Question
stringlengths 39
1.33k
| Tag
stringlengths 3
46
| label
int64 0
206
|
|---|---|---|
Box P has 2 red balls and 3 blue balls and box Q has 3 red balls and 1 blue ball A ball is selected as follows i select a box ii choose a ball from the selected box such that each ball in the box is equally likely to be chosen The probabilities of selecting boxes P and Q are dfrac 1 3 and dfrac 2 3 respectively Given that a ball selected in the above process is a red ball the probability that it came from the box P is dfrac 4 19 dfrac 5 19 dfrac 2 9 dfrac 19 30 | Bayes Theorem | 7 |
An automobile plant contracted to buy shock absorbers from two suppliers X and Y X supplies 60 and Y supplies 40 of the shock absorbers All shock absorbers are subjected to a quality test The ones that pass the quality test are considered reliable Of X u2019s shock absorbers 96 are reliable Of Y u2019s shock absorbers 72 are reliable The probability that a randomly chosen shock absorber which is found to be reliable is made by Y is A 0 288 B 0 334 C 0 667 D 0 720 | Bayes Theorem | 7 |
A binary operation oplus on a set of integers is defined as x oplus y x 2 y 2 Which one of the following statements is TRUE about oplus A Commutative but not associative B Both commutative and associative C Associative but not commutative D Neither commutative nor associative | Binary Operation | 8 |
The number of possible commutative binary operations that can be defined on a set of n elements for a given n is ___________ | Binary Operation | 8 |
On the set N of non negative integers the binary operation ______ is associative and non commutative | Binary Operation | 8 |
The number of possible commutative binary operations that can be defined on a set of n elements for a given n is ___________ | Binary Operation | 8 |
The number of possible commutative binary operations that can be defined on a set of n elements for a given n is ___________ | Binary Operation | 8 |
A binary operation oplus on a set of integers is defined as x oplus y x 2 y 2 Which one of the following statements is TRUE about oplus A Commutative but not associative B Both commutative and associative C Associative but not commutative D Neither commutative nor associative | Binary Operation | 8 |
The binary operator u2260 is defined by the following truth table p q p u2260 q 0 0 0 0 1 1 1 0 1 1 1 0 Which one of the following is true about the binary operator u2260 Both commutative and associative Commutative but not associative Not commutative but associative Neither commutative nor associative | Binary Operation | 8 |
Suppose that we have numbers between 1 and 100 in a binary search tree and want to search for the number 55 Which of the following sequences CANNOT be the sequence of nodes examined 10 75 64 43 60 57 55 90 12 68 34 62 45 55 9 85 47 68 43 57 55 79 14 72 56 16 53 55 | Binary Search Tree | 9 |
A Binary Search Tree BST stores values in the range 37 to 573 Consider the following sequence of keys 81 537 102 439 285 376 305 52 97 121 195 242 381 472 142 248 520 386 345 270 307 550 149 507 395 463 402 270 Suppose the BST has been unsuccessfully searched for key 273 Which all of the above sequences list nodes in the order in which we could have encountered them in the search II and III only I and III only III and IV only III only | Binary Search Tree | 9 |
For each element in a set of size 2n an unbiased coin is tossed The 2n coin tosses are independent An element is chosen if the corresponding coin toss was a head The probability that exactly n elements are chosen is frac 2n mathrm C _n 4 n frac 2n mathrm C _n 2 n frac 1 2n mathrm C _n frac 1 2 | Binomial Distribution | 11 |
An unbiased coin is tossed repeatedly until the outcome of two successive tosses is the same Assuming that the trials are independent the expected number of tosses is 3 4 5 6 | Binomial Distribution | 11 |
For each element in a set of size 2n an unbiased coin is tossed The 2n coin tosses are independent An element is chosen if the corresponding coin toss was a head The probability that exactly n elements are chosen is frac 2n mathrm C _n 4 n frac 2n mathrm C _n 2 n frac 1 2n mathrm C _n frac 1 2 | Binomial Distribution | 11 |
