Lr grammars
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LR-Grammars. LR(0), LR(1), and LR(K). Deterministic Context-Free Languages. DCFL A family of languages that are accepted by a Deterministic Pushdown Automaton (DPDA) Many programming languages can be described by means of DCFLs. Prefix and Proper Prefix. Prefix (of a string)

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Lr grammars


LR(0), LR(1), and LR(K)

Deterministic context free languages
Deterministic Context-Free Languages

  • DCFL

  • A family of languages that are accepted by a Deterministic Pushdown Automaton (DPDA)

  • Many programming languages can be described by means of DCFLs

Prefix and proper prefix
Prefix and Proper Prefix

  • Prefix (of a string)

    • Any number of leading symbols of that string

    • Example: abc

      • Prefixes: , a, ab, abc

  • Proper Prefix (of a string)

    • A prefix of a string, but not the string itself

    • Example: abc

      • Proper prefixes: , a, ab

Prefix property
Prefix Property

  • Context-Free Language (CFL) L is said to have the prefix property whenever w is in L and no proper prefix of w is in L

  • Not considered a serve restriction

    • Why?

      • Because we can easily convert a DCFL to a DCFL with the prefix property by introducing an endmarker

Suffix and proper suffix
Suffix and Proper Suffix

  • Suffix (of a string)

    • Any number of trailing symbols

  • Proper Suffix

    • A suffix of a string, but not the string itself

Example grammar
Example Grammar

  • This is the grammar that will be used in many of the examples:

    • S’  Sc

    • S  SA | A

    • A  aSb | ab

Lr grammar

  • Left-to-right scan of the input producing a rightmost derivation

  • Simply:

    • L stands for Left-to-right

    • R stands for rightmost derivation

Lr items

  • An item (for a given CFG)

    • A production with a dot anywhere in the right side (including the beginning and end)

    • In the event of an -production: B  

      • B · is an item

Example items
Example: Items

  • Given our example grammar:

    • S’  Sc, S  SA|A, A  aSb|ab

  • The items for the grammar are:

    S’·Sc, S’S·c, S’Sc·

    S·SA, SS·A, SSA·, S·A, SA·

    A·aSb, Aa·Sb, AaS·b, AaSb·, A·ab, Aa·b, Aab·

Some notation
Some Notation

  • * = 1 or more steps in a derivation

  • *rm = rightmost derivation

  • rm = single step in rightmost derivation

Right sentential form
Right-Sentential Form

  • A sentential form that can be derived by a rightmost derivation

    • A string of terminals and variables  is called a sentential form if S* 

More terms
More terms

  • Handle

    • A substring which matches the right-hand side of a production and represents 1 step in the derivation

    • Or more formally:

      • (of a right-sentential form  for CFG G)

      • Is a substring  such that:

        • S *rm w

        • w = 

    • If the grammar is unambiguous:

      • There are no useless symbols

      • The rightmost derivation (in right-sentential form) and the handle are unique


  • Given our example grammar:

    • S’  Sc, S  SA|A, A  aSb|ab

  • An example right-most derivation:

    • S’  Sc  SAc  SaSbc

  • Therefore we can say that: SaSbc is in right-sentential form

    • The handle is aSb

More terms1
More terms

  • Viable Prefix

    • (of a right-sentential form for )

    • Is any prefix of  ending no farther right than the right end of a handle of .

  • Complete item

    • An item where the dot is the rightmost symbol


  • Given our example grammar:

    • S’  Sc, S  SA|A, A  aSb|ab

  • The right-sentential form abc:

    • S’ *rm Ac  abc

  • Valid prefixes:

    • A  ab for prefix ab

    • A  ab for prefix a

    • A  ab for prefix 

  • Aab is a complete item,  Ac is the right-sentential form for abc


  • Left-to-right scan of the input producing a rightmost derivation with a look-ahead (on the input) of 0 symbols

  • It is a restricted type of CFG

  • 1st in the family of LR-grammars

  • LR(0) grammars define exactly the DCFLs having the prefix property

Computing sets of valid items
Computing Sets of Valid Items

  • The definition of LR(0) and the method of accepting L(G) for LR(0) grammar G by a DPDA depends on:

    • Knowing the set of valid items for each prefix 

  • For every CFG G, the set of viable prefixes is a regular set

    • This regular set is accepted by an NFA whose states are the items for G


  • Given an NFA (whose states are the items for G) that accepts the regular set

    • We can apply the subset construction to this NFA and yield a DFA

    • The DFA whose state is the set of valid items for 

Nfa m

  • NFA M recognizes the viable prefixes for CFG

    • M = (Q, V  T, , q0, Q)

      • Q = set of items for G plus state q0

    • G = (V, T, P, S)

  • Three Rules

    • (q0,) = {S| S is a production}

    • (AB,) = {B| B is a production}

      • Allows expansion of a variable B appearing immediately to the right of the dot

    • (AX, X) = {AX}

      • Permits moving the dot over any grammar symbol X if X is the next input symbol

  • Theorem 10 9
    Theorem 10.9

    • The NFA M has property that (q0, ) contains A iff A is valid for 

    • This theorem gives a method for computing the sets of valid items for any viable prefix

      • Note: It is an NFA. It can be converted to a DFA. Then by inspecting each state it can be determine if it is a valid LR(0) grammar

