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Graph Theory » Extremal G.T.

What is the smallest number of disjoint spanning trees made a graph Hamiltonian ★★

Author(s): Goldengorin

We are given a complete simple undirected weighted graph $ G_1=(V,E) $ and its first arbitrary shortest spanning tree $ T_1=(V,E_1) $. We define the next graph $ G_2=(V,E\setminus E_1) $ and find on $ G_2 $ the second arbitrary shortest spanning tree $ T_2=(V,E_2) $. We continue similarly by finding $ T_3=(V,E_3) $ on $ G_3=(V,E\setminus \cup_{i=1}^{2}E_i) $, etc. Let k be the smallest number of disjoint shortest spanning trees as defined above and let $ T^{k}=(V,\cup_{i=1}^{k}E_i) $ be the graph obtained as union of all $ k $ disjoint trees.

Question 1. What is the smallest number of disjoint spanning trees creates a graph $ T^{k} $ containing a Hamiltonian path.

Question 2. What is the smallest number of disjoint spanning trees creates a graph $ T^{k} $ containing a shortest Hamiltonian path?

Questions 3 and 4. Replace in questions 1 and 2 a shortest spanning tree by a 1-tree. What is the smallest number of disjoint 1-trees creates a Hamiltonian graph? What is the smallest number of disjoint 1-trees creates a graph containing a shortest Hamiltonian cycle?

Keywords: 1-trees; cycle; Hamitonian path; spanning trees

Posted by boris
updated September 10th, 2007
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