scholarly journals Geodetic Number of Powers of Cycles

Symmetry ◽  
2018 ◽  
Vol 10 (11) ◽  
pp. 592
Author(s):  
Mohammad Abudayah ◽  
Omar Alomari ◽  
Hassan Ezeh

The geodetic number of a graph is an important graph invariant. In 2002, Atici showed the geodetic set determination of a graph is an NP-Complete problem. In this paper, we compute the geodetic set and geodetic number of an important class of graphs called the k-th power of a cycle. This class of graphs has various applications in Computer Networks design and Distributed computing. The k-th power of a cycle is the graph that has the same set of vertices as the cycle and two different vertices in the k-th power of this cycle are adjacent if the distance between them is at most k.

1982 ◽  
Vol 34 (3) ◽  
pp. 519-524 ◽  
Author(s):  
Svatopluk Poljak ◽  
Daniel Turzík

Let G be a symmetric connected graph without loops. Denote by b(G) the maximum number of edges in a bipartite subgraph of G. Determination of b(G) is polynomial for planar graphs ([6], [8]); in general it is an NP-complete problem ([5]). Edwards in [1], [2] found some estimates of b(G) which give, in particular,for a connected graph G of n vertices and m edges, whereand ﹛x﹜ denotes the smallest integer ≧ x.We give an 0(V3) algorithm which for a given graph constructs a bipartite subgraph B with at least f(m, n) edges, yielding a short proof of Edwards’ result.Further, we consider similar methods for obtaining some estimates for a particular case of the satisfiability problem. Let Φ be a Boolean formula of variables x1, …, xn.


2019 ◽  
Author(s):  
Vitor Ponciano ◽  
Romulo Oliveira

For a nontrivial connected and simple graphs G= (V(G), E(G)), a set S E(G) is called edge geodetic set of G if every edge of G it’s in S or is contained in a geodesic joining some pair of edges in S. The edge geodetic number eds(G) of G is the minimum order of its edge geodetic sets. We prove that it is NP-complete to decide for a given bipartiti graphs G and a given integer k whether G has a edge geodetic set of cardinality at most k. A set M V(G) is called P3 set of G if all vertices of G have two neighbors in M. The P3 number of G is the minimum order of its P3 sets. We prove that it is NP-complete to decide for a given graphs G(diamond,odd-hole) free and a given integer k whether G has a P3 set of cardinality at most k.


2001 ◽  
Vol 34 (44) ◽  
pp. 9555-9567 ◽  
Author(s):  
Tomohiro Sasamoto ◽  
Taro Toyoizumi ◽  
Hidetoshi Nishimori

2012 ◽  
Vol Vol. 14 no. 2 (Graph Theory) ◽  
Author(s):  
Laurent Gourvès ◽  
Adria Lyra ◽  
Carlos A. Martinhon ◽  
Jérôme Monnot

Graph Theory International audience In this paper we deal from an algorithmic perspective with different questions regarding properly edge-colored (or PEC) paths, trails and closed trails. Given a c-edge-colored graph G(c), we show how to polynomially determine, if any, a PEC closed trail subgraph whose number of visits at each vertex is specified before hand. As a consequence, we solve a number of interesting related problems. For instance, given subset S of vertices in G(c), we show how to maximize in polynomial time the number of S-restricted vertex (resp., edge) disjoint PEC paths (resp., trails) in G(c) with endpoints in S. Further, if G(c) contains no PEC closed trails, we show that the problem of finding a PEC s-t trail visiting a given subset of vertices can be solved in polynomial time and prove that it becomes NP-complete if we are restricted to graphs with no PEC cycles. We also deal with graphs G(c) containing no (almost) PEC cycles or closed trails through s or t. We prove that finding 2 PEC s-t paths (resp., trails) with length at most L > 0 is NP-complete in the strong sense even for graphs with maximum degree equal to 3 and present an approximation algorithm for computing k vertex (resp., edge) disjoint PEC s-t paths (resp., trails) so that the maximum path (resp., trail) length is no more than k times the PEC path (resp., trail) length in an optimal solution. Further, we prove that finding 2 vertex disjoint s-t paths with exactly one PEC s-t path is NP-complete. This result is interesting since as proved in Abouelaoualim et. al.(2008), the determination of two or more vertex disjoint PEC s-t paths can be done in polynomial time. Finally, if G(c) is an arbitrary c-edge-colored graph with maximum vertex degree equal to four, we prove that finding two monochromatic vertex disjoint s-t paths with different colors is NP-complete. We also propose some related problems.


2021 ◽  
Vol 76 (4) ◽  
Author(s):  
Marta Borowiecka-Olszewska ◽  
Ewa Drgas-Burchardt ◽  
Nahid Yelene Javier-Nol ◽  
Rita Zuazua

AbstractWe consider arc colourings of oriented graphs such that for each vertex the colours of all out-arcs incident with the vertex and the colours of all in-arcs incident with the vertex form intervals. We prove that the existence of such a colouring is an NP-complete problem. We give the solution of the problem for r-regular oriented graphs, transitive tournaments, oriented graphs with small maximum degree, oriented graphs with small order and some other classes of oriented graphs. We state the conjecture that for each graph there exists a consecutive colourable orientation and confirm the conjecture for complete graphs, 2-degenerate graphs, planar graphs with girth at least 8, and bipartite graphs with arboricity at most two that include all planar bipartite graphs. Additionally, we prove that the conjecture is true for all perfect consecutively colourable graphs and for all forbidden graphs for the class of perfect consecutively colourable graphs.


Author(s):  
Lance Fortnow

This chapter demonstrates several approaches for dealing with hard problems. These approaches include brute force, heuristics, and approximation. Typically, no single technique will suffice to handle the difficult NP problems one needs to solve. For moderate-sized problems one can search over all possible solutions with the very fast computers available today. One can use algorithms that might not work for every problem but do work for many of the problems one cares about. Other algorithms may not find the best possible solution but still a solution that's good enough. Other times one just cannot get a solution for an NP-complete problem. One has to try to solve a different problem or just give up.


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