857 lines
18 KiB
TeX
857 lines
18 KiB
TeX
\chapter{Matrices}
|
|
|
|
\index{matrix}
|
|
|
|
A \key{matrix} is a mathematical concept
|
|
that corresponds to a two-dimensional array
|
|
in programming. For example,
|
|
\[
|
|
A =
|
|
\begin{bmatrix}
|
|
6 & 13 & 7 & 4 \\
|
|
7 & 0 & 8 & 2 \\
|
|
9 & 5 & 4 & 18 \\
|
|
\end{bmatrix}
|
|
\]
|
|
is a matrix of size $3 \times 4$, i.e.,
|
|
it has 3 rows and 4 columns.
|
|
The notation $[i,j]$ refers to
|
|
the element in row $i$ and column $j$
|
|
in a matrix.
|
|
For example, in the above matrix,
|
|
$A[2,3]=8$ and $A[3,1]=9$.
|
|
|
|
\index{vector}
|
|
|
|
A special case of a matrix is a \key{vector}
|
|
that is a one-dimensional matrix of size $n \times 1$.
|
|
For example,
|
|
\[
|
|
V =
|
|
\begin{bmatrix}
|
|
4 \\
|
|
7 \\
|
|
5 \\
|
|
\end{bmatrix}
|
|
\]
|
|
is a vector that contains three elements.
|
|
|
|
\index{transpose}
|
|
|
|
The \key{transpose} $A^T$ of a matrix $A$
|
|
is obtained when the rows and columns in $A$
|
|
are swapped, i.e., $A^T[i,j]=A[j,i]$:
|
|
\[
|
|
A^T =
|
|
\begin{bmatrix}
|
|
6 & 7 & 9 \\
|
|
13 & 0 & 5 \\
|
|
7 & 8 & 4 \\
|
|
4 & 2 & 18 \\
|
|
\end{bmatrix}
|
|
\]
|
|
|
|
\index{square matrix}
|
|
|
|
A matrix is a \key{square matrix} if it
|
|
has the same number of rows and columns.
|
|
For example, the following matrix is a
|
|
square matrix:
|
|
|
|
\[
|
|
S =
|
|
\begin{bmatrix}
|
|
3 & 12 & 4 \\
|
|
5 & 9 & 15 \\
|
|
0 & 2 & 4 \\
|
|
\end{bmatrix}
|
|
\]
|
|
|
|
\section{Operations}
|
|
|
|
The sum $A+B$ of matrices $A$ and $B$
|
|
is defined if the matrices are of the same size.
|
|
The result is a matrix where each element
|
|
is the sum of the corresponding elements
|
|
in $A$ and $B$.
|
|
|
|
For example,
|
|
\[
|
|
\begin{bmatrix}
|
|
6 & 1 & 4 \\
|
|
3 & 9 & 2 \\
|
|
\end{bmatrix}
|
|
+
|
|
\begin{bmatrix}
|
|
4 & 9 & 3 \\
|
|
8 & 1 & 3 \\
|
|
\end{bmatrix}
|
|
=
|
|
\begin{bmatrix}
|
|
6+4 & 1+9 & 4+3 \\
|
|
3+8 & 9+1 & 2+3 \\
|
|
\end{bmatrix}
|
|
=
|
|
\begin{bmatrix}
|
|
10 & 10 & 7 \\
|
|
11 & 10 & 5 \\
|
|
\end{bmatrix}.
|
|
\]
|
|
|
|
Multiplying a matrix $A$ by a value $x$ means
|
|
that each element of $A$ is multiplied by $x$.
|
|
For example,
|
|
\[
|
|
2 \cdot \begin{bmatrix}
|
|
6 & 1 & 4 \\
|
|
3 & 9 & 2 \\
|
|
\end{bmatrix}
|
|
=
|
|
\begin{bmatrix}
|
|
2 \cdot 6 & 2\cdot1 & 2\cdot4 \\
|
|
2\cdot3 & 2\cdot9 & 2\cdot2 \\
|
|
\end{bmatrix}
|
|
=
|
|
\begin{bmatrix}
|
|
12 & 2 & 8 \\
|
|
6 & 18 & 4 \\
|
|
\end{bmatrix}.
