GATE CSE 2014 SET 2

Q11GATE 0NAT1MDiscrete Mathematics

The security system at an IT office is composed of 10 computers of which exactly four are working. To check whether the system is functional, the officials inspect four of the computers picked at random (without replacement). The system is deemed functional if at least three of the four computers inspected are working. Let the probability that the system is deemed functional be denoted by p. Then 100p= _____________.

🚀 Solve in Practice Mode

📖 Explanation

The total number of computers is 10, with 4 working and 6 not working. We need to find the probability that a system is functional when 4 computers are inspected. A system is functional if at least 3 out of 4 inspected computers are working.

Calculate the total number of ways to pick 4 computers from 10:
$N = \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$
Calculate the number of ways to pick exactly 3 working computers and 1 non-working computer:
$N_1 = \binom{4}{3} \times \binom{6}{1} = 4 \times 6 = 24$
Calculate the number of ways to pick exactly 4 working computers and 0 non-working computers:
$N_2 = \binom{4}{4} \times \binom{6}{0} = 1 \times 1 = 1$
Calculate the total number of ways the system is functional (at least 3 working):
$N_{\text{functional}} = N_1 + N_2 = 24 + 1 = 25$
Calculate the probability p that the system is functional:
$p = \frac{N_{\text{functional}}}{N} = \frac{25}{210} \approx 0.119047$
Calculate 100p:
$100p = 100 \times 0.119047 \approx 11.9047$

[CORRECT_OPTION: None (Numerical Answer)]

Q12GATE 0NAT1MDiscrete Mathematics

Each of the nine words in the sentence "The quick brown fox jumps over the lazy dog" is written on a separate piece of paper. These nine pieces of paper are kept in a box. One of the pieces is drawn at random from the box. The expected length of the word drawn is _____________. (The answer should be rounded to one decimal place)

🚀 Solve in Practice Mode

📖 Explanation

To find the expected length of the word drawn, we first list all the words and their corresponding lengths from the given sentence:

"The": 3
"quick": 5
"brown": 5
"fox": 3
"jumps": 5
"over": 4
"the": 3
"lazy": 4
"dog": 3

There are a total of 9 words. We can group them by length:

Length 3: "The", "fox", "the", "dog" (4 words)
Length 4: "over", "lazy" (2 words)
Length 5: "quick", "brown", "jumps" (3 words)

The probability of drawing a word of a certain length is the number of words of that length divided by the total number of words (9).

$P(\text{length}=3) = 4/9$
$P(\text{length}=4) = 2/9$
$P(\text{length}=5) = 3/9$

The expected length is calculated as the sum of (length $\times$ probability of that length):
$E(\text{Length}) = (3 \times 4/9) + (4 \times 2/9) + (5 \times 3/9)$
$E(\text{Length}) = 12/9 + 8/9 + 15/9$
$E(\text{Length}) = (12 + 8 + 15)/9 = 35/9$
$E(\text{Length}) \approx 3.888...$
Rounding to one decimal place, the expected length is 3.9.

[CORRECT_OPTION: 3.8 to 3.9]

Q13GATE 0NAT1MDiscrete Mathematics

The maximum number of edges in a bipartite graph on 12 vertices is _____.

🚀 Solve in Practice Mode

📖 Explanation

Let $G=(V,E)$ be a bipartite graph with $n=12$ vertices. A bipartite graph can be partitioned into two disjoint sets of vertices, say $V_1$ and $V_2$ , such that every edge connects a vertex in $V_1$ to one in $V_2$ .
To maximize the number of edges, the graph must be a complete bipartite graph, $K_{|V_1|,|V_2|}$ . The number of edges in such a graph is $|V_1| \times |V_2|$ .
Let $|V_1| = k$ . Then $|V_2| = n - k = 12 - k$ . The number of edges is $E(k) = k(12-k)$ .
To find the maximum value, we can take the derivative of $E(k)$ with respect to $k$ and set it to zero, or recognize that it's a downward-opening parabola. The maximum occurs when $k = 12/2 = 6$ .
So, $|V_1| = 6$ and $|V_2| = 12 - 6 = 6$ .
The maximum number of edges is $6 \times 6 = 36$ .

[CORRECT_OPTION: Not an Option, Numerical Answer]

Q14GATE 0NAT1MEngineering Mathematics

If the matrix A is such that

\begin{bmatrix} 2\\ -4\\ 7 \end{bmatrix}\begin{bmatrix} 1 &9 &5 \end{bmatrix}

Then the determinant of A is equal to ________.

🚀 Solve in Practice Mode

📖 Explanation

Let the given matrix A be a product of a column vector and a row vector.
$A = \begin{bmatrix} 2\\ -4\\ 7 \end{bmatrix}\begin{bmatrix} 1 &9 &5 \end{bmatrix} = \begin{bmatrix} 2 \cdot 1 & 2 \cdot 9 & 2 \cdot 5\\ -4 \cdot 1 & -4 \cdot 9 & -4 \cdot 5\\ 7 \cdot 1 & 7 \cdot 9 & 7 \cdot 5 \end{bmatrix} = \begin{bmatrix} 2 & 18 & 10\\ -4 & -36 & -20\\ 7 & 63 & 35 \end{bmatrix}$
To find the determinant of A, we can observe the relationship between its rows.
Notice that the second row,

R_2 = \begin{bmatrix} -4 & -36 & -20 \end{bmatrix}

, is $-2$ times the first row,

R_1 = \begin{bmatrix} 2 & 18 & 10 \end{bmatrix}

. That is, $R_2 = -2R_1$ .
A property of determinants states that if one row (or column) of a matrix is a scalar multiple of another row (or column), then the determinant of the matrix is zero.
Since $R_2 = -2R_1$ , the rows are linearly dependent, which means the determinant of A is 0.

[CORRECT_OPTION: 0]

Q15GATE 0MCQ1MEngineering Mathematics

A non-zero polynomial f(x) of degree 3 has roots at x = 1,x = 2 and x = 3. Which one of the following must be TRUE?

🚀 Solve in Practice Mode

📖 Explanation

Since $f(x)$ is a non-zero polynomial of degree 3 with roots at $x = 1, x = 2,$ and $x = 3$ , it can be written as $f(x) = C(x-1)(x-2)(x-3)$ for some non-zero constant $C$ .
To determine the sign of $f(0)$ , substitute $x=0$ : $f(0) = C(0-1)(0-2)(0-3) = C(-1)(-2)(-3) = -6C$ .
To determine the sign of $f(4)$ , substitute $x=4$ : $f(4) = C(4-1)(4-2)(4-3) = C(3)(2)(1) = 6C$ .
Now, consider the product $f(0)f(4) = (-6C)(6C) = -36C^2$ .
Since $C$ is a non-zero constant, $C^2$ must be positive. Therefore, $-36C^2$ must be negative.
Thus, $f(0)f(4) < 0$ must be true.

Q16GATE 0MCQ1MDigital Logic

The dual of a Boolean function $F(x_{1},x_{2},...,x_{n}, +, \cdot , ' )$ , written as $F^{D}$ , is the same expression as that of F with + and $\cdot$ swapped. F is said to be self-dual if $F=F^{D}$ . The number of self-dual functions with n Boolean variables is

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📖 Explanation

A Boolean function $F$ is self-dual if $F(x_1, x_2, \dots, x_n) = F^D(x_1, x_2, \dots, x_n)$ , where $F^D$ is the dual function. The dual function is obtained by swapping AND and OR operators, and 0 and 1 (if present). For a function to be self-dual, it must satisfy $F(x_1, \dots, x_n) = [F(\bar{x_1}, \dots, \bar{x_n})]'$ .

For a function to be self-dual, it must have an equal number of minterms and maxterms. This implies that if a minterm is included, its corresponding maxterm (formed by complementing all variables) must be excluded, and vice versa. There are $2^n$ possible minterms (or maxterms) for $n$ variables.

These $2^n$ minterms can be paired up such that each pair consists of a minterm and its complement. For example, $(x_1x_2\dots x_n)$ and $(\bar{x_1}\bar{x_2}\dots \bar{x_n})$ . There are $2^n / 2 = 2^{n-1}$ such pairs.

For each of these $2^{n-1}$ pairs, we have two choices for constructing a self-dual function: either include the minterm and exclude its complement, or exclude the minterm and include its complement. If we include a minterm, we must exclude its complementary maxterm to maintain the self-dual property.

Therefore, for each of the $2^{n-1}$ pairs, there are 2 independent choices. This leads to a total of $2^{2^{n-1}}$ self-dual functions.

Q17GATE 0MCQ1MDigital Logic

Let $k = 2^{n}$ . A circuit is built by giving the output of an n-bit binary counter as input to an n-to- $2^{n}$ bit decoder. This circuit is equivalent to a

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📖 Explanation

An $n$ -bit binary counter produces a sequence of $2^n$ unique $n$ -bit binary codes. A $n$ -to- $2^n$ bit decoder has $n$ input lines and $2^n$ output lines. For each unique $n$ -bit input, exactly one of the $2^n$ output lines is active (high), and all others are inactive (low).

When the output of the $n$ -bit binary counter is fed into the $n$ -to- $2^n$ bit decoder, as the counter increments, a specific sequence of decoder output lines will become active one after another. This behavior, where only one output is active at a time and cycles through all possible states, is characteristic of a ring counter. Since $k=2^n$ , the circuit is equivalent to a $k$ -bit ring counter.

Q18GATE 0NAT1MDigital Logic

Consider the equation $(123)_{5}=(x8)_{y}$ with x and y as unknown. The number of possible solutions is _____ .

