GATE-CS-2000

Q.1

The minimum number of cards to be dealt from an arbitrarily shuffled deck of 52 cards to guarantee that three cards are from some same suit is

Ans .

Explanation :
Total number of suits is 4 We need minimum 9 cards to make sure that there are 3 cards of same suit. For example, With 8 cards, we can have 2 cards of each suit.

Q.2

Let L denotes the language generated by the grammar S -> 0S0/00. Which of the following is true?

L = 0+

L is regular but not 0+

L is context free but not regular

L is not context free

Ans .

Explanation :
Option A : L is not 0+ , because 0+ will contain any arbitrary string over alphabet 0 with any no of 0's ( except empty string ), for ex: {0, 00, 000,00000}, but L will only have the strings as { 00, 0000, 000000,...}, i.e only even no of 0's ( excluding empty string}. Option D : L is a Context Free Language, because the Grammar G which generates the language L is Context Free Grammar. A Grammar G is CFG if all of its productions are of the form A->α, where A is a single non-terminal and α belongs to (V∪ T)* , i.e α can be a string of terminals and/or Non-terminals. (V represents a non-terminal and T represents a terminal) Option C : L is a Regular Language, Because we are able to write a regular expression for it ( and also able to make a Finite Automaton), which is (00)+. Option B : Hence This option is Correct, because L is Regular but not 0+, as we proved above.

Q.3

The number 43 in 2's complement representation is

01010101

11010101

00101011

10101011

Ans .

Explanation :
In2's complement representation, positive numbers are represented as their representation and negative numbers are represented by first doing 1's complement, then adding 1 to the result. So 43 is represented as 00101011. Note that option represents -43.

Q.4

To put the 8085 microprocessor in the wait state

lower the-HOLD input

lower the READY input

raise the HOLD input

raise the READY input

Ans .

Explanation :
If ready pin is high the microprocessor will complete the operation and proceeds for the next operation. If ready pin is low the microprocessor will wait until it goes high. Thus, option (B) is the answer. Please comment below if you find anything wrong in the above post.

Q.5

Comparing the time T1 taken for a single instruction on a pipelined CPU with time T2 taken on a non pipelined but identical CPU, we can say that

T1 <= T2

T1 >= T2

T1 < T2

T1 is T2 plus the time taken for one instruction fetch cycle

Ans .

Explanation :
TPipelining does not increase the execution time of a single instruction. It increases the overall performance by executing instructions in multiple pipeline stages. We assume that each stage takes ‘T’ unit of time both in pipelined and non-pipelined CPU. Let total stages in pipelined CPU = Total stages in non-pipelined CPU = K and number of Instructions = N = 1 Pipelined CPU : Total time (T1) = (K + (N - 1)) * T = KT Non-Pipelined CPU : Total time (T2) = KNT = KT Considering buffer delays in pipelined CPU, T1 >= T2 Thus, option (B) is the answer. Please comment below if you find anything wrong in the above post.

Q.6

The Palghat Gap (or Palakkad Gap), a region about 30 km wide in the southern part of the Western Ghats in India, is lower than the hilly terrain to its north and south. The exact reasons for the formation of this gap are not clear. It results in the neighbouring regions of Tamil Nadu getting more rainfall from the South West monsoon and the neighbouring regions of Kerala having higher summer temperatures. What can be inferred from this passage? <-/div>

The Palghat gap is caused by high rainfall and high temperatures in southern Tamil Nadu and Kerala

The regions in Tamil Nadu and Kerala that are near the Palghat Gap are low-lying

The low terrain of the Palghat Gap has a significant impact on weather patterns in neighbouring parts of Tamil Nadu and Kerala

Higher summer temperatures result in higher rainfall near the Palghat Gap area

Ans .

Explanation :
Here the passage is asking, " what can be inferred ? ", i.e. it's asking about the hidden conclusion of the passage. Why not Option A ? - because it is clearly given in the passage : "The exact reasons for the formation of this gap are not clear ". Why not Option B ? - because the passage no where talks about the low-lying regions of Kerala and Tamil Nadu near the Palghat Gap. Why not option D ? - Because the passage only says : "neighbouring regions of Kerala having higher summer temperatures", it doesn't say that this high temperature results in rainfall. Note: We have to think and conclude only from what is given in the passage. We can't make our own conclusions. Why Option C ? - because the passage says : " It results in the neighbouring regions of Tamil Nadu getting more rainfall from the South West monsoon and the neighbouring regions of Kerala having higher summer temperatures", hence we can conclude here that the weather is getting affected due to the Gap. Therefore, option C is correct.

Q.7

Geneticists say that they are very close to confirming the genetic roots of psychiatric illnesses such as depression and schizophrenia, and consequently, that doctors will be able to eradicate these diseases through early identification and gene therapy. On which of the following assumptions does the statement above rely?

Strategies are now available for eliminating psychiatric illnesses

Certain psychiatric illnesses have a genetic basis

All human diseases can be traced back to genes and how they are expressed

In the future, genetics will become the only relevant field for identifying psychiatric illnesses

Ans .

Explanation :
- A says that strategies are now available for eliminating psychiatric illnesses but it is mentioned in the very first line that the geneticists are very close but the strategy is not available till now. So, A is incorrect.
- The given data says that the geneticists are working on the genetic roots of psychiatric illnesses such as depression and schizophrenia, which implies that these two diseases have genetic basis. Thus, B is the correct option.
- C says that all human diseases can be traced back to genes and how they are expressed, but the data given in the question talks about psychiatric illnesses such as depression and schizophrenia only. So, C is also incorrect.
- D says that in future, genetics will be the only relevant field for identifying psychiatric illnesses, which cannot be inferred from the given data. So, D is also incorrect.
So, B is the correct option

Q.8

Round-trip tickets to a tourist destination are eligible for a discount of 10% on the total fare. In addition, groups of 4 or more get a discount of 5% on the total fare. If the one way single person fare is Rs 100, a group of 5 tourists purchasing round-trip tickets will be charged Rs _________.

850

900

800

1000

Ans .

Explanation :
One way single person fare is 100. Therefore 2 way is 200.
For 5 persons it is 200 x 5 = 1000
Now 1st discount is of 10 % on total fare, which is 100( 10 % of 1000). And as the group is of more than 4, an additional discount of 5 % on total fare, which is 50( 5 % of 1000).
Hence total discount is equal to 100 + 50 = 150.
Therefore tickets are charged at Rs 1000 - 150 = 850.

Q.9

Suppose you break a stick of unit length at a point chosen uniformly at random. Then the expected length of the shorter stick is ________

0.24 to 0.27

0.15 to 0.30

0.20 to 0.30

0.10 to 0.15

Ans .

Explanation :
The smaller sticks will range in length from almost 0 unit up to a maximum of 0.5 unit, with each length equally possible. Therefore, the average length will be about (0 + 0.5)/2 = 0.25 unit, which is answer. Another approach - See math.stackexchange.com/questions/350679/we-break-a-unit-length-rod-into-two-pieces-at-a-uniformly-chosen-point-find-the/350863#350863 for explanation

Q.10

When a point inside of a tetrahedron (a solid with four triangular surfaces) is connected by straight lines to its corners, how many (new) internal planes are created with these lines? _____________

Ans .

