Bottom-up parser SLR PARSER Parsing techniques

1. SLR PARSER
Steps:
1. create augment grammar
2. generate kernel items
3. find closure
4. compute goto()
5. construct parsing table
6. parse the string
Let us consider grammar:
S->a
S->(L)
L->S
L->L,S

2. • Step :1 Create augment grammar
S is start symbol in the given grammar
Augment grammar is S’-> S
• Step :2 Generate kernel items
Introduce dot in RHS of the production``
S’-> .S
• Step :3 Find closure
(Rule: A -> α.Xβ i.e if there is nonterminal next to dot then Include X production))
S’-> .S
S-> . a
S->.(L)
I0

3. Step :4 compute goto()
S’-> S .
I1
Goto (I0,a) S-> a .
I2
Goto (I0,( )
S’-> .S
S-> . a
S->.(L) I0
S-> (.L)
L->.S
L>.L,S
S->.a
S->.(L)
I3
Goto (I3,L )
S-> (L .)
L->L . ,S
I4
Goto (I3,S ) L->S . I5
Goto (I3,a ) S->a . I2
Goto (I3, ( )
S-> (. L )
L->.S
L>.L,S
S->.a
S->.(L)
I3
Goto (I0,S)

4. Goto (I4, ) ) S-> (L) . I6
Goto (I4, , ) L->L , . S
S-> . a
S->.(L)
I7
Goto (I7,S ) L->L , S . I8
Goto (I7,a ) S-> a . I2
Goto (I7,( )
S-> (.L)
L->.S
L>.L,S
S->.a
S->.(L)
I3
label Production Ending
iteration
R0 S’-> S . I1
R1 S->a . I2
R2 S->(L) . I6
R3 L->S . I5
R4 L->L,S . I8

5. Step :5 construct parsing table
Label the rows of table with no. of iterations
Divide the Column into two parts
• Action part: terminal symbols
• Goto part: Nonterminal symbols
Terminals Non terminals
a ( ) , $ S L
0
1
2
3
4
5
6
7
8

6. Step :5 construct parsing table from step 4:
Label the rows of table with no. of iterations
Divide the Column into two parts
• Action part: terminal symbols
• Goto part: Nonterminal symbols
ACTION GOTO
a ( ) , $ S L
0 S2 S3 1
1
2
3 S2 S3 5 4
4 S6 S7
5
6
7 S2 S3 8
8
Goto (I0, S) 1 Goto (I3, ( ) 3
Goto (I0, a) 2 Goto (I4, ) ) 6
Goto (I0, ( ) 3
Goto (I4, ,)
7
Goto (I3, L) 4 Goto (I7, S) 8
Goto (I3, S) 5 Goto (I7, a) 2
Goto (I3, a) 2 Goto (I7, ( ) 3

7. Step :5 construct parsing table
Label the rows of table with no. of iterations
Divide the Column into two parts
• Action part: terminal symbols
• Goto part: Nonterminal symbols
ACTION GOTO
a ( ) , $ S L
0 S2 S3 1
1 accept
2 R1 R1 R1
3 S2 S3 5 4
4 S6 S7
5 R3 R3
6 R2 R2 R2
7 S2 S3 8
8 R4 R4
label Production
Ending
iteration
Follow(LHS)
R0 S’-> S . I1 Follow(S’)={ $ }
R1 S-> a . I2
Follow(S)={ $ ) , }
R2 S-> ( L ) . I6
Follow(S)={ $ ) , }
R3 L-> S . I5
Follow(L)={ ) , }
R4 L-> L , S . I8
Follow(L)={ ) , }

