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@ -10,28 +10,25 @@ but worse than $O(\log n)$.
Still, many square root algorithms are fast in practice
and have small constant factors.
As an example, let's consider the problem of
handling sum queries in an array.
The required operations are:
\begin{itemize}
\item change the value at index $x$
\item calculate the sum in the range $[a,b]$
\end{itemize}
As an example, let us consider the problem of
creating a data structure that supports
two operations in an array:
modifying an element at a given position
and calculating the sum of elements in the given range.
We have previously solved the problem using
a binary indexed tree and a segment tree,
that support both operations in $O(\log n)$ time.
However, now we will solve the problem
in another way using a square root structure
so that we can calculate sums in $O(\sqrt n)$ time
and modify values in $O(1)$ time.
that allows us to modify elements in $O(1)$ time
and calculate sums in $O(\sqrt n)$ time.
The idea is to divide the array into blocks
of size $\sqrt n$ so that each block contains
the sum of elements inside the block.
The following example shows an array and the
corresponding segments:
For example, an array of 16 elements will be
divided into blocks of 4 elements as follows:
\begin{center}
\begin{tikzpicture}[scale=0.7]
@ -67,9 +64,15 @@ corresponding segments:
\end{tikzpicture}
\end{center}
When a value in the array changes,
we have to calculate the sum of the corresponding
block again:
Using this structure,
it is easy to modify the array,
because it is only needed to calculate
the sum of a single block again
after each modification,
which can be done in $O(1)$ time.
For example, the following picture shows
how the value of an element and
the sum of the corresponding block change:
\begin{center}
\begin{tikzpicture}[scale=0.7]
@ -107,9 +110,12 @@ block again:
\end{tikzpicture}
\end{center}
Any sum in the array can be calculated as a combination
of single values in the array and the sums of the
blocks between them:
Calculating the sum of elements in a range is
a bit more difficult.
It turns out that we can always divide
the range into three parts such that
the sum consists of values of single elements
and sums of blocks between them:
\begin{center}
\begin{tikzpicture}[scale=0.7]
@ -152,35 +158,23 @@ blocks between them:
\end{tikzpicture}
\end{center}
We can change a value in $O(1)$ time,
because we only have to change the sum of a single block.
A sum in a range consists of three parts:
Since the number of single elements is $O(\sqrt n)$
and also the number of blocks is $O(\sqrt n)$,
the time complexity of the sum query is $O(\sqrt n)$.
Thus, the parameter $\sqrt n$ balances two things:
the array is divided into $\sqrt n$ blocks,
each of which contains $\sqrt n$ elements.
\begin{itemize}
\item first, there are $O(\sqrt n)$ single values
\item then, there are $O(\sqrt n)$ consecutive blocks
\item finally, there are $O(\sqrt n)$ single values
\end{itemize}
Calculating each sum takes $O(\sqrt n)$ time,
so the total complexity for calculating the sum
of values in any range is $O(\sqrt n)$.
The reason why we use the parameter $\sqrt n$ is that
it balances two things:
for example, an array of $n$ elements is divided
into $\sqrt n$ blocks, each of which contains
$\sqrt n$ elements.
In practice, it is not needed to use exactly
the parameter $\sqrt n$ in algorithms, but it may be better to
In practice, it is not needed to use the
exact parameter $\sqrt n$, but it may be better to
use parameters $k$ and $n/k$ where $k$ is
larger or smaller than $\sqrt n$.
The optimal parameter depends on the problem and input.
The best parameter depends on the problem
and input.
For example, if an algorithm often goes through
blocks but rarely iterates elements inside
blocks, it may be good to divide the array into
For example, if an algorithm often goes
through the blocks but rarely inspects
single elements inside the blocks,
it may be a good idea to divide the array into
$k < \sqrt n$ blocks, each of which contains $n/k > \sqrt n$
elements.
@ -188,24 +182,19 @@ elements.
\index{batch processing}
In \key{batch processing}, the operations of an
algorithm are divided into batches,
and each batch will be processed separately.
Between the batches some precalculation is done
to process the future operations more efficiently.
Sometimes the operations of an algorithm
can be divided into batches so that
each batch can be processed separately.
Some precalculation is done
between the batches
in order to process the future operations more efficiently.
If there are $O(\sqrt n)$ batches of size $O(\sqrt n)$,
this results in a square root algorithm.
In a square root algorithm, $n$ operations are
divided into batches of size $O(\sqrt n)$,
and the number of both batches and operations in each
batch is $O(\sqrt n)$.
This balances the precalculation time between
the batches and the time needed for processing
the batches.
