Identify the class of the language
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Given a description of a language <math>L</math>, how to find the class of <math>L</math>?

Find the set of strings generated by <math>L</math>

  1. If the set of strings in <math>L</math> is finite, <math>L</math> is regular since all finite languages are regular
  2. 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
  3. 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>.
  4. If the moves of <math>PDA</math> are all deterministic, then <math>L</math> is a <math>DCFL</math>
  5. If <math>PDA</math> is not possible for <math>L</math>, see if we can get a <math>TM</math> for <math>L</math>
  6. 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
  7. 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).
  8. If <math>TM</math> can say both "yes" and "no" then <math>L</math> is recursive
  9. If no <math>TM</math> is possible for <math>L</math>, then <math>L</math> is undecidable

Some Facts

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.

Some Twisted Examples

1. $L = \{ww \mid w ∈ (a+b)^*\}$

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>.

2. $L = \{ww\mid w ∈ (a+b)^+\}$

Same explanation as above, <math>L</math> is <math>CSL</math>.

3. $L = \{ww_R\mid w ∈ ({a+b})^*\}$

<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>.

4. $L = \{ww_R\mid w ∈ ({a+b})^+\}$

Same explanation as above. <math>L</math> is <math>CFL</math>.

5. $L = \{wxw \mid 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>

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.

6. $L = \{wxw\mid w,x ∈ ({a+b})^+\}$

<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>.

7. $L = \{wxw_R\mid w,x ∈ ({a+b})^*\}$

<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 | 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.

8. $L = \{wxw_R\mid w,x ∈ ({a+b})^+\}$

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>.

9. $L = \{wxwy \mid w,x,y ∈ ({a+b})^+\}$

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 $a(a+b)^+a(a+b)^+ + b(a+b)^+b(a+b)^+$

10. $L = \{xwyw \mid w,x,y ∈ ({a+b})^+\}$

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 $(a+b)^+a(a+b)^+a + (a+b)^+b(a+b)^+b$


11. $L = \{wxyw \mid w,x,y ∈ ({a+b})^+\}$

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 $w = ab, x,y = a$. In short we need to compare the substring at the beginning of the string with that at the end, making this a CSL.




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Given a description of a language <math>L</math>, how to find the class of <math>L</math>?[edit]

Find the set of strings generated by <math>L</math>

  1. If the set of strings in <math>L</math> is finite, <math>L</math> is regular since all finite languages are regular
  2. 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
  3. 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>.
  4. If the moves of <math>PDA</math> are all deterministic, then <math>L</math> is a <math>DCFL</math>
  5. If <math>PDA</math> is not possible for <math>L</math>, see if we can get a <math>TM</math> for <math>L</math>
  6. 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
  7. 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).
  8. If <math>TM</math> can say both "yes" and "no" then <math>L</math> is recursive
  9. If no <math>TM</math> is possible for <math>L</math>, then <math>L</math> is undecidable

Some Facts[edit]

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.

Some Twisted Examples[edit]

1. $L = \{ww \mid w ∈ (a+b)^*\}$[edit]

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>.

2. $L = \{ww\mid w ∈ (a+b)^+\}$[edit]

Same explanation as above, <math>L</math> is <math>CSL</math>.

3. $L = \{ww_R\mid w ∈ ({a+b})^*\}$[edit]

<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>.

4. $L = \{ww_R\mid w ∈ ({a+b})^+\}$[edit]

Same explanation as above. <math>L</math> is <math>CFL</math>.

5. $L = \{wxw \mid w,x ∈ ({a+b})^*\} $[edit]

<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.

6. $L = \{wxw\mid w,x ∈ ({a+b})^+\}$[edit]

<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>.

7. $L = \{wxw_R\mid w,x ∈ ({a+b})^*\}$[edit]

<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 | 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.

8. $L = \{wxw_R\mid w,x ∈ ({a+b})^+\}$[edit]

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>.

9. $L = \{wxwy \mid w,x,y ∈ ({a+b})^+\}$[edit]

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 $a(a+b)^+a(a+b)^+ + b(a+b)^+b(a+b)^+$

10. $L = \{xwyw \mid w,x,y ∈ ({a+b})^+\}$[edit]

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 $(a+b)^+a(a+b)^+a + (a+b)^+b(a+b)^+b$


11. $L = \{wxyw \mid w,x,y ∈ ({a+b})^+\}$[edit]

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 $w = ab, x,y = a$. In short we need to compare the substring at the beginning of the string with that at the end, making this a CSL.




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