Syntax error Eliminate epsilon, unit and useless symbols and rewrite into CNF

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

Updated on: 2021-06-12T12:15:02+05:30

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