From a2675d0a6e3c9979f081078681617db1f83b2f3f Mon Sep 17 00:00:00 2001
From: Camil Staps
Date: Wed, 2 Sep 2015 14:45:28 +0200
Subject: 1.2d, 1.4

---
 assignment1.tex | 9 +++++++--
 1 file changed, 7 insertions(+), 2 deletions(-)

(limited to 'assignment1.tex')

diff --git a/assignment1.tex b/assignment1.tex
index ea0151e..ebda083 100644
--- a/assignment1.tex
+++ b/assignment1.tex
@@ -49,7 +49,7 @@
             \item Yes.
             \item Yes.
             \item No.
-            \item %Todo
+            \item No.
             \item Yes.
             \item Yes.
             \item No.
@@ -76,7 +76,12 @@
                 \end{proof}
         \end{enumerate}
 
-    \item %Todo
+    \item I'm assuming constant time for basic instructions (declaration, comparison, logical operations, branches , function calls and returns, arithmetic instructions).
+
+        \begin{enumerate}
+            \item In this case \texttt{v[i] == x} is never true, meaning we execute the \texttt{while} loop as long as possible (that is, for $i\in\{k\in\mathbb{N}\mid k<n\}$). For the rest, all operations take constant time, giving a worst case time of $c_1\cdot n+c_2$ for some constants $c_1,c_2$.
+            \item In this case \texttt{v[i] == x} for $i=0$, meaning we loop the \texttt{while} loop exactly once, giving it a constant time like all other operations. This gives us best case time $c$ for some constant $c$.
+        \end{enumerate}
 
     \item %Todo
 \end{enumerate}
-- 
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