LR-Grammars
LR(0), LR(1), and LR(K)
Deterministic Context-Free
Languages
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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
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Prefix (of a string)
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Any number of leading symbols of that
string
Example: abc
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Prefixes: , a, ab, abc
Proper Prefix (of a string)
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A prefix of a string, but not the string itself
Example: abc
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Proper prefixes: , a, ab
Prefix Property
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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
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Why?
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Because we can easily convert a DCFL to a
DCFL with the prefix property by introducing an
endmarker
Suffix and Proper Suffix
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Suffix (of a string)
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Any number of trailing symbols
Proper Suffix
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A suffix of a string, but not the string itself
Example Grammar
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This is the grammar that will be used in
many of the examples:
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S’  Sc
S  SA | A
A  aSb | ab
LR-Grammar
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Left-to-right scan of the input producing
a rightmost derivation
Simply:
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L stands for Left-to-right
R stands for rightmost derivation
LR-Items
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An item (for a given CFG)
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A production with a dot anywhere in the
right side (including the beginning and end)
In the event of an -production: B  
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B  · is an item
Example: Items
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Given our example grammar:
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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
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* = 1 or more steps in a derivation
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*rm = rightmost derivation
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rm = single step in rightmost
derivation
Right-Sentential Form
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A sentential form that can be derived by
a rightmost derivation
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A string of terminals and variables  is
called a sentential form if S* 
More terms
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Handle
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A substring which matches the right-hand side of a
production and represents 1 step in the derivation
Or more formally:
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(of a right-sentential form  for CFG G)
Is a substring  such that:
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S *rm w
w = 
If the grammar is unambiguous:
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There are no useless symbols
The rightmost derivation (in right-sentential
form) and the handle are unique
Example
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Given our example grammar:
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An example right-most derivation:
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S’  Sc, S  SA|A, A  aSb|ab
S’  Sc  SAc  SaSbc
Therefore we can say that: SaSbc is in
right-sentential form
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The handle is aSb
More terms
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Viable Prefix
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(of a right-sentential form for )
Is any prefix of  ending no farther right
than the right end of a handle of .
Complete item
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An item where the dot is the rightmost
symbol
Example
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Given our example grammar:
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The right-sentential form abc:
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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
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S’  Sc, S  SA|A, A  aSb|ab
LR(0)
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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
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The definition of LR(0) and the method
of accepting L(G) for LR(0) grammar G
by a DPDA depends on:
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Knowing the set of valid items for each
prefix 
For every CFG G, the set of viable
prefixes is a regular set
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This regular set is accepted by an NFA
whose states are the items for G
Continued
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Given an NFA (whose states are the
items for G) that accepts the regular set
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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
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NFA M recognizes the viable prefixes for CFG
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M = (Q, V  T, , q0, Q)
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Q = set of items for G plus state q0
G = (V, T, P, S)
Three Rules
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(q0,) = {S| S is a production}
(AB,) = {B| B is a production}
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Allows expansion of a variable B appearing
immediately to the right of the dot
(AX, X) = {AX}
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Permits moving the dot over any grammar symbol X
if X is the next input symbol
Theorem 10.9
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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
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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
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G is an LR(0) grammar if
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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
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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:
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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
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We trace out the rightmost derivation in
reverse
The stack holds a viable prefix (in rightsentential form) and the current state (of
the DFA)
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Viable prefixes: X1X2…Xk
States: s1, s2,…,sk
Stack: s0X1s1…Xksk
Reduction
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If sk contains A
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Then A is valid for X1X2…Xk
 = suffix of X1X2…Xk
Let
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 = Xi+1…Xk
w such that X1…Xkw is a right-sentential
form.
Reduction Continued
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There is a derivation:
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S *rm X1…XiAw rm X1…Xkw
To obtain the right-sentential form
(X1…Xkw) in a right derivation we
reduce  to A
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Therefore, we pop Xi+1…Xk from the stack
and push A onto the stack
Shift
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If sk contains only incomplete items
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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
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If L is L(G) for an LR(0) grammar G,
then L is N(M) for a DPDA M
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N(M) = the language accepted by empty
stack or null stack
Proof
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Construct from G the DFA D
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Stack Symbols of M are
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Transition function: recognizes G’s prefixes
Grammar Symbols of G
States of D
M has start state q and other states
used to perform reduction
We know that:
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If G is LR(0) then
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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:
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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)
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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
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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
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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
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Every LR(0) grammar is unambiguous
Why?
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The rightmost derivation of w is unique
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(Given the construction we provided)
LR(1) Grammars
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LR grammar with 1 look-ahead
All and only deterministic CFL’s have
LR(1) grammars
Are greatly important to compiler design
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Why?
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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
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Consists of an LR(0) item followed by a
look-ahead set consisting of terminals
and/or the special symbol $
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General Form:
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$ = the right end of the string
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
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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
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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:
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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
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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
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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
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Rows: sets of items
Columns: terminals and $