typset algorithm better

streaming/streaming.tex

@@ -42,18 +42,16 @@ 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}$(\alpha, k)$
+\algo{FrequencyEstimate} \algalias{Misra/Gries Algorithm}
 \algin the data stream $\alpha$, the target for the estimator $k$
-\:\em{Init}: $A \= \emptyset$. (an empty map)
+\:\em{Init}: $A \= \emptyset$. \cmt{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
+\::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.
+\algout $\hat{f}_a = A[a]$ If $a \in $~keys($A$), and $\hat{f}_a = 0$ otherwise.
 \endalgo
 
 Let us show that $\hat{f}_a$ is a good estimate for the frequency $f_a$.
@@ -95,6 +93,10 @@ $\vert C \vert = \vert$keys($A$)$\vert \leq k - 1$, and a key-value pair can
 be stored in $\O(\log n + \log m)$ bits.
 \qed
 
+\subsection{The Count-Min sketch}
+
+We will now look at a randomized streaming algorithm that performs the same task 
+
 \endchapter