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Let G be a simple connected graph on 5 vertices all of degree at least 3. Must G have a Hamiltonian circuit? Explain.
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Start by assuming each vertex has a degree of exactly 3. For any given vertex, you have \(\dbinom43=4\) possible ways of connecting to three other vertices. Here's one such arrangement: |dw:1430965672570:dw|
Let's consider the extreme case, in which we connect one of the currently 1-degree vertices to two others. We have \(\dbinom32=3\) ways of doing so. Here's one: |dw:1430965854041:dw|
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