Chapter 4: Problem 30
Prove that, if \(G\) is a 3-connected plane graph, then its geometric dual is a simple graph.
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Chapter 4: Problem 30
Prove that, if \(G\) is a 3-connected plane graph, then its geometric dual is a simple graph.
These are the key concepts you need to understand to accurately answer the question.
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(i) Use Euler's formula to prove that, if \(G\) is a connected planar graph of girth 5 with \(n\) vertices and \(m\) edges, then \(m \leq 5 / 3(n-2)\). Deduce that the Petersen graph is non-planar. (ii) Obtain an inequality, generalizing that in part (i), for connected planar graphs of girth \(r\).
Let \(G\) be a polyhedron (or polyhedral graph), each of whose faces is bounded by a pentagon or a hexagon. (i) Use Euler's formula to show that \(G\) must have at least 12 pentagonal faces. (ii) Prove, in addition, that if \(G\) is such a polyhedron with exactly three faces meeting at each vertex (such as a football), then \(G\) has exactly 12 pentagonal faces.
Let \(G\) be a simple plane graph with fewer than 12 faces, in which each vertex has degree at least 3 . (i) Use Euler's formula to prove that \(G\) has a face bounded by at most four edges. (ii) Give an example to show that the result of part (i) is false if \(G\) has 12 faces.
A planar graph \(G\) is outerplanar if \(G\) can be drawn in the plane so that all of its vertices lie on the exterior boundary. (i) Show that \(K_{4}\) and \(K_{2,3}\) are not outerplanar. (ii) Deduce that, if \(G\) is an outerplanar graph, then \(G\) contains no subgraph homeomorphic or contractible to \(K_{4}\) or \(K_{23}\). (The converse result also holds, yielding a Kuratowski-type criterion for a graph to be outerplanar.)
Show that the dual of the cube graph is the octahedron graph, and that the dual of the dodecahedron graph is the icosahedron graph. What is the dual of the tetrahedron graph?
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