If L is a language defined by :
L = { awa | w ∈ {a, b}* },
is aa a string of the language L? (notice that w is being null string here)
According to you definition it seems L consists only of two words aba and aaa.
EDIT: after you have edited the question I can say yes "aa" is a word of this language
w ∈ {a, b}* means zero or more chars of this alphabet and thus w may have zero chars and be empty.
Your Problem can be restated as this.
L= a(a|b)*a
Which translates into intuition as "Strings that begin with and end with 'a'". So naturally, 'aa' is a valid string.
I hope this answers your question.
Related
I have a question regarding this question:
L= empty where the alphabet is {a,b}
how to create a grammar for this ? how can be the production rule ?
thanks in advance
A grammar G is an ordered 4-tuple {S, N, E, e, P} where:
N is a set of non-terminal symbols
E is a set of terminal symbols
N and E are disjoint
E is a superset of the alphabet of L(G)
e is the empty string
P is a set of ordered pairs of elements of (N U E U e); that is, P is a subset of (N U E U e) X (N U E U e)*.
S, the start symbol, is in N
A derivation in G is a sequence of elements of (N U E U e)* such that:
The first element is S
Adjacent elements w[i] and w[i+1] can be written as w[i] = uxv and w[i+1] = uyv such that (x, y) is in P
If there is a derivation in G whose last element is a string w[n] over (E U e)*, we say G generates w[n]; that is, w[n] is in L(G).
Now, we want to define a grammar G such that L(G) is the empty set. We fix the alphabet E = {a, b}. We must still define:
N, the set of nonterminals
S, the start symbol
P, the productions
We might as well take S as our start symbol. So N contains at least S; N is a superset of {S}. We will only add more nonterminals if we determine we need them. Let us turn our attention to the condition that L(G) is empty.
If L(G) is empty, that means there is no derivation in G that leads to a string of only terminal symbols. We can accomplish this easily be ensuring all our productions produce at least one nonterminal with any terminal. Or produce no terminals at all. So the following grammars would all work:
S := S
or
S := aSb
or
S := aXb | XXSSX
X := aabbXbbaaS
etc. All of these grammars have L(G) empty since none of them can derive a string of nonterminals.
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 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
Using pumping lemma, we can easily prove that the language L1 = {WcW^R|W ∈ {a,b}*} is not a regular language. (the alphabet is {a,b,c}; W^R represents the reverse string W)
However, If we replace character c with "x"(x ∈ {a,b}+), say, L2 = {WxW^R| x, W ∈ {a,b}^+}, then L2 is a regular language.
Could you give me some ideas?
If we replace character c with x where (x ∈ {a,b}+), say, L2 = {WXWR| x, W ∈ {a,b}+}, then L2 is a regular language.
Yes, L2 is Regular Language :).
You can write regular expression for L2 too.
Language L2 = {WXWR| x, W ∈ {a,b}+} means:
string should start any string consist of a and b that is W and end with reverse string WR.
notice: because W and WR are reverse of each other so string start and end with same symbol (that can be either a or b)
And contain any string of a and b in middle that is X. (because of +, length of X becomes greater than one |X| >= 1)
Example of this kind of strings can be following:
aabababa, as follows:
a ababab a
-- -------- --
w X W^R
or it can be also:
babababb, as follows:
b ababab b
-- -------- --
w X W^R
See length of W is not a constraint in language definition.
so any string WXWR can be assume equals to a(a + b)+a or b(a + b)+b
a (a + b)+ a
--- -------- ---
W X W^R
or
b (a + b)+ b
--- -------- ---
W X W^R
And Regular Expression for this language is: a(a + b)+a + b(a + b)+b
Don't mix WXWR with WCWR, its X with + that makes language regular. Think by including X that is (a + b)* we can have finite choice for W that is a and b (finite is regular).
Language WXWR can be say: if start with a ends with a and if start with b end with b. so correspondingly we need two final states.
Q6 if W is a
Q5 if W is b
ITs DFA is as given below.
Any string in the language with |W| > 1 can be interpreted as a string in the language where |W| = 1. Thus, a string is in the language if it begins and ends with the same symbol. There are two symbols: a and b. So that language is equivalent to the language a(a+b)(a+b)*a + b(a+b)(a+b)*b. To prove this, you should formalize the argument that "if y is in WxW, then y is in a(a+b)(a+b)*a + b(a+b)(a+b)*b; and if y is in a(a+b)(a+b)*a + b(a+b)(a+b)*b, then y is in WxW".
It doesn't work in the other case since c is a fixed symbol, and can't include all but the characters on the ends. As soon as you bound the length of "x" in your example, the language becomes non-regular.
The question says W ∈ {a,b}^+ , so a^n(a+b)a^n should be in the language L2. Now there is no such DFA that will accept the string a^n(a+b)a^n because, after accepting n number of a and (a+b)^+, there is no way for the dfa to remember exactly how many a it accepted in the begining, so L2 should not be regular.........But every where i search for this answer it says it is regular.....this bugs me
Here's a few questions I had on a quiz in a class and just want to verify their correctness.
Grammar:
S -> ABC
A -> df | epsilon
B -> f | epsilon
C -> g | epsilon
1.) The Follow set of B contains g and epsilon (T/F)? Ans: F.
There is no epsilon in the Follow sets, correct? (Only $ aka end of input)
2.) The First set of S contains d, f, g, and epsilon (T/F)? Ans: T.
I said false for this because I thought First(S) = First(A), which g is not a part of. Who is correct?
You are correct. If epsilon is involved, it will be accounted for in the First set, not the Follow set. If it's possible for the production to end the string, then $ goes in the Follow set, not epsilon.
The quiz is correct. The production S can indeed start with any of d, f, and g, and it can also be started by the empty string. Consider the input string g. It matches S, right? A is satisfied by the empty string, B is satisfied by the empty string, and C is satisfied by g. Since A, B, and C are all satisfied, S is satisfied. The first character consumed by S is g, so g must be in First(S).