Given a grammar : S1→Sc, S→SA|A, A→aSb|ab, there is a rightmost derivation S1=>Sc =>SAC=>SaSbc. Thus, SaSbc is a right sentential form, and its handle is (A) SaS (B) be (C) Sbe (D) aSb

Given a grammar : S1→Sc, S→SA|A, A→aSb|ab, there is a rightmost derivation S1=>Sc =>SAC=>SaSbc. Thus, SaSbc is a right sentential form, and its handle is (A) SaS (B) be (C) Sbe (D) aSb 
by

1 Answer

Answer :

(D) aSb
by

Related questions

Which of the following derivations does a top-down parser use while parsing an input string ? The input is scanned from left to right. (A) Leftmost derivation (B) Leftmost derivation traced out in reverse (C) Rightmost derivation traced out in reverse (D) Rightmost derivation

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Last Answer : (A) Leftmost derivation

Given the following statements: S1 : The grammars S→asb | bsa | ss | a and S→asb | bsa | a are not equivalent. S2: The grammars S→ss | sss | asb | bsa | λ and S→ss | asb | bsa | λ are equivalent. Which of the following is true? (A) S1 is correct and S2 is not correct. (B) Both S1 and S2 are correct (C) S1 is not correct and S2 is correct (D) Both S1 and S2 are not correct.

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The grammar with production rules S → aSb |SS|λ generates language L given by: (A) L = {w∈{a, b}* | na(w) = nb(w) and na(v) ≥ nb(v) where v is any prefix of w} (B) L = {w∈{a, b}* | na(w) = nb(w) and na(v) ≤ nb(v) where v is any prefix of w} (C) L = {w∈{a, b}* | na(w) ≠ nb(w) and na(v) ≥ nb(v) where v is any prefix of w} (D) L = {w∈{a, b}* | na(w) ≠ nb(w) and na(v) ≤ nb(v) where v is any prefix of w}

Description : The grammar with production rules S → aSb |SS|λ generates language L given by: (A) L = {w∈{a, b}* | na(w) = nb(w) and na(v) ≥ nb(v) where v is any prefix of w} (B) L = {w∈{a, b}* | na(w) = nb(w) and na(v ... (D) L = {w∈{a, b}* | na(w) ≠ nb(w) and na(v) ≤ nb(v) where v is any prefix of w}

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Last Answer : (D) All the above 

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Last Answer : (D) Both S1 and S2 are correct.

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Last Answer : (D) Both S1 and S2 are not true.

If S1 is total number of modules defined in the program architecture, S3 is the number of modules whose correct function depends on prior processing then the number of modules not dependent on prior processing is: (A) 1 + S3/S1 (B) 1 - S3/S1 (C) 1 + S1/S3 (D) 1 - S1/S3

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Last Answer : (B) 1 - S3/S1

Consider the following statements : S1: A queue can be implemented using two stacks. S2: A stack can be implemented using two queues. Which of the following is correct ? (A) S1 is correct and S2 is not correct. (B) S1 is not correct and S2 is correct. (C) Both S1 and S2 are correct. (D) Both S1 and S2 are not correct.

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Last Answer : S1 is not conflict serializable and S2 is conflict serializable.

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Last Answer : Answer: 2

Given L1 = L(a*baa*) and L2 = L(ab*) The regular expression corresponding to language L3 = L1/L2 (right quotient) is given by (A) a*b (B) a*baa* (C) a*ba* (D) None of the above

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Last Answer : (C) a*ba* 

Given the production rules of a grammar G1 as S1→AB | aaB A→a | Aa B→b and the production rules of a grammar G2 as S2→aS2bS2 | bS2aS2 | λ Which of the following is correct statement? (A) G1 is ambiguous and G2 is not ambiguous. (B) G1 is ambiguous and G2 is ambiguous. (C) G1 is not ambiguous and G2 is ambiguous. (D) G1 is not ambiguous and G2 is not ambiguous.

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Last Answer : (B) G1 is ambiguous and G2 is ambiguous.

A context free grammar for L={w|n0(w)>n1(w)} is given by: (A) S→0|0S|1SS (B) S→0S|1S|0SS|1SS|0|1 (C) S→0|0S|1SS|S1S|SS1 (D) S→0S|1S|0|1

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Last Answer : (C) S→0|0S|1SS|S1S|SS1 

Inductive logic is also known as A) Formal logic B) Model logic C) Material logic D) Sentential logic

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Last Answer : C) Material logic 

The Greibach normal form grammar for the language L={an bn+1|n≥0} is (A) S→a SB, B→bB|λ (B) S→a SB, B→bB|b (C) S→a SB|b, B→b (D) S→a Sb|b

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Last Answer : (C) S→a SB|b, B→b 

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Last Answer : Answer: D

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Community Service ideas for elementary aged ASB kids?

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Description : The context free grammar for the language L = {an bm | n≤m+3, n≥0, m≥0} is (A) S→aaaA; A→aAb|B, B→Bb|λ (B) S→aaaA|λ, A→aAb|B, B→Bb|λ (C) S→aaaA|aaA|λ, A→aAb|B, B→Bb|λ (D) S→aaaA|aaA|aA|λ, A→aAb|B, B→Bb|λ

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If all the production rules have single non - terminal symbol on the left side, the grammar defined is : (A) context free grammar (B) context sensitive grammar (C) unrestricted grammar (D) phrase grammar

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Last Answer : (C) Four Explanation:

What is the full form of 'SAS' ? -How To ?

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What is the full form of 'SAS' ? -How To ?

Description : What is the full form of 'SAS' ? -How To ?

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Description : What is the full form of SAS ?

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You have just joined the collective in the past few weeks, either from Ab or from wherever. What has your experience been like thus far?

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SAS importance in recent and past conflicts?

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What is 'SSS Congruence ' , 'C.P.C.T' ,'SAS' ?? -Maths 9th

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What is 'SSS Congruence ' , 'C.P.C.T' ,'SAS' ?? -Maths 9th

Description : What is 'SSS Congruence ' , 'C.P.C.T' ,'SAS' ?? -Maths 9th

Last Answer : Side-Side-Side (SSS) Rule Side-Side-Side is a rule used to prove whether a given set of triangles are congruent. The SSS rule states that If three sides of one triangle are ... ASA, SSS and RHS congruence rules then remaining angles and sides corresponding in both triangles are equal.Example:

The simplified form of a Boolean equation (AB’+AB’C+AC)(A’C’+B’) is : (A) AB’ (B) AB’C (C) A’B (D) ABC

Description : The simplified form of a Boolean equation (AB’+AB’C+AC)(A’C’+B’) is : (A) AB’ (B) AB’C (C) A’B (D) ABC

Last Answer : (A) AB’

What is the output when following statement is executed ? >>>"abcd"[2:] a) a b) ab c) cd d) dc

Description : What is the output when following statement is executed ? >>>"abcd"[2:] a) a b) ab c) cd d) dc

Last Answer : c) cd

What is the output of the following? x = ['ab', 'cd'] for i in x: x.append(i.upper()) print(x) a) [‘AB’, ‘CD’]. b) [‘ab’, ‘cd’, ‘AB’, ‘CD’]. c) [‘ab’, ‘cd’]. d) none of the mentioned

Description : What is the output of the following? x = ['ab', 'cd'] for i in x: x.append(i.upper()) print(x) a) [‘AB’, ‘CD’]. b) [‘ab’, ‘cd’, ‘AB’, ‘CD’]. c) [‘ab’, ‘cd’]. d) none of the mentioned

Last Answer : d) none of the mentioned