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Return to Identify the class of the language.
Find the set of strings generated by <math>L</math>
Partially undecidable or semi-undecidable is considered undecidable. For example halting problem is considered undecidable but is 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 \mid 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 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>
This language is different from the $L = \{wcw_R \mid w ∈ \{{a,b}\}^*\} $ which is clearly a DCFL. Here, we cannot do any reduction and hence there is no way to accept a string without checking that the string after c is the reverse of the string before c, which requires a DPDA.
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 a finite automata making <math>L</math> regular . Regular expression for <math>L</math> is <math>a(a+b)^+a + b(a+b)^+b</math>.
For any string to be in $L$, the beginning part of the string ($w$) must repeat at some other point (between the second and last characters) of the string (next $w$). Since $y$ is there at the end which can generate any string, we can make $w$ as small as possible as per the given condition. So, $w$ can be either $a$ or $b$. We can thus write regular expression for $L$ as $a(a+b)^+a(a+b)^+ + b(a+b)^+b(a+b)^+$
Similar explanation for example 9, except that instead of first character being $a$ or $b$ we have the last character. So, regular expression for $L$ will be $(a+b)^+a(a+b)^+a + (a+b)^+b(a+b)^+b$
Here, $w$ is coming at the beginning and also at the end. Unlike as in example 8 or 9, we cannot restrict $w$ to be $a$ or $b$ as a string starting with $a$ can end in $b$ and still be in $L$- example $abaaab$, where $w = ab$ and $x,y = a$. In short we need to compare the substring at the beginning of the string with that at the end, making this a CSL.
<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>
Here, $w$ cannot be $\epsilon$ and hence to accept the string we do need the power of an LBA making $L$ a CSL.
Here, $w$ cannot be $\epsilon$ and hence to accept the string we do need the power of a PDA making $L$ a NCFL (non-determinism is required to guess the start of $w$).
GATECSE