\magnification 2400 \parindent 0pt \parskip 8pt \nopagenumbers \hsize 7.5truein \vsize 9.9 truein \hoffset -0.45truein \def\u{\vskip -7pt} \def\v{\vskip -5pt} \input ../../../PAPERS/GOLD/psfig Suppose $D^*$ is the condensation of $D= (V, A)$. Let $K_1, ..., K_p$ be the strong components of $D$. Then $\{K_1, ..., K_p\}$ are the vertices of $D^*$. Thm 2.7: $D^*$ is acyclic. Proof 1: Suppose $D^*$ contains a cycle $K_{i_1}, ..., K_{i_m}$ where $K_{i_1} = K_{i_m}$. Then there exists $v_{j,b} \in K_{i_j}$, $v_{j+ 1,e} \in K_{i_{j+1}$ such that $(v_{j,b}, v_{j+ 1,e}) \in A$. Since $v_{j, e}, v_{j,b}$ are in the same strong component, $K_{i_j}$, there exists a path $v_{j, e}, u_{j, 1}, ..., u_{j, n_j}, v_{j,b}$. Similarly $v_{m, e}, v_{1, b}$ are in the same strong component, $K_{i_1}= K_{i, m}$, so there exists a path $v_{m, e}, u_{1, 1}, ..., u_{1, n_1}, v_{1,b}$ Thus $v_{1, b}, v_{2, e}, u_{2, 1}$, ..., $u_{2, n_2}$, $v_{2, b}$, $v_{3, e}$, $u_{3, 1}$, ..., $u_{3, n_2}$, $v_{3, b}$, ..., $v_{m-1, b}$, $v_{m, e}$, $u_{1, 1}, ..., u_{1, n_1}, v_{1,b}$ is a closed path in $D$. Hence there exists a path from $v_{1, b}$ to $v_{2, e}$ (namely $v_{1, b}, v_{2, e}$) and there is a path from $v_{2, e}, v_{1, b}$ (namely $v_{2, e}, u_{2, 1}, ..., u_{2, n_2}, v_{2, b}, v_{3, e}, u_{3, 1}, ...,$ $u_{3, n_2}$, $v_{3, b},$ ..., $v_{m-1, b}$, $v_{m, e}$, u_{1, 1}, ..., u_{1, n_1}, v_{1,b}$). Thus $v_{2, e} \in K_{i_1}$. But this is a contradiction by thm 2.6 since $v_{2, e} \in K_{i_2}$ and $K_{i_1} \not= K_{i_2}$ (since $K_{i_1}, ..., K_{i_m}$ is a cycle). Note: using a lot of notation helped us to write the above proof. Often notation can help us in writing a proof and can also help our understanding. But proofs using less notation and more English are also valid and sometimes are easier to understand. I will first simplify the above proof by using different notation. Instead of giving the path in terms of vertices, I can just give it a name, say ${\bf p_j}$. Thus in the following ${\bf p_j}$ will refer to the path $v_{j, e}, u_{j, 1}, ..., u_{j, n_j}, v_{j,b}$ ${\bf p_1}$ will refer to the path $v_{m, e}, u_{1, 1}, ..., u_{1, n_1}, v_{1,b}$. Proof 2: Suppose $D^*$ contains a cycle $K_{i_1}, ..., K_{i_m}$ where $K_{i_1} = K_{i_m}$. Then there exists $v_{j,b} \in K_{i_j}$, $v_{j+ 1,e} \in K_{i_{j+1}$ such that $(v_{j,b}, v_{j+ 1,e}) \in A$. Since $v_{j, e}, v_{j,b}$ are in the same strong component, $K_{i_j}$, there exists a path ${\bf p_j}$ from $v_{j, e}$ to $v_{j,b}$. Similarly $v_{m, e}, v_{1, b}$ are in the same strong component, $K_{i_1} = K_{i, m}$, so there exists a path ${\bf p_1}$ from $v_{m, e}$ to $v_{1,b}$. Thus ${\bf p_2}{\bf p_3}, ..., {\bf p_m}{\bf p_1}$ is a closed path in $D$ starting and ending at $v_{1, b}$. Since $v_{2, e}$ is a vertex in this closed path, there exists a path from $v_{1, b}$ to $v_{2, e}$ and there is a path from $v_{2, e}, v_{1, b}$ Thus $v_{2, e} \in K_{i_1}$. But this is a contradiction by thm 2.6 since $v_{2, e} \in K_{i_2}$ and $K_{i_1} \not= K_{i_2}$ (since $K_{i_1}, ..., K_{i_m}$ is a cycle). Now with more English: Proof 3: Suppose $D^*$ contains a cycle $K_{i_1}, ..., K_{i_m}$ where $K_{i_1} = K_{i_m}$. Then there exists $v_{j,b} \in K_{i_j}$, $v_{j+ 1,e} \in K_{i_{j+1}$ such that $(v_{j,b}, v_{j+ 1,e}) \in A$. Since $v_{j, e}, v_{j,b}$ are in the same strong component, $K_{i_j}$, there exists a path from $v_{j, e}$ to $v_{j,b}$. We can now form a closed path by joining these paths together. Since