As an example, let's consider a problem
As an example, let us consider a problem
where a grid of size $k \times k$
initially consists of white squares.
Our task is to perform $n$ operations,
initially consists of white squares,
and our task is to perform $n$ operations,
each of which is one of the following:
\begin{itemize}
\item
@ -217,7 +206,8 @@ between squares $(y_1,x_1)$ and $(y_2,x_2)$
is $|y_1-y_2|+|x_1-x_2|$
\end{itemize}
The solution is to divide the operations into
We can solve the problem by dividing
the operations into
$O(\sqrt n)$ batches, each of which consists
of $O(\sqrt n)$ operations.
At the beginning of each batch,
@ -227,81 +217,99 @@ This can be done in $O(k^2)$ time using breadth-first search.
When processing a batch, we maintain a list of squares
that have been painted black in the current batch.
Now, the distance from a square to the nearest black
The list contains $O(\sqrt n)$ elements,
because there are $O(\sqrt n)$ operations in each batch.
Thus, the distance from a square to the nearest black
square is either the precalculated distance or the distance
to a square that has been painted black in the current batch.
The algorithm works in
$O((k^2+n) \sqrt n)$ time.
First, between the batches,
there are $O(\sqrt n)$ searches that each take
$O(k^2)$ time.
Second, the total number of processed
squares is $O(n)$, and at each square,
we go through a list of $O(\sqrt n)$ squares
in a batch.
First, there are $O(\sqrt n)$ breadth-first searches
and each search takes $O(k^2)$ time.
Second, the total number of
squares processed during the algorithm
is $O(n)$, and at each square,
we go through a list of $O(\sqrt n)$ squares.
If the algorithm would perform a breadth-first search
at each operation, the complexity would be
at each operation, the time complexity would be
$O(k^2 n)$.
And if the algorithm would go through all painted
squares at each operation,
the complexity would be $O(n^2)$.
The square root algorithm combines these complexities,
and turns the factor $n$ into $\sqrt n$.
the time complexity would be $O(n^2)$.
Thus, the time complexity of the square root algorithm
is a combination of these time complexities,
but in addition, a factor $n$ is replaced by $\sqrt n$.
\section{Case processing}
\section{Subalgorithms}
\index{case processing}
Some square root algorithms consists of
subalgorithms that are specialized for different
input parameters.
Typically, there are two subalgorithms:
one algorithm is efficient when
some parameter is smaller than $\sqrt n$,
and another algorithm is efficient
when the parameter is larger than $\sqrt n$.
In \key{case processing}, an algorithm has
specialized subalgorithms for different cases that
may appear during the algorithm.
Typically, one part is efficient for
small parameters, and another part is efficient
for large parameters, and the turning point is
about $\sqrt n$.
As an example, let's consider a problem where
we are given a tree that contains $n$ nodes,
As an example, let us consider a problem where
we are given a tree of $n$ nodes,
each with some color. Our task is to find two nodes
that have the same color and the distance
between them is as large as possible.
that have the same color and whose distance
is as large as possible.
The problem can be solved by going through all
colors one after another, and for each color,
finding two nodes of that color whose distance is
maximum.
For a fixed color, a subalgorithm will be used
that depends on the number of nodes of that color.
Let's assume that the current color is $x$
and there are $c$ nodes whose color is $x$.
There are two cases:
For example, in the following tree,
the maximum distance is 4 between
the red nodes 3 and 4:
\subsubsection*{Case 1: $c \le \sqrt n$}
\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle, fill=green!40] (1) at (1,3) {$2$};
\node[draw, circle, fill=red!40] (2) at (4,3) {$3$};
\node[draw, circle, fill=red!40] (3) at (1,1) {$5$};
\node[draw, circle, fill=blue!40] (4) at (4,1) {$6$};
\node[draw, circle, fill=red!40] (5) at (-2,1) {$4$};
\node[draw, circle, fill=blue!40] (6) at (-2,3) {$1$};
\path[draw,thick,-] (1) -- (2);
\path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (3) -- (4);
\path[draw,thick,-] (3) -- (6);
\path[draw,thick,-] (5) -- (6);
\end{tikzpicture}
\end{center}
The problem can be solved by going through
all colors and calculating
the maximum distance of two nodes for each color
separately.
Assume that the current color is $x$ and
there are $c$ nodes whose color is $x$.
There are two subalgorithms
that are specialized for small and large
values of $c$:
\emph{Case 1}: $c \le \sqrt n$.
If the number of nodes is small,
we go through all pairs of nodes whose
color is $x$ and select the pair that
has the maximum distance.
