4. Prove the following theorem: If a graph has more than two vertices of odd degree, then it does not have an Euler Path. (Do a proof by contradiction please)
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4. Prove the following theorem: If a graph has more than two vertices of odd degree, then it does not have an Euler Path.
(Do a proof by contradiction please)
Step by step
Solved in 3 steps
- The following conditions must be met for a graph to include an Euler circuit: (1) it must be linked, and (2) each vertex must have an even degree. What are the conditions for having an Euler path?Provide a sufficient but not required condition for a graph to lack an Eulerian cycle even if it is undirected. Defend your position.Use Euler's inequality (m≤ 3n-6) to prove that each planar graph has at least two vertices of degree at most 5. (Can you do better than 2?)
- 5.1.2. (-) Prove that the chromatic number of a graph equals the maximum of the chromatic numbers of its components.please type only***Graph Complement and Planarity,(a) Let G be a 4-regular plane graph with 10 faces. Determine how many vertices and edges G hasand give a plane drawing of such a graph. for this problem, you must writecomplete sentences, include all details, show all of your work, and clarify all of your reasoning.
- Determine whether the graph is planar or not. To show planarity, give aplanar embedding. To show that a graph is not planar, use Kuratowski’s theorem5. Another Proof Consider an undirected, unweighted, simple graph G. Suppose the longest simple path in G has length 10 and goes from node u to node v. Prove that the degree of node u is at most 10.In each part, either draw a graph with the given specifications or explain why no such graph exists. A graph with 4 vertices that has a Euler circuit but no Hamiltonian circuits.