8. Step :5 construct parsing table
Label the rows of table with no. of iterations
Divide the Column into two parts
• Action part: terminal symbols
• Goto part: Nonterminal symbols
ACTION GOTO
a ( ) , $ S L
0 S2 S3 1
1 accept
2 R1 R1 R1
3 S2 S3 5 4
4 S6 S7
5 R3 R3
6 R2 R2 R2
7 S2 S3 8
8 R4 R4
label Production
Ending
iteration
Follow(LHS)
R0 S’-> S . I1 Follow(S’)={ $ }
R1 S-> a . I2
Follow(S)={ $ ) , }
R2 S-> ( L ) . I6
Follow(S)={ $ ) , }
R3 L-> S . I5
Follow(L)={ ) , }
R4 L-> L , S . I8
Follow(L)={ ) , }
Goto (I0, S) 1 Goto (I3, ( ) 3
Goto (I0, a) 2 Goto (I4, ) ) 6
Goto (I0, ( ) 3
Goto (I4, ,)
7
Goto (I3, L) 4 Goto (I7, S) 8
Goto (I3, S) 5 Goto (I7, a) 2
Goto (I3, a) 2 Goto (I7, ( ) 3

9. Step :6 String parsing
Let string w=(a,a) derived from the given grammar
S – Shift action
* shift input symbol to the stack
* shift number to the stack
R – Reduce action
* pop 2 times of RHS symbols
* shift LHS to the stack
* find the number in the parsing table for
the last two elements in the stack
ACTION GOTO
a ( ) , $ S L
0 S2 S3 1
1 accept
2 R1 R1 R1
3 S2 S3 5 4
4 S6 S7
5 R3 R3
6 R2 R2 R2
7 S2 S3 8
8 R4 R4
Stack Input Action
$0 (a,a)$ S3
label Production
R0 S’-> S .
R1 S-> a .
R2 S-> ( L ) .
R3 L-> S .
R4 L-> L , S .

10. Step :6 String parsing
Let string w=(a,a) derived from the given grammar
S – Shift action
* shift input symbol to the stack
* shift number to the stack
R – Reduce action
* pop 2 times of RHS symbols
* shift LHS to the stack
* find the number in the parsing table for
the last two elements in the stack
ACTION GOTO
a ( ) , $ S L
0 S2 S3 1
1 accept
2 R1 R1 R1
3 S2 S3 5 4
4 S6 S7
5 R3 R3
6 R2 R2 R2
7 S2 S3 8
8 R4 R4
Stack Input Action
$0 (a,a)$ S3
$0(3 a,a)$ S2
label Production
R0 S’-> S .
R1 S-> a .
R2 S-> ( L ) .
R3 L-> S .
R4 L-> L , S .

11. Step :6 String parsing
Let string w=(a,a) derived from the given grammar
S – Shift action
* shift input symbol to the stack
* shift number to the stack
R – Reduce action
* pop 2 times of RHS symbols
* shift LHS to the stack
* find the number in the parsing table for
the last two elements in the stack
ACTION GOTO
a ( ) , $ S L
0 S2 S3 1
1 accept
2 R1 R1 R1
3 S2 S3 5 4
4 S6 S7
5 R3 R3
6 R2 R2 R2
7 S2 S3 8
8 R4 R4
Stack Input Action
$0 (a,a)$ S3
$0(3 a,a)$ S2
$0(3a2 ,a)$ R1
label Production
R0 S’-> S .
R1 S-> a .
R2 S-> ( L ) .
R3 L-> S .
R4 L-> L , S .

12. Step :6 String parsing
Let string w=(a,a) derived from the given grammar
S – Shift action
* shift input symbol to the stack
* shift number to the stack
R – Reduce action
* pop 2 times of RHS symbols
* shift LHS to the stack
* find the number in the parsing table for
the last two elements in the stack
ACTION GOTO
a ( ) , $ S L
0 S2 S3 1
1 accept
2 R1 R1 R1
3 S2 S3 5 4
4 S6 S7
5 R3 R3
6 R2 R2 R2
7 S2 S3 8
8 R4 R4
Stack Input Action
$0 (a,a)$ S3
$0(3 a,a)$ S2
$0(3a2 ,a)$ R1
$0(3S5 ,a)$ R3
label Production
R0 S’-> S .
R1 S-> a .
R2 S-> ( L ) .
R3 L-> S .
R4 L-> L , S .

