1. Bhaskar Bagchi (11CS10058) Lecture Slides( 9 th Sept. 2013)

2. In LR Parser L stands for Left to right and R stands for Rightmost derivation LR Parsers are also known as Shift-Reduce Parser Whenever we have a transition in the input symbol we either perform a SHIFT or perform a REDUCE. For the shift and reduce operations we need an FSM, to construct which we need 1. State 2. Transition 3. A small augmentation in grammar

3. There are various types of LR parsers, namely 1. SLR (Simple LR) Parser 2. CLR (Canonical LR) Parser 3. LALR (look-Ahead LR) Parser For all these parsers we need State transition machine The construction of STM will be further discussed

4. Augment the grammar G to G’ 1. Start symbol S 2. Introduce new non-terminal S’  S Find CLOSURE() Find GOTO()

5. Let there be a production A  XYZ We can construct 4 items from this production, as listed below 1. A .XYZ 2. A  X.YZ 3. A  XY.Z 4. A  XYZ.

6. Each of the item tell us to how much the input string has been recognized. A  X.YZ implies that the prefix of the input parsed has been reduced to non-terminal X and the remaining string may be expected to match with YZ. Similarly A  XY.Z says that XY has been recognized and Z is expected. Also, A .XYZ means that no prefix has been matched yet but it may match to XYZ.

7. In LR parser to know whether to shift or to reduce we need a parsing table, and in order to construct this parsing table we first need a State Transition Machine, the LR(0) automaton. Each state of this automaton is a set of items As mentioned earlier, to construct this automaton we need two functions: 1. CLOSURE() 2. GOTO()

8. Helps us find state of the automaton. Algorithm to find CLOSURE 1. Put all items of I 0 in CLOSURE I 0 2. If there is a production of the form A  B.CD in I 0 and there is a production of the form C  E then add the production C .E to the CLOSURE I 0 Let we have an example grammar as follows  E  E + T | T  T  T * F | F  F  id

9. First augment the grammar by introducing a new non terminal as E’  E To find the state S 0 add the item E’ .E to the state. Now for all productions in the augmented grammar G’ with E as its head, say, E  E + T | T, add the corresponding items E .E + T and E .T to the state S 0 Again we get an item with T appearing to the right of the dot, add the corresponding items. Proceed similarly…

10. GOTO(I i, X)  I i, where I i is the current state(set of items), X is the terminal/non-terminal symbol and I j is the next state. If there is a production of the form A  A.XB in I i and a production of the form A  AX.B in I j then if the next input matches X in the state I i then the GOTO() function returns I j Let the input string is of the form x 1 x 2 …x m x m=1 …x n-1 x n then x 1 x 2 …x m has already been matched to B and the remaining x m=1 …x n-1 x n can be matched to X.

11. Let I 0 be a start state for the above mentioned example grammar, then the GOTO() functions will be as follows: GOTO(I 0, E)  I 1 Item E’  E is available so we add the item E’  E. in I 1 and take the CLOSURE of the items. So we get state I 1 as E’  E. and E’  E. + T GOTO(I 0, T)  I2 For the item E .T and T .T* F we add T  T. * F to I 2 and find its CLOSURE. So, we get I 2 as E  T. and T  T. * F

12. Similarly we get  GOTO(I 0, F)  I 3 : T  F.  GOTO(I 0, id)  I 4 : F  id. Each of the above set of items I 1, I 2, I 3, I 4 corresponding to new states of the LR(0) automaton. For each of these states we use GOTO() function for the terminals and non-terminals appearing in the items in the states and right shift the dot, i.e. match one symbol in at least one of the items present in the state.