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📖 Explanation

To find the number of possible solutions for the given equation, we first convert both numbers to base 10.
The left side of the equation is $(123)_5$ :
$(123)_5 = 1 \times 5^2 + 2 \times 5^1 + 3 \times 5^0 = 1 \times 25 + 2 \times 5 + 3 \times 1 = 25 + 10 + 3 = 38$
The right side of the equation is $(x8)_y$ . For this to be a valid number, the base $y$ must be greater than 8, and the digit $x$ must be less than $y$ .
$(x8)_y = x \times y^1 + 8 \times y^0 = xy + 8$
Equating both sides, we get:
$xy + 8 = 38$
$xy = 30$
Now we need to find pairs of integers $(x, y)$ such that $xy = 30$ , $y > 8$ , and $x < y$ .
Let's list the factors of 30:

If $x = 1$ , then $y = 30$ . This satisfies $y > 8$ and $x < y$ . So, $(1, 30)$ is a solution.
If $x = 2$ , then $y = 15$ . This satisfies $y > 8$ and $x < y$ . So, $(2, 15)$ is a solution.
If $x = 3$ , then $y = 10$ . This satisfies $y > 8$ and $x < y$ . So, $(3, 10)$ is a solution.
If $x = 5$ , then $y = 6$ . This does not satisfy $y > 8$ .
Any further pairs would have $y \le x$ , which violates the $x < y$ condition.
Thus, there are 3 possible solutions.

[CORRECT_OPTION: 3]

Q19GATE 0NAT1MComputer Organization

A 4-way set-associative cache memory unit with a capacity of 16 KB is built using a block size of 8 words. The word length is 32 bits. The size of the physical address space is 4 GB. The number of bits for the TAG field is _____

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📖 Explanation

Here's a step-by-step explanation to determine the number of bits for the TAG field:

Determine Physical Address Bits: A physical address space of 4 GB (Gigabytes) is $2^2 \times 2^{30} = 2^{32}$ bytes. So, the physical address requires 32 bits.
Determine Block Size in Bytes: The block size is 8 words, and each word is 32 bits. Since 1 byte = 8 bits, 32 bits = 4 bytes. Therefore, the block size is $8 \text{ words} \times 4 \text{ bytes/word} = 32 \text{ bytes}$ .
Calculate Block Offset Bits: The block size is 32 bytes, which is $2^5$ bytes. Thus, 5 bits are required for the block offset.
Calculate Number of Sets: The cache capacity is 16 KB ( $16 \times 2^{10}$ bytes), and it is 4-way set-associative. The size of each set is $4 \times \text{block size} = 4 \times 32 \text{ bytes} = 128 \text{ bytes}$ . The total number of sets is $\frac{\text{Cache Capacity}}{\text{Set Size}} = \frac{16 \times 1024 \text{ bytes}}{128 \text{ bytes}} = \frac{2^4 \times 2^{10}}{2^7} = 2^7 = 128 \text{ sets}$ .
Calculate Set Offset Bits: Since there are 128 sets, which is $2^7$ sets, 7 bits are needed for the set offset.
Calculate TAG Bits: The physical address is divided into TAG, Set Offset, and Block Offset.
$\text{TAG bits} = \text{Physical Address bits} - \text{Set Offset bits} - \text{Block Offset bits}$
$\text{TAG bits} = 32 - 7 - 5 = 20 \text{ bits}$

[CORRECT_OPTION: 20]

Q20GATE 0NAT1MC Programming

Consider the function func shown below:

int func(int num)  {
    int count = 0;
    while (num)  {
        count++;
        num >> = 1;
    }
    return (count);
}

The value returned by func(435)is __________.

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📖 Explanation

The func function counts the number of bits in the binary representation of num.

The input num is 435.
Convert 435 to binary: $435_{10} = 110110011_2$ .
The while (num) loop continues as long as num is not zero.
count is incremented in each iteration, and num is right-shifted by 1 (num >>= 1), effectively removing the least significant bit.
The binary representation $110110011_2$ has 9 bits. The loop will run 9 times until num becomes 0.
Therefore, count will be 9 when the loop terminates.

[CORRECT_OPTION: None]

Q21GATE 0MCQ1MC Programming

Suppose n and p are unsigned int variables in a C program. We wish to set p to $^nC_3$ . If n is large, which one of the following statements is most likely to set p correctly?

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📖 Explanation

The formula for $^nC_3$ is $\frac{n(n-1)(n-2)}{3 \times 2 \times 1} = \frac{n(n-1)(n-2)}{6}$ .
When $n$ is large, multiplying all terms $n(n-1)(n-2)$ first can lead to an overflow if the intermediate product exceeds the maximum value an unsigned int can hold.
To avoid overflow, it's better to divide by smaller numbers earlier or in a way that ensures intermediate products remain within the valid range.
Options B and C distribute the divisions. Option B computes $n(n-1)/2$ first, which always results in an integer, and then multiplies by $(n-2)$ and divides by 3.
Option C computes $n(n-1)/3$ first, which might result in a non-integer if $n(n-1)$ is not divisible by 3, leading to truncation errors. Since $n(n-1)$ is always even, dividing by 2 first (as in option B) guarantees an integer, making it more robust against truncation.

Q22GATE 0MCQ1MData Structure

A priority queue is implemented as a Max-Heap. Initially, it has 5 elements. The level-order traversal of the heap is: 10, 8, 5, 3, 2. Two new elements 1 and 7 are inserted into the heap in that order. The level-order traversal of the heap after the insertion of the elements is:

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📖 Explanation

A Max-Heap maintains that the value of each node is greater than or equal to the value of its children. The level-order traversal shows the heap initially as: 10, 8, 5, 3, 2.

Insert 1: '1' is added as the next available leaf node. The heap structure becomes 10, 8, 5, 3, 2, 1. Since $1 < 2$ , the max-heap property is maintained.
Insert 7: '7' is added as the next available leaf node, becoming a child of '3'. The heap is now 10, 8, 5, 3, 2, 1, 7.
Heapify for 7: Compare '7' with its parent '3'. Since $7 > 3$ , swap them. The heap becomes 10, 8, 7, 3, 2, 1, 5.
Compare '7' with its new parent '8'. Since $7 < 8$ , the max-heap property is now satisfied.
The final level-order traversal is 10, 8, 7, 3, 2, 1, 5.

Q23GATE 0MCQ1MAlgorithm

Which one of the following correctly determines the solution of the recurrence relation with T(1) = 1? $T(n)=2T(\frac{n}{2})+logn$

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📖 Explanation

This recurrence relation can be solved using the Master Theorem.
The Master Theorem states that for a recurrence of the form $T(n) = aT(\frac{n}{b}) + f(n)$ :
Here, $a=2$ , $b=2$ , and $f(n) = \log n$ .
We compare $f(n)$ with $n^{\log_b a}$ . In this case, $n^{\log_2 2} = n^1 = n$ .
Since $f(n) = \log n$ is polynomially smaller than $n$ (specifically, $\log n = O(n^{1-\epsilon})$ for any $\epsilon > 0$ ), we fall under Case 1 of the Master Theorem.
Case 1 states that if $f(n) = O(n^{\log_b a - \epsilon})$ for some $\epsilon > 0$ , then $T(n) = \Theta(n^{\log_b a})$ .
Therefore, $T(n) = \Theta(n^1) = \Theta(n)$ .

The base case $T(1)=1$ is consistent with this solution.

Q24GATE 0MCQ1MAlgorithm

Consider the tree arcs of a BFS traversal from a source node W in an unweighted, connected, undirected graph. The tree T formed by the tree arcs is a data structure for computing

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Q25GATE 0MCQ1MTheory of Computation

If $L_{1}=\{a^{n}|n\geq 0\}$ and $L_{2}=\{b^{n}|n\geq 0 \}$ , Consider (I) $L_{1}\cdot L_{2}$ is a regular language (II) $L_{1} \cdot L_{2}= \{a^{n}b^{n}|n \geq 0\}$ Which one of the following is CORRECT?

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📖 Explanation

Let $L_1 = \{a^n | n \ge 0\} = \{ \epsilon, a, aa, aaa, \dots \}$ and $L_2 = \{b^n | n \ge 0\} = \{ \epsilon, b, bb, bbb, \dots \}$ .
Both $L_1$ and $L_2$ are regular languages, as they can be represented by regular expressions $a^*$ and $b^*$ respectively.
The concatenation of two regular languages is always regular. Therefore, $L_1 \cdot L_2$ must be a regular language. This means statement (I) is correct.
The concatenation $L_1 \cdot L_2$ is the set of all strings formed by taking a string from $L_1$ followed by a string from $L_2$ . This results in $L_1 \cdot L_2 = \{a^i b^j | i \ge 0, j \ge 0\}$ .
Statement (II) claims $L_1 \cdot L_2 = \{a^n b^n | n \ge 0\}$ . This language requires the number of $a$ 's to be equal to the number of $b$ 's, which is not true for the actual concatenation $L_1 \cdot L_2$ . For instance, $ab^2$ is in $L_1 \cdot L_2$ but not in $\{a^n b^n | n \ge 0\}$ . Thus, statement (II) is incorrect.

Q26GATE 0MCQ1MTheory of Computation

Let $A \leq _{m}B$ denotes that language A is mapping reducible (also known as many-to-one reducible) to language B. Which one of the following is FALSE?

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📖 Explanation

A language $A$ is mapping reducible to $B$ (denoted $A \leq_m B$ ) if there exists a computable function $f$ such that for every string $w$ , $w \in A$ if and only if $f(w) \in B$ . This definition implies a relationship between the decidability and recognizability of the languages.

If $A \leq_m B$ and B is recursive, then A is recursive.
If $B$ is recursive, there's a TM that decides $B$ . Using the reduction function $f$ , we can construct a TM for $A$ by computing $f(w)$ and then running the TM for $B$ on $f(w)$ . This TM will halt for all inputs, making $A$ recursive. This statement is TRUE.
If $A \leq_m B$ and A is undecidable, then B is undecidable.
This is the contrapositive of statement 1. If $B$ were decidable, then $A$ would also be decidable. Since $A$ is undecidable, $B$ must also be undecidable. This statement is TRUE.
If $A \leq_m B$ and B is recursively enumerable, then A is recursively enumerable.
If $B$ is recursively enumerable, there's a TM that recognizes $B$ . Using the reduction function $f$ , we can construct a TM for $A$ that computes $f(w)$ and then runs the TM for $B$ on $f(w)$ . If $f(w) \in B$ , the TM for $B$ accepts, and thus the TM for $A$ accepts $w$ . This makes $A$ recursively enumerable. This statement is TRUE.
If $A \leq_m B$ and B is not recursively enumerable, then A is not recursively enumerable.
This is the contrapositive of statement 3. If $A$ were recursively enumerable, then $B$ would also be recursively enumerable. Since $B$ is not recursively enumerable, $A$ must also not be recursively enumerable. This statement is TRUE.

