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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-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
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
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
example
Example
  • 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
example1
Example
  • 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
slide16
LR(0)
  • 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
continued
Continued
  • 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
    • 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
reduction
Reduction
  • 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
shift
Shift
  • 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
proof
Proof
  • 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 $
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