|
|
\]
|
|
|
|
\subsubsection{Matrix multiplication}
|
|
|
|
\index{matrix multiplication}
|
|
|
|
The product $AB$ of matrices $A$ and $B$
|
|
is defined if $A$ is of size $a \times n$
|
|
and $B$ is of size $n \times b$, i.e.,
|
|
the width of $A$ equals the height of $B$.
|
|
The result is a matrix of size $a \times b$
|
|
whose elements are calculated using the formula
|
|
\[
|
|
AB[i,j] = \sum_{k=1}^n A[i,k] \cdot B[k,j].
|
|
\]
|
|
|
|
The idea is that each element of $AB$
|
|
is a sum of products of elements of $A$ and $B$
|
|
according to the following picture:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.5]
|
|
\draw (0,0) grid (4,3);
|
|
\draw (5,0) grid (10,3);
|
|
\draw (5,4) grid (10,8);
|
|
|
|
\node at (2,-1) {$A$};
|
|
\node at (7.5,-1) {$AB$};
|
|
\node at (11,6) {$B$};
|
|
|
|
\draw[thick,->,red,line width=2pt] (0,1.5) -- (4,1.5);
|
|
\draw[thick,->,red,line width=2pt] (6.5,8) -- (6.5,4);
|
|
\draw[thick,red,line width=2pt] (6.5,1.5) circle (0.4);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
For example,
|
|
|
|
\[
|
|
\begin{bmatrix}
|
|
1 & 4 \\
|
|
3 & 9 \\
|
|
8 & 6 \\
|
|
\end{bmatrix}
|
|
\cdot
|
|
\begin{bmatrix}
|
|
1 & 6 \\
|
|
2 & 9 \\
|
|
\end{bmatrix}
|
|
=
|
|
\begin{bmatrix}
|
|
1 \cdot 1 + 4 \cdot 2 & 1 \cdot 6 + 4 \cdot 9 \\
|
|
3 \cdot 1 + 9 \cdot 2 & 3 \cdot 6 + 9 \cdot 9 \\
|
|
8 \cdot 1 + 6 \cdot 2 & 8 \cdot 6 + 6 \cdot 9 \\
|
|
\end{bmatrix}
|
|
=
|
|
\begin{bmatrix}
|
|
9 & 42 \\
|
|
21 & 99 \\
|
|
20 & 102 \\
|
|
\end{bmatrix}.
|
|
\]
|
|
|
|
Matrix multiplication is associative,
|
|
so $A(BC)=(AB)C$ holds,
|
|
but it is not commutative,
|
|
so $AB = BA$ does not usually hold.
|
|
|
|
\index{identity matrix}
|
|
|
|
An \key{identity matrix} is a square matrix
|
|
where each element on the diagonal is 1
|
|
and all other elements are 0.
|
|
For example, the following matrix
|
|
is the $3 \times 3$ identity matrix:
|
|
\[
|
|
I = \begin{bmatrix}
|
|
1 & 0 & 0 \\
|
|
0 & 1 & 0 \\
|
|
0 & 0 & 1 \\
|
|
\end{bmatrix}
|
|
\]
|
|
|
|
\begin{samepage}
|
|
Multiplying a matrix by an identity matrix
|
|
does not change it. For example,
|
|
\[
|
|
\begin{bmatrix}
|
|
1 & 0 & 0 \\
|
|
0 & 1 & 0 \\
|
|
0 & 0 & 1 \\
|
|
\end{bmatrix}
|
|
\cdot
|
|
\begin{bmatrix}
|
|
1 & 4 \\
|
|
3 & 9 \\
|
|
8 & 6 \\
|
|
\end{bmatrix}
|
|
=
|
|
\begin{bmatrix}
|
|
1 & 4 \\
|
|
3 & 9 \\
|
|
8 & 6 \\
|
|
\end{bmatrix} \hspace{10px} \textrm{and} \hspace{10px}
|
|
\begin{bmatrix}
|
|
1 & 4 \\
|
|
3 & 9 \\
|
|
8 & 6 \\
|
|
\end{bmatrix}
|
|
\cdot
|
|
\begin{bmatrix}
|
|
1 & 0 \\
|
|
0 & 1 \\
|
|
\end{bmatrix}
|
|
=
|
|
\begin{bmatrix}
|
|
1 & 4 \\
|
|
3 & 9 \\
|
|
8 & 6 \\
|
|
\end{bmatrix}.