Explanation :
The Tetrahedron has 4 triangular surfaces with 4 vertices/corners ( say A,B,C and D) as can be seen here. http://www.sjsu.edu/faculty/wa... Now if you take a point inside a tetrahedron (suppose O) and connect it with any two of its corners which are nothing but vertices( suppose A and B), you will get 1 internal plane as OAB. So we can see from here that, no of new internal planes = no of different pair of corners or vertices Similarly you can take any other 2 corners like (A,C) or (A,D) or (B,C) or (B,D) or (C,D), hence total possible pair of corners are 6. Therefore 6 new internal planes possible. We could also calculate the possible corners by using combinations formula, which is nCr, i.e. no of ways to select a combination of r things from a given set of n things. here n = 4 ( as total 4 vertices, A,B,C and D) and r =2 ( as we need two corners at a time) Thus, 4C2 = 6.

Q.11

Consider the statement “Not all that glitters is gold” Predicate glitters(x) is true if x glitters and predicate gold(x) is true if x is gold. Which one of the following logical formulae represents the above statement?

Ans .

Explanation :
The statement “Not all that glitters is gold” can be expressed as follows :
¬(∀x(glitters(x)⇒gold(x)) ... (1)
Where ∀x(glitters(x)⇒gold(x) refers that all glitters is gold. Now ,
∃x¬(glitters(x)⇒gold(x)) ... (2) , Since we know ¬∀x() = ∃x¬()
(Where ∀ refers to -> All and ∃x refers to -> There exists some).
As we know, A⇒B is true only in the case that either A is false or B is true. It can also defined in the other way :
A⇒B=¬A∨B (negationA or B ) ... (3)
From equation (2) and (3) , we have ∃x(¬(¬glitters(x)∨gold(x))
⇒∃x(glitters(x)∧¬gold(x)) ... (4) , Negation cancellation ¬(¬) = () : and ¬(()∨()) = (¬()∧¬()) .
So Answer is (D) .

Q.12

In a survey, 300 respondents were asked whether they own a vehicle or not. If yes, they were further asked to mention whether they own a car or scooter or both. The irresponses are tabulated below. What percent of respondents do not own a scooter?

Ans .

Explanation :
No. of respondents do not own scooter = No. of persons who have car alone + No. of persons who don't have both.
In men, 40 + 20 = 60 and
In women 34 + 50 = 84.
Total 60 + 84 = 144 .
Now the percentage is, (144/300) *100 = 48

Q.13

Let G=(V,E) be a directed graph where V is the set of vertices and E the set of edges. Then which one of the following graphs has the same strongly connected components as G ?

Ans .

Explanation :
If we reverse directions of all arcs in a graph, the new graph has same set of strongly connected components as the original graph. Se http://www.geeksforgeeks.org/strongly-connected-components/ for more details.

Q.20

Let G be a graph with n vertices and m edges. What is the tightest upper bound on the running time on Depth First Search of G? Assume that the graph is represented using adjacency matrix.

O(n)

O(m+n)

O(n²)

O(mn)

Ans .

Explanation :
Depth First Search of a graph takes O(m+n) time when the graph is represented using adjacency list. In adjacency matrix representation, graph is represented as an "n x n" matrix. To do DFS, for every vertex, we traverse the row corresponding to that vertex to find all adjacent vertices (In adjacency list representation we traverse only the adjacent vertices of the vertex). Therefore time complexity becomes O(n²)

Q.21

Consider a rooted Binary tree represented using pointers. The best upper bound on the time required to determine the number of subtrees having having exactly 4 nodes O(na Logn b). Then the value of a + 10b is ________

Ans .

Explanation :
We can find the subtree with 4 nodes in O(n) time. Following can be a simple approach.
1) Traverse the tree in bottom up manner and find size of subtree rooted with current node
2) If size becomes 4, then print the current node.

Following is C implementation

#include
#include

struct Node
{
int data;
struct Node *left, *right;
};

// A utility function to create a new Binary Tree Node
struct Node *newNode(int item)
{
struct Node *temp = (struct Node *)malloc(sizeof(struct Node));
temp->data = item;
temp->left = temp->right = NULL;
return temp;
}

int print4Subtree(struct Node *root)
{
if (root == NULL)
return 0;
int l = print4Subtree(root->left);
int r = print4Subtree(root->right);
if ((l + r + 1) == 4)
printf("%d ", root->data);
return (l + r + 1);
}

// Driver Program to test above functions
int main()
{
struct Node *root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
root->right->left = newNode(6);
root->right->right = newNode(7);
root->right->left->right = newNode(8);

print4Subtree(root);

return 0;
}

Q.22

Consider the directed graph given below. Which one of the following is TRUE? gate paper solved

The graph doesn't have any topological ordering

Both PQRS and SRPQ are topological ordering

Both PSRQ and SPRQ are topological ordering

PSRQ is the only topological ordering

Ans .

Explanation :
The graph doesn't contain any cycle, so there exist topological ordering. P and S must appear before R and Q because there are edges from P to R and Q, and from S to R and Q. See Topological Ordering for more details.

Q.23

Let P be a QuickSort Program to sort numbers in ascending order using the first element as pivot. Let t1 and t2 be the number of comparisons made by P for the inputs {1, 2, 3, 4, 5} and {4, 1, 5, 3, 2} respectively. Which one of the following holds?

t1 = 5

t1 < t2

t1 > t2

t1 = t2

Ans .

Explanation :
When first element or last element is chosen as pivot, Quick Sort's worst case occurs for the sorted arrays. In every step of quick sort, numbers are divided as per the following recurrence. T(n) = T(n-1) + O(n)

Q.24

GATECS2014Q25

Ans .

Explanation :
(A) L = {a n b n |n >= 0} is not regular because there does not exists a finite automaton that can
derive this grammar. Intuitively, finite automaton has finite memory, hence it can’t track
number of as. It is a standard CFL though.

(B) L = {a n b n |n is prime} is again not regular because there is no way to remember/check if
current n is prime or not. Hence, no finite automaton exists to derive this grammar, thus it
is not regular.

(C) L = {w|w has 3k+1 bs} is a regular language because k is a fixed constant and we can easily
emulate L as a ∗ ba ∗ …..ba ∗ such that there are exactly 3k + 1 bs and a ∗ s surrounding each b in
the grammar.

v_24
(D) L = {ww| w ∈ Σ ∗ } is again not a regular grammar, infact it is not even a CFG. There is no
way to remember and derive double word using finite automaton.

Hence, correct answer would be (C).

Q.25

GATECS2014Q26

{q0, q1, q2}

{q0, q1}

{q0, q1, q2, q3}

{q3}

Ans .

Explanation :
So, q0, q1 and q2 are reachable states for the input string 0011, but q3 is not. So, option (A) is answer.

