Results 1 to 4 of 4

Math Help - Graph Theory

  1. #1
    Member Jason Bourne's Avatar
    Joined
    Nov 2007
    Posts
    132

    Graph Theory

    every planar connected simple graph G has at least 3 vertices of degree no more than 5.

    every planar connected simple graph G with 11 or fewer vertices has at least 1 vertex of degree no more than 5.

    Question: show that every planar simple graph with 11 or fewer vertices is 5 colourable.

    ~~~~~~~~~~~#
    im not sure how to prove the first statment for the 3 vertices but I need to know how to prove the last statement.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    For the first question we work by contradiction.

    Assume there are at most 2 vertices of degrees less or equal than 5. Then <br />
e = \tfrac{1}<br />
{2} \cdot \left( {\sum\limits_{x \in V\left( G \right)} {\deg \left( x \right)} } \right) \geqslant \tfrac{1}<br />
{2} \cdot \left( {\deg(u)+\deg(w) + 6 \cdot \left( {v - 2} \right)} \right)=<br />
\tfrac{{\deg \left( u \right) + \deg \left( w \right)}}<br />
{2}<br />
+3(v-2)<br />
(1) - u and w are the vertices that may have degree less or equal than 5 -

    However, if G is planar and v=|V(G)|\geq{3} we have: <br />
3v - 6 \geqslant e <br />
(2) NOw note that (1) and (2) cannot be true simultaneously. -since G is connected-
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Jason Bourne's Avatar
    Joined
    Nov 2007
    Posts
    132
    how can I prove it's 5 - colourable?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    May 2009
    From
    Tokyo, Japan
    Posts
    46
    Just use induction to the number of vertices. Remove a vertice with 5 or less edges, make a 5-coloring and then inspect the colors. if the node had less than 5 edges you are done, if the vertice had 5 edges of less than 5 colors ur done. If the vertice had 5 edges with 5 colors the fact that the K_5 isn't planar saves you, you can smelt two non-connected nodes together, make a 5-coloring, then remove the nodes again and color the removed node in the remaining color.

    That's how the 5 coloring proof goes, if you need a proper one with mathematical notation etc. feel free to ask and I'll cook one up.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Graph theory, bipartite Graph
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: March 10th 2012, 05:47 AM
  2. Graph Theory / Chromatic Number of a Complete Graph
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 15th 2011, 09:59 AM
  3. (Graph Theory)Prove that graph X is a tree.
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: August 1st 2011, 03:30 AM
  4. Replies: 0
    Last Post: September 25th 2010, 05:59 AM
  5. Graph Theory - Size of a Line Graph
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: July 25th 2010, 11:15 PM

Search Tags


/mathhelpforum @mathhelpforum