An unbiased coin is tossed repeatedly until the outcome of two successive tosses is the same Assuming that the trials are independent the expected number of tosses is 3 4 5 6 | Binomial Distribution | 11 |
A random bit string of length n is constructed by tossing a fair coin n times and setting a bit to 0 or 1 depending on outcomes head and tail respectively The probability that two such randomly generated strings are not identical is frac 1 2 n 1 frac 1 n frac 1 n 1 frac 1 2 n | Binomial Distribution | 11 |
A random bit string of length n is constructed by tossing a fair coin n times and setting a bit to 0 or 1 depending on outcomes head and tail respectively The probability that two such randomly generated strings are not identical is frac 1 2 n 1 frac 1 n frac 1 n 1 frac 1 2 n | Binomial Distribution | 11 |
An unbiased coin is tossed repeatedly until the outcome of two successive tosses is the same Assuming that the trials are independent the expected number of tosses is 3 4 5 6 | Binomial Distribution | 11 |
For each element in a set of size 2n an unbiased coin is tossed The 2n coin tosses are independent An element is chosen if the corresponding coin toss was a head The probability that exactly n elements are chosen is frac 2n mathrm C _n 4 n frac 2n mathrm C _n 2 n frac 1 2n mathrm C _n frac 1 2 | Binomial Distribution | 11 |
When a coin is tossed the probability of getting a Head is p 0 lt p lt 1 Let N be the random variable denoting the number of tosses till the first Head appears including the toss where the Head appears Assuming that successive tosses are independent the expected value of N is dfrac 1 p dfrac 1 1 p dfrac 1 p 2 dfrac 1 1 p 2 | Binomial Distribution | 11 |
When a coin is tossed the probability of getting a Head is p 0 lt p lt 1 Let N be the random variable denoting the number of tosses till the first Head appears including the toss where the Head appears Assuming that successive tosses are independent the expected value of N is dfrac 1 p dfrac 1 1 p dfrac 1 p 2 dfrac 1 1 p 2 | Binomial Distribution | 11 |
For each element in a set of size 2n an unbiased coin is tossed The 2n coin tosses are independent An element is chosen if the corresponding coin toss was a head The probability that exactly n elements are chosen is frac 2n mathrm C _n 4 n frac 2n mathrm C _n 2 n frac 1 2n mathrm C _n frac 1 2 | Binomial Distribution | 11 |
Let x_ 1 u2295 x_ 2 u2295 x_ 3 u2295 x_ 4 0 where x_ 1 x_ 2 x_ 3 x_ 4 are Boolean variables and u2295 is the XOR operator Which one of the following must always be TRUE x_ 1 x_ 2 x_ 3 x_ 4 0 x_ 1 x_ 3 x_ 2 0 bar x _ 1 u2295 bar x _ 3 bar x _ 2 u2295 bar x _ 4 x_ 1 x_ 2 x_ 3 x_ 4 0 | Boolean Algebra | 12 |
For x in 0 1 let lnot x denote the negation of x that is lnot x begin cases 1 amp mbox iff x 0 0 amp mbox iff x 1 end cases If x in 0 1 n then lnot x denotes the component wise negation of x that is lnot x _i biggl lnot x_i mid i in 1 n biggr Consider a circuit C computing a function f 0 1 n o 0 1 using AND land OR lor and NOT lnot gates Let D be the circuit obtained from C by replacing each AND gate by an OR gate and replacing each OR gate by an AND Suppose D computes the function g Which of the following is true for all inputs x g x lnot f x g x f x land f lnot x g x f x lor f lnot x g x lnot f lnot x None of the above | Boolean Algebra | 12 |
Let x_ 1 u2295 x_ 2 u2295 x_ 3 u2295 x_ 4 0 where x_ 1 x_ 2 x_ 3 x_ 4 are Boolean variables and u2295 is the XOR operator Which one of the following must always be TRUE x_ 1 x_ 2 x_ 3 x_ 4 0 x_ 1 x_ 3 x_ 2 0 bar x _ 1 u2295 bar x _ 3 bar x _ 2 u2295 bar x _ 4 x_ 1 x_ 2 x_ 3 x_ 4 0 | Boolean Algebra | 12 |
The simultaneous equations on the Boolean variables x y z and w x y z 1 xy 0 xz w 1 xy bar z bar w 0 have the following solution for x y z and w respectively 0 1 0 0 1 1 0 1 1 0 1 1 1 0 0 0 | Boolean Algebra | 12 |
For x in 0 1 let lnot x denote the negation of x that is lnot x begin cases 1 amp mbox iff x 0 0 amp mbox iff x 1 end cases If x in 0 1 n then lnot x denotes the component wise negation of x that is lnot x _i biggl lnot x_i mid i in 1 n biggr Consider a circuit C computing a function f 0 1 n o 0 1 using AND land OR lor and NOT lnot gates Let D be the circuit obtained from C by replacing each AND gate by an OR gate and replacing each OR gate by an AND Suppose D computes the function g Which of the following is true for all inputs x g x lnot f x g x f x land f lnot x g x f x lor f lnot x g x lnot f lnot x None of the above | Boolean Algebra | 12 |