Q.26

A machine has a 32-bit architecture, with 1-word long instructions. It has 64 registers, each of which is 32 bits long. It needs to support 45 instructions, which have an immediate operand in addition to two register operands. Assuming that the immediate operand is an unsigned integer, the maximum value of the immediate operand is ____________.

Ans .

16383

Explanation :
1 Word = 32 bits

Each instruction has 32 bits

To support 45 instructions, opcode must contain 6-bits

Register operand1 requires 6 bits, since the total registers are 64.

Register operand 2 also requires 6 bits.

14-bits are left over for immediate Operand Using 14-bits, we can give maximum 16383,

Since 2^14=16384(from 0 to 16383)

Q.27

Which one of the following is FALSE?

A basic block is a sequence of instructions where control enters the sequence at the beginning and exits at the end.

Available expression analysis can be used for common subexpression elimination.

Live variable analysis can be used for dead code elimination.

x = 4 ∗ 5 => x = 20 is an example of common subexpression elimination

Ans .

Explanation :
A) A basic block is a sequence of instructions where control enters the sequence at the beginning and exits at the end is TRUE. (B) Available expression analysis can be used for common subexpression elimination is TRUE. Available expressions is an analysis algorithm that determines for each point in the program the set of expressions that need not be recomputed. Available expression analysis is used to do global common subexpression elimination (CSE). If an expression is available at a point, there is no need to re-evaluate it. (C)Live variable analysis can be used for dead code elimination is TRUE. (D) x = 4 ∗ 5 => x = 20 is an example of common subexpression elimination is FALSE. Common subexpression elimination (CSE) refers to compiler optimization replaces identical expressions (i.e., they all evaluate to the same value) with a single variable holding the computed value when it is worthwhile to do so. Below is an example
In the following code:

a = b * c + g;
d = b * c * e;

it may be worth transforming the code to:

tmp = b * c;
a = tmp + g;
d = tmp * e;

Q.28

1) Waterfall model        a) Specifications can be
                              developed incrementally

2) Evolutionary model     b) Requirements compromises 
                             are inevitable

3) Component-based        c) Explicit recognition of risk
software engineering

4) Spiral development     d) Inflexible partitioning of 
                             the project into stages

1-a, 2-b, 3-c, 4-d

1-d, 2-a, 3-b, 4-c

1-d, 2-b, 3-a, 4-c

1-c, 2-a, 3-b, 4-d

Ans .

Explanation :
Waterfall Model: We can not go back in previous project phase as soon as as we proceed to next phase ,So inflexible

Evolutionary: It keeps changing with evolution so incremental in nature

Component based: Reuse-based approach to defining, implementing and composing loosely coupled independent components into systems

Spiral: Spiral model is the most advanced .It includes four faces one of which is Risk.

Phases: Planning, Risk Analysis, Engineering and Evaluation

Q.29

Suppose a disk has 201 cylinders, numbered from 0 to 200. At some time the disk arm is at cylinder 100, and there is a queue of disk access requests for cylinders 30, 85, 90, 100, 105, 110, 135 and 145. If Shortest-Seek Time First (SSTF) is being used for scheduling the disk access, the request for cylinder 90 is serviced after servicing ____________ number of requests.

Ans .

Explanation :
In Shortest-Seek-First algorithm, request closest to the current position of the disk arm and head is handled first. In this question, the arm is currently at cylinder number 100. Now the requests come in the queue order for cylinder numbers 30, 85, 90, 100, 105, 110, 135 and 145. The disk will service that request first whose cylinder number is closest to its arm. Hence 1st serviced request is for cylinder no 100 ( as the arm is itself pointing to it ), then 105, then 110, and then the arm comes to service request for cylinder 90. Hence before servicing request for cylinder 90, the disk would had serviced 3 requests. Hence option C.

Q.30

Which one of the following is FALSE?

User level threads are not scheduled by the kernel.

When a user level thread is blocked, all other threads of its process are blocked.

Context switching between user level threads is faster than context switching between kernel level threads.

Kernel level threads cannot share the code segment

Ans .

Explanation :

User level thread Kernel level thread

User thread are implemented by user processes. kernel threads are implemented by OS.

OS doesn’t recognized user level threads. Kernel threads are recognized by OS.

Implementation of User threads is easy. Implementation of Kernel thread is complicated.

Context switch time is less. Context switch time is more.

Context switch requires no hardware support. Hardware support is needed.

If one user level thread perform blocking operation then entire process will be blocked. If one kernel thread perform blocking operation then another thread can continue execution.

Example : Java thread, POSIX threads. Example : Window Solaris.

User level thread	Kernel level thread
User thread are implemented by user processes.	kernel threads are implemented by OS.
OS doesn’t recognized user level threads.	Kernel threads are recognized by OS.
Implementation of User threads is easy.	Implementation of Kernel thread is complicated.
Context switch time is less.	Context switch time is more.
Context switch requires no hardware support.	Hardware support is needed.
If one user level thread perform blocking operation then entire process will be blocked.	If one kernel thread perform blocking operation then another thread can continue execution.
Example : Java thread, POSIX threads.	Example : Window Solaris.

Q.31

Consider the relation scheme R = {E, F, G, H, I, J, K, L, M, M} and the set of functional dependencies {{E, F} -> {G}, {F} -> {I, J}, {E, H} -> {K, L}, K -> {M}, L -> {N} on R. What is the key for R?

{E, F}

{E, F, H}

{E, F, H, K, L}

{E}

Ans .

Explanation :
All attributes can be derived from {E, F, H} To solve these kind of questions that are frequently asked in GATE paper, try to solve it by using shortcuts so that enough amount of time can be saved. Fist Method: Using the given options try to obtain closure of each options. The solution is the one that contains R and also minimal Super Key, i.e Candidate Key.

A) {EF}+ = {EFGIJ} ≠ R(The given relation)

B) {EFH}+ = {EFGHIJKLMN} = R (Correct since each member of the given relation is determined)

C) {EFHKL}+ = {EFGHIJKLMN} = R (Not correct although each member of the given relation can be determined but it is not minimal, since by the definition of Candidate key it should be minimal Super Key)

D) {E}+ = {E} ≠ R

Second Method:
Since, {EFGHIJKLMN}+ = {EFGHIJKLMN}

{EFGHIJKLM}+ = {EFGHIJKLMN} ( Since L -> {N}, hence can replace N by L)

In a similar way K -> {M} hence replace M by K

{EFGHIJKL}+ = {EFGHIJKLMN}

Again {EFGHIJ}+ = {EFGHIJKLMN} (Since {E, H} -> {K, L}, hence replace KL by EH)

{EFGH}+ = {EFGHIJKLMN} (Since {F} -> {I, J} )

{EFH}+ = {EFGHIJKLMN} (Since {E, F} -> {G} )

Q.32

Given the following statements:

    S1: A foreign key declaration can always 

        be replaced by an equivalent check

        assertion in SQL.

    S2: Given the table R(a,b,c) where a and

        b together form the primary key, the 

        following is a valid table definition.

