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Design an unambiguous CFG in CNF that generates E?
Problem
Define the language, E={aibj|i not equal to j and i not equal to 2j} and design an unambiguous context free grammar (CFG) in Chomsky normal form (CNF) that generates E.
Solution
The unambiguous CFG for the given language is as follows −
S->AC|CB
A->aA|a
B->Bb|b
C->aCb|aaCb|epsilon
Now, convert this CFG into CNF. You can follow the below mentioned steps for the successful conversion.
Step 1
First add a new start symbol S0
S0->S
S->AC|CB
A->aA|a
B->Bb|b
C->aCb|aaCb|epsilon
Step 2
Next eliminate the epsilon symbol in the production other than the start symbol.
C->epsilon is a null production
After eliminating null production, the new productions are as follows −
S0->S
S->AC|CB
A->aA|a
B->Bb|b
C->aCb|aaCb|ab|aab
Step 3
Now remove the unit production, if it exists
S0->S is a unit production
Replace S, with S production
S0->AC|CB
S->AC|CB
A->aA|a
B->Bb|b
C->aCb|aaCb|ab|aab
Step 4
There are no useless productions to remove.
Step 5
Now check the productions that are not in CNF, If not, let's convert into CNF
A context free grammar is in CNF, if the production rules satisfy one of the following conditions −
- If there is start Symbol generating ε. Example − A->
- If a non-terminal generates two non-terminals. Example − S->AB
- If a non-terminal generates a terminal. Example − S->a
Now, our grammar that doesn't follow the rules given below −
A->aA
B->Bb
C->aCb|aaCb
Convert this production into CNF as shown below −
A->XA
B->BY
C->XCY|XXCY
X->a
Y->b
Still C->XCY|XXCY violating rules because RHS there are more than two symbols So, let's decompose the production
C->XCY
C->RY
R->XC
And
C->XXCY
C->LM
L->XX
M->CY
The CNF that generates E is as follows −
S0->AC|CB
S->AC|CB
A->XA|a
B->BY|b
C->RY
C->LM
X->a
Y->b
R->XC
L->XX
M->CY
Now, each production is generating either a terminal or two non-terminals.
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