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Orthogonal Drawing Kees Visser Overview Introduction Orthogonal representation Flow network Bend optimal drawing Introduction Orthogonal drawing Given embedding Only vertical or horizontal line segments Maximum degree of 4 Introduction Uses in VLSI circuits Minimize bends Finding an orthogonal drawing of graph G with the minimum number of bends without a fixed embedding is NP-Complete. Introduction Tamassia and Garg and Tamassia found algorithms to find a minimal bend orthogonal drawing of graph G in polynomial time. Orthogonal Representation Planar representation Set of circularly ordered edge list P(F) One for each face P(F1) = (e8, e7, e11, e6, e3) P(F2) = (e1, e2, e10, e9, e11, e7, e8) P(F3) = (e6, e9, e10, e5, e4) P(F0) = (e1, e2, e5, e4, e3) Orthogonal representation Definitions: An edge appearing in the same list twice is a bridge An angle formed by to edges is a vertex- angle An angle formed by two line segments at a bend is a bend-angle Both vertex angles and bend angles are 90, 180, 270 or 360 degrees Orthogonal representation Facts: The sum of vertex-angles around any vertex is 360 degrees The sum or the angles inside any facial polygon is (2p-4)90 degree, and the sum of the angles of the outer facial polygon is (2p+4)90 degrees, where p is the number of edges Orthogonal representation Orthogonal representation 1 list for each face Each element in the list has 3 values, the edge number (Er), a bit string and the angle formed in the face from this edge to the next edge (Ar). The bit string indicated the bends in the line. A 0 indicates a 90 degree bend, 1 means 270 and ε means a straight line Orthogonal representation R(F1) = ((e8, ε, 180),(e7, ε, 90),(e11, ε, 90),(e6, ε, 180),(e3, 0, 90)) R(F2) = ((e1, 0, 90), (e2, 00, 90), (e10, ε, 180), (e9, ε, 180), (e11, ε, 270), (e7, ε, 180), (e8, ε, 90)) Orthogonal representation Necessary properties of a orthogonal representation: (p1) There is some planer graph whose planer representation is given by the e- fields of the list in R (p2) For each pair of elements r and r’ in R with Er=Er’, string Sr’ can be obtained by applying bitwise negation to the reversion of Sr’ Orthogonal representation (p3) For each element r in R, define the rotation p(r) as follows: p(r) = |Sr|0 - |Sr|1 + (2-Ar/90) (p4) For each vertex v, the sum of the vertex angles around v given by the A- fields in R is equal to 360 degrees Flow network A flow network N is a directed graph with two disjoint non empty sets of sources and sinks Each arc e of N is labelled with 3 nonnegative integers A lower bound λ(e) A capacity μ(e) A cost c(e) Flow network The flow Ф(e) must equal or higher then the lower bound and lower of equal then the capacity. σ(u) is the production of source u -σ(u) is the consumption of sink u Bend-optimal drawing 1 unit of flow represents an angle of 90 degrees Each vertex and face is a node Each vertex Node is a source (Uv), σ(Uv) =4 Each face Node is a sink -σ(UF), -σ(UF) = 2p(F)-4 (if F is an inner face), -σ(UF) = 2p(F)+4 (if F is the outer face). p(F) is the number of vertex-angles in face F Bend-optimal drawing Bend-optimal drawing The arc set of network N consists of the vertex arcs (Av) and face arcs (AF) The set of Av consists of arcs from every vertex node to every face node the vertex borders. Flow Ф(Uv, UF) is the sum of vertex angles at vertex v inside face F. λ(Uv, UF) is equal to the number of vertex- angles at v inside face F. μ(Uv, UF) = 4 c(Uv, UF) = 0 Bend-optimal drawing Bend-optimal drawing The set of AF consists of arcs from every face F that shares an edge with face F’ Flow Ф(UF, UF’) is the number of bends with an angle of 90 degrees inside face F along the edges which are common to F and F’ λ(UF, UF’) = 0 μ(UF, UF’) = +∞ c(UF, UF) = 1 Bend-optimal drawing Bend-optimal drawing Theorem: Let G be a plane graph, let N be the network constructed from G. For each integer flow Ф in network N, there is an orthogonal representation R that represents an orthogonal drawing D of G and whose number of bends is equal to the cost of the flow Ф. In particular, the minimum cost flow can be used to construct a bend-optimal orthogonal drawing of G. Bend-optimal drawing Proof We construct a orthogonal representation R of G by computing the a- and s-fields from the flow values in Av and AF. L = |(R(v,F)| and R(v,F) ={r1,r2,…,rL}, where 1 =< L =< 4 Ar1 = 90(Ф(Uv, UF)- L + 1) and Ari = 90 for each i, 2 =< i =< L Bend-optimal drawing Sr1 = 0 Ф(UF, UF’)1 Ф(UF’, UF) Sr’1 = 0 Ф(UF’, UF)1 Ф(UF, UF’) Sri = Sr’i = ε for each i 2 =< i =< L Example: L = 3, Ф(UF, UF’) = 2 and Ф(UF’, UF) = 1 Sr1 = 001 and Sr2 = Sr3 = ε Bend-optimized drawing R(F) needs to satisfy Properties (p1)-(p4) (p1) There is some planer graph whose planer representation is given by the e-fields of the list in R (p2) For each pair of elements r and r’ in R with Er=Er’, string Sr’ can be obtained by applying bitwise negation to the reversion of Sr’ (p3) For each element r in R, define the rotation p(r) as follows: p(r) = |Sr|0 - |Sr|1 + (2-Ar/90) (p4) For each vertex v, the sum of the vertex angles around v given by the A-fields in R is equal to 360 degrees Bend-optimized drawing (p1) is automatically satisfied since we have build R from a planar representation (p2) easily follows from the assignment of the S-values. (p4) follows from the definition that every source produces 4 units. (p3) Solving the equation by replacing with the flow equations makes this satisfied. Look in the book for a more precise proof of this. Conclusion Garg and Tamassia have shown that the minimum cost flow problem in this specific network can be solved in time O(n7/4√(log n)) Questions?