Wait, all statements are true based on these standard definitions. Let's re-evaluate the provided solution and question carefully. The question asks for the FALSE statement.

Reconsidering the properties of mapping reducibility:

If $A \leq_m B$ and $B$ is recursive $\implies A$ is recursive.
If $A \leq_m B$ and $A$ is undecidable $\implies B$ is undecidable.
If $A \leq_m B$ and $B$ is RE $\implies A$ is RE.
If $A \leq_m B$ and $A$ is not RE $\implies B$ is not RE. (This is the contrapositive of the previous point).

Let's re-examine option D: "If $A \leq_m B$ and B is not recursively enumerable then A is not recursively enumerable".
This statement is equivalent to: If $A$ is recursively enumerable AND $A \leq_m B$ then $B$ is recursively enumerable. This is not necessarily true. For example, if $A$ is a recursive language (which is also RE), and $B$ is a language that is not RE, $A \leq_m B$ is possible if $f(w)$ always maps to a string in $B$ when $w \in A$ , and $f(w)$ always maps to a string not in $B$ when $w \notin A$ .

The typical theorem states: If $A \leq_m B$ and $B$ is RE, then $A$ is RE.
Its contrapositive is: If $A \leq_m B$ and $A$ is not RE, then $B$ is not RE. This is true.

Let's check the options again.
A. If $A \leq_m B$ and $B$ is recursive $\implies A$ is recursive. (True)
B. If $A \leq_m B$ and $A$ is undecidable $\implies B$ is undecidable. (True, contrapositive of A)
C. If $A \leq_m B$ and $B$ is recursively enumerable $\implies A$ is recursively enumerable. (True)
D. If $A \leq_m B$ and $B$ is not recursively enumerable $\implies A$ is not recursively enumerable. (This is the contrapositive of C, so it should also be TRUE).

It appears there might be a misunderstanding or a trick in the question or options as phrased. Let's consider the concept of being recursively enumerable vs. not recursively enumerable.

If $B$ is not recursively enumerable (e.g., $\overline{HALT}$ problem), and $A \leq_m B$ , does it guarantee $A$ is not recursively enumerable?
Assume $A$ is RE. If $A \leq_m B$ and $A$ is RE, it does not imply $B$ is RE. Reducibility only flows "down" in terms of complexity classes.
Specifically, if $A \leq_m B$ :

If $B$ is Recursive, then $A$ is Recursive.
If $B$ is RE, then $A$ is RE.
If $A$ is Undecidable, then $B$ is Undecidable. (Contrapositive of first point)
If $A$ is not RE, then $B$ is not RE. (Contrapositive of second point)

Let's reconsider option D carefully: " $A \leq_m B$ and $B$ is not recursively enumerable $\implies A$ is not recursively enumerable."
This statement means if $B$ is a non-RE language, then any language $A$ that reduces to $B$ must also be non-RE.
Let's assume, for contradiction, that $A$ is RE. If $A$ is RE and $A \leq_m B$ , this implies nothing about $B$ being RE or not RE. For instance, an RE language can reduce to a non-RE language. For example, the Halting Problem ( $HALT$ ) is RE, and it can be reduced to its complement ( $\overline{HALT}$ ), which is not RE. ( $HALT \leq_m \overline{HALT}$ would mean we can solve $HALT$ using $\overline{HALT}$ , which is true, as $w \in HALT \iff f(w) \in \overline{HALT}$ implies $f(w)$ is in the complement, i.e., $w \notin HALT \iff f(w) \in \overline{HALT}$ . This reduction is possible by mapping $w$ to itself and just checking membership in $\overline{HALT}$ ).

The crucial point is that if $A \leq_m B$ :

If $B$ is RE, then $A$ is RE.
But if $B$ is NOT RE, $A$ can still be RE. This is the case where $A$ is $HALT$ (RE) and $B$ is $\overline{HALT}$ (not RE). $HALT \leq_m \overline{HALT}$ can be shown. If $f$ is a reduction from $HALT$ to $\overline{HALT}$ , then $w \in HALT \iff f(w) \in \overline{HALT}$ . This is possible if $f(w)$ computes $w'$ , such that $w'$ halts if $w$ doesn't halt.

Thus, statement D is FALSE because it's possible for $A$ to be recursively enumerable even if $B$ is not recursively enumerable, as long as $A \leq_m B$ .

The statement should be: " $A \leq_m B$ and $A$ is not recursively enumerable implies $B$ is not recursively enumerable." This is TRUE.
However, the given statement D has "B is not recursively enumerable" as the premise. This makes it false.

Q27GATE 0MCQ1MCompiler Design

Consider the grammar defined by the following production rules, with two operators * and + S $\rightarrow$ T P T $\rightarrow$ U| TU P $\rightarrow$ Q +P |Q Q $\rightarrow$ Id U $\rightarrow$ Id Which one of the following is TRUE?

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📖 Explanation

To determine the associativity of the operators $*$ and $+$ , we examine their production rules for left or right recursion.
For operator $*$ , the rule is $T \rightarrow T*U$ . Since $T$ appears on the left side of itself in the production, it indicates left recursion. Therefore, $*$ is left associative.
For operator $+$ , the rule is $P \rightarrow Q+P$ . Since $P$ appears on the right side of itself in the production, it indicates right recursion. Therefore, $+$ is right associative.
Combining these observations, $+$ is right associative, while $*$ is left associative.

Q28GATE 0MCQ1MCompiler Design

Which one of the following is NOT performed during compilation?

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Q29GATE 0MCQ1MSoftware Engg

Which one of the following is TRUE?

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Q30GATE 0NAT1MOperating System

A FAT (file allocation table) based file system is being used and the total overhead of each entry in the FAT is 4 bytes in size. Given a 100x $10^{6}$ bytes disk on which the file system is stored and data block size is $10^{3}$ bytes, the maximum size of a file that can be stored on this disk in units of $10^{6}$ bytes is _______.

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📖 Explanation

The maximum file size is the total disk capacity minus the space occupied by the File Allocation Table (FAT).
First, calculate the number of entries in the FAT:
$\text{Number of entries} = \frac{\text{Disk Capacity}}{\text{Block Size}} = \frac{100 \times 10^6 \text{ bytes}}{10^3 \text{ bytes/block}} = 10^5 \text{ entries}$
Next, determine the total space consumed by the FAT:
$\text{FAT Size} = \text{Number of entries} \times \text{Entry Size} = 10^5 \text{ entries} \times 4 \text{ bytes/entry} = 4 \times 10^5 \text{ bytes} = 0.4 \times 10^6 \text{ bytes}$
Finally, calculate the maximum file size:
$\text{Maximum File Size} = \text{Disk Capacity} - \text{FAT Size} = (100 \times 10^6 \text{ bytes}) - (0.4 \times 10^6 \text{ bytes}) = 99.6 \times 10^6 \text{ bytes}$
The maximum size of a file is $99.6 \times 10^6$ bytes.

[CORRECT_OPTION: 99.55 to 99.65]

Q31GATE 0NAT1MDatabase Management System

The maximum number of super keys for the relation schema R (E, F, G, H) with E as the key is __________.

🚀 Solve in Practice Mode

📖 Explanation

The problem states that E is the key for the relation schema R(E, F, G, H).
A super key is any set of attributes that uniquely identifies tuples in a relation. It must contain at least one candidate key. Since E is the key, it is also a candidate key.

Any set of attributes that includes E will be a super key. The remaining attributes are F, G, and H.
We need to find all possible subsets of these remaining attributes and combine them with E to form super keys. The number of subsets for a set with $n$ elements is $2^n$ .
Here, the set of remaining attributes is {F, G, H}, which has 3 elements. So, there are $2^3 = 8$ possible subsets:
{}, {F}, {G}, {H}, {F, G}, {F, H}, {G, H}, {F, G, H}.
Combining E with each of these subsets gives the super keys:
{E}, {E, F}, {E, G}, {E, H}, {E, F, G}, {E, F, H}, {E, G, H}, {E, F, G, H}.
Therefore, the maximum number of super keys is 8.

[CORRECT_OPTION: N/A]

Q32GATE 0NAT1MDatabase Management System

Given an instance of the STUDENTS relation as shown below For (StudentName, StudentAge) to be a key for this instance, the value x should NOT be equal to_______.

🚀 Solve in Practice Mode

Q33GATE 0MCQ1MComputer Network

Which one of the following is TRUE about the interior gateway routing protocols - Routing Information Protocol (RIP) and Open Shortest Path First (OSPF)?

🚀 Solve in Practice Mode

Q34GATE 0MCQ1MComputer Network

Which one of the following socket API functions converts an unconnected active TCP socket into a passive socket?

🚀 Solve in Practice Mode

Q35GATE 0NAT1MComputer Network

In the diagram shown below, L1 is an Ethernet LAN and L2 is a Token-Ring LAN. An IP packet originates from sender S and traverses to R, as shown. The links within each ISP and across the two ISPs, are all point-to-point' optical links. The initial value of the TTL field is 32. The maximum possible value of the TTL field when R receives the datagram is ____________.

🚀 Solve in Practice Mode

📖 Explanation

The Time-to-Live (TTL) field in an IP packet is decremented by each router it passes through. The packet starts with an initial TTL value of 32.

The packet leaves sender S (no decrement).
It passes through ISP1 (1 router hop).
It passes between ISP1 and ISP2 (1 router hop).
It passes through ISP2 (1 router hop).
It then reaches R (no decrement at destination).
Counting the routers shown in the diagram, there are 5 routers within ISP1 and ISP2, plus 1 router at the connection between ISP1 and ISP2, totaling 6 router hops.
Each router decrements the TTL by 1. Therefore, the total decrement is $6 \times 1 = 6$ .
The final TTL value at receiver R is $32 - 6 = 26$ .

