Below is the grammar for L = {a^n b^n c^n | n>=1}
S → abc | aSAc
cA → Ac
bA → bb
Let’s try deriving string w = aaabbbccc
S → aSAc [S → aSAc]
→ aaSAcAc [S → aSAc]
→ aaSAAcc [cA → Ac]
→ aaabcAAcc [S → abc]
→ aaabAcAcc [cA → Ac]
→ aaabAAccc [cA → Ac]
→ aaabbAccc [bA → bb]
→ aaabbbccc [bA → bb]
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