CHOMSKY NORMAL FORM(CNF)

&LEFTMOST

DERIVATION (LMD)

M A D E B Y : D I M P Y C H U G H ( 1 8 3 3 )

D R I S H T I B H A L L A ( 1 8 3 8 )

DEFINITIONIf a Context Free Grammar or CFG has only productions of

the formNonterminal -> string of exactly two non-terminals

or Nonterminal -> one terminal then, it is said to be in Chomsky Normal Form, CNF.

Example: S → AS | a A → SA | b

STEPS FOR CONVERSION

• The conversion to Chomsky Normal Form has four main steps:

1. Get rid of all λ-Productions .2. Eliminate Non-terminal Unit Productions. A → B , B → b A → b3. Elimination of more than one terminals by a non- terminal in a production.4. Replace every production that is too long by shorter productions. A → BCD A → BE E → CD

THEOREM-1STATEMENT- If L is a language generated by some CFG, then there

is another CFG that generates all the non- λ words of L, all of whose productions are of one of two basic forms:

Nonterminal-> string of only Non-terminals or

Nonterminal -> one terminalPROOF- Let us start with the CFG-

S -> X1 | X2aX2 | aSb | bXl -> X2X2 | bX2 -> aX2 | aaX1

Now, in order to convert it into the above form , we introduce new productions : A->a and B->b

After conversion we have,

S -> X1 X1 -> X2X2S -> X2AX2 X1 -> BS -> ASB X2 -> AX2S -> B X2 -> AAX1

Thus, we have converted the original CFG into the desired form.

STATEMENT- For any context- free language L, the non- λ words of L can be generated by a grammar in which all the productions are in CNF.

PROOF - Let our initial CFG is S -> XI X2 X3 X8 S -> X3 X5 S -> b

Now, in order to get the given CFG in CNF form we’ll add certain new non-terminals , i.e. S -> XI X2 X3 X8 S -> XI R1 R1 -> X2 X3 X8 R1 -> X2 R2 R2 -> X3 X8

THEOREM-2

Consider another CFG X -> X1aX2bX3Apply Theorem 1 and add productions A->a and B->b to get string of solid non-terminals.Now, X -> X1AX2BX3 X -> X1R1 R1 -> AX2BX3 R1 -> AR2 R2 -> X2BX3 R2 -> X2R3 R3 -> BX3

In this way, we have converted the long productions with long strings of non-terminals into sequences of productions with exactly 2 non-terminals on RHS. Thus, the new grammar in CNF generates the same language as the old grammar.

Ques16 i). The following CFG has unit productions. Convert the following CGF into CNF.S -> X

X -> Y Y-> Z Z ->aa

Sol16). S -> X -> Y => S -> Y S -> Y -> Z => S -> Z S -> Z -> aa =>S ->aa Since S->aa is not in CNF , we introduce a new production A->a to transform it to CNF. Thus, the new grammar in CNF form is : A -> a S -> AA

QUESTIONS

Ques14 v). Convert the following CGF into CNF. S -> ABABAB A -> a|λ B -> b|λSol13). Removing λ productions, A -> a ---------1 B -> b ---------2

S -> ABABAB Using Theorem 2, S -> AR1 ---------3 R1 -> BABAB R1 -> BR2 ---------4 R2 -> ABAB R2 -> AR3 ---------5 R3 -> BAB R3 -> BR4 ---------6 R4 -> AB ---------7Thus, the new grammar in CNF form comprises of (1,2,3,4,5,6,7).

Ques14 iv). Convert the following CGF into CNF. E -> E + E E -> E*E E -> (E) E -> 7 The terminals here are +, *, (, ), 7.Sol12). E -> 7 --------------1 Using Theorem 1, we add new productions that will help in converting CGF to CNF. F -> + --------------2 G -> * --------------3 H-> ( --------------4 I -> ) --------------5 Now , we have E -> EFE E -> ER1 --------------6 R1 -> FE --------------7 E -> EGE E -> ER2 --------------8 R2 -> GE --------------9 E -> HEI E -> HR3 -------------10 R3 -> EI -------------11Thus, the new grammar in CNF form comprises of (1,2,3,4,5,6,7,8,9,10,11).

LEFT MOST DERIVATIONS• The leftmost non-terminal in a working string is the first

non-terminal that we encounter when we scan the string from left to right.

For example : In the string abNbaXYa , the leftmost nonterminal is N.

• If a word w is generated by a CFG by a certain derivation and at each step in the derivation, a rule of production is applied to the leftmost non-terminal in the working string, then this derivation is called a leftmost derivation.

Ques17 i)Find the leftmost derivation for the word “abba” in the grammar:

S -> AA A -> aB B -> bB | λ Sol18)(i). The following is a leftmost derivation:S -> AAS -> aBAS -> abBAS -> abbBAS -> abbλAS -> abbaBS -> abbaλS -> abba

QUESTIONS

Ques17 ii) Find the leftmost derivation for the word “abbabaabbbabbab” in the CFG:

S->SSS|aXbX->ba|bba|abb

Sol18)(ii). The following is a leftmost derivation: S -> SSS S -> aXbSS S -> abbabSS S -> abbabaXbS S -> abbabaabbbS S -> abbabaabbbaXb S -> abbabaabbbabbab

ADVANTAGES..• CNF Grammar can be used to determine whether a string is a

member of the language associated with it or not .• It can be used to find the no. of production steps required to

generate a string. The key advantage is that in Chomsky Normal Form, every derivation of a string of n letters has exactly 2 n − 1 steps.

Example : S -> AX|AB A -> a X -> SB B -> b

The string to be generated is “ aabb” having n(no. of letters in a string ) =4. The following is a leftmost derivation:1) S -> AX2) S -> aX3) S -> aSB4) S -> aABB5) S -> aaBB6) S -> aabB7) S -> aabbHere, A string of n(n=4) letters has exactly 2n-1(2*4-1=7) steps .

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