Commit ad01c805 authored by Parth Mittal's avatar Parth Mittal
Browse files

wrote misra/gries algorithm and analysis

parent cdaacc99
......@@ -38,4 +38,60 @@ Algorithm does.
\subsection{Misra/Gries Algorithm}
TODO: Typeset the algorithm better.
\proc{FrequencyEstimate}$(\alpha, k)$
\algin the data stream $\alpha$, the target for the estimator $k$
\:Init: $A \= \emptyset$
\:For $j$ a number from the stream:
\:If $j$ is a key in $A$, $A[j] \= A[j] + 1$.
\:Else If $\vert A \vert < k - 1$, add the key $j$ to $A$ and set $A[j] \= 1$.
\:Else For each key $\ell$ in $A$, reduce $A[\ell] \= A[\ell] - 1$.
Delete $\ell$ from $A$ if $A[\ell] = 0$.
\:After processing the entire stream, return A.
Let us show that $A[j]$ is a good estimate for the frequency $f_j$.
$f_j - m / k \leq A[j] \leq f_j$
Suppose that $A$ maintains the value for each key $j \in [n]$ (instead of
just $k - 1$ of them). We can recast \alg{FrequencyEstimate} in this setting:
We always increment $A[j]$ on seeing $j$ in the stream, but if there are
$\geq k$ positive values $A[\ell]$ after this step, we decrease each of them
by 1.
In particular, this reduces the value of the most recently added key $A[j]$
back to $0$.
Now, we see immediately that $A[j] \leq f_j$, since it is only incremented when
we see $j$ in the stream. To see the other inequality, consider the potential
function $\Phi = \sum_{\ell} A[\ell]$. Note that $\Phi$ increases by exactly
$m$ (since the stream contains $m$ elements), and is decreased by $k$ every
time $A[j]$ decreases. Since $\Phi = 0$ initially and $\Phi \geq 0$, we get
that $A[j]$ is decreased at most $m / k$ times.
Now, for $j \in F_k$, we know that $f_j > m / k$, which implies that $A[j] > 0$.
Hence $F_k \subseteq C = \{ j \mid A[j] > 0 \}$, and we have a $C$ of size
$k - 1$ ready for the second pass over the input.
There exists a deterministic 2-pass algorithm that finds $F_k$ in
$\O(k(\log n + \log m))$ space.
The correctness of the algorithm follows from the discussion above, we show
the bound on the space used below.
In the first pass, we only need to store $k - 1$ key-value pairs for $A$
(for example, as an unordered-list),
and the key and the value need $\lfloor\log_2 n \rfloor + 1$ and
$\lfloor \log_2 m \rfloor + 1$ bits respectively.
In the second pass, we have one key-value pair for each element of $C$, and
they take the same amount of space as above.
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