For each node, we have calculate the distance
For each node, it is needed to calculate the distance
to $O(\sqrt n)$ other nodes (see 18.3),
so the total time needed for processing all
nodes in case 1 is $O(n \sqrt n)$.
\subsubsection*{Case 2: $c > \sqrt n$}
nodes is $O(n \sqrt n)$.
\emph{Case 2}: $c > \sqrt n$.
If the number of nodes is large,
we traverse through the whole tree
we go through the whole tree
and calculate the maximum distance between
two nodes with color $x$.
The time complexity of the tree traversal is $O(n)$,
and this will be done at most $O(\sqrt n)$ times,
so the total time needed for case 2 is
$O(n \sqrt n)$.\\\\
\noindent
so the total time needed is $O(n \sqrt n)$.
The time complexity of the algorithm is $O(n \sqrt n)$,
because both case 1 and case 2 take $O(n \sqrt n)$ time.
because both cases take $O(n \sqrt n)$ time.
\section{Mo's algorithm}
@ -312,54 +320,53 @@ that require processing range queries in
a \emph{static} array.
Before processing the queries, the algorithm
sorts them in a special order which guarantees
that the algorithm runs efficiently.
that the algorithm works efficiently.
At each moment in the algorithm, there is an active
subarray and the algorithm maintains the answer
for a query to that subarray.
The algorithm processes the given queries one by one,
and always changes the active subarray
by inserting and removing elements
so that it corresponds to the current query.
range and the algorithm maintains the answer
to a query related to that range.
The algorithm processes the queries one by one,
and always updates the endpoints of the
active range by inserting and removing elements.
The time complexity of the algorithm is
$O(n \sqrt n f(n))$ when there are $n$ queries
and each insertion and removal of an element
takes $O(f(n))$ time.
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 $k=O(\sqrt n)$
The trick in Mo's algorithm is the order
in which the queries are processed:
The array is divided into blocks of $O(\sqrt n)$
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.
the number of the block that contains the first element
in the range, and secondarily by the position of the
last element in the range.
It turns out that using this order, the algorithm
only performs $O(n \sqrt n)$ operations,
because the left border of the subarray moves
because the left endpoint of the range moves
$n$ times $O(\sqrt n)$ steps,
and the right border of the subarray moves
$\sqrt n$ times $O(n)$ steps. Thus, both the
borders move a total of $O(n \sqrt n)$ steps.
and the right endpoint of the range moves
$\sqrt n$ times $O(n)$ steps. Thus, both
endpoints move a total of $O(n \sqrt n)$ steps during the algorithm.
\subsubsection*{Example}
As an example, let's consider a problem
where we are given a set of subarrays in an array,
and our task is to calculate for each subarray
the number of distinct elements in the subarray.
As an example, consider a problem
where we are given a set of queries,
each of them corresponding to a range in an array,
and our task is to calculate for each query
the number of distinct elements in the range.
In Mo's algorithm, the queries are always sorted
in the same way, but the way the answer for the query
is maintained depends on the problem.
in the same way, but it depends on the problem
how the answer to the query is maintained.
In this problem, we can maintain an array
\texttt{c} where $\texttt{c}[x]$
indicates how many times an element $x$
occurs in the active subarray.
occurs in the active range.
When we move from a query to another query,
the active subarray changes.
For example, if the current subarray is
When we move from one query to another query,
the active range changes.
For example, if the current range is
\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (1,0) rectangle (5,1);
@ -375,7 +382,7 @@ For example, if the current subarray is
\node at (8.5, 0.5) {4};
\end{tikzpicture}
\end{center}
and the next subarray is
and the next range is
\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (2,0) rectangle (7,1);
@ -392,21 +399,21 @@ and the next subarray is
\end{tikzpicture}
\end{center}
there will be three steps:
the left border moves one step to the left,
and the right border moves two steps to the right.
the left endpoint moves one step to the left,
and the right endpoint moves two steps to the right.
After each step, we update the
array \texttt{c}.
If an element $x$ is added to the subarray,
If an element $x$ is added to the range,
the value
$\texttt{c}[x]$ increases by one,
and if an element $x$ is removed from the subarray,
and if an element $x$ is removed from the range,
the value $\texttt{c}[x]$ decreases by one.
If after an insertion
$\texttt{c}[x]=1$,
the answer for the query increases by one,
the answer to the query will be increased by one,
and if after a removal $\texttt{c}[x]=0$,
the answer for the query decreases by one.
the answer to the query will be decreased by one.
In this problem, the time needed to perform
each step is $O(1)$, so the total time complexity