2,502 research outputs found

    On Hardness of the Joint Crossing Number

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    The Joint Crossing Number problem asks for a simultaneous embedding of two disjoint graphs into one surface such that the number of edge crossings (between the two graphs) is minimized. It was introduced by Negami in 2001 in connection with diagonal flips in triangulations of surfaces, and subsequently investigated in a general form for small-genus surfaces. We prove that all of the commonly considered variants of this problem are NP-hard already in the orientable surface of genus 6, by a reduction from a special variant of the anchored crossing number problem of Cabello and Mohar

    Planarity of Streamed Graphs

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    In this paper we introduce a notion of planarity for graphs that are presented in a streaming fashion. A streamed graph\textit{streamed graph} is a stream of edges e1,e2,...,eme_1,e_2,...,e_m on a vertex set VV. A streamed graph is ω\omega-stream planar\textit{stream planar} with respect to a positive integer window size ω\omega if there exists a sequence of planar topological drawings Γi\Gamma_i of the graphs Gi=(V,{ej∣i≤j<i+ω})G_i=(V,\{e_j \mid i\leq j < i+\omega\}) such that the common graph G∩i=Gi∩Gi+1G^{i}_\cap=G_i\cap G_{i+1} is drawn the same in Γi\Gamma_i and in Γi+1\Gamma_{i+1}, for 1≤i<m−ω1\leq i < m-\omega. The Stream Planarity\textit{Stream Planarity} Problem with window size ω\omega asks whether a given streamed graph is ω\omega-stream planar. We also consider a generalization, where there is an additional backbone graph\textit{backbone graph} whose edges have to be present during each time step. These problems are related to several well-studied planarity problems. We show that the Stream Planarity\textit{Stream Planarity} Problem is NP-complete even when the window size is a constant and that the variant with a backbone graph is NP-complete for all ω≥2\omega \ge 2. On the positive side, we provide O(n+ωm)O(n+\omega{}m)-time algorithms for (i) the case ω=1\omega = 1 and (ii) all values of ω\omega provided the backbone graph consists of one 22-connected component plus isolated vertices and no stream edge connects two isolated vertices. Our results improve on the Hanani-Tutte-style O((nm)3)O((nm)^3)-time algorithm proposed by Schaefer [GD'14] for ω=1\omega=1.Comment: 21 pages, 9 figures, extended version of "Planarity of Streamed Graphs" (9th International Conference on Algorithms and Complexity, 2015

    Orthogonal Graph Drawing with Inflexible Edges

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    We consider the problem of creating plane orthogonal drawings of 4-planar graphs (planar graphs with maximum degree 4) with constraints on the number of bends per edge. More precisely, we have a flexibility function assigning to each edge ee a natural number flex(e)\mathrm{flex}(e), its flexibility. The problem FlexDraw asks whether there exists an orthogonal drawing such that each edge ee has at most flex(e)\mathrm{flex}(e) bends. It is known that FlexDraw is NP-hard if flex(e)=0\mathrm{flex}(e) = 0 for every edge ee. On the other hand, FlexDraw can be solved efficiently if flex(e)≥1\mathrm{flex}(e) \ge 1 and is trivial if flex(e)≥2\mathrm{flex}(e) \ge 2 for every edge ee. To close the gap between the NP-hardness for flex(e)=0\mathrm{flex}(e) = 0 and the efficient algorithm for flex(e)≥1\mathrm{flex}(e) \ge 1, we investigate the computational complexity of FlexDraw in case only few edges are inflexible (i.e., have flexibility~00). We show that for any ε>0\varepsilon > 0 FlexDraw is NP-complete for instances with O(nε)O(n^\varepsilon) inflexible edges with pairwise distance Ω(n1−ε)\Omega(n^{1-\varepsilon}) (including the case where they induce a matching). On the other hand, we give an FPT-algorithm with running time O(2k⋅n⋅Tflow(n))O(2^k\cdot n \cdot T_{\mathrm{flow}}(n)), where Tflow(n)T_{\mathrm{flow}}(n) is the time necessary to compute a maximum flow in a planar flow network with multiple sources and sinks, and kk is the number of inflexible edges having at least one endpoint of degree 4.Comment: 23 pages, 5 figure

