raph P2 Hamiltonian. One approach: Show that any Hamilton path must begin and end at even-numbered vertices. Why does this prevent forming a Hamilton cycle?

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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The graph \( P_2 \times P_2 \) is depicted below. Show that this graph is not Hamiltonian. One approach: Show that any Hamilton path must begin and end at even-numbered vertices. Why does this prevent forming a Hamilton cycle?

**Graph Description:**

The graph is a 3x3 grid with vertices labeled from 0 to 8. It is structured as follows:

- The bottom row has vertices 0, 1, and 2 connected by horizontal edges.
- The middle row has vertices 3, 4, and 5 connected horizontally, directly above the bottom row, with vertical edges connecting vertices 0 to 3, 1 to 4, and 2 to 5.
- The top row has vertices 6, 7, and 8 connected horizontally, directly above the middle row, with vertical edges connecting vertices 3 to 6, 4 to 7, and 5 to 8.

**Analysis:**

The vertices form a grid with each vertex connected to its adjacent vertices directly beside, above, or below it.

**Task:**

The problem asks to demonstrate why the graph is not Hamiltonian by showing any Hamilton path must begin and end at even-numbered vertices, and explaining why this prevents forming a Hamilton cycle. A Hamilton cycle would require the path to return to its starting point, which is not possible under these conditions.
Transcribed Image Text:The graph \( P_2 \times P_2 \) is depicted below. Show that this graph is not Hamiltonian. One approach: Show that any Hamilton path must begin and end at even-numbered vertices. Why does this prevent forming a Hamilton cycle? **Graph Description:** The graph is a 3x3 grid with vertices labeled from 0 to 8. It is structured as follows: - The bottom row has vertices 0, 1, and 2 connected by horizontal edges. - The middle row has vertices 3, 4, and 5 connected horizontally, directly above the bottom row, with vertical edges connecting vertices 0 to 3, 1 to 4, and 2 to 5. - The top row has vertices 6, 7, and 8 connected horizontally, directly above the middle row, with vertical edges connecting vertices 3 to 6, 4 to 7, and 5 to 8. **Analysis:** The vertices form a grid with each vertex connected to its adjacent vertices directly beside, above, or below it. **Task:** The problem asks to demonstrate why the graph is not Hamiltonian by showing any Hamilton path must begin and end at even-numbered vertices, and explaining why this prevents forming a Hamilton cycle. A Hamilton cycle would require the path to return to its starting point, which is not possible under these conditions.
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