|
|
\]
|
|
\end{samepage}
|
|
|
|
Using a straightforward algorithm,
|
|
we can calculate the product of
|
|
two $n \times n$ matrices
|
|
in $O(n^3)$ time.
|
|
There are also more efficient algorithms
|
|
for matrix multiplication\footnote{The first such
|
|
algorithm was Strassen's algorithm,
|
|
published in 1969 \cite{str69},
|
|
whose time complexity is $O(n^{2.80735})$;
|
|
the best current algorithm \cite{gal14}
|
|
works in $O(n^{2.37286})$ time.},
|
|
but they are mostly of theoretical interest
|
|
and such algorithms are not necessary
|
|
in competitive programming.
|
|
|
|
|
|
\subsubsection{Matrix power}
|
|
|
|
\index{matrix power}
|
|
|
|
The power $A^k$ of a matrix $A$ is defined
|
|
if $A$ is a square matrix.
|
|
The definition is based on matrix multiplication:
|
|
\[ A^k = \underbrace{A \cdot A \cdot A \cdots A}_{\textrm{$k$ times}} \]
|
|
For example,
|
|
|
|
\[
|
|
\begin{bmatrix}
|
|
2 & 5 \\
|
|
1 & 4 \\
|
|
\end{bmatrix}^3 =
|
|
\begin{bmatrix}
|
|
2 & 5 \\
|
|
1 & 4 \\
|
|
\end{bmatrix} \cdot
|
|
\begin{bmatrix}
|
|
2 & 5 \\
|
|
1 & 4 \\
|
|
\end{bmatrix} \cdot
|
|
\begin{bmatrix}
|
|
2 & 5 \\
|
|
1 & 4 \\
|
|
\end{bmatrix} =
|
|
\begin{bmatrix}
|
|
48 & 165 \\
|
|
33 & 114 \\
|
|
\end{bmatrix}.
|
|
\]
|
|
In addition, $A^0$ is an identity matrix. For example,
|
|
\[
|
|
\begin{bmatrix}
|
|
2 & 5 \\
|
|
1 & 4 \\
|
|
\end{bmatrix}^0 =
|
|
\begin{bmatrix}
|
|
1 & 0 \\
|
|
0 & 1 \\
|
|
\end{bmatrix}.
|
|
\]
|
|
|
|
The matrix $A^k$ can be efficiently calculated
|
|
in $O(n^3 \log k)$ time using the
|
|
algorithm in Chapter 21.2. For example,
|
|
\[
|
|
\begin{bmatrix}
|
|
2 & 5 \\
|
|
1 & 4 \\
|
|
\end{bmatrix}^8 =
|
|
\begin{bmatrix}
|
|
2 & 5 \\
|
|
1 & 4 \\
|
|
\end{bmatrix}^4 \cdot
|
|
\begin{bmatrix}
|
|
2 & 5 \\
|
|
1 & 4 \\
|
|
\end{bmatrix}^4.
|
|
\]
|
|
|
|
\subsubsection{Determinant}
|
|
|
|
\index{determinant}
|
|
|
|
The \key{determinant} $\det(A)$ of a matrix $A$
|
|
is defined if $A$ is a square matrix.
|
|
If $A$ is of size $1 \times 1$,
|
|
then $\det(A)=A[1,1]$.
|
|
The determinant of a larger matrix is
|
|
calculated recursively using the formula \index{cofactor}
|
|
\[\det(A)=\sum_{j=1}^n A[1,j] C[1,j],\]
|
|
where $C[i,j]$ is the \key{cofactor} of $A$
|
|
at $[i,j]$.
|
|
The cofactor is calculated using the formula
|
|
\[C[i,j] = (-1)^{i+j} \det(M[i,j]),\]
|
|
where $M[i,j]$ is obtained by removing
|
|
row $i$ and column $j$ from $A$.
|
|
Due to the coefficient $(-1)^{i+j}$ in the cofactor,
|
|
every other determinant is positive
|
|
and negative.
