1. LR Parsing Table Costruction
Lecture 6
Syntax Analysis

2. LR parsing example Grammar: E -> E + T E -> T T -> T * F
F -> ( E )
F -> id

3. LR parsing example CONFIGURATIONS STACK INPUT ACTION
id * id + id $ shift 5

4. Fig. 4.32. Moves of LR parser on id * id + id.
STACK
INPUT
ACTION
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
0 id 5
0 F 5
0 T 2
0 T 2 * 7
0 T 2 * 7 id 5
0 T 2 * 7 F 10
0 E 1
0 E 1 + 6
0 E id 5
0 E F 3
0 E T 9
id * id + id$
* id + id$
id + id$
+ id$
id$ $
shift
reduced by F  id
reduced by T  F
reduced by T  T*F
reduced by E  T
E  E + T
accept
Fig Moves of LR parser on id * id + id.

5. LR grammars
If it is possible to construct an LR parse table for G, we say “G is an LR grammar”.
LR parsers DO NOT need to parse the entire stack to decide what to do (other shift-reduce parsers might).
Instead, the STATE symbol summarizes all the information needed to make the decision of what to do next.
The GOTO function corresponds to a DFA that knows how to find the HANDLE by reading the top of the stack downwards.
In the example, we only looked at 1 input symbol at a time. This means the grammar is LR(1).

6. How to construct an LR parse table?
We will look at 3 methods:
Simple LR (SLR): simple but not very powerful
Canonical LR: very powerful but too many states
LALR: almost as powerful with many fewer states
yacc uses the LALR algorithm.

7. SLR (Simple LR) Parse Table Construction

8. SLR parse tables
The SLR parse table is easy to construct, but the resulting parser is a little weak.
The table is based on LR(0) ITEMS, or just plain ITEMS.
A LR(0) item is a production G with a dot at some position on the RHS.
The production A -> XYZ could generate the following LR(0) items:
A -> .XYZ
A -> X.YZ
A -> XY.Z
A -> XYZ.
The production A -> ε only generates 1 LR(0) item:
A -> .

9. LR(0) items An item indicates how far we are in parsing the RHS.
A -> .XYZ means we think we’re at the beginning of an A production, but haven’t seen an X yet.
A -> X.YZ means we think we’re in the middle of an A production, have seen an X, and should see a Y soon.

10. Augmenting the grammar G
Before we can produce an SLR parse table, we have to AUGMENT the input grammar, G.
Given G, we produce G’, the AUGMENTED GRAMMAR for G:
Add a new symbol S’
Add a new production S’ -> S (where S is the old start symbol)
Make S’ the new start symbol

11. Item set closure
We need a new concept: the CLOSURE of a set of LR(0) items.
If I is a set of items for grammar G’, then the CLOSURE of I is defined recursively:
Initially, every item in I is added to closure(I)
If A -> α . B β is in closure(I) and B -> γ is a production, then add the item B -> . γ to I, if not already there.

12. Itemset closure example
E’ -> E Closure(I) = { E’ -> . E
E -> E + T | T E -> . E + T
T -> T * F | F E -> . T
F -> ( E ) | id T -> . T * F
T -> . F
Initial itemset I is { E’ -> .E } F -> . ( E )
F -> . id }

13. The goto table
We also need the function goto(I,X) that takes an itemset I and a grammar symbol X, and returns the closure of the set of all items [ A -> α X . β ] such that [ A -> α . X β ] is in I.
Example: I = { [E’ -> E.], [E -> E. + T] } goto(I,+) =

14. Fig. 4.35. Canonical LR(0) collection for grammar (4.19)
E'  · E
E  · E + T
E  · T
T  · T * F
T  · F
F  · (E)
F  · id
I5:
F  id ·
I6:
E  E + · T
I7:
T  T * · F
I1:
E'  E ·
E  E · + T
I2:
E  T ·
T  T · * F
I8:
F  ( E · )
I9:
E  E + T ·
I3:
T  F ·
I4:
F  (· E )
I10:
T  T * F ·
I11:
F  ( E ) ·
Fig Canonical LR(0) collection for grammar (4.19)

15. Fig. 4.36. Transition diagram of DFA D form viable prefixes.

16. Canonical LR(0) itemsets
The CANONICAL LR(0) ITEMSETS can be used to create the states in the SLR parse table.
We begin with an initial set C = {closure({ [S’->.S] })}.
Then, foreach I in C and each grammar symbol X such that goto(I,X) is not empty and not in C already, do
Add goto(I,X) to C
Example: canonical LR(0) itemsets for the same grammar.
Each set in C corresponds to a state in a DFA.

17. How to build the SLR parse table
Take the augmented grammar G’
Construct the canonical LR(0) itemsets C for G’
Associate a state with each itemset Ii in C
Construct the parse table as follows:
If A -> α . a β is in Ii and goto(Ii,a) = Ij, then set action[i,a] to “shift j” (“a” here is a terminal)
If A -> α . is in Ii then set action[i,a] to “reduce A -> α” for all a in FOLLOW(A)
If S’ -> S . is in Ii then set action[i,$] to “accept”
If any of the actions in the table conflict, then G is NOT SLR.

