exercise-clustering: Improved the text regarding points that are adde…

…d to the candidate set to better reflect the algorithm

exercise/5-Clustering.tex

@@ -1347,7 +1347,7 @@ \subsection*{Task 2: Application of DBSCAN}
 
 			      \item \textbf{Add Points in the $\varepsilon$-Neighborhood of $(1,2)$ to the Candidate Set $N$:}
 
-			            Only $(1,1)$ is an unvisited point in the $\varepsilon$-neighborhood of $(1,2)$:
+			            Only $(1,1)$ is within the does $\varepsilon$-Neighborhood of $(1,2)$ and does not belong to a cluster yet:
 
 			            \begin{alignat*}{2}
 				            Distance_{(1,2)\leftrightarrow(1,1)} & = \sqrt{(1-1)^2+(2-1)^2} = 1 \\
@@ -1405,7 +1405,7 @@ \subsection*{Task 2: Application of DBSCAN}
 
 					                  \item \textbf{Add Points in the $\varepsilon$-Neighborhood of $(1,1)$ to the Candidate Set $N$:}
 
-					                        Only $(2,1)$ is an unvisited point in the $\varepsilon$-neighborhood of $(1,1)$:
+					                        Only $(2,1)$ is within the $\varepsilon$-neighborhood of $(1,1)$ and does not belong to a cluster yet:
 
 					                        \begin{alignat*}{2}
 						                        Distance_{(1,1)\leftrightarrow(2,1)} & = \sqrt{(1-2)^2+(1-1)^2} = 1 \\
@@ -1503,7 +1503,7 @@ \subsection*{Task 2: Application of DBSCAN}
 
 		\item \textbf{Select $(4,4)$ as Random Point:}
 
-		      Every unvisited point can be selected as the next point to visit. In this sample solution, we  decided to use $(4,4)$ as the next point:
+		      Every unvisited point can be selected as the next point to visit. In this sample solution, we decided to use $(4,4)$ as the next point:
 
 		      \begin{enumerate}
 			      \item \textbf{Mark $(4,4)$ as Visited:}
@@ -1539,7 +1539,7 @@ \subsection*{Task 2: Application of DBSCAN}
 
 			      \item \textbf{Add Points in the $\varepsilon$-Neighborhood of $(4,4)$ to the Candidate Set $N$:}
 
-			            Both $(3,4)$ and $(4,3)$ are unvisited points in the $\varepsilon$-neighborhood of $(4,4)$:
+			            Both $(3,4)$ and $(4,3)$ are within the $\varepsilon$-neighborhood of $(4,4)$ and do not belong to a cluster yet:
 
 			            \begin{alignat*}{2}
 				            Distance_{(4,4)\leftrightarrow(3,4)} & = \sqrt{(4-3)^2+(4-4)^2} = 1 \\