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Matrices and spanning trees

  1. Sep 30, 2011 #1
    1. The problem statement, all variables and given/known data
    The costs (in millions of dollars) of connecting any two of the four cities A,B,C and D by telephone lines are given in the following matrix:

    0 3 5 4
    3 0 2 3
    5 2 0 6
    4 3 6 0

    a) Draw a diagram of the complete graph
    b) find a minimal spanning tree

    3. The attempt at a solution

    I honestly have no idea what the question is even asking.

    my best guess is that I need to draw a cost vs city bar graph so 2-6 million dollars on the y axis, then fill in for A-B, A-C etc.
    is that right?

    And I am not 100% sure about minimal spanning trees either. Would you draw 4 vertices labelled A to D then on the connecting lines fill in the proper cost values?
     
  2. jcsd
  3. Oct 2, 2011 #2
  4. Oct 2, 2011 #3
    A graph is a collection of vertices and edges. Your matrix describes the http://en.wikipedia.org/wiki/Graph_%28mathematics%29" [Broken]. Algorithms for calculating spanning trees are well known and can be found in any introductory book on graph theory.
     
    Last edited by a moderator: May 5, 2017
  5. Oct 2, 2011 #4
    Note that the wiki-link I gave you on spanning trees also hyperlinks to minimal spanning trees. Links to algorithms are supplied therein.
     
  6. Oct 2, 2011 #5
    Notice that trees satisfy the relation V=E+1 , where V,E are the number of vertices, edges
    resp in the tree. First, make sure your tree is connected (A tree is a graph that is
    connected, but has no loops. You need to find a subset of your graph that goes
    thru every vertex, but has no loops).

    So start at any vertex v , and join v to some edge v1 so that v1 is " 1 away" from
    v, i.e., there is a path vev1 of length 1 from v to v1 . Then you have a subgraph
    {v,v1,e} , with 2 vertices and 1 edge, i.e., a tree. Continue with the idea until the
    tree spans, i.e., until the graph uses all the vertices (again, make sure your graph is
    connected.) Can you take it from there?
     
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