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Notebook[{
Cell[CellGroupData[{
Cell["Modelling with Directed Graphs", "Title",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell[BoxData[{
\(Author : \ Thomas\ Shores\), "\[IndentingNewLine]",
\(University\ of\ Nebraska\), "\[IndentingNewLine]",
\(Send\ comments\ \(to : \ tshores@math . unl . edu\)\)}], "Input",
FontFamily->"Helvetica",
FontWeight->"Plain",
FontVariations->{"CompatibilityType"->0}],
Cell[CellGroupData[{
Cell["Definition of a Graph", "Subsection",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell[TextData[{
"A (",
StyleBox["directed",
FontSlant->"Italic"],
") ",
StyleBox["graph",
FontSlant->"Italic"],
" is simply a set of points V, whose elements are called ",
StyleBox["vertices",
FontSlant->"Italic"],
", together with a set E of ordered pairs from V",
"\[Cross]",
"V, whose elements are called (",
StyleBox["directed",
FontSlant->"Italic"],
") ",
StyleBox["edges",
FontSlant->"Italic"],
". In the edges are not directed in the sense that whenever (a,b) is an \
edge, so is (b,a), then the graph is called ",
StyleBox["undirected",
FontSlant->"Italic"],
" or simply, a graph. All the graphs we examine will be directed.\n\
Throughout the following discussion unless otherwise stated, the graph G is \
understood to have ",
StyleBox["n",
FontSlant->"Italic"],
" vertices in the set V = {1, 2, ... , n} and ",
StyleBox["m",
FontSlant->"Italic"],
" edges in the set E = {e",
StyleBox["1",
FontVariations->{"CompatibilityType"->"Subscript"}],
", e",
StyleBox["2",
FontVariations->{"CompatibilityType"->"Subscript"}],
", ... , e",
StyleBox["m",
FontVariations->{"CompatibilityType"->"Subscript"}],
"}. Here each edge e",
StyleBox["i",
FontVariations->{"CompatibilityType"->"Subscript"}],
" is an ordered pair."
}], "Text",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell[TextData[{
"A picture is worth a thousand words. So use your Browser to look in the \
subdirectory \n/Users/Math314Files for the file called GraphExample.eps Now \
activate the figures by double clicking on the file name. You will see the \
Example used for the following discussion. Next click inside the Mathematica \
window to get back to this tutorial. Notice that the Example graph G has \
vertices V = {1, 2, 3, 4} and edges E = {e",
StyleBox["1",
FontVariations->{"CompatibilityType"->"Subscript"}],
", e",
StyleBox["2",
FontVariations->{"CompatibilityType"->"Subscript"}],
", e",
StyleBox["3",
FontVariations->{"CompatibilityType"->"Subscript"}],
", e",
StyleBox["4",
FontVariations->{"CompatibilityType"->"Subscript"}],
", e",
StyleBox["5",
FontVariations->{"CompatibilityType"->"Subscript"}],
"}. To be exact, e",
StyleBox["1",
FontVariations->{"CompatibilityType"->"Subscript"}],
" = (1,2), e",
StyleBox["2",
FontVariations->{"CompatibilityType"->"Subscript"}],
" = (2,3), e",
StyleBox["3",
FontVariations->{"CompatibilityType"->"Subscript"}],
" = (1,3), e",
StyleBox["4",
FontVariations->{"CompatibilityType"->"Subscript"}],
" = (3,4) and e",
StyleBox["5",
FontVariations->{"CompatibilityType"->"Subscript"}],
" = (1,4). We will sometimes abuse the language and refer to the edges as \
simply edges 1, 2, 3 ,4 and 5. Now for some terminology we need to talk \
about graphs :"
}], "Text",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell[TextData[{
"A ",
StyleBox["path",
FontSlant->"Italic"],
" in a graph G is a sequence of edges (v",
StyleBox["0",
FontVariations->{"CompatibilityType"->"Subscript"}],
", v",
StyleBox["1",
FontVariations->{"CompatibilityType"->"Subscript"}],
"), (v",
StyleBox["1",
FontVariations->{"CompatibilityType"->"Subscript"}],
", v",
StyleBox["2",
FontVariations->{"CompatibilityType"->"Subscript"}],
"), ... , (v",
StyleBox["m-1",
FontVariations->{"CompatibilityType"->"Subscript"}],
", v",