this path is closed, all the vertices in this path must be in the same strong component. Hence $v_{1,b} \in K_{1_j}$ and $v_{2,e} \in K_{i_{2}$ are in the same strong component. Thus $K_{i_1} = K_{i_2}$, But this is a contradiction since $K_{i_1}, ..., K_{i_m}$ is a cycle. Note the first proof has the most detail. If anyone questions your proof, you need to be able to provide more details. For example, in the 2nd and 3rd proof, I state that if two vertices are in the same closed path, then these vertices must be in the same strong component. This is true, but we don't have a theorem stating it. Thus it would be valid for someone to be unsure if it were true and to ask me to provide more detail. Sometimes additional notation is necessary to provide the detail. Sometimes more notation is not necessary. {\bf NOTE:} Sometimes things that seem to be obviously true are NOT true. Hence anything you state, you should be able to prove or refer to a theorem or definition. A general rule for what needs to be written in a proof is that is you can prove it, you don't need to (since you can prove it, you know it is true), but if you don't know how to prove it, then you must prove it (in case it is not true). \eject 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 been given a particular order. Defn: The adjacency matrix of $D = (V, A)$ is \centerline{$A(D) = (a_{ij})$ 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{$R = B(I + A + A^2 + ... + A^{n-1}) = B( (I + A)^{n-1} )$} \vskip 6pt \hrule $I =$ 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 $n =$ \# of vertices of $D$. Then a.) $D$ is strongly connected iff $R = J$. b.) $D$ is unilaterally connected iff $B(R + R') = J$. c.) $D$ is weakly connected iff $B( I + A + A' )^{n-1}) = J$. \eject Thm 2.14. $R =$ 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.) \# of vertices in the strong component $u_i$ = $s_{ii}$. \vskip 8pt \hrule Defn: The distance matrix of the graph $G$ is 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 $B(A^k) = 1$. \vskip 8pt \hrule HW\#4 (due 2/20): 2.3: 1, 2, 3, 4, 6, 10, 11, 18 and give rough estimate of computational complexity of TSP. HW\#5 (due 2/27): 2.4: 1, 2, 3, 4, 5, 7, 8, 12 AND 13, 14, 15. \end Thm 2.8: An acyclic digraph, $D$ has a unique vertex basis consisting of all vertices with no incoming arcs. Proof: Let $B$ be the set of all vertices with no incoming arcs. If $v \in B$. Then since $v$ has no incoming arcs, $v$ is only reachable from itself. Thus $B$ must be a subset of any vertex base. Suppose $u \not\in B$. Let $u_1 = u$. Then $u_1$ has an incoming arc $(u_2, u_1)$. If $u_2 \in B$, then $u_1$ is reachable from a vertex in $B$. If $u_2 \not\in B$ then $u_2$ has an incoming arc $(u_3, u_2)$. Suppose the path $u_n, ..., u_1$ is defined such that all vertices are distinct. If $u_n \in B$, then $u$ is reachable from a vertex in $B$. If $u_n \not\in B$ then $u_n$ has an incoming arc $(u_{n+1}, u_n)$. If $u_{n+1} = u_i$ for some $i = 1, ..., n$, then $u_{n+1}, u_n, ..., u_i$ is a cycle, a contradiction. Hence all the vertices of $u_{n+1}, u_n, ..., u_1$ are distinct. Since the number of vertices of $D$ is finite, this process must eventually end, say with the path $u_t, ..., u_1$. Since we cannot continue this process, $u_t$ must not have any incoming arcs. Hence $u_t \in B$, and hence $u$ is reachable from a vertex in $B$. Thus any vertex basis must be contained in $B$. Hence $B$ is the unique vertex basis of $D$.