For x in 0 1 let lnot x denote the negation of x that is lnot x begin cases 1 amp mbox iff x 0 0 amp mbox iff x 1 end cases If x in 0 1 n then lnot x denotes the component wise negation of x that is lnot x _i biggl lnot x_i mid i in 1 n biggr Consider a circuit C computing a function f 0 1 n o 0 1 using AND land OR lor and NOT lnot gates Let D be the circuit obtained from C by replacing each AND gate by an OR gate and replacing each OR gate by an AND Suppose D computes the function g Which of the following is true for all inputs x g x lnot f x g x f x land f lnot x g x f x lor f lnot x g x lnot f lnot x None of the above | Boolean Algebra | 12 |
The total number of Boolean functions which can be realised with four variables is 4 17 256 65 536 | Boolean Algebra | 12 |
For x in 0 1 let lnot x denote the negation of x that is lnot x begin cases 1 amp mbox iff x 0 0 amp mbox iff x 1 end cases If x in 0 1 n then lnot x denotes the component wise negation of x that is lnot x _i biggl lnot x_i mid i in 1 n biggr Consider a circuit C computing a function f 0 1 n o 0 1 using AND land OR lor and NOT lnot gates Let D be the circuit obtained from C by replacing each AND gate by an OR gate and replacing each OR gate by an AND Suppose D computes the function g Which of the following is true for all inputs x g x lnot f x g x f x land f lnot x g x f x lor f lnot x g x lnot f lnot x None of the above | Boolean Algebra | 12 |
The simultaneous equations on the Boolean variables x y z and w x y z 1 xy 0 xz w 1 xy bar z bar w 0 have the following solution for x y z and w respectively 0 1 0 0 1 1 0 1 1 0 1 1 1 0 0 0 | Boolean Algebra | 12 |
The total number of Boolean functions which can be realised with four variables is 4 17 256 65 536 | Boolean Algebra | 12 |
The simultaneous equations on the Boolean variables x y z and w x y z 1 xy 0 xz w 1 xy bar z bar w 0 have the following solution for x y z and w respectively 0 1 0 0 1 1 0 1 1 0 1 1 1 0 0 0 | Boolean Algebra | 12 |
What values of A B C and D satisfy the following simultaneous Boolean equations overline A AB 0 AB AC AB A overline C CD overline C D A 1 B 0 C 0 D 1 A 1 B 1 C 0 D 0 A 1 B 0 C 1 D 1 A 1 B 0 C 0 D 0 | Boolean Expressions | 13 |
Choose the correct alternatives more than one may be correct and write the corresponding letters only The operation which is commutative but not associative is AND OR EX OR NAND | Boolean Operations | 14 |
Choose the correct alternatives more than one may be correct and write the corresponding letters only The operation which is commutative but not associative is AND OR EX OR NAND | Boolean Operations | 14 |
Choose the correct alternatives more than one may be correct and write the corresponding letters only The operation which is commutative but not associative is AND OR EX OR NAND | Boolean Operations | 14 |
Choose the correct alternatives more than one may be correct and write the corresponding letters only The operation which is commutative but not associative is AND OR EX OR NAND | Boolean Operations | 14 |
Choose the correct alternatives more than one may be correct and write the corresponding letters only The operation which is commutative but not associative is AND OR EX OR NAND | Boolean Operations | 14 |
Choose the correct alternatives more than one may be correct and write the corresponding letters only The operation which is commutative but not associative is AND OR EX OR NAND | Boolean Operations | 14 |
Choose the correct alternatives more than one may be correct and write the corresponding letters only The operation which is commutative but not associative is AND OR EX OR NAND | Boolean Operations | 14 |
Choose the correct alternatives more than one may be correct and write the corresponding letters only The operation which is commutative but not associative is AND OR EX OR NAND | Boolean Operations | 14 |
Choose the correct alternatives more than one may be correct and write the corresponding letters only The operation which is commutative but not associative is AND OR EX OR NAND | Boolean Operations | 14 |
