🚧 Fin 4.9

appendix_b.txt

@@ -1 +1 @@
-928
+930

chapter_4/exercises.md

@@ -8835,14 +8835,50 @@ Check that the number of edges equals one-half of the total degree.
 
 1. See page 265.
 
+$\text{deg(v_1)} = 3$
+
+$\text{deg(v_2)} = 2$
+
+$\text{deg(v_3)} = 4$
+
+$\text{deg(v_4)} = 2$
+
+$\text{deg(v_5)} = 1$
+
+$\text{deg(v_6)} = 0$
+
+$\text{deg}(\text{total}) = 12$
+
+$\text{total edges} = 6$
+
 2. See page 265.
 
+$\text{deg(v_1)} = 1$
+
+$\text{deg(v_2)} = 5$
+
+$\text{deg(v_3)} = 4$
+
+$\text{deg(v_4)} = 4$
+
+$\text{deg(v_5)} = 1$
+
+$\text{deg(v_6)} = 3$
+
+$\text{deg}(\text{total}) = 18$
+
+$\text{total edges} = 9$
+
 3. A graph has vertices of degrees 0, 2, 2, 3, and 9. How many edges does the
    graph have?
 
+$$ \frac{1}{2}(0 + 2 + 2 + 3 + 9) = 8 \text{ edges} $$
+
 4. A graph has vertices of degrees 1, 1, 4, 4, and 6. How many edges does the
    graph have?
 
+$$ \frac{1}{2}(1 + 1 + 4 + 4 + 6) = 8 \text{ edges} $$
+
 In each of 5-13 either draw a graph with the specified properties or explain why
 no such graph exists.
 
@@ -8850,14 +8886,27 @@ no such graph exists.
 
 6. Graph of four vertices of degrees 1, 2, 3, and 3.
 
+Not possible. It has an odd number of total degrees, 9. By 4.9.2, the total
+degree of a graph must be even.
+
 7. Graph with four vertices of degrees 1, 1, 1, and 4.
 
+Not possible. It has an odd number of total degrees, 7. By 4.9.2, the total
+degree of a graph must be even.
+
 8. Graph with four vertices of degrees 1, 2, 3, and 4.
 
 9. Simple graph with four vertices of degrees 1, 2, 3, and 4.
 
+No such graph. The vertex of degree 4 would have to be connected by edges to 4
+distinct vertices other than itself. This is not possible in a simple graph
+since it cannot loop back on itself.
+
 10. Simple graph with five vertices of degrees 2, 3, 3, 3, and 5.
 
+Not possible, as the vertex of degree 5 would have to loop back on itself in a
+graph of 5 vertices, which contradicts the definition of a simple graph.
+
 11. Simple graph with five vertices of degrees 1, 1, 1, 2, and 3.
 
 12. Simple graph with six edges and all vertices of degree 3.
@@ -8871,8 +8920,20 @@ no such graph exists.
 
 a. How many people attending the party knew three other people before the party?
 
+$$ (2 \cdot 1) + (5 \cdot 2) + 3x = 2 + 10 + 3x = 12 + 3x $$
+
+$$ 12 + 3x = 2(15) $$
+
+$$ 12 + 3x = 30 $$
+
+$$ 3x = 18 $$
+
+$$ \boxed{x = 6} $$
+
 b. How many people attended the party?
 
+$$ 2 + 5 + 6 = \boxed{13} $$
+
 15. A small social network contains three people who are network friends with
     six other people in the network, one person who is network friend with five
     other people in the network, and five people who are network friends with
@@ -8882,29 +8943,84 @@ b. How many people attended the party?
 
 a. How many people are network friends with three other people in the network?
 
+$$ (3 \cdot 6) + (1 \cdot 5) + (5 \cdot 4) + 3x = 41(2) $$
+
+$$ 43 + 3x = 82 $$
+
+$$ 3x = 39 $$
+
+$$ \boxed{x = 13} $$
+
 b. How many people are in the network?
 
+$$ 3 + 1 + 5 + 13 = \boxed{22} $$
+
 16.
 
 a. In a group of 15 people, is it possible for each person to have exactly 3
 friends? Justify your answer. (Assume that friendship is a symmetric
 relationship: If $x$ is a friend of $y$, then $y$ is a friend of $x$.)
 
