Bhaskar Bagchi (11CS10058)

Lecture Slides( 9th Sept. 2013)

In LR Parser L stands for Left to right and R stands for Rightmost derivation

LR Parsers are also known as Shift-Reduce ParserWhenever 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 need1.State2.Transition3.A small augmentation in grammar

There are various types of LR parsers, namely1. SLR (Simple LR) Parser2. CLR (Canonical LR) Parser3. LALR (look-Ahead LR) Parser

For all these parsers we need State transition machineThe construction of STM will be further discussed

Augment the grammar G to G’1. Start symbol S2. Introduce new non-terminal S’ S

Find CLOSURE()Find GOTO()

Let there be a production A XYZWe can construct 4 items from this

production, as listed below1. A .XYZ2. A X.YZ3. A XY.Z4. A XYZ.

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.

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 itemsAs mentioned earlier, to construct this

automaton we need two functions:1. CLOSURE()2. GOTO()

Helps us find state of the automaton.Algorithm to find CLOSURE

1. Put all items of I0 in CLOSURE I0

2. If there is a production of the form A B.CD in I0 and there is a production of the form C E then add the production C .E to the CLOSURE I0

Let we have an example grammar as follows E E + T | T T T * F | F F id

First augment the grammar by introducing a new non terminal asE’ E

To find the state S0 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 S0

Again we get an item with T appearing to the right of the dot, add the corresponding items.

Proceed similarly…

GOTO(Ii , X) Ii , where Ii is the current state(set of items), X is the terminal/non-terminal symbol and Ij is the next state.

If there is a production of the form A A.XB in Ii and a production of the form A AX.B in Ij then if the next input matches X in the state Ii then the GOTO() function returns Ij

Let the input string is of the form x1 x2 …xm xm=1 …xn-1 xn then x1 x2 …xm has already been matched to B and the remaining xm=1 …xn-1 xn can be matched to X.

Let I0 be a start state for the above mentioned example grammar, then the GOTO() functions will be as follows: GOTO(I0 , E) I1

Item E’ E is available so we add the item E’ E. in I1 and take the CLOSURE of the items. So we get state I1 as E’E. and E’E. + T

GOTO(I0 , T) I2

For the item E .T and T .T* F we add T T. * F to I2 and find its CLOSURE. So, we get I2 as E T. and TT. * F

Similarly we getGOTO(I0 , F) I3 : T F.GOTO(I0 , id) I4 : F id.

Each of the above set of items I1 , I2 , I3 , I4 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.