1. NORMAL FORMS FDP ON THEORY OF COMPUTINGBy
G Sudha Sadasivam
Assistant Professor, CSE

2. CONTEXT FREE GRAMMAR Formal languages, NFA and DFA describe grammars
English Sentence rules
<Sentence>  <Noun Phrase> <Verb Phrase>
<Noun Phrase>  <Article> <Noun> | <Noun>
<Verb Phrase>  <Verb> | <Verb> <Noun Phrase>
<Article>  a | the
<Noun>  Sita | boy | girl | ball | dog | ...
<Verb>  caught | saw | took | ...

3. CONTENTS CFG Chomsky hierarchy Normal forms Unit productions
Useless productions
CNF
Applications

4. Sentence NP VP
Noun
Verb
Article
Sita
took
the
ball

5. Chomsky Hierarchy- Avram Noam Chomsky
Grammar G = ( VN, VT, P, S), where VN is set of NT/ Variables, VT is alphabet set / terminals (∑), P is set of productions , S start symbol
In CFG, rules are of the form A  w, w ε ( VT U VN)
Type
Name
Productions (P)
Unrestricted
a ® b with a Î (VT È VN)+ and b Î (VT È VN)* 1
Context-Sensitive
a1Aa2 ® a1b a2 with A Î VN and
a1, a2 Î (VT È VN)* and b Î (VT È VN)+ 2
Context-Free
A ® b with A Î VN and b Î (VT È VN)* 3
Regular, Finite
A ® bB or A ® b with A, B Î VN and b Î V*T

6. CFG Context-free means a variable can be replaced with w.
They are powerful to describe Programming languages.
CFG are simple to construct efficient parsing algorithms – LR & LL parsers.
BNF (Backus-Naur form is used to represent CFG.
S  aSb | λ
Panini described Sanskrit using CFG
Venpa is governed by CFG
Text Mining in Bio-medicine

7. Normal Forms
A NF for a grammar has additional conditions imposed upon its productions and is equivalent to the given grammar.
Two types
Chomsky NF (CNF)
Greibach NF (GNF)
Simple form of productions.
Rules in CNF has both theoretical and practical implications.

8. T0 T1 T2
GNF T3
CNF

9. Examples
CNF:
CYK membership algorithm to find if a string is in the language represented by the grammar.
GNF:
is used for conversion from CFG to NDPA and vice versa.

10. CNF CFG, RHS can be a combination of V & T Eg NP  the N is reduced to
DET  the NP  DET N
A λ-free CFG is said to be in CNF if prod. are
A  a B  CD , with A, B, C, D εVN and a ε VT.
If CFG is not λ-free then include S λ
As 2nd prod has two variables – binary grammars

11. GNF Productions are of the form Aax , where a is a VT and x ε VN*.
They are long

12. 1. λ-free languages Let L be any CF language,
G (with λ proiductions) has prod S0  S | λ
λ - productions are A  λ;
A variable A for A * λ is nullable
In this case λ-prod can be removed
If G is a λ-free CFG, then there is G1 having no λ-prod

13. STEPS Find the set VN of all nullable variables A  λ, put A in VN
Repeat to Add variables to VN
For prod B  A1 A2 A3 A4
A1 A2 A3 A4 are in VN, add B to VN
2. For A  x1 x2 … xm with m >=1, put into P1, that production and prod generated by replacing nullable variables with λ in all combinations

14. S  ABaC
A  BC
B  b | λ
C  D | λ
D  d
Nullable is { A, B, C}
S  ABaC | BaC | AaC | ABa | aC | Ba | Aa | a
A  BC | C | B
B  b
C  D
D  d

15. 2. Substitution Rule G : A  x1 B x2 and B  y1 | y2 |.. | yn
G1: A  x1 y1 x2 | x1 y2 x2 | x1 y3 x2 |…| x1 yn x2
B  y1 | y2 |.. | yn
Then G = G1
Example:
A  a | aaA | abBc
B  abbA | b
Then
A  a | aaA | ababbAc | abbc

16. 3. Removing Useless Productions
Prod that do not take part in any derivations
S * xAy * w is useful
Useless variables
Cannot be reached from S
S  A; A  aA | λ;
B  bA
2. Cannot derive a terminal string
S  A | b
A  aA

17. Identify the variables that can lead to a terminal string
Set V1(G1=(V1,T1,P1,S1) to NULL
Repeat
For A  x1x2x3……x n for xi in V1 U T, add A to V1
Add to P1 all prod in P whose symbols are in A U V1
Eliminate var that cannot be reached from S
Dependency graph
Useful-var is reached from start (S).

