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===5. $L = \{wxw | w,x ∈ \{{a,b}\}^*\} $=== | ===5. $L = \{wxw | w,x ∈ \{{a,b}\}^*\} $=== | ||
<math>L</math> is regular since <math> L = \Sigma^*</math>, by making <math>x = (a+b)^*</math> and <math>w = \epsilon</math>. i.e.; the set of strings generated by <math>L</math> is <math>\{ \epsilon, a, b, aa, ab, ba, bb, aaa, ...\} = \Sigma^*</math> | <math>L</math> is regular since <math> L = \Sigma^*</math>, by making <math>x = (a+b)^*</math> and <math>w = \epsilon</math>. i.e.; the set of strings generated by <math>L</math> is <math>\{ \epsilon, a, b, aa, ab, ba, bb, aaa, ...\} = \Sigma^*</math> | ||
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| + | This language is different from the $L = \{wcw | w ∈ \{{a,b}\}^*\} $ which is clearly a CSL. Here, we cannot do any reduction and hence there is no way to accept a string without checking w before the c and w after c are the same which requires an LBA. | ||
===6. $L = \{wxw| w,x ∈ \{{a,b}\}^+\}$=== | ===6. $L = \{wxw| w,x ∈ \{{a,b}\}^+\}$=== | ||
Find the set of strings generated by <math>L</math> If the set of strings in <math>L</math> is finite, <math>L</math> is regular since all finite languages are regular If the set of strings in <math>L</math> is infinite, check if we can draw an NFA for recognizing <math>L</math>. If so, <math>L</math> is regular If <math>NFA</math> is not possible for <math>L</math>, check if we can recognize <math>L</math> using a <math>PDA</math>, that is with a stack in addition to set of states. If so, <math>L</math> is <math>CFL</math>. If the moves of <math>PDA</math> are all deterministic, then <math>L</math> is a <math>DCFL</math> If <math>PDA</math> is not possible for <math>L</math>, see if we can get a <math>TM</math> for <math>L</math> If <math>TM</math> takes only a linear space (in terms of length of input string), then <math>L</math> is <math>CSL</math>. otherwise its just recursive If <math>L</math> is a decision problem and TM can just say "yes" and may not halt in case of "no", then <math>L</math> is recursively enumerable (partially decidable). If <math>TM</math> can say both "yes" and "no" then <math>L</math> is recursive If no <math>TM</math> is possible for <math>L</math>, then <math>L</math> is undecidable
Partially undecidable or semi-undecidable is under undecidable class. For example halting problem is undecidable but is also semi-decidable. <math>P</math>, <math>NP</math> and <math>NPC</math> problems can all be decided by a <math>TM</math> and hence are recursive. All undecidable problems are NP-Hard, but all NP-Hard problems are not undecidable. Turing decidable problems are recursive but Turing recognizable (Turing acceptable) problems are only recursively enumerable.
The set of strings in <math>L</math> are <math>\{aa, bb, aaaa, abab, baba, bbbb, aaaaaa, ...\}</math>. We cannot accept these strings using an <math>NFA</math>. Now, even a <math>PDA</math> is not possible as once we store <math>w</math> on stack, it can only be read back in reverse order. Thus, we require 2 stacks to recognize <math>L</math>. Now, <math>L</math> can be accepted by a <math>TM</math> in linear space and hence <math>L</math> is <math>CSL</math>.
Same explanation as above, <math>L</math> is <math>CSL</math>.
<math>ww_R</math> can be accepted by a <math>PDA</math> and hence is <math>CFL</math>. But we need a <math>NPDA</math> for this as there is no deterministic way to identify where <math>w</math> ends and <math>w_R</math> starts. <math>wcw_R, w\in(a+b)^*</math> is accepted by a <math>DPDA</math> and hence is <math>DCFL</math>.
Same explanation as above. <math>L</math> is <math>CFL</math>.
<math>L</math> is regular since <math> L = \Sigma^*</math>, by making <math>x = (a+b)^*</math> and <math>w = \epsilon</math>. i.e.; the set of strings generated by <math>L</math> is <math>\{ \epsilon, a, b, aa, ab, ba, bb, aaa, ...\} = \Sigma^*</math>
This language is different from the $L = \{wcw | w ∈ \{{a,b}\}^*\} $ which is clearly a CSL. Here, we cannot do any reduction and hence there is no way to accept a string without checking w before the c and w after c are the same which requires an LBA.
<math>L</math> doesn't contain all strings in <math>\Sigma^*</math> as the strings like <math>abab</math> are not contained in <math>L</math>. All words starting and ending in <math>a</math> or starting and ending in <math>b</math> are in <math>L</math>. But <math>L</math> also contains words starting with <math>a</math> and ending in <math>b</math> like <math>abbab, aabbbabaab</math> etc where the starting sub-string exactly matches the ending sub-string and at least a letter separates them. To accept such strings we need a <math>TM</math> with linear space (this is at least as hard as accepting <math>ww, w \in (a+b)^*</math>), making <math>L</math>, a <math>CSL</math>.
