block contains a critical corner implying path. A maximal block of a graph G is a biconnected subgraph of G which is not contained inany other block. v v v v v v v v v v v v v v v Figure 3: A counter example contradicting Theorem 1.1.
Although the given graph G in Fig. 3 satisfies all the conditions given inTheorem 1.1, it is not an RDG. Using the existing algorithm [8], one canfind an RFP for each of its blocks. Then, an RFP for G can be obtainedby gluing them in a rectangular area which is not possible because of theadjacency relation of cut vertices v and v . Corresponding to a cut vertex,there always associate a through rectangle [30] in an RFP for G . But inFig. 3, the cut-vertices are adjacent. Hence, it is not possible to maintainrectangular enclosure while keeping R and R as through rectangles.ii. Except Theorem 1.1, there does not exist a theorem to check the existenceof an RFP for separable connected planar graphs in the literature.iii. Lai and Leinward [6] derived the following necessary and sufficient condi-tion for an extended plane triangulated graph (EPTG) to be an RDG: Theorem 1.2. [6, Theorem 3] An EPTG is an RDG if and only if eachof its triangular regions can be assigned to one of its corner vertices suchthat each vertex v i has exactly d ( v i ) − Results
In this paper, we find a necessary and sufficient condition for a given planegraph to be an RDG. We show that the unbounded region of an RFP can be real-ized by an unbounded rectangle in the extended Euclidean plane. Equivalently, A through rectangle shares two sides to the exterior, but they are opposite sides.
5e can say that an RFP can be seen as a quadrangulation of the Euclideanplane for which we find a stereographic projection of an RDG.A brief description of our contribution is as follows: In Section 2, we discussexisting facts about RDGs. Section 3 describes the extended RDG constructionprocess. In Section 4, we find stenographic projection of the dual of an RDGin order to extract some result pertaining to the exterior (unbounded) regionof the dual. In Section 5, we derive a necessary and sufficient condition for anEPTG to be an RDG. Finally, we conclude our contribution and discuss futurescope in Section 6.A list of notations used in this paper can be seen in Table 1.Symbol DescriptionRFP rectangular floorplanRDG rectangularly dualizable graphPTG plane triangulated graphEPTG extended plane triangulated graph G a simple connected planar triangulated graph G ∗ Extended plane triangulated graph v i i th vertex of a graph d ( v i ) degree of v i ( v i , v j ) an edge incident to vertices v i and v j R i i th rectangle (region) of an RFP (RDG) corresponding to v i Table 1: List of Notations
2. Preliminaries
In this section, we survey several facts about RFPs that would be helpful toprove our results.A graph is called planar if it can be drawn in the Euclidean plane withoutcrossing its edges except endpoints. A plane graph is a planar graph with a fixedplanar drawing. It splits the Euclidean plane into connected regions called faces;the unbounded region is the exterior face (the outermost face) and all other facesare interior faces. The vertices lying on the exterior face are exterior verticesand all other vertices are interior vertices. A graph is said to be k − connectedif it has at least k vertices and the removal of fever than k vertices does notdisconnect the graph. If a connected graph has a cut vertex, then it is calleda separable graph, otherwise it is called a nonseparable graph. Since floorplansare concerned with connectivity, we only consider nonseparable (biconnected)and separable connected graphs in this paper. A plane graph is called planetriangulated graph (PTG) if it has triangular faces. A TPG may or may nothave exterior face triangular. In this paper, an PTG represents a plane graphwith interior triangular faces. Definition 2.1.
A graph is said to be rectangular graph if each of its edgescan be oriented horizontally or vertically such that it encloses a rectangular6rea. If the dual graph of a planar graph is a rectangular graph, then thegraph is said to be a rectangularly dualizable graphs (RDG). In other words,a planar graph is rectangular dualizable (RDG) if its dual can be realized asa rectangular floorplan (RFP). An RFP is a partition of a rectangle R into n rectangles R , R , . . . , R n provided that no four of them meet at a point. Definition 2.2. [1] The block neighborhood graph (BNG) of a plane graph G is a graph where vertices are represented by biconnected components of G suchthat there is an edge between two vertices if and only if the two biconnectedcomponents they represent, have a vertex in common. Definition 2.3. [1] A shortcut in a plane block G is an edge that is incidentto two vertices on the outermost cycle C of G and is not a part of C . A cornerimplying path (CIP) in G is a v − v k path on the outermost cycle of G suchthat it does not contain any vertices of a shortcut other than v and v k and theshortcut ( v , v k ) is called a critical shortcut.For a better understanding to Definition 2.3, consider a graph shown in Fig.4. Edges ( v , v ), ( v , v ) and ( v , v ) are shortcuts. Paths v v v v and v v v are CIPs while path v v v v v is is not a CIP since it contains the endpointsof other shortcut ( v , v ) and hence ( v , v ) is not a critical shortcut. Bothshortcuts ( v , v ) and ( v , v ) are of length 2. v v v v v v v v v v v v v v v v v v v v v a b Figure 4: (a) Presence of CIPs v v v and (b) a separating triangle v v v v . A component rectangle in an RFP is called a corner rectangle [30] if its twoadjacent sides are adjacent to the exterior while a through rectangle shares itstwo opposite sides to the exterior. A component rectangle in an RFP is calledan end rectangle if its three sides are adjacent to the exterior.