|
|
For example,
|
|
\[
|
|
\det(
|
|
\begin{bmatrix}
|
|
3 & 4 \\
|
|
1 & 6 \\
|
|
\end{bmatrix}
|
|
) = 3 \cdot 6 - 4 \cdot 1 = 14
|
|
\]
|
|
and
|
|
\[
|
|
\det(
|
|
\begin{bmatrix}
|
|
2 & 4 & 3 \\
|
|
5 & 1 & 6 \\
|
|
7 & 2 & 4 \\
|
|
\end{bmatrix}
|
|
) =
|
|
2 \cdot
|
|
\det(
|
|
\begin{bmatrix}
|
|
1 & 6 \\
|
|
2 & 4 \\
|
|
\end{bmatrix}
|
|
)
|
|
-4 \cdot
|
|
\det(
|
|
\begin{bmatrix}
|
|
5 & 6 \\
|
|
7 & 4 \\
|
|
\end{bmatrix}
|
|
)
|
|
+3 \cdot
|
|
\det(
|
|
\begin{bmatrix}
|
|
5 & 1 \\
|
|
7 & 2 \\
|
|
\end{bmatrix}
|
|
) = 81.
|
|
\]
|
|
|
|
\index{inverse matrix}
|
|
|
|
The determinant of $A$ tells us
|
|
whether there is an \key{inverse matrix}
|
|
$A^{-1}$ such that $A \cdot A^{-1} = I$,
|
|
where $I$ is an identity matrix.
|
|
It turns out that $A^{-1}$ exists
|
|
exactly when $\det(A) \neq 0$,
|
|
and it can be calculated using the formula
|
|
|
|
\[A^{-1}[i,j] = \frac{C[j,i]}{det(A)}.\]
|
|
|
|
For example,
|
|
|
|
\[
|
|
\underbrace{
|
|
\begin{bmatrix}
|
|
2 & 4 & 3\\
|
|
5 & 1 & 6\\
|
|
7 & 2 & 4\\
|
|
\end{bmatrix}
|
|
}_{A}
|
|
\cdot
|
|
\underbrace{
|
|
\frac{1}{81}
|
|
\begin{bmatrix}
|
|
-8 & -10 & 21 \\
|
|
22 & -13 & 3 \\
|
|
3 & 24 & -18 \\
|
|
\end{bmatrix}
|
|
}_{A^{-1}}
|
|
=
|
|
\underbrace{
|
|
\begin{bmatrix}
|
|
1 & 0 & 0 \\
|
|
0 & 1 & 0 \\
|
|
0 & 0 & 1 \\
|
|
\end{bmatrix}
|
|
}_{I}.
|
|
\]
|
|
|
|
\section{Linear recurrences}
|
|
|
|
\index{linear recurrence}
|
|
|
|
A \key{linear recurrence}
|
|
is a function $f(n)$
|
|
whose initial values are
|
|
$f(0),f(1),\ldots,f(k-1)$
|
|
and larger values
|
|
are calculated recursively using the formula
|
|
\[f(n) = c_1 f(n-1) + c_2 f(n-2) + \ldots + c_k f (n-k),\]
|
|
where $c_1,c_2,\ldots,c_k$ are constant coefficients.
|
|
|
|
Dynamic programming can be used to calculate
|
|
any value of $f(n)$ in $O(kn)$ time by calculating
|
|
all values of $f(0),f(1),\ldots,f(n)$ one after another.
|
|
However, if $k$ is small, it is possible to calculate
|
|
$f(n)$ much more efficiently in $O(k^3 \log n)$
|
|
time using matrix operations.
|
|
|
|
\subsubsection{Fibonacci numbers}
|
|
|
|
\index{Fibonacci number}
|
|
|
|
A simple example of a linear recurrence is the
|
|
following function that defines the Fibonacci numbers:
|
|
\[
|
|
\begin{array}{lcl}
|
|
f(0) & = & 0 \\
|
|
f(1) & = & 1 \\
|
|
f(n) & = & f(n-1)+f(n-2) \\
|
|
\end{array}
|
|
\]
|
|
In this case, $k=2$ and $c_1=c_2=1$.
