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Assume G = (V,E) is a complete graph.

Let the vertices be a set of points in the plane and let the edges be line segments between the points. Let the weight of each edge [a, b] be the length of the segment 'ab'.

After reading about Prim's Algorithm and Kruskal's Algorithm, I have some sound knowledge that these greedy algorithms output the minimum spanning tree of a graph.

My Question is: After obtaining a minimum spanning tree of G, Is there a way to prove that the minimum spanning tree of G is a plane graph?

arsalunic612
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  • My gut says all trees are planar. Pick a root. Put it on row zero column zero. Add its children on row one in separate columns. Repeat, each time shifting parents right to align with the leftmost of its direct children in the next row. Now none of the edges intersect. – Patrick87 Nov 30 '17 at 01:03
  • all trees are two colourable - so by the four colour theorem, it must be planar. – Rusty Rob Nov 30 '17 at 01:09
  • @Patrick87 The OP is considering a set of points in the plane with specific starting coordinates, where the lengths of the edges are the actual distances between points. It's certainly possible to construct a tree where the lines do cross if the points are fixed in space. The question is, given a set of points embedded in spaces with specific coordinates, do any of the edges of the MST intersect (other than at endpoints). – Dave Nov 30 '17 at 01:20

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You can check if the minimum spanning tree is planar as any graph. There are a simple way to check if a graph is planar. The very known Euler formula

“If G is a connected planar graph with e edges and v vertices, where v >= 3, then e <= 3v - 6. Also G cannot have a vertex of degree exceeding 5.”

or you can rely on the following method:

Theorem – “Let G be a connected simple planar graph with e edges and v vertices. Then the number of faces f in the graph is equal to f = e-v+2.”

Euler also showed that for any connected planar graph, the following relationship holds:

v - e + f = 2.

Good lucky

enyard
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