Corrections
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Antti H S Laaksonen
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luku23.tex
103
luku23.tex
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@ -73,7 +73,7 @@ The sum $A+B$ of matrices $A$ and $B$
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is defined if the matrices are of the same size.
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The result is a matrix where each element
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is the sum of the corresponding elements
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in matrices $A$ and $B$.
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in $A$ and $B$.
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For example,
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\[
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@ -99,8 +99,7 @@ For example,
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\]
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Multiplying a matrix $A$ by a value $x$ means
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that we multiply each element in $A$ by $x$.
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that each element of $A$ is multiplied by $x$.
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For example,
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\[
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2 \cdot \begin{bmatrix}
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@ -180,18 +179,18 @@ For example,
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\end{bmatrix}.
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\]
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Matrix multiplication is not commutative,
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so $AB = BA$ doesn't hold.
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However, it is associative,
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so $A(BC)=(AB)C$ holds.
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Matrix multiplication is associative,
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so $A(BC)=(AB)C$ holds,
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but it is not commutative,
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so $AB = BA$ does not hold.
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\index{identity matrix}
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An \key{identity matrix} is a square matrix
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where each element on the diagonal is 1,
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where each element on the diagonal is 1
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and all other elements are 0.
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For example, the $3 \times 3$ identity matrix
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is as follows:
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For example, the following matrix
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is the $3 \times 3$ identity matrix:
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\[
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I = \begin{bmatrix}
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1 & 0 & 0 \\
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@ -202,7 +201,7 @@ is as follows:
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\begin{samepage}
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Multiplying a matrix by an identity matrix
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doesn't change it. For example,
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does not change it. For example,
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\[
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\begin{bmatrix}
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1 & 0 & 0 \\
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@ -220,7 +219,7 @@ doesn't change it. For example,
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1 & 4 \\
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3 & 9 \\
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8 & 6 \\
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\end{bmatrix} \hspace{10px} \textrm{ja} \hspace{10px}
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\end{bmatrix} \hspace{10px} \textrm{and} \hspace{10px}
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\begin{bmatrix}
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1 & 4 \\
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3 & 9 \\
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@ -328,13 +327,11 @@ where $C[i,j]$ is the \key{cofactor} of $A$
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at $[i,j]$.
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The cofactor is calculated using the formula
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\[C[i,j] = (-1)^{i+j} \det(M[i,j]),\]
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where $M[i,j]$ is a copy of matrix $A$
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where row $i$ and column $j$ are removed.
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Because of the multiplier $(-1)^{i+j}$ in the cofactor,
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where $M[i,j]$ is obtained by removing
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row $i$ and column $j$ from $A$.
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Due to the multiplier $(-1)^{i+j}$ in the cofactor,
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every other determinant is positive
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and negative.
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\begin{samepage}
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For example,
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\[
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\det(
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@ -344,10 +341,7 @@ For example,
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\end{bmatrix}
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) = 3 \cdot 6 - 4 \cdot 1 = 14
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\]
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\end{samepage}
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and
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\[
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\det(
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\begin{bmatrix}
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@ -381,9 +375,9 @@ and
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\index{inverse matrix}
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The determinant indicates if matrix $A$
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has an \key{inverse matrix}
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$A^{-1}$ for which $A \cdot A^{-1} = I$,
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The determinant of $A$ tells us
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whether there is an \key{inverse matrix}
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$A^{-1}$ such that $A \cdot A^{-1} = I$,
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where $I$ is an identity matrix.
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It turns out that $A^{-1}$ exists
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exactly when $\det(A) \neq 0$,
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@ -426,16 +420,16 @@ For example,
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A \key{linear recurrence}
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can be represented as a function $f(n)$
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with initial values
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$f(0),f(1),\ldots,f(k-1)$,
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whose values for $k$ and larger parameters
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are calculated recursively using a formula
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such that the initial values are
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$f(0),f(1),\ldots,f(k-1)$
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and the larger values
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are calculated recursively using the formula
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\[f(n) = c_1 f(n-1) + c_2 f(n-2) + \ldots + c_k f (n-k),\]
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where $c_1,c_2,\ldots,c_k$ are constant multipliers.
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We can use dynamic programming to calculate
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any value $f(n)$ in $O(kn)$ time by calculating
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all values $f(0),f(1),\ldots,f(n)$ one after another.
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any value of $f(n)$ in $O(kn)$ time by calculating
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all values of $f(0),f(1),\ldots,f(n)$ one after another.
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However, if $k$ is small, it is possible to calculate
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$f(n)$ much more efficiently in $O(k^3 \log n)$
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time using matrix operations.
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@ -445,7 +439,7 @@ time using matrix operations.
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\index{Fibonacci number}
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A simple example of a linear recurrence is the
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function that calculates Fibonacci numbers:
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following function that defines the Fibonacci numbers:
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\[
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\begin{array}{lcl}
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f(0) & = & 0 \\
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@ -456,9 +450,9 @@ f(n) & = & f(n-1)+f(n-2) \\
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In this case, $k=2$ and $c_1=c_2=1$.
