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a b s t r a c tSuppose we are given a graph G together with two proper vertex k-colourings of G, α and β. How easily can we decide whether it is possible to transform α into β by recolouring vertices of G one at a time, making sure we always have a proper k-colouring of G? This decision problem is trivial for k = 2, and decidable in polynomial time for k = 3. Here we prove it is PSPACE-complete for all k ≥ 4. In particular, we prove that the problem remains PSPACE-complete for bipartite graphs, as well as for: (i) planar graphs and 4 ≤ k ≤ 6, and (ii) bipartite planar graphs and k = 4. Moreover, the values of k in (i) and (ii) are tight, in the sense that for larger values of k, it is always possible to recolour α to β.We also exhibit, for every k ≥ 4, a class of graphs {G N,k : N ∈ N * }, together with two k-colourings for each G N,k , such that the minimum number of recolouring steps required to transform the first colouring into the second is superpolynomial in the size of the graph: the minimum number of steps is Ω(2 N ), whereas the size of G N is O(N 2 ). This is in stark contrast to the k = 3 case, where it is known that the minimum number of recolouring steps is at most quadratic in the number of vertices. We also show that a class of bipartite graphs can be constructed with this property, and that: (i) for 4 ≤ k ≤ 6 planar graphs and (ii) for k = 4 bipartite planar graphs can be constructed with this property. This provides a remarkable correspondence between the tractability of the problem and its underlying structure.

The Shortest Path Reconfiguration problem has as input a graph G with unit edge lengths, with vertices s and t, and two shortest st-paths P and Q . The question is whether there exists a sequence of shortest st-paths that starts with P and ends with Q , such that subsequent paths differ in only one vertex. This is called a rerouting sequence. This problem is shown to be PSPACE-complete. For claw-free graphs and chordal graphs, it is shown that the problem can be solved in polynomial time, and that shortest rerouting sequences have linear length. For these classes, it is also shown that deciding whether a rerouting sequence exists between all pairs of shortest st-paths can be done in polynomial time. Finally, a polynomial time algorithm for counting the number of isolated paths is given.

Abstract. We present a polynomial-time algorithm that, given two independent sets in a claw-free graph G, decides whether one can be transformed into the other by a sequence of elementary steps. Each elementary step is to remove a vertex v from the current independent set S and to add a new vertex w (not in S) such that the result is again an independent set. We also consider the more restricted model where v and w have to be adjacent.

In the unsplittable flow problem on a path, we are given a capacitated path P and n tasks, each task having a demand, a profit, and start and end vertices. The goal is to compute a maximum profit set of tasks, such that for each edge e of P , the total demand of selected tasks that use e does not exceed the capacity of e. This is a well-studied problem that has been studied under alternative names, such as resource allocation, bandwidth allocation, resource constrained scheduling, temporal knapsack and interval packing.We present a polynomial time constant-factor approximation algorithm for this problem. This improves on the previous best known approximation ratio of O(log n). The approximation ratio of our algorithm is 7 + ǫ for any ǫ > 0.We introduce several novel algorithmic techniques, which might be of independent interest: a framework which reduces the problem to instances with a bounded range of capacities, and a new geometrically inspired dynamic program which solves a special case of the maximum weight independent set of rectangles problem to optimality. In the setting of resource augmentation, wherein the capacities can be slightly violated, we give a (2 + ǫ)-approximation algorithm. In addition, we show that the problem is strongly NP-hard even if all edge capacities are equal and all demands are either 1, 2, or 3.