7 The Mathematics of Networks

2. 7 The Mathematics of Networks 7.1 Trees 7.2 Spanning Trees 7.3 Kruskal’s Algorithm 7.4 The Shortest Network Connecting Three Points 7.5 Shortest Networks for Four or More Points

3. Reformulate Property 3 to . . . PROPERTY 4 ■If a network has N vertices and M edges, then M ≥ N – 1.[For convenience, we will refer to the difference R = M – (N – 1) as the redundancyof the network.] ■If M = N – 1, the network is a tree; ifM > N – 1,the network has circuits and is not a tree. (In other words, a tree is a network with zero redundancy, and a network with positive redundancy is not a tree.)

4. Spanning Trees In the case of a network with positive redundancy, there are many treeswithin the network that connect its vertices–these are the spanning trees of thenetwork.

5. Example 7.4 Counting Spanning Trees The network in the figure has N = 8vertices and M = 8 edges. The redundancyof the network is R = M – (N – 1)= 1, so to find a spanning tree we will have to“discard” one edge.

6. Example 7.4 Counting Spanning Trees Five of these edges are bridges of the network, and they willhave to be part of any spanning tree. The other three edges (BC, CG, and GB) forma circuit of length 3, and if we exclude any one of the three edges, then we will havea spanning tree.

7. Example 7.4 Counting Spanning Trees Thus, the network has three different spanning trees.

8. Example 7.4 Counting Spanning Trees The network in the Figure has M = 9 edges and N = 8 vertices. The redundancy of the network is R = 2,so to find a spanning tree we will have to “discard”two edges. Edges AB and AH are bridges of the network, so they will have to bepart of any spanning tree.

9. Example 7.4 Counting Spanning Trees The other seven edges are split into two separate circuits(B, C, G, B of length 3 and C, D, E, F, C of length 4). A spanning tree can be foundby “busting” each of the two circuits. This means excluding any one of the threeedges of circuit B, C, G, B and any one of the four edges of circuit C, D, E, F, C.

10. Example 7.4 Counting Spanning Trees Forexample, if we exclude BC and CD, we get the spanning tree shown.

11. Example 7.4 Counting Spanning Trees We could also exclude BC and DE and get the spanning tree shown.

12. Example 7.4 Counting Spanning Trees Given that there are 3  4 = 12different ways to choose an edge fromthe circuit of length 3 and an edge from the circuit of length 4, there are 12 spanning trees. We have already shown two, here’s one more.

13. Example 7.5 Counting Spanning Trees This network has M = 9 edges and N = 8 vertices. Here the circuits B, C, G, B and C, D, E, G, C share a common edge CG. Determiningwhich pairs of edges can be excluded in this case is a bit more complicated.

14. Example 7.6 Minimum Spanning Trees This example builds on the ideas introduced in Example 7.2. The network shownin is the same network as Example 7.2 now with weights added tothe edges.

15. Example 7.6 Minimum Spanning Trees Vertices represent computer labs, and edges are potential Ethernet connections. The weights represent the cost in K’s(1K = $1000) of installing the Ethernet connections.

16. Example 7.6 Minimum Spanning Trees Using the terminology we introduced in this section, the weighted network has a redundancy ofR = 3(the network hasM = 14vertices andN = 12edges).

17. Example 7.6 Minimum Spanning Trees The network has many possible spanning trees, and our job is to find one with least weight, that is, a minimum spanning tree (MST) for the network.We know that we can find a spanning tree byexcluding certain edges of the network (those that close circuits) from the spanning tree and that there are many different ways in which this can be done. Giventhat the edges now have weights, a reasonable strategy for sorting through theoptions would be to always try to exclude from the network the most expensive edges.

18. Example 7.6 Minimum Spanning Trees To illustrate the point, let’s look at circuit B, C, D, E, B on the left side of the network. It makes sense to exclude CD(which costs $95,000 to build).

19. Example 7.6 Minimum Spanning Trees Likewise, on the right side of the network wehave a configuration of two circuits K, H, I, Kand K,J,I,K that share the common edgeKI.

20. Example 7.6 Minimum Spanning Trees We exclude the two most expensive edges from these two circuits (KJand KI).

21. Example 7.6 Minimum Spanning Trees At this point, all the circuits of theoriginal network are “busted,” and we endup with the red spanning tree shown.

22. Minimum Spanning Trees Is the red spanning tree obtained in Example 7.6 the MST of the originalnetwork? We would like to think it is, but how can we be sure, especially after thelessons learned in Chapter 6? And even if it is, what assurances do we have that oursimple strategy will work in more complicated graphs? These are the questionswe will answer next.