        CREATE TABLE S (

            a INTEGER,

            d INTEGER,

            e INTEGER,

            PRIMARY KEY (d),

            FOREIGN KEY (a) references R) 

Which one of the following statements is CORRECT?

S1 is TRUE and S2 is FALSE.

Both S1 and S2 are TRUE.

S1 is FALSE and S2 is TRUE.

Both S1 and S2 are FALSE

Ans .

Explanation :
S1: A foreign key declaration can always
be replaced by an equivalent checkS1: A foreign key declaration can always
be replaced by an equivalent check
assertion in SQL.
False: Check assertions are not sufficient to replace foreign key. Foreign key declaration may have cascade delete which is not possible by just check insertion.
S2: Given the table R(a,b,c) where a and
b together form the primary key, the
following is a valid table definition.
CREATE TABLE S (
a INTEGER,
d INTEGER,
e INTEGER,
PRIMARY KEY (d),
FOREIGN KEY (a) references R)
False: Foreign key in one table should uniquely identifies a row of other table. In above table definition, table S has a foreign key that refers to field 'a' of R. The field 'a' in table S doesn't uniquely identify a row in table R.
assertion in SQL.
False: Check assertions are not sufficient to replace foreign key. Foreign key declaration may have cascade delete which is not possible by just check insertion.
S2: Given the table R(a,b,c) where a and
b together form the primary key, the
following is a valid table definition.
CREATE TABLE S (
a INTEGER,
d INTEGER,
e INTEGER,
PRIMARY KEY (d),
FOREIGN KEY (a) references R)
False: Foreign key in one table should uniquely identifies a row of other table. In above table definition, table S has a foreign key that refers to field 'a' of R. The field 'a' in table S doesn't uniquely identify a row in table R.

Q.35

Which of the following are used to generate a message digest by the network security protocols? (P) RSA
(Q) SHA-1
(R) DES
(S) MD5

P and R only

Q and R only

Q and S only

R and S only

Ans .

Explanation :
RSA – It is an algorithm used to encrypt and decrypt messages.
SHA 1 – Secure Hash Algorithm 1, or SHA 1 is a cryptographic hash function. It produces a 160 bit (20 byte) hash value (message digest).
DES – Data Encryption Standard, or DES is a symmetric key algorithm for encryption of electronic data.
MD5 – Message Digest 5, or MD5 is a widely used cryptographic hash function that produces a 128 bit hash value (message digest).
Q and S i.e SHA 1 and MD5 are used to generate a message digest by the network security protocols. So, C is the correct choice.

Q.36

Which of the following are used to generate a message digest by the network security protocols? (P) RSA
(Q) SHA-1
(R) DES
(S) MD5

P and R only

Q and R only

Q and S only

R and S only

Ans .

Explanation :
RSA – It is an algorithm used to encrypt and decrypt messages.
SHA 1 – Secure Hash Algorithm 1, or SHA 1 is a cryptographic hash function. It produces a 160 bit (20 byte) hash value (message digest).
DES – Data Encryption Standard, or DES is a symmetric key algorithm for encryption of electronic data.
MD5 – Message Digest 5, or MD5 is a widely used cryptographic hash function that produces a 128 bit hash value (message digest).
Q and S i.e SHA 1 and MD5 are used to generate a message digest by the network security protocols. So, C is the correct choice.

Q.37

Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of the connection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.

1100 to 1300

800 to 1000

1400 to 1600

1500 to 1700

Ans .

Explanation :

Current size of congestion window in terms of number of segments
                            = (Size in Bytes)/(Maximum Segment Size)
                            = 32KB / 2KB 
                            = 16 MSS
          
When timeout occurs, in TCP's Slow Start algorithm, threshold is 
reduced to half which is 16KB or 8MSS. Also, slow start phase begins 
where congestion window is increased twice. 
So from 1MSS to 8 MSS window size will grow exponentially. 
Congestion window becomes 2MSS after one RTT and becomes 4MSS after
2 RTTs and 8MSS after 3 RTTs.  At 8MSS, threshold is reached and
congestion avoidance phase begins.  In congestion avoidance phase,
window is increased linearly. So to cover from 8MSS to 16MSS, it needs
8 RTTs

Together, 11RTTs are needed (3 in slow start phase and 8 in congestion
avoidance phase).

Q.38

Consider a selective repeat sliding window protocol that uses a frame size of 1 KB to send data on a 1.5 Mbps link with a one-way latency of 50 msec. To achieve a link utilization of 60%, the minimum number of bits required to represent the sequence number field is ________.

Ans .

Explanation :

Transmission delay = Frame Size/bandwidth
                   = (1*8*10^3)/(1.5 * 10^6)=5.33ms
Propagation delay = 50ms
Efficiency = Window Size/(1+2a) = .6

a = Propagation delay/Transmission delay
So, window size = 11.856(approx)
min sequence number = 2*window size = 23.712
bits required in Min sequence number = log2(23.712)
Answer is 4.56

Ceil(4.56) = 5

Q.39

Consider the following four schedules due to three transactions (indicated by the subscript) using read and write on a data item x, denoted by r(x) and w(x) respectively. Which one of them is conflict serializable. gate paper solved

Ans .

Explanation :
In option D, there is no interleaving of operations. The option D has first all operations of transaction 2, then 3 and finally 1 There can not be any conflict as it is a serial schedule with sequence 2 --> 3 -- > 1

Q.40

S1: Every table with two single-valued 
      attributes is in 1NF, 2NF, 3NF and BCNF.

  S2: AB->C, D->E, E->C is a minimal cover for 
      the set of functional dependencies 
      AB->C, D->E, AB->E, E->C.

Which one of the following is CORRECT?

S1 is TRUE and S2 is FALSE.

Both S1 and S2 are TRUE.

S1 is FALSE and S2 is TRUE.

Both S1 and S2 are FALSE.

Ans .

Explanation :

S1: Every table with two single-valued 
      attributes is in 1NF, 2NF, 3NF and BCNF.
A relational schema R is in BCNF iff in Every non-trivial Functional Dependency X->Y, X is Super Key. If we can prove the relation is in BCNF then by default it would be in 1NF, 2NF, 3NF also. Let R(AB) be a two attribute relation, then
If {A->B} exists then BCNF since {A}+ = AB = R
If {B->A} exists then BCNF since {B}+ = AB = R
If {A->B,B->A} exists then BCNF since A and B both are Super Key now.
If {No non trivial Functional Dependency} then default BCNF.
Hence it's proved that a Relation with two single - valued attributes is in BCNF hence its also in 1NF, 2NF, 3NF. Hence S1 is true.
S2: AB->C, D->E, E->C is a minimal cover for 
      the set of functional dependencies 
      AB->C, D->E, AB->E, E->C.
As we know Minimal Cover is the process of eliminating redundant Functional Dependencies and Extraneous attributes in Functional Dependency Set. So each dependency of F = {AB->C, D->E, AB->E, E->C} should be implied in minimal cover. As we can see AB->E is not covered in minimal cover since {AB}+ = ABC in the given cover {AB->C, D->E, E->C} Hence, S2 is false.

