Commit c23efdce authored by Martin Mareš's avatar Martin Mareš
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Merge branch 'pm-streaming'

parents 983e1af4 9882fc7d
include ../Makerules
\input adsmac.tex
\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:
\: 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.
\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 knew some small set
$C$ which contains $F_k$. Then, with a pass over the input, we can count the
occurrences of each element of $C$, and hence find $F_k$ in
$\O(\vert C \vert \log m)$ space.
\subsection{The Misra/Gries Algorithm}
We will now see a deterministic one-pass algorithm that estimates the frequency
of each element in a stream of integers. We shall see that it also provides
us with a small set $C$ containing $F_k$, and hence lets us solve the frequent
elements problem efficiently.
\algo{FrequencyEstimate} \algalias{Misra/Gries Algorithm}
\algin the data stream $\alpha$, the target for the estimator $k$.
\:\em{Init}: $A \= \emptyset$. \cmt{an empty map}
\::If $x \in$ keys($A$), $A[x] \= A[x] + 1$.
\::Else If $\vert$keys($A$)$\vert < k - 1$, $A[x] \= 1$.
\forall $a \in $~keys($A$): $A[a] \= A[a] - 1$,
delete $a$ from $A$ if $A[a] = 0$.
\algout $\hat{f}_a = A[a]$ If $a \in $~keys($A$), and $\hat{f}_a = 0$ otherwise.
Let us show that $\hat{f}_a$ is a good estimate for the frequency $f_a$.
$f_a - m / k \leq \hat{f}_a \leq f_a$
We see immediately that $\hat{f}_a \leq f_a$, since it is only incremented when
we see $a$ in the stream.
To see the other inequality, suppose that we have a counter for each
$a \in [n]$ (instead of just $k - 1$ keys at a time). Whenever we have at least
$k$ non-zero counters, we will decrease all of them by $1$; this gives exactly
the same estimate as the algorithm above.
Now consider the potential
function $\Phi = \sum_{a \in [n]} A[a]$. Note that $\Phi$ increases by
exactly $m$ (since $\alpha$ contains $m$ elements), and is decreased by $k$
every time any $A[x]$ decreases. Since $\Phi = 0$ initially and $\Phi \geq 0$,
we get that $A[x]$ decreases at most $m / k$ times.
There exists a deterministic 2-pass algorithm that finds $F_k$ in
$\O(k(\log n + \log m))$ space.
In the first pass, we obtain the frequency estimate $\hat{f}$ by the
Misra/Gries algorithm.
We set $C = \{ a \mid \hat{f}_a > 0 \}$. For $a \in F_k$, we have
$f_a > m / k$, and hence $\hat{f}_a > 0$ by the previous Lemma.
In the second pass, we count $f_c$ exactly for each $c \in C$, and hence know
$F_k$ at the end.
To see the bound on space used, note that
$\vert C \vert = \vert$keys($A$)$\vert \leq k - 1$, and a key-value pair can
be stored in $\O(\log n + \log m)$ bits.
\subsection{The Count-Min Sketch}
We will now look at a randomized streaming algorithm that solves the
frequency estimation problem. While this algorithm can fail with some
probability, it has the advantage that the output on two different streams
can be easily combined.
\algo{FrequencyEstimate} \algalias{Count-Min Sketch}
\algin the data stream $\alpha$, the accuracy $\varepsilon$,
the error parameter $\delta$.
$C[1\ldots t][1\ldots k] \= 0$, where $k \= \lceil 2 / \varepsilon \rceil$
and $t \= \lceil \log(1 / \delta) \rceil$.
\:: Choose $t$ independent hash functions $h_1, \ldots , h_t : [n] \to [k]$, each
from a 2-independent family.
\::For $i \in [t]$: $C[i][h_i(x)] \= C[i][h_i(x)] + 1$.
\algout Report $\hat{f}_a = \min_{i \in t} C[i][h_i(a)]$.
