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  1. An identifier in a programming language consists of up to six letters and digits of which the first character must be a letter. Derive a regular expression for the identifier.

  2. Build an $LL(1)$ parsing table for the language defined by the $LL(1)$ grammar with productions

    $\text{Program} \rightarrow \text{ begin } d \text{ semi } X\text{ end}$

    $X \rightarrow d \text{ semi } X \mid sY$

    $Y \rightarrow \text{ semi } s Y \mid \epsilon$

asked in Compiler Design by Veteran (52k points)
edited by | 940 views

2 Answers

+18 votes
Best answer

a.  $(letter)(letter+digit+\epsilon)^5$

b. 

  1. Program $\rightarrow$ begin d semi $X$ end 
  2. $X \rightarrow d \ semi X$  
  3. $X \rightarrow sY$ 
  4. $Y \rightarrow semi \ sY$
  5. $Y \rightarrow \epsilon$

$$\begin{array}{|l|l|l|}\hline \textbf{Variable} & \textbf{First} & \textbf{Follow} \\\hline \text{Program} & \text{begin} & \text{\$} \\\hline \text{X} & \text{d, s} & \text{end} \\\hline \text{Y} & \text{semi, } \epsilon & \text{end}\\\hline \end{array}$$
Here, First$(Y)$ contains $\epsilon$ so we need to add   $Y \rightarrow \epsilon$ to Follow$(Y)$ 
$$\begin{array}{|l|l|l|l|l|l|l|}\hline \textbf{Variable} & \textbf{begin} & \textbf{d} & \textbf{semi} & \textbf{s} & \text{end} & \textbf{\$} \\\hline \text{Program} & \text{A} & \text{} & \text{}& \text{}& \text{}& \text{}\\\hline \text{X} & \text{} & \text{B} & \text{} & \text{C} & \text{} & \text{} \\\hline \text{Y} & \text{}& \text{} & \text{D} & \text{} & \text{Y} \rightarrow \epsilon \\\hline \end{array}$$

answered by Boss (25.3k points)
edited by
0
How did you decide "semi" is variable or terminal?
+2
"semi" cannot be a variable as we do not have any productions from it.
+3

for part a) (letter)(letter+digit+epsilon)5

0
please edit part a.
+7 votes

a.
  $(letter)(letter + digit + epsilon)^5$
b.
1.program ---> begin d semi X end      
2.    X -----> d semi x
3.              | sY
4.    Y ----->  semi sY
5.              | epsilon

                   begin       d       semi      s       end     $
program       1
     X                            2                      3
     Y                                       4                                5

answered by Veteran (59.7k points)
edited by
+1

Given is, Programs →  begin d semi X end

 while u considered program ---> begin s end 

why?

0
Given

 Programs →  begin d semi X end

then why program ---> begin s end  is considered here
+2

@Digvijay Pandey  entry '5' should be in 'y' row and 'end' column

0
How did you decide "semi" is variable or terminal?
+1
terminals are given in small letters.
0
if any symbol(semi in this case) is a non terminal, then we must have it at the LHS of a production.So semi is a terminal.

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