Problem on LR(0) parser

Prerequisite: LR Parser.

The LR parser is an efficient bottom-up syntax analysis technique that can be used for a large class of context-free grammar. This technique is also called LR(0) parsing.
L stands for the left to right scanning
R stands for rightmost  derivation in reverse
0 stands for no. of input symbols of lookahead.

Augmented grammar :
If G is a grammar with starting symbol S, then G’ (augmented grammar for G) is a grammar with a new starting symbol S^‘ and productions  S’-> .S . The purpose of this new starting production is to indicate to the parser when it should stop parsing. The ‘ . ‘ before S indicates the left side of ‘ . ‘  has been read by a compiler and the right side of ‘ . ‘ is yet to be read by a compiler.

Steps for constructing the LR parsing table :

1. Writing augmented grammar
2. LR(0) collection of items to be found
3. Defining 2 functions: goto(list of non-terminals) and action(list of terminals) in the parsing table.

Q. Construct an LR parsing table for the given context-free grammar –

S–>AA
      A–>aA|b

Solution :
STEP 1- Find augmented grammar –

The augmented grammar of the given grammar is:-

      S’–>.S      [0th production]
      S–>.AA    [1st production]
      A–>.aA    [2nd production]
      A–>.b      [3rd production]

STEP 2 –  Find LR(0) collection of items
Below is the figure showing the LR(0) collection of items. We will understand everything one by one.

The terminals of this grammar are {a,b}
The non-terminals of this grammar are {S,A}

RULE – if any nonterminal has ‘ . ‘ preceding it, we have to write all its production and add ‘ . ‘ preceding each of its-production.
RULE –  from each state to the next state, the ‘ . ‘ shifts to one place to the right.

• In the figure, I₀ consists of augmented grammar.
• I_o goes to I₁ when  ‘ . ‘ of 0th production is shifted towards the right of S(S’->S.). This state is the accepted state. 
  S is seen by the compiler
• I_o goes to I₂ when  ‘ . ‘ of 1st production is shifted towards the right (S->A.A) . A is seen by the compiler
• I₀ goes to I₃ when  ‘ . ‘ of the 2nd production is shifted towards the right (A->a.A) . a is seen by the compiler.
• I₀ goes to I₄ when  ‘ . ‘ of the 3rd production is shifted towards the right (A->b.) . b is seen by the compiler.
• I₂ goes to I₅ when  ‘ . ‘ of 1st production is shifted towards the right (S->AA.) . A is seen by the compiler
• I₂ goes to I₄ when  ‘ . ‘ of 3rd production is shifted towards the right (A->b.) . b is seen by the compiler.
• I₂ goes to I₃ when  ‘ . ‘ of the 2nd production is shifted towards the right (A->a.A) . a is seen by the compiler.
• I₃ goes to I₄ when  ‘ . ‘ of the 3rd production is shifted towards the right (A->b.) . b is seen by the compiler.
• I₃ goes to I₆ when  ‘ . ‘ of 2nd production is shifted towards the right (A->aA.) . A is seen by the compiler
• I₃ goes to I₃ when  ‘ . ‘ of the 2nd production is shifted towards the right (A->a.A) . a is seen by the compiler.
  

STEP3 – defining 2 functions: goto[list of non-terminals] and action[list of terminals] in the parsing table

• $ is by default a nonterminal that takes the accepting state.
• 0,1,2,3,4,5,6 denotes I₀,I₁,I₂,I₃,I₄,I₅,I₆
• I₀ gives A in I₂, so 2 is added to the A column and 0 rows.
• I₀ gives S in I1, so 1 is added to the S column and 1 row.
• similarly, 5 is written in  A column and 2nd row, 6 is written in A column and 3 rows.
• I₀ gives an in I₃ to .so S3(shift 3) is added to a column and 0 rows.
• I₀ gives b in I4, so S4(shift 4) is added to the b column and 0 rows.
• Similarly, S3(shift 3) is added on a column and 2,3 rows, S4(shift 4) is added on b column and 2,3 rows.
• I₄ is reduced state as ‘ . ‘ is at the end. I₄ is the 3rd production of grammar. So write r3(reduce 3) in terminals.
• I₅ is reduced state as ‘ . ‘ is at the end. I₅ is the 1st production of grammar. So write r1(reduce 1) in terminals.
• I₆ is reduced state as ‘ . ‘ is at the end. I₆ is the 2nd production of grammar. So write r2(reduce 2) in terminals.

As each cell has only 1 value in it, hence, the given grammar is LR(0).