Note that the algorithm needs $\O(tk \log m)$ bits to store the table $C$, and
$\O(t \log n)$ bits to store the hash functions $h_1, \ldots , h_t$, and hence
uses $\O(1/\varepsilon \cdot \log (1 / \delta) \cdot \log m
+ \log (1 / \delta)\cdot \log n)$ bits. It remains to show that it computes
a good estimate.
$f_a \leq \hat{f}_a \leq f_a + \varepsilon m$ with probability $\delta$.
Clearly $\hat{f}_a \geq f_a$ for all $a \in [n]$; we will show that
$\hat{f}_a \leq f_a + \varepsilon m$ with probability at least $\delta$.
For a fixed element $a$, define the random variable
$$X_i := C[i][h_i(a)] - f_a$$
For $j \in [n] \setminus \{ a \}$, define the
indicator variable $Y_{i, j} := [ h_i(j) = h_i(a) ]$. Then we can see that
$$X_i = \sum_{j \neq a} f_j\cdot Y_{i, j}$$
Note that $\E[Y_{i, j}] = 1/k$ since each $h_i$ is from a 2-independent family,
and hence by linearity of expectation:
$$\E[X_i] = {\vert\vert f \vert\vert_1 - f_a \over k} =
{\vert\vert f_{-a} \vert\vert_1 \over k}$$
And by applying Markov's inequality we obtain a bound on the error of a single
$$ \Pr[X_i > \varepsilon \cdot m ] \geq
\Pr[ X_i > \varepsilon \cdot \vert\vert f_{-a} \vert\vert_1 ] \leq
{1 \over k\varepsilon} \leq 1/2$$
Finally, since we have $t$ independent counters, the probability that they
are all wrong is:
$$ \Pr\left[\bigcap_i X_i > \varepsilon \cdot m \right] \leq 1/2^t \leq \delta $$
The main advantage of this algorithm is that its output on two different
streams (computed with the same set of hash functions $h_i$) is just the sum
of the respective tables $C$. It can also be extended to support events
which remove an occurence of an element $x$ (with the caveat that upon
termination the ``frequency'' $f_x$ for each $x$ must be non-negative).
(TODO: perhaps make the second part an exercise?).
\section{Counting Distinct Elements}
We continue working with a stream $\alpha[1 \ldots m]$ of integers from $[n]$,
and define $f_a$ (the frequency of $a$) as before. Let
$d = \vert \{ j : f_j > 0 \} \vert$. Then the distinct elements problem is
to estimate $d$.
\subsection{The AMS Algorithm}
Suppose we map our universe $[n]$ to itself via a random permutation $\pi$.
Then if the number of distinct elements in a stream is $d$, we expect
$d / 2^i$ of them to be divisible by $2^i$ after applying $\pi$. This is the
core idea of the following algorithm.
Define ${\tt tz}(x) := \max\{ i \mid 2^i $~divides~$ x \}$
(i.e. the number of trailing zeroes in the base-2 representation of $x$).
\algo{DistinctElements} \algalias{AMS}
\algin the data stream $\alpha$.
\:\em{Init}: Choose a random hash function $h : [n] \to [n]$ from a 2-independent
\:: $z \= 0$.
\::If ${\tt tz}(h(x)) > z$: $z \= {\tt tz}(h(x))$.
\algout $\hat{d} \= 2^{z + 1/2}$
The AMS algorithm is a $(3, \delta)$-estimator for some constant
For $j \in [n]$, $r \geq 0$, let $X_{r, j} := [ {\tt tz}(h(j)) \geq r ]$, the
indicator that is true if $h(j)$ has at least $r$ trailing $0$s.
Now define $$ Y_r = \sum_{j : f_j > 0} X_{r, j} $$
How is our estimate related to $Y_r$? If the algorithm outputs
$\hat{d} \geq 2^{a + 1/2}$, then we know that $Y_a > 0$. Similarly, if the
output is smaller than $2^{a + 1/2}$, then we know that $Y_a = 0$. We will now
bound the probabilities of these events.
For any $j \in [n]$, $h(j)$ is uniformly distributed over $[n]$ (since $h$
is $2$-independent). Hence $\E[X_{r, j}] = 1 / 2^r$. By linearity of
expectation, $\E[Y_{r}] = d / 2^r$.