Choose the correct alternatives more than one may be correct and write the corresponding letters only The operation which is commutative but not associative is AND OR EX OR NAND | Boolean Operations | 14 |
Choose the correct alternatives more than one may be correct and write the corresponding letters only The operation which is commutative but not associative is AND OR EX OR NAND | Boolean Operations | 14 |
Choose the correct alternatives more than one may be correct and write the corresponding letters only The operation which is commutative but not associative is AND OR EX OR NAND | Boolean Operations | 14 |
Choose the correct alternatives more than one may be correct and write the corresponding letters only The operation which is commutative but not associative is AND OR EX OR NAND | Boolean Operations | 14 |
Choose the correct alternatives more than one may be correct and write the corresponding letters only The operation which is commutative but not associative is AND OR EX OR NAND | Boolean Operations | 14 |
When multiplicand Y is multiplied by multiplier X x_ n 1 x_ n 2 dots x_0 using bit pair recoding in Booth s algorithm partial products are generated according to the following table Row xi 1 xi xi 1 Partial Product 1 0 0 0 0 2 0 0 1 Y 3 0 1 0 Y 4 0 1 1 2Y 5 1 0 0 6 1 0 1 Y 7 1 1 0 Y 8 1 1 1 The partial products for rows 5 and 8 are 2Y and Y 2Y and 2Y 2Y and 0 0 and Y | Booth Recoding | 15 |
When multiplicand Y is multiplied by multiplier X x_ n 1 x_ n 2 dots x_0 using bit pair recoding in Booth s algorithm partial products are generated according to the following table Row xi 1 xi xi 1 Partial Product 1 0 0 0 0 2 0 0 1 Y 3 0 1 0 Y 4 0 1 1 2Y 5 1 0 0 6 1 0 1 Y 7 1 1 0 Y 8 1 1 1 The partial products for rows 5 and 8 are 2Y and Y 2Y and 2Y 2Y and 0 0 and Y | Booth Recoding | 15 |
When multiplicand Y is multiplied by multiplier X x_ n 1 x_ n 2 dots x_0 using bit pair recoding in Booth s algorithm partial products are generated according to the following table Row xi 1 xi xi 1 Partial Product 1 0 0 0 0 2 0 0 1 Y 3 0 1 0 Y 4 0 1 1 2Y 5 1 0 0 6 1 0 1 Y 7 1 1 0 Y 8 1 1 1 The partial products for rows 5 and 8 are 2Y and Y 2Y and 2Y 2Y and 0 0 and Y | Booth Recoding | 15 |
When multiplicand Y is multiplied by multiplier X x_ n 1 x_ n 2 dots x_0 using bit pair recoding in Booth s algorithm partial products are generated according to the following table Row xi 1 xi xi 1 Partial Product 1 0 0 0 0 2 0 0 1 Y 3 0 1 0 Y 4 0 1 1 2Y 5 1 0 0 6 1 0 1 Y 7 1 1 0 Y 8 1 1 1 The partial products for rows 5 and 8 are 2Y and Y 2Y and 2Y 2Y and 0 0 and Y | Booth Recoding | 15 |
When multiplicand Y is multiplied by multiplier X x_ n 1 x_ n 2 dots x_0 using bit pair recoding in Booth s algorithm partial products are generated according to the following table Row xi 1 xi xi 1 Partial Product 1 0 0 0 0 2 0 0 1 Y 3 0 1 0 Y 4 0 1 1 2Y 5 1 0 0 6 1 0 1 Y 7 1 1 0 Y 8 1 1 1 The partial products for rows 5 and 8 are 2Y and Y 2Y and 2Y 2Y and 0 0 and Y | Booth Recoding | 15 |
When multiplicand Y is multiplied by multiplier X x_ n 1 x_ n 2 dots x_0 using bit pair recoding in Booth s algorithm partial products are generated according to the following table Row xi 1 xi xi 1 Partial Product 1 0 0 0 0 2 0 0 1 Y 3 0 1 0 Y 4 0 1 1 2Y 5 1 0 0 6 1 0 1 Y 7 1 1 0 Y 8 1 1 1 The partial products for rows 5 and 8 are 2Y and Y 2Y and 2Y 2Y and 0 0 and Y | Booth Recoding | 15 |
When multiplicand Y is multiplied by multiplier X x_ n 1 x_ n 2 dots x_0 using bit pair recoding in Booth s algorithm partial products are generated according to the following table Row xi 1 xi xi 1 Partial Product 1 0 0 0 0 2 0 0 1 Y 3 0 1 0 Y 4 0 1 1 2Y 5 1 0 0 6 1 0 1 Y 7 1 1 0 Y 8 1 1 1 The partial products for rows 5 and 8 are 2Y and Y 2Y and 2Y 2Y and 0 0 and Y | Booth Recoding | 15 |
When multiplicand Y is multiplied by multiplier X x_ n 1 x_ n 2 dots x_0 using bit pair recoding in Booth s algorithm partial products are generated according to the following table Row xi 1 xi xi 1 Partial Product 1 0 0 0 0 2 0 0 1 Y 3 0 1 0 Y 4 0 1 1 2Y 5 1 0 0 6 1 0 1 Y 7 1 1 0 Y 8 1 1 1 The partial products for rows 5 and 8 are 2Y and Y 2Y and 2Y 2Y and 0 0 and Y | Booth Recoding | 15 |