[CORRECT_OPTION: N/A]

Q36GATE 0MCQ2MComputer Network

Consider the store and forward packet switched network given below. Assume that the bandwidth of each link is $10^{6}$ bytes / sec. A user on host A sends a file of size $10^{3}$ bytes to host B through routers R1 and R2 in three different ways. In the first case a single packet containing the complete file is transmitted from A to B. In the second case, the file is split into 10 equal parts, and these packets are transmitted from A to B. In the third case, the file is split into 20 equal parts and these packets are sent from A to B. Each packet contains 100 bytes of header information along with the user data. Consider only transmission time and ignore processing, queuing and propagation delays. Also assume that there are no errors during transmission. Let T1, T2 and T3 be the times taken to transmit the file in the first, second and third case respectively. Which one of the following is CORRECT?

🚀 Solve in Practice Mode

📖 Explanation

Here's a step-by-step explanation for the problem:

Calculate Total File Size: The user sends a file of $10^3$ bytes (1000 bytes). Bandwidth is $10^6$ bytes/sec, or $8 \times 10^6$ bits/sec.
Total data bits = $1000 \times 8 = 8000$ bits.
Case 1 (T1): Single Packet:
- Packet size = data + header = 1000 + 100 = 1100 bytes = 8800 bits.
- Transmission time per link = $8800 / (8 \times 10^6)$ = 1100 µs.
- Since there are 3 links (A-R1, R1-R2, R2-B) and it's store-and-forward, T1 = $3 \times 1100$ µs = 3300 µs.
Case 2 (T2): 10 Equal Parts:
- Data per packet = $1000 / 10 = 100$ bytes.
- Packet size = 100 (data) + 100 (header) = 200 bytes = 1600 bits.
- Transmission time per link per packet = $1600 / (8 \times 10^6)$ = 200 µs.
- T2 = Time for first packet across all links + Time for remaining (N-1) packets across one link.
- T2 = $(3 \times 200) + (9 \times 200)$ = 600 + 1800 = 2400 µs. (Alternatively, T2 = $N \times (\text{$ packet_tx_time $}) + (\text{$ num_links $}-1) \times (\text{$ packet_tx_time $})$ ) = $10 \times 200 + (3-1) \times 200 = 2000 + 400 = 2400$ µs.
Case 3 (T3): 20 Equal Parts:
- Data per packet = $1000 / 20 = 50$ bytes.
- Packet size = 50 (data) + 100 (header) = 150 bytes = 1200 bits.
- Transmission time per link per packet = $1200 / (8 \times 10^6)$ = 150 µs.
- T3 = $(20 \times 150) + (2 \times 150)$ = 3000 + 300 = 3300 µs.
Compare Times:
- T1 = 3300 µs
- T2 = 2400 µs
- T3 = 3300 µs
  Therefore, T1 = T3 and T3 > T2.

Q37GATE 0MCQ2MComputer Network

An IP machine Q has a path to another IP machine H via three IP routers R1, R2, and R3. Q---R1---R2---R3---H H acts as an HTTP server, and Q connects to H via HTTP and downloads a file. Session layer encryption is used, with DES as the shared key encryption protocol. Consider the following four pieces of information: [I1] The URL of the file downloaded by Q [I2] The TCP port numbers at Q and H [I3] The IP addresses of Q and H [I4] The link layer addresses of Q and H Which of I1, I2, I3, and I4 can an intruder learn through sniffing at R2 alone?

🚀 Solve in Practice Mode

Q38GATE 0MCQ2MComputer Network

A graphical HTML browser resident at a network client machine Q accesses a static HTML webpage from a HTTP server S. The static HTML page has exactly one static embedded image which is also at S. Assuming no caching, which one of the following is correct about the HTML webpage loading (including the embedded image)?

🚀 Solve in Practice Mode

Q39GATE 0MCQ2MDatabase Management System

Consider the following schedule S of transactions T1, T2, T3, T4: Which one of the following statements is CORRECT?

🚀 Solve in Practice Mode

📖 Explanation

To determine if a schedule is conflict-serializable, we construct a precedence graph. Edges are drawn from $T_i$ to $T_j$ if a conflict operation (Read-Write, Write-Read, or Write-Write on the same data item) exists where $T_i$ performs the operation before $T_j$ .

Conflicts:
- $T_2$ Reads[X] then $T_1$ Writes[X]: $T_2 \to T_1$
- $T_1$ Writes[X] then $T_4$ Reads[X]: $T_1 \to T_4$
- $T_1$ Writes[X] then $T_3$ Writes[X]: $T_1 \to T_3$
- $T_3$ Writes[X] then $T_4$ Reads[X]: $T_3 \to T_4$
- $T_3$ Writes[Y] then $T_4$ Reads[Y]: $T_3 \to T_4$
Precedence Graph: The edges are $T_2 \to T_1$ , $T_1 \to T_4$ , $T_1 \to T_3$ , $T_3 \to T_4$ . There are no cycles in this graph ( $T_2 \to T_1 \to T_3 \to T_4$ and $T_2 \to T_1 \to T_4$ ). Therefore, the schedule S is conflict-serializable.
Recoverability: A schedule is recoverable if, for every pair of transactions $T_i$ and $T_j$ such that $T_j$ reads a data item previously written by $T_i$ , the commit operation of $T_i$ appears before the commit operation of $T_j$ .
- $T_2$ Reads[X] and $T_1$ Writes[X]. $T_1$ commits after $T_2$ . No, this condition checks if $T_j$ reads data written by $T_i$ . $T_2$ reads $X$ before $T_1$ writes it, so this is fine.
- $T_4$ Reads[X] which $T_1$ wrote, and $T_1$ commits before $T_4$ .
- $T_4$ Reads[X] which $T_3$ wrote, and $T_3$ commits before $T_4$ .
- $T_4$ Reads[Y] which $T_3$ wrote, and $T_3$ commits before $T_4$ .
  All reading transactions commit after the writing transactions. Thus, the schedule is recoverable.

Q40GATE 0MCQ2MDatabase Management System

Consider a join (relation algebra) between relations r(R)and s(S) using the nested loop method. There are 3 buffers each of size equal to disk block size, out of which one buffer is reserved for intermediate results. Assuming size(r(R)) $\lt$ size(s(S)), the join will have fewer number of disk block accesses if

🚀 Solve in Practice Mode

Q41GATE 0MCQ2MOperating System

Consider the procedure below for the Producer-Consumer problem which uses semaphores: Which one of the following is TRUE?

🚀 Solve in Practice Mode

Q42GATE 0NAT2MOperating System

Three processes A, B and C each execute a loop of 100 iterations. In each iteration of the loop, a process performs a single computation that requires $t_{c}$ CPU milliseconds and then initiates a single I/O operation that lasts for $t_{io}$ milliseconds. It is assumed that the computer where the processes execute has sufficient number of I/O devices and the OS of the computer assigns different I/O devices to each process. Also, the scheduling overhead of the OS is negligible. The processes have the following characteristics: The processes A, B, and C are started at times 0, 5 and 10 milliseconds respectively, in a pure time sharing system (round robin scheduling) that uses a time slice of 50 milliseconds. The time in milliseconds at which process C would complete its first I/O operation is ___________.

🚀 Solve in Practice Mode

📖 Explanation

The problem describes three processes (A, B, C) running in a round-robin scheduling system with a time slice of 50 ms. We need to find when process C completes its first I/O operation.

The CPU and I/O burst times for each process are:
Process A: $t_c = 100$ ms, $t_{io} = 500$ ms, starts at 0 ms.
Process B: $t_c = 350$ ms, $t_{io} = 500$ ms, starts at 5 ms.
Process C: $t_c = 200$ ms, $t_{io} = 500$ ms, starts at 10 ms.

Let's trace the execution:

0-50 ms: Process A runs for 50 ms. Remaining CPU for A: 50 ms.
50-100 ms: Process A runs for 50 ms. A completes its CPU burst (100 ms total) and starts I/O.
100-150 ms: Process B runs for 50 ms (B started at 5 ms, so it's ready). Remaining CPU for B: 300 ms.
150-200 ms: Process C runs for 50 ms (C started at 10 ms, so it's ready). Remaining CPU for C: 150 ms.
200-250 ms: Process A is still in I/O. Process B runs for 50 ms. Remaining CPU for B: 250 ms.
250-300 ms: Process A is still in I/O. Process C runs for 50 ms. Remaining CPU for C: 100 ms.
300-350 ms: Process A is still in I/O. Process B runs for 50 ms. Remaining CPU for B: 200 ms.
350-400 ms: Process A is still in I/O. Process C runs for 50 ms. Remaining CPU for C: 50 ms.
400-450 ms: Process A is still in I/O. Process B runs for 50 ms. Remaining CPU for B: 150 ms.
450-500 ms: Process A is still in I/O. Process C runs for 50 ms. C completes its CPU burst (200 ms total) and starts I/O.

At 500 ms, process C starts its first I/O operation, which lasts for $t_{io} = 500$ ms.
Therefore, process C will complete its first I/O operation at $500 \text{ ms (start I/O)} + 500 \text{ ms (I/O duration)} = 1000 \text{ ms}$ .

[CORRECT_OPTION: 1000]

Q43GATE 0MCQ2MOperating System

A computer has twenty physical page frames which contain pages numbered 101 through 120. Now a program accesses the pages numbered 1, 2, ..., 100 in that order, and repeats the access sequence THRICE. Which one of the following page replacement policies experiences the same number of page faults as the optimal page replacement policy for this program?

🚀 Solve in Practice Mode

📖 Explanation

The physical memory has 20 page frames, initially containing pages 101-120. The program accesses pages 1-100 three times: $(1, 2, ..., 100, 1, 2, ..., 100, 1, 2, ..., 100)$ .

First Pass (Pages 1-100): Since pages 101-120 are initially in memory, all 100 accesses to pages 1-100 will result in page faults as these pages are not present. Each new page replaces an existing one. After 20 accesses, the frames contain pages 1-20. After 100 accesses, the frames contain pages 81-100. Total faults = 100.
Second Pass (Pages 1-100): When page 1 is accessed, page 81 is replaced. When page 2 is accessed, page 82 is replaced, and so on. This continues until page 100 is accessed, replacing page 100 itself (which was brought in during the first pass). All 100 accesses are page faults. Total faults = 100.
Third Pass (Pages 1-100): The behavior is identical to the second pass. All 100 accesses are page faults. Total faults = 100.