Definition 2.4. A separating cycle is a cycle in a plane graph G that enclosesvertices inside as well as outside. If a separating cycle is of length 3, it is called a separating triangle or a complex triangle. We say a separating triangle in a planegraph is critical separating triangle if it does not contain any other separatingtriangle in its interior.For instance, in Fig. 4b, the cycle v v v v is a separating triangle while thecycle v v v v is a critical separating triangle.7 heorem 2.1. [1, Theorem 3] A nonseperable plane graph G with triangularinterior faces except exterior one is an RDG if and only if it has atmost 4 CIPsand has no separating triangle. Theorem 2.2. [31] A graph G is 4-connected if and only if there exist atleast4 vertex-disjoint path between any two vertices of G .
3. Extended RDG Construction
In a graph described by a VLSI system, vertices and edges correspond tocomponent modules and required interconnections respectively. Communica-tion with units outside the given system are modeled by edges having one endincident to a vertex at the infinity (denoted by v ∞ , see Fig. 5). The vertex v ∞ of an RDG corresponds to the unbounded region of its rectangular dual (RFP).Only planar graphs can be dualized. Whenever the graph structure is notplanar, it can be made planar by adding crossover vertices until the resultantgraph is planar. Such vertices are inserted at crossings of edges in order tosplit the edges of a nonplanar graph. In general, maximum interconnectionsby abutment and minimum through channels is used as an objective function.Without loss of generality, we consider simple connected planar graphs in thispaper.In a floorplan, meeting k -component rectangles at a point is called k -joints.Since an RFP has three joints or four joints only, its dual has triangular orquadrangle regions only. Abiding by common design practice, we consider RFPswith three joints only. In fact, a quadrangle region can be partitioned into twotriangular regions. In such cases, some extra adjacency requests allow unrelatedcomponents in the RDG to connect, but these connections are not used forinterconnection.Furthermore, a rectangular graph needs to be fitted in a rectangular enclo-sure while connecting to the outside world. Vertices that correspond to regionsnext to the enclosure are called enclosure vertices [9] and those vertices corre-spond to corner regions are called corner enclosure vertices. In figure 5, vertices v , v , v , v , v , v , v are enclosure vertices and v , v , v , v are corner enclosurevertices. Since the enclosure has 4 sides, out of these enclosure vertices, the en-closure corner vertices correspond to corner rectangles or end rectangles of anRFP where a corner rectangle shares its two sides to the unbounded (exterior)region and end rectangle shares three sides to the exterior. Therefore, we needto consider atmost 4 extra edges between the selected enclosure corner verticesand v ∞ . These atmost 4 extra edges are known as construction edges [6]. APTG where enclosure vertices are connected to v ∞ together with 4 additionalconstruction edges is called an EPTG (extended planar triangulated graph). AnEPTG is depicted in Fig. 5 by red edges.It is interesting to note that the regions including unbounded region aretriangulated in EPTG so that every region including unbounded region of thedual of RDG is quadrangle. This permits the enclosure to be rectangular. Adetailed description of unbounded quadrangle region of the dual can be seen in8ection 4. Since there is one to one corresponding between the edges of a planegraph and its dual, an enclosure corner vertex has parallel edges to v ∞ . In thispaper, we consider a simple connected plane triangulated graph, i.e., there is noloops or parallel edges. However, some minor changes ( parallel edges betweenenclosure corner vertices and v ∞ only) in the EPTG is done in order to choosefour construction edges. v v v v v v v v ∞ v Figure 5: Construction of an extended RDG (red edges) and its corresponding RDG (darkedges).