Q.41

An operating system uses the Banker’s algorithm for deadlock avoidance when managing the allocation of three resource types X, Y, and Z to three processes P0, P1, and P2. The table given below presents the current system state. Here, the Allocation matrix shows the current number of resources of each type allocated to each process and the Max matrix shows the maximum number of resources of each type required by each process during its execution. GATECS2014Q42 There are 3 units of type X, 2 units of type Y and 2 units of type Z still available. The system is currently in a safe state. Consider the following independent requests for additional resources in the current state:

REQ1: P0 requests 0 units of X,  
      0 units of Y and 2 units of Z
REQ2: P1 requests 2 units of X, 
      0 units of Y and 0 units of Z

Which one of the following is TRUE?

Only REQ1 can be permitted.

Only REQ2 can be permitted

Both REQ1 and REQ2 can be permitted.

Neither REQ1 nor REQ2 can be permitted

Ans .

Explanation :
This is the current safe state.

AVAILABLE X=3, Y=2, Z=2

MAX ALLOCATION

X Y Z X Y Z

P0 8 4 3 0 0 1

P1 6 2 0 3 2 0

P2 3 3 3 2 1 1

Now, if the request REQ1 is permitted, the state would become :

AVAILABLE X=3, Y=2, Z=0

MAX ALLOCATION NEED

X Y Z X Y Z X Y Z

P0 8 4 3 0 0 3 8 4 0

P1 6 2 0 3 2 0 3 0 0

P2 3 3 3 2 1 1 1 2 2

Now, with the current availability, we can service the need of P1. The state would become :

AVAILABLE X=6, Y=4, Z=0

MAX ALLOCATION NEED

X Y Z X Y Z X Y Z

P0 8 4 3 0 0 3 8 4 0

P1 6 2 0 3 2 0 0 0 0

P2 3 3 3 2 1 1 1 2 2

With the resulting availability, it would not be possible to service the need of either P0 or P2, owing to lack of Z resource. Therefore, the system would be in a deadlock. ⇒ We cannot permit REQ1. Now, at the given safe state, if we accept REQ2 :

AVAILABLE X=1, Y=2, Z=2

MAX ALLOCATION NEED

X Y Z X Y Z X Y Z

P0 8 4 3 0 0 1 8 4 2

P1 6 2 0 5 2 0 1 0 0

P2 3 3 3 2 1 1 1 2 2

With this availability, we service P1 (P2 can also be serviced). So, the state is :

AVAILABLE X=7, Y=4, Z=2

MAX ALLOCATION NEED

X Y Z X Y Z X Y Z

P0 8 4 3 0 0 1 8 4 2

P1 6 2 0 5 2 0 0 0 0

P2 3 3 3 2 1 1 1 2 2

With the current availability, we service P2. The state becomes :

AVAILABLE X=10, Y=7, Z=5

MAX ALLOCATION NEED

X Y Z X Y Z X Y Z

P0 8 4 3 0 0 1 8 4 2

P1 6 2 0 5 2 0 0 0 0

P2 3 3 3 2 1 1 0 0 0

Finally, we service P0. The state now becomes :

AVAILABLE X=18, Y=11, Z=8

MAX ALLOCATION NEED

X Y Z X Y Z X Y Z

P0 8 4 3 0 0 1 0 0 0

P1 6 2 0 5 2 0 0 0 0

P2 3 3 3 2 1 1 0 0 0

The state so obtained is a safe state. ⇒ REQ2 can be permitted. So, only REQ2 can be permitted. Hence, B is the correct choice. Please comment below if you find anything wrong in the above post.

Q.42

Consider the following set of processes that need to be scheduled on a single CPU. All the times are given in milliseconds.

Process Name      Arrival Time      Execution Time
    A                  0                  6
    B                  3                  2
    c                  5                  4
    D                  7                  6
    E                  10                 3

Using the shortest remaining time first scheduling algorithm, the average process turnaround time (in msec) is ____________________

7.2

7.5.

Ans .

Explanation :
Turn around time of a process is total time between submission of the process and its completion. Shortest remaining time (SRT) scheduling algorithm selects the process for execution which has the smallest amount of time remaining until completion. Solution: Let the processes be A, ,C,D and E. These processes will be executed in following order. Gantt chart is as follows: First 3 sec, A will run, then remaining time A=3, B=2,C=4,D=6,E=3 Now B will get chance to run for 2 sec, then remaining time. A=3, B=0,C=4,D=6,E=3 Now A will get chance to run for 3 sec, then remaining time. A=0, B=0,C=4,D=6,E=3 By doing this way, you will get above gantt chart. Scheduling table: As we know, turn around time is total time between submission of the process and its completion. i.e turn around time=completion time-arrival time. i.e. TAT=CT-AT Turn around time of A = 8 (8-0) Turn around time of B = 2 (5-3) Turn around time of C = 7 (12-5) Turn around time of D = 14 (21-7) Turn around time of E = 5 (15-10) Average turn around time is (8+2+7+14+5)/5 = 7.2. Answer is 7.2.
Reference: https://www.cs.uic.edu/~jbell/CourseNotes/OperatingSystems/5_CPU_Scheduling.html This solution is contributed by Nitika Bansal Alternate Explanatio:
```
After drawing Gantt Chart

Completion Time for processes A, B, C, D 
and E are 8, 5, 12, 21 and 15 respectively.

Turnaround Time = Completion Time - Arrival Time
```

Q.43

Assume that there are 3 page frames which are initially empty. If the page reference string is 1, 2, 3, 4, 2, 1, 5, 3, 2, 4, 6, the number of page faults using the optimal replacement policy is__________.

Ans .

Explanation :

In optimal page replacement replacement policy, we replace the place which is not used for longest duration in future.

Given three page frames.

Reference string is 1, 2, 3, 4, 2, 1, 5, 3, 2, 4, 6

Initially, there are three page faults and entries are
1  2  3

Page 4 causes a page fault and replaces 3 (3 is the longest
distant in future), entries become
1  2  4
Total page faults =  3+1 = 4

Pages 2 and 1 don't cause any fault.

5 causes a page fault and replaces 1, entries become
5  2  4
Total page faults =  4 + 1 = 5

3 causes a page fault and replaces 1, entries become
3  2  4
Total page faults =  5 + 1 = 6

3, 2 and 4 don't cause any page fault.

6 causes a page fault.
Total page faults =  6 + 1 = 7

Q.44

A canonical set of items is given below

S --> L. > R
Q --> R.

On input symbol < the set has

a shift-reduce conflict and a reduce-reduce conflict.

a shift-reduce conflict but not a reduce-reduce conflict.

a reduce-reduce conflict but not a shift-reduce conflict.

neither a shift-reduce nor a reduce-reduce conflict.

Ans .

Explanation :
The question is asked with respect to the symbol ' < ' which is not present in the given canonical set of items. Hence it is neither a shift-reduce conflict nor a reduce-reduce conflict on symbol '<'. Hence D is the correct option. But if the question would have asked with respect to the symbol ' > ' then it would have been a shift-reduce conflict.