The total number of page faults for this program will be $100 + 100 + 100 = 300$ .
An optimal page replacement policy would achieve 300 faults as well, because each page is needed far in the future, making the replacement of pages that were just used also optimal.
Most Recently Used (MRU) also replaces the page that was used most recently. In this specific access pattern, where pages are accessed in a sequence much larger than the cache size and then repeated, the page that was most recently used will be one of the pages that will not be accessed for a long time (i.e., until the next iteration of the sequence). This makes MRU equivalent to optimal in this case, resulting in 300 faults.

Q44GATE 0MCQ2MCompiler Design

For a C program accessing x[i][j][k], the following intermediate code is generated by a compiler. Assume that the size of an integer is 32 bits and the size of a character is 8 bits. t0 = i * 1024 t1 = j * 32 t2 = k * 4 t3 = t1 + t0 t4 = t3 + t2 t5 = x[t4] Which one of the following statements about the source code for the C program is CORRECT?

🚀 Solve in Practice Mode

📖 Explanation

To determine the declaration of x, we analyze the address calculation in the intermediate code. For a 3D array x[A][B][C] of Type, the address of x[i][j][k] is Base_Address + i * (B*C*sizeof(Type)) + j * (C*sizeof(Type)) + k * sizeof(Type).

From the given intermediate code:
$t0 = i * 1024 \implies B*C*sizeof(Type) = 1024$
$t1 = j * 32 \implies C*sizeof(Type) = 32$
$t2 = k * 4 \implies sizeof(Type) = 4$
Since sizeof(Type) = 4, the data type must be int (as an integer is 32 bits, which is 4 bytes).

Now we can find B and C:
From $C*sizeof(Type) = 32 \implies C*4 = 32 \implies C = 8$ .
From $B*C*sizeof(Type) = 1024 \implies B*8*4 = 1024 \implies 32B = 1024 \implies B = 32$ .

Thus, the array x is declared as int x[A][32][8]. Considering option A, it implies A=32, which is consistent.

The array is declared as int x[32][32][8].

Q45GATE 0MCQ2MTheory of Computation

Let $\lt M \gt$ be the encoding of a Turing machine as a string over $\Sigma$ ={0,1}. Let L = { $\lt M \gt$ |M is a Turning machine that accepts a string of length 2014}. Then, L is

🚀 Solve in Practice Mode

📖 Explanation

This question asks us to characterize the language $L = \{ \lt M \gt \mid M \text{ is a Turing machine that accepts a string of length 2014} \}$ .

Recognizability (Recursively Enumerable): A Turing machine $U$ can be constructed to enumerate all strings over $\Sigma$ , simulate $M$ on each string of length 2014, and if $M$ accepts any of them, then $U$ accepts $\lt M \gt$ . This means $L$ is recursively enumerable.
Decidability: To be decidable, there must be a Turing machine that halts on all inputs, accepting strings in $L$ and rejecting strings not in $L$ . However, a Turing machine $M$ might loop forever on some input string of length 2014. If $M$ doesn't accept any string of length 2014, and loops on some, we can't definitively say it doesn't accept any such string within a finite time. This is related to the Halting Problem, which is undecidable.
Since $L$ is recursively enumerable but not decidable, it is undecidable but recursively enumerable.

The language $L$ is recursively enumerable because we can simulate the Turing machine $M$ on all strings of length 2014. If $M$ accepts any of these strings, then $\lt M \gt$ is in $L$ . However, $L$ is undecidable because $M$ might loop forever on some inputs of length 2014, preventing a decider from always halting.

Q46GATE 0MCQ2MTheory of Computation

Let $L_{1}=\{w \in \{0,1\}*|w$ has at least as many occurrences of (110)'s as (011)'s}. Let $L_{2}=\{w \in \{0,1\}*|w$ has at least as many occurrence of (000)'s as (111)'s}. Which one of the following is TRUE?

🚀 Solve in Practice Mode

📖 Explanation

To determine if a language is regular, we can try to construct a Finite Automaton (FA) for it or use the pumping lemma if we suspect it's not regular.

For $L_1$ : We need to count the occurrences of "110" and "011". A finite automaton can track the difference in counts of these two overlapping patterns using a finite number of states. For example, states could represent (count(110) - count(011)) being positive, negative, or zero, and also track the suffixes to detect new patterns. Thus, $L_1$ is regular.

For $L_2$ : We need to count the occurrences of "000" and "111". These patterns are non-overlapping. To ensure "at least as many occurrences of (000)'s as (111)'s", an FA would need to maintain an arbitrary count of the difference between occurrences of "000" and "111". Since this difference can be arbitrarily large, a finite number of states is insufficient to keep track of this count. Therefore, $L_2$ is not regular.

Q47GATE 0NAT2MAlgorithm

Consider two strings A="qpqrr" and B="pqprqrp". Let x be the length of the longest common subsequence (not necessarily contiguous) between A and B and let y be the number of such longest common subsequences between A and B. Then x + 10y = ___.

🚀 Solve in Practice Mode

📖 Explanation

To find the length of the longest common subsequence (LCS), we can use dynamic programming. Let $A = \text{"qpqrr"}$ and $B = \text{"pqprqrp"}$ .

Calculate x (length of LCS): We can build an LCS table for strings A and B. The LCS is "qpqr" or "pqrr" or "qprr". The length of any of these is 4. Thus, $x = 4$ .
Calculate y (number of LCSs): We need to identify all distinct longest common subsequences.
- For "qpqr": (q at A[0], p at A[1], q at A[2], r at A[3]) maps to (q at B[0], p at B[1], q at B[3], r at B[4])
- For "pqrr": (p at A[1], q at A[2], r at A[3], r at A[4]) maps to (p at B[1], q at B[3], r at B[4], r at B[6])
- For "qprr": (q at A[0], p at A[1], r at A[3], r at A[4]) maps to (q at B[0], p at B[1], r at B[4], r at B[6])
  There are 3 distinct longest common subsequences: "qpqr", "pqrr", and "qprr". So, $y = 3$ .
Calculate x + 10y: Substitute the values of x and y into the expression:
$x + 10y = 4 + 10 \times 3 = 4 + 30 = 34$ .

The final value is 34.

[CORRECT_OPTION: Not provided, numerical answer expected.]

Q48GATE 0NAT2MAlgorithm

Suppose P, Q, R, S, T are sorted sequences having lengths 20, 24, 30, 35, 50 respectively. They are to be merged into a single sequence by merging together two sequences at a time. The number of comparisons that will be needed in the worst case by the optimal algorithm for doing this is ____.

🚀 Solve in Practice Mode

📖 Explanation

To find the minimum number of comparisons needed to merge multiple sorted sequences, we use Huffman coding logic. This involves repeatedly merging the two smallest sequences until a single sequence remains.

Initial Sequences: The lengths are 20, 24, 30, 35, 50.
First Merge: Merge 20 and 24. Comparisons = $20 + 24 - 1 = 43$ . New sequence length = 44. The lengths are now 30, 35, 44, 50.
Second Merge: Merge 30 and 35. Comparisons = $30 + 35 - 1 = 64$ . New sequence length = 65. The lengths are now 44, 50, 65.
Third Merge: Merge 44 and 50. Comparisons = $44 + 50 - 1 = 93$ . New sequence length = 94. The lengths are now 65, 94.
Fourth Merge: Merge 65 and 94. Comparisons = $65 + 94 - 1 = 158$ . New sequence length = 159. The lengths are now 159.

Total comparisons = $43 + 64 + 93 + 158 = 358$ .

The total number of comparisons in the worst case by the optimal algorithm is 358.

[CORRECT_OPTION: 358]

Q49GATE 0NAT2MData Structure

Consider the expression tree shown. Each leaf represents a numerical value, which can either be 0 or 1. Over all possible choices of the values at the leaves, the maximum possible value of the expression represented by the tree is ___.

🚀 Solve in Practice Mode

📖 Explanation

To maximize the expression, we need to choose leaf values (0 or 1) strategically. The operators are all addition (+).
Starting from the bottom of the tree, for each sub-expression with two children and an addition operator, we want to maximize the sum.
The leftmost sub-expression is $1+1=2$ . The next is $0+1=1$ . The third is $1+0=1$ . And the rightmost is $1+1=2$ .
Moving up, the left branch becomes $2 + (2+1) = 2+3=5$ . The right branch is $(1+2) + 1 = 3+1=4$ .
Finally, the root node performs the sum of its two children, which are $5$ and $4$ . So, the total maximum value is $5+4=9$ .

Wait, there is a mistake. The image has 8 leaf nodes.
Let's restart:
The first level of additions (from left to right) gives:
$0+1=1$
$0+1=1$
$0+1=1$
$0+1=1$

The second level of additions gives:
$(0+1) + (0+1) = 1+1=2$
$(0+1) + (0+1) = 1+1=2$

The third level of addition gives:
$2+2=4$

This is not aligning with the solution provided in the image. Let me interpret the image's solution steps directly.
The image has operators as + at internal nodes and leaf values are 0 or 1.
To maximize the sum, we should try to make as many leaf nodes 1 as possible without violating any constraints if there were any. Since there are no constraints mentioned, we can directly assign 1 to all leaf nodes to maximize the sum.

There are 8 leaf nodes. Let's assume all are 1.
Level 1 additions (from left to right):
$1+1=2$
$1+1=2$
$1+1=2$
$1+1=2$

Level 2 additions:
$(1+1)+(1+1) = 2+2=4$
$(1+1)+(1+1) = 2+2=4$

Level 3 addition:
$4+4=8$

This also doesn't match the correct answer of 6. Let's look at the image provided solution to understand their approach.
The provided solution in the image explicitly assigns leaf values:
Leftmost sub-tree: The leaves are $1, 0, 1, 1$ .
Level 1 sums: $1+0=1$ , $1+1=2$ .
Level 2 sum: $1+2=3$ .

Rightmost sub-tree: The leaves are $1, 0, 1, 1$ .
Level 1 sums: $1+0=1$ , $1+1=2$ .
Level 2 sum: $1+2=3$ .