4. RDG Stereographic Projection
In this section, we describe stereographic projection of a rectangular graph.A small difference between a rectangular dual graph and an RFP is thatRFP is an edge-oriented graph such that each of its edges is either horizontallyor vertically aligned together with rectangular enclosure whereas the edges of arectangular dual graph may not be oriented either horizontally or vertically, butthey can always be oriented horizontally or vertically. Thus, for transforminga rectangular dual graph into be an RFP, we need to know the orientations ofthe edges. Once it is known, a rectangular dual graph can be converted into anRFP by orienting edges aligned along horizontal or vertical axis of the Euclideanplane. Thus, an RFP can be seen an embedding of a rectangular dual graph.Further it is interesting to note that an RFP is always a rectangular dual graph,but converse it not true.Let D be the rectangular dual graph of an RDG G . Note that a connectedplane graph is a single piece made up of continuous curves (called edges) joiningtheir ends to pairs of the specified points (called vertices) in the Euclidean plane.Consider a sphere S centered at (0 , , /
2) having radius 1 /
2, and a fixed planeembedding D ∗ of D in the Euclidean plane passing through z = 0 ( xy -plane).Let (0 , ,
1) be the north pole N and p be a point of an edge of D ∗ . Draw a linesegment joining the points N and p . Let t be a point where it intersects thesurface of S . Thus we see that the point p is mapped to the point t . In this way,the image of each of its points is a curved line on the surface of S and hence9ach edge of D ∗ is mapped to a curved line on the surface of S . This results anembedding of D ∗ on the surface of a sphere.Now, it is important to identify why the edge of D ∗ is mapped to the edgeson the surface of S ? In fact, a connected graph is carried to a connected graphby a continuous map. Thus being the mapping continuous, the image of D ∗ isagain a plane graph on the surface of S with its exterior bounded. Note that theunbounded region is now mapped into a bounded region on S passing through N . This process is known as stereographic projection and sphere is known asRiemann sphere. But D ∗ is a rectangular dual graph. Its exterior is a four sidedrectangular enclosure. This results the unbounded region of D ∗ corresponds toa four sided bounded region of the corresponding plane graph embedded on thesurface of S . Consequently, when we assign horizontal or vertical orientationsto the edges of D ∗ to transform into an RFP, the unbounded region of D ∗ corresponds to an unbounded rectangle (region) R ∞ passing through ∞ . Thuswe see that the exterior of an RFP is a rectangle R ∞ passing through ∞ . Notethat R ∞ is not a part of an RFP, but is a rectangle that shares its two adjacentsides to each of its enclosure corner rectangles. Recall that a rectangle is afour-sided region with 4 right interior angles formed by its sides. Although incase of R ∞ , these interior angles can be realized to be 90 ◦ by looking at it froma point at ∞ , otherwise we realize every interior angle to be 270 ◦ . The roleof the point at ∞ is played by N and hence an alternative way is to realizeright angle between two sides of the four-sided region passing through N in thestereographic projection of the rectangular dual graph is the angle between theintersection of their tangents to the sides of this region. This discussion realizesus that an RFP is quadrangulation of the Euclidean plane.
5. RDG Existence Theory
In this section, we describe the theory of RDGs.
Theorem 5.1.
A necessary and sufficient condition for an EPTG G ∗ to be anRDG is that it is 4-connected and has atmost 4 critical separating trianglespassing through v ∞ . Proof.
Necessary Condition.
