Thursday 10-27

    Last class we applied Euler's Formula for the plane or the sphere:
    Theorem: For any connected graph in the plane,

V+R = E + 2.
OR  V -E+R =2  

    We'll continue to examine more applications of this topological property of graphs (networks) in the plane.

    Last class we applied "Euler" to analysis of the Utilities problem.

      Complete graphs. A complete graph is a graph in which each vertex of the graph is connected by an edge to every other vertex.

      For example:

      2 vertices: 1 edge - a line segment
      3 vertices: 3 edges -  a triangle
      4 vertices: 6  edges - a tetrahedron
      Notice these all have planar representations.
      The technical way this is decribed is to say these graphs can be imbedded in the plane.
      What about 5 vertices?
      Draw a complete graph using 5 vertices.

    Discussion: Analysis. Suppose this were possible and consider the consequences.
    If the graph were drawn it would have V = 5 vertices and E= ?_    __ edges.
    Based on Euler's formula, the graph would therefore have to have R=?                regions in the plane.

    By design each region would have to have at least 3 edges.
    Thus if we count the edges by the regions we would have at least 3R=21 edges counting each edge twice- because an edge is a boundary for exactly two regions.
    Thus the number of edges must be more than 3/2R=10 1/2.
    Is this possible? What can we conclude?

    Coloring problems: Suppose we have a map that consists of a graph that deteremines R regions ( countries) with E edges (borders).  How many colors are needed to color the regions so that countries that share a border (not merely a vertex) have different colors.
    The Junkyard on the color problem.

From Serendip at Bryn Mawr College
The Shockwave movie below makes it possible for you to play with the four color problem yourself. Create a "map of countries" of any number, shape, and size by drawing in the space, or let Serendip create a map for you by clicking on the Generate Map button. The question is how many colors are required to color such a map, using the rule that no two countries with adjacent borders may be the same color (countries meeting at a point may be the same color). 

Once generated, either by you or by Serendip, click on Submit Map. After a processing delay, Serendip will then present the map to you in grey, and you can color individual countries by clicking on them (or let Serendip color the map for you). See if you can create a map that requires two colors, or three colors, or four colors. If you create one that requires five colors, you will upset mathematicians (and a few others) worldwide.Warning: Serendip has a small bug- so that sometimes it will give a less than best coloring of the map or will evaluate your work incorrectly.

Created by Paul Grobstein and Sasha Schwartz. Shockwave programming by Sasha Schwartz. Reactions, comments, suggestions gratefully received.

Send us your comments at Serendip

© by Serendip 1994-2003 - Last Modified: Wednesday, 05-Mar-2003 17:11:21 EST

Examples: A map using only circles (possibly interesecting) needs only two colors!

    Draw a map that needs 3 colors.
    Draw a map that needs 4 colors.
    Draw a map that needs 5 colors.

    The Four Color Theorem: Any map on the plane or the sphere can be colored with at most 4 colors.
    The Color problems: Discussion and proof of the five color theorem.

    Some References for "proofs":
    summary of a new proof of the Four Color Theorem and a four-coloring algorithm found by Neil Robertson, Daniel P. Sanders, Paul Seymour and Robin Thomas.
    article by Eric W. Weisstein © 1999 CRC Press LLC, © 1999-2003 Wolfram Research, Inc.

From the Math History web site:

Article by: J J O'Connor and E F Robertson JOC/EFR September 1996

However the final ideas necessary for the solution of the Four Colour Conjecture had been introduced before these last two results. Heesch in 1969 introduced the method of discharging. This consists of assigning to a vertex of degree i the charge 6-i. Now from Euler's formula we can deduce that the sum of the charges over all the vertices must be 12. A given set S of configurations can be proved unavoidable if for a triangulation T which does not contain a configuration in S we can redistribute the charges (without changing the total charge) so that no vertex ends up with a positive charge.

Heesch thought that the Four Colour Conjecture could be solved by considering a set of around 8900 configurations. There were difficulties with his approach since some of his configurations had a boundary of up to 18 edges and could not be tested for reducibility. The tests for reducibility used Kempe chain arguments but some configurations had obstacles to prevent reduction.

The year 1976 saw a complete solution to the Four Colour Conjecture when it was to become the Four Colour Theorem for the second, and last, time. The proof was achieved by Appel and Haken, basing their methods on reducibility using Kempe chains. They carried through the ideas of Heesch and eventually they constructed an unavoidable set with around 1500 configurations. They managed to keep the boundary ring size down to less than or equal to 14 making computations easier that for the Heesch case. There was a long period where they essentially used trial and error together with unbelievable intuition to modify their unavoidable set and their discharging procedure. Appel and Haken used 1200 hours of computer time to work through the details of the final proof. Koch assisted Appel and Haken with the computer calculations.

The Four Colour Theorem was the first major theorem to be proved using a computer, having a proof that could not be verified directly by other mathematicians. Despite some worries about this initially, independent verification soon convinced everyone that the Four Colour Theorem had finally been proved. Details of the proof appeared in two articles in 1977. Recent work has led to improvements in the algorithm.