Vít Jelínek

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We study the following problem: Given a planar graph <i>G</i> and a planar drawing (embedding) of a subgraph of <i>G</i>, can such a drawing be extended to a planar drawing of the entire graph <i>G</i>? This problem fits the paradigm of extending a partial solution to a complete one, which has been studied before in many different settings. Unlike many(More)
It is known that every planar graph has a planar embedding where edges are represented by non-crossing straight-line segments. We study the planar slope number, i.e., the minimum number of distinct edge-slopes in such a drawing of a planar graph with maximum degree Δ. We show that the planar slope number of every series-parallel graph of maximum degree(More)
A set partition of size n is a collection of disjoint blocks B1, B2, . . . , Bd whose union is the set [n] = {1, 2, . . . , n}. We choose the ordering of the blocks so that they satisfy minB1 < minB2 < · · · < minBd. We represent such a set partition by a canonical sequence π1, π2, . . . , πn, with πi = j if i ∈ Bj. We say that a partition π contains a(More)
A partially embedded graph (or PEG) is a triple (G,H,EH), where G is a graph, H is a subgraph of G, and EH is a planar embedding of H. We say that a PEG (G,H,EH) is planar if the graph G has a planar embedding that extends the embedding EH. We introduce a containment relation of PEGs analogous to graph minor containment, and characterize the minimal(More)
In this paper we present a bijection between composition matrices and (2+ 2)free posets. This bijection maps partition matrices to factorial posets, and induces a bijection from upper triangular matrices with non-negative entries having no rows or columns of zeros to unlabeled (2+ 2)-free posets. Chains in a (2+ 2)-free poset are shown to correspond to(More)
In this paper we study properties of intersection graphs of kbend paths in the rectangular grid. A k-bend path is a path with at most k 90 degree turns. The class of graphs representable by intersections of k-bend paths is denoted by Bk-VPG. We show here that for every fixed k, Bk-VPG ( Bk+1-VPG and that recognition of graphs from Bk-VPG is NP-complete even(More)