StyleBox["m",
FontVariations->{"CompatibilityType"->"Subscript"}],
") which goes from vertex v",
StyleBox["0",
FontVariations->{"CompatibilityType"->"Subscript"}],
" to vertex v",
StyleBox["m",
FontVariations->{"CompatibilityType"->"Subscript"}],
", where all vertices except possibly v",
StyleBox["0",
FontVariations->{"CompatibilityType"->"Subscript"}],
" and v",
StyleBox["m",
FontVariations->{"CompatibilityType"->"Subscript"}],
" are distinct. The length of the path is ",
StyleBox["m",
FontSlant->"Italic"],
". For instance, you can see that the sequence of edges 1, 2, and 4 is a \
path from vertex 1 to vertex 4 of length 3. "
}], "Clipboard",
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Cell[TextData[{
"The graphs we shall consider are all ",
StyleBox["connected",
FontSlant->"Italic"],
": this means that between any two vertices ",
StyleBox[" i",
FontSlant->"Italic"],
" and ",
StyleBox["j",
FontSlant->"Italic"],
" there is a sequence of vertices, each of which is connected by an edge, \
though these edges may run in different directions. Note this is not the \
same as saying that there is a path between any two vertices. Our Example \
graph is connected, yet there is not a path from vertex 4 to vertex 2, for \
instance; yet we see that there is an edge between vertex 4 and 3, and then \
one between vertex 3 and 2. This notion of connecting vertices by edges not \
all running in the same direction is a useful one. We can represent such an \
object as a vector whose coordinates are indexed by the edges of the graph \
and whose entries are 0, -1 or 1. For instance, in our Example, the \
connection from vertex 4 to 2 may be represented by the vector {0, -1, 0, -1, \
0}. The minus signs indicate that edges 2 and 4 are traversed in the \
negative direction, i.e., from head to tail. "
}], "Text",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell[TextData[{
"A sequence of edges which goes from a vertex back to itself without \
passing through any vertex except the initial one more than once, is called a \
",
StyleBox["loop ",
FontSlant->"Italic"],
"(or",
StyleBox[" ",
FontSlant->"Italic"],
"a",
StyleBox[" cycle",
FontSlant->"Italic"],
"). As in the previous paragraph, loops can be represented by a vector \
indexed by the edges. For instance, in our Example graph, {0,0,1,1,-1} \
represents the loop starting at vertex 1, passing through vertex 3 by way of \
edge 3, through vertex 4 by way of edge 4, and finally back to 1 by way of \
edge 5. Notice, the -1 in the 5th place means that edge 5 is traversed in \
the negative direction. There are other loops in our graph; for example, \
{1,1,-1,0,0} and {1,1,0,1,-1}. Interestingly enough, {1,1,0,1,-1} = \
{1,1,-1,0,0} + {0,0,1,1,-1}. As a matter of fact, all loops are built from \
the smallest loops {1,1,-1,0,0} and {0,0,1,1,-1}."
}], "Text",
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Cell[CellGroupData[{
Cell["The Adjacency Matrix", "Subsection",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell[TextData[{
"Adjacency Matrix: The adjacency matrix of a graph G is an n",
"\[Cross]",
"n matrix whose ",
StyleBox["(i,j)",
FontSlant->"Italic"],
"th entry is 1 if there is an edge from v",
StyleBox["i",
FontVariations->{"CompatibilityType"->"Subscript"}],
" to v",
StyleBox["j",
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", i.e., if (v",
StyleBox["i",
FontVariations->{"CompatibilityType"->"Subscript"}],
",v",
StyleBox["j",
FontVariations->{"CompatibilityType"->"Subscript"}],
") ",
"\[Element]",
" E, and is 0 otherwise. For example, the adjacency matrix A of our \
Example is a 4",
"\[Cross]",
"4 matrix defined in the cell below. Activate this cell by putting the \
insertion point in it and pressing enter or by marking it and choosing the \
option Action, then Evaluate Selection."
