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

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**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

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**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 Proper prefixes: , a, ab

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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

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**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

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Example Grammar This is the grammar that will be used in many of the examples: S’ Sc S SA | A A aSb | ab

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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

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**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

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**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, SS·A, SSA·, S·A, SA· A·aSb, Aa·Sb, AaS·b, AaSb·, A·ab, Aa·b, Aab·

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**Some Notation * = 1 or more steps in a derivation**

*rm = rightmost derivation rm = single step in rightmost derivation

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**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*

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**More terms Handle If the grammar is unambiguous:**

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

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**Example Given our example grammar: An example right-most derivation:**

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

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**More terms Viable Prefix Complete item**

(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

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**Example Given our example grammar: The right-sentential form abc:**

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 ab for prefix a A ab for prefix Aab is a complete item, Ac is the right-sentential form for abc

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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

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**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

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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

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**NFA M Three Rules 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} (AB,) = {B| B is a production} Allows expansion of a variable B appearing immediately to the right of the dot (AX, X) = {AX} Permits moving the dot over any grammar symbol X if X is the next input symbol

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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

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**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

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**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

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**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

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**Reduction If sk contains A Let 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.

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**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

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**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

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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

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**Proof Construct from G the DFA D Stack Symbols of M are**

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

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**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

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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)

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**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

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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)

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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)

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**Corollary of Theorem 10.10 Every LR(0) grammar is unambiguous Why?**

The rightmost derivation of w is unique (Given the construction we provided)

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**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)

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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

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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

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**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

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**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 AaB, T, and a is the current input symbol, then shift

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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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