18. Example SLR table construction
For the first LR(0) itemset in our favorite grammar:
I0: E’ -> .E
E -> .E + T
E -> .T
T -> .T * F
T -> .F
F -> .(E) This gives us action[0,(] = shift 4
F -> .id This gives us action[0,id] = shift 5

19. Using Ambiguous Grammars

20. What to do with ambiguity?
Sometimes it is convenient to leave ambiguity in G
For instance, G1: is simpler than G2:
E -> E + E E -> E + T | T
| E * E E -> T * F | F
| ( E ) F -> ( E ) | id
| id
But SLR(1), LR(1), and LALR(1) parsers will all have a
shift/reduce conflict for G1.

21. What to do with ambiguity?
Sometimes it is convenient to leave ambiguity in G
For instance, G1: is simpler than G2:
E -> E + E E -> E + T | T
| E * E E -> T * F | F
| ( E ) F -> ( E ) | id
| id
But SLR(1), LR(1), and LALR(1) parsers will all have a
shift/reduce conflict for G1.

22. LR(0) itemsets for G1

23. Ambiguity leads to conflicts
G1 is ambiguous, so we are guaranteed to get conflicts.
For example, in I7:
We will add rules to “shift 4” on ‘+’ and “shift 5” on ‘*’.
For the item E -> E+E. we will add the rule “reduce E->E+E” to the parse table for each terminal in FOLLOW(E).
But! FOLLOW(E) contains + and * -- shift/reduce conflict.
LR(1) and LALR(1) tables will have the same problems.

24. Resolving the conflicts
Knowing about operator precedence and associativity, we can resolve the conflicts.
Example: for input “id + id * id”, we will be in state 7 after processing “id + id”
STACK INPUT
0 E E * id $ since * has higher precedence than +, we should really shift, not reduce.
With a + next in the input, we should reduce, to enforce left-associativity.
See Fig in text for a complete SLR(1) table.

25. If-else ambiguity
The ambiguity of the “dangling else” creates a shift-reduce conflict in parsers for most languages.
Since the else is normally associated with the nearest if, we resolve the conflict by shifting, instead of educing, when we see “else” in the input.
See the LR(0) states and parse table on page 251.
This method is much simpler than writing an unambiguous grammar.

26. Non-SLR grammars Consider the assignment grammar
S’ -> S generating, e.g. S =*> id = * id
S -> L = R
S -> R
L -> * R
L -> id
R -> L

27. Non-SLR grammars Construct the initial canonical LR(0) itemset I0.
Compute I2 = goto(I0,L) and I6 = goto(I2,=).
Compute FOLLOW(L)
Compute parse table entries for I2: shift/reduce conflict!
This means in state I2, with ‘=’ in the input, we do not know whether to shift and go to state I6 or reduce with R -> L, since ‘=’ is in FOLLOW(L).
To correct this, we need to know more about the context of the L we just parsed.
“Canonical LR(1)” and “LALR(1)” are powerful enough.

28. Canonical LR Parse Table Construction

29. Fig. 4.37. Canonical LR(0) collection for grammar (4.20).
S'  · S
S  · L = R
S  · R
L  · * R
L  · id
R  · L
I5:
L  id ·
I6:
S  L = · R
I1:
S'  S ·
I7:
L  * R ·
I2:
S  L · = R
R  L ·
I8:
I9:
S  L = R ·
I3:
S  R ·
I4:
L  * · R
Fig Canonical LR(0) collection for grammar (4.20).

30. More states means more memory
In SLR, we said in state i we should reduce by A -> α if the itemset contains the item [A -> α .] and a is in FOLLOW(A).
However, sometimes when state i is on top of the stack, and a is next in the input, what comes BEFORE α on the stack might invalidate the reduction A -> α.
Example from previous grammar: sentential form “R = …” is impossible, but “* R =” is possible.
So actually, we really want to reduce by L -> * R when we see R on stack and “=” in the input.

31. LR(1) idea
Our parser needs to keep track of more state information. How can it?
Idea: use canonical LR(0) states, but split states as needed by adding a terminal symbol to each item.
LR(1) ITEMS take the form [A-> α.β,a], where A-> αβ is a production in G and a is a terminal symbol or $.
The “1” refers to the length of a, the LOOKAHEAD for each item. If length = k, we would have an LR(k) item.
In parsing, we will now only reduce αβ. to A if an item’s lookahead symbol agrees with the next input.

32. LR(1) parse table construction
We need to redefine closure(I) for a set of LR(1) items:
for each
item [A-> α.B β,a] in I
production B -> γ in G’
terminal b in FIRST(β a)
such that [B->. γ,b] is not already in I, do:
add [B->. γ,b] to I
repeat until no more items can be added to I
goto(I,X) is the same as for SLR(1).