Q.45

Let L be a language and L' be its complement. Which one of the following is NOT a viable possibility?

Neither L nor L' is recursively enumerable (r.e.).

One of L and L' is r.e. but not recursive; the other is not r.e.

Both L and L' are r.e. but not recursive.

Both L and L' are recursive

Ans .

Explanation :
A) It is possible if L itself is NOT RE. Then L' will also not be RE. B) Suppose there is a language such that turing machine halts on the input. The given language is RE but not recursive and its complement is NOT RE. C) This is not possible because if we can write enumeration procedure for both languages and it's complement, then the language becomes recursive. D) It is possible because L is closed under complement if it is recursive. Thus, C is the correct choice.

Q.46

Which of the regular expressions given below represent the following DFA? gate paper solved

I) 0*1(1+00*1)*
II) 0*1*1+11*0*1
III) (0+1)*1

I and II only

I and III only

II and III only

I, II, and III

Ans .

Explanation :

I) 0*1(1+00*1)*
II) 0*1*1+11*0*1
III) (0+1)*1 

(I) and  (III) represent DFA.

(II) doesn't represent as the DFA accepts strings like 11011,
    but the regular expression doesn't accept.

Q.47

There are 5 bags labeled 1 to 5. All the coins in a given bag have the same weight. Some bags have coins of weight 10 gm, others have coins of weight 11 gm. I pick 1, 2, 4, 8, 16 coins respectively from bags 1 to 5. Their total weight comes out to 323 gm. Then the product of the labels of the bags having 11 gm coins is ___.

Ans .

Explanation :

There are 5 bags numbered 1 to 5.

We don't know how many bags contain 10 gm and 
11 gm coins.

We only know that the total weights of coins is 323.

Now the idea here is to get 3 in the place of total 
sum's unit digit.

Mark no 1 bag as having 11 gm coins.
Mark no 2 bag as having 10 gm coins.
Mark no 3 bag as having 11 gm coins.
Mark no 4 bag as having 11 gm coins.
Mark no 5 bag as having 10 gm coins.

Note: The above marking is done after getting false 
results for some different permutations, the permutations
which were giving 3 in the unit place of the total sum.

Now, we have picked 1, 2, 4, 8, 16 coins respectively
from bags 1 to 5.

Hence total sum coming from each bag from 1 to 5 is 11,
20, 44, 88, 160 gm respectively.

For the above combination we are getting 3 as unit digit
in sum.

Lets find out the total sum, it's 11 + 20 + 44 + 88 + 160 = 323.

So it's coming right.

Now 11 gm coins containing bags are 1, 3 and 4.
Hence, the product is : 1 x 3 x 4 = 12.

Q.48

Suppose a polynomial time algorithm is discovered that correctly computes the largest clique in a given graph. In this scenario, which one of the following represents the correct Venn diagram of the complexity classes P, NP and NP Complete (NPC)?

Ans .

Explanation :
Clique is an NP complete problem. If one NP complete problem can be solved in polynomial time, then all of them can be. So NPC set becomes equals to P.

Q.49

The minimum number of comparisons required to find the minimum and the maximum of 100 numbers is _________________.

148

147

146

140

Ans .

Explanation :
To find minimum and maximum element out of n numbers, we need to have at least (3n/2 -2) comparisons. By putting the value n=100, (3*100/2)-2 = 148.

Q.50

Consider the following C function in which size is the number of elements in the array E: The value returned by the function MyX is the

int MyX(int *E, unsigned int size)
{
    int Y = 0;
    int Z;
    int i, j, k;
    for(i = 0; i < size; i++)
        Y = Y + E[i];
    for(i = 0; i < size; i++)
        for(j = i; j < size; j++)
        {
            Z = 0;
            for(k = i; k <= j; k++)
                Z = Z + E[k];
            if (Z > Y)
                Y = Z;
        }
    return Y;
}

maximum possible sum of elements in any sub-array of array E.

maximum element in any sub-array of array E.

sum of the maximum elements in all possible sub-arrays of array E

the sum of all the elements in the array E.

Ans .

Explanation :
The function does following Y is used to store maximum sum seen so far and Z is used to store current sum 1) Initialize Y as sum of all elements 2) For every element, calculate sum of all subarrays starting with arr[i]. Store the current sum in Z. If Z is greater than Y, then update Y.

Q.51

Consider the following pseudo code. What is the total number of multiplications to be performed?

D = 2
for i = 1 to n do
   for j = i to n do
      for k = j + 1 to n do
           D = D * 3

Half of the product of the 3 consecutive integers.

One-third of the product of the 3 consecutive integers.

One-sixth of the product of the 3 consecutive integers.

None of the above

Ans .

Explanation :
The statement “D = D * 3” is executed n*(n+1)*(n-1)/6 times. Let us see how. For i = 1, the multiplication statement is executed (n-1) + (n-2) + .. 2 + 1 times. For i = 2, the statement is executed (n-2) + (n-3) + .. 2 + 1 times ……………………….. ………………………. For i = n-1, the statement is executed once. For i = n, the statement is not executed at all So overall the statement is executed following times [(n-1) + (n-2) + .. 2 + 1] + [(n-2) + (n-3) + .. 2 + 1] + … + 1 + 0 The above series can be written as S = [n*(n-1)/2 + (n-1)*(n-2)/2 + ….. + 1] The sum of above series can be obtained by trick of subtraction the series from standard Series S1 = n2 + (n-1)2 + .. 12. The sum of this standard series is n*(n+1)*(2n+1)/6 S1 – 2S = n + (n-1) + … 1 = n*(n+1)/2 2S = n*(n+1)*(2n+1)/6 – n*(n+1)/2 S = n*(n+1)*(n-1)/6

Q.52

Consider a hash table with 9 slots. The hash function is ℎ(k) = k mod 9. The collisions are resolved by chaining. The following 9 keys are inserted in the order: 5, 28, 19, 15, 20, 33, 12, 17, 10. The maximum, minimum, and average chain lengths in the hash table, respectively, are

3, 0, and 1

3, 3, and 3

4, 0, and 1

3, 0, and 2

Ans .

Explanation :

Following are values of hash function for all keys
 5 --> 5
28 --> 1
19 --> 1  [Chained with 28]
15 --> 6
20 --> 2
33 --> 6  [Chained with 15]
12 --> 3
17 --> 8
10 --> 1 [Chained with 28 and 19]
The maximum chain length is 3. The keys 28, 19 and 10 go to same slot 1, and form a chain of length 3. The minimum chain length 0, there are empty slots (0, 4 and 7). Average chain length is (0 + 3 + 1 + 1 + 0 + 1 + 2 + 0 + 1)/9 = 1

Q.53

Consider a 6-stage instruction pipeline, where all stages are perfectly balanced.Assume that there is no cycle-time overhead of pipelining. When an application is executing on this 6-stage pipeline, the speedup achieved with respect to non-pipelined execution if 25% of the instructions incur 2 pipeline stall cycles is

Ans .

Explanation :

It was a numerical digit type question so answer must be 4.

