Arjun Suresh (talk | contribs) |
Arjun Suresh (talk | contribs) |
||
| Line 1: | Line 1: | ||
===1. $L = \{wxw | w,x ∈ \{{a,b}\}^*\} $=== | ===1. $L = \{wxw | w,x ∈ \{{a,b}\}^*\} $=== | ||
| + | Here, L can generate all strings in <math>\Sigma</math>*, by making x = (a+b)* and w = \epsilon. Hence, L is regular. | ||
| + | ===2. $L = \{wxw| w,x ∈ \{{a,b}\}^+\}$=== | ||
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). | 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. | 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, L can generate all strings in <math>\Sigma</math>*, by making x = (a+b)* and w = \epsilon. Hence, L 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, L can generate all strings in <math>\Sigma</math>*, by making x = (a+b)* and w = \epsilon. Hence, L 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