Context-Free-Language
Question 1 |
Consider the following languages:
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I. {ambncpdq ∣ m + p = n + q, where m, n, p, q ≥ 0}
II. {ambncpdq ∣ m = n and p = q, where m, n, p, q ≥ 0}
III. {ambncpdq ∣ m = n = p and p ≠ q, where m, n, p, q ≥ 0}
IV. {ambncpdq ∣ mn = p + q, where m, n, p, q ≥ 0}
Which of the above languages are context-free?
I and IV only | |
I and II only
| |
II and III only | |
II and IV only |
Question 1 Explanation:
i) am bn cp dq | m+p = n+q,
m-n = q-p
First we will push a’s in the stack then we will pop a’s after watching b’s.
Then some of a’s might left in the stack.
Now we will push c’s in the stack and then pop c’s by watching d’s.
And then the remaining a’s will be popped off the stack by watching some more d’s.
And if no d’s is remaining and the stack is empty then the language is accepted by CFG.
ii) am bn cp dq | m=n, p=q
Push a’s in stack. Pop a’s watching b’s.
Now push c’s in stack.
Pop c’s watching d’s.
So it is context free language.
iii) am bn cp dq | m=n=p & p≠q
Here three variables are dependent on each other. So not context free.
iv) Not context free because comparison in stack can’t be done through multiplication operation.
m-n = q-p
First we will push a’s in the stack then we will pop a’s after watching b’s.
Then some of a’s might left in the stack.
Now we will push c’s in the stack and then pop c’s by watching d’s.
And then the remaining a’s will be popped off the stack by watching some more d’s.
And if no d’s is remaining and the stack is empty then the language is accepted by CFG.
ii) am bn cp dq | m=n, p=q
Push a’s in stack. Pop a’s watching b’s.
Now push c’s in stack.
Pop c’s watching d’s.
So it is context free language.
iii) am bn cp dq | m=n=p & p≠q
Here three variables are dependent on each other. So not context free.
iv) Not context free because comparison in stack can’t be done through multiplication operation.
Question 2 |
Consider the context-free grammars over the alphabet {a, b, c} given below. S and T are non-terminals.
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G1: S → aSb|T, T → cT|ϵ
G2: S → bSa|T, T → cT|ϵ
The language L(G1) ∩ L(G2) is
Finite | |
Not finite but regular | |
Context-Free but not regular | |
Recursive but not context-free |
Question 2 Explanation:
Strings generated by G1:
{ϵ, c, cc, ccc, … ab, aabb, aaabbb….acb, accb… aacbb, aaccbb, …}
Strings generated by G2:
{ϵ, c, cc, ccc, … ba, bbaa, bbbaaa….bca, bcca… bbcaa, bbccaa, …}
The strings common in L (G1) and L (G2) are:
{ϵ, c, cc, ccc…}
So, L (G1) ∩ L (G2) can be represented by regular expression: c*
Hence the language L (G1) ∩ L (G2) is “Not finite but regular”.
{ϵ, c, cc, ccc, … ab, aabb, aaabbb….acb, accb… aacbb, aaccbb, …}
Strings generated by G2:
{ϵ, c, cc, ccc, … ba, bbaa, bbbaaa….bca, bcca… bbcaa, bbccaa, …}
The strings common in L (G1) and L (G2) are:
{ϵ, c, cc, ccc…}
So, L (G1) ∩ L (G2) can be represented by regular expression: c*
Hence the language L (G1) ∩ L (G2) is “Not finite but regular”.
Question 3 |
Consider the following languages over the alphabet Σ = {a, b, c}.
Let L1 = {an bn cm│m,n ≥ 0} and L2 = {am bn cn│m,n ≥ 0}
Which of the following are context-free languages?
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I. L1 ∪ L2
II. L1 ∩ L2
I only | |
II only | |
I and II | |
Neither I nor II |
Question 3 Explanation:
Strings in L1 = {ϵ, c, cc, …., ab, aabb,…., abc, abcc,…, aabbc, aabbcc, aabbccc,..}
Strings in L2 = {ϵ, a, aa, …., bc, bbcc,…., abc, aabc,…, abbcc, aabbcc, aaabbcc,..}
Strings in L1 ∪ L2 ={ϵ, a, aa, .., c, cc,.. ab, bc, …, aabb, bbcc,.., abc, abcc, aabc,…}
Hence (L1 ∪ L2) will have either (number of a’s = equal to number of b’s) OR (number of b’s = number of c’s).
