1. Section 4.8 Aggelos Kiayias Computer Science & Engineering Department
The University of Connecticut
371 Fairfield Road, Box U-155
Storrs, CT

2. Ambiguous Grammars Ambiguous Grammars produce “conflicts”
Shift/reduce, or reduce/reduce.
Three “typical examples”
The dangling else
The “expression” grammar
Eqn pre-processor of Troff

3. Dangling Else Ambiguity
Recall Grammar
Can be rewritten as:
S  iSeS | iS | a
We compute LR(0) items + Goto and the SLR
Parsing table.
stmt  if expr then stmt
| if expr then stmt else stmt
| other (any other statement)

4. Canonical sets of LR(0) items
I0 I3
S’  .S S  a.
S  .iSeS
S  .iS I4
S  .a S  iS.eS
S  iS. I1
S’ S. I5
S  iSe.S
I2 S  .iSeS
S  i.SeS S  .iS
S  i.S S  .a
S  .iS I6
S  .a S  iSeS.
Follow(S)= $,e

5. Parsing Table i e a $ S 0 s2 s3 1 1 acc 2 s2 s3 4 3 r3 r3 4 s5/r2 r2
1. S iSeS
2. S iS
3. S a
i e a $ S
0 s2 s3 1
1 acc
2 s2 s3 4
3 r3 r3
4 s5/r2 r2
5 s2 s3 6
6 r1 r1
Resolve: choose s5
e.g. parse iiaea to understand the conflict resolution.

6. Tracing $0
iiaea$
STACK
INPUT
Remark

7. Expressions Recall grammar: Originally written as:
E  E + E | E *E | (E) | id
In fact we can try to construct the SLR parsing table and see what will happen
E’  E E  E + T | T
T  T * F | F
F  ( E ) | id

8. Canonical Sets I0 I2 I5 I8 I1 I3 I6 I9 Follow(E’)=$ Follow(E)=+*)$
E’  .E E  .( E ) E  E * . E E  E * E .
E  .E + E E  .E + E E  .E + E E  E . + E
E  .E * E E  .E * E E  .E * E E  E . * E
E  .( E ) E  .( E ) E  .( E )
E  .id E  .id E  .id
I1 I3 I6 I9
E’  E. E  id. E  ( E . ) E  ( E ) .
E  E . + E E  E . + E
E  E . * E I4 E  E . * E
E  E + . E
E  .E + E I7
E  .E * E E  E + E .
E  .( E ) E  E . + E
E  .id E  E . * E
Follow(E’)=$
Follow(E)=+*)$

9. The Parsing Table id + * ( ) $ E 0 s3 s2 1 1 s4 s5 acc 2 s3 s2 6
1. E E+E
2. E E*E
3. E (E)
4. E id
id + * ( ) $ E
0 s3 s2 1
1 s4 s5 acc
2 s3 s2 6
3 r4 r4 r4 r4
4 s3 s2 7
5 s3 s2 8
6 s4 s5 s9
7 s4/r1 s5/r1 r1 r1
8 s4/r2 s5/r2 r2 r2
9 r3 r3 r3 r3

10. Resolving Ambiguity
If the state contains E  E op E . and we find op on the input we
Reduce (left-association)
Shift (right-association)
If the state contains E  E op E . and we find op’ on the input we
Reduce (op’ has lower precedence than op)
Shift (op’ has higher precedence than op)

11. Subscript/Superscript
1. E EsubEsupE
2. E EsubE
3. E EsupE
4. E {E}
5. E c
Even if we resolve conflicts using precedence and associations
Rule 1 will still produce problems,
i.e. when seeing a } or $
Being at the state that contains the item E EsubEsupE.
We will have two possibilities for reduce:
Either r3 or r1. (reduce/reduce conflict)
Resolving for r1 allows a special meaning for it.