When multiplicand Y is multiplied by multiplier X x_ n 1 x_ n 2 dots x_0 using bit pair recoding in Booth s algorithm partial products are generated according to the following table Row xi 1 xi xi 1 Partial Product 1 0 0 0 0 2 0 0 1 Y 3 0 1 0 Y 4 0 1 1 2Y 5 1 0 0 6 1 0 1 Y 7 1 1 0 Y 8 1 1 1 The partial products for rows 5 and 8 are 2Y and Y 2Y and 2Y 2Y and 0 0 and Y | Booth Recoding | 15 |
When multiplicand Y is multiplied by multiplier X x_ n 1 x_ n 2 dots x_0 using bit pair recoding in Booth s algorithm partial products are generated according to the following table Row xi 1 xi xi 1 Partial Product 1 0 0 0 0 2 0 0 1 Y 3 0 1 0 Y 4 0 1 1 2Y 5 1 0 0 6 1 0 1 Y 7 1 1 0 Y 8 1 1 1 The partial products for rows 5 and 8 are 2Y and Y 2Y and 2Y 2Y and 0 0 and Y | Booth Recoding | 15 |
When multiplicand Y is multiplied by multiplier X x_ n 1 x_ n 2 dots x_0 using bit pair recoding in Booth s algorithm partial products are generated according to the following table Row xi 1 xi xi 1 Partial Product 1 0 0 0 0 2 0 0 1 Y 3 0 1 0 Y 4 0 1 1 2Y 5 1 0 0 6 1 0 1 Y 7 1 1 0 Y 8 1 1 1 The partial products for rows 5 and 8 are 2Y and Y 2Y and 2Y 2Y and 0 0 and Y | Booth Recoding | 15 |
When multiplicand Y is multiplied by multiplier X x_ n 1 x_ n 2 dots x_0 using bit pair recoding in Booth s algorithm partial products are generated according to the following table Row xi 1 xi xi 1 Partial Product 1 0 0 0 0 2 0 0 1 Y 3 0 1 0 Y 4 0 1 1 2Y 5 1 0 0 6 1 0 1 Y 7 1 1 0 Y 8 1 1 1 The partial products for rows 5 and 8 are 2Y and Y 2Y and 2Y 2Y and 0 0 and Y | Booth Recoding | 15 |
When multiplicand Y is multiplied by multiplier X x_ n 1 x_ n 2 dots x_0 using bit pair recoding in Booth s algorithm partial products are generated according to the following table Row xi 1 xi xi 1 Partial Product 1 0 0 0 0 2 0 0 1 Y 3 0 1 0 Y 4 0 1 1 2Y 5 1 0 0 6 1 0 1 Y 7 1 1 0 Y 8 1 1 1 The partial products for rows 5 and 8 are 2Y and Y 2Y and 2Y 2Y and 0 0 and Y | Booth Recoding | 15 |
When multiplicand Y is multiplied by multiplier X x_ n 1 x_ n 2 dots x_0 using bit pair recoding in Booth s algorithm partial products are generated according to the following table Row xi 1 xi xi 1 Partial Product 1 0 0 0 0 2 0 0 1 Y 3 0 1 0 Y 4 0 1 1 2Y 5 1 0 0 6 1 0 1 Y 7 1 1 0 Y 8 1 1 1 The partial products for rows 5 and 8 are 2Y and Y 2Y and 2Y 2Y and 0 0 and Y | Booth Recoding | 15 |
Booth s coding in 8 bits for the decimal number 57 is 0 100 1000 0 100 100 1 0 1 100 10 1 00 10 100 1 | Booths Algorithm | 16 |
Booth u2019s algorithm for integer multiplication gives worst performance when the multiplier pattern is 101010 u2026 1010 100000 u2026 0001 111111 u2026 1111 011111 u2026 1110 | Booths Algorithm | 16 |
Booth u2019s algorithm for integer multiplication gives worst performance when the multiplier pattern is 101010 u2026 1010 100000 u2026 0001 111111 u2026 1111 011111 u2026 1110 | Booths Algorithm | 16 |
Booth s coding in 8 bits for the decimal number 57 is 0 100 1000 0 100 100 1 0 1 100 10 1 00 10 100 1 | Booths Algorithm | 16 |
Using Booth s Algorithm for multiplication the multiplier 57 will be recoded as 0 1 0 0 1 0 0 1 1 1 0 0 0 1 1 1 0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 1 | Booths Algorithm | 16 |
The two numbers given below are multiplied using the Booth s algorithm Multiplicand 0101 1010 1110 1110 Multiplier 0111 0111 1011 1101 How many additions Subtractions are required for the multiplication of the above two numbers 6 8 10 12 | Booths Algorithm | 16 |
Using Booth s Algorithm for multiplication the multiplier 57 will be recoded as 0 1 0 0 1 0 0 1 1 1 0 0 0 1 1 1 0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 1 | Booths Algorithm | 16 |
Booth u2019s algorithm for integer multiplication gives worst performance when the multiplier pattern is 101010 u2026 1010 100000 u2026 0001 111111 u2026 1111 011111 u2026 1110 | Booths Algorithm | 16 |
Booth s coding in 8 bits for the decimal number 57 is 0 100 1000 0 100 100 1 0 1 100 10 1 00 10 100 1 | Booths Algorithm | 16 |
Booth u2019s algorithm for integer multiplication gives worst performance when the multiplier pattern is 101010 u2026 1010 100000 u2026 0001 111111 u2026 1111 011111 u2026 1110 | Booths Algorithm | 16 |
