In the Chomsky classification of formal languages, I need some examples of Non-Linear, Unambiguous and also Non-Deterministic Context-Free-Language(N-CFL)?
Linear Language: For which Linear grammar is possible( ⊆ CFG) e.g.
L1 = {anbn | n ≥ 0 }
Deterministic Context Free Language(D-CFG): For which Deterministic Push-Down-Automata(D-PDA) is possible e.g.
L2 = {anbncm | n ≥ 0, m ≥ 0 }
L2 is unambiguous.
A CF grammar that is not linear is nonlinear.
Lnl = {w: na(w) = nb(w)} is also a Non-Linear CFG.
--
3.
Non-Deterministic Context Free Language(N-CFG): For which only Non-Deterministic Push-Down-Automata(N-PDA) is possible e.g.
L3 = {wwR | w ∈ {a, b}* }
L3 is also Linear CFG.
--4. Ambiguous CFL: CFL for which only ambiguous CFG is possible
L4 = {anbncm | n ≥ 0, m ≥ 0 } U {anbmcm | n ≥ 0, m ≥ 0 }
L4 is both non-linear and Ambiguous CFG And Every Ambigous CFL \subseteq N-CFL.
My Question is:
Whether all non-linear, Non-Deterministic CFL are Ambiguous? If not then
I need a example that is non-linear, non-deterministic CFL and also unambiguous?
Given Venn-diagram below:
Also asked here
"IN CONTEXT OF CHOMSHY CLASSIFICATION OF FORMAL LANGUAGE"
(1) L3 = {wwR | w ∈ {a, b}* }
Language L3 is a Non-Deterministic Context Free Language, its also Unambiguous and Liner language.
(2) Lp is language of parenthesis matching. There are two terminal symbols "(" and ")".
Grammar for Lp is:
S → SS
S → (S)
S → ()
This language is nonlinear, deterministic and unambiguous too.
Language L that is union of Lp and L3 is unambiguous, nonlinear (due to Lp), and non-deterministic (due to L3) (Assuming language symbols for both languages are different).
This Language is an example of language in Venn-diagram for which I marked ??.
Also the correct diagram is below:
An ambiguous context free language also be a liner context free
Related
I find difficulties in constructing a Grammar for the language especially with linear grammar.
Can anyone please give me some basic tips/methodology where i can construct the grammar for any language ? thanks in advance
I have a doubt whether the answer for this question "Construct a linear grammar for the language: is right
L ={a^n b c^n | n belongs to Natural numbers}
Solution:
Right-Linear Grammar :
S--> aS | bA
A--> cA | ^
Left-Linear Grammar:
S--> Sc | Ab
A--> Aa | ^
As pointed out in the comments, these grammars are wrong since they generate strings not in the language. Here's a derivation of abcc in both grammars:
S -> aS -> abA -> abcA -> abccA -> abcc
S -> Sc -> Scc -> Abcc -> Aabcc -> abcc
Also as pointed out in the comments, there is a simple linear grammar for this language, where a linear grammar is defined as having at most one nonterminal symbol in the RHS of any production:
S -> aSc | b
There are some general rules for constructing grammars for languages. These are either obvious simple rules or rules derived from closure properties and the way grammars work. For instance:
if L = {a} for an alphabet symbol a, then S -> a is a gammar for L.
if L = {e} for the empty string e, then S -> e is a grammar for L.
if L = R U T for languages R and T, then S -> S' | S'' along with the grammars for R and T are a grammar for L if S' is the start symbol of the grammar for R and S'' is the start symbol of the grammar for T.
if L = RT for languages R and T, then S = S'S'' is a grammar for L if S' is the start symbol of the grammar for R and S'' is the start symbol of the grammar for T.
if L = R* for language R, then S = S'S | e is a grammar for L if S' is the start symbol of the grammar for R.
Rules 4 and 5, as written, do not preserve linearity. Linearity can be preserved for left-linear and right-linear grammars (since those grammars describe regular languages, and regular languages are closed under these kinds of operations); but linearity cannot be preserved in general. To prove this, an example suffices:
R -> aRb | ab
T -> cTd | cd
L = RT = a^n b^n c^m d^m, 0 < a,b,c,d
L' = R* = (a^n b^n)*, 0 < a,b
Suppose there were a linear grammar for L. We must have a production for the start symbol S that produces something. To produce something, we require a string of terminal and nonterminal symbols. To be linear, we must have at most one nonterminal symbol. That is, our production must be of the form
S := xYz
where x is a string of terminals, Y is a single nonterminal, and z is a string of terminals. If x is non-empty, reflection shows the only useful choice is a; anything else fails to derive known strings in the language. Similarly, if z is non-empty, the only useful choice is d. This gives four cases:
x empty, z empty. This is useless, since we now have the same problem to solve for nonterminal Y as we had for S.
x = a, z empty. Y must now generate exactly a^n' b^n' b c^m d^m where n' = n - 1. But then the exact same argument applies to the grammar whose start symbol is Y.
x empty, z = d. Y must now generate exactly a^n b^n c c^m' d^m' where m' = m - 1. But then the exact same argument applies to the grammar whose start symbol is Y.
x = a, z = d. Y must now generate exactly a^n' b^n' bc c^m' d^m' where n' and m' are as in 2 and 3. But then the exact same argument applies to the grammar whose start symbol is Y.
