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I know the grammar for the language $\{ xy : |x| = |y|, x ≠ y \}$ if $\Sigma=\{a,b\}$:

$$ \begin{align*} &S→AB∣BA \\ &A→a∣aAa∣aAb∣bAa∣bAb \\ &B→b∣aBa∣aBb∣bBa∣bBb \end{align*} $$

I know this is a grammar, but I need a PDA for this language, and intuition how $\{xy: |x|=|y|,x \neq y\}$ is a CFL while $\{xy: |x|=|y|,x=y\}$ is a CSL but not a CFL. How is this possible?

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  • $\begingroup$ How do you derive $aaab$? Oh, damn my description was imprecise. I meant what if $x$ and $y$ have even length? Also, you can generate words of non-equal length using $A$ and $B$. $\endgroup$ – ThreeFx Dec 2 '16 at 15:33
  • $\begingroup$ Regarding the distinction between your problem (which is a CFL) and the similar problem (which is a CSL): In your case you accept a word when you found a non-equal pair of characters. For a non-valid input the PDA will potentially loop forever. If you require the words to be equal, you have to check the condition for every possible pair $x_i$ and $y_i$ and remember if you have already checked $i$. Since a PDA cannot do this, it will have to loop forever on correct inputs and you cannot distinguish between a valid input looping and an invalid input looping. $\endgroup$ – ThreeFx Dec 2 '16 at 16:09
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    $\begingroup$ Please ask only one question per post. For your first question, what have you tried? A construction for converting a grammar to a PDA is found in almost every textbook on automata theory; there is little reason for us to laboriously repeat it here. We expect you to do research on your own to try to answer your own question before asking here. $\endgroup$ – D.W. Dec 2 '16 at 16:54
  • $\begingroup$ Yeah i know the algorithm and there are many simulators too. . . But i just want the intuition like the language of equal number of a's and b's can be done by pushing a symbol for a's and popping symbol for b's. . so for this type of intuition i posted it here. And i will ask only one question per post $\endgroup$ – Pavan Kumar Munnam Dec 2 '16 at 17:42
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    $\begingroup$ I second the point made by Yuval and D.W.: going from CFG to NPDA is an algorithmic task with well-established solutions. That does, however, not provide any insight. Here are two hints: 1) Check even length and $x \neq y$ separately. 2) Use non-determinism to "find" a position where $x$ and $y$ differ. $\endgroup$ – Raphael Dec 3 '16 at 20:00
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You are asking two questions: how to construct a PDA for the language, and why this language is context-free while the same language with the condition $x \neq y$ replaced by the condition $x = y$, is not. I will only answer the second, since for the first question there are known algorithms.

The reason that your language is context-free is that we can rewrite it as follows: $$ \{ \Sigma^i a \Sigma^i \Sigma^j b \Sigma^j : i,j \neq 0 \} \cup \{ \Sigma^i b \Sigma^i \Sigma^j a \Sigma^j : i,j \neq 0 \}. $$ This gives a different description of the language as the union of concatenations of context-free languages. A similar trick just doesn't work for the language $\{ xy : x=y \}$.

Here is a different example. Consider the following two collections of natural numbers (without zero):

  • $A = \{ x \cdot y : x \neq y \}$.
  • $B = \{ x \cdot y : x = y \}$.

The set $A$ consists of all natural numbers other than $1$, whereas the set $B$ consists of all squares. Even though we defined them in a similar way, the set $A$ has a much simpler description, while $B$ doesn't.

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  • $\begingroup$ i was thinking like this , correct it if i am thinking wrong. I was assuming a deterministic machine with different states so for example of input size 3 there will be 3 different states where x(i) will be compared to y(i) for 0<i<4 so these three states cannot be simulated at the same instant, but all can be simulated using one stack at different instances , so if one case fails we can say unequal , but for it to be equal it should be match in all cases , so equality cannot be generated . am i thinking right? $\endgroup$ – Pavan Kumar Munnam Dec 2 '16 at 17:49
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    $\begingroup$ Showing that a language isn't context-free requires proof. Intuition isn't enough. $\endgroup$ – Yuval Filmus Dec 2 '16 at 17:52
  • $\begingroup$ I know that it is cfl , I was talking intuition about how npda works $\endgroup$ – Pavan Kumar Munnam Dec 2 '16 at 17:56

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