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Fix grammar and indices

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Antti H S Laaksonen 2017-04-19 20:44:51 +03:00
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chapter09.tex
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@ -82,7 +82,7 @@ the answer for any possible query.
\index{prefix sum array} \index{prefix sum array}
It turns out that we can easily process We can easily process
sum queries on a static array, sum queries on a static array,
because we can use a data structure called because we can use a data structure called
a \key{prefix sum array}. a \key{prefix sum array}.
@ -145,12 +145,12 @@ The corresponding prefix sum array is as follows:
Let $\textrm{sum}(a,b)$ denote the sum of elements Let $\textrm{sum}(a,b)$ denote the sum of elements
in the range $[a,b]$. in the range $[a,b]$.
Since the prefix sum array contains all values Since the prefix sum array contains all values
of the form $\textrm{sum}(0,k)$, of $\textrm{sum}(0,k)$,
we can calculate any value of we can calculate any value of
$\textrm{sum}(a,b)$ in $O(1)$ time, because $\textrm{sum}(a,b)$ in $O(1)$ time, because
\[ \textrm{sum}(a,b) = \textrm{sum}(0,b) - \textrm{sum}(0,a-1).\] \[ \textrm{sum}(a,b) = \textrm{sum}(0,b) - \textrm{sum}(0,a-1).\]
By defining $\textrm{sum}(0,-1)=0$, By defining $\textrm{sum}(0,-1)=0$,
the above formula also holds when $a=1$. the above formula also holds when $a=0$.
For example, consider the range $[3,6]$: For example, consider the range $[3,6]$:
\begin{center} \begin{center}
@ -442,7 +442,7 @@ we can conclude that $\textrm{rmq}(1,6)=1$.
\index{binary indexed tree} \index{binary indexed tree}
\index{Fenwick tree} \index{Fenwick tree}
A \key{binary indexed tree} or \key{Fenwick tree}\footnote{The A \key{binary indexed tree} or a \key{Fenwick tree}\footnote{The
binary indexed tree structure was presented by P. M. Fenwick in 1994 \cite{fen94}.} binary indexed tree structure was presented by P. M. Fenwick in 1994 \cite{fen94}.}
can be seen as a dynamic version of a prefix sum array. can be seen as a dynamic version of a prefix sum array.
This data structure supports two $O(\log n)$ time operations: This data structure supports two $O(\log n)$ time operations:
@ -450,18 +450,18 @@ calculating the sum of elements in a range
and modifying the value of an element. and modifying the value of an element.
The advantage of a binary indexed tree is The advantage of a binary indexed tree is
that it allows us to efficiently update that it allows us to efficiently \emph{update}
the array elements between the sum queries. array elements between sum queries.
This would not be possible using a prefix sum array, This would not be possible using a prefix sum array,
because after each update, we should build the because after each update, it would be necessary to build the
whole array again in $O(n)$ time. whole prefix sum array again in $O(n)$ time.
\subsubsection{Structure} \subsubsection{Structure}
In this section we assume that one-based indexing In this section we assume that one-based indexing
is used in all arrays. is used, because it makes the implementation easier.
A binary indexed tree can be represented as an array A binary indexed tree is as an array
where the value at position $x$ whose value at position $x$
equals the sum of elements in the range $[x-k+1,x]$ equals the sum of elements in the range $[x-k+1,x]$
of the original array, of the original array,
where $k$ is the largest power of two that divides $x$. where $k$ is the largest power of two that divides $x$.
@ -578,9 +578,9 @@ corresponds to a range in the array:
\subsubsection{Sum queries} \subsubsection{Sum queries}
The values in the binary indexed tree The values in a binary indexed tree
can be used to efficiently calculate can be used to efficiently calculate
the sum of elements in any range $[1,k]$, the sum of array elements in any range $[1,k]$,
because such a range because such a range
can be divided into $O(\log n)$ ranges can be divided into $O(\log n)$ ranges
whose sums are available in the binary indexed tree. whose sums are available in the binary indexed tree.
@ -745,7 +745,7 @@ takes $O(1)$ time using bit operations.
\index{segment tree} \index{segment tree}
A \key{segment tree}\footnote{Quite similar structures were used A \key{segment tree}\footnote{Quite similar structures were used
in the late 1970's to solve geometric problems \cite{ben80}. in late 1970's to solve geometric problems \cite{ben80}.