None of the possible choices for a useful production for S is actually useful in getting us closer to a string in the language. Therefore, no strings are derived, a contradiction, meaning that the grammar for L cannot be linear.
Suppose there were a grammar for L'. Then that grammar has to generate all the strings in (a^n b^n)R(a^m b^m), plus those in e + R. But it can't generate the ones in the former by the argument used above: any production useful for that purpose would get us no closer to a string in the language.
I'm in a Formal languages class and have a grammar quiz coming up. I'm assuming something like this will appear.
Consider the alphabet ∑ = {a, b, c}. Construct a grammar that generates the language L = {bab^nabc^na^p : n ≥ 0, p ≥ 1}. Assume that the start variable is S.
It was a very long time since I worked with formal languages for the last time, so, please, forgive my rustyness, but this would be the language: We divide S to a prefix variable (A) and a suffix variable (B). Then, we handle the prefix and the suffix separately, both of them have a possible rule of further recursion, and an end sign of empty, where no occurrence is required and the constant where at least an occurrence is required.
{bab^nabc^na^p : n ≥ 0, p ≥ 1}
S -> ASB
A -> babAabc
A -> {empty}
B -> Ba
B -> a
I'm trying to understand a concept with respect to grammar and Production Rules.
According to most material on this subject:
1) Epsilon production rules are only allowable if they do not appear on the RHS of any other production rule.
However, taking a grammar:
G = { T,N,P,S }
Where:
T = {a,b}
N = {S,S1}
S = {S}
P {
S -> aSb
S -> ab
S1 -> SS1
S1 -> E //Please note, using E to represent Epsilon.
}
Where, the language of the grammar is:
L(G) = { a^n, b^n | n >= 1 }
In this case, a production rule containing Epsilon exists (derived from S1) but S1 also forms part of a RHS of another production rule (S1 -> SS1).
Doesn't this violate point 1?
Your statement:
Epsilon production rules are only allowable if they do not appear on the RHS of any other production rule.
would be better stated as
A non-terminal may have an epsilon production rules if that non-terminal does not appear on the right-hand side of any production rule.
In Chomsky's original hierarchy, epsilon productions were banned for all but Type 0 (unrestricted) grammars. If all epsilon productions are banned, then it is impossible for the grammar to produce the empty string. I believe this was not a concern for Chomsky; consequently, most modern formulations allow the start symbol to have an empty right-hand side as long as the start symbol itself does not appear on the right-hand side of any production.
As it happens, the restriction on epsilon-productions is somewhat stronger than is necessary. In the case of both context-free grammars and regular grammars (Chomsky type 2 and type 3 grammars), it is always possible to create a weakly-equivalent grammar without epsilon productions (except possibly the single production S → ε if the grammar can produce the empty string.) It is also possible to remove a number of other anomalies which complicate grammar analysis: unreachable symbols, unproductive symbols, and cyclic productions. The result of the combination of all these eliminations is a "proper context-free grammar".
Consequently, most modern formulations of context-free grammars do not require the right-hand sides to be non-empty.
Your grammar G = {T, N, S, P} with
T = {a, b}
N = {S, S1}
S = {S}
P {
S → a S b
S → a b
S1 → S S1
S1 → ε
}
contains an unreachable symbol, S1. We can easily eliminate it, producing the equivalent grammar G' = { T, N', S, P' }:
N' = {S}
P' {
S → a S b
S → a b
}
G' does not contain any epsilon productions (but even if it had, they could have been eliminated).
I'm trying to get my head around context free grammars and I think I'm close. What is baffling me is this one question (I'm doing practise questions as I have an exam in a month's time):
I've come up with this language but I believe it's wrong.
S --> aSb | A | B
A --> aA | Σ
B --> bB | Σ
Apparently this is the correct solution:
S --> aSb | aA | bB
A --> aA | Σ
B --> bB | Σ
What I don't quite understand is why we have S --> aSb | aA | bB and not just S --> aSb | A | B. What is the need for the terminals? Can't I just call A instead and grab my terminals that way?
