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@ -27,9 +27,9 @@ in another way using a square root structure
so that we can calculate sums in $O(\sqrt n)$ time so that we can calculate sums in $O(\sqrt n)$ time
and modify values in $O(1)$ time. and modify values in $O(1)$ time.
The idea is to divide the array into segments The idea is to divide the array into blocks
of size $\sqrt n$ so that each segment contains of size $\sqrt n$ so that each block contains
the sum of values inside the segment. the sum of elements inside the block.
The following example shows an array and the The following example shows an array and the
corresponding segments: corresponding segments:
@ -68,8 +68,8 @@ corresponding segments:
\end{center} \end{center}
When a value in the array changes, When a value in the array changes,
we have to calculate the sum in the corresponding we have to calculate the sum of the corresponding
segment again: block again:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.7] \begin{tikzpicture}[scale=0.7]
@ -109,7 +109,7 @@ segment again:
Any sum in the array can be calculated as a combination Any sum in the array can be calculated as a combination
of single values in the array and the sums of the of single values in the array and the sums of the
segments between them: blocks between them:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.7] \begin{tikzpicture}[scale=0.7]
@ -153,12 +153,12 @@ segments between them:
\end{center} \end{center}
We can change a value in $O(1)$ time, We can change a value in $O(1)$ time,
because we only have to change the sum of a single segment. because we only have to change the sum of a single block.
A sum in a range consists of three parts: A sum in a range consists of three parts:
\begin{itemize} \begin{itemize}
\item first, there are $O(\sqrt n)$ single values \item first, there are $O(\sqrt n)$ single values
\item then, there are $O(\sqrt n)$ consecutive segments \item then, there are $O(\sqrt n)$ consecutive blocks
\item finally, there are $O(\sqrt n)$ single values \item finally, there are $O(\sqrt n)$ single values
\end{itemize} \end{itemize}
@ -169,7 +169,7 @@ of values in any range is $O(\sqrt n)$.
The reason why we use the parameter $\sqrt n$ is that The reason why we use the parameter $\sqrt n$ is that
it balances two things: it balances two things:
for example, an array of $n$ elements is divided for example, an array of $n$ elements is divided
into $\sqrt n$ segments, each of which contains into $\sqrt n$ blocks, each of which contains
$\sqrt n$ elements. $\sqrt n$ elements.
In practice, it is not needed to use exactly In practice, it is not needed to use exactly
the parameter $\sqrt n$ in algorithms, but it may be better to the parameter $\sqrt n$ in algorithms, but it may be better to
@ -179,16 +179,16 @@ larger or smaller than $\sqrt n$.
The best parameter depends on the problem The best parameter depends on the problem
and input. and input.
For example, if an algorithm often goes through For example, if an algorithm often goes through
segments but rarely iterates the elements inside blocks but rarely iterates elements inside
the segments, it may be good to divide the array into blocks, it may be good to divide the array into
$k < \sqrt n$ segments, each of which contains $n/k > \sqrt n$ $k < \sqrt n$ blocks, each of which contains $n/k > \sqrt n$
elements. elements.
\section{Batch processing} \section{Batch processing}
\index{batch processing} \index{batch processing}
In \key{batch processing}, the operations in the In \key{batch processing}, the operations of an
algorithm are divided into batches, algorithm are divided into batches,
and each batch will be processed separately. and each batch will be processed separately.
Between the batches some precalculation is done Between the batches some precalculation is done
@ -308,57 +308,58 @@ because both case 1 and case 2 take $O(n \sqrt n)$ time.
\index{Mo's algorithm} \index{Mo's algorithm}
\key{Mo's algorithm} can be used in many problems \key{Mo's algorithm} can be used in many problems
where we are asked to process range queries in that require processing range queries in
a \emph{static} array. a \emph{static} array.
The algorithm handles the queries in a special order Before processing the queries, the algorithm
so that it is efficient to process them. sorts them in a special order which guarantees
that the algorithm runs efficiently.
The algorithm maintains a range in the array, At each moment in the algorithm, there is an active
and the answer for a query for that range. subarray and the algorithm maintains the answer
When moving from a range to another range, for a query to that subarray.
the algorithm modifies the range step by step The algorithm processes the given queries one by one,
so that the answer for the next range can be and always changes the active subarray
calculated. by inserting and removing elements
so that it corresponds to the current query.