As for 6 stages, non-pipelining takes 6 cycles.

There were 2 stall cycles for pipelining for 25% of the instructions

So pipe line time = (1+(25/100)*2) = 1.5

Speed up = Non pipeline time/Pipeline time = 6/1.5 = 4

Q.54

An access sequence of cache block address is of length N and contains n unique block addresses. The number of unique block addresses between two consecutive accesses to the same block address is bounded above by k. What is the miss ration is the access sequence is passed through a cache of associativity A >= k exercising least-recently used replacement policy.

n/N

1/N

1/A

k/n

Ans .

Explanation :

Their are N access request for the cache blocks out this n
blocks are unique .

In between two access of the same block their are request of 
(k-1) other block block.

And if their associativity >=k and use LRU, then
there will be only one cache miss for every unique block i.e.,
n and it will be the time when the enter the cahe for the first 
time.  Therefore Miss ratio =(Cache miss)/(No. of request) = n/N

Q.55

Consider a 4-to-1 multiplexer with two select lines S1 and S0, given below GATECS2014Q55 The minimal sum-of-products form of the Boolean expression for the output F of the multiplexer is

P'Q + QR' + PQ'R

P'Q + P'QR' + PQR' + PQ'R

P'QR + P'QR' + QR' + PQ'R

PQR'

Ans .

Explanation :

For 4 to 1 mux =p’q’(0)+p’q(1)+pq’r+pqr’ 
                    =p’q+pq’r+pqr’ 
					=q(p’+pr’)+pq’r
					=q(p’+r’)+pq’r 
					=p’q+qr’+pq’r 
					Ans (a)

Q.56

The function f(x) = x sinx satisfies the following equation f''(x) + f(x) + t cosx = 0. The value of t is

-1

-2

Ans .

Explanation :

We have f(x) = x sin x

⇒ f'(x) = x cos x + sin x

⇒ f′′(x) = x (− sin x ) + cos x + cos x = (−x sin x ) + 2 cos x

Now, it is given that f(x) = x sin x satisfies the equation f′′(x) + f(x) + t cos x = 0

⇒ (−x sin x ) + 2 cos x + x sin x + t cos x = 0

⇒ 2 cos x + t cos x = 0

⇒ cos x ( t + 2 ) = 0

⇒ t + 2 = 0

⇒ t = −2

Q.57

A function is continuous in the interval [0, 2]. It is known that f(0) = f(2) = -1 and f(1) = 1. Which one of the following statements must be true.

There exist a y in the interval (0, 1), such that f(y) = f(y+1)

For every y in the interval (0, 1) , f(y) = f(2-y)

The maximum value of the function in the interval (0, 2) is 1

There exist a y in the interval (0, 1), such that f(y) = -f(2-y)

Ans .

Explanation :
Since the function is continuous, there must be point between 0 and 1 where it becomes 0 and there must also be a point between 1 and 2 where it becomes 0.

Q.58

Four fair six-sided dice are rolled. The probability that the sum being 22 is X/1296. The value of X is

Ans .

Explanation :

In general, Probability (of an event ) 
= No of favorable outcomes to the event / Total number of possible outcomes in the random experiment.
 Here, 4 six-faces dices are tossed, for one dice there can be 6 equally likely and mutually exclusive outcomes. 
Taking 4 together, there can be total number of 6*6*6*6 = 1296 possible outcomes.
 Now, No of favorable cases to the event : here event is getting sum as 22.
 So, there can be only 2 cases possible.
 Case 1: Three 6's and one 4, for example: 6,6,6,4 ( sum is 22)
 Hence, No of ways we can obtain this = 4!/3! = 4 ways 
( 3! is for removing those cases where all three 6 are swapping among themselves) 
Case 2: Two 6's and two 5's,for example: 6,6,5,5 ( sum is 22) 
Hence, No of ways we can obtain this = 4! /( 2! * 2!) = 6 ways 
( 2! is for removing those cases where both 6 are swapping between themselves, similarly for both 5 also) 
Hence total no of favorable cases = 4 + 6 = 10. 
Hence probability = 10/1296. 
Therefore option D.

Q.59

A pennant is a sequence of numbers, each number being 1 or 2. An n-pennant is a sequence of numbers with sum equal to n. For example, (1,1,2) is a 4-pennant. The set of all possible 1-pennants is {(1)}, the set of all possible 2-pennants is {(2), (1,1)} and the set of all 3-pennants is {(2,1), (1,1,1), (1,2)}. Note that the pennant (1,2) is not the same as the pennant (2,1). The number of 10- pennants is ______________.

Ans .

88.9 to 89.1

Explanation :

1-pennant {(1)} - #1

2-pennant {(1,1),(2)} - #2

3-pennant {(1,1,1),(1,2),(2,1)} - #3

4-pennant {(1,1,1,1),(2,2),(1,1,2),(1,2,1),(2,1,1)} - #5

5-pennant {(1,1,1,1,1),(2,1,1,1),(1,2,1,1),(1,1,2,1),
            (1,1,1,2),(2,2,1),(2,1,2),(1,2,2)} - #8

If one observe carefully they are the terms(part of) a Fibonacci series. (0,1,1,2,3,5,8,13 ....). Hence the # of 10-pennant is the 12th term of the series ie. 89

Q.60

Let S denote the set of all functions f: {0,1}⁴ -> {0,1}. Denote by N the number of functions from S to the set {0,1}. The value of Log₂Log₂N is ______.

Ans .

Explanation :

The given mapping S is defined by f:{0,1}^4 -> {0,1} .
So, number of functions from S will be 2^16.
Now N is defined by f : S-> {0,1}.
So Number of functions from S to {0,1} will be 2^S.
Hence log₂log₂N = log₂S = 16

Q.61

Consider an undirected graph G where self-loops are not allowed. The vertex set of G is {(i, j): 1 <= i <= 12, 1 <= j <= 12}. There is an edge between (a, b) and (c, d) if |a − c| <= 1 and |b − d| <= 1. The number of edges in this graph is __________.

500

502

506

510

Ans .

Explanation :

Given:
The vertex set of G is {(i, j): 1 <= i <= 12, 1 <= j <= 12}. 
There is an edge between (a, b) and (c, d) if |a − c| <= 1 and 
                                              |b − d| <= 1.



There can be total 12*12 possible vertices. The vertices are (1, 1), 
(1, 2) ....(1, 12) (2, 1), (2, 2), ....

The number of edges in this graph?
Number of edges is equal to number of pairs of vertices that satisfy
above conditions. For example, vertex pair {(1, 1), (1, 2)} satisfy
above condition.

For (1, 1), there can be an edge to (1, 2), (2, 1), (2, 2). Note that
there can be self-loop as mentioned in the question.
Same is count for (12, 12), (1, 12) and (12, 1)

For (1, 2), there can be an edge to (1, 1), (2, 1), (2, 2), (2, 3),
                                    (1, 3)
Same is count for (1, 3), (1, 4)....(1, 11), (12, 2), ....(12, 11)

For (2, 2), there can be an edge to (1, 1), (1, 2), (1, 3), (2, 1),
                                    (2, 3), (3, 1), (3, 2), (3, 3)
Same is count for remaining vertices.

