From dcd4774bfc5d3962a981c7b2ffbadf9a65ad83d6 Mon Sep 17 00:00:00 2001
From: Sepehr Bayat <bayats1@mcmaster.ca>
Date: Mon, 29 Jan 2018 07:12:31 -0500
Subject: [PATCH] Update T4.tex

---
 Tutorials/T04-A2Example/slides/T4.tex | 11 +++++++----
 1 file changed, 7 insertions(+), 4 deletions(-)

diff --git a/Tutorials/T04-A2Example/slides/T4.tex b/Tutorials/T04-A2Example/slides/T4.tex
index 9da20965..f8cc63f6 100644
--- a/Tutorials/T04-A2Example/slides/T4.tex
+++ b/Tutorials/T04-A2Example/slides/T4.tex
@@ -117,7 +117,10 @@ McMaster University\\ }
                 \begin{itemize} \item A file with encapsulated code to implement a specific functionality
                                        \item Ex: For designing a website with a login system, we may have a module that deals with logging out
                                        \item A module comes with an "interface"
-			\item An interface includes things like functions and arguments of the function
+	          \item An interface includes things like functions and arguments of the function
+	          \item Interface connects two system/modules, like a touch screen between people and phone
+	          \item Implementation and Interface are separated, no matter how the code changes, the output should always be the same
+
                 \end{itemize}
 \end{frame}
 
@@ -193,7 +196,7 @@ McMaster University\\ }
         	     Consider the following triangle:
                \begin{figure}[h]
     \centering
-    \includegraphics[width=0.5\textwidth]{triangle.png}
+%    \includegraphics[width=0.5\textwidth]{triangle.png}
   \end{figure}
      We are using the following inequality which is called inequality equation to know the possibility of having triangle with three points A, B and C:
              
@@ -456,8 +459,8 @@ $out := ((self.sides[0] + self.slide[1]) > self.sides[2]$
     $\wedge (self.sides[1] + self.slide[2]) > self.sides[0]$
 $\wedge (self.sides[0] + self.slide[2]) > self.sides[1]$)
 \item exception: 
-\\ ex := ((p1.xcoord()==p2.xcoord() ==p3.xcoord() $\vee$
-\\ p1.ycoord()==p2.ycoord() ==p3.ycoord()) $\Rightarrow \mbox{LINE})$
+\\ $ex := ((p1.xcoord()==p2.xcoord() ==p3.xcoord()$ $\vee$
+\\ $p1.ycoord()==p2.ycoord() ==p3.ycoord())$ $\Rightarrow \mbox{LINE})$
 \end{itemize}
 
 \end{frame}
-- 
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