Skip links

Main navigation

GATE CSE

Resource Portal of GATE Overflow

Identify the Class of a Given Language

by 4 Comments

Given a description of a language $L$, how to find the class of $L$?

Find the set of strings generated by $L$

  1. If the set of strings in $L$ is finite, $L$ is regular since all finite languages are regular
  2. If the set of strings in $L$ is infinite, check if we can draw an NFA for recognizing $L$. If so, $L$ is regular
  3. If $NFA$ is not possible for $L$, check if we can recognize $L$ using a $PDA$, that is with a stack in addition to set of states. If so, $L$ is $CFL$.
  4. If the moves of $PDA$ are all deterministic, then $L$ is a $DCFL$
  5. If $PDA$ is not possible for $L$, see if we can get a $TM$ for $L$
  6. If $TM$ takes only a linear space (in terms of length of input string), then $L$ is $CSL$. otherwise its just recursive
  7. If $L$ is a decision problem and $TM$ can just say “yes” and may not halt in case of “no”, then $L$ is recursively enumerable (partially decidable).
  8. If $TM$ can say both “yes” and “no” then $L$ is recursive
  9. If no $TM$ is possible for $L$, then $L$ is undecidable

Some Facts

  • Partially decidable or semi-decidable is broadly under the category of undecidable. For example halting problem is considered undecidable but is semi-decidable.
  • $P$, $NP$ and $NPC$ problems can all be decided by a $TM$ 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 $L$ are $\{aa, bb, aaaa, abab, baba, bbbb, aaaaaa, …\}$. We cannot accept these strings using an $NFA$. Now, even a $PDA$ is not possible as once we store $w$ on stack, it can only be read back in reverse order. Thus, we require 2 stacks to recognize $L$. Now, $L$ can be accepted by a $TM$ in linear space and hence $L$ is $CSL$.

2. $L = \{ww\mid w ∈ (a+b)^+\}$

Same explanation as above, $L$ is $CSL$.

3. $L = \{ww_R\mid w ∈ ({a+b})^*\}$

$ww_R$ can be accepted by a $PDA$ and hence is $CFL$. But we need a $NPDA$ for this as there is no deterministic way to identify where $w$ ends and $w_R$ starts. $wcw_R, w\in(a+b)^*$ is accepted by a $DPDA$ and hence is $DCFL$.

4. $L = \{ww_R\mid w ∈ ({a+b})^+\}$

Same explanation as above. $L$ is $CFL$.

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

$L$ is regular since $ L = \Sigma^*$, by making $x = (a+b)^*$ and $w = \epsilon$. i.e.; the set of strings generated by $L$ is $\{ \epsilon, a, b, aa, ab, ba, bb, aaa, …\} = \Sigma^*$

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. $L = \{wxw\mid w,x ∈ ({a+b})^+\}$

$L$ doesn’t contain all strings in $\Sigma^*$ as the strings like $abab$ are not contained in $L$. All words starting and ending in $a$ or starting and ending in $b$ are in $L$. But $L$ also contains words starting with $a$ and ending in $b$ like $abbab, aabbbabaab$ 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 $TM$ with linear space (this is at least as hard as accepting $ww, w \in (a+b)^*$), making $L$, a $CSL$.

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

$L$ is regular. Since, $w$ can be $\epsilon$ and $x \in (a+b)^*$, making $L =\Sigma^*$. i.e.; the set of strings generated by $L$ is $\{ \epsilon, a, b, aa, ab, ba, bb, aaa, …\} = \Sigma^*$

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.

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

The set of strings in $L$ are $\{aaa, aba, aaaa, aaba, abaa, abba, baab, …\}$ i.e.; $L$ contains all strings starting and ending with $a$ or starting and ending with $b$ and containing at least 3 letters. Moreover, $L$ doesn’t contain any other strings. Thus $L$ can be accepted by a finite automata making $L$ regular . Regular expression for $L$ is $a(a+b)^+a + b(a+b)^+b$.

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 $w = ab$ 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.

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

$L$ is regular. Since, $w$ can be $\epsilon$ and $x \in (a+b)^*$, making $L =\Sigma^*$. i.e.; the set of strings generated by $L$ is $\{ \epsilon, a, b, aa, ab, ba, bb, aaa, …\} = \Sigma^*$

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

Here, $w$ cannot be $\epsilon$ and hence to accept the string we do need the power of an LBA making $L$ a CSL.

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

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 $w$).

 

 

Loading