نتایج جستجو برای: hamiltonian graphs
تعداد نتایج: 127472 فیلتر نتایج به سال:
The Heawood graph and K3;3 have the property that all of their 2-factors are Hamilton circuits. We call such graphs 2-factor hamiltonian. We prove that if G is a k-regular bipartite 2factor hamiltonian graph then either G is a circuit or k 1⁄4 3: Furthermore, we construct an infinite family of cubic bipartite 2-factor hamiltonian graphs based on the Heawood graph and K3;3 and conjecture that th...
A graph G is a combined k-fault-tolerant Hamiltonian graph (also called a combined k-Hamiltonian graph) if G − F is Hamiltonian for every subset F ⊂ (V(G) ∪ E(G)) with |F| = k. A combined k-Hamiltonian graph G with |V(G)| = n is optimal if it has the minimum number of edges among all n-node k-Hamiltonian graphs. Using the concept of node expansion, we present a powerful construction scheme to c...
Motivated by multi-robot construction systems, we introduce the problem of finding squeeze-free Hamiltonian paths in grid graphs. A Hamiltonian path is squeeze-free if it does not pass between two previously visited vertices lying on opposite sides. We determine necessary and sufficient conditions for the existence of squeeze-free Hamiltonian paths in staircase grid graphs. Our proofs are const...
The Cartesian product of two hamiltonian graphs is always hamiltonian. For directed graphs, the analogous statement is false. We show that the Cartesian product C,,, x C,, of directed cycles is hamiltonian if and only if the greatest common divisor (g.c.d.) d of n, and n, is a t least two and there exist positive integers d,, d, so that d, + d, = d and g.c.d. (n,, d,) = g.c.d. (n,, d,) = 1. We ...
Algorithm tests if a Hamiltonian cycle exists in directed graphs, if it is exists algorithm can show found Hamiltonian cycle. If you want to test an undirected graph, such a graph should be converted to the form of directed graph. Previously known algorithm solving Hamiltonian cycle problem brute-force search can’t handle relatively small graphs. Algorithm presented here is referred to simply a...
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