In the following graph, it is possible to travel from one vertex to any other vertex. In the graph, a vertex should have edges with all other vertices, then it called a complete graph. In other words, if a vertex is connected to all other vertices in a graph, then it is called a complete graph.
A vertex or node of a graph is one of the objects that are connected together. The connections between the vertices are called edges or links. A graph with 10 vertices or nodes and 11 edges links. For more information about graph vertices, see the network introduction. An acyclic graph is a graph having no graph cycles. Acyclic graphs are bipartite. A connected acyclic graph is known as a tree, and a possibly disconnected acyclic graph is known as a forest i.
A graph with a single cycle is known as a unicyclic graph. Make a conjecture about the relation between the degrees of a graph and its number of edges? Does that work for any graph? How would you prove it? Why do you think that this is? However, the graph in Figure A7 has the sequence of degrees: 1, 1, 1, 1, 1, 1. Experiment with this.
What conjectures guesses do you have? Can you prove them? Does the Theorem see after Q This is not true for graphs that are not simple though. Find an example. The one with each vertex joined to all of the other vertices. Make sure that you can prove that result in 2 ways. There seem to be three non-isomorphic simple graphs with every vertex of degree 2.
We've listed them in Figure A8. Does this have anything to do with the number of edges? Does it have anything to do with the degrees of vertices? Does Q16 have anything to do with this? All of these things help to simplify things, but in the last resort you have to go back to the definition to be sure if two graphs are isomorphic.
How about the number of edges or the structure of the graphs? This is a repeat of Q. For 3 vertices the maximum number of edges is 3; for 4 it is 6; for 5 it is 10 and for 6 it is There are two ways at least to prove this.
We'll show you one way and then you find the another. This continues until the last vertex has no new edges. So we have a one-to-one link here. And this will be easier if you use Q Or did you use it without knowing? Also refer forward to Q So the complement is regular.
We'll list all of the graphs with 0, 1, 2, 3, and 4 edges. The graphs on 10, 9, 8, 7 and 6 edges follow by complementation. We'll then list all of the graphs with 5 edges some of which are self-complementary. This is done in Figure A On 5 vertices: there are two of them; one with every vertex of degree 2 and the other with degrees of 1, 1, 2, 3, 3.
On 6 vertices: there are none. Why not? This problem is a non-trivial exercise. Work on graphs on up to 13 vertices and see if you can generalise from there. You might find the graph in Figure A12 useful.
The simple graph on 1 vertex is regular of degree 0. The graphs on 2 vertices are either regular of degree 0 or regular of degree 1. On 3 vertices there is one regular graph of degree 0 and one of degree 2. On 4 vertices there is one regular graph of degree 0, one of degree 1, one of degree 2, and one of degree 3.
We have shown the regular graphs of degree 2 on 8 vertices in Q21; there are no others. There are no graphs that are regular of degree 3 on 9 vertices. How many edges would such a graph have? Can we go further than this? What can be said about the number of vertices of odd degree in any graph? We'll prove the general result here. By the theorem, the sum of the degrees of all of the vertices is even.
But this sum is also the sum of the even degree vertices and the sum of the odd degree ones. Now the sum of the even degree vertices is even. So the sum of the odd degrees has to be even too. This means that there have to be an even number of vertices of odd degree. On 3 vertices the graphs with 2 or 3 edges will do the trick. On 4 vertices graphs 5, 6, 8, 9, 10, 11 in the answer to Q On the 5 vertices graphs show in Figure A12, the 9th, 10th, 12th, 15th, 16th, 18th, 19th and 20th in the answer to Q The complements of some of the graphs there are also connected.
Which ones? There are several of these. We give three in Figure A13, but you might try to find them all. Be systematic! On 3 vertices: the graphs with none and one edge. On 4 vertices: from Q 11, these are graphs 1, 2, 3, 4, and 7. There is still a vertex not joined to any other vertex, so this graph is disconnected. If you don't see how this works, try the idea out on some specific disconnected graphs.
In general, at least one of a graph or its complement is connected. This follows from the argument above. All the graphs in Figure A13 have a connected complement.
So a graph and its complement can both be connected. On 3 vertices: 1 graph with 2 edges; on 4 vertices: 2 graphs with 3 edges; on 5 vertices: 3 graphs with 4 edges. You can find them easily from the graphs in Q11 and Q Your proof has to take into account all the notes above. We'll give a proof of all the conjectures later in the text.
Suppose we have a connected graph. If there is only one way to get from any vertex to any other then the graph has no cycles. This can be done for all connected graphs, but you have to be careful to take off edges that don't disconnect the graph. How do you know when to stop though? It looks as if they all have at least two vertices of degree 1. Are there some trees with only 2 vertices of degree 1? But how to prove these things? We'll take this up in the text later when we prove a series of results about trees.
There are a lot of them There are no cycles in any tree. Suppose there were. Then you could delete an edge from that tree and it would still be connected. But this would contradict the condition of trees having the minimum number of edges in any connected graph on a given number of vertices. The answer to the uniqueness is 'sometimes'. What do you think of the following conjecture?
If a graph has a spanning path, then it is possible to go from any vertex to any other vertex by edges of the graph. Hence the graph is connected.
The graphs in Figure A14 the answer to Q48 are connected graphs that don't have spanning paths. In Figure 4, the path 1, 2, 3, 4, 5, 10, 8, 6, 9, 7 is one possible spanning path. How many more are there?
See also Q Yes, again there are generally lots of them. The thing that distinguishes connected graphs from disconnected graphs is that connected graphs have spanning trees. So we'll conjecture that a graph is connected if and only if it contains a spanning tree. We can colour it with 3 colours though. One way to do this it to colour the vertices 1, 4, 7, 8 with colour A; 2, 5, 6 with colour B; and 3, 9, 10 with colour C. Our original definition assumes that trees are connected and Theorem T1 shows that they are acyclic.
If a graph is connected and acyclic we only have to prove that it has the fewest edges of any connected graph to show that it is a tree in our original definition.
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