+**Proof by contradiction:**
+
+Suppose that, in a group of 15 people, each person had exactly three friends.
+Then you could draw a graph representing each person by a vertex and connecting
+two vertices by an edge if the corresponding people were friends. But such a
+graph would have 15 vertices, each of degree 3, for a total of 45. This would
+contradict the fact that the total degree of any graph is even. Hence the
+supposition must be false, and in a group of 15 people it is not possible for
+each to have exactly three friends.
+
 b. In a group of 4 people, is it possible for each person to have exactly 3
 friends? Justify your answer.
 
+**Proof:**
+
+Suppose that, in a group of 4 people, each person has exactly 3 friends. Then
+you could draw a graph representing each person by a vertex and connecting two
+vertices by an edge if the corresponding people were friends. Such a graph would
+have 4 vertices, each of degree 3, for a total of 12. The total degree is even,
+and therefore it is possible for each person to have exactly 3 friends.
+
 17. In a group of 25 people, is it possible for each to shake hands with exactly
     3 other people? Justify your answer.
 
+No $25 \cdot 3 = 75$ is odd number of total degrees.
+
 18. Is there a simple graph, each of whose vertices has even degree? Justify
     your answer.
 
+Yes, a minimum number of vertices would be 3. Each vertex would have a degree of
+2 that would reach out to the other two vertices.
+
 19. Suppose that $G$ is a graph with $v$ vertices and $e$ edges and that the
     degree of each vertex is at least $d_{\text{min}}$ and at most
     $d_{\text{max}}$. Show that
 
 $$ \frac{1}{2}d_{\text{min}} \cdot v \leq e \leq \frac{1}{2}d_{\text{max}} \cdot v $$
 
+**Proof:**
+
+Let $e$ be the total number of edges, let $t$ be the total degree of the graph,
+let $d_{\text{min}}$ be the minimum degree of any vertex in $G$, and let
+$d_{\text{max}}$ be the maximum degree of any vertex in $G$.
+
+The total degree of $G$ is greater than or equal to the minimum degree times the
+total amount of vertices.
+
+$$ d_{\text{min}} \cdot v \leq t  $$
+
+Also the total degree of $G$ is less than or equal to the maximum degree times
+the total amount of vertices.
+
+$$ d_{\text{min}} \cdot v \leq t \leq d_{\text{max}} \cdot v $$
+
+The total degree of $G$ is $2e$.
+
+$$ d_{\text{min}} \cdot v \leq 2e \leq d_{\text{max}} \cdot v $$
+
+$$ \frac{1}{2}d_{\text{min}} \cdot v \leq e \leq \frac{1}{2}d_{\text{max}} \cdot v $$
+
 20.
 
 a. Draw $K_6$, a complete graph on six vertices.
@@ -8912,20 +9028,90 @@ a. Draw $K_6$, a complete graph on six vertices.
 b. Use the result of Example 4.9.9 to show that the number of edges of a simple
 graph with $n$ vertices is less than or equal to $\dfrac{n(n - 1)}{2}$.
 
+Prove that for any positive integer $n$, the number of edges of a simple graph
+with $n$ vertices is less than or equal to $\dfrac{n(n - 1)}{2}$.
+
+**Proof:**
+
+Suppose $n$ and $e$ are any positive integers such that a simple graph $K_n$ has
+$n$ vertices and $e$ edges.
+
+By Example 4.9.9, we know the number of edges of a complete graph, $K_m$ is:
+
+$$ \text{the number of edges of } K_m = \frac{m(m - 1)}{2} $$
+
+A simple graph is a graph that does not have any loops or parallel edges, while
+a complete graph is a simple graph with $m$ vertices and exactly one edge
+connecting each pair of distinct vertices.
+
+It follows then that the total number of edges for the simple graph $K_n$ could
+only have at most the total number of edges for a complete graph of $n$
+vertices.
+
+Therefore we have proven that:
+
+$$ e \text{ for } K_n \leq \frac{n(n - 1)}{2} $$
+
+Q.E.D.
+
 21.
 
 a. In a simple graph, must every vertex have degree that is less than the number
 of vertices in the graph? Why?
 
+Yes. Let $G$ be a simple graph with $n$ vertices and let $v$ be a vertex of $G$.
+Since $G$ has no parallel edges, $v$ can be joined by at most a single edge to
+each of the $n - 1$ other vertices of $G$, and since $G$ has no loops, $v$
+cannot be joined to itself. Therefore, the maximum degree of $v$ is $n - 1$.
+
 b. Can there be a simple graph that has four vertices all of different degrees?
 Why?
 
