Arjun Suresh (talk | contribs) |
Arjun Suresh (talk | contribs) |
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| − | == $w,x ∈ \{{a,b}\}^* $== | + | ==1. $w,x ∈ \{{a,b}\}^* $== |
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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. | ||
| − | ==2 | + | ==2. $w,x ∈ \{{a,b\}}^+$== |
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). 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, 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