Math 2.0

Combinatorics Proposition

K5 and K3,3 are not planar

Builds on Complete graphComplete bipartite graphPlanar graphEdge bound for planar graphs

Statement

The complete graph K5K_5 and the complete bipartite graph K3,3K_{3,3} are not planar.

Remarks

The graph K5K_5 has 55 vertices and 1010 edges. The edge bound for planar graphs allows at most 3⋅5−6=93 \cdot 5 - 6 = 9 edges. The graph K3,3K_{3,3} has 66 vertices and 99 edges, and it has no cycle of length 33. The bound e≤2v−4e \le 2v - 4 allows at most 88 edges. The statement for K3,3K_{3,3} is the three utilities problem: three houses cannot each be joined to three utilities without crossings.

Reading path

To understand this page from the ground up, read these 10 pages in order. The foundations come first.

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Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.
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  1. Complete bipartite graph example
  2. Complete graph example
  3. Graph definition
  4. Planar graph definition
  5. Walk definition
  6. Reachability definition
  7. Connected graph definition
  8. Euler's formula for planar graphs statement
  9. Edge bound for planar graphs statement
  10. K5 and K3,3 are not planar statement

Where this leads

1 page builds on this one.

Further reading

  1. Web https://en.wikipedia.org/wiki/Three_utilities_problem, Unsolvability
About this page
Tag030A3FHN14
URIm2://030A3FHN14
Kindstatement, proposition
WritingW0 stub
MSC05C10
StatusThis copy of the site has no build data. Run m2 status 030A3FHN14.
UsesComplete graph, Complete bipartite graph, Planar graph, Edge bound for planar graphs

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