+No. Suppose there is a simple graph with four vertices, all of which have
+different degrees. By part (a), no vertex can have a degree greater than three,
+and of course, no vertex can have a degree less than $0$. Therefore, the only
+possible degrees of the vertices are 0, 1, 2, and 3. Since all four vertices
+have different degrees, there is one vertex with each degree. But then the
+vertex of degree 3 is connected to all other vertices, which contradicts the
+fact that one of the vertices has degree 0. Hence the supposition is false, and
+there is no simple graph with four vertices each of which has a different
+degree.
+
 c. For any integer $n \geq 5$, can there be a simple graph that has $n$ vertices
 all of different degrees? Why?
 
+No, let $n$ be an integer such that $n \geq 5$, and let $G$ be a simple graph
+with $n$ vertices. By part (a) no vertex can have a degree greater than $n - 1$.
+Since $n \geq 5$, then $n - 1 \geq 4$. Therefore the only possible degrees of
+the vertices are $0, 1, \dots, n - 1$.
+
+Since there are $n$ vertices and $n$ possible degree values, each degree must
+occur exactly once. In particular there must be a vertex of degree $0$ and a
+vertex of degree $n - 1$.
+
+However, a vertex of degree $n - 1$ is adjacent to every other vertex in the
+graph, including the vertex of degree $0$. This contradicts the fact that a
+vertex of degree $0$ is adjacent to no vertices.
+
 22. In a group of two or more people, must there always be at least two people
     who are acquainted with the same number of people within the group? Why?
 
+Yes.
+
+Let the group contain $n \geq 2$ people. Model the situation with a simple graph
+$G$ where each person is represented by a vertex. Two vertices are connected by
+an edge if the corresponding people are acquainted.
+
+Then the degree of a vertex is the number of people in the group that person is
+acquainted with.
+
+By Exercise 21, a simple graph with $n$ vertices cannot have all $n$ vertices of
+different degrees. Equivalently, there must be at least two vertices with the
+same degree.
+
+Therefore, there must be at least two people who are acquainted with the same
+number of people within the group.
+
 23. Recall that $K_{m, n}$ denotes a complete bipartite graph on $(m, n)$
     vertices.
 
@@ -8937,11 +9123,21 @@ c. Draw $K_{3, 4}$.
 
 d. How many vertices of $K_{m, n}$ have degree $m$? degree $n$?
 
+Vertices that have degree $m$ are all vertices in $\{n\}$.
+
+Vertices that have degree $n$ are all vertices in $\{m\}$.
+
 e. What is the total degree of $K_{m, n}$?
 
+$$ (m \cdot n) + (n \cdot m) = 2mn $$
+
 f. Find a formula in terms of $m$ and $n$ for the number of edges of $K_{m, n}$.
 Justify your answer.
 
+$$ e = mn $$
+
+Justification omitted.
+
 24. A (general) **bipartite graph** $G$ is a simple graph whose vertex set can
     be partitioned into two disjoint nonempty subsets $V_1$ and $V_2$ such that
     vertices in $V_1$ may be connected to vertices in $V_2$, but no vertices in
@@ -8963,3 +9159,5 @@ See Page 266.
 25. Suppose $r$ and $s$ are any positive integers. Does there exist a graph $G$
     with the property that $G$ has vertices of degrees $r$ and $s$ and no other
     degrees? Explain.
+
+Omitted.

chapter_4/test_yourself.md

@@ -258,17 +258,30 @@ Page 265
 
 1. The total degree of a graph is defined as ______.
 
+The sum of the degrees of all the vertices of the graph
+
 2. The handshake theorem says that the total degree of a graph is ______.
 
+equal to the number of edges of the graph
+
 3. In any graph the number of vertices of odd degree is ______.
 
+an even number
+
 4. A simple graph is ______.
 
+a graph with no loops or parallel edges
+
 5. A complete graph on $n$ vertices is a ______.
 
+a simple graph with $n$ vertices whose set of edges contains exactly one edge
+for each pair of vertices
+
 6. A complete bipartite graph on $(m, n)$ vertices is a simple graph whose
    vertices can be divided into two distinct, non-overlapping sets, say $V$ with
    $m$ vertices and $W$ with $m$ vertices, in such a way that (1) there is
    ______ from each vertex of $V$ to each vertex of $W$, (2) there is ______
    from any one vertex of $V$ to any other of $V$, and (3) there is ______ from
    any one vertex of $W$ to any other vertex of $W$.
+
+one edge; no edge; no edge

leftoff.txt

@@ -1 +1 @@
-258
+267