Assume that G ∗ is an RDG. Then it has arectangular dual graph D . Let v i be a vertex of G ∗ dual to some interior region R i of D . Since every region of D is four-sided, atleast 4 regions are requiredto fully enclose an interior region of D . This implies that R i is surrounded byatleast 4 regions of D and hence v i is adjacent to atleast 4 vertices of G ∗ , i.e., d ( v i ) ≥
4. Let v e be a vertex of G ∗ dual to an enclosure (exterior) region R e of D . There arise two possibilities: • R e surrounds exactly its two sides with R ∞ if it is an enclosure cornerregion, • R e surrounds exactly its one side with R ∞ if it is not an enclosure cornerregion. 10n the first case, R ∞ surrounds the two sides of R e . There are two edges between v ∞ and v e where v ∞ corresponds to R ∞ . The remaining two sides of R e aresurrounded by atleast two interior regions other than R ∞ . This implies that d ( v e ) ≥
4. In the second case, only one side of R e is surrounded by R ∞ and theremaining sides are surrounded by atleast three interior regions. This impliesthat d ( v e ) ≥
4. Since v e and v i are arbitrary vertices of G ∗ , G ∗ is 4-connected.This proves the first condition.As discussed in Section 4, R ∞ surrounds exactly its two adjacent sides toeach of the enclosure corner regions of D and exactly one side to the remainingexterior regions of D . Let v c be a vertex of G ∗ dual to an enclosure cornerregion of D . We have already shown that G ∗ is 4-connected, i.e., d ( v c ) ≥ ∀ v c ∈ G ∗ . If d ( v c ) = 4, then two adjacent sides of R c are surrounded by R ∞ whereas the remaining two sides of R c are surrounded by two regions R a and R b . Clearly, R a and R b are the enclosure regions. Since G ∗ is an EPTG,every region of G ∗ is triangular. This implies that R a and R b are adjacent.Consequently, there is a separating triangle passing through v ∞ and verticesthat are dual to R a and R b . Clearly, it encloses exactly one vertex v c . Thisimplies that there is no separating triangle inside this separating triangle andhence it is a critical separating triangle. This situation is depicted in Fig.6a. If d ( v c ) >
4, there are atleast three interior regions that surround R c .Vertices that are dual to these interior regions together with v ∞ is a cycleof length atleast 4. Only possibility for the existence of a critical separatingtriangle passing through v ∞ and enclosing v c is depicted in Fig. 6b. Now it isevident that there is atmost one critical separating triangle passing through v ∞ corresponding to each enclosure corner region. Since a rectangular graph hasatmost four enclosure corner regions, there can be atmost 4 critical separatingtriangles passing through v ∞ . This proves the second condition. v a v b R b R a R c v c v ∞ R b v b v ∞ R c v c v a R a a b Figure 6: Two possibilities of a critical separating triangle enclosing an enclosure corner vertex.
Sufficient Condition.
Assume that the given conditions hold. We provethe result by applying the induction method on the vertices of G ∗ . Recall thatan EPTG contains atleast two vertices. Let n be the number of vertices of G ∗ . If n = 2, then it is a graph consisting of a single edge and hence it is anRDG. Let us assume that n > n − n − n -vertex EPTG H satisfying thegiven conditions is an RDG. Since there can be atmost four critical separatingtriangles in H , there arise two possibilities: (1) there are exactly three edgesbetween v ∞ and atleast one of the enclosure vertices, (2) there are exactly twoedges between v ∞ and each enclosure corner vertex. Let v i be an enclosurecorner vertex of H and A = { v , v , . . . , v t } be the set of vertices adjacent to v i .Consider the first case, i.e., there exist edges ( v i , v ∞ ), ( v i , v p ), ( v i , v q ) wherevertices v p and v q are incident to v ∞ as shown in Fig. 7a. Construct a newEPTG H by deleting v i together with the incident edges and introducing newedges ( v ∞ , v ), ( v ∞ , v ) . . . ( v ∞ , v t ) (see Fig. 7b). We prove that H satisfiesthe given conditions stated in the theorem.Consider two vertices v a and v b of H such that i (cid:54) = a, b . As H is 4-connected,by Menger’s theorem, there exist four vertex-disjoint paths between v a and v b .Choose each path of the shortest possible length. If none of these paths usesthe edges ( v i , v p ) and ( v i , v q ), then the same path would exist in H with theedge ( v ∞ , v k ), (1 ≤ k ≤ t ) substituted in the place of ( v k , v i ) ∪ ( v i , v ∞ ) if theyoccur in the path. Otherwise suppose that one of the four paths passes through( v i , v p ). Being the shortest possible path, it can not pass through v ∞ or v k ,(1 ≤ k ≤ t ). Consequently, it must use the edge ( v i , v q ). If a path passesthrough v ∞ , it would pass through v p or v q , contradicting to the facts that pathis the shortest. Thus vertex v ∞ is not used by any of the four paths. Now bysubstituting the part ( v i , v p ) ∪ ( v i , v q ) of the path in H by ( v p , v ∞ ) ∪ ( v ∞ , v q )in H , we can obtain 4 vertex-disjoint paths in H also. Then by Menger’stheorem, H is 4-connected.Next we claim that the number of critical separating triangles in H can notbe more than the number of critical separating triangles in H . As discussed inthe necessary part that there is atmost one critical separating triangle enclosingan enclosure corner vertex and H has three enclosure corner vertices, there areatmost three critical separating triangles in H . Then the only possibility ofoccurring a separating triangle in H is as follows. If an enclosure vertex v l isincident to both v p , v k where v k ∈ A , then there exists a separating triangle in H passing through v k , v p and v l . Similarly, there can be another separatingtriangle in H passing through v t , v q and v s ∈ A . thus there can be atmosttwo new separating triangles in H . If there exists a critical separating triangle T c containing v i in H , then there are three possibilities: (1) there no longerremains T c in H , (2) T c is contained in one of the new created separatingtriangles in H , and (3) One of the new created separating triangle is containedin T c . All these possibilities show that there can not be more than four criticalseparating triangles in H . This shows that H has atmost 4 critical separatingtriangles. Thus, H has n − H is an RDG and hence admits an RFP. This RFP canbe transformed to another RFP by adjoining a region R i (corresponding to v i )as shown in Fig. 7c. Then the resultant RFP corresponds to H . Hence H is anRDG.Consider the second case. In this case, H appears as shown in Fig. 8a with12tleast four more vertices v , v , v and v . Consider the four enclosure cornervertices v , v , v and v as shown in Fig. 8a. Now we show that there isa separating cycle C passing through v i , v ∞ and an enclosure vertex v d butnot passing through v or v such that the removal of vertices of C from H disconnects it into two connected pieces, each containing atleast one vertex.If there is an edge ( v , v ) in H , there is a separating cycle passing through v , v and v ∞ . In this case, H is separated into two parts, one of which containsatleast v and another contains atleast v . Remainder v i v p v q v ∞ Remainder of H v p v q v ∞ a b R i RFP for H cof H v =v v v k v t = Figure 7: (a) Sketch of the graph H when there are three edges between v ∞ and enclosurevertex v i , (b) sketch of the graph H when there are exactly two edges between each enclosurecorner vertex and v ∞ , and (c) the construction of an rectangular dual for H . If there is no edge ( v , v ) in H . All vertices adjacent to v lie on a path y y . . . y k where y and y k are the enclosure vertices. Let y k x x . . . v bea path of the enclosure vertices starting from y k and ending with v . Then C = ty y . . . y k x x . . . v is a separating cycle which separates H into twoparts, one of which atleast contains v and another contains atleast v . v v v v v ∞ v v v v ∞ a bz z i-1 z i z i+1 z m z m+1 =Remainder of H Remainder of H Figure 8: (a) A separating cycle shown by red edges and (b) the appearance of H u . Once a separating cycle exists, there also exists the shortest separating cy-cle C s = v ∞ z z . . . z m z m +1 . This situation is depicted in 8b. Without lossof generality, suppose C s separates v and v . Construct an EPTG H u fromthe subgraph contained in the interior of C by adding a vertex v ∞ and edgesbetween v ∞ and enclosure vertices of this subgraph. The new edges in this13onstruction are ( v (cid:48)∞ , z ), ( v (cid:48)∞ , z ), . . . ( v (cid:48)∞ , z m +1 ). Now we show that H u sat-isfies the given conditions. Only possibility for creating a separating triangle isa triangle z i z i +1 v ∞ for 1 ≤ i ≤ n . If there would exist an edge ( z i , z i +1 ) in H u ,then it contradicts that C s is the shortest separating cycle. Therefore, any cyclein H u is of length atleast 4 and consequently, H u is 4-connected and can nothave more than 4 separating triangles. By induction hypothesis, H u is an RDG.Similarly, we can show that the EPTG H b constructed from the the remainingpart of H is an RDG. Then the corresponding RFP can be placed one abovethe other and can be merged after applying homeomorphic transformation so asto preserve orthogonal directions of the edges such that the resultant floorplanis an RFP of H as shown in Fig. 9. This completes the induction process andhence completes the proof. a b c Figure 9: (a) Merging two RDGs of H u and H b into an RDG for H Now we turn our attention to derive a necessary and sufficient conditionfor a PTG to be an RDG. A plane graph can be either nonseparable graph(block) or a separable connected graph. A disconnected graph is also a separablegraph. However, we are not considering this case since RFP are concerned withconnectivity.
Theorem 5.2.
A necessary and sufficient condition for a nonseparable PTG G to an RDG is that it is 4-connected and has atmost 4 critical shortcuts. Proof.
Necessary Condition.
