why do we need to add two times S in example 2 time when we can do this using single s as well . Suppose 0S1 where S Should contain equal number of zeroes and ones . approximately 20:00 Please reply .
Whatever you are suggesting is going to generate the language 0^n 1^n. In Example 2: we are only bothering about no of 0s and no of 1s order doesn't matter say for example string: 0100101101 or 10100011 is valid for above said condition.
yes @@mehulkhatiwala is correct. To be consise there are strings which match the question definition but cannot be produced by your rule @rajendrakujur2078
is it incorrect to write S->(S)S | e for L3 with the intuition of L2
what you have written is equivalent to what the professor has written. It is after one step of substitution.
SS --- set S = (S) for 1st S ---> (S)S
why do we need to add two times S in example 2 time when we can do this using single s as well . Suppose 0S1 where S Should contain equal number of zeroes and ones . approximately 20:00 Please reply .
Whatever you are suggesting is going to generate the language 0^n 1^n. In Example 2: we are only bothering about no of 0s and no of 1s order doesn't matter say for example string: 0100101101 or 10100011 is valid for above said condition.
yes @@mehulkhatiwala is correct. To be consise there are strings which match the question definition but cannot be produced by your rule @rajendrakujur2078