The bottom-up-implementation in this chapter corresponds to The bottom-up-implementation in this chapter corresponds to
that in \cite{sta06}.} is a data structure that in \cite{sta06}.} is a data structure
that supports two operations: that supports two operations:
@ -1153,7 +1153,7 @@ and the operations move one level forward in the tree at each step.
Segment trees can support any queries Segment trees can support any queries
as long as we can divide a range into two parts, as long as we can divide a range into two parts,
calculate the answer for both parts calculate the answer separately for both parts
and then efficiently combine the answers. and then efficiently combine the answers.
Examples of such queries are Examples of such queries are
minimum and maximum, greatest common divisor, minimum and maximum, greatest common divisor,
@ -1207,7 +1207,7 @@ In this segment tree, every node in the tree
contains the smallest element in the corresponding contains the smallest element in the corresponding
range of the array. range of the array.
The top node of the tree contains the smallest The top node of the tree contains the smallest
element in the whole array. element of the whole array.
The operations can be implemented like previously, The operations can be implemented like previously,
but instead of sums, minima are calculated. but instead of sums, minima are calculated.
@ -1351,14 +1351,14 @@ For example, consider the following array:
\footnotesize \footnotesize
\node at (0.5,1.4) {$1$}; \node at (0.5,1.4) {$0$};
\node at (1.5,1.4) {$2$}; \node at (1.5,1.4) {$1$};
\node at (2.5,1.4) {$3$}; \node at (2.5,1.4) {$2$};
\node at (3.5,1.4) {$4$}; \node at (3.5,1.4) {$3$};
\node at (4.5,1.4) {$5$}; \node at (4.5,1.4) {$4$};
\node at (5.5,1.4) {$6$}; \node at (5.5,1.4) {$5$};
\node at (6.5,1.4) {$7$}; \node at (6.5,1.4) {$6$};
\node at (7.5,1.4) {$8$}; \node at (7.5,1.4) {$7$};
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
@ -1378,28 +1378,28 @@ The difference array for the above array is as follows:
\footnotesize \footnotesize
\node at (0.5,1.4) {$1$}; \node at (0.5,1.4) {$0$};
\node at (1.5,1.4) {$2$}; \node at (1.5,1.4) {$1$};
\node at (2.5,1.4) {$3$}; \node at (2.5,1.4) {$2$};
\node at (3.5,1.4) {$4$}; \node at (3.5,1.4) {$3$};
\node at (4.5,1.4) {$5$}; \node at (4.5,1.4) {$4$};
\node at (5.5,1.4) {$6$}; \node at (5.5,1.4) {$5$};
\node at (6.5,1.4) {$7$}; \node at (6.5,1.4) {$6$};
\node at (7.5,1.4) {$8$}; \node at (7.5,1.4) {$7$};
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
For example, the value 5 at position 6 in the original array For example, the value 2 at position 6 in the original array
corresponds to the sum $3-2+4=5$. corresponds to the sum $3-2+4-3=2$.
The advantage of the difference array is The advantage of the difference array is
that we can update a range that we can update a range
in the original array by changing just in the original array by changing just
two elements in the difference array. two elements in the difference array.
For example, if we want to For example, if we want to
increase the elements in the range $2 \ldots 5$ by 5, increase the elements in the range $1 \ldots 4$ by 5,
it suffices to increase the value at position 2 by 5 it suffices to increase the value at position 1 by 5
and decrease the value at position 6 by 5. and decrease the value at position 5 by 5.
The result is as follows: The result is as follows:
\begin{center} \begin{center}
@ -1416,14 +1416,14 @@ The result is as follows:
\node at (7.5,0.5) {$0$}; \node at (7.5,0.5) {$0$};
\footnotesize \footnotesize
\node at (0.5,1.4) {$1$}; \node at (0.5,1.4) {$0$};
\node at (1.5,1.4) {$2$}; \node at (1.5,1.4) {$1$};
\node at (2.5,1.4) {$3$}; \node at (2.5,1.4) {$2$};
\node at (3.5,1.4) {$4$}; \node at (3.5,1.4) {$3$};
\node at (4.5,1.4) {$5$}; \node at (4.5,1.4) {$4$};
\node at (5.5,1.4) {$6$}; \node at (5.5,1.4) {$5$};
\node at (6.5,1.4) {$7$}; \node at (6.5,1.4) {$6$};
\node at (7.5,1.4) {$8$}; \node at (7.5,1.4) {$7$};
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}