Problem

Solutions For Applied Combinatorics Chapter 1.2 Problem 2E

Step-by-Step Solution

Solution 1

To list all nonisomorphic directed graphs with 3 vertices, do the following. First note that there are 6 possible edges. Now consider the cases where number of edges range from 0 to 6. For each case, draw all possible graphs, and check if any are isomorphic to any other. Keep only one graph from each set of isomorphic graphs.

Step #2 of 7

Case 1: 0 edges. In this case, only one graph is possible. It is:

$D:\Dropbox\Chegg TBS\514969-Applied Combinatorics\Solutions\1.2-2E-1.png$

Case 2: 1 edge. In this case, only one graph is possible, as there is only one edge. It is:

$D:\Dropbox\Chegg TBS\514969-Applied Combinatorics\Solutions\1.2-2E-2.png$

Step #3 of 7

Case 3: 2 edges. As there are only 3 vertices, the edges have to share an endpoint. Thus, they may share one or two endpoints. When they share two endpoints, only one case is possible.

$D:\Dropbox\Chegg TBS\514969-Applied Combinatorics\Solutions\1.2-2E-3.png$

When they share 1 endpoint, there are several cases to consider. The shared vertex might be the incoming vertex for both, outgoing vertex for both, or incoming for one and outgoing for the other. Hence the three possible graphs are

$D:\Dropbox\Chegg TBS\514969-Applied Combinatorics\Solutions\1.2-2E-4.png$ $D:\Dropbox\Chegg TBS\514969-Applied Combinatorics\Solutions\1.2-2E-5.png$ $D:\Dropbox\Chegg TBS\514969-Applied Combinatorics\Solutions\1.2-2E-6.png$

Step #4 of 7

Case 4: 3 edges. The three edges can either form a triangle, or two of the edges can share two endpoints.

When the three edges form a triangle, there are two possible cases. Either the edges can share the same orientation, or one edge can have opposite orientation to the other. The graphs are:

$D:\Dropbox\Chegg TBS\514969-Applied Combinatorics\Solutions\1.2-2E-8.png$ $D:\Dropbox\Chegg TBS\514969-Applied Combinatorics\Solutions\1.2-2E-9.png$

When two edges share two endpoints, then the third edge has two possible orientations, towards the edges or away from them. The two graphs are:

$D:\Dropbox\Chegg TBS\514969-Applied Combinatorics\Solutions\1.2-2E-10.png$ $D:\Dropbox\Chegg TBS\514969-Applied Combinatorics\Solutions\1.2-2E-11.png$

Step #5 of 7

Case 5: 4 edges. This is similar to Case 3, as the non-edges can be considered to be edges from Case 3, hence there are 4 graphs. The graphs are:

$D:\Dropbox\Chegg TBS\514969-Applied Combinatorics\Solutions\1.2-2E-12.png$

Step #6 of 7

$D:\Dropbox\Chegg TBS\514969-Applied Combinatorics\Solutions\1.2-2E-13.png$ $D:\Dropbox\Chegg TBS\514969-Applied Combinatorics\Solutions\1.2-2E-14.png$ $D:\Dropbox\Chegg TBS\514969-Applied Combinatorics\Solutions\1.2-2E-15.png$

Step #7 of 7

Case 6: 5 edges. This is similar to Case 2, as non-edges can be considered as edges from Case 2. Hence there is only one graph which is:

$D:\Dropbox\Chegg TBS\514969-Applied Combinatorics\Solutions\1.2-2E-16.png$

Case 7: 6 edges. As all possible edges are present, there is only one possible case. It is:

$D:\Dropbox\Chegg TBS\514969-Applied Combinatorics\Solutions\1.2-2E-17.png$

These are all the possible cases.

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Solutions For Problems in Chapter 1.2

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