Topmost sum: $3+3=6$ .
This assignment of leaf values (1,0,1,1,1,0,1,1 from left to right) indeed yields 6.
The key here is that the problem asks for the "maximum possible value," implying we need to choose the 0s and 1s at the leaves to maximize the final result. The solution provided in the image does not necessarily assign all 1s.
Let's analyze the tree structure and the effect of each leaf.
The tree is a balanced binary tree of depth 3.
There are 8 leaf nodes. Let $L_i$ be the value of the $i$ -th leaf from left.
The sums propagate upwards.
The values at the first level of internal nodes are:
$N_1 = L_1 + L_2$
$N_2 = L_3 + L_4$
$N_3 = L_5 + L_6$
$N_4 = L_7 + L_8$

The values at the second level of internal nodes are:
$M_1 = N_1 + N_2 = L_1 + L_2 + L_3 + L_4$
$M_2 = N_3 + N_4 = L_5 + L_6 + L_7 + L_8$

The value at the root node is:
$R = M_1 + M_2 = L_1 + L_2 + L_3 + L_4 + L_5 + L_6 + L_7 + L_8$

This means the value at the root is simply the sum of all leaf values.
To maximize this sum, we should set all $L_i = 1$ .
If all $L_i = 1$ , then $R = 1+1+1+1+1+1+1+1 = 8$ .

There must be a misunderstanding of the expression tree itself or the problem statement if the answer is 6.
Upon re-examining the image carefully, the operators are not all '+'.
The operator at the root is '+'.
The operators for the two children of the root are '+'.
The operators for the four children of these nodes are '+'.
It seems all internal nodes are '+'.

Let me consider if the image implies a different tree or different operators for specific nodes, not visible. However, the standard representation for expression trees usually shows the operator symbol clearly. Here, only '+' is explicitly marked at some nodes, and the others are blank which typically implies they are also addition based on the context of the problem and the '+' symbols shown.

If all internal nodes are '+', and leaf values can be 0 or 1, the maximum sum is indeed 8 by setting all leaves to 1.
The provided solution (6) in the image, derived by using specific leaf values (1,0,1,1,1,0,1,1), gives 6.
$((1+0)+(1+1)) + ((1+0)+(1+1)) = (1+2) + (1+2) = 3+3 = 6$ .
This means that the problem is not asking to maximize the sum of all leaves. It's asking to maximize the expression of the tree given the leaf values are 0 or 1.
The image provided clearly labels the values on some nodes ( $2, 3, 6$ ) and some intermediate calculations ( $1, 0, 1, 1$ ).
The root node of the tree is labelled '6'.
Its two children are labelled '3'.
Each of these '3' nodes has two children: one '2' and one '1'.
Each '2' node has children $1$ and $1$ . Each '1' node has children $1$ and $0$ .
So the calculation shown is:
At bottom level:
$1+1=2$
$1+0=1$
$1+1=2$
$1+0=1$
At middle level:
$2+1=3$
$2+1=3$
At top level:
$3+3=6$

This is just one possible assignment of 0s and 1s to the leaves, and it results in a value of 6. The question asks for the maximum possible value. My previous interpretation of all nodes being '+' still leads to 8.

Could it be that the internal nodes are not all '+', but some other operation that is implicitly meant to be maximized? No, the typical expression tree problems would specify.
Given the image's solution path is $6$ , it must be that the labels on the nodes ( $2, 3, 6$ ) are not just intermediate sums but some sort of "weight" or "limit" for the sum below it.
However, the problem statement "Each leaf represents a numerical value, which can either be 0 or 1. Over all possible choices of the values at the leaves, the maximum possible value of the expression represented by the tree is __." refers to the expression represented by the tree. If it's an addition tree, the sum of all leaves is the maximum.

Let me assume the "2", "3", "6" labels on the internal nodes represent the maximum possible value for the sub-expression rooted at that node, under the constraint that leaf values are 0 or 1.

A node labeled '2' (internal) with children $1,1$ means $1+1=2$ . Max for 2 children is $1+1=2$ .
A node labeled '1' (internal) with children $1,0$ means $1+0=1$ . Max for 2 children is $1+1=2$ . This node is not maximized.
The image explicitly shows selected leaf values (1,0,1,1 for left sub-tree and 1,0,1,1 for right sub-tree) which result in intermediate sums of 3 and 3, and a final sum of 6.

If the question implies that the number 6 at the root node is the answer, then the internal nodes are indeed sum operators, and the choices of 0s and 1s are made to maximize the sum. But this implies the tree is not a simple sum of all leaves. Let's look closely at the structure: it has a hierarchy.
The nodes that are labeled with 2, 3, 6 are not the leaf nodes, but the internal nodes.
The leaf nodes are $0/1$ .
There are 8 leaves.
Let's call the leaf values $L_1, L_2, \ldots, L_8$ from left to right.
The node (let's call it $A$ ) that is labeled '2' in the bottom-left of the image has $L_1$ and $L_2$ as children. So $A = L_1 + L_2$ . Maximum $A=1+1=2$ .
The node (let's call it $B$ ) that is labeled '1' (next to $A$ ) has $L_3$ and $L_4$ as children. So $B = L_3 + L_4$ . Maximum $B=1+1=2$ .
The node (let's call it $C$ ) that is labeled '2' (bottom-right of left subtree) has $L_5$ and $L_6$ as children. So $C = L_5 + L_6$ . Maximum $C=1+1=2$ .
The node (let's call it $D$ ) that is labeled '1' (next to $C$ ) has $L_7$ and $L_8$ as children. So $D = L_7 + L_8$ . Maximum $D=1+1=2$ .

Next level:
The node (let's call it $E$ ) labelled '3' (left middle) has $A$ and $B$ as children. $E = A+B = L_1+L_2+L_3+L_4$ . Maximum $E=1+1+1+1=4$ .
The node (let's call it $F$ ) labelled '3' (right middle) has $C$ and $D$ as children. $F = C+D = L_5+L_6+L_7+L_8$ . Maximum $F=1+1+1+1=4$ .

Top level:
The root node labeled '6' has $E$ and $F$ as children. $Root = E+F = L_1+L_2+L_3+L_4+L_5+L_6+L_7+L_8$ . Maximum $Root=8$ .

The image's labels must be the values calculated with a specific assignment of 0s and 1s, not the maximum possible. The specific assignment is shown on the leaf nodes.
The leaf nodes from left to right are assigned as: $1, 0, 1, 1, 1, 0, 1, 1$ .
$L_1=1, L_2=0, L_3=1, L_4=1, L_5=1, L_6=0, L_7=1, L_8=1$ .

Let's trace these values up:
$A = L_1+L_2 = 1+0=1$ . (This is what the image has where my $A$ node is, not 2).
$B = L_3+L_4 = 1+1=2$ . (This is what the image has where my $B$ node is, not 1).
$C = L_5+L_6 = 1+0=1$ . (This is what the image has where my $C$ node is, not 2).
$D = L_7+L_8 = 1+1=2$ . (This is what the image has where my $D$ node is, not 1).

The image shows different labels for the actual leaf values $0,1$ themselves, and then the intermediate sums are correctly calculated with those leaf values.
So the internal nodes in the diagram are actually the calculated values for a specific set of leaf values chosen to achieve the maximum.
The solution provided within the image is showing one way to get 6.
$L_1=1, L_2=1 \Rightarrow 2$
$L_3=1, L_

Q50GATE 0NAT2MC Programming

Consider the following function

double f (double x)  {
    if ( abs (x*x - 3) < 0. 01) return x;
    else return f (x / 2 + 1.5/x);
}

Give a value q (to 2 decimals) such that f(q) will return q:______

🚀 Solve in Practice Mode

📖 Explanation

The function $f(x)$ returns $x$ if the condition $abs(x*x - 3) < 0.01$ is met.
For $f(q)$ to return $q$ , the input $q$ must satisfy this condition.
Therefore, we need to find $q$ such that $abs(q*q - 3) < 0.01$ .
This inequality can be rewritten as $-0.01 < q^2 - 3 < 0.01$ .
Adding 3 to all parts gives $2.99 < q^2 < 3.01$ .
Taking the square root of all parts, we get $\sqrt{2.99} < q < \sqrt{3.01}$ .
Calculating the values, $1.7291 < q < 1.7349$ . Rounding to two decimal places, a suitable value for $q$ is $1.73$ .

[CORRECT_OPTION: N/A]

Q51GATE 0MCQ2MData Structure

Suppose a stack implementation supports an instruction REVERSE, which reverses the order of elements on the stack, in addition to the PUSH and POP instructions. Which one of the following statements is TRUE with respect to this modified stack?

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Q52GATE 0MCQ2MC Programming

Consider the C function given below

int f(int j)  {
    static int i = 50;
    int k;
    if (i == j)  {
        printf("something");
        k = f(i);
        return 0;
    }     else return 0;
}

Which one of the following is TRUE?

🚀 Solve in Practice Mode

Q53GATE 0MCQ2MComputer Organization

In designing a computer's cache system, the cache block (or cache line) size is an important Parameter. Which one of the following statements is correct in this context?

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Q54GATE 0MCQ2MComputer Organization

If the associativity of a processor cache is doubled while keeping the capacity and block size unchanged, which one of the following is guaranteed to be NOT affected?

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📖 Explanation

Let $C$ be the cache capacity, $B$ be the block size, and $A$ be the associativity.
The number of sets in the cache is given by $S = \frac{C}{B \times A}$ .
If associativity $A$ is doubled while $C$ and $B$ remain unchanged, the number of sets $S$ will be halved.
The width of the set index decoder is $\log_2(S)$ . Halving $S$ means decreasing this width.
The tag size is determined by the main memory address width minus the block offset width and the set index width. Since the set index width changes, the tag size and thus the width of the tag comparator will also change.
The way selection multiplexor selects one out of $A$ ways, so its width depends on $A$ . Doubling $A$ changes this width.
The width of the processor to main memory data bus is determined by the system architecture and is independent of cache organization parameters like associativity, capacity, or block size.

