Context-Free-Language
Question 1 |
L 1 and L 2 are regular. | |
L 1 and L 2 are context-free. | |
L 1 is regular and L 2 is context-free. | |
L 1 and L 2 are context-free but not regular. |
Question 1 Explanation:
Both L1 and L2 are regular.
Since no condition on the value of “n” is mentioned so for a particular case we can assume n=0, now when n=0 the language L1= w =(a+b)*
So if one case is (a+b)* then now even we assume any value of “n” the generated string will already present in (a+b)* thus L1=(a+b)*.
L2: In L2 the middle X can expand and consume all symbols of w except the first symbol and symbols of wr except the last symbol so L2 will be equivalent to language all strings start and end with the same symbol, hence L2 is regular.
Since no condition on the value of “n” is mentioned so for a particular case we can assume n=0, now when n=0 the language L1= w =(a+b)*
So if one case is (a+b)* then now even we assume any value of “n” the generated string will already present in (a+b)* thus L1=(a+b)*.
L2: In L2 the middle X can expand and consume all symbols of w except the first symbol and symbols of wr except the last symbol so L2 will be equivalent to language all strings start and end with the same symbol, hence L2 is regular.
Question 2 |
Consider the following languages:
L 1 is not context-free but L 2 and L 3 are deterministic context-free. | |
Neither L 1 nor L 2 is context-free. | |
 | |
Neither L 1 nor its complement is context-free. |
Question 2 Explanation:
L1=ww is a well-known CSL (non-CFL) and its complement is CFL.
L2={an bn cm | m,n >=0} this contains only one comparison (number of a’s = number of b’s) so this is DCFL.
L3={am bn cn | m,n >=0} this contains only one comparison (number of b’s = number of c’s) so this is DCFL.
Intersection of L2 and L3 will have (number of a’s= number of b’s = number of c’s) i.e., {an bn cn | n >=0} so this is CSL (non-CFL).
So A is a true statement and rest all are false statements.
L2={an bn cm | m,n >=0} this contains only one comparison (number of a’s = number of b’s) so this is DCFL.
L3={am bn cn | m,n >=0} this contains only one comparison (number of b’s = number of c’s) so this is DCFL.
Intersection of L2 and L3 will have (number of a’s= number of b’s = number of c’s) i.e., {an bn cn | n >=0} so this is CSL (non-CFL).
So A is a true statement and rest all are false statements.
Question 3 |
Define for a context free language L ⊆ {0,1}*, init(L) = {u ∣ uv ∈ L for some v in {0,1}∗} (in other words, init(L) is the set of prefixes of L)
Let L = {w ∣ w is nonempty and has an equal number of 0’s and 1’s}
Then init(L) is
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 3 Explanation:
(B) is the answer. Because for any binary string 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 4 |
The grammar whose productions are
→ if id then → if id then else → id := id
is ambiguous because
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 4 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 5 |
Consider the context-free grammars over the alphabet {a, b, c} given below. S and T are non-terminals.
G1: S → aSb|T, T → cT|ϵ
G2: S → bSa|T, T → cT|ϵ
The language L(G1) ∩ L(G2) is
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 5 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 6 |
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?
I. L1 ∪ L2
II. L1 ∩ L2
Let L1 = {an bn cm│m,n ≥ 0} and L2 = {am bn cn│m,n ≥ 0}
Which of the following are context-free languages?
I. L1 ∪ L2
II. L1 ∩ L2
I only | |
II only | |
I and II | |
Neither I nor II |
Question 6 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 7 |
Context-free languages are closed under:
Union, intersection | |
Union, Kleene closure | |
Intersection, complement | |
Complement, Kleene closure |
Question 7 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 8 |
Show that the language L = {xcx| x ∈ {0,1}* and c is a terminal symbol} is not context free, c is not 0 or 1.
Theory Explanation. |
Question 9 |
For a string w, we define wR to be the reverse of w. For example, if w = 01101 then wR= 10110.
Which of the following languages is/are context-free?
L={w x wR xR | w, x ∈ {0,1}* } | |
L={w wR x xR | w, x ∈ {0,1}* } | |
L={w x xR wR | w, x ∈ {0,1}* } | |
L={w x wR | w, x ∈ {0,1}* } |
Question 9 Explanation:
Option A: L={w x wR xR | w, x ∈ {0,1}* }
This is not CFL as if we push “w” then “x” then we cannot match wR with “w” as top of stack contains x.
Option B: L={w wR x xR | w, x ∈ {0,1}* }
This is CFL. We non deterministically guess the middle of the string. So we push “w” then match with wR and again push x and match with xR
Option C: L={w x xR wR | w, x ∈ {0,1}* }
This is also CFL. We non deterministically guess the middle of the string. So we push “w” then push x and then match with xR and again match with wR
Option D: L={w x wR | w, x ∈ {0,1}* }
This is a regular language (hence CFL). In this language every string start and end with same symbol (as x can expand).
Question 10 |
Let L1be a regular language and L2be a context-free language. Which of the following languages is/are context free
 | |
 | |
 | |
 |
Question 10 Explanation:
Question 11 |
The intersection of two CFL's is also a CFL.
True | |
False |
Question 11 Explanation:
Context free language is not closed under intersection.
Question 12 |
Consider the following languages:
L1 = {an bm cn+m: m,n ≥ 1}
L2= {an bn c2n: n ≥ 1}
Which one of the following is TRUE?
L1 = {an bm cn+m: m,n ≥ 1}
L2= {an bn c2n: n ≥ 1}
Which one of the following is TRUE?
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 12 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 13 |
Consider the languages:
L1 = {anbncm |n,m > 0} and 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 13 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 14 |
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 14 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 15 |
Let L1 = {0n+m1n0m|n,m ≥ 0}, L2 = {0n+m1n+m0m|n,m ≥ 0}, and L3 = {0n+m1n+m0n+m|n,m ≥ 0}. Which of these languages are NOT context free?
L1 only | |
L3 only | |
L1 and L2 | |
L2 and L3 |
Question 15 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.
There are 15 questions to complete.