    Constructions of Large Graphs on Surfaces

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    We consider the degree/diameter problem for graphs embedded in a surface, namely, given a surface Σ\Sigma and integers Δ\Delta and kk, determine the maximum order N(Δ,k,Σ)N(\Delta,k,\Sigma) of a graph embeddable in Σ\Sigma with maximum degree Δ\Delta and diameter kk. We introduce a number of constructions which produce many new largest known planar and toroidal graphs. We record all these graphs in the available tables of largest known graphs. Given a surface Σ\Sigma of Euler genus gg and an odd diameter kk, the current best asymptotic lower bound for N(Δ,k,Σ)N(\Delta,k,\Sigma) is given by 38gΔ⌊k/2⌋.\sqrt{\frac{3}{8}g}\Delta^{\lfloor k/2\rfloor}. Our constructions produce new graphs of order \begin{cases}6\Delta^{\lfloor k/2\rfloor}& \text{if $\Sigma$ is the Klein bottle}\\ \(\frac{7}{2}+\sqrt{6g+\frac{1}{4}}\)\Delta^{\lfloor k/2\rfloor}& \text{otherwise,}\end{cases} thus improving the former value by a factor of 4.Comment: 15 pages, 7 figure

    Compact Drawings of 1-Planar Graphs with Right-Angle Crossings and Few Bends

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    We study the following classes of beyond-planar graphs: 1-planar, IC-planar, and NIC-planar graphs. These are the graphs that admit a 1-planar, IC-planar, and NIC-planar drawing, respectively. A drawing of a graph is 1-planar if every edge is crossed at most once. A 1-planar drawing is IC-planar if no two pairs of crossing edges share a vertex. A 1-planar drawing is NIC-planar if no two pairs of crossing edges share two vertices. We study the relations of these beyond-planar graph classes (beyond-planar graphs is a collective term for the primary attempts to generalize the planar graphs) to right-angle crossing (RAC) graphs that admit compact drawings on the grid with few bends. We present four drawing algorithms that preserve the given embeddings. First, we show that every nn-vertex NIC-planar graph admits a NIC-planar RAC drawing with at most one bend per edge on a grid of size O(n)×O(n)O(n) \times O(n). Then, we show that every nn-vertex 1-planar graph admits a 1-planar RAC drawing with at most two bends per edge on a grid of size O(n3)×O(n3)O(n^3) \times O(n^3). Finally, we make two known algorithms embedding-preserving; for drawing 1-planar RAC graphs with at most one bend per edge and for drawing IC-planar RAC graphs straight-line

    Relating Graph Thickness to Planar Layers and Bend Complexity

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    The thickness of a graph G=(V,E)G=(V,E) with nn vertices is the minimum number of planar subgraphs of GG whose union is GG. A polyline drawing of GG in R2\mathbb{R}^2 is a drawing Γ\Gamma of GG, where each vertex is mapped to a point and each edge is mapped to a polygonal chain. Bend and layer complexities are two important aesthetics of such a drawing. The bend complexity of Γ\Gamma is the maximum number of bends per edge in Γ\Gamma, and the layer complexity of Γ\Gamma is the minimum integer rr such that the set of polygonal chains in Γ\Gamma can be partitioned into rr disjoint sets, where each set corresponds to a planar polyline drawing. Let GG be a graph of thickness tt. By F\'{a}ry's theorem, if t=1t=1, then GG can be drawn on a single layer with bend complexity 00. A few extensions to higher thickness are known, e.g., if t=2t=2 (resp., t>2t>2), then GG can be drawn on tt layers with bend complexity 2 (resp., 3n+O(1)3n+O(1)). However, allowing a higher number of layers may reduce the bend complexity, e.g., complete graphs require Θ(n)\Theta(n) layers to be drawn using 0 bends per edge. In this paper we present an elegant extension of F\'{a}ry's theorem to draw graphs of thickness t>2t>2. We first prove that thickness-tt graphs can be drawn on tt layers with 2.25n+O(1)2.25n+O(1) bends per edge. We then develop another technique to draw thickness-tt graphs on tt layers with bend complexity, i.e., O(2t⋅n1−(1/β))O(\sqrt{2}^{t} \cdot n^{1-(1/\beta)}), where β=2⌈(t−2)/2⌉\beta = 2^{\lceil (t-2)/2 \rceil }. Previously, the bend complexity was not known to be sublinear for t>2t>2. Finally, we show that graphs with linear arboricity kk can be drawn on kk layers with bend complexity 3(k−1)n(4k−2)\frac{3(k-1)n}{(4k-2)}.Comment: A preliminary version appeared at the 43rd International Colloquium on Automata, Languages and Programming (ICALP 2016
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