\chapter{Square root algorithms}

\index{square root algorithm}

A \key{square root algorithm} is an algorithm
that has a square root in its time complexity.
A square root can be seen as a ''poor man's logarithm'':
the complexity $O(\sqrt n)$ is better than $O(n)$
but worse than $O(\log n)$.
Still, many square root algorithms are fast in practice
and have small constant factors.

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
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.
For example, an array of 16 elements will be
divided into blocks of 4 elements as follows:

\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,0) grid (16,1);

\draw (0,1) rectangle (4,2);
\draw (4,1) rectangle (8,2);
\draw (8,1) rectangle (12,2);
\draw (12,1) rectangle (16,2);

\node at (0.5, 0.5) {5};
\node at (1.5, 0.5) {8};
\node at (2.5, 0.5) {6};
\node at (3.5, 0.5) {3};
\node at (4.5, 0.5) {2};
\node at (5.5, 0.5) {7};
\node at (6.5, 0.5) {2};
\node at (7.5, 0.5) {6};
\node at (8.5, 0.5) {7};
\node at (9.5, 0.5) {1};
\node at (10.5, 0.5) {7};
\node at (11.5, 0.5) {5};
\node at (12.5, 0.5) {6};
\node at (13.5, 0.5) {2};
\node at (14.5, 0.5) {3};
\node at (15.5, 0.5) {2};

\node at (2, 1.5) {21};
\node at (6, 1.5) {17};
\node at (10, 1.5) {20};
\node at (14, 1.5) {13};

\end{tikzpicture}
\end{center}

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]
\fill[color=lightgray] (5,0) rectangle (6,1);
\draw (0,0) grid (16,1);

\fill[color=lightgray] (4,1) rectangle (8,2);
\draw (0,1) rectangle (4,2);
\draw (4,1) rectangle (8,2);
\draw (8,1) rectangle (12,2);
\draw (12,1) rectangle (16,2);

\node at (0.5, 0.5) {5};
\node at (1.5, 0.5) {8};
\node at (2.5, 0.5) {6};
\node at (3.5, 0.5) {3};
\node at (4.5, 0.5) {2};
\node at (5.5, 0.5) {5};
\node at (6.5, 0.5) {2};
\node at (7.5, 0.5) {6};
\node at (8.5, 0.5) {7};
\node at (9.5, 0.5) {1};
\node at (10.5, 0.5) {7};
\node at (11.5, 0.5) {5};
\node at (12.5, 0.5) {6};
\node at (13.5, 0.5) {2};
\node at (14.5, 0.5) {3};
\node at (15.5, 0.5) {2};

\node at (2, 1.5) {21};
\node at (6, 1.5) {15};
\node at (10, 1.5) {20};
\node at (14, 1.5) {13};

\end{tikzpicture}
\end{center}

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]
\fill[color=lightgray] (3,0) rectangle (4,1);
\fill[color=lightgray] (12,0) rectangle (13,1);
\fill[color=lightgray] (13,0) rectangle (14,1);
\draw (0,0) grid (16,1);

\fill[color=lightgray] (4,1) rectangle (8,2);
\fill[color=lightgray] (8,1) rectangle (12,2);
\draw (0,1) rectangle (4,2);
\draw (4,1) rectangle (8,2);
\draw (8,1) rectangle (12,2);
\draw (12,1) rectangle (16,2);

\node at (0.5, 0.5) {5};
\node at (1.5, 0.5) {8};
\node at (2.5, 0.5) {6};
\node at (3.5, 0.5) {3};
\node at (4.5, 0.5) {2};
\node at (5.5, 0.5) {5};
\node at (6.5, 0.5) {2};
\node at (7.5, 0.5) {6};
\node at (8.5, 0.5) {7};
\node at (9.5, 0.5) {1};
\node at (10.5, 0.5) {7};
\node at (11.5, 0.5) {5};
\node at (12.5, 0.5) {6};
\node at (13.5, 0.5) {2};
\node at (14.5, 0.5) {3};
\node at (15.5, 0.5) {2};

\node at (2, 1.5) {21};
\node at (6, 1.5) {15};
\node at (10, 1.5) {20};
\node at (14, 1.5) {13};

\draw [decoration={brace}, decorate, line width=0.5mm] (14,-0.25) -- (3,-0.25);

\end{tikzpicture}
\end{center}

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.

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.

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.

\section{Batch processing}

\index{batch processing}

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.

As an example, let us consider a problem
where a grid of size $k \times k$
initially consists of white squares,
and our task is to perform $n$ operations,
each of which is one of the following:
\begin{itemize}
\item
paint square $(y,x)$ black
\item
find the nearest black square to
square $(y,x)$ where the distance
between squares $(y_1,x_1)$ and $(y_2,x_2)$
is $|y_1-y_2|+|x_1-x_2|$
\end{itemize}

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,
we calculate for each square in the grid
the smallest distance to a black square.
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.
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, 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 time complexity would be
$O(k^2 n)$.
And if the algorithm would go through all painted
squares at each operation,
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{Subalgorithms}

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 $\sqrt n$.

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 whose distance
is as large as possible.

For example, in the following tree,
the maximum distance is 4 between
the red nodes 3 and 4:

\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, it is needed to calculate the distance
to $O(\sqrt n)$ other nodes (see Chapter 18.3),
so the total time needed for processing all
nodes is $O(n \sqrt n)$.

\emph{Case 2}: $c > \sqrt n$.
If the number of nodes is large,
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 is $O(n \sqrt n)$.

The time complexity of the algorithm is $O(n \sqrt n)$,
because both cases take $O(n \sqrt n)$ time.

\section{Mo's algorithm}

\index{Mo's algorithm}

\key{Mo's algorithm} can be used in many problems
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 works efficiently.

At each moment in the algorithm, there is an active
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 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 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 endpoint of the range moves
$n$ times $O(\sqrt n)$ 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, 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 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 range.

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);
\draw (0,0) grid (9,1);
\node at (0.5, 0.5) {4};
\node at (1.5, 0.5) {2};
\node at (2.5, 0.5) {5};
\node at (3.5, 0.5) {4};
\node at (4.5, 0.5) {2};
\node at (5.5, 0.5) {4};
\node at (6.5, 0.5) {3};
\node at (7.5, 0.5) {3};
\node at (8.5, 0.5) {4};
\end{tikzpicture}
\end{center}
and the next range is
\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (2,0) rectangle (7,1);
\draw (0,0) grid (9,1);
\node at (0.5, 0.5) {4};
\node at (1.5, 0.5) {2};
\node at (2.5, 0.5) {5};
\node at (3.5, 0.5) {4};
\node at (4.5, 0.5) {2};
\node at (5.5, 0.5) {4};
\node at (6.5, 0.5) {3};
\node at (7.5, 0.5) {3};
\node at (8.5, 0.5) {4};
\end{tikzpicture}
\end{center}
there will be three steps:
the left endpoint moves one step to the left,
and the right endpoint moves two steps to the right.

After each step, the array \texttt{c}
needs to be updated.
After adding an element $x$,
we increase the value of 
$\texttt{c}[x]$ by one
and if $\texttt{c}[x]=1$ after this,
we also increase the answer to the query by one.
Similarly, after removing an element $x$,
we decrease the value of 
$\texttt{c}[x]$ by one
and if $\texttt{c}[x]=0$ after this,
we also decrease the answer to the query by one.

In this problem, the time needed to perform
each step is $O(1)$, so the total time complexity
of the algorithm is $O(n \sqrt n)$.