Ex review: fixes from discussion session

info/exercises/src/ex-review/ex/ll1.tex

@@ -26,17 +26,30 @@
     \end{solution}
     \item Explain why the grammar is not LL(1).
     \begin{solution}
-      \(IDs\) has two rules with the same \(\first\) set.
+      \(decl\), \(IDs\), and \(type\) each have two rules with the same
+      \(\first\) set. 
     \end{solution}
     \item Give an LL(1) grammar describing the same sequences of tokens as the
     previous grammar.
     \begin{solution}
+      We can remove the common prefix from \(decl\):
+      \begin{gather*}
+        decl = type~\lstinline|ID|~decl' \\
+        decl' = \lstinline|;| \mid \lstinline|(| ~optIDs~\lstinline|);|
+      \end{gather*}
+
       We can remove the left recursion from the \(IDs\) rule:
       \begin{gather*}
         IDs = \lstinline|ID| ~IDs' \\
         IDs' = \epsilon \mid \lstinline|ID| ~IDs'
       \end{gather*}
 
+      We can also remove the left recursion from the \(type\) rule:
+      \begin{gather*}
+        type = \lstinline|int| ~type' \\
+        type' = \epsilon \mid \lstinline|*| ~type'
+      \end{gather*}
+
       Note that \(IDs\) can be followed only by a closing parenthesis.
     \end{solution}
   \end{enumerate}

info/exercises/src/ex-review/ex/regex.tex

@@ -5,7 +5,7 @@
   concatenation is given by:
   %%
   \begin{gather*}
-    L_1L_1 = \{u_1u_2 \mid u_1 \in L_1 \land u_2 \in L_2\}
+    L_1L_2 = \{u_1u_2 \mid u_1 \in L_1 \land u_2 \in L_2\}
   \end{gather*}
   %%
   We that a language \(L\) left-cancels if and only if for every \(L_1, L_2\):
@@ -15,7 +15,7 @@
   \end{gather*}
 
   \begin{enumerate}
-    \item Does \(L = \empty\) left-cancel? \ifAns{No}
+    \item Does \(L = \emptyset\) left-cancel? \ifAns{No}
     \item Does \(L = \epsilon\) left-cancel? \ifAns{Yes}
     \item Give a regular expression describing an infinite language \(L\) that left-cancels. \ifAns{\(a^*b\)}
     \item Give a context-free grammar for another language \(L\) that