13. Step :6 String parsing
Let string w=(a,a) derived from the given grammar
S – Shift action
* shift input symbol to the stack
* shift number to the stack
R – Reduce action
* pop 2 times of RHS symbols
* shift LHS to the stack
* find the number in the parsing table for
the last two elements in the stack
ACTION GOTO
a ( ) , $ S L
0 S2 S3 1
1 accept
2 R1 R1 R1
3 S2 S3 5 4
4 S6 S7
5 R3 R3
6 R2 R2 R2
7 S2 S3 8
8 R4 R4
Stack Input Action
$0 (a,a)$ S3
$0(3 a,a)$ S2
$0(3a2 ,a)$ R1
$0(3S5 ,a)$ R3
$0(3L4 ,a)$ S7
label Production
R0 S’-> S .
R1 S-> a .
R2 S-> ( L ) .
R3 L-> S .
R4 L-> L , S .

14. Step :6 String parsing
Let string w=(a,a) derived from the given grammar
S – Shift action
* shift input symbol to the stack
* shift number to the stack
R – Reduce action
* pop 2 times of RHS symbols
* shift LHS to the stack
* find the number in the parsing table for
the last two elements in the stack
ACTION GOTO
a ( ) , $ S L
0 S2 S3 1
1 accept
2 R1 R1 R1
3 S2 S3 5 4
4 S6 S7
5 R3 R3
6 R2 R2 R2
7 S2 S3 8
8 R4 R4
Stack Input Action
$0 (a,a)$ S3
$0(3 a,a)$ S2
$0(3a2 ,a)$ R1
$0(3S5 ,a)$ R3
$0(3L4 ,a)$ S7
label Production
R0 S’-> S .
R1 S-> a .
R2 S-> ( L ) .
R3 L-> S .
R4 L-> L , S .

15. Step :6 String parsing
Let string w=(a,a) derived from the given grammar
S – Shift action
* shift input symbol to the stack
* shift number to the stack
R – Reduce action
* pop 2 times of RHS symbols
* shift LHS to the stack
* find the number in the parsing table for
the last two elements in the stack
ACTION GOTO
a ( ) , $ S L
0 S2 S3 1
1 accept
2 R1 R1 R1
3 S2 S3 5 4
4 S6 S7
5 R3 R3
6 R2 R2 R2
7 S2 S3 8
8 R4 R4
Stack Input Action
$0 (a,a)$ S3
$0(3 a,a)$ S2
$0(3a2 ,a)$ R1
$0(3S5 ,a)$ R3
$0(3L4 ,a)$ S7
$0(3L4 , 7 a)$ S2
label Production
R0 S’-> S .
R1 S-> a .
R2 S-> ( L ) .
R3 L-> S .
R4 L-> L , S .

16. Step :6 String parsing
Let string w=(a,a) derived from the given grammar
S – Shift action
* shift input symbol to the stack
* shift number to the stack
R – Reduce action
* pop 2 times of RHS symbols
* shift LHS to the stack
* find the number in the parsing table for
the last two elements in the stack
ACTION GOTO
a ( ) , $ S L
0 S2 S3 1
1 accept
2 R1 R1 R1
3 S2 S3 5 4
4 S6 S7
5 R3 R3
6 R2 R2 R2
7 S2 S3 8
8 R4 R4
Stack Input Action
$0 (a,a)$ S3
$0(3 a,a)$ S2
$0(3a2 ,a)$ R1
$0(3S5 ,a)$ R3
$0(3L4 ,a)$ S7
$0(3L4 , 7 a)$ S2
$0(3L4 , 7a2 )$ R1
label Production
R0 S’-> S .
R1 S-> a .
R2 S-> ( L ) .
R3 L-> S .
R4 L-> L , S .

