From e70e45fd4b5720520608d97c07c3adc511aad92f Mon Sep 17 00:00:00 2001
From: gareth12 <gareth12@users.noreply.github.com>
Date: Tue, 7 Oct 2014 23:04:01 +0100
Subject: [PATCH] simplify proof of lemma 5.3 {hashing-mapping}

the proof of lemma 5.3 {hashing-mapping} can be significantly simplified by noticing that k must be zero. the second half of the proof is then not necessary. also, move the paragraph introducing the star notation until after the proof of lemma 5.3 as the proof no longer uses that notation.
---
 latex/hashing.tex | 42 +++++++++---------------------------------
 1 file changed, 9 insertions(+), 33 deletions(-)

diff --git a/latex/hashing.tex b/latex/hashing.tex
index 6dbf7119..7c20318c 100644
--- a/latex/hashing.tex
+++ b/latex/hashing.tex
@@ -184,13 +184,6 @@ \subsection{Multiplicative Hashing}
   as required.
 \end{proof}
 
-Now, we want to prove \lemref{universal-hashing}, but first we need a
-result from number theory.  In the following proof, we use the notation
-$(b_r,\ldots,b_0)_2$ to denote $\sum_{i=0}^r b_i2^i$, where each $b_i$
-is a bit, either 0 or 1.  In other words, $(b_r,\ldots,b_0)_2$ is
-the integer whose binary representation is given by $b_r,\ldots,b_0$.
-We use $\star$ to denote a bit of unknown value.
-
 \begin{lem}\lemlabel{hashing-mapping}
   Let $S$ be the set of odd integers in $\{1,\ldots,2^{#w#}-1\}$; let $q$
   and $i$ be any two elements in $S$.  Then there is exactly one value
@@ -215,32 +208,8 @@ \subsection{Multiplicative Hashing}
   \begin{equation}
     (#z#-#z#')q = k 2^{#w#} \eqlabel{factors} 
   \end{equation}
-  for some integer $k$.  Thinking in terms of binary numbers, we have 
-  \[
-    (#z#-#z#')q = k\cdot(1,\underbrace{0,\ldots,0}_{#w#})_2 \enspace ,
-  \]
-  so that the #w# trailing bits in the binary representation of
-  $(#z#-#z#')q$ are all 0's.
-
-  Furthermore $k\neq 0$, since $q\neq 0$ and $#z#-#z#'\neq 0$.  Since $q$
-  is odd, it has no trailing 0's in its binary representation:
-  \[
-    q = (\star,\ldots,\star,1)_2 \enspace .
-  \]
-  Since $|#z#-#z#'| < 2^{#w#}$, $#z#-#z#'$ has fewer than #w# trailing
-  0's in its binary representation:
-  \[
-    #z#-#z#' = (\star,\ldots,\star,1,\underbrace{0,\ldots,0}_{<#w#})_2
-      \enspace .
-  \]
-  Therefore, the product $(#z#-#z#')q$ has fewer than #w# trailing 0's in
-  its binary representation:
-  \[
-   (#z#-#z#')q = (\star,\cdots,\star,1,\underbrace{0,\ldots,0}_{<#w#})_2 
-    \enspace .
-  \]
-  Therefore $(#z#-#z#')q$ cannot satisfy \myeqref{factors}, yielding a
-  contradiction and completing the proof.
+  for some integer $k$. Since $z$ is in $S$, $k$ is necessarily zero, 
+  which means that $z=z'$, a contradiction. This completes the proof.
 \end{proof}
 
 The utility of \lemref{hashing-mapping} comes from the following
@@ -249,6 +218,13 @@ \subsection{Multiplicative Hashing}
 to think of the binary representation of #z#, which consists of $#w#-1$
 random bits followed by a 1.
 
+Now, we want to prove \lemref{universal-hashing}, but first we need a
+result from number theory.  In the following proof, we use the notation
+$(b_r,\ldots,b_0)_2$ to denote $\sum_{i=0}^r b_i2^i$, where each $b_i$
+is a bit, either 0 or 1.  In other words, $(b_r,\ldots,b_0)_2$ is
+the integer whose binary representation is given by $b_r,\ldots,b_0$.
+We use $\star$ to denote a bit of unknown value.
+
 \begin{proof}[Proof of \lemref{universal-hashing}]
   First we note that the condition $#hash(x)#=#hash(y)#$ is equivalent to
   the statement ``the highest-order #d# bits of $#z# #x#\bmod2^{#w#}$