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  1. Let $G_1 = (N, T, P, S_1)$ be a CFG where, $N=\{S_1, A, B\},T=\{a, b\}$ and $P$ is given by$$\begin{array}{l|l}
    S_1 \rightarrow a S_1 b &S_1 \rightarrow a B b \\
    S_1 \rightarrow a A b & B \rightarrow Bb\\
    A \rightarrow a A & B \rightarrow b\\
    A \rightarrow a
    \end{array}$$What is $L(G_1)$?
  2. Use the grammar in Part(a) to give a CFG for $L_2 = \{a^ib^ja^kb^l \mid i, j, k, l \geq 1, i=j \text{ or } k=l\}$ by adding not more than $5$ production rules.

  3. Is $L_2$ inherently ambiguous?

in Compiler Design by Veteran (52.1k points)
edited by | 769 views

3 Answers

+8 votes
Best answer

$(a)$ $L(G_1) = \{a^nb^m \mid n \neq m\}$

$(b)$

  • $S\rightarrow S_1S_2\mid S_2S_1 \qquad (2)$    
  • $S_2\rightarrow aS_2b \mid \epsilon \qquad (2)$
  • $S_1 \rightarrow \epsilon\qquad (1)$ 

So, totally $2+2+1 = 5$ extra productions. 

$(c)$ If grammar is ambiguous and no unambiguous grammar is possible for the language, then language is inherently  ambiguous.

Non-determinism of language (if it means, $L$ is not DCFL) + ambiguity for all possible grammars implies inherent ambiguity. Or if a language is deterministic, it is surely not inherently ambiguous. But if a language is not deterministic, it may or may not be inherently ambiguous. 

The given language $L_2$ is the set of strings where number of $a's$ is followed by an equal number of $b's$ and if not, for the remaining part, the number of $a's$ is followed by an equal number of $b's$. This is actually a deterministic language as we do not need to do multiple counts here. Only after the first condition is violated we need to start doing the count for the second part. This makes the language deterministic and hence it cannot be inherently ambiguous. 

by Veteran (56.4k points)
selected by
+3
Yes, non-determinism does not imply "inherent ambiguity". But determinism do imply no "inherent ambiguity". So, how to formally prove a language to be inherently ambiguous?
0

First case : language is not deterministic.

Second case: Grammar for language is Ambiguous, (that is undecidable).

0
But is it enough? I mean if an ambiguous grammar exist for a CFL which is not deterministic, that makes it inherently ambiguous?
+5

No, it is not enough. as per definition , we need to say every possible grammar for language is ambiguous. Intuitively we can say, that every grammar will contain some set of productions that will be lead to ambiguity. 

Afaik, no formal way to prove that. 

0

@Praveen Saini  as mentioned in the question, for language L2, $i,j,k,l>=1$ but the grammar given by you can generate null string, right?

+1
Instead of $S_1 \to \epsilon$ and  $S_2 \to \epsilon$, productions should be  $S_1 \to ab$ and  $S_2 \to ab$ as $i,j,k,l >= 1$

The smallest string in the language will be $abab$.
0
how to solve such questions. Please tell any source for learning and practising
0

@Manoja Rajalakshmi A ma'am same doubt

+2 votes

a)L(G1)={aibj |i,j>1, i>j or j>i}

b)S ->S1S2 | S2S1

S1 -> aS1b | ab

S2 -> S3S4

S3 -> aS3 | a

S4 -> bS4 | b

c) Inherently Ambiguous grammar means there will not be any unambiguity in that grammar

But here we can generate aabbaaab ,aabbbaabb..........these grammar which are unambiguous in nature

So, it is not inherently ambiguous grammar

by Veteran (113k points)
edited by
0
how do we prove the c) part like is there any shortcut for it??
0
also in b) production rules are more than 5
+1 vote
b)

S-->  S1S2 / S2S1

S1--> aS1b / ϵ

S2-->AB

A-->aA/a

B-->bB/b

just added 4 new extra  productions
by Boss (16.5k points)

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