For all pairs (i, j) there can total 8 vertices connected to them if 
i and j are not in {1, 12}

There are total 100 vertices without a 1 or 12.  So total 800 edges.  

For vertices with 1, total edges = (Edges where 1 is first part) +
                    (Edges where 1 is second part and not first part)   
                              
                                =  (3 + 5*10 + 3) + (5*10) edges

Same is count for vertices with 12

Total number of edges:
 = 800 + [(3 + 5*10 + 3) + 5*10]  + [(3 + 5*10 + 3) + 5*10]
 = 800 + 106 + 106
 = 1012

Since graph is undirected, two edges from v1 to v2 and v2 to v1 
should be counted as one.

So total number of undirected edges = 1012/2 = 506.

Q.62

An ordered n-tuple (d1, d2, … , dn) with d1 >= d2 >= ⋯ >= dn is called graphic if there exists a simple undirected graph with n vertices having degrees d1, d2, … , dn respectively. Which of the following 6-tuples is NOT graphic?

(1, 1, 1, 1, 1, 1)

(2, 2, 2, 2, 2, 2)

(3, 3, 3, 1, 0, 0)

(3, 2, 1, 1, 1, 0)

Ans .

Explanation :

The required graph is not possible with the given degree set of (3, 3, 3, 1, 0, 0). Using this 6-tuple the graph formed will be a Disjoint undirected graph, where the two vertices of the graph should not be connected to any other vertex ( i.e. degree will be 0 for both the vertices ) of the graph. And for the remaining 4 vertices the graph need to satisfy the degrees of (3, 3, 3, 1).

Let's see this with the help of a logical structure of the graph :

Let's say vertices labelled as <ABCDEF> should have their degree as <3, 3, 3, 1, 0, 0> respectively.

Now E and F should not be connected to any vertex in the graph. And A, B, C and D should have their degree as <3, 3, 3, 1> respectively. Now to fulfill the requirement of A, B and C, the node D will never be able to get its degree as 1. It's degree will also become as 3. This is shown in the above diagram.

Hence tuple <3, 3, 3, 1, 0, 0> is not graphic.

Q.63

Which one of the following propositional logic formulas is TRUE when exactly two of p, q, and r are TRUE? gate paper solved

Ans .

Explanation :

Draw the truth table of three variables and output will be 1 (true) only when exactly two variables are 1 (true) else output will be 0 (false). Therefore,
p    q     r   Output (f)
0	0	0	0
0	0	1	0
0	1	0	0
0	1	1	1
1	0	0	0
1	0	1	1
1	1	0	1
1	1	1	0

f = ( ~p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) ∨ (p ∧ q ∧ ~r)
f = {( ~p ∧ q ) ∨ (p ∧ ~q )} ∧ r  ∨ (p ∧ q ∧ ~r)
f = {(p ⊕ q )} ∧ r  ∨ (p ∧ q ∧ ~r)
f = {~ (p ↔ q )} ∧ r  ∨ (p ∧ q ∧ ~r)
Hence, option (b) is true.

Q.63

Given the following schema:

	 employees(emp-id, first-name, last-name, hire-date, dept-id, salary)
     departments(dept-id, dept-name, manager-id, location-id)

You want to display the last names and hire dates of all latest hires in their respective departments in the location ID 1700. You issue the following query:

SQL> SELECT last-name, hire-date
     FROM employees
     WHERE (dept-id, hire-date) IN
     (SELECT dept-id, MAX(hire-date)
     FROM employees JOIN departments USING(dept-id)
     WHERE location-id = 1700
     GROUP BY dept-id); 
What is the outcome?

It executes but does not give the correct result.

It executes and gives the correct result.

It generates an error because of pairwise comparison.

It generates an error because the GROUP BY clause cannot be used with table joins in a subquery

Ans .

Explanation :
The given query uses below inner query.
```
SELECT dept-id, MAX(hire-date)
     FROM employees JOIN departments USING(dept-id)
     WHERE location-id = 1700
     GROUP BY dept-id
	 
```
The inner query produces last max hire-date in every department located at location id 1700. The outer query simply picks all pairs of inner query. Therefore, the query produces correct result.
```
SELECT last-name, hire-date
     FROM employees
     WHERE (dept-id, hire-date) IN
     (Inner-Query);
```

Q.65

Consider two processors P1 and P2 executing the same instruction set. Assume that under identical conditions, for the same input, a program running on P2 takes 25% less time but incurs 20% more CPI (clock cycles per instruction) as compared to the program running on P1. If the clock frequency of P1 is 1GHz, then the clock frequency of P2 (in GHz) is _________.

1.6

3.2

1.2

0.8

Ans .

Explanation :

For P1 clock period = 1ns 

Let clock period for P2 be t.

Now consider following equation based on specification
7.5 ns = 12*t ns

We get t and inverse of t will be 1.6GHz

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	AVAILABLE	X=3, Y=2, Z=2

	MAX	ALLOCATION
	X Y Z	X Y Z
P0	8 4 3	0 0 1
P1	6 2 0	3 2 0
P2	3 3 3	2 1 1

	AVAILABLE	X=3, Y=2, Z=0

	MAX	ALLOCATION	NEED
	X Y Z	X Y Z	X Y Z
P0	8 4 3	0 0 3	8 4 0
P1	6 2 0	3 2 0	3 0 0
P2	3 3 3	2 1 1	1 2 2

	AVAILABLE	X=6, Y=4, Z=0

	MAX	ALLOCATION	NEED
	X Y Z	X Y Z	X Y Z
P0	8 4 3	0 0 3	8 4 0
P1	6 2 0	3 2 0	0 0 0
P2	3 3 3	2 1 1	1 2 2

	AVAILABLE	X=1, Y=2, Z=2

	MAX	ALLOCATION	NEED
	X Y Z	X Y Z	X Y Z
P0	8 4 3	0 0 1	8 4 2
P1	6 2 0	5 2 0	1 0 0
P2	3 3 3	2 1 1	1 2 2

	AVAILABLE	X=7, Y=4, Z=2

	MAX	ALLOCATION	NEED
	X Y Z	X Y Z	X Y Z
P0	8 4 3	0 0 1	8 4 2
P1	6 2 0	5 2 0	0 0 0
P2	3 3 3	2 1 1	1 2 2

	AVAILABLE	X=10, Y=7, Z=5

	MAX	ALLOCATION	NEED
	X Y Z	X Y Z	X Y Z
P0	8 4 3	0 0 1	8 4 2
P1	6 2 0	5 2 0	0 0 0
P2	3 3 3	2 1 1	0 0 0

	AVAILABLE	X=18, Y=11, Z=8

	MAX	ALLOCATION	NEED
	X Y Z	X Y Z	X Y Z
P0	8 4 3	0 0 1	0 0 0
P1	6 2 0	5 2 0	0 0 0
P2	3 3 3	2 1 1	0 0 0