Commit 5d2b3d02 by Parth Mittal

### rewrote misra/gries

 ... ... @@ -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