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Eliminate epsilon, unit and useless symbols and rewrite into CNF
Problem
Eliminate epsilon, unit and the useless symbols for the given grammar and rewrite it into CNF.
S->0E0|1FF| ε
E->G
F->S|E
G->S| ε
Solution
In the given grammar, we will first remove the null production. There are two null productions in the grammar, as given below −
S ==> ε
G ==> ε
So, remove null production and rewrite all the other rules containing G by epsilon there, along with old productions. We do not remove S ==> epsilon as it is the start symbol.
Remove G ==> epsilon, we get the following −
S ==> 0E0 | 1FF | ε
E ==> G | ε
F ==> S | E
G ==> S
Now remove E ==> epsilon, we get the following −
S ==> 0E0 | 1FF | 00 | ε
E ==> G
F ==> S | E | ε
G ==> S
Now remove F ==> epsilon, we get the following −
S ==> 0E0 | 1FF | 00 | 1S | 1E | 1 | ε
E ==> G
F ==> S | E
G ==> S
Now, we have to remove unit production, there is only one unit production E ==> G and G ==> S,
By removing G ==> S, we get the following −
S ==> 0E0 | 1FF | 00 | 1S | 1E | 1 | ε
E ==> S
F ==> S | E
Remove E ==> S, we get the following −
S ==> 0S0 | 1FF | 00 | 1S | 1 | ε
F ==> S
Removing F ==> S, we get the following −
S ==> 0S0 | 1SS | 00 | 1S | 1 | ε
Now we have to convert it to Chomsky normal form (CNF). So, we do the following −
Add production A ==> 0, B==>1, C ==> AS and D ==> BS,
We get final grammar as follows −
S ==> CA | DS | BB | AS | 1 | ε
A ==> 0
B ==> 1
C ==> AS
D ==> BS
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