Assume that G is an RDG. Then it admits anRFP F . Let v i be an interior vertex of G dual to a rectangular region R i of F .Recall that there require atleast 4 component rectangular regions to surround arectangular region in an RFP. Therefore, there exist atleast 4 rectangular regionsin F enclosing R i . Then v i is adjacent to atleast 4 vertices, i.e., d ( v i ) ≥
4. Since v i is an arbitrary interior vertex of G , G is 4-connected.To the contrary, if there exist 5 critical shortcuts in G , the correspondingEPTG G ∗ would contain 5 critical separating triangles, each passing throughexactly one critical shortcut. This is a contradiction to Theorem 5.1. Thisshows that G can not have more than 4 critical shortcuts. Sufficient Condition.
Assume that the given conditions hold. Choose 4enclosure corner vertices, each on the path joining the endpoints of the criticalshortcut lying on its outermost cycle but not as the endpoints of these paths. If14he number of critical shortcuts are less than 4, choose the remaining enclosurecorner vertices randomly among enclosure vertices. Join each of these 4 verticesto v ∞ by two parallel edges and join each of the remaining n − v ∞ by a single edge. This constructs an EPTG G ∗ satisfying allthe conditions given in Theorem 5.1. Hence G is an RDG. This completes theproof. Theorem 5.3.
A necessary and sufficient condition for a separable connectedPTG G to be an RDG is that:i. each of its blocks is 4-connected,ii. BNG is a path,iii. both endpoints of an exterior edge of each of its blocks are not cut vertices,iv. each maximal blocks corresponding to the endpoints of the BNG containsat most 2 critical shortcuts, not passing through cut vertices,v. Other remaining maximal blocks do not contain a critical shortcut, notpassing through a cut vertex. Proof.
Necessary Condition.
Assume that G is an RDG. The proof of thefirst condition is a direct consequence followed by Theorem 5.1. The BNG of G has the following possibilities:i. it can be path,ii. it can be a cycle of length ≥ v c of G . The construction of anEPTG G ∗ create more than 4 critical separating triangles, each passing through v c , v ∞ , and a vertex adjacent to v c that belongs to the outermost cycle of eachblock. This situation can be depicted in Fig. 10a. Then by Theorem 5.1, G nolonger is an RDG. A similar argument can be applied when it is a tree. Thissituation can be depicted in Fig. 11a. Thus, the BNG is left with one possibility,i.e., the BNG is a path.To the contrary, suppose that both the endpoints of an exterior edge ( v i , v j )of a block are cut vertices, then there are more than 4 critical separating trianglespassing through v i , v j and v ∞ in G ∗ , which is a contradiction to Theorem 5.1.Hence both the endpoints of an exterior edge of a block can not be cut verticessimultaneously.Let M i be a maximal block corresponding to the endpoints of the BNG. Since G is an RDG, each of its block is an RDG. Suppose that M i is an RDG. Then itadmits an RFP F i . It can be easily noted that out of 4 corner rectangular regionsof F i , only two can be the corner rectangular regions of F . Then there can beatmost two critical separating triangles in G ∗ and hence there can be atmosttwo critical shortcuts in each M i . This implies that the second condition holds.Also, any other maximal block of the BNG can not share critical separatingtriangles since any corner rectangular region in F is an RFP. This implies that15o other maximal block has a critical separating triangle in G ∗ and hence thereis no critical shortcut in the remaining maximal blocks. AB C AB Ca b
Figure 10: (a) A separable connected graph constituted by three blocks A, B and C, and (b)its BNG. Here only the outermost cycles of the blocks are shown.
Sufficient Condition.
Assume that the given conditions hold. The firstcondition shows that G ∗ is 4-connected. The remaining conditions show thatthere are atmost four critical separating triangles in G ∗ . By Theorem 5.1, G isan RDG. Hence the proof. AB C D ACB Da b
Figure 11: A separable connected graph constituted by three blocks A, B, C and D, and (b)its BNG. Here only the outermost cycles of the blocks are shown.
6. Concluding remarks and future task
We developed graph theoretic characterization of RFPs. We reported thatthe existing RDG theory may fail in some cases. Hence, we proposed a new RDGtheory which is easily implementable and it simplifies the floorplan constructionprocess of the VLSI circuits as well architectural buildings.In future, it would be interesting to transform a nonRDG into an RDG byremoving those edges which violates the RDG property and then adding newedges (maintaining RDG property) in such a way that the distances of endpointsof the deleted edges can be minimized. This idea would be useful in reducingthe interconnection wire-lengths as well as in complex buildings, it gives the16hortest possible paths for those pairs of rooms which is impossible to directlyconnect.