Booth s coding in 8 bits for the decimal number 57 is 0 100 1000 0 100 100 1 0 1 100 10 1 00 10 100 1 | Booths Algorithm | 16 |
A prime attribute of a relation scheme R is an attribute that appears in all candidate keys of R in some candidate key of R in a foreign key of R only in the primary key of R | Candidate Keys | 17 |
An instance of a relational scheme R A B C has distinct values for attribute A Can you conclude that A is a candidate key for R | Candidate Keys | 17 |
An instance of a relational scheme R A B C has distinct values for attribute A Can you conclude that A is a candidate key for R | Candidate Keys | 17 |
An instance of a relational scheme R A B C has distinct values for attribute A Can you conclude that A is a candidate key for R | Candidate Keys | 17 |
The maximum number of superkeys for the relation schema R E F G H with E as the key is _____ | Candidate Keys | 17 |
The maximum number of superkeys for the relation schema R E F G H with E as the key is _____ | Candidate Keys | 17 |
Consider a relational table with a single record for each registered student with the following attributes ext Registration_Num Unique registration number for each registered student ext UID Unique identity number unique at the national level for each citizen ext BankAccount_Num Unique account number at the bank A student can have multiple accounts or joint accounts This attribute stores the primary account number ext Name Name of the student ext Hostel_Room Room number of the hostel Which of the following options is INCORRECT A ext BankAccount_Num is a candidate key B ext Registration_Num can be a primary key C ext UID is a candidate key if all students are from the same country D If S is a super key such that S cap ext UID is ext NULL then S cup ext UID is also a superkey | Candidate Keys | 17 |
An instance of a relational scheme R A B C has distinct values for attribute A Can you conclude that A is a candidate key for R | Candidate Keys | 17 |
A prime attribute of a relation scheme R is an attribute that appears in all candidate keys of R in some candidate key of R in a foreign key of R only in the primary key of R | Candidate Keys | 17 |
Consider a relational table with a single record for each registered student with the following attributes ext Registration_Num Unique registration number for each registered student ext UID Unique identity number unique at the national level for each citizen ext BankAccount_Num Unique account number at the bank A student can have multiple accounts or joint accounts This attribute stores the primary account number ext Name Name of the student ext Hostel_Room Room number of the hostel Which of the following options is INCORRECT A ext BankAccount_Num is a candidate key B ext Registration_Num can be a primary key C ext UID is a candidate key if all students are from the same country D If S is a super key such that S cap ext UID is ext NULL then S cup ext UID is also a superkey | Candidate Keys | 17 |
A Boolean expression is an expression made out of propositional letters such as p q r and operators wedge vee and eg e g p wedge eg q vee eg r An expression is said to be in sum of product form also called disjunctive normal form if all eg occur just before letters and no vee occurs in scope of wedge e g p wedge eg q vee eg p wedge q The expression is said to be in product of sum form also called conjunctive normal form if all negations occur just before letters and no wedge occurs in the scope of vee e g p vee eg q wedge eg p vee q Which of the following is not correct Every Boolean expression is equivalent to an expression is sum of product form Every Boolean expression is equivalent to an expression in product of sum form Every Boolean expression is equivalent to an expression without vee operator Every Boolean expression is equivalent to an expression without wedge operator Every Boolean expression is equivalent to an expression without eg operator | Canonical Normal Form | 18 |
Given f_1 f_3 and f in canonical sum of products form in decimal for the circuit f_1 Sigma m 4 5 6 7 8 f_3 Sigma m 1 6 15 f Sigma m 1 6 8 15 hen f_2 is Sigma m 4 6 Sigma m 4 8 Sigma m 6 8 Sigma m 4 6 8 | Canonical Normal Form | 18 |