<math>L</math> is regular. Since, <math>w</math> can be <math>\epsilon</math> and <math>x \in (a+b)^*, making L =\Sigma^*</math>. i.e.; the set of strings generated by <math>L</math> is <math>\{ \epsilon, a, b, aa, ab, ba, bb, aaa, ...\} = \Sigma^*</math>
The set of strings in <math>L</math> are <math>\{aaa, aba, aaaa, aaba, abaa, abba, baab, ...\}</math> i.e.; <math>L</math> contains all strings starting and ending with <math>a</math> or starting and ending with <math>b</math> and containing at least 3 letters. Moreover, <math>L</math> doesn't contain any other strings. Thus <math>L</math> can be accepted by an <math>NFA</math> making <math>L</math> regular . Regular expression for <math>L</math> is <math>a(a+b)^+a + b(a+b)^+b</math>.
Find the set of strings generated by <math>L</math> If the set of strings in <math>L</math> is finite, <math>L</math> is regular since all finite languages are regular If the set of strings in <math>L</math> is infinite, check if we can draw an NFA for recognizing <math>L</math>. If so, <math>L</math> is regular If <math>NFA</math> is not possible for <math>L</math>, check if we can recognize <math>L</math> using a <math>PDA</math>, that is with a stack in addition to set of states. If so, <math>L</math> is <math>CFL</math>. If the moves of <math>PDA</math> are all deterministic, then <math>L</math> is a <math>DCFL</math> If <math>PDA</math> is not possible for <math>L</math>, see if we can get a <math>TM</math> for <math>L</math> If <math>TM</math> takes only a linear space (in terms of length of input string), then <math>L</math> is <math>CSL</math>. otherwise its just recursive If <math>L</math> is a decision problem and TM can just say "yes" and may not halt in case of "no", then <math>L</math> is recursively enumerable (partially decidable). If <math>TM</math> can say both "yes" and "no" then <math>L</math> is recursive If no <math>TM</math> is possible for <math>L</math>, then <math>L</math> is undecidable
Partially undecidable or semi-undecidable is under undecidable class. For example halting problem is undecidable but is also semi-decidable. <math>P</math>, <math>NP</math> and <math>NPC</math> problems can all be decided by a <math>TM</math> and hence are recursive. All undecidable problems are NP-Hard, but all NP-Hard problems are not undecidable. Turing decidable problems are recursive but Turing recognizable (Turing acceptable) problems are only recursively enumerable.
The set of strings in <math>L</math> are <math>\{aa, bb, aaaa, abab, baba, bbbb, aaaaaa, ...\}</math>. We cannot accept these strings using an <math>NFA</math>. Now, even a <math>PDA</math> is not possible as once we store <math>w</math> on stack, it can only be read back in reverse order. Thus, we require 2 stacks to recognize <math>L</math>. Now, <math>L</math> can be accepted by a <math>TM</math> in linear space and hence <math>L</math> is <math>CSL</math>.
Same explanation as above, <math>L</math> is <math>CSL</math>.
<math>ww_R</math> can be accepted by a <math>PDA</math> and hence is <math>CFL</math>. But we need a <math>NPDA</math> for this as there is no deterministic way to identify where <math>w</math> ends and <math>w_R</math> starts. <math>wcw_R, w\in(a+b)^*</math> is accepted by a <math>DPDA</math> and hence is <math>DCFL</math>.
Same explanation as above. <math>L</math> is <math>CFL</math>.
<math>L</math> is regular since <math> L = \Sigma^*</math>, by making <math>x = (a+b)^*</math> and <math>w = \epsilon</math>. i.e.; the set of strings generated by <math>L</math> is <math>\{ \epsilon, a, b, aa, ab, ba, bb, aaa, ...\} = \Sigma^*</math>
<math>L</math> doesn't contain all strings in <math>\Sigma^*</math> as the strings like <math>abab</math> are not contained in <math>L</math>. All words starting and ending in <math>a</math> or starting and ending in <math>b</math> are in <math>L</math>. But <math>L</math> also contains words starting with <math>a</math> and ending in <math>b</math> like <math>abbab, aabbbabaab</math> etc where the starting sub-string exactly matches the ending sub-string and at least a letter separates them. To accept such strings we need a <math>TM</math> with linear space (this is at least as hard as accepting <math>ww, w \in (a+b)^*</math>), making <math>L</math>, a <math>CSL</math>.
<math>L</math> is regular. Since, <math>w</math> can be <math>\epsilon</math> and <math>x \in (a+b)^*, making L =\Sigma^*</math>. i.e.; the set of strings generated by <math>L</math> is <math>\{ \epsilon, a, b, aa, ab, ba, bb, aaa, ...\} = \Sigma^*</math>
The set of strings in <math>L</math> are <math>\{aaa, aba, aaaa, aaba, abaa, abba, baab, ...\}</math> i.e.; <math>L</math> contains all strings starting and ending with <math>a</math> or starting and ending with <math>b</math> and containing at least 3 letters. Moreover, <math>L</math> doesn't contain any other strings. Thus <math>L</math> can be accepted by an <math>NFA</math> making <math>L</math> regular . Regular expression for <math>L</math> is <math>a(a+b)^+a + b(a+b)^+b</math>.
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