How to Construct a Tree with Ten Vertices of Degree 1 or 5

Introduction

In graph theory, a tree with n vertices has n-1 edges. This adds a unique constraint on the sum of the degrees of all vertices, given by 2n-2. This article explores how to construct a tree with ten vertices such that each vertex has a degree of either 1 or 5. We will also discuss the underlying principles and equivalencies that make this construction possible.

Theoretical Background

Let's start by examining the conditions under which a tree can be constructed with a specific set of vertex degrees. Consider a tree with n vertices, where ell vertices have degree 1 and the remaining n-ell vertices have degree 5.

The sum of the degrees of the vertices in such a tree must equal 2n-1. Given the specified degrees, we can write the equation:

1*ell 5*(n-ell) 2n-2

This equation simplifies to:

5n - ell 2n-2

Solving for ell, we get:

ell 3n - 2

For this equation to have a positive integer solution for ell, n must be even. If n is even, say n 2m, then:

ell 3m - 2

This suggests that for n to be of the form 4k 2, ell 3k - 1, indicating the number of vertices of degree 1. Consequently, the number of vertices of degree 5 is k.

Existence and Uniqueness of Such a Tree

From the theory of graph construction, a tree with a given degree sequence exists if and only if the sum of the degrees is even. Specifically, for our case, the degree sequence is 1, 5, 1, 5, ..., 1, 5 over n vertices. We can apply the following result to confirm the existence:

For any sequence of positive integers a_1, ..., a_n, there exists a tree with degree sequence a_1, ..., a_n if and only if the sum of the degrees is 2n-2.

Given that the sum of the degrees is indeed 2n-2, we can proceed to construct such a tree.

Constructing a Tree with Ten Vertices

Given n 10, we need to satisfy the condition that the sum of the degrees equals 18. Therefore:

5*10 - ell 2*10 - 2

Solving for ell, we get:

ell 30 - 18 12

This means we need 12 vertices of degree 5, which is not possible since we only have 10 vertices. Thus, we need to adjust our approach to ensure the sum of the degrees matches 2n-2.

By setting n 10 and ell 2, we satisfy:

1*2 5*(10-2) 18

Hence, we can construct a tree with 2 vertices of degree 5 and 8 vertices of degree 1. This configuration ensures that the sum of the degrees is 18, which is 2*10 - 2.

Inductive Proof and Further Insights

The existence of such a tree can be proven using induction. Given a sequence of positive integers a_1, ..., a_n, the following holds:

The sequence represents the degree sequence of a tree if and only if the sum of the degrees is 2n-2.

For the set S {1, 5}, the equation:

5x_1 - 1x_2 10 - 2

must have a solution in positive integers. Here, x_1 and x_2 represent the number of vertices of degree 1 and 5, respectively. Solving for x_1 and x_2, we find the specific configuration that satisfies the sum condition.

By solving the equation, we determine that the unique solution is two vertices of degree 5 and eight vertices of degree 1, forming a connected tree. This configuration is the only possible solution for a tree with ten vertices where each vertex has a degree of either 1 or 5.

In conclusion, the construction of a tree with ten vertices of degree 1 or 5 is feasible if and only if we have a specific configuration that meets the sum of degrees condition. This example illustrates the constraints and possibilities in graph theory and demonstrates the application of degree sequence conditions in constructing trees.