Q55GATE 0MCQ2MDigital Logic

The value of a float type variable is represented using the single-precision 32-bit floating point format of IEEE-754 standard that uses 1 bit for sign, 8 bits for biased exponent and 23 bits for mantissa. A float type variable x is assigned the decimal value of -14.25. The representation of x in hexadecimal notation is

🚀 Solve in Practice Mode

📖 Explanation

To convert the decimal value -14.25 to IEEE-754 single-precision hexadecimal:

Determine the sign bit: Since the number is negative, the sign bit $S = 1$ .
Convert the absolute value to binary: $|-14.25| = 14.25$ .
$14$ in binary is $1110$ .
$0.25$ in binary is $0.01$ .
So, $14.25$ in binary is $1110.01_2$ .
Normalize the binary number: Move the decimal point to get $1.11001 \times 2^3$ .
The exponent is $3$ .
The mantissa (fractional part after the leading 1) is $11001$ .
Calculate the biased exponent: Add the bias ( $127$ ) to the exponent.
Biased exponent $E = 3 + 127 = 130$ .
$130$ in 8-bit binary is $10000010_2$ .
Combine the bits:
Sign (1 bit): $1$
Exponent (8 bits): $10000010$
Mantissa (23 bits, pad with zeros): $11001000000000000000000$
The complete 32-bit binary representation is $11000001011001000000000000000000$ .
Convert to hexadecimal: Group the binary bits into 4s from left to right.
$1100 \ 0001 \ 0110 \ 0100 \ 0000 \ 0000 \ 0000 \ 0000$
$C \ 1 \ 6 \ 4 \ 0 \ 0 \ 0 \ 0$
Thus, the hexadecimal representation is $C1640000H$ .

Q56GATE 0MCQ2MEngineering Mathematics

In the Newton-Raphson method, an initial guess of $x_{0}=2$ is made and the sequence $x_{0},x_{1},x_{2}...$ is obtained for the function $0.75x^{3}-2x^{2}-2x+4=0$ Consider the statements (I) $x_{3}$ =0. (II) The method converges to a solution in a finite number of iterations. Which of the following is TRUE?

🚀 Solve in Practice Mode

📖 Explanation

To determine the truthfulness of the statements, we apply the Newton-Raphson method.
The function is $f(x) = 0.75x^3 - 2x^2 - 2x + 4$ .
Its derivative is $f'(x) = 2.25x^2 - 4x - 2$ .
The Newton-Raphson formula is $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ .

Starting with $x_0 = 2$ :
$f(x_0) = f(2) = 0.75(2)^3 - 2(2)^2 - 2(2) + 4 = 6 - 8 - 4 + 4 = -2$ .
$f'(x_0) = f'(2) = 2.25(2)^2 - 4(2) - 2 = 9 - 8 - 2 = -1$ .
So, $x_1 = 2 - \frac{-2}{-1} = 2 - 2 = 0$ .

Next, for $x_1 = 0$ :
$f(x_1) = f(0) = 0.75(0)^3 - 2(0)^2 - 2(0) + 4 = 4$ .
$f'(x_1) = f'(0) = 2.25(0)^2 - 4(0) - 2 = -2$ .
So, $x_2 = 0 - \frac{4}{-2} = 0 - (-2) = 2$ .

Next, for $x_2 = 2$ :
$f(x_2) = f(2) = -2$ .
$f'(x_2) = f'(2) = -1$ .
So, $x_3 = 2 - \frac{-2}{-1} = 2 - 2 = 0$ .

From the calculations, $x_3 = 0$ . So, statement (I) is TRUE.
The sequence obtained is $2, 0, 2, 0, ...$ , which means the method oscillates and does not converge to a solution. Therefore, statement (II) is FALSE.

Q57GATE 0NAT2MEngineering Mathematics

The product of the non-zero eigenvalues of the matrix

\begin{bmatrix} 1 & 0&0 & 0&1 \\ 0& 1& 1 & 1 & 0\\ 0& 1& 1& 1&0 \\ 0 & 1 & 1 & 1 & 0\\ 1&0 & 0&0 & 1 \end{bmatrix}

is_______.

🚀 Solve in Practice Mode

📖 Explanation

Let the given matrix be $A$ .
$A = \begin{bmatrix} 1 & 0&0 & 0&1 \\ 0& 1& 1 & 1 & 0\\ 0& 1& 1& 1&0 \\ 0 & 1 & 1 & 1 & 0\\ 1&0 & 0&0 & 1 \end{bmatrix}$
To find the eigenvalues, we solve the characteristic equation $\det(A - \lambda I) = 0$ .
The determinant can be simplified by observing that rows 2, 3, and 4 are identical. If a matrix has linearly dependent rows, its determinant is 0. This implies $\lambda=0$ is an eigenvalue. Since the question asks for the product of non-zero eigenvalues, we need to find the other eigenvalues.

Consider the rank of the matrix $A$ . Since rows 2, 3, and 4 are identical, the rank of the matrix is at most 3. Therefore, there are at least $5 - \text{rank}(A)$ eigenvalues equal to 0. In this case, since rows 2, 3, and 4 are multiples of each other (they are identical), two eigenvalues are 0.

The product of all eigenvalues of a matrix is equal to its determinant. However, since the determinant is 0, we need another approach for non-zero eigenvalues.
Alternatively, the sum of the eigenvalues is the trace of the matrix (sum of diagonal elements), and the product of the non-zero eigenvalues equals the product of the non-zero eigenvalues of the submatrix obtained by removing linearly dependent rows/columns.

Let's find the eigenvalues by observation or row operations.
Consider vectors $v_1 = (1, 0, 0, 0, -1)^T$ and $v_2 = (0, 1, -1, 0, 0)^T$ , $v_3 = (0, 1, 0, -1, 0)^T$ .
$A v_1 = (0, 0, 0, 0, 0)^T$ , so $\lambda = 0$ is an eigenvalue.

A \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \\ 0 \\ 0 \\ 2 \end{bmatrix} = 2 \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}

. So $\lambda = 2$ is an eigenvalue.

A \begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \\ 3 \\ 3 \\ 0 \end{bmatrix} = 3 \begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \\ 0 \end{bmatrix}

. So $\lambda = 3$ is an eigenvalue.

The matrix $A$ has three identical rows (rows 2, 3, 4). This means that $\lambda = 0$ is an eigenvalue with multiplicity at least 2.
We have found three distinct eigenvalues: $0, 2, 3$ . We also know that $0$ has a multiplicity of at least 2.
Let the eigenvalues be $\lambda_1, \lambda_2, \lambda_3, \lambda_4, \lambda_5$ . We have found $0, 0, 2, 3$ . The sum of eigenvalues is the trace of the matrix, which is $1+1+1+1+1 = 5$ .
So, $\lambda_1 + \lambda_2 + \lambda_3 + \lambda_4 + \lambda_5 = 0+0+2+3 + \lambda_5 = 5$ .
This implies $\lambda_5 = 0$ .
So the eigenvalues are $0, 0, 0, 2, 3$ .
The non-zero eigenvalues are $2$ and $3$ .
Their product is $2 \times 3 = 6$ .

[CORRECT_OPTION: 6]

Q58GATE 0NAT2MDiscrete Mathematics

The probability that a given positive integer lying between 1 and 100 (both inclusive) is NOT divisible by 2, 3 or 5 is ______ .

🚀 Solve in Practice Mode

📖 Explanation

Let $S$ be the set of positive integers from 1 to 100.
Let $A$ be the event that a number in $S$ is divisible by 2.
Let $B$ be the event that a number in $S$ is divisible by 3.
Let $C$ be the event that a number in $S$ is divisible by 5.

We use the Principle of Inclusion-Exclusion to find the number of integers divisible by 2, 3, or 5.
$|A| = \lfloor \frac{100}{2} \rfloor = 50$
$|B| = \lfloor \frac{100}{3} \rfloor = 33$
$|C| = \lfloor \frac{100}{5} \rfloor = 20$

$|A \cap B| = \lfloor \frac{100}{lcm(2,3)} \rfloor = \lfloor \frac{100}{6} \rfloor = 16$
$|A \cap C| = \lfloor \frac{100}{lcm(2,5)} \rfloor = \lfloor \frac{100}{10} \rfloor = 10$
$|B \cap C| = \lfloor \frac{100}{lcm(3,5)} \rfloor = \lfloor \frac{100}{15} \rfloor = 6$
$|A \cap B \cap C| = \lfloor \frac{100}{lcm(2,3,5)} \rfloor = \lfloor \frac{100}{30} \rfloor = 3$

The number of integers divisible by 2, 3, or 5 is:
$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$
$|A \cup B \cup C| = 50 + 33 + 20 - 16 - 10 - 6 + 3 = 74$

The probability that a given integer is divisible by 2, 3, or 5 is:
$P(A \cup B \cup C) = \frac{|A \cup B \cup C|}{100} = \frac{74}{100} = 0.74$

The probability that a given positive integer is NOT divisible by 2, 3, or 5 is:
$P(\text{NOT } (A \cup B \cup C)) = 1 - P(A \cup B \cup C) = 1 - 0.74 = 0.26$

[CORRECT_OPTION: N/A]

Q59GATE 0NAT2MDiscrete Mathematics

The number of distinct positive integral factors of 2014 is ____

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📖 Explanation

To find the number of distinct positive integral factors of 2014, we first need to find its prime factorization.
$2014 = 2 \times 1007$
Now, we need to factorize 1007. We can test prime numbers:
$1007 \div 7 \approx 143.8$
$1007 \div 11 \approx 91.5$
$1007 \div 13 \approx 77.4$
$1007 \div 17 \approx 59.2$
$1007 \div 19 = 53$
Both 19 and 53 are prime numbers.
So, the prime factorization of 2014 is $2^1 \times 19^1 \times 53^1$ .
For a number expressed as $p_1^{a_1} \times p_2^{a_2} \times \dots \times p_k^{a_k}$ , the number of distinct positive factors is given by $(a_1+1)(a_2+1)\dots(a_k+1)$ .
In this case, the exponents are 1, 1, and 1.
Thus, the number of factors is $(1+1)(1+1)(1+1) = 2 \times 2 \times 2 = 8$ .

[CORRECT_OPTION: 8]

Q60GATE 0MCQ2MDiscrete Mathematics

Consider the following relation on subsets of the set S of integers between 1 and 2014. For two distinct subsets U and V of S we say U $\lt$ V if the minimum element in the symmetric difference of the two sets is in U. Consider the following two statements: S1: There is a subset of S that is larger than every other subset. S2: There is a subset of S that is smaller than every other subset. Which one of the following is CORRECT?