|
|
|
|
\begin{samepage}
|
|
To efficiently calculate Fibonacci numbers,
|
|
we represent the
|
|
Fibonacci formula as a
|
|
square matrix $X$ of size $2 \times 2$,
|
|
for which the following holds:
|
|
\[ X \cdot
|
|
\begin{bmatrix}
|
|
f(i) \\
|
|
f(i+1) \\
|
|
\end{bmatrix}
|
|
=
|
|
\begin{bmatrix}
|
|
f(i+1) \\
|
|
f(i+2) \\
|
|
\end{bmatrix}
|
|
\]
|
|
Thus, values $f(i)$ and $f(i+1)$ are given as
|
|
''input'' for $X$,
|
|
and $X$ calculates values $f(i+1)$ and $f(i+2)$
|
|
from them.
|
|
It turns out that such a matrix is
|
|
|
|
\[ X =
|
|
\begin{bmatrix}
|
|
0 & 1 \\
|
|
1 & 1 \\
|
|
\end{bmatrix}.
|
|
\]
|
|
\end{samepage}
|
|
\noindent
|
|
For example,
|
|
\[
|
|
\begin{bmatrix}
|
|
0 & 1 \\
|
|
1 & 1 \\
|
|
\end{bmatrix}
|
|
\cdot
|
|
\begin{bmatrix}
|
|
f(5) \\
|
|
f(6) \\
|
|
\end{bmatrix}
|
|
=
|
|
\begin{bmatrix}
|
|
0 & 1 \\
|
|
1 & 1 \\
|
|
\end{bmatrix}
|
|
\cdot
|
|
\begin{bmatrix}
|
|
5 \\
|
|
8 \\
|
|
\end{bmatrix}
|
|
=
|
|
\begin{bmatrix}
|
|
8 \\
|
|
13 \\
|
|
\end{bmatrix}
|
|
=
|
|
\begin{bmatrix}
|
|
f(6) \\
|
|
f(7) \\
|
|
\end{bmatrix}.
|
|
\]
|
|
Thus, we can calculate $f(n)$ using the formula
|
|
\[
|
|
\begin{bmatrix}
|
|
f(n) \\
|
|
f(n+1) \\
|
|
\end{bmatrix}
|
|
=
|
|
X^n \cdot
|
|
\begin{bmatrix}
|
|
f(0) \\
|
|
f(1) \\
|
|
\end{bmatrix}
|
|
=
|
|
\begin{bmatrix}
|
|
0 & 1 \\
|
|
1 & 1 \\
|
|
\end{bmatrix}^n
|
|
\cdot
|
|
\begin{bmatrix}
|
|
0 \\
|
|
1 \\
|
|
\end{bmatrix}.
|
|
\]
|
|
The value of $X^n$ can be calculated in
|
|
$O(k^3 \log n)$ time,
|
|
so the value of $f(n)$ can also be calculated
|
|
in $O(k^3 \log n)$ time.
|
|
|
|
\subsubsection{General case}
|
|
|
|
Let us now consider the general case where
|
|
$f(n)$ is any linear recurrence.
|
|
Again, our goal is to construct a matrix $X$
|
|
for which
|
|
|
|
\[ X \cdot
|
|
\begin{bmatrix}
|
|
f(i) \\
|
|
f(i+1) \\
|
|
\vdots \\
|
|
f(i+k-1) \\
|
|
\end{bmatrix}
|
|
=
|
|
\begin{bmatrix}
|
|
f(i+1) \\
|
|
f(i+2) \\
|
|
\vdots \\
|
|
f(i+k) \\
|
|
\end{bmatrix}.
|
|
\]
|
|
Such a matrix is
|
|
\[
|
|
X =
|
|
\begin{bmatrix}
|
|
0 & 1 & 0 & 0 & \cdots & 0 \\
|
|
0 & 0 & 1 & 0 & \cdots & 0 \\
|
|
0 & 0 & 0 & 1 & \cdots & 0 \\
|
|
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
|
|
0 & 0 & 0 & 0 & \cdots & 1 \\
|
|
c_k & c_{k-1} & c_{k-2} & c_{k-3} & \cdots & c_1 \\
|
|
\end{bmatrix}.
|
|
\]
|
|
In the first $k-1$ rows, each element is 0
|
|
except that one element is 1.