Testing to see if I can generate the string: aaabbbb
S --> aSb --> aaSbb --> aaaSbbb --> aaaBbbb --> aaabbbb
I believe I generate the string correctly, but I'm not quite sure. I'm telling myself that the reason for S --> aSb | aA | bB is that if we start with aA and then replace A with a, we have two a's which gives us our correct string as they're not equal, this can be done with b as well. Any advice is greatly appreciated.
Into the Tuple (G-4-tuple)
V (None terminals) = {A, B}
Σ (Terminals) = {a, b}
P = { } // not quite sure how to express my solution in R? Would I have to use a test string to do so?
S = A
First:
Σ means language symbols. in your language Σ = {a, b}
^ means null symbols (it is theoretical, ^ is not member of any language symbol)
ε means empty string (it is theoretical, ε can be a member of some language)
See ^ symbol means nothing but we use it just for theoretical purpose, like ∞ infinity symbol we uses in mathematics(really no number is ∞ but we use it to understand, to proof some theorems) similarly ^ is nothing but we use it.
this point not written in any book, I am writing it to explain/for understanding point of view. The subject more belongs to theoretical and math and I am from computer science.
As you says your grammar is L = {am bn | m != n}. Suppose if productions are as follows:
First:
S --> aSb | A | B
A --> aA | Σ
B --> bB | Σ
It means.(very rare book may use Σ in grammar rules)
S --> aSb | A | B
A --> aA | a | b
B --> bB | a | b
I replaced Σ by a | b (a, b language symbols).
This grammar can generates a string of equal numbers of symbols a and symbol b(an bn). How it can generate an bn? See below an example derivation:
S ---> aSb ---> aAb ---> aaAb ---> aabb
^ ^ ^ ^
rule-1 S-->A A--> aA A --> b
But these kind of strings are not possible in language L because m != n.
Second:
For the same reason production rules S --> aSb | aA | bB is also not correct grammar if A --> aA | Σ or B --> bB | Σ are in grammar.
I think in second grammar you mean:
S --> aSb | aA | bB
A --> aA | ^
B --> bB | ^
Then this is correct grammar for language L = {am bn | m != n}. Because using:
S --> aSb
you can only generate equal numbers of a' and b and by replacing S either by aA or by bB you make a sentential form in which unequal numbers of a and b symbols are present and that can't convert back to generate a string of type an bn. (since A doesn't generates b and B doesn't generates a).
Third:
But usually we write grammar rules like:
S --> aSb | A | B
A --> aA | a
B --> bB | b
Both forms are equivalent (generate same language L = {am bn | m != n}) because once you convert S into either A or B you have to generate at-least one a or b (or more) respectively and thus constraint m != n holds.
Remember proofing, whether two grammars are equivalent or not is undecidable problem. We can't prove it by algorithm (but logically possible, that works because we are human being having brain better then processor :P :) ).
Fourth:
At the end I would also like to add, Grammar:
S --> aSb | A | B
A --> aA | ^
B --> bB | ^
doesn't produces L = {am bn | m != n} because we can generate an bn for example:
S ---> aSb ---> aAb ---> ab
^
A --> ^
Grammar in formal languages
Any class of formal languages can be represented by a formal Grammar consisting of the four-tuple (S, V, Σ, P). (note a Grammar or an automata both are finite representation weather language is finite or infinite: Check figures one & two).
Σ: Finite set of language symbols.
In grammar we commonly call it finite set of terminals (in contrast of variables V). Language symbols or terminals are thing, using which language strings (sentences) are constructed. In your example set of terminals Σ is {a, b}. In natural language you can correlate terminals with vocabulary or dictionary words.
Natural language means what we speak Hindi, English
V: Finite set of Non-terminals.
Non-terminal or say 'variable', should always participate in grammar production rules. (otherwise the variable counts in useless variables, that is a variable that doesn't derives terminals or nothing).
See: 'ultimate aim of grammar is to produce language's strings in correct form hence every variable should be useful in some way.
In natural language you can correlate variable set with Noun/Verbs/Tens that defined a specific semantical property of an language (like Verb means eating/sleeping, Noun means he/she/Xiy etc).
Note: One can find in some books V ∩ Σ = ∅ that means variables are not terminals.
S: Start Variable. (S ∈ V)
S is a special variable symbol, that is called 'Start Symbol'. We can only consider a string in language of grammar L(G) if it can be derived from Start variable S. If a string can not be derived from S (even if its consist of language symbols Σ) then string will not be consider in the language of grammar( actually that string belongs to 'complement language' of L(G), we writes complement language L' = Σ* - L(G) , Check: "the complement language in case of regular language")
P: Finite set of Production Rules.
Production Rules defines replacement rules in the from α --> β, that means during the derivation of a string from S, from grammar rules at any time α (lhs) can be replaced by β (rhs).(this is similar to Noun can be replace by he,she or Xiy, and Verb can be replace by eating, sleeping etc in natural language.