33. Example LR(1) parser construction
Begin with augmented grammar G’:
S’ -> S
S -> C C [ what is L(G’)?? ]
C -> c C | d
The first itemset I0 = closure({S’->.S,$}) = {
S’ -> .S,$
S -> .CC,$ [ from S’->.S,$ and S->CC, B=S, α=ε, β= ε ]
C -> .cC,c/d [ from S’->.CC,$ and C->cC, B=C, α= ε, β=C ]
C -> .d,c/d [ from S’->.CC,$ and C->d, B=C, α= ε, β=C ] }

34. Fig. 4.39. The goto graph for grammar (4.21).

35. LR(1) parsers: the good news
LR(1) is quite similar to SLR(1), with one main difference:
We only add reduce rules to the parse table when the input matches the LOOKAHEAD for the item
SLR(1) adds reduce rules for any terminal in the FOLLOW set.
This means LR(1) will have fewer shift/reduce and reduce/reduce conflicts, because it tries to reduce in fewer situations.

36. LR(1) parsers: the bad news
LR(1) parsers are powerful, able to parse almost any unambiguous CFG used for real programming languages.
But there is a price: the number of states is huge.
For the very simple c*dc*d language with 4 productions, we already needed 10 LR(1) states.
For a typical PL like Pascal, the LR(1) table would contain a few THOUSAND states!
Is there a technique as powerful with fewer states?

37. Fig. 4.40. Canonical parsing table for grammar (4.21).
STATE
action
goto c d $ S C s3 s4 1 2
acc s6 s7 5 3 8 4 r3 r1 6 9 7 r2
Fig Canonical parsing table for grammar (4.21).

38. LALR Parse Table Construction

39. LALR parse tables
LALR makes smaller parse tables than canonical LR, but still covers most common programming language constructs.
LALR has the same number of states as the SLR parser for the same grammar, but is more picky about when to reduce, so fewer conflicts come up.
yacc actually constructs a LALR(1) table, not a canonical LR(1) table.

40. LALR idea
Usually, in a LR parser, there will be many states that are identical, except for the lookahead symbol.
LALR takes these identical states and MERGES them, forming the UNION of the lookahead symbols for the merged items.
Algorithm: build the LR(1) itemsets, then merge itemsets with the same CORES.

41. LALR example I0: S’ -> .S,$ I3: C -> c.C,c/d
Which LR(1) itemsets
can be merged?
I0: S’ -> .S,$ I3: C -> c.C,c/d
S -> .CC,$ C -> .cC,c/d
C -> .cC,c/d C -> .d,c/d
C -> .d,c/d
I5: S -> CC.,$
I1: S’ -> S.,$
I6: C -> c.C,$
I2: S -> C.C,$ C -> .cC,$
C -> .cC,$ C -> .d,$
C -> .d,$
I7: C -> d.,$
I4: C -> d.,c/d
I8: C -> cC.,c/d
I9: C -> cC.,$

42. Fig. 4.41. LALR parsing table for grammar (4.21).
STATE
action
goto c d $ S C
s36
s47 1 2
acc 5 36 89 47 r3 r1 r2
Fig LALR parsing table for grammar (4.21).

43. Efficient Construction of LALR Parsing Tables
Example Let us again consider the augmented grammar
S'  S
S  L = R | R
A  * R | id
B  L
The kernels of the sets of LR(0) items for this grammar are shown in Fig
I0:
S'  · S
I1:
S'  S ·
I2:
S  L · = R
R  L ·
I3:
S  R ·
I4:
L  * · R
I5:
L  id ·
I6:
S  L = · R
I7:
L  * R ·
I8:
R  L ·
I9:
S  L = R ·
Fig Kernels of the sets of LR(0) items for grammar (4.20).

44. Efficient Construction of LALR Parsing Tables
Example Let us construct the kernels of the LALR(1) items for the grammar in the previous example. The kernels of the LR(0) items were shown in Fig When we apply Algorithm 4.12 to the kernel of set of items I0, we compute closure ({[S'  · S, #]}), which is
S'  · S, #
S  · L = R, #
S  · R, #
L  · * R, #/=
L  · id, #/=
R  · L, #

45. Fig.4.44. Propagation of lookaheads.
FROM TO
I0:
S'  · S
I1:
I2:
I3:
I4:
I5:
S'  S ·
S  L · = R
R  L ·
S  R ·
L  * · R
L  id ·
I6:
S  L = · R
I7:
I8:
L  * R ·
I9:
S  L = R ·
Fig Propagation of lookaheads.

46. Fig. 4.45. Computation of lookaheads.
SET
ITEM
LOOKAHEADS
INIT
PASS1
PASS2
PASS3
I0:
S'  · S $
I1:
S'  S ·
I2:
S  L · = R
R L ·
I3:
S  R ·
I4:
L  * · R =
=/$
I5:
L  id ·
I6:
S  L = · R
I7:
L  * R ·
I8:
R  L ·
I9:
S  L = R ·
Fig Computation of lookaheads.

47. Next time
- Yacc 사용법은 조교가 설명
- Semantic 처리 (Yacc에서 배운 것 구현 방법)