\magnification 2050 \parindent 0pt \parskip 5pt \nopagenumbers \hsize 7.5truein \vsize 9.9 truein \hoffset -0.45truein \def\u{\vskip -7pt} \def\v{\vskip -5pt} \input ../../../PAPERS/GOLD/psfig 2.3: Communication Network Question: What (and who) are the minimum number of people you need to give a message in order for the message to reach everyone one else in the network. Answer: Find a vertex basis. Defn: A vertex basis of $D = (V, A)$ is a set of vertices $B \subset V$ such that (1) Every vertex in $V$ is reachable form some vertex in $B$. (2) If $C$ proper $B$ then there exists a $v \in V$ such that $v$ is not reachable from any vertex in $C$. Defn: A strong component of $D$ is a strongly connected subgraph generated by a set of vertices such that this set of vertices is maximal with respect to strong connectedness. Thm 2.6: Every vertex is in exactly one strong component. Defn: $D^* = (V^*, A^*)$ is the condensation of $D = (V, A)$ if (1) $V^* = \{K_1, ..., K_p\} =$ the set of strong components of $D$. (2) There exists and arc $(K_i, K_j)$ iff there exists $u \in K_i$ and $v \in K_j$ such that $(u, v) \in A$. Thm 2.7: $D^*$ is acyclic (i.e. $D^*$ does not contain any cycles). Thm 2.8: An acyclic digraph has a unique vertex base consisting of all vertices with no incoming arcs. Cor: In an acyclic digraph, there exists a vertex with no incoming arcs. Thm 2.9: Let $B^*= \{K_{i_1}, ...,K_{i_n}\} $ be the vertex basis of $D^*$ where $D^*$ is the condensation of $D$. Then the set $\{v_1, ..., v_n\}$ is a vertex basis of $D$ if $v_j \in K_{i_j}$. Furthermore, every vertex basis of $D$ can be formed in this way. Find a vertex basis for the following digraph: \centerline{\psfig{figure=graph2_3.eps,width=5truein}} \vskip 10pt \hrule \vskip 10pt \centerline{\psfig{figure=graph2_3.eps,width=5truein}} \vskip 10pt \hrule \vskip 10pt \centerline{\psfig{figure=graph2_3.eps,width=5truein}} \end 2.4: Digraphs and matrices Recall if $A = (a_{ij}) $ and $B = (b_{ij}) $ are two $n \times n$ matrices, then \centerline{$A + B = ( a_{ij} + b_{ij} )$ } \centerline{$A \times B = ( a_{ij} b_{ij} )$ } \centerline{$AB = ( c_{ij} ) $ where $c_{ij} = \Sigma_{k=1}^n a_{ik}b_{kj}$ = $row(i) \cdot column(j)$} The following refer to the digraph $D = (V, A)$ where $V = (v_1, ..., v_n)$ where the vertices have not been given a particular order. Defn: The adjacency matrix of $D = (V, A) = A(D) = (a_{ij})$ where $a_{ij} = \cases{1 & if $(v_i, v_j) \in A$ \cr 0 & otherwise}$. Thm 2.11: If $A$ is the adjacency matrix of $D$, then the $i, j$th entry of $A^k$ is the number of paths of length $k$ from $v_i$ to $v_j$. Defn: The reachability matrix of $D = R(D) = (r_{ij})$ where $r_{ij} = \cases{1 & if $v_j$ is reachable from $v_i$ \cr 0 & otherwise}$. Defn: The Boolean function $b: {\cal M} \rightarrow \{0, 1\}$ where ${\cal M} = \{0, 1, 2, ... \}$ is defined by $b(x) = \cases{0 & if $x = 0$ \cr 1 & otherwise}$. $B:$ set of matrices with nonnegative integer entries $ \rightarrow $ set of matrices with entries 0 and 1 is defined by $B(A) = ( b(a_{ij} ) )$. Thm 2.12: Suppose $A$ is the adjacency matrix and $R$ is the reachability matrix of the digraph $D$. If $D$ has $n$ vertices, then \centerline{$R = B(I + A + A^2 + ... _ A^{n-1}) = B( (I + A)^{n-1} )$} Defn: $I =$ the identity matrix = $(i_{kp})$ where $i_{kp} = \cases{0 & if $k \not= p$ \cr 1 & if $k = p$}$ $J = (j_{kp})$ where $j_{kp} = 1$ for all $k, p$. $R' =(r_{ji})$ is the transpose of $R = (r_{ij})$. Thm 2.13 (Ross and Harary 1959). Suppose $A$ is the adjacency matrix and $R$ is the reachability matrix of the digraph $D$. Let $n =$ the number of vertices of $D$. Then (a.) $D$ is strongly connected if and only if $R = J$. (b.) $D$ is unilaterally connected if and only if $B(R + R') = J$. (c.) $D$ is weakly connected if and only if $B( I + A + A' )^{n-1}) = J$. Thm 2.14. Suppose $R$ is the reachability matrix of the digraph $D$. Let $R \times R' = (t_{ij})$ Let $R^2 = (s_{ij})$. Then (a.) $v_j$ is in the strong component containing the vertex $v_i$ if and only if $t_{ij} = 1$. (b.) The number of vertices in the strong component $u_i$ is $s_{ii}$. Defn: The distance matrix of $D$ the matrix $(d_{ij})$ where $d_{ij}$ is the distance from vertex $v_i$ to $v_j$. Thm 2.15: $d_{ij}$ is the smallest number $k$ such that $B(A^k) = 1$. HW: 2.3: 1, 2, 3, 4, 6, 10, 11, 18; 2.4: 1, 2, 3, 4, 5, 7, 8, 12 (13 or 14 or 15). \end Thm: If $X$ is Hausdorff, then the diagonal, $D$, is closed in $X \times X$. Need to prove $D$ is closed in $X \times X$. Question: How do you prove a set is closed? a.) Show $X \times X - D$ is open. b.) Show $D' \subset D$. c.) Show $D = \overline{D}$. d.) Show $D$ is a finite union of closed sets or an arbitrary union of open sets. e.) etc. Question: What is given? If $X$ is Hausdorff. Hence for every pair of points in $X$ there exists a pair of disjoint open sets containing them (on exams give the precise definition). Plus we also have some theorems from $X$ Hausdorff. Question: How can we relate what is given to what we need to prove. Which definitions/theorems involving what we need to prove are most related to which definitions/theorems of what is given. The definition of Hausdorff gives open sets (in X). Hence we will try focusing on the definition $D$ closed in $X \times X$ iff $X \times X - D$ is open in $X \times X$. Concern: $X$ Hausdorff gives open sets in $X$, but we need open set in $X \times X$. We will ignore this concern for either of the following two reasons: 1.) We have to try something. Might as well give this a try. Working blindly with little motivation as to why this might work may be disconcerting, BUT if often results in the correct answer. If you don't know what to do, try something (hopefully relating what is given to what you need to prove even if weakly related). When successful, look back over your proof and see if you can figure out why it worked and how you could have come up with is with more motivation. 2.) $X \times X$ has the product topology, thus we can create the needed open set in $X \times X$ by taking the product of two open sets in $X$. Since $X$ Hausdorff gives us two open sets in $X$, this is excellent motivation. Thus we will try to show $X \times X - D$ is open in $X \times X$. There are many ways to show a set is open. We'll try the definition. Take $(x, y) \in X \times X - D$ (note since it often helps to be specific, I didn't take a point $p$ in $X \times X - D$, but instead took an ordered pair $(x, y)$). \end Dear Viveca, Just got your letter and am excited that you and your family are planning to visit Iowa. You are, of course, welcome to stay with us for as long as your time allows. I will probably be visiting Texas in June and perhaps the beginning of July, so after July ? would be best for me. If you can only make it before July , mother will stay in Iowa, so you are welcome anytime. My schedule is not yet set, but one of my students is defending her thesis this summer so it will not be very flexible, but I will definitely be free after July . Last summer we bought a badminton set, but have not had time to use it as we have spent our time unpacking boxes and replacing the upstairs carpet with laminate flooring. Mother and I are doing the work ourselves so it is taking longer than expected. But I think it will look good in a couple of months when we have finished the upstairs.