Commit cdaacc99 authored by Parth Mittal's avatar Parth Mittal
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wrote intro to streaming / frequent elements

parent 66411d7c
TOP=..
include ../Makerules
\ifx\chapter\undefined
\input adsmac.tex
\singlechapter{20}
\fi
\chapter[streaming]{Streaming Algorithms}
For this chapter, we will consider the streaming model. In this
setting, the input is presented as a ``stream'' which we can read
\em{in order}. In particular, at each step, we can do some processing,
and then move forward one unit in the stream to read the next piece of data.
We can choose to read the input again after completing a ``pass'' over it.
There are two measures for the performance of algorithms in this setting.
The first is the number of passes we make over the input, and the second is
the amount of memory that we consume. Some interesting special cases are:
\tightlist{o}
\: 1 pass, and $O(1)$ memory: This is equivalent to computing with a DFA, and
hence we can recognise only regular languages.
\: 1 pass, and unbounded memory: We can store the entire stream, and hence this
is just the traditional computing model.
\endlist
\section{Frequent Elements}
For this problem, the input is a stream $\alpha[1 \ldots m]$ where each
$\alpha[i] \in [n]$.
We define for each $j \in [n]$ the \em{frequency} $f_j$ which counts
the occurences of $j$ in $\alpha[1 \ldots m]$. Then the majority problem
is to find (if it exists) a $j$ such that $f_j > m / 2$.
We consider the more general frequent elements problem, where we want to find
$F_k = \{ j \mid f_j > m / k \}$. Suppose that we (magically) knew some small set
$C$ which contains $F_k$. Then we can pass over the input once, keeping track of
how many times we see each member of $C$, and then find $F_k$ easily.
The challenge is to find a small $C$, which is precisely what the Misra/Gries
Algorithm does.
\subsection{Misra/Gries Algorithm}
\endchapter
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