We will also use the variance of these variables -- note that
$$\Var[X_{r, j}] \leq \E[X_{r, j}^2] = \E[X_{r, j}] = 1/2^r$$
And because $h$ is $2$-independent, the variables $X_{r, j}$ and $X_{r, j'}$
are independent for $j \neq j'$, and hence:
$$\Var[Y_{r}] = \sum_{j : f_j > 0} \Var[X_{r, j}] \leq d / 2^r $$
Now, let $a$ be the smallest integer such that $2^{a + 1/2} \geq 3d$. Then we
$$ \Pr[\hat{d} \geq 3d] = \Pr[Y_a > 0] = \Pr[Y_a \geq 1] $$
Using Markov's inequality we get:
$$ \Pr[\hat{d} \geq 3d] \leq \E[Y_a] = {d \over 2^a} \leq {\sqrt{2} \over 3} $$
For the other side, let $b$ be the smallest integer so that
$2^{b + 1/2} \leq d/3$. Then we have:
$$ \Pr[\hat{d} \leq d / 3] = \Pr[ Y_{b + 1} = 0] \leq
\Pr[ \vert Y_{b + 1} - \E[Y_{b + 1}] \vert \geq d / 2^{b + 1} ]$$
Using Chebyshev's inequality, we get:
$$ \Pr[\hat{d} < d / 3] \leq {\Var[Y_b] \over (d / 2^{b + 1})^2} \leq
{2^{b + 1} \over d} \leq {\sqrt{2} \over 3}$$
The previous algorithm is not particularly satisfying -- by our analysis it
can make an error around $94\%$ of the time (taking the union of the two bad
events). However we can improve the success probability easily; we run $t$
independent estimators simultaneously, and print the median of their outputs.
By a standard use of Chernoff Bounds one can show that the probability that
the median is more than $3d$ is at most $2^{-\Theta(t)}$ (and similarly also
the probability that it is less than $d / 3$).
Hence it is enough to run $\O(\log (1/ \delta))$ copies of the AMS estimator
to get a $(3, \delta)$ estimator for any $\delta > 0$. Finally, we note that
the space used by a single estimator is $\O(\log n)$ since we can store $h$ in
$\O(\log n)$ bits, and $z$ in $\O(\log \log n)$ bits, and hence a $(3, \delta)$
estimator uses $\O(\log (1/\delta) \cdot \log n)$ bits.
\subsection{The BJKST Algorithm}
We will now look at another algorithm for the distinct elements problem.
Note that unlike the AMS algorithm, it accepts an accuracy parameter
\algo{DistinctElements} \algalias{BJKST}
\algin the data stream $\alpha$, the accuracy $\varepsilon$.
\:\em{Init}: Choose a random hash function $h : [n] \to [n]$ from a 2-independent
\:: $z \= 0$, $B \= \emptyset$.
\::If ${\tt tz}(h(x)) \geq z$:
\:::$B \= B \cup \{ (x, {\tt tz}(h(x)) \}$
\:::While $\vert B \vert \geq c/\varepsilon^2$:
\::::$z \= z + 1$.
\::::Remove all $(a, b)$ from $B$ such that $b = {\tt tz}(h(a)) < z$.
\algout $\hat{d} \= \vert B \vert \cdot 2^{z}$.
For any $\varepsilon > 0$, the BJKST algorithm is an
$(\varepsilon, \delta)$-estimator for some constant $\delta$.
We setup the random variables $X_{r, j}$ and $Y_r$ as before. Let $t$ denote
the value of $z$ when the algorithm terminates, then $Y_t = \vert B \vert$,
and our estimate $\hat{d} = \vert B \vert \cdot 2^t = Y_t \cdot 2^t$.
