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)
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 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 (of a string)
Any number of trailing symbols Proper Suffix
A suffix of a string, but not the string itself
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 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·
Some Notation * = 1 or more steps in a derivation
*rm = rightmost derivation
rm = single step in rightmost derivation
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 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 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 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
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 ab for prefix a A ab for prefix
Aab is a complete item, Ac is the right-sentential form for abc
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 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 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} (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
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 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: 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 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
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 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
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 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 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: 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: 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)
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 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 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 Every LR(0) grammar is unambiguous Why?
The rightmost derivation of w is unique (Given the construction we provided)
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 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 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: 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 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