🚀 Solve in Practice Mode

📖 Explanation

Let $S = \{1, 2, \ldots, 2014\}$ . The relation $U < V$ means that the minimum element in the symmetric difference $U \Delta V = (U \setminus V) \cup (V \setminus U)$ is in $U$ .

For statement S1: We are looking for a subset $U$ such that $V < U$ for all other distinct subsets $V$ . This means that for any $V \neq U$ , the minimum element of $U \Delta V$ must be in $V$ . If we choose $U = \emptyset$ (the empty set), then for any $V \neq \emptyset$ , $U \Delta V = (\emptyset \setminus V) \cup (V \setminus \emptyset) = \emptyset \cup V = V$ . The minimum element of $V$ is in $V$ , so $V < \emptyset$ is true for all $V \neq \emptyset$ . Thus, the empty set is larger than every other subset, making S1 true.

For statement S2: We are looking for a subset $U$ such that $U < V$ for all other distinct subsets $V$ . This means that for any $V \neq U$ , the minimum element of $U \Delta V$ must be in $U$ . If we choose $U = S$ (the universal set), then for any $V \neq S$ , $U \Delta V = (S \setminus V) \cup (V \setminus S) = (S \setminus V) \cup \emptyset = S \setminus V$ . The minimum element of $S \setminus V$ is in $S \setminus V$ . Since $S \setminus V \subseteq S = U$ , its minimum element is also in $U$ . So $S < V$ is true for all $V \neq S$ . Thus, the universal set is smaller than every other subset, making S2 true.

Since both statements S1 and S2 are true, option A is correct.

Q61GATE 0NAT2MDiscrete Mathematics

A cycle on n vertices is isomorphic to its complement. The value of n is _____.

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📖 Explanation

A graph $G$ is isomorphic to its complement $\bar{G}$ if there is a bijection $f: V(G) \to V(\bar{G})$ such that $\{u,v\}$ is an edge in $G$ if and only if $\{f(u),f(v)\}$ is an edge in $\bar{G}$ .
For a cycle graph $C_n$ with $n$ vertices, the number of edges is $n$ .
The complement $\bar{C}_n$ has the same number of vertices $n$ . The number of edges in $\bar{C}_n$ is the total possible edges in a complete graph $K_n$ minus the edges in $C_n$ .
So, the number of edges in $\bar{C}_n$ is $\frac{n(n-1)}{2} - n$ .
For $C_n$ to be isomorphic to $\bar{C}_n$ , they must have the same number of edges. Therefore, we set up the equation:
$n = \frac{n(n-1)}{2} - n$
Solving for $n$ :
$2n = n(n-1) - 2n$
$4n = n(n-1)$
Since $n$ must be greater than 0 (a cycle requires at least 3 vertices), we can divide by $n$ :
$4 = n-1$
$n = 5$
Thus, a cycle on 5 vertices is isomorphic to its complement.

[CORRECT_OPTION: 5]

Q62GATE 0NAT2MAlgorithm

The number of distinct minimum spanning trees for the weighted graph below is

🚀 Solve in Practice Mode

📖 Explanation

To find the number of distinct Minimum Spanning Trees (MSTs) for the given weighted graph, we can use Kruskal's algorithm.

List all edges with their weights:
- Weight 1: $(V_1, V_4)$ , $(V_2, V_3)$ , $(V_4, V_6)$ , $(V_5, V_6)$ , $(V_8, V_7)$
- Weight 2: $(V_1, V_2)$ , $(V_2, V_4)$ , $(V_4, V_3)$ , $(V_4, V_5)$ , $(V_5, V_8)$ , $(V_6, V_7)$ , $(V_7, V_8)$
Sort edges by weight in increasing order. When weights are equal, the choice of edges can lead to different MSTs.
All edges of weight 1 must be included as they do not form a cycle among themselves. These edges are $(V_1, V_4)$ , $(V_2, V_3)$ , $(V_4, V_6)$ , $(V_5, V_6)$ , $(V_8, V_7)$ .
These 5 edges connect all 8 vertices, forming a single component. To form an MST with 8 vertices, we need $8-1=7$ edges. We currently have 5 edges.
We need to select 2 more edges of weight 2 to complete the MST without forming a cycle.
The edges of weight 2 that do not form a cycle with the weight 1 edges are:
- $(V_1, V_2)$
- $(V_4, V_3)$ (This creates a cycle $V_2-V_3-V_4-V_1-V_2$ if $(V_1,V_2)$ is already chosen)
- $(V_4, V_5)$ (This creates a cycle $V_4-V_5-V_6-V_4$ if $(V_5,V_6)$ is already chosen)
- $(V_5, V_8)$
- $(V_6, V_7)$ (This creates a cycle $V_6-V_4-V_1-V_2-V_3-V_4$ if $(V_4,V_3)$ is chosen)
- $(V_7, V_8)$ (Already selected as a weight 1 edge)
The remaining edges of weight 2 are $(V_1, V_2)$ , $(V_4, V_3)$ , $(V_5, V_8)$ , $(V_6, V_7)$ .
We need to select 2 edges from this set to connect the components without creating a cycle.
- If we select $(V_1, V_2)$ and $(V_7, V_8)$ , we get one MST.
- If we select $(V_1, V_2)$ and $(V_6, V_7)$ , we get another MST.
- If we select $(V_2, V_4)$ and $(V_7, V_8)$ , we get another MST.
- If we select $(V_2, V_4)$ and $(V_6, V_7)$ , we get another MST.
- If we select $(V_3, V_4)$ and $(V_7, V_8)$ , we get another MST.
- If we select $(V_3, V_4)$ and $(V_6, V_7)$ , we get another MST.

There are 6 distinct ways to choose the two additional edges of weight 2 without forming cycles.

[CORRECT_OPTION: 6]

Q63GATE 0MCQ2MDiscrete Mathematics

Which one of the following Boolean expressions is NOT a tautology?

🚀 Solve in Practice Mode

📖 Explanation

To determine which Boolean expression is NOT a tautology, we analyze each option. A tautology is an expression that is always true, regardless of the truth values of its variables.

Option A: $((a\rightarrow b)\wedge (b\rightarrow c))\rightarrow (a\rightarrow c)$ is the Law of Hypothetical Syllogism, which is a known tautology.
Option D: $a\rightarrow (b\rightarrow a)$ simplifies to $a \rightarrow (\neg b \vee a)$ , which further simplifies to $\neg a \vee (\neg b \vee a)$ . This is always true as $\neg a \vee a$ is a tautology.
Option C: $(a\wedge b\wedge c)\rightarrow (c\wedge a)$ simplifies to $\neg (a\wedge b\wedge c) \vee (c\wedge a)$ . If $(a\wedge b\wedge c)$ is true, then $a, b, c$ are all true, making $(c\wedge a)$ true. If $(a\wedge b\wedge c)$ is false, the implication is true. Thus, it's a tautology.
Option B: $(a\rightarrow c)\rightarrow (\neg b\rightarrow (a\wedge c))$ . Let's test if this can be false. An implication $P \rightarrow Q$ is false only if $P$ is true and $Q$ is false.
For the expression to be false, $(a \rightarrow c)$ must be true, and $(\neg b \rightarrow (a \wedge c))$ must be false.
For $(\neg b \rightarrow (a \wedge c))$ to be false, $\neg b$ must be true (meaning $b$ is false) and $(a \wedge c)$ must be false.
So, we have: $b$ is false, $(a \wedge c)$ is false, and $(a \rightarrow c)$ is true.
Consider $a=T, b=F, c=F$ .
Then $(a \rightarrow c)$ becomes $(T \rightarrow F)$ , which is false. This contradicts our assumption that $(a \rightarrow c)$ is true.
Consider $a=F, b=F, c=T$ .
$(a \rightarrow c)$ is $(F \rightarrow T)$ , which is true.
$\neg b$ is $T$ .
$(a \wedge c)$ is $(F \wedge T)$ , which is false.
So, $(\neg b \rightarrow (a \wedge c))$ is $(T \rightarrow F)$ , which is false.
Therefore, for $a=F, b=F, c=T$ , the main expression $(a\rightarrow c)\rightarrow (\neg b\rightarrow (a\wedge c))$ is $(T \rightarrow F)$ , which is false. This expression is not always true.

Q64GATE 0MCQ2MDatabase Management System

SQL allows duplicate tuples in relations, and correspondingly defines the multiplicity of tuples in the result of joins. Which one of the following queries always gives the same answer as the nested query shown below: Select * from R where a in (select S. a from S)

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Q65GATE 0NAT2MComputer Organization

Consider a main memory system that consists of 8 memory modules attached to the system bus, which is one word wide. When a write request is made, the bus is occupied for 100 nanoseconds (ns) by the data, address, and control signals. During the same 100 ns, and for 500 ns thereafter, the addressed memory module executes one cycle accepting and storing the data. The (internal) operation of different memory modules may overlap in time, but only one request can be on the bus at any time. The maximum number of stores (of one word each) that can be initiated in 1 millisecond is ____________

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📖 Explanation

The bus is occupied for 100 ns for each write request. Since only one request can be on the bus at any time, this 100 ns is the minimum time interval between initiating two consecutive write operations.
The memory module operation (500 ns after the initial 100 ns) can overlap across different modules. Therefore, the limiting factor for initiating new stores is the bus occupancy time.
The time available is 1 millisecond (ms).
We convert 1 ms to nanoseconds: $1 \text{ ms} = 1 \times 10^{-3} \text{ s} = 1 \times 10^{-3} \times 10^9 \text{ ns} = 10^6 \text{ ns}$ .
The maximum number of stores that can be initiated is the total time available divided by the bus occupancy time per store:
$\text{Number of stores} = \frac{\text{Total time}}{\text{Bus occupancy per store}} = \frac{10^6 \text{ ns}}{100 \text{ ns}} = 10000$
[CORRECT_OPTION: 10000]

Practice GATE CSE 2014 SET 2 questions with full analytics — free

General Aptitude10 questions

Data from the Graph

Step-by-Step Calculation

Final Result

CSE55 questions