}], "Text",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell["A = {{0,1,1,1},{0,0,1,0},{0,0,0,1},{0,0,0,0}}", "Input",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell["\<\
Next, you might prefer to see the matrix in standard form. \
Activate the next cell:\
\>", "Text",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell["MatrixForm[A]", "Input",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell[TextData[{
"Here is the connection between adjacency matrices and paths: the number \
of distinct paths of length 1 from vertex ",
StyleBox["i",
FontSlant->"Italic"],
" to vertex ",
StyleBox["j",
FontSlant->"Italic"],
" is simply the ",
StyleBox["(i,j)",
FontSlant->"Italic"],
"th entry of A. Likewise, the number of distinct paths from vertex ",
StyleBox["i",
FontSlant->"Italic"],
" to ",
StyleBox["j",
FontSlant->"Italic"],
" of length 2 is the ",
StyleBox["(i,j)",
FontSlant->"Italic"],
"th entry of A A. And so on! The ",
StyleBox["(i,j)",
FontSlant->"Italic"],
"th entry of A times itself ",
StyleBox["m",
FontSlant->"Italic"],
" times gives the number of paths from vertex ",
StyleBox["i",
FontSlant->"Italic"],
" to vertex ",
StyleBox["j",
FontSlant->"Italic"],
" of length ",
StyleBox["m",
FontSlant->"Italic"],
". It follows that the number of paths from vertex ",
StyleBox["i",
FontSlant->"Italic"],
" to",
StyleBox[" j",
FontSlant->"Italic"],
" of length less than or equal to ",
StyleBox["m",
FontSlant->"Italic"],
" is the ",
StyleBox["(i,j)",
FontSlant->"Italic"],
"th entry of the sum of matrices A + A A + ... + A A ... A, where the last \
product is A multiplied by itself ",
StyleBox["m",
FontSlant->"Italic"],
" times. Try experimenting with the following two cells and compare with \
the picture of our Example A to see how things work. Activate them, then \
change them to see the effects of computing longer paths by adding more \
terms. For example, the following product has (1,4)th entry equal to 1 and, \
sure enough, there is only one path of length two from vertex 1 to vertex 4 \
(see it?). Also notice that the next cell has (1,4)th entry equal to 2, \
which is exactly the number of paths of length at most 2 from vertex 1 to \
vertex 4."
}], "Text",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell["MatrixForm[A.A]", "Input",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell["MatrixForm[A + A.A]", "Input",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell[TextData[{
"There are several ideas that are very important in applications of graph \
theory to the social sciences. One of the idea of a \
\[OpenCurlyDoubleQuote]",
StyleBox["dominance-directed",
FontSlant->"Italic"],
"\[CloseCurlyDoubleQuote] graph. This is a graph that has no paths from a \
vertex to itself, and for any pair of distinct vertices, ",
StyleBox["i",
FontSlant->"Italic"],
" and ",
StyleBox["j",
FontSlant->"Italic"],
", there may be a path from ",
StyleBox["i",
FontSlant->"Italic"],
" to ",
StyleBox["j",
FontSlant->"Italic"],
", or from",
StyleBox[" j",
FontSlant->"Italic"],
" to ",
StyleBox["i",
FontSlant->"Italic"],
", but not both. For such a graph, the \[OpenCurlyDoubleQuote]power\
\[CloseCurlyDoubleQuote] of a vertex is defined to be the number of paths of \
length 1 or 2 from that vertex to any other vertex. For instance, you can \
see in our Example, which is a dominance-directed graph, that the power of \
vertex 2 is 2 (there is a path of length 1 to vertex 3 and one of length 2 to \
vertex 4). In general, the power of the ",
StyleBox["i",
FontSlant->"Italic"],
"th vertex is simply the sum of all the entries in row ",
StyleBox["i",
FontSlant->"Italic"],
" of A + A A, in view of the discussion of the previous paragraph. Here \
is an illustration of the idea: suppose a group of teams play a round-robin \
tournament of games with no ties. Build a graph as follows. The vertices of \
the graph are the teams. There is an edge from ",
StyleBox["i",
FontSlant->"Italic"],
" to",
StyleBox[" j",
FontSlant->"Italic"],
" if team ",
StyleBox["i",
FontSlant->"Italic"],
" defeated team",
StyleBox[" j.",
FontSlant->"Italic"],
" Thus, the power of a team is the number of teams that it has defeated or \
that have been defeated by a team it defeated. Add up the entries of each \
row of the matrix A + A A in our Example and you see that vertex 1 is the \
most powerful of the vertices, while vertex 4 is the least powerful. \n \
This idea of dominance-directed graphs is used in sociological studies of \
group interactions, where \[OpenCurlyDoubleQuote]dominance\
\[CloseCurlyDoubleQuote] means cooperation between groups, which are \
represented as vertices of a graph. In voting analysis studies, dominance \
represents influence with respect to voting choice. In communications, \
dominance can mean the ability of one vertex to communicate with another."