The total number of prime implicants of the function f w x y z sum 0 2 4 5 6 10 is __________ | Canonical Normal Form | 18 |
The minterm expansion of f P Q R PQ Q bar R P bar R is m_2 m_4 m_6 m_7 m_0 m_1 m_3 m_5 m_0 m_1 m_6 m_7 m_2 m_3 m_4 m_5 | Canonical Normal Form | 18 |
A Boolean expression is an expression made out of propositional letters such as p q r and operators wedge vee and eg e g p wedge eg q vee eg r An expression is said to be in sum of product form also called disjunctive normal form if all eg occur just before letters and no vee occurs in scope of wedge e g p wedge eg q vee eg p wedge q The expression is said to be in product of sum form also called conjunctive normal form if all negations occur just before letters and no wedge occurs in the scope of vee e g p vee eg q wedge eg p vee q Which of the following is not correct Every Boolean expression is equivalent to an expression is sum of product form Every Boolean expression is equivalent to an expression in product of sum form Every Boolean expression is equivalent to an expression without vee operator Every Boolean expression is equivalent to an expression without wedge operator Every Boolean expression is equivalent to an expression without eg operator | Canonical Normal Form | 18 |
The total number of prime implicants of the function f w x y z sum 0 2 4 5 6 10 is __________ | Canonical Normal Form | 18 |
The total number of prime implicants of the function f w x y z sum 0 2 4 5 6 10 is __________ | Canonical Normal Form | 18 |
Given the function F P QR where F is a function in three Boolean variables P Q and R and P P consider the following statements S1 F sum 4 5 6 S2 F sum 0 1 2 3 7 S3 F Pi 4 5 6 S4 F Pi 0 1 2 3 7 Which of the following is true S1 False S2 True S3 True S4 False S1 True S2 False S3 False S4 True S1 False S2 False S3 True S4 True S1 True S2 True S3 False S4 False | Canonical Normal Form | 18 |
Given two three bit numbers a_ 2 a_ 1 a_ 0 and b_ 2 b_ 1 b_ 0 and c the carry in the function that represents the carry generate function when these two numbers are added is a_ 2 b_ 2 a_ 2 a_ 1 b_ 1 a_ 2 a_ 1 a_ 0 b_ 0 a_ 2 a_ 0 b_ 1 b_ 0 a_ 1 b_ 2 b_ 1 a_ 1 a_ 0 b_ 2 b_ 0 a_ 0 b_ 2 b_ 1 b_ 0 a_ 2 b_ 2 a_ 2 b_ 1 b_ 0 a_ 2 a_ 1 b_ 1 b_ 0 a_ 1 a_ 0 b_ 2 b_ 1 a_ 1 a_ 0 b_ 2 a_ 1 a_ 0 b_ 2 b_ 0 a_ 2 a_ 0 b_ 1 b_ 0 a_ 2 b_ 2 a_ 2 oplus b_ 2 a_ 1 b_ 1 a_ 1 oplus b_ 1 a_ 0 b_ 0 a_ 2 b_ 2 overline a_ 2 a_ 1 b_ 1 overline a_ 2 a_ 1 a_ 0 b_ 0 overline a_ 2 a_ 0 overline b_ 1 b_ 0 a_ 1 overline b_ 2 b_ 1 overline a_ 1 a_ 0 overline b_ 2 b_ 0 a_ 0 overline b_ 2 b_ 1 b_ 0 | Carry Generator | 19 |
In a look ahead carry generator the carry generate function G_i and the carry propagate function P_i for inputs A_i and B_i are given by P_i A_i oplus B_i ext and G_i A_iB_i The expressions for the sum bit S_i and the carry bit C_ i 1 of the look ahead carry adder are given by S_i P_i oplus C_i ext and C_ i 1 G_i P_iC_i ext where C_0 ext is the input carry Consider a two level logic implementation of the look ahead carry generator Assume that all P_i and G_i are available for the carry generator circuit and that the AND and OR gates can have any number of inputs The number of AND gates and OR gates needed to implement the look ahead carry generator for a 4 bit adder with S_3 S_2 S_1 S_0 and C_4 as its outputs are respectively 6 3 10 4 6 4 10 5 | Carry Generator | 19 |
Given two three bit numbers a_ 2 a_ 1 a_ 0 and b_ 2 b_ 1 b_ 0 and c the carry in the function that represents the carry generate function when these two numbers are added is a_ 2 b_ 2 a_ 2 a_ 1 b_ 1 a_ 2 a_ 1 a_ 0 b_ 0 a_ 2 a_ 0 b_ 1 b_ 0 a_ 1 b_ 2 b_ 1 a_ 1 a_ 0 b_ 2 b_ 0 a_ 0 b_ 2 b_ 1 b_ 0 a_ 2 b_ 2 a_ 2 b_ 1 b_ 0 a_ 2 a_ 1 b_ 1 b_ 0 a_ 1 a_ 0 b_ 2 b_ 1 a_ 1 a_ 0 b_ 2 a_ 1 a_ 0 b_ 2 b_ 0 a_ 2 a_ 0 b_ 1 b_ 0 a_ 2 b_ 2 a_ 2 oplus b_ 2 a_ 1 b_ 1 a_ 1 oplus b_ 1 a_ 0 b_ 0 a_ 2 b_ 2 overline a_ 2 a_ 1 b_ 1 overline a_ 2 a_ 1 a_ 0 b_ 0 overline a_ 2 a_ 0 overline b_ 1 b_ 0 a_ 1 overline b_ 2 b_ 1 overline a_ 1 a_ 0 overline b_ 2 b_ 0 a_ 0 overline b_ 2 b_ 1 b_ 0 | Carry Generator | 19 |