Production rules defines formation rules of language sentences. Formal language are similar to Natural language having a pattern that is certain thing can occurs in certain form--that we call syntax in programming language. And because of this ability of grammar, grammar use for syntax checking called parse).
Note: In α --> β, α and β both are consists of language symbols and terminals (V U Σ)* with a constraint that in α their must be at-least one variable. (as we can replace only a string contain variable by rhs of rule. a terminal can't replace by other terminal or we can say a sentence can't be replaced by other sentence)
Remember: There is two form Sentential Form and Sentence of a string:
Sentence: if all symbols are terminals (sentence can be either in L(G) or in complement language L' = Σ* - L)
Sentential: if any symbol is variable (not a language string but derivation string)
From #MAV (Thanks!!):
To represent grammar of above language L = {am bn | m != n}, 4-tuple are :
V = {S, A, B}
Σ = {a, b}
P = {S --> aSb | A | B, A --> aA | a, B --> bB | a}
S = S
note: Generally I use P for Production rules, your book may use R for rules
Terminology uses in theory of formal languages and automate
Capital letters are uses for variables e.g. S, A, B in grammar construction.
Small letter from start uses for terminals(language symbols) for example a, b.
(some time numbers like 0, 1 uses. Also ^ is null symbol).
Small letters form last uses for string of terminals z, y, w, x (for example you can find these notations in pumping lemma,
symbols use for language string or sub strings).
α, β, γ for Sentential forms.
Σ for language symbols.
Γ for input or output tap symbol, other then language symbols.
^ for null symbol, # or ☐ Symbol for blank symbol in Turing machine and PDA (^, #, ☐ are other then language symbols.
ε uses for empty string (can be a part of language string for example { } is empty body in C language, you can write while(1); or
while(1){ } both are valid see here I have defined a valid program
with empty sentences).
∅ means empty set in set theory.
Φ, Ψ uses for substring in Sentential forms.
Note: ∅ means set is empty, ε means string is empty, ^ means none symbol (don't mix in theory, all are different in semantic)
There is no rules I know about symbol notation, but these are commonly used terminology once can find in most standard books I observed during study.
Next post: Tips for writing Context free grammar
Is it possible to come up with a linear grammar with unequal number of 0s and 1s?
Such as 0100, 01100, 111,1,0, 100101001...
I know there is a context-free grammar for this, but is there a linear grammar?
Thanks.
A grammar is regular if and only if it is either left regular or right regular. The left regular grammars are equivalent to the left linear grammars. The right regular grammars are equivalent to the right linear grammars. Therefore, if a regular grammar exists that generates the indicated language, then it is either right or left regular, and hence equivalent to either a left or right linear grammar.
edit1:
Note that there's no regular grammar generating the indicated language LUNEQ. To see this, consider the fact that LEQ = { w : na(w) = nb(w)} is the complement of LUNEQ. Because the regular languages are closed under complementation and LEQ is not a regular language, LUNEQ is not a regular language.
edit2:
I believe the pumping lemma for linear languages can be used to show that the indicated language LUNEQ is not linear. Here is what I've come up with. I'm fairly confident it's correct. My primary concern is that you were asked - presumably - for a linear language generating the indicated language; however, I came to the conclusion that there is no such grammar.
Assume LUNEQ is linear. By the pumping lemma for linear languages, there exists an n > 0 depending on LUNEQ such that for all z ∈ LUNEQ, z can be written uvwxy where:
|vx| > 0,
|uvxy| ≤ n, and
uviwxiy ∈ LUNEQ for all i ≥ 0.
Let n be the constant guaranteed be the pumping lemma. Consider the string
z = anb(n! + 2n)an
Since z ∈ LUNEQ, it can be decomposed into substrings uvwxy satisfying the constraints of the pumping lemma such that, for all i ≥ 0, the string
uviwxiy = a|u|ai|v|a(n - |u| - |v|)b(n! + 2n)a(n - |x| - |y|)ai|x|a|y|
is a member of LUNEQ. Since 1 ≤ |vx| ≤ n, |vx| divides n!. Hence, (n!|vx|-1 + 1) is a natural number. Setting i to (n!|vx|-1 + 1) gives the string
z' = uv(n!|vx|-1 + 1)wx(n!|vx|-1 + 1)y = a|u|a(n!|vx|-1 + 1)|v|a(n - |u| - |v|)b(n! + 2n)a(n - |x| - |y|)a(n!|vx|-1 + 1)|x|a|y|
Simplifying the pumped string gives us an equal number of a's and b's:
na(z') = 2n - |vx| + (n!|vx|-1 + 1)|vx| = 2n + n!
Since (2n + n!) is equivalent to the number of b's in the pumped string, z' ∉ LUNEQ. But this contradicts the assumption that LUNEQ is a linear language. Hence, LUNEQ is not a linear language.