}], "Text",
ImageRegion->{{-0, 1}, {0, 1}}]
}, Closed]],
Cell[CellGroupData[{
Cell["The Incidence Matrix", "Subsection",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell[TextData[{
"The next important idea about graphs that we will discuss is that the the \
",
StyleBox["edge-vertex incidence matrix",
FontSlant->"Italic"],
" (",
StyleBox["incidence matrix",
FontSlant->"Italic"],
" for short) of a graph. Here is the idea: form a matrix B whose rows \
are indexed by the edges of the graph, and whose columns are indexed by the \
vertices of the graph. Now if edge e",
StyleBox["i",
FontVariations->{"CompatibilityType"->"Subscript"}],
" is the ordered pair (j,k), put a -1 in the ",
StyleBox["j",
FontSlant->"Italic"],
"th column of row",
StyleBox[" i",
FontSlant->"Italic"],
", a +1 in the ",
StyleBox["k",
FontSlant->"Italic"],
"th column of row ",
StyleBox["i",
FontSlant->"Italic"],
" and 0 in every other entry of row ",
StyleBox["i. ",
FontSlant->"Italic"],
"The result is the incidence matrix B of the graph.\n For instance, the \
incidence matrix of our Example is given below. Activate this cell to define \
the incidence matrix and the subsequent cell to get a nice form for the \
matrix."
}], "Text",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell["B = {{-1,1,0,0},{0,-1,1,0},{-1,0,1,0},{0,0,-1,1},{-1,0,0,1}}", "Input",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell["MatrixForm[B]", "Input",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell["\<\
Notice that the sum of the entries in each row is 0. Hence the \
vector {1,1,1,1} belongs to the null space of B. Now it is not hard to see \
that the rank of B is 3. In fact, it in instructive to see the row-echelon \
form of B. Try activating the next cell to see this.\
\>", "Text",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell["MatrixForm[RowReduce[B]]", "Input",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell["Next, have a look at the null space of B:", "Text",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell["NullSpace[B]", "Input",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell[TextData[{
"Is this an accident? No. Here is what happens in general: if G is a \
connected graph and B is the incidence matrix of G, then the rank of B is \
exactly ",
StyleBox["n",
FontSlant->"Italic"],
" - 1, where G has ",
StyleBox["n",
FontSlant->"Italic"],
" vertices, and the vector {1,1,...,1} spans the null space of B."
}], "Text",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell["\<\
An even more interesting object is the null space of the transpose \
of B. Try activating the next cell.\
\>", "Text",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell["NullSpace[Transpose[B]]", "Input",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell["\<\
Now compare this output to the discussion of loops at the end of \
the introductory section of these notes and note that these amount to two of \
the three loops (up to a factor of \[PlusMinus]) of our Example graph. \
Moreover, we could get the third loop (up to a sign) by subtracting these two \
loops. In other words, all the information about loops is contained in the \
null space of the transpose of B.\
\>", "Text",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell[TextData[{
"There is one more context in which the incidence matrix turns out to be \
interesting. Suppose our graph G represents an electrical curcuit. The \
edges might represent locations in the circuit where objects like resistors \
or capacitors are placed. At each vertex we would have a potential assigned; \
say at vertex ",
StyleBox["i",
FontSlant->"Italic"],
" we have potential value x",
StyleBox["i",
FontVariations->{"CompatibilityType"->"Subscript"}],
". Then the jump in potential across the ",
StyleBox["k",
FontSlant->"Italic"],
"th edge e = (i, j) would simply be x",
StyleBox["j",
FontVariations->{"CompatibilityType"->"Subscript"}],
" - x",
StyleBox["i",
FontVariations->{"CompatibilityType"->"Subscript"}],
". Interestingly enough, this is exactly the ",
StyleBox["k",
FontSlant->"Italic"],
"th entry of Bx. So suppose we wanted to design a circuit where the \
potential differences across the edges is given by the vector ",
StyleBox["f",
FontSlant->"Italic"],
". Is it possible for any vector ",
StyleBox["f",
FontSlant->"Italic"],
"? The answer is no. For ",
StyleBox["f",
FontSlant->"Italic"],
" represents the potential differences across edges corresponding to some \
distribution of potential values ",
StyleBox["x",
FontSlant->"Italic"],
" if and only if the equation Bx = ",
StyleBox["f",
FontSlant->"Italic"],
" is consistent. Since the rank of B is ",
StyleBox["n",
FontSlant->"Italic"],
" - 1, this will not always happen, because the system will not be \
consistent for every choice of ",
StyleBox["b",
FontSlant->"Italic"],
". When will it happen?"
