2010 **Abstract:** Abstract. We consider the problem of periodic graph exploration in which a mobile entity with (at most) constant memory, an agent, has to visit all n nodes of an arbitrary undirected graph G in a periodic manner. Graphs are supposed to be anonymous, that is, nodes are unlabeled. However, while visiting a node, the robot has to distinguish between edges incident to it. For each node v the endpoints of the edges incident to v are uniquely identi ed by di erent integer labels called port numbers. We are intereste…

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“…While the stated problem and several combinatorial observations, in particular merging and exchanging contents of cycles, attracted attention in the community, the bound of 10n on their port labeling turned out to be incorrect. Very recently Czyzowicz et al in [27] proposed a polished version of the previous argument supported by a new combinatorial structure of a three-layer partition of graphs. This led to the first provably correct port labeling inducing a linear tour of length 4 1 3 n. Moreover, the labeling based on the three-layer partition can be performed in the optimal O(|E|)−time.…”

confidence: 99%

“…Unfortunately the problem of deciding whether the input graph has a spanning tree with the required property is NP-hard since this problem corresponds to selection of a Hamiltonian path in 3-regular graphs, which is known to be hard. The authors in [27] give also an n-node graph, shown in Figure 5, in which the witness cycle must contain all arcs in ← → G and therefore cannot be shorter than 2.8n.…”

confidence: 99%

“…At any node v with the degree d, the incoming edge incident via a port i becomes the predecessor of the outgoing arc incident via port (i + 1) mod d. A certain arrangement of the edge of the nodes of the graph leads to a cycle that visits every node. In [20] the authors show that there exists an edge order that leads to a cycle of length 4.33n that visits every node.…”

confidence: 99%