From 126288502f64e1959e82bcab21c107f7fbee8f68 Mon Sep 17 00:00:00 2001
From: Camil Staps
Date: Tue, 15 Dec 2015 19:53:30 +0000
Subject: Finish report practical 2

---
 Practical2/report/analysis.tex | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

(limited to 'Practical2/report/analysis.tex')

diff --git a/Practical2/report/analysis.tex b/Practical2/report/analysis.tex
index d989fe6..ac5d6a9 100644
--- a/Practical2/report/analysis.tex
+++ b/Practical2/report/analysis.tex
@@ -1,11 +1,11 @@
 \section{Complexity analysis}
 \label{sec:analysis}
 
-We iterate $p$ in $\pazocal O(|\bar a'|) = \pazocal O(|\bar a|)$. In every iteration we do two things:
+We iterate over $p$ in $\pazocal O(|\langle a_i\in\bar a\mid 5\nmid a_i\rangle|) = \pazocal O(|\bar a|)$. In every iteration we do two things:
 
 \begin{itemize}
     \item We build $S$ in $\pazocal O(p)=\pazocal O(|\bar a|)$.
-    \item We iterate $d$ in $\pazocal O(k)$. Apart from the special cases that $p=0$ or $d=0$, we then have to iterate over $i$ in $\pazocal O(p)=\pazocal O(|\bar a|)$.
+    \item We iterate over $d$ in $\pazocal O(k)$. Apart from the special cases that $p=0$ or $d=0$, we then have to iterate over $i$ in $\pazocal O(p)=\pazocal O(|\bar a|)$.
         
         The computations in this last loop take constant time: $\PR$ can be implemented in constant time as has been seen in \autoref{sec:implementation}, and computing the maximum can be done while iterating over $i$.
 \end{itemize}
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