WebNov 11, 2024 · Approach : Let us understand the approach by taking the example “aabb”. Scan the input from the left. First, replace an ‘a’ with ‘X’ and move right. Then skip all the a’s and b’s and move right. When the pointer reaches Blank (B) Blank will remain Blank (B) and the pointer turns left. Now it scans the input from the right and ... WebLet L = {a m b m m ≥ 1}. Then L is not regular. Proof: Let n be as in Pumping Lemma. Let w = a n b n. Let w = xyz be as in Pumping Lemma. Thus, xy 2 z ∈ L, however, xy 2 z contains more a’s than b’s. Share Improve this answer Follow edited Mar 26, 2024 at 18:17 Lucas 518 2 12 18 answered Feb 22, 2010 at 8:53 cletus 612k 166 906 942 12
FORMAL LANGUAGES, AUTOMATA AND COMPUTATION
WebJun 15, 2024 · The shortest word I was able to produce using this grammar is abdd which does not conform to your language. It should have been possible to construct an empty word for n=0 and the word abbd for n=1. But: The proposed language is not context free and cannot be described by a context free grammar. See this answer for proof. Share … WebDec 8, 2024 · The first rule guarantees, that for every a in the beginning there are two f in the end. It enforces at least one a. The second half enforces the sequence d e ff.... The second rule enforces the correct number of b and d and also that the single c is between the b s and the c s Share Improve this answer Follow answered Dec 8, 2024 at 13:03 notes on competition act 2002
Grammar for a^N b^N c^N
WebOct 11, 2016 · Option (4) is correct as first part has #a = #b+#c and second part has #b = #a+#c, which is required for given language. First part, for n = k + m : S 1 → a S 1 c S 2 λ, S 2 → a S 2 b λ Second part, for m = k + n : S 3 → a S 3 b S 4 λ, S 4 → b S 4 c λ Thus, or : Language of above grammar would be inherently ambiguous. Share Cite Follow WebMay 11, 2024 · 1 Answer Sorted by: 0 Consider the regular language R = a*b*cd. The intersection of two regular languages must be a regular language. The intersection of L and R is a^n b^n cd. However, this is easily shown not to be regular using the pumping lemma or Myhill-Nerode theorem. This is a contradiction, so L must not be regular. Share Follow notes on company law