Hence (L1 ∪ L2) is CFL.
Strings in L1 ∩ L2 = {ϵ, abc, aabbcc, aaabbbccc,…}
Hence (L1 ∩ L2) will have (number of a’s = number of b’s = number of c’s)
i.e., (L1 ∩ L2) = {anbncn | n ≥ 0} which is CSL.
Strings in L2 = {ϵ, a, aa, …., bc, bbcc,…., abc, aabc,…, abbcc, aabbcc, aaabbcc,..}
Strings in L1 ∪ L2 ={ϵ, a, aa, .., c, cc,.. ab, bc, …, aabb, bbcc,.., abc, abcc, aabc,…}
Hence (L1 ∪ L2) will have either (number of a’s = equal to number of b’s) OR (number of b’s = number of c’s).
Hence (L1 ∪ L2) is CFL.
Strings in L1 ∩ L2 = {ϵ, abc, aabbcc, aaabbbccc,…}
Hence (L1 ∩ L2) will have (number of a’s = number of b’s = number of c’s)
i.e., (L1 ∩ L2) = {anbncn | n ≥ 0} which is CSL.
Question 4 |
Let L1, L2 be any two context-free languages and R be any regular language. Then which of the following is/are CORRECT?
I, II and IV only | |
I and III only | |
II and IV only | |
I only |
Question 4 Explanation:
Since CFL is closed under UNION so L1 ∪ L2 is CFL, is a true statement.
CFL is not closed under complementation.
So L1 compliment may or may not be CFL. Hence
is Context free, is a false statement.
L1 – R means
and Regular language is closed under compliment, so
is also a regular language, so we have now L1 ∩ R .
Regular language is closed with intersection with any language, i.e. L∩R is same type as L.
So L1∩R is context free.
CFL is not closed under INTERSECTION, so L1 ∩ L2 may or may not be CFL and hence IVth is false.
CFL is not closed under complementation.
So L1 compliment may or may not be CFL. Hence
is Context free, is a false statement.L1 – R means
and Regular language is closed under compliment, so is also a regular language, so we have now L1 ∩ R .Regular language is closed with intersection with any language, i.e. L∩R is same type as L.
So L1∩R is context free.
CFL is not closed under INTERSECTION, so L1 ∩ L2 may or may not be CFL and hence IVth is false.
Question 5 |
Consider the following languages:
-
L1 = {ap│p is a prime number}
L2 = {an bm c2m | n ≥ 0, m ≥ 0}
L3 = {an bn c2n │ n ≥ 0}
L4 = {an bn │ n ≥ 1}
Which of the following are CORRECT?
-
I. L1 is context-free but not regular.
II. L2 is not context-free.
III. L3 is not context-free but recursive.
IV. L4 is deterministic context-free.
I, II and IV only | |
II and III only | |
I and IV only | |
III and IV only |
Question 5 Explanation:
L1 = {ap│p is a prime number} is a context sensitive language. So I is false.
L2 = {an bm c2m│n ≥ 0, m ≥ 0} is a context free as we have to do only one comparison (between b’s and c’s) which can be done by using PDA, so L2 is Context free and II is true.
L3 = {an bn c2n│n≥0} is context sensitive.
The reason it has more than one comparison (at a time) as we have to compare number of a’s, b’s and c’s.
So this cannot be done using PDA. Hence III is CSL and every CSL is recursive, so III is True
L4 = {an bn│n ≥ 1} is Context free (as well as Deterministic context free).
We can define the transition of PDA in a deterministic manner.
In beginning push all a’s in stack and when b’s comes pop one “a” for one “b”.
If input and stack both are empty then accept.
L2 = {an bm c2m│n ≥ 0, m ≥ 0} is a context free as we have to do only one comparison (between b’s and c’s) which can be done by using PDA, so L2 is Context free and II is true.
L3 = {an bn c2n│n≥0} is context sensitive.
The reason it has more than one comparison (at a time) as we have to compare number of a’s, b’s and c’s.
So this cannot be done using PDA. Hence III is CSL and every CSL is recursive, so III is True
L4 = {an bn│n ≥ 1} is Context free (as well as Deterministic context free).