    Definition of lr 0 grammar
    Definition of LR(0) Grammar

    • G is an LR(0) grammar if

      • The start symbol does not appear on the right side of any productions

      •  prefixes  of G where A is a complete item, then it is unique

        • i.e., there are no other complete items (and there are no items with a terminal to the right of the dot) that are valid for 

    Facts we now know
    Facts we now know:

    • Every LR(0) grammar generates a DCFL

    • Every DCFL with the prefix property has a LR(0) grammar

    • Every language with LR(0) grammar have the prefix property

    • L is DCFL iff L has a LR(0) grammar

    Dpda s from lr 0 grammars
    DPDA’s from LR(0) Grammars

    • We trace out the rightmost derivation in reverse

    • The stack holds a viable prefix (in right-sentential form) and the current state (of the DFA)

      • Viable prefixes: X1X2…Xk

      • States: s1, s2,…,sk

      • Stack: s0X1s1…Xksk


    • If sk contains A

      • Then A is valid for X1X2…Xk

      •  = suffix of X1X2…Xk

    • Let

      •  = Xi+1…Xk

      • w such that X1…Xkw is a right-sentential form.

    Reduction continued
    Reduction Continued

    • There is a derivation:

      • S *rm X1…XiAw rm X1…Xkw

    • To obtain the right-sentential form (X1…Xkw) in a right derivation we reduce  to A

      • Therefore, we pop Xi+1…Xk from the stack and push A onto the stack


    • If sk contains only incomplete items

      • Then the right-sentential form (X1…Xkw) cannot be formed using a reduction

    • Instead we simply “shift” the next input symbol onto the stack

    Theorem 10 10
    Theorem 10.10

    • If L is L(G) for an LR(0) grammar G, then L is N(M) for a DPDA M

      • N(M) = the language accepted by empty stack or null stack


    • Construct from G the DFA D

      • Transition function: recognizes G’s prefixes

    • Stack Symbols of M are

      • Grammar Symbols of G

      • States of D

    • M has start state q and other states used to perform reduction

    We know that
    We know that:

    • If G is LR(0) then

      • Reductions are the only way to get the right-sentential form when the state of the DFA (on the top of the stack) contains a complete item

    • When M starts on input w it will construct a right-most derivation for w in reverse order

    What we need to prove
    What we need to prove:

    • When a shift is called for and the top DFA state on the stack has only incomplete items then there are no handles

    • (Note: if there was a handle, then some DFA state on the stack would have a complete item)

    Suppose state a complete item
    Suppose  state A (complete item)

    • Each state is put onto the top of the stack

    • It would then immediately be reduced to A

    • Therefore, a complete item cannot possibly become buried on the stack

    Proof continued
    Proof continued

    • The acceptance of G occurs when the top of the stack contains the start symbol

    • The start symbol by definition of LR(0) grammars cannot appear on the right side of a production

    • L(G) always has a prefix property if G is LR(0)

    Conclusion of proof
    Conclusion of Proof

    • Thus, if w is in L(G), M finds the rightmost derivation of w, reduces w to S, and accepts

    • If M accepts w, then the sequence of right-sentential forms provides a derivation of w from S

    • N(M) = L(G)

    Corollary of theorem 10 10
    Corollary of Theorem 10.10

    • Every LR(0) grammar is unambiguous

    • Why?

      • The rightmost derivation of w is unique

        • (Given the construction we provided)

    Lr 1 grammars
    LR(1) Grammars

    • LR grammar with 1 look-ahead

    • All and only deterministic CFL’s have LR(1) grammars

    • Are greatly important to compiler design

      • Why?

        • Because they are broad enough to include the syntax of almost all programming languages

        • Restrictive enough to have efficient parsers (that are essentially DPDAs)

    Lr 1 item
    LR(1) Item

    • Consists of an LR(0) item followed by a look-ahead set consisting of terminals and/or the special symbol $

      • $ = the right end of the string

    • General Form:

      • A  , {a1, a2, …, an}

    • The set of LR(1) items forms the states of a viable prefix by converting the NFA to a DFA

    A grammar is lr 1 if
    A grammar is LR(1) if

    • The start symbol does not appear on the right side of any productions

    • The set of items, I, valid for some viable prefix includes some complete item A, {a1,…,an} then

      • No ai appears immediately to the right of the dot in any item of I

      • If B, {b1,…,bk} is another complete item in I, then ai  bj for any 1  i  n and 1  j  k

    Accepting lr 1 language
    Accepting LR(1) language:

    • Similar to the DPDA used with LR(0) grammars

    • However, it is allowed to use the next input symbol during it’s decision making

    • This is accomplished by appending a $ to the end of the input and the DPDA keeps the next input symbol as part of the state

    Lr 1 rules for reduce shift
    LR(1) Rules for Reduce/Shift

    • If the top set of items has a complete item A, {a1, a2, …, an}, where A  S, reduce by A if the current input symbol is in {a1, a2, …, an}

    • If the top set of items has an item S, {$}, then reduce by S and accept if the current symbol is $ (i.e., the end of the input is reached)

    • If the top set of items has an item AaB, T, and a is the current input symbol, then shift

    Regarding the rules
    Regarding the Rules

    • Guarantees that at most one of the rules will be applied for any input symbol or $

    • Often for practicality the information is summarized into a table

      • Rows: sets of items

      • Columns: terminals and $