}], "Text",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell[TextData[{
"Both null spaces can be informative in this context of electrical \
curcuits. For example, we get a necessary condition for Bx =",
StyleBox[" f",
FontSlant->"Italic"],
" to have a solution. Let ",
StyleBox["y",
FontSlant->"Italic"],
" be any element of the null space of the transpose of B. Then yB = 0. \
Given that Bx =",
StyleBox[" f",
FontSlant->"Italic"],
", we can multiply on the left by (the transpose of) ",
StyleBox["y",
FontSlant->"Italic"],
" to obtain that y ",
StyleBox["f",
FontSlant->"Italic"],
" = yBx = 0 x = 0 . In particular, if ",
StyleBox["y",
FontSlant->"Italic"],
" represents a loop, then the condition y ",
StyleBox["f ",
FontSlant->"Italic"],
"= 0",
StyleBox[" ",
FontSlant->"Italic"],
"really says that the total potential drop around the loop should be 0.\n\
The null space of B is also relevant. If x",
StyleBox["0",
FontVariations->{"CompatibilityType"->"Subscript"}],
" is any particular solution to the equation Bx = ",
StyleBox["f",
FontSlant->"Italic"],
", then so is x",
StyleBox["0",
FontVariations->{"CompatibilityType"->"Subscript"}],
" + z, where ",
StyleBox["z",
FontSlant->"Italic"],
" is any element of the null space of B. We have seen above that ",
StyleBox["z",
FontSlant->"Italic"],
" will simply be a multiple of {1,1,...,1}. So what does this say? It \
says that if we add the same constant to the potential at each vertex of a \
circuit, we obtain a new vector of potentials whose potential differences is \
the same as before."
}], "Text",
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Cell["Exercises", "Subsection",
ImageRegion->{{-0, 1}, {0, 1}}],
Cell[TextData[{
"You are given that the (directed) graph G has vertex set V = {1,2,3,4,5} \
and edge set \nE = {(2,1), (1,5), (2,5), (5,4), (4,2), (4,3), (3,2)}. \
Answer the following questions about the graph G:\n\n(1) What does the \
graph look like? You may draw this by hand or, if you prefer, you may use \
the Draw application (it has the icon with a compass on it) to draw the \
picture. \n\n(2) Next view the graph as representing an electrical circuit \
with potentials x",
StyleBox["1",
FontVariations->{"CompatibilityType"->"Subscript"}],
",...,x",
StyleBox["5",
FontVariations->{"CompatibilityType"->"Subscript"}],
" to be assigned to the vertices. Find N(A) and N(A",
StyleBox["T",
FontSize->11,
FontVariations->{"CompatibilityType"->"Superscript"}],
") using the Mathematica command NullSpace. What does the latter tell you \
about the loop structure of the circuit? Finally, use that fact that Ax = b \
implies that for all ",
StyleBox["y",
FontSlant->"Italic"],
" in N(A",
StyleBox["T",
FontSize->11,
FontVariations->{"CompatibilityType"->"Superscript"}],
"), y",
StyleBox["T",
FontSize->11,
FontVariations->{"CompatibilityType"->"Superscript"}],
" b = 0 to find conditions that a vector ",
StyleBox["b",
FontSlant->"Italic"],
" must satisfy in order for it to be a vector of potential differences for \
some potential distribution on the vertices.\n\nMath 314H students : this \
last exercise is not part of the assignment. Do not turn it in.\n\n(3) \
View this graph as representing a communications network. First find the \
adjacency matrix of the graph. Notice that G is a dominance-directed \
graph. Which point(s) have the greatest power? Which point(s) would be the \
best from which to initiate communications, based on this information?\n"
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*)
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End of Mathematica Notebook file.
***********************************************************************)