We can define the transition of PDA in a deterministic manner.
In beginning push all a’s in stack and when b’s comes pop one “a” for one “b”.
If input and stack both are empty then accept.
Question 6 |
Consider the following languages:
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L1 = {an bm cn+m : m,n ≥ 1}
L2 = {an bn c2n : n ≥ 1}
Which one of the following isTRUE?
Both L1 and L2 are context-free. | |
L1 is context-free while L2 is not context-free. | |
L2 is context-free while L1 is not context-free. | |
Neither L1 nor L2 is context-free. |
Question 6 Explanation:
L1 can be recognized by PDA, we have to push a’s and b’s in stack and when c’s comes then pop every symbol from stack for each c’s.
At the end if input and stack is empty then accept.
Hence, it is CFL.
But L2 can’t be recognized by PDA, i.e. by using single stack.
The reason is, it has two comparison at a time,
1st comparison:
number of a’s = number of b’s
2nd comparison:
number of c’s must be two times number of a’s (or b’s)
It is CSL.
At the end if input and stack is empty then accept.
Hence, it is CFL.
But L2 can’t be recognized by PDA, i.e. by using single stack.
The reason is, it has two comparison at a time,
1st comparison:
number of a’s = number of b’s
2nd comparison:
number of c’s must be two times number of a’s (or b’s)
It is CSL.
Question 7 |
Which of the following languages are context-free?
L1 = {ambnanbm ⎪ m, n ≥ 1}
L2 = {ambnambn ⎪ m, n ≥ 1}
L3 = {ambn ⎪ m = 2n + 1}
L1 and L2 only | |
L1 and L3 only | |
L2 and L3 only | |
L3 only |
Question 7 Explanation:
L1: First push all the a's in the stack then push all the b's in the stack. Now pop all the b's from the stack watching next no. of a's. And then pop all the a's from the stack watching next no. of b's. So can be done by PDA, hence CFL.
L2: First push all the a's in the stack then push all the b's in the stack. Now again a's come which cannot be compared by previous a's in the stack because at top of the stack's there are b's which is also needed to be pushed for further comparision with the next b's. So not CFL.
L3: First simply read one 'a', then push one 'a' in the stack after reading two a's and then pop all the a's by reading the b's. Since can be done by PDA hence CFL.
L2: First push all the a's in the stack then push all the b's in the stack. Now again a's come which cannot be compared by previous a's in the stack because at top of the stack's there are b's which is also needed to be pushed for further comparision with the next b's. So not CFL.
L3: First simply read one 'a', then push one 'a' in the stack after reading two a's and then pop all the a's by reading the b's. Since can be done by PDA hence CFL.
Question 8 |
Consider the languages
L1 = {0i1j | i != j}.
L2 = {0i1j | i = j}.
L3 = {0i1j | i = 2j+1}.
L4 = {0i1j | i != 2j}.
Which one of the following statements is true?
Only L2 is context free | |
Only L2 and L3 are context free | |
Only L1 and L2 are context free | |
All are context free |
Question 8 Explanation:
All languages viz L1, L2, L3 and L4 has only one comparison and it can be accepted by PDA (single stack), hence all are Context Free Languages.
Question 9 |
S → aSa|bSb|a|b; The language generated by the above grammar over the alphabet {a,b} is the set of
All palindromes. | |
All odd length palindromes. | |
Strings that begin and end with the same symbol. | |
All even length palindromes. |
Question 9 Explanation:
From the grammar, we can easily infer that it generates all the palindromes of odd length, for ex: strings {aba, bab, aaa, bbb, ….}
Question 10 |
L1 only | |
L3 only
| |
L1 and L2
| |
L2 and L3
|
Question 10 Explanation:
L1 can be accepted by PDA, we need to push all 0’s before 1’s and when 1’s comes in input string we need to pop every 0’s from stack for every 1’s and then for every 0’s. If stack and input string is empty at the same time then the string belongs to L1.
But for L2 and L3 PDA implementation is not possible. The reason is, in L2 there are two comparison at a time, first the number of 0’s in beginning should be equal to 1’s and then 0’s after 1’s must be less than or equal to number of 1’s. Suppose for initial 0’s and 1’s are matched by using stack then after matching stack will become empty and then we cannot determine the later 0’s are equal to or less than number of 1’s. Hence PDA implementation is not possible. Similarly L3 also has the similar reason.