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\begin{samepage}
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The idea is to represent the formula for
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calculating Fibonacci numbers as a
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square matrix $X$ of size $2 \times 2$
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The idea is to represent the
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Fibonacci formula as a
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square matrix $X$ of size $2 \times 2$,
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for which the following holds:
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\[ X \cdot
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\begin{bmatrix}
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@ -473,7 +467,7 @@ for which the following holds:
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\]
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Thus, values $f(i)$ and $f(i+1)$ are given as
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''input'' for $X$,
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and $X$ constructs values $f(i+1)$ and $f(i+2)$
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and $X$ calculates values $f(i+1)$ and $f(i+2)$
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from them.
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It turns out that such a matrix is
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@ -540,14 +534,14 @@ X^n \cdot
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1 \\
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\end{bmatrix}.
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\]
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The power $X^n$ on can be calculated in
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The power $X^n$ can be calculated in
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$O(k^3 \log n)$ time,
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so the value $f(n)$ can also be calculated
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so the value of $f(n)$ can also be calculated
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in $O(k^3 \log n)$ time.
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\subsubsection{General case}
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Let's now consider a general case where
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Let us now consider the general case where
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$f(n)$ is any linear recurrence.
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Again, our goal is to construct a matrix $X$
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for which
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@ -583,8 +577,8 @@ In the first $k-1$ rows, each element is 0
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except that one element is 1.
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These rows replace $f(i)$ with $f(i+1)$,
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$f(i+1)$ with $f(i+2)$, etc.
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The last row contains the multipliers in the recurrence,
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and it calculates the new value $f(i+k)$.
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The last row contains the multipliers of the recurrence
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to calculate the new value $f(i+k)$.
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\begin{samepage}
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Now, $f(n)$ can be calculated in
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@ -674,8 +668,8 @@ Using a similar idea in a weighted graph,
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we can calculate for each pair of nodes the shortest
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path between them that contains exactly $n$ edges.
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To calculate this, we have to define matrix multiplication
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in another way, so that we don't calculate the number
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of paths but minimize the length of a path.
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in a new way, so that we do not calculate the numbers
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of paths but minimize the lengths of paths.
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\begin{samepage}
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As an example, consider the following graph:
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@ -700,8 +694,8 @@ As an example, consider the following graph:
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\end{center}
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\end{samepage}
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Let's construct an adjacency matrix where
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$\infty$ means that an edge doesn't exist,
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Let us construct an adjacency matrix where
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$\infty$ means that an edge does not exist,
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and other values correspond to edge weights.
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The matrix is
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\[
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@ -721,15 +715,16 @@ AB[i,j] = \sum_{k=1}^n A[i,k] \cdot B[k,j]
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\]
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we now use the formula
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\[
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AB[i,j] = \min_{k=1}^n A[i,k] + B[k,j],
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AB[i,j] = \min_{k=1}^n A[i,k] + B[k,j]
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\]
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for matrix multiplication, so we calculate
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a minimum instead of a sum,
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and a sum of elements instead of a product.
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After this modification,
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matrix powers can be used for calculating
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shortest paths in the graph:
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matrix powers correspond to
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shortest paths in the graph.
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For example, as
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\[
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V^4= \begin{bmatrix}
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\infty & \infty & 10 & 11 & 9 & \infty \\
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@ -738,9 +733,9 @@ V^4= \begin{bmatrix}
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\infty & 8 & \infty & \infty & \infty & \infty \\
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\infty & \infty & \infty & \infty & \infty & \infty \\
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\infty & \infty & 12 & 13 & 11 & \infty \\
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\end{bmatrix}
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\end{bmatrix},
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\]
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For example, the shortest path of 4 edges
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we can conclude that the shortest path of 4 edges
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from node 2 to node 5 has length 8.
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This path is
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$2 \rightarrow 1 \rightarrow 4 \rightarrow 2 \rightarrow 5$.
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@ -750,9 +745,9 @@ $2 \rightarrow 1 \rightarrow 4 \rightarrow 2 \rightarrow 5$.
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\index{Kirchhoff's theorem}
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\index{spanning tree}
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\key{Kirchhoff's theorem} provides us a way
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\key{Kirchhoff's theorem} provides a way
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to calculate the number of spanning trees
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in a graph as a determinant of a special matrix.
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of a graph as a determinant of a special matrix.
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For example, the graph
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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@ -802,12 +797,12 @@ has three spanning trees:
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\end{tikzpicture}
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\end{center}
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\index{Laplacean matrix}
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We construct a \key{Laplacean matrix} $L$,
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To calculate the number of spanning trees,
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we construct a \key{Laplacean matrix} $L$,
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where $L[i,i]$ is the degree of node $i$
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and $L[i,j]=-1$ if there is an edge between
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nodes $i$ and $j$, and otherwise $L[i,j]=0$.
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In this graph, the matrix is as follows:
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The Laplacean matrix for the above graph is as follows:
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\[
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L= \begin{bmatrix}
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3 & -1 & -1 & -1 \\
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@ -817,7 +812,7 @@ L= \begin{bmatrix}
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\end{bmatrix}
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\]
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The number of spanning trees is the same as
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The number of spanning trees equals
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the determinant of a matrix that is obtained
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when we remove any row and any column from $L$.
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For example, if we remove the first row
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