|
|
These rows replace $f(i)$ with $f(i+1)$,
|
|
$f(i+1)$ with $f(i+2)$, and so on.
|
|
The last row contains the coefficients of the recurrence
|
|
to calculate the new value $f(i+k)$.
|
|
|
|
\begin{samepage}
|
|
Now, $f(n)$ can be calculated in
|
|
$O(k^3 \log n)$ time using the formula
|
|
\[
|
|
\begin{bmatrix}
|
|
f(n) \\
|
|
f(n+1) \\
|
|
\vdots \\
|
|
f(n+k-1) \\
|
|
\end{bmatrix}
|
|
=
|
|
X^n \cdot
|
|
\begin{bmatrix}
|
|
f(0) \\
|
|
f(1) \\
|
|
\vdots \\
|
|
f(k-1) \\
|
|
\end{bmatrix}.
|
|
\]
|
|
\end{samepage}
|
|
|
|
\section{Graphs and matrices}
|
|
|
|
\subsubsection{Counting paths}
|
|
|
|
The powers of an adjacency matrix of a graph
|
|
have an interesting property.
|
|
When $V$ is an adjacency matrix of an unweighted graph,
|
|
the matrix $V^n$ contains the numbers of paths of
|
|
$n$ edges between the nodes in the graph.
|
|
|
|
For example, for the graph
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\node[draw, circle] (1) at (1,3) {$1$};
|
|
\node[draw, circle] (2) at (1,1) {$4$};
|
|
\node[draw, circle] (3) at (3,3) {$2$};
|
|
\node[draw, circle] (4) at (5,3) {$3$};
|
|
\node[draw, circle] (5) at (3,1) {$5$};
|
|
\node[draw, circle] (6) at (5,1) {$6$};
|
|
|
|
\path[draw,thick,->,>=latex] (1) -- (2);
|
|
\path[draw,thick,->,>=latex] (2) -- (3);
|
|
\path[draw,thick,->,>=latex] (3) -- (1);
|
|
\path[draw,thick,->,>=latex] (4) -- (3);
|
|
\path[draw,thick,->,>=latex] (3) -- (5);
|
|
\path[draw,thick,->,>=latex] (3) -- (6);
|
|
\path[draw,thick,->,>=latex] (6) -- (4);
|
|
\path[draw,thick,->,>=latex] (6) -- (5);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
the adjacency matrix is
|
|
\[
|
|
V= \begin{bmatrix}
|
|
0 & 0 & 0 & 1 & 0 & 0 \\
|
|
1 & 0 & 0 & 0 & 1 & 1 \\
|
|
0 & 1 & 0 & 0 & 0 & 0 \\
|
|
0 & 1 & 0 & 0 & 0 & 0 \\
|
|
0 & 0 & 0 & 0 & 0 & 0 \\
|
|
0 & 0 & 1 & 0 & 1 & 0 \\
|
|
\end{bmatrix}.
|
|
\]
|
|
Now, for example, the matrix
|
|
\[
|
|
V^4= \begin{bmatrix}
|
|
0 & 0 & 1 & 1 & 1 & 0 \\
|
|
2 & 0 & 0 & 0 & 2 & 2 \\
|
|
0 & 2 & 0 & 0 & 0 & 0 \\
|
|
0 & 2 & 0 & 0 & 0 & 0 \\
|
|
0 & 0 & 0 & 0 & 0 & 0 \\
|
|
0 & 0 & 1 & 1 & 1 & 0 \\
|
|
\end{bmatrix}
|
|
\]
|
|
contains the numbers of paths of 4 edges
|
|
between the nodes.
|
|
For example, $V^4[2,5]=2$,
|
|
because there are two paths of 4 edges
|
|
from node 2 to node 5:
|
|
$2 \rightarrow 1 \rightarrow 4 \rightarrow 2 \rightarrow 5$
|
|
and
|
|
$2 \rightarrow 6 \rightarrow 3 \rightarrow 2 \rightarrow 5$.
|
|
|
|
\subsubsection{Shortest paths}
|
|
|
|
Using a similar idea in a weighted graph,
|
|
we can calculate for each pair of nodes the minimum
|
|
length of a path
|
|
between them that contains exactly $n$ edges.
