CS101
532
CS201
225
CS301
232
CS302
174
CS304
192
CS401
224
CS402
258
CS403
228
CS408
113
CS411
121
CS502
249
CS504
268
CS601
679
CS604
381
CS605
261
CS607
184
CS609
230
CS610
300
CS614
100
CS703
65
Code | Question | Option A | Option B | Option C | Option D | Answer | |
---|---|---|---|---|---|---|---|
CS402 | Kleene Star Closure Can Be Defined | Over Any Set Of String | Over Specific Type Of String | Over Any Set Of Language | Over Specific Type Of Language | A | |
CS402 | L Is A Regular Language So By Kleene S Theorem There Exists An __________ | Fa | Gtg | Tg | Cnf | A | |
CS402 | L= Language Of Words Containing Even Number Of A’S. Regular Expression Is | (A+B)*Aa(A+B)* | (B+Ab*A)* | A+Bb*Aab*A | (A+B)*Ab(A+B)* | B | |
CS402 | Language | Must Be Finite | Must Be Infinite | Can Be Finite Or Infinite | Must Be Finite And Cannot Be Infinite | B | |
CS402 | Language Can Be Expressed By More Than One Fa”. This Statement Is | True | False | Some Times True & Sometimes False | None Of These | A | |
CS402 | Language Which Are Context –Free Are Called Non –Cfl | True | False | Na | Na | B | |
CS402 | Languages Are Proved To Be Regular Or Non Regular Using Pumping Lemma | True | False | Na | Na | A | |
CS402 | Languages Generated By Kleene Star Are Always __________ | Finite | Infinite | Sometimes Finite & Sometimes Infinite | None Of The These | B | |
CS402 | Left Hand Side Of A Production In Cfg Consists Of | One Terminal | More Than One Terminal | One Non-Terminal | Terminals And Non-Terminals | D | |
CS402 | Left Hand Side Of A Production In Cfg Consists Of: | One Terminal | More Than One Terminal | One Non-Terminal | Terminals And Non-Terminals | D | |
CS402 | Left Hand Side Of Cfg May Consist Of | One Terminal | More Than One Terminal | One Non-Terminal | Terminals And Non-Terminals | C | |
CS402 | Length Of Null String Is | Always Not Equal To 0 | Always Equal To 0 | It Has Variable Length | All Are True | B | |
CS402 | Length Of Strings Generated By Infinite Language Is | Finite | Infinite | Depends On The Language | None Of These | A | |
CS402 | Let A = {0 1}. The Number Of Possible Strings Of Length ‘N’ That Can Be Formed By The Elements Of The Set A Is | N! | N^2 | N^M | 2^N | B | |
CS402 | Let Fa 3 Be An Fa Accepting L1nL2 Then The Initial Stat Of Fa3 Must Be Correspond To The Initial State Of__________ And Initial State Of__________ | Fa3 Fa2 | Fa1 Fa2 | Fa1 Fa3 | None Of The Given | B | |
CS402 | Let L Be A Language Defined Over An Alphabet ? Then The Language Of Strings Defined Over ? Not Belonging To L Denoted By Lc Or L. Is Called | Non Regular Language Of L | Complement Of The Language L | Non Of The Given | All Of Above | B | |
CS402 | Let L Be A Language Defined Over An Alphabet S Then The Language Of Strings Defined Over S Not Belonging To L Is Called Complement Of The Language L Denoted By Lc Or L’ | True | False | Na | Na | T | |
CS402 | Let L Be Any Infinite Regular Language Defined Over An Alphabet S Then There Exist Three Strings X Y And Z Belonging To S* Such That All The Strings Of The Form Xy Z N For N=1 2 3 … Are The Words In L. Called Complement Of L | Pumping Lemma | Kleene’S Theorem | None Of The Given | 1 2 Both | A | |
CS402 | Let Q = {Aa Abaaabb Bbaaaaa Bbbbbbbbbb} And R = {B Bbbb Bbbaaa Bbbaaaaa} Pref (Q In R) Is Equal To | {B Bbba Bbbaaa} | {B Bba Bbaaa} | {Ab Bba Bbbaa} | {B Bba Bbba} | A | |
CS402 | Let Q And R Be Expressed By Ab*A And (Ba)* Respectively I.E Q={Aa Aba Abba ……}And R={? Ba Baba Bababa…….}..Aba Is The Only Word In Q Which Can Make A Word In R Because The Words In R Don T Contain The | Single Letter | Double Letter | String | Null String | B |