Reputation: 85
S → aSab | ABB | BB
A → baab | BABB | AaA
B → bAaBa | bb | b
I started solving it but it seems I just kept on adding new rules to keep up, can someone please explain how to do this one.
Upvotes: 0
Views: 125
Reputation: 114
This is the resulting grammar in Chomsky normal form:
defined by a 4-tuple of set of non-terminal symbols, set of terminal symbols, rules and the start symbol, respectively.
G = ({S,A,B,<Sab>,<ab>,<BB>,<aab>,<ABB>,<aA>,<AaBa>,<aBa>,<Ba>,a',b'},{a,b},R,S)
R =
S → a'<Sab> | A<BB> | BB
A → b'<aab> | B<ABB> | A<aA>
B → b'<AaBa> | b'b' | b
<Sab> → S<ab>
<ab> → a'b'
<BB> → BB
<aab> → a'<ab>
<ABB> → A<BB>
<aA> → a'A
<AaBa> → A<aBa>
<aBa> → a'<Ba>
<Ba> → Ba'
a' → a
b' → b
Feel free to rename the non-terminals in angle brackets <> by one-character capital symbols as you feel appropriate (they do mean a single non-terminal character), I find this more suitable.
Upvotes: 1