But for L2 and L3 PDA implementation is not possible. The reason is, in L2 there are two comparison at a time, first the number of 0’s in beginning should be equal to 1’s and then 0’s after 1’s must be less than or equal to number of 1’s. Suppose for initial 0’s and 1’s are matched by using stack then after matching stack will become empty and then we cannot determine the later 0’s are equal to or less than number of 1’s. Hence PDA implementation is not possible. Similarly L3 also has the similar reason.
Question 11 |
Consider the languages:
L1 = {wwR |w ∈ {0, 1}*}
L2 = {w#wR | w ∈ {0, 1}*}, where # is a special symbol
L3 = {ww | w ∈ (0, 1}*)
Which one of the following is TRUE?L1 is a deterministic CFL
| |
L2 is a deterministic CFL
| |
L3 is a CFL, but not a deterministic CFL
| |
L3 is a deterministic CFL |
Question 11 Explanation:
Given: L1 = {wwR | w ∈ {0,1}*}
→ Given L1 is CFL but not DCFL.
→ Because, we can't predict where w ends and where it reverse is starts.
→ L2 = {w#wR | w ∈ (0,1)*}
→ Given L2 is CFL and also DCFL.
→ The string w and wR are separated by special symbol '#'.
→ L3 = {ww | w ∈ (0,1)*}
This is not even a CFL. This can be proved by using pumping lemma. So, L2 is DCFL. (✔️)
→ Given L1 is CFL but not DCFL.
→ Because, we can't predict where w ends and where it reverse is starts.
→ L2 = {w#wR | w ∈ (0,1)*}
→ Given L2 is CFL and also DCFL.
→ The string w and wR are separated by special symbol '#'.
→ L3 = {ww | w ∈ (0,1)*}
This is not even a CFL. This can be proved by using pumping lemma. So, L2 is DCFL. (✔️)
Question 12 |
Consider the languages:
L1 = {anbncm | n, m > 0}
L2 = {anbmcm | n, m > 0}
Which one of the following statements is FALSE?
L1 ∩ L2 is a context-free language | |
L1 ∪ L2 is a context-free language
| |
L1 and L2 are context-free languages
| |
L1 ∩ L2 is a context sensitive language
|
Question 12 Explanation:
CFL is closed under Union.
CFL is not closed under Intersection.
L1 = {anbncm | n>0 & m>0}
L2 = {ambncn | n>0 & m>0}
L3 = L1 ∩ L2
={anbncn | n>0} It is not context-free.
CFL is not closed under Intersection.
L1 = {anbncm | n>0 & m>0}
L2 = {ambncn | n>0 & m>0}
L3 = L1 ∩ L2
={anbncn | n>0} It is not context-free.
Question 13 |
Context-free languages are closed under:
Union, intersection | |
Union, Kleene closure | |
Intersection, complement | |
Complement, Kleene closure
|
Question 13 Explanation:
Context free languages are not closed under Intersection and complementation.
By checking the options only option B is correct.
By checking the options only option B is correct.
Question 14 |
the set of all binary strings with unequal number of 0’s and 1’s | |
the set of all binary strings including the null string
| |
the set of all binary strings with exactly one more 0’s than the number of 1’s or one more 1 than the number of 0’s
| |
None of the above |
Question 14 Explanation:
(B) is the answer. Because for any binarystring of 0's and 1's we can append another string to make it contain equal no. of 0's and 1's, i.e., any string over {0,1} is a prefix of a string in L.
Question 15 |
the sentence if a then if b then c:=d | |
the left most and right most derivations of the sentence if a then if b then c:=d give rise top different parse trees | |
the sentence if a then if b then c:=d else c:=f has more than two parse trees | |
the sentence if a then if then c:=d else c:=f has two parse trees |
Question 15 Explanation:
We have to generate
"if a then if b then c:=d else c:=f".
Parse tree 1:
Parse tree 2:
"if a then if b then c:=d else c:=f".
Parse tree 1:
Parse tree 2:
Question 16 |
The intersection of two CFL's is also a CFL.
True | |
False |
Question 16 Explanation:
Context free language is not closed under intersection.
There are 16 questions to complete.