18. For eg1:
S  aSb | λ | A
A  aA
A – is useless
Eg2:
S  A
A  aA | λ
B  bA
A cannot derive a terminal string
B is not reachable from start

19. 1) C is useless since it does not derive a terminal string
Eg3:
1: S  aS | A | C
2: A  a
3: B  aa
4: C  aCb
1) C is useless since it does not derive a terminal string
2) Reachability graph – B is not reachable S A B

20. 4. Removing Unit Productions
Prod of the form A  B are unit productions
Find A * B from a dependency graph
Add to P1 all non-unit prod from P
If AB and B  y1 | y2 |.. | yn then
A  y1 | y2 |.. | yn

21. S  Aa | B S  Aa | bb A  a | bc | B A  a | bc | bb B  A | bb
Non-unit rules
S  Aa
A  a | bc
B  bb
2. As S * B , S bb is added
As A * B, A bb is added
As B * A, B  a|bc is added
S  Aa | bb
A  a | bc | bb
B  bb | a | bc S A B

22. Altogether Remove λ productions Remove Unit Productions.
Remove useless productions and non-terminals.

23. CNF CNF -- Rules are of the form A  BC or A  a
Eg: S  AS | a; A  SA | b
Steps
Eliminate λ, unit and useless productions
Add production of form A a or A  BC to P1
Consider A  x1x2x3……x n
If n = 1 then x1 should be a Terminal (T)
If n >=2, introduce Ba for a ε T and
Convert A  x1x2x3……x n to A  B1 B2 … B3 and add B1  x1 … to P1
For n>2 introduce new Var D1,D2…
Eg. A BCDE is written as
A BD1 D1  C D2 D2  D D3 D3 E D4

24. G:
S  ABa
A  aab
B  Ac
Result:
Na  a
Nb  b
Nc  c
S  AX1
X1  BNa
A  NaX2
X2  NaNb
B  ANc

25. Remove useless symbols
Eg1:
Remove useless symbols
Eg 2:
Convert to CNF
A  BD
A  a
S  λ
S  AA|CD|bB
A  aA|a
B  bB|bC
C  cB
D  dD|d

26. Eg3:
S  A|ABa|AbA
A  Aa|λ
B  Bb|BC
C  CB|CA|bB

27. THANK YOU

28. NORMAL FORMS FDP ON THEORY OF COMPUTINGBy
G Sudha Sadasivam
Assistant Professor, CSE

29. GNF Productions are of the form Aax , where a is a VT and x ε VN*.
They are long
Can be used to construct PDA to recognise CFG
Both GNF and s-grammars require that rules have the form A  ax.
s-grammars requires that the first VN of all the A-rules , be distinct. GNF does not impose such a restriction.

30. Remove λ productions
Remove Unit Productions.
Remove useless productions and non-terminals.
Convert to CNF
Convert to GNF

31. GNF S  aAB | bBB | bB S  AB A  aA | bB | b A  aA | bB | b B  b
S  aY SY | aX
X  a
Y  b
Eg1:
S  AB
A  aA | bB | b
B  b
Eg2:
S  abSb | aa

32. Removing direct left recursions
A  A α | β, the equivalent is
A  β A’ ; A’  α A’ | λ
Eg:
S  Sa | b is equivalent to
S  b S’ ; S’  a S’ | λ

33. Answer:
S  A | C
A  B A’ | a A’
A’  aBA’ | aCA’
B  Cb B’
B’  bB’
C  cC | c

34. 1. Grammar G:
2. Removing left recursion:
3. BAA is out of order
4. Take each rule that does not have a terminal at start and follow the derivation until a terminal is produced

35. answer

36. Construct GNF for
Answer

37. CYK algorithm
Cocke-Younger-Kasami algorithm - To recognise CF language
To prove membership of strings to CFL
To construct a possible parse tree.
Can also process Stochastic CFG, where probabilities a stored in a table.
Asymptotic time complexity is θ(n3),
where, n -- strlen

38. I/P string a 1 ... a n of length ‘n’. G has ‘r’ terminals.
Grammar has nonterminals R R r & R 1 is start symbol.
P[n n r] is an array of booleans initialized to false.
For each i = 1 to n
For each unit production R j → a i set P[i 1 j] = true.
For each i = 2 to n -- Length of span
For each j = 1 to n-i+1 -- Start of span
For each k = 1 to i-1 -- Partition of span
For each production R A -> R B R C
if P[j k B] and P[j+k i-k C]
then set P[j i A] = true
if P[1 n 1] is true
then string is member of language
else string is not member of language

39. CYK Parsers i/p str: w = w1w2 w3……wn;
wij = wi wi+1……wj and Vij = { A ε V | A * wij }
W belongs to L iff S ε V1n.
Vii. Is found by examining RHS of rules
O(n3) – n is string length

40. S  AB
A  BB | a
B  AB | b
To generate string aabbb

41. String: aaabbb

42. String : baaa