Arjun Suresh (talk | contribs) |
Arjun Suresh (talk | contribs) |
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===1. $L = \{wxw | w,x ∈ \{{a,b}\}^*\} $=== | ===1. $L = \{wxw | w,x ∈ \{{a,b}\}^*\} $=== | ||
| − | Here, <math>L</math> can generate all strings in <math>\Sigma</math> | + | Here, <math>L</math> can generate all strings in <math>\Sigma^*</math>, by making <math>x = (a+b)^*</math> and <math>w = \epsilon</math>. Hence, <math>L</math> is regular. |
Here, <math>L</math> can generate all strings in <math>\Sigma^*</math>, by making <math>x = (a+b)^*</math> and <math>w = \epsilon</math>. Hence, <math>L</math> is regular.
Here, our problem is this- given a word s, whether it belongs to L or not. I say that a word belongs to L, iff it starts and end with a or starts and ends with b (and contains at least 3 letters in total). Now, in case of wxw, the same logic won't work. As you told if s ∈ {a (a+b)^+ a} U {b (a+b)^+ b}, then its of the form wxw. But if s ∉ {a (a+b)^+ a} U {b (a+b)^+ b}, we can't say its not of the form wxw. For example, take s as abbab, its of the form wxw with w = "ab" and x = "b". Thus the reduction will work only one way and hence it cannot be used.
Here, <math>L</math> can generate all strings in <math>\Sigma</math>*, by making <math>x = (a+b)</math>* and <math>w = \epsilon</math>. Hence, <math>L</math> is regular.
Here, our problem is this- given a word s, whether it belongs to L or not. I say that a word belongs to L, iff it starts and end with a or starts and ends with b (and contains at least 3 letters in total). Now, in case of wxw, the same logic won't work. As you told if s ∈ {a (a+b)^+ a} U {b (a+b)^+ b}, then its of the form wxw. But if s ∉ {a (a+b)^+ a} U {b (a+b)^+ b}, we can't say its not of the form wxw. For example, take s as abbab, its of the form wxw with w = "ab" and x = "b". Thus the reduction will work only one way and hence it cannot be used.
GATECSE