Note that if $t = 0$, the algorithm computes $d$ exactly (since we never remove
any elements from $B$, and $\hat{d} = \vert B \vert$). For $t \geq 1$, we
say that the algorithm \em{fails} iff
$\vert Y_t \cdot 2^t - d \vert > \varepsilon d$. Rearranging, we have that the
algorithm fails iff:
$$ \left\vert Y_t - {d \over 2^t} \right\vert \geq {\varepsilon d \over 2^t} $$
To bound the probability of this event, we will sum over all possible values
$r \in [\log n]$ that $t$ can take. Note that for \em{small} values of $r$,
a failure is unlikely when $t = r$, since the required deviation $d / 2^t$ is
large. For \em{large} values of $r$, simply achieving $t = r$ is difficult.
More formally, let $s$ be the unique integer such that:
$$ {12 \over \varepsilon^2} \leq {d \over 2^s} \leq {24 \over \varepsilon^2}$$
Then we have:
$$ \Pr[{\rm fail}] = \sum_{r = 1}^{\log n}
\Pr\left[ \left\vert Y_r - {d \over 2^r} \right\vert \geq {\varepsilon d \over 2^r}
\land t = r \right] $$
After splitting the sum around $s$, we bound small and large values by different
methods as described above to get:
$$ \Pr[{\rm fail}] \leq \sum_{r = 1}^{s - 1}
\Pr\left[ \left\vert Y_r - {d \over 2^r} \right\vert \geq {\varepsilon d \over 2^r} \right] +
\sum_{r = s}^{\log n}
\Pr\left[t = r \right] $$
Recall that $\E[Y_r] = d / 2^r$, so the terms in the first sum can be bounded
using Chebyshev's inequality. The second sum is equal to the probability of
the event $[t \geq s]$, that is, the event $Y_{s - 1} \geq c / \varepsilon^2$
(since $z$ is only increased when $B$ becomes larger than this threshold).
We will use Markov's inequality to bound the probability of this event.
Putting it all together, we have:
\Pr[{\rm fail}] &\leq \sum_{r = 1}^{s - 1}
{\Var[Y_r] \over (\varepsilon d / 2^r)^2} + {\E[Y_{s - 1}] \over c / \varepsilon^2}
\leq \sum_{r = 1}^{s - 1}
{d / 2^r \over (\varepsilon d / 2^r)^2} + {d / 2^{s - 1} \over c / \varepsilon^2}\cr
&= \sum_{r = 1}^{s - 1} {2^r \over \varepsilon^2 d} + {\varepsilon^2 d \over c2^{s - 1}}
\leq {2^{s} \over \varepsilon^2 d} + {\varepsilon^2 d \over c2^{s - 1}}
Recalling the definition of $s$, we have $2^s / d \leq \varepsilon^2 / 12$, and
$d / 2^{s - 1} \leq 48 / \varepsilon^2$, and hence:
$$ \Pr[{\rm fail}] \leq {1 \over 12} + {48 \over c} $$
which is smaller than (say) $1 / 6$ for $c > 576$. Hence the algorithm is an
$(\varepsilon, 1 / 6)$-estimator.
As before, we can run $\O(\log \delta)$ independent copies of the algorithm,
and take the median of their estimates to reduce the probability of failure
to $\delta$. The only thing remaining is to look at the space usage of the
The counter $z$ requires only $\O(\log \log n)$ bits, and $B$ has
$\O(1 / \varepsilon^2)$ entries, each of which needs $\O( \log n )$ bits.
Finally, the hash function $h$ needs $\O(\log n)$ bits, so the total space
used is dominated by $B$, and the algorithm uses $\O(\log n / \varepsilon^2)$
space. As before, if we use the median trick, the space used increases to
$\O(\log\delta \cdot \log n / \varepsilon^2)$.
(TODO: include the version of this algorithm where we save space by storing
$(g(a), {\tt tz}(h(a)))$ instead of $(a, {\tt tz}(h(a)))$ in $B$ for some
hash function $g$ as an exercise?)
......@@ -170,6 +170,7 @@
\def\E{{\bb E}}
\def\Pr{{\rm Pr}\mkern0.5mu}
\def\Prsub#1{{\rm Pr}_{#1}}
\def\Var{{\rm Var}\mkern0.5mu}
% Vektory
\def\t{{\bf t}}
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