|
|
To calculate this, we have to define matrix multiplication
|
|
in a new way, so that we do not calculate the numbers
|
|
of paths but minimize the lengths of paths.
|
|
|
|
\begin{samepage}
|
|
As an example, consider the following graph:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\node[draw, circle] (1) at (1,3) {$1$};
|
|
\node[draw, circle] (2) at (1,1) {$4$};
|
|
\node[draw, circle] (3) at (3,3) {$2$};
|
|
\node[draw, circle] (4) at (5,3) {$3$};
|
|
\node[draw, circle] (5) at (3,1) {$5$};
|
|
\node[draw, circle] (6) at (5,1) {$6$};
|
|
|
|
\path[draw,thick,->,>=latex] (1) -- node[font=\small,label=left:4] {} (2);
|
|
\path[draw,thick,->,>=latex] (2) -- node[font=\small,label=left:1] {} (3);
|
|
\path[draw,thick,->,>=latex] (3) -- node[font=\small,label=north:2] {} (1);
|
|
\path[draw,thick,->,>=latex] (4) -- node[font=\small,label=north:4] {} (3);
|
|
\path[draw,thick,->,>=latex] (3) -- node[font=\small,label=left:1] {} (5);
|
|
\path[draw,thick,->,>=latex] (3) -- node[font=\small,label=left:2] {} (6);
|
|
\path[draw,thick,->,>=latex] (6) -- node[font=\small,label=right:3] {} (4);
|
|
\path[draw,thick,->,>=latex] (6) -- node[font=\small,label=below:2] {} (5);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
\end{samepage}
|
|
|
|
Let us construct an adjacency matrix where
|
|
$\infty$ means that an edge does not exist,
|
|
and other values correspond to edge weights.
|
|
The matrix is
|
|
\[
|
|
V= \begin{bmatrix}
|
|
\infty & \infty & \infty & 4 & \infty & \infty \\
|
|
2 & \infty & \infty & \infty & 1 & 2 \\
|
|
\infty & 4 & \infty & \infty & \infty & \infty \\
|
|
\infty & 1 & \infty & \infty & \infty & \infty \\
|
|
\infty & \infty & \infty & \infty & \infty & \infty \\
|
|
\infty & \infty & 3 & \infty & 2 & \infty \\
|
|
\end{bmatrix}.
|
|
\]
|
|
|
|
Instead of the formula
|
|
\[
|
|
AB[i,j] = \sum_{k=1}^n A[i,k] \cdot B[k,j]
|
|
\]
|
|
we now use the formula
|
|
\[
|
|
AB[i,j] = \min_{k=1}^n A[i,k] + B[k,j]
|
|
\]
|
|
for matrix multiplication, so we calculate
|
|
a minimum instead of a sum,
|
|
and a sum of elements instead of a product.
|
|
After this modification,
|
|
matrix powers correspond to
|
|
shortest paths in the graph.
|
|
|
|
For example, as
|
|
\[
|
|
V^4= \begin{bmatrix}
|
|
\infty & \infty & 10 & 11 & 9 & \infty \\
|
|
9 & \infty & \infty & \infty & 8 & 9 \\
|
|
\infty & 11 & \infty & \infty & \infty & \infty \\
|
|
\infty & 8 & \infty & \infty & \infty & \infty \\
|
|
\infty & \infty & \infty & \infty & \infty & \infty \\
|
|
\infty & \infty & 12 & 13 & 11 & \infty \\
|
|
\end{bmatrix},
|
|
\]
|
|
we can conclude that the minimum length of a path
|
|
of 4 edges
|
|
from node 2 to node 5 is 8.
|
|
Such a path is
|
|
$2 \rightarrow 1 \rightarrow 4 \rightarrow 2 \rightarrow 5$.
|
|
|
|
\subsubsection{Kirchhoff's theorem}
|
|
|
|
\index{Kirchhoff's theorem}
|
|
\index{spanning tree}
|
|
|
|
\key{Kirchhoff's theorem}
|
|
%\footnote{G. R. Kirchhoff (1824--1887) was a German physicist.}
|
|
provides a way
|
|
to calculate the number of spanning trees
|
|
of a graph as a determinant of a special matrix.
