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We study the following problem: Given a planar graph G and a planar drawing (embedding) of a subgraph of G, can such a drawing be extended to a planar drawing of the entire graph G? 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 cases, in which the presence of a partial solution in the input makes hard an otherwise easy problem, we show that the planarity question remains polynomial-time solvable. Our algorithm is based on several combinatorial lemmata which show that the planarity of partially embedded graphs meets the "oncas" behaviour -obvious necessary conditions for planarity are also sufficient. These conditions are expressed in terms of the interplay between (a) rotation schemes and containment relationships between cycles and (b) the decomposition of a graph into its connected, biconnected, and triconnected components. This implies that no dynamic programming is needed for a decision algorithm and that the elements of the decomposition can be processed independently.Further, by equipping the components of the decomposition with suitable data structures and by carefully splitting the problem into simpler subproblems, we improve our algorithm to reach linear-time complexity.Finally, we consider several generalizations of the problem, e.g. minimizing the number of edges of the partial embedding that need to be rerouted to extend it, and argue that they are NP-hard. Also, we show how our algorithm can be applied to solve related Graph Drawing problems.

Abstract. The b-chromatic number b(G) of a graph G = (V, E) is the largest integer k such that G admits a vertex partition into k independent sets Xi (i = 1, . . . , k) such that each Xi contains a vertex xi adjacent to at least one vertex of each Xj, j = i. We discuss on the tightness of some bounds on b(G) and on the complexity of determining b(G). We also determine the asymptotic behavior of b(Gn,p) for the random graph, within the accuracy of a multiplicative factor 2 + o(1) as n → ∞.

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