$2$ LL$(1)$ Grammar

$```$

::$=$ BEG :

END EOF

::$=$ f $|$ g

::$=()$

::$=,|\backslash$epsilon

::$=$

::$=\backslash$t

$<mtext></mtext><mi>\backslash </mi>$text $\{$ morestmts$>$ ::$=$

$\backslash |\}$

::$=||$

::$==$

$```$

$```$

::$=$ if :

$\backslash$t else :

$<mtext></mtext><mi>\backslash </mi>$text $\{$ returnstmt$>$ ::$=$ return $\}$

$```$

$\backslash (<mtext></mtext><mi>\backslash </mi><mi>)</mi>$ expr $\backslash (>$::$=<mtext></mtext><mi>\backslash </mi><mi>)</mi>$ term $\backslash (>+<mtext></mtext><mi>\backslash </mi><mi>)</mi>$ term $\backslash (>\backslash )$

$\backslash (<mtext></mtext><mi>\backslash </mi><mi>)</mi>$ term $\backslash (>$::$=<mtext></mtext><mi>\backslash </mi><mi>)</mi>$ variable $\backslash (>\backslash$mid$<mtext></mtext><mi>\backslash </mi><mi>)</mi>$ digit $\backslash (>\backslash )$

$\backslash (<mtext></mtext><mi>\backslash </mi><mi>)</mi>$ variable $\backslash (>$::$=\backslash$mathbf$\{$a$\}|\backslash$mathbf$\{$b$\}|\backslash$mathbf$\{$c$\}\backslash )$

$\backslash (<mtext></mtext><mi>\backslash </mi><mi>)</mi>$ digit$>$ :: $\backslash (=\backslash$mathbf$\{0\}|\backslash$mathbf$\{1\}|\backslash$mathbf$\{2\}\backslash )$

$\backslash (\backslash$backslash $\backslash$mathbf$\{$n$\}\backslash )$ represents the "new line" terminal. $\backslash (\backslash$backslash $\backslash$mathbf$\{$t$\}\backslash )$ represents the "tab" terminal.

$($a$)$ Write an example program that can be derived from this grammar.

$($b$)$ Show that the grammar above is LL$(1).$ Use a formal argument based on the definition of the LL$(1)$ grammar.

$($c$)$ Show the LL$(1)$ parse table.

$($d$)$ Write a recursive descent parser in pseudo$-$code. You may assume that the next$\_$token$()$ function is already implemented for reading the next token in the remaining input string, and you can directly use it in your pseudocode.