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Design Context Free Grammar Examples 2 In this class, We discuss Design Context Free Grammar Examples 2. The reader should have prior knowledge of previous examples. Click Here. Example 1: Language l1 = {0^n1^(n+1) where n = 0}. L1= { 1,011, 00111, . . .} The language L1 will have a strings sequence of zeros followed by a sequence of one’s. The number of one’s must be one greater than the number of zero’s. The below grammar shows the CFG for the language L1. S – A1 A – 0A1 A – ε The production S is checking for an extra 1. First, we call the non-terminal A. The production A will check for zero’s followed by an equal number of one’s. After finding an equal number of one’s, the production S will check for an extra 1. Example 2: Language L2 = {a^mb^n where m not equal to n, m = 0, n=0} the language L2 will not accept strings containing a’s followed by an equal number of b’s. The below grammar shows the CFG for language L2. S – aSb S – A S – B A – aA | a B – bB | b We need to find strings containing a’s followed by b’s. But not equal a’s and b’s. We should find atleast one extra a or one extra b. The production S will check for a’s followed by an equal number of b’s. We should not end the production S. We need to check for atleast one extra a or one extra b. Tho check for extra a’s, we are using production A. To check for extra b’s, we use production B. Example 3: Language L3 = {a^mb^n where nm, m = 1, n=1} The language L3 similar to language L2. The condition nm says we need more number of b’s. Similar to example 2, we need to check for extra b’s. The below grammar shows the CFG for language L3. S – aSb | aAb A – bA | b Link for playlists: / @wisdomerscse Link for our website: https://learningmonkey.in Follow us on Facebook @ / learningmonkey Follow us on Instagram @ / learningmonkey1 Follow us on Twitter @ / _learningmonkey Mail us @ learningmonkey01@gmail.com