|
|
For example, the graph
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\node[draw, circle] (1) at (1,3) {$1$};
|
|
\node[draw, circle] (2) at (3,3) {$2$};
|
|
\node[draw, circle] (3) at (1,1) {$3$};
|
|
\node[draw, circle] (4) at (3,1) {$4$};
|
|
|
|
\path[draw,thick,-] (1) -- (2);
|
|
\path[draw,thick,-] (1) -- (3);
|
|
\path[draw,thick,-] (3) -- (4);
|
|
\path[draw,thick,-] (1) -- (4);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
has three spanning trees:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\node[draw, circle] (1a) at (1,3) {$1$};
|
|
\node[draw, circle] (2a) at (3,3) {$2$};
|
|
\node[draw, circle] (3a) at (1,1) {$3$};
|
|
\node[draw, circle] (4a) at (3,1) {$4$};
|
|
|
|
\path[draw,thick,-] (1a) -- (2a);
|
|
%\path[draw,thick,-] (1a) -- (3a);
|
|
\path[draw,thick,-] (3a) -- (4a);
|
|
\path[draw,thick,-] (1a) -- (4a);
|
|
|
|
\node[draw, circle] (1b) at (1+4,3) {$1$};
|
|
\node[draw, circle] (2b) at (3+4,3) {$2$};
|
|
\node[draw, circle] (3b) at (1+4,1) {$3$};
|
|
\node[draw, circle] (4b) at (3+4,1) {$4$};
|
|
|
|
\path[draw,thick,-] (1b) -- (2b);
|
|
\path[draw,thick,-] (1b) -- (3b);
|
|
%\path[draw,thick,-] (3b) -- (4b);
|
|
\path[draw,thick,-] (1b) -- (4b);
|
|
|
|
\node[draw, circle] (1c) at (1+8,3) {$1$};
|
|
\node[draw, circle] (2c) at (3+8,3) {$2$};
|
|
\node[draw, circle] (3c) at (1+8,1) {$3$};
|
|
\node[draw, circle] (4c) at (3+8,1) {$4$};
|
|
|
|
\path[draw,thick,-] (1c) -- (2c);
|
|
\path[draw,thick,-] (1c) -- (3c);
|
|
\path[draw,thick,-] (3c) -- (4c);
|
|
%\path[draw,thick,-] (1c) -- (4c);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
\index{Laplacean matrix}
|
|
To calculate the number of spanning trees,
|
|
we construct a \key{Laplacean matrix} $L$,
|
|
where $L[i,i]$ is the degree of node $i$
|
|
and $L[i,j]=-1$ if there is an edge between
|
|
nodes $i$ and $j$, and otherwise $L[i,j]=0$.
|
|
The Laplacean matrix for the above graph is as follows:
|
|
\[
|
|
L= \begin{bmatrix}
|
|
3 & -1 & -1 & -1 \\
|
|
-1 & 1 & 0 & 0 \\
|
|
-1 & 0 & 2 & -1 \\
|
|
-1 & 0 & -1 & 2 \\
|
|
\end{bmatrix}
|
|
\]
|
|
|
|
It can be shown that
|
|
the number of spanning trees equals
|
|
the determinant of a matrix that is obtained
|
|
when we remove any row and any column from $L$.
|
|
For example, if we remove the first row
|
|
and column, the result is
|
|
|
|
\[ \det(
|
|
\begin{bmatrix}
|
|
1 & 0 & 0 \\
|
|
0 & 2 & -1 \\
|
|
0 & -1 & 2 \\
|
|
\end{bmatrix}
|
|
) =3.\]
|
|
The determinant is always the same,
|
|
regardless of which row and column we remove from $L$.
|
|
|
|
Note that Cayley's formula in Chapter 22.5 is
|
|
a special case of Kirchhoff's theorem,
|
|
because in a complete graph of $n$ nodes
|
|
|
|
\[ \det(
|
|
\begin{bmatrix}
|
|
n-1 & -1 & \cdots & -1 \\
|
|
-1 & n-1 & \cdots & -1 \\
|
|
\vdots & \vdots & \ddots & \vdots \\
|
|
-1 & -1 & \cdots & n-1 \\
|
|
\end{bmatrix}
|
|
) =n^{n-2}.\]
|
|
|
|
|
|
|