The time complexity of the algorithm is The time complexity of the algorithm is
$O(n \sqrt n f(n))$ when there are $n$ queries $O(n \sqrt n f(n))$ when there are $n$ queries
and each step takes $f(n)$ time. and each insertion and removal of an element
takes $O(f(n))$ time.
The algorithm processes the queries in a special
order which makes the algorithm efficient.
When the queries correspond to ranges of the form $[a,b]$,
they are primarily sorted according to
the value $\lfloor a/\sqrt n \rfloor$,
and secondarily according to the value $b$.
Hence, all queries whose starting index
is in a fixed segment
are processed after each other.
The essential trick in Mo's algorithm is that
the queries are processed in a special order,
which makes the algorithm efficient.
The array is divided into blocks of $k=O(\sqrt n)$
elements, and the queries are sorted primarily by
the index of the block that contains the first element
of the query, and secondarily by the index of the
last element of the query.
It turns out that using this order, the algorithm It turns out that using this order, the algorithm
only performs $O(n \sqrt n)$ steps. only performs $O(n \sqrt n)$ operations,
The reason for this is that the left border of because the left border of the subarray moves
the range moves $n$ times $O(\sqrt n)$ steps, $n$ times $O(\sqrt n)$ steps,
and the right border of the range moves and the right border of the subarray moves
$\sqrt n$ times $O(n)$ steps. Thus, both the $\sqrt n$ times $O(n)$ steps. Thus, both the
borders move a total of $O(n \sqrt n)$ steps. borders move a total of $O(n \sqrt n)$ steps.
\subsubsection*{Example} \subsubsection*{Example}
As an example, let's consider a problem As an example, let's consider a problem
where we are given a set of ranges in an array, where we are given a set of subarrays in an array,
and our task is to calculate for each range and our task is to calculate for each subarray
the number of distinct elements in the range. the number of distinct elements in the subarray.
In Mo's algorithm, the queries are always sorted In Mo's algorithm, the queries are always sorted
in the same way, but it depends on the problem in the same way, but the way the answer for the query
how the answer for queries is maintained. is maintained depends on the problem.
In this problem, we can maintain an array In this problem, we can maintain an array
\texttt{c} where $\texttt{c}[x]$ \texttt{c} where $\texttt{c}[x]$
indicates how many times an element $x$ indicates how many times an element $x$
occurs in the active range. occurs in the active subarray.
When we move from a query to another query, When we move from a query to another query,
the active range changes. the active subarray changes.
For example, if the current range is For example, if the current subarray is
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.7] \begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (1,0) rectangle (5,1); \fill[color=lightgray] (1,0) rectangle (5,1);
@ -374,7 +375,7 @@ For example, if the current range is
\node at (8.5, 0.5) {4}; \node at (8.5, 0.5) {4};
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
and the next range is and the next subarray is
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.7] \begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (2,0) rectangle (7,1); \fill[color=lightgray] (2,0) rectangle (7,1);
@ -394,21 +395,19 @@ there will be three steps:
the left border moves one step to the left, the left border moves one step to the left,
and the right border moves two steps to the right. and the right border moves two steps to the right.
After each step, we should update the After each step, we update the
array \texttt{c}. array \texttt{c}.
If an element $x$ is added to the range, If an element $x$ is added to the subarray,
the value the value
$\texttt{c}[x]$ increases by one, $\texttt{c}[x]$ increases by one,
and if a value $x$ is removed from the range, and if an element $x$ is removed from the subarray,
the value $\texttt{c}[x]$ decreases by one. the value $\texttt{c}[x]$ decreases by one.
If after an insertion If after an insertion
$\texttt{c}[x]=1$, $\texttt{c}[x]=1$,
the answer for the query increases by one, the answer for the query increases by one,
and if after a removel $\texttt{c}[x]=0$, and if after a removal $\texttt{c}[x]=0$,
the answer for the query decreases by one. the answer for the query decreases by one.
In this problem, the time needed to perform In this problem, the time needed to perform
each step is $O(1)$, so the total time complexity each step is $O(1)$, so the total time complexity
of the algorithm is $O(n \sqrt n)$. of the algorithm is $O(n \sqrt n)$.