17. Step :6 String parsing
Let string w=(a,a) derived from the given grammar
S – Shift action
* shift input symbol to the stack
* shift number to the stack
R – Reduce action
* pop 2 times of RHS symbols
* shift LHS to the stack
* find the number in the parsing table for
the last two elements in the stack
ACTION GOTO
a ( ) , $ S L
0 S2 S3 1
1 accept
2 R1 R1 R1
3 S2 S3 5 4
4 S6 S7
5 R3 R3
6 R2 R2 R2
7 S2 S3 8
8 R4 R4
Stack Input Action
$0 (a,a)$ S3
$0(3 a,a)$ S2
$0(3a2 ,a)$ R1
$0(3S5 ,a)$ R3
$0(3L4 ,a)$ S7
$0(3L4 , 7 a)$ S2
$0(3L4 , 7a2 )$ R1
$0(3L4 , 7S8 )$ R4
label Production
R0 S’-> S .
R1 S-> a .
R2 S-> ( L ) .
R3 L-> S .
R4 L-> L , S .

18. Step :6 String parsing
Let string w=(a,a) derived from the given grammar
S – Shift action
* shift input symbol to the stack
* shift number to the stack
R – Reduce action
* pop 2 times of RHS symbols
* shift LHS to the stack
* find the number in the parsing table for
the last two elements in the stack
ACTION GOTO
a ( ) , $ S L
0 S2 S3 1
1 accept
2 R1 R1 R1
3 S2 S3 5 4
4 S6 S7
5 R3 R3
6 R2 R2 R2
7 S2 S3 8
8 R4 R4
Stack Input Action
$0 (a,a)$ S3
$0(3 a,a)$ S2
$0(3a2 ,a)$ R1
$0(3S5 ,a)$ R3
$0(3L4 ,a)$ S7
$0(3L4 , 7 a)$ S2
$0(3L4 , 7a2 )$ R1
$0(3L4 , 7S8 )$ R4
$0(3L4 ) $ S6
label Production
R0 S’-> S .
R1 S-> a .
R2 S-> ( L ) .
R3 L-> S .
R4 L-> L , S .

19. Step :6 String parsing
Let string w=(a,a) derived from the given grammar
S – Shift action
* shift input symbol to the stack
* shift number to the stack
R – Reduce action
* pop 2 times of RHS symbols
* shift LHS to the stack
* find the number in the parsing table for
the last two elements in the stack
ACTION GOTO
a ( ) , $ S L
0 S2 S3 1
1 accept
2 R1 R1 R1
3 S2 S3 5 4
4 S6 S7
5 R3 R3
6 R2 R2 R2
7 S2 S3 8
8 R4 R4
Stack Input Action
$0 (a,a)$ S3
$0(3 a,a)$ S2
$0(3a2 ,a)$ R1
$0(3S5 ,a)$ R3
$0(3L4 ,a)$ S7
$0(3L4 , 7 a)$ S2
$0(3L4 , 7a2 )$ R1
$0(3L4 , 7S8 )$ R4
$0(3L4 ) $ S6
$0(3L4)6 $ R2
label Production
R0 S’-> S .
R1 S-> a .
R2 S-> ( L ) .
R3 L-> S .
R4 L-> L , S .

20. Step :6 String parsing
Let string w=(a,a) derived from the given grammar
S – Shift action
* shift input symbol to the stack
* shift number to the stack
R – Reduce action
* pop 2 times of RHS symbols
* shift LHS to the stack
* find the number in the parsing table for
the last two elements in the stack
ACTION GOTO
a ( ) , $ S L
0 S2 S3 1
1 accept
2 R1 R1 R1
3 S2 S3 5 4
4 S6 S7
5 R3 R3
6 R2 R2 R2
7 S2 S3 8
8 R4 R4
Stack Input Action
$0 (a,a)$ S3
$0(3 a,a)$ S2
$0(3a2 ,a)$ R1
$0(3S5 ,a)$ R3
$0(3L4 ,a)$ S7
$0(3L4 , 7 a)$ S2
$0(3L4 , 7a2 )$ R1
$0(3L4 , 7S8 )$ R4
$0(3L4 ) $ S6
$0(3L4)6 $ R2
$0S1 $ accept
Hence the string parsed successfully !!!!!!
label Production
R0 S’-> S .
R1 S-> a .
R2 S-> ( L ) .
R3 L-> S .
R4 L-> L , S .