In a look ahead carry generator the carry generate function G_i and the carry propagate function P_i for inputs A_i and B_i are given by P_i A_i oplus B_i ext and G_i A_iB_i The expressions for the sum bit S_i and the carry bit C_ i 1 of the look ahead carry adder are given by S_i P_i oplus C_i ext and C_ i 1 G_i P_iC_i ext where C_0 ext is the input carry Consider a two level logic implementation of the look ahead carry generator Assume that all P_i and G_i are available for the carry generator circuit and that the AND and OR gates can have any number of inputs The number of AND gates and OR gates needed to implement the look ahead carry generator for a 4 bit adder with S_3 S_2 S_1 S_0 and C_4 as its outputs are respectively 6 3 10 4 6 4 10 5 | Carry Generator | 19 |
Given two three bit numbers a_ 2 a_ 1 a_ 0 and b_ 2 b_ 1 b_ 0 and c the carry in the function that represents the carry generate function when these two numbers are added is a_ 2 b_ 2 a_ 2 a_ 1 b_ 1 a_ 2 a_ 1 a_ 0 b_ 0 a_ 2 a_ 0 b_ 1 b_ 0 a_ 1 b_ 2 b_ 1 a_ 1 a_ 0 b_ 2 b_ 0 a_ 0 b_ 2 b_ 1 b_ 0 a_ 2 b_ 2 a_ 2 b_ 1 b_ 0 a_ 2 a_ 1 b_ 1 b_ 0 a_ 1 a_ 0 b_ 2 b_ 1 a_ 1 a_ 0 b_ 2 a_ 1 a_ 0 b_ 2 b_ 0 a_ 2 a_ 0 b_ 1 b_ 0 a_ 2 b_ 2 a_ 2 oplus b_ 2 a_ 1 b_ 1 a_ 1 oplus b_ 1 a_ 0 b_ 0 a_ 2 b_ 2 overline a_ 2 a_ 1 b_ 1 overline a_ 2 a_ 1 a_ 0 b_ 0 overline a_ 2 a_ 0 overline b_ 1 b_ 0 a_ 1 overline b_ 2 b_ 1 overline a_ 1 a_ 0 overline b_ 2 b_ 0 a_ 0 overline b_ 2 b_ 1 b_ 0 | Carry Generator | 19 |
In a look ahead carry generator the carry generate function G_i and the carry propagate function P_i for inputs A_i and B_i are given by P_i A_i oplus B_i ext and G_i A_iB_i The expressions for the sum bit S_i and the carry bit C_ i 1 of the look ahead carry adder are given by S_i P_i oplus C_i ext and C_ i 1 G_i P_iC_i ext where C_0 ext is the input carry Consider a two level logic implementation of the look ahead carry generator Assume that all P_i and G_i are available for the carry generator circuit and that the AND and OR gates can have any number of inputs The number of AND gates and OR gates needed to implement the look ahead carry generator for a 4 bit adder with S_3 S_2 S_1 S_0 and C_4 as its outputs are respectively 6 3 10 4 6 4 10 5 | Carry Generator | 19 |
Given two three bit numbers a_ 2 a_ 1 a_ 0 and b_ 2 b_ 1 b_ 0 and c the carry in the function that represents the carry generate function when these two numbers are added is a_ 2 b_ 2 a_ 2 a_ 1 b_ 1 a_ 2 a_ 1 a_ 0 b_ 0 a_ 2 a_ 0 b_ 1 b_ 0 a_ 1 b_ 2 b_ 1 a_ 1 a_ 0 b_ 2 b_ 0 a_ 0 b_ 2 b_ 1 b_ 0 a_ 2 b_ 2 a_ 2 b_ 1 b_ 0 a_ 2 a_ 1 b_ 1 b_ 0 a_ 1 a_ 0 b_ 2 b_ 1 a_ 1 a_ 0 b_ 2 a_ 1 a_ 0 b_ 2 b_ 0 a_ 2 a_ 0 b_ 1 b_ 0 a_ 2 b_ 2 a_ 2 oplus b_ 2 a_ 1 b_ 1 a_ 1 oplus b_ 1 a_ 0 b_ 0 a_ 2 b_ 2 overline a_ 2 a_ 1 b_ 1 overline a_ 2 a_ 1 a_ 0 b_ 0 overline a_ 2 a_ 0 overline b_ 1 b_ 0 a_ 1 overline b_ 2 b_ 1 overline a_ 1 a_ 0 overline b_ 2 b_ 0 a_ 0 overline b_ 2 b_ 1 b_ 0 | Carry Generator | 19 |
Given two three bit numbers a_ 2 a_ 1 a_ 0 and b_ 2 b_ 1 b_ 0 and c the carry in the function that represents the carry generate function when these two numbers are added is a_ 2 b_ 2 a_ 2 a_ 1 b_ 1 a_ 2 a_ 1 a_ 0 b_ 0 a_ 2 a_ 0 b_ 1 b_ 0 a_ 1 b_ 2 b_ 1 a_ 1 a_ 0 b_ 2 b_ 0 a_ 0 b_ 2 b_ 1 b_ 0 a_ 2 b_ 2 a_ 2 b_ 1 b_ 0 a_ 2 a_ 1 b_ 1 b_ 0 a_ 1 a_ 0 b_ 2 b_ 1 a_ 1 a_ 0 b_ 2 a_ 1 a_ 0 b_ 2 b_ 0 a_ 2 a_ 0 b_ 1 b_ 0 a_ 2 b_ 2 a_ 2 oplus b_ 2 a_ 1 b_ 1 a_ 1 oplus b_ 1 a_ 0 b_ 0 a_ 2 b_ 2 overline a_ 2 a_ 1 b_ 1 overline a_ 2 a_ 1 a_ 0 b_ 0 overline a_ 2 a_ 0 overline b_ 1 b_ 0 a_ 1 overline b_ 2 b_ 1 overline a_ 1 a_ 0 overline b_ 2 b_ 0 a_ 0 overline b_ 2 b_ 1 b_ 0 | Carry Generator | 19 |
In a look ahead carry generator the carry generate function G_i and the carry propagate function P_i for inputs A_i and B_i are given by P_i A_i oplus B_i ext and G_i A_iB_i The expressions for the sum bit S_i and the carry bit C_ i 1 of the look ahead carry adder are given by S_i P_i oplus C_i ext and C_ i 1 G_i P_iC_i ext where C_0 ext is the input carry Consider a two level logic implementation of the look ahead carry generator Assume that all P_i and G_i are available for the carry generator circuit and that the AND and OR gates can have any number of inputs The number of AND gates and OR gates needed to implement the look ahead carry generator for a 4 bit adder with S_3 S_2 S_1 S_0 and C_4 as its outputs are respectively 6 3 10 4 6 4 10 5 | Carry Generator | 19 |
Subsets and Splits