rewrote misra/gries

streaming/streaming.tex

@@ -30,68 +30,73 @@ 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.
+$F_k = \{ j \mid f_j > m / k \}$. 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{Misra/Gries Algorithm}
+\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.
 
 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.
+\:\em{Init}: $A \= \emptyset$. (an empty map)
+\:\em{Process}($x$): 
+\:  If $x \in$ keys($A$), $A[x] \= A[x] + 1$.
+\:  Else If $\vert$keys($A$)$\vert < k - 1$, $A[x] \= 1$.
+\:  Else
+      \forall $a \in $~keys($A$): $A[a] \= A[a] - 1$,
+      delete $a$ from $A$ if $A[a] = 0$.
+\:\em{Output}: $\hat{f}_a = A[a]$ If $a \in $~keys($A$), and $\hat{f}_a = 0$ otherwise.
 \endalgo
 
-Let us show that $A[j]$ is a good estimate for the frequency $f_j$.
+Let us show that $\hat{f}_a$ is a good estimate for the frequency $f_a$.
 
 \lemma{
-$f_j - m / k \leq A[j] \leq f_j$
+$f_a - m / k \leq \hat{f}_a \leq f_a$
 }
 
 \proof
-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.
+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.
 \qed
 
-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.
-
 \theorem{
     There exists a deterministic 2-pass algorithm that finds $F_k$ in
     $\O(k(\log n + \log m))$ space.
 }
 \proof
-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.
-
+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.
 \qed
 
 \endchapter
+
+
+
+