CSC 3130: Automata theory and formal languages Andrej Bogdanov http://www.cse.cuhk.edu.hk/ ~andrejb/csc3130 The Chinese University of Hong Kong Normal forms and parsing Fall 2009
CSC 3130: Automata theory and formal languages
Andrej Bogdanovhttp://www.cse.cuhk.edu.hk/~andrejb/
csc3130
The Chinese University of Hong Kong
Normal forms and parsing
Fall 2009
Testing membership and parsing• Given a grammar
• How can we know if a string x is in its language?
• If so, can we obtain a parse tree for x?• Can we tell if the parse tree is unique?
S → 0S1 | 1S0S1 | TT → S | e
First attempt
• Maybe we can try all possible derivations:
S → 0S1 | 1S0S1 | TT → S | x = 00111
S 0S1
1S0S1
T
00S1101S0S110T1
S
10S10S1...
when do we stop?
Problems
• How do we know when to stop?
S → 0S1 | 1S0S1 | TT → S | x = 00111
S 0S1
1S0S1
00S1101S0S110T1
10S10S1...
when do we stop?
Problems
• Idea: Stop derivation when length exceeds |x|• Not right because of -productions
• We might want to eliminate -productions too
S → 0S1 | 1S0S1 | TT → S | x = 01011
S 0S1 01S0S11 01S011 010111 3 7 6 5
Problems
• Loops among the variables (S → T → S) might make us go forever
• We want to eliminate such loops
S → 0S1 | 1S0S1 | TT → S | x = 00111
Removal of -productions• A variable N is nullable if there is a derivation
• How to remove -productions (except from S)Find all nullable variables N1, ..., Nk
For every production of the form A → Ni,
add another production A → If Ni → is a production, remove itIf S is nullable, add the special production S →
N *
Example• Find the nullable variables
S ACDA aB C ED | D BC | bE b
B C D
nullable variablesgrammar
Find all nullable variables N1, ..., Nk
Finding nullable variables• To find nullable variables, we work backwards
– First, mark all variables A s.t. A as nullable– Then, as long as there are productions of the form
where all of A1,…, Ak are marked as nullable, mark A as nullable
A → A1… Ak
Eliminating -productionsS ACDA aB C ED | D BC | bE b
nullable variables: B, C, D
For every production of the form A → Ni,add another production A →
If Ni → is a production, remove it
D CS ADD BD S ACS AC E
Dealing with loops• A unit production is a production of the
form
where A1 and A2 are both variables• Example
A1 → A2
S → 0S1 | 1S0S1 | TT → S | R | R → 0SR
grammar: unit productions:
S T
R
Removal of unit productions• If there is a cycle of unit productions
delete it and replace everything with A1
• Example
A1 → A2 → ... → Ak → A1
S → 0S1 | 1S0S1 | TT → S | R | R → 0SR
S T
R
S → 0S1 | 1S0S1S → R | R → 0SR
T is replaced by S in the {S, T} cycle
Removal of unit productions• For other unit productions, replace every
chain
by productions A1 → ,... , Ak →
• Example
A1 → A2 → ... → Ak →
S → R → 0SR is replaced by S → 0SR, R → 0SR
S → 0S1 | 1S0S1 | R | R → 0SR
S → 0S1 | 1S0S1 | 0SR | R → 0SR
Recap• After eliminating -productions and unit
productions, we know that every derivation
doesn’t shrink in length and doesn’t go into cycles
• Exception: S → – We will not use this rule at all, except to check if L
• Note -productions must be eliminated before unit
productions
S a1…ak where a1, …, ak are terminals*
Example: testing membershipS → 0S1 | 1S0S1 | TT → S |
x = 00111
S → | 01 | 101 | 0S1 |10S1 | 1S01 | 1S0S1
S 01, 101
10S11S011S0S1
10011, strings of length ≥ 610101, strings of length ≥ 6
unit, -prod
eliminate
only strings of length ≥ 6
0S1 0011, 0101100S11strings of length ≥ 6
only strings of length ≥ 6
Algorithm 1 for testing membership• How to check if a string x ≠ is in L(G)
Eliminate all -productions and unit productionsLet X := SWhile some new rule R can be applied to X
Apply R to XIf X = x, you have found a
derivation for xIf |X| > |x|, backtrack
If no more rules can be applied to X, x is not in L
Practical limitations of Algorithm I• This method can be very slow if x is long
• There is a faster algorithm, but it requires that we do some more transformations on the grammar
G = CFG of the java programming languagex = code for a 200-line java program
algorithm might take about 10200 steps!
Chomsky Normal Form• A grammar is in Chomsky Normal Form if every
production (except possibly S → ) is of the type
• Conversion to Chomsky Normal Form is easy:
A → BC A → aor
A → BcDEreplace terminalswith new variables
A → BCDEC → c break up
sequenceswith new variables
A → BX1
X1 → CX2
X2 → DEC → c
Exercise• Convert this CFG into Chomsky Normal Form:
S |ADDAA aC cD bCb
Algorithm 2 for testing membership
S AB | BCA BA | aB CC | bC AB | a
x = baaba
Idea: We generate each substring of x bottom up
ab b aaACB B ACACBSA SASCB– B
SAC–SAC
Parse tree reconstruction
S AB | BCA BA | aB CC | bC AB | a
x = baabaab b aa
ACB B ACACBSA SASCB– B
SAC–SAC
Tracing back the derivations, we obtain the parse tree
Cocke-Younger-Kasami algorithm
For cells in last row If there is a production A xi
Put A in table cell iiFor cells st in other rows If there is a production A BC where B is in cell sj and C is in cell jt Put A in cell st
x1 x2 … xk
11 22 kk12 23
… …1k
tablecells
s j t k1
Input: Grammar G in CNF, string x = x1…xk
Cell ij remembers all possible derivations of substring xi…xj