Converting grammar to Chomsky Normal Form?

Question

Convert the grammar below into Chomsky Normal Form. Give all the intermediate steps.

S -> AB | aB
A -> aab|lambda
B -> bbA

Ok so t

Answer 1

Alternative answer: The grammar can only produce a finite number of strings, namely 6.

 S -> aabbbaab | aabbb | bbaab | bb | abbaab | abb.

You can now condense this back to Chomsky Normal Form by hand.

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By substitution, we can find the set of all strings produced. Your initial rules:

S -> AB | aB.
A -> aab | lambda.
B -> bbA.

First split up the S rule:

S -> AB.
S -> aB.

Now substitute what A and B expand into:

S -> AB
  -> (aab | lambda) bbA
  -> (aab | lambda) bb (aab | lambda).
S -> aB
  -> abbA
  -> abb (aab | lambda).

Expand these again to get:

S -> aabbbaab.
S -> aabbb.
S -> bbaab.
S -> bb.
S -> abbaab.
S -> abb.

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To change this finite set to Chomsky Normal Form, it suffices to do it by brute force without any intelligent factoring. First we introduce two terminal rules:

X -> a.
Y -> b.

Now for each string, we consume the first letter with a terminal variable and the remaining letters with a new variables. For example, like this:

S -> aabbb. (initial rule, not in Chomsky Normal Form)

S -> XC, where X->a and C->abbb.
C -> XD, where X->a and D->bbb.
D -> YE, where Y->b and E->bb.
E -> YY, where Y->b and Y->b.

We just go through this process for all 6 strings, generating a lot of new intermediate variables.