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Unformatted text preview: 232 Chapter 9 Graphs CHAPTER 9
Graphs SECTION 9.1 Graphs and Graph Models 2. a) A simple graph would be the model here, since there are no parallel edges or loops, and the edges are
undirected.
b) A multigraph would, in theory, be needed here, since there may be more than one interstate highway
between the same pair of cities. c) A pseudograph is needed here, to allow for loops.
4. This is a multigraph; the edges are undirected, and there are no loops, but there are parallel edges.
6. This is a multigraph; the edges are undirected, and there are no loops, but there are parallel edges.
8. This is a directed multigraph; the edges are directed, and there are parallel edges. 10. The graph in Exercise 3 is simple. The multigraph in Exercise 4 can be made simple by removing one of the
edges between a and b, and two of the edges between b and d. The pseudograph in Exercise 5 can be made
simple by removing the three loops and one edge in each of the three pairs of parallel edges. The multigraph
in Exercise 6 can be made simple by removing one of the edges between a and c, and one of the edges between
I; and d. The other three are not undirected graphs. (Of course removing any superscts of the answers given
here are equally valid answers; in particular, we could remove all the edges in each case.) 12. If u RU, then there is an edge joining vertices 'u and v, and since the graph is undirected, this is also an edge
joining vertices c and v. This means that @1311. Thus the relation is symmetric. The relation is reﬂexive
because the loops guarantee that aRu. for each vertex u. 14. Since there are edges from Hawk to Crow, Owl, and Raccoon, the graph is telling us that the hawk competes with these three animals. 16. Each person is represented by a vertex, with an edge between two vertices if and only if the people are
acquainted. Hanks 5am; Patric!) Jeff Har“
Am; lion 18. Fred inﬂuences Brian, since there is an edge from Fred to Brian. Yvonne and Deborah influence Fred, since
there are edges from these vertices to Fred . 20. Team four beat the vertices to which there are edges from Team four, namely only Team three. The other
teamsiTeam one, Team two, Team ﬁve, and Team sixﬁall beat Team four, since there are edges from them
to Team four. Section 9.1 Graphs and Graph illodels 233 ' 22. 24. 26. 28. 30. 32. 34. This is a directed multigraph with one edge from a to b for each call made by e to l). Rather than draw
the parallel edges with parallel lines, we have indicated what is intended by writing a numeral on the edge to
indicate how many calls were made, if it was more than one. 1333 1200 2222 0091 This is similar to the use of directed graphs to model telephone calls. a) We can have a vertex for each mailbox or email address in the network, with a directed edge between two
vertices if a message is sent from the tail of the edge to the head. b) As in part (a) we use a directed edge for each message sent during the week.
Vertiees with thousands 01' millions of edges going out from them could be such lists. We make the subway stations the vertices, with an edge from station “u to station n if there is a train going
from u to 1) without stopping. It is quite possible that some segments are one—way, so we should use directed
edges. (If there are no onesway segments, then we could use undirected edges.) There would be no need for
multiple edges, unless we had two kinds of edges, maybe with different. colors, to represent local and express
trains. In that case, there could be parallel edges of different colors between the same vertices, because both a
local and an express train might travel the same segment. There would be no point in having loops, because
no passenger would want. to travel from a station back to the same station without stopping. The model says that the statements for which there are edges to 85 must be executed before 36, namely the
statements 31, SQ, S3, and 3.1. The vertices in the directed graph represent cities. \Vhenever there is a nonstop flight from city A to city B,
we put a directed edge into our directed graph from vertex A to vertex B, and furthermore we label that
edge with the ﬂight time. Let us see how to incorporate this into the mathematical deﬁnition. Let us call
such a thing a directed graph with weighted edges. It is deﬁned to be a triple (V. E, W), where (V, E) is
a. directed graph (i.e., V is a set of vertices and E is a set of ordered pairs of elements of V) and W is a
function from E to the set of Ilonnegative real numbers. Here we are simply thinking of ill/(e) as the weight
of edge e, which in this case is the flight time. We can let. the vertices represent people; an edge from n. to '1; would indicate that u can send a message to e.
We would need a directed multigraph in which the edges have labels, where the label on each edge indicates
the form of communication (cell phone audio, text meSsaging, and so on). 234 Chapter 9 Graphs SECTION 9.2 Graph Terminology and Special Types of Graphs 2. In this pseudograph there are 5 vertices and 13 edges. The degree of vertex 0. is 6, since in addition to
the 4 nonloops incident to a, there is a loop contributing 2 to the degree. The degrees of the other vertices
are deg(b) 2 (i, deg(c) 2 6, deg(r1) = 5, and deg(e) 2 3. There are no pendant or isolated vertices in this
pseudograph. 4. For the graph in Exercise 1, the sum is 2+4 +1 +0+2 +3 2 l2 2 2 . 6; there are 6 edges. For the pseudograph
in Exercise 2, the sum is 0+ 6 +6 +5+3 2 26 2 2 13; there are 13 edges. For the pseudograph in Exercise 3,
the sum is 3+2+4+0+6+0+4+2+3 2 24 2 212; there are 12 edges. 6. Model this problem by letting the vertices of a graph be the people at the party. with an edge between two
people if they shake hands. Then the degree of each vertex is the number of people the person that vertex
represents shakes hands with. By Theorem 1 the sum of the degrees is even (it is 26). 8. In this directed multigraph there are 4 vertices and 8 edges. The degrees are deg—(a) = 21 deg+(e) 2 2,
deg—(b) 2 3, deg+[b) 2 4, deg"(e) 2 2, deg+(e) 2 1, deg—(d) 2 1, and deg+{(i) 2 1. 10. For Exercise 7 the sum of the in~degrees is 3+ 1 +2+1 2 7, and the sum of the outdegrees is 1+2+1 +3 2 7;
there are 7 edges. For Exercise 8 the sum of the in—degrees is 2 + 3+ 2 + l 2 8, and the sum of the outdegrees
is 2 + 4 + 1 + 1 2 8; there are 8 edges. For Exercise 9 the sum of the inndegrees is 6 + l + 2 + 4 + 0 2 13,
and the sum of the out—degrees is 1 + 5 + 5 + 2 + 0 = 13; there are 13 edges. 12. Since there is an edge from a person to each of his or her acquaintances, the degree of u is the number of
people v knows. An isolated vertex would be a person who knew on one, and a pendant vertex would be a
person who knew just one other person (it is doubtful that there are many, if any, isolated or pendant vertices).
lithe average degree is 1000, then the average person knows 1000 other people. 14. Since there is an edge from a person to each 01' the other actors that person has appeared with in a movie,
the degree of I} is the number of other actors that person has appeared with. An isolated vertex Would be a
person who has appeared only in movies in which he or she was the only actor, and a pendant vertex would
be a person who has appeared with only one other actor in any movie (it is doubtful that there are many, if
any, isolated or pendant vertices). 16. Since there is an edge from a page to each page that it links to, the outdegree of a vertex is the number of
links on that page, and the iiidegree of a vertex is the number of other pages that have a link to it. 18. This is essentially the same as Exercise 36 in Section 5.2. where the graph models the “know each other”
relation on the people at the party. See the solution given for that exercise. The number of people a person knows is the degree of the corresponding vertex in the graph. 20. a) This graph has 7 vertices, with an edge joining each pair of distinct vertices. ...
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 Spring '09
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