Solution

Solution

Since the numbers are given in the form of ratio that means, their common factor have been cancelled.

Each one’s common factor is HCF, here H.C.F = 12

The numbers are 12, 24, 36.

Solution

Other than 72 remaining all are perfect squares.

Solution

Solution

Solution

Cost price = Rs. 750

Profit = 20% of 750 = Rs. 150

Selling price = 750 + 150 = Rs. 900

Solution

Application is the action of putting something into action.

Supplication is the action of asking something humbly.

Germination is the development of a plant from a seed.

Segregation is the action of setting something or someone apart from others.

Solution

Beating around the bush means not speaking precisely or to the point.

Over the moon means extremely happy or

delighted.

Bite the bullet means to get something over because it's inevitable.

Break a leg means good luck.

Solution

In present tense the sentence is ' The house upon the rock stands firm.'

In the past tense the sentence is ' The house upon the rock stood firm.'

Solution

Solution

Finding the Kth smallest element, the predecessor, may require traveling down the height of the tree, making the running time O(h).

Solution

Finding the Kth smallest element, the predecessor, may require traveling down the height of the tree, making the running time O(h).

Solution

In-order is one type of DFS.

There are three types of depth-first traversal: pre-order, in-order, and post-order.

Solution

Based on the example graph, we can claim that both statements are correct.

Solution

A directed graph is strongly connected if there is a path between all pairs of vertices. A strongly connected component (SCC) of a directed graph is a maximal strongly connected subgraph.

The strongly connected components are

1. Time complexity is O(V+E) because using DFS
2. We can find strongly connected components using the Kosaraju algorithm and Trojan’s algorithm. The kosaraju algorithm and Trojan’s algorithm are not required for GATE.

Solution

As per the inversion definition, we have to consider the

21 > 7, 21 > 15 , 21 > 18 → 3 inversions

41 > 22, 41 > 15, 41> 18, 41 > 25 → 4 inversions

22 > 15, 22 > 18 → 2 inversions.

15, 18, 25, 58, 103, 203 are in ascending order, so we won’t get single inversion from these elements.

Total= 3 + 4 + 2

= 9

Key points in inversion array:

1. If the array is already sorted then the inversion count is 0.

2. If the array is sorted in reverse order that inversion count is the maximum.

Time Complexity: Using divide and conquer we will get O(n log n).

Space Complexity: No extra space is required for this algorithm.

Solution

The recurrence relation to get minimum number of scalar multiplications are

We can get a maximum of 5 possible parentheses using standard dynamic programming method (or) using catalan numbers.

Given the matrices A,B,C,D

Assume the dimensions of A=d₀×d₁, etc..,

Below are the five possible parenthesizations of these arrays, along with the number of multiplications:

(AB)(CD) → d₀d₁d₂ + d₂d₃d₄ + d₀d₂d₄ → 300 + 378 + 540 = 1218

((AB)C)D → d₀d₁d₂ + d₀d₂d₃ + d₀d₃d₄ → 300 + 420 + 630 = 1350

(A(BC))D → d₁d₂d₃ + d₀d₁d₃ + d₀d₃d₄ → 210 + 350 + 630 = 1190

A((BC)D) → d₁d₂d₃ + d₁d₃d₄ + d₀d₁d₄ → 210 + 315 + 450 = 975

A(B(CD)) → d₂d₃d₄ + d₁d₂d₄ + d₀d₁d₄ → 378 + 270 + 450 = 1098

→ The minimum number of scalar multiplications are 975 and optimal parenthesization is A((BC)D)

→ The maximum number of scalar multiplications are 1350 and optimal parenthesization is ((AB)C)D

Solution

Step 1:

Step-2:

Step-3:

110100 → M

1100 → I

11011 → T

110101 → H

Solution

The quicksort gives worst case time complexity for two scenarios.

1. If all the elements are in sorted order
2. All elements are having the same number.

According to above scenarios t₁ gives average case time complexity and t₂ also gives worst case time complexity.

→ It would be t₁< t₂

The best and average case time complexity of quick sort is O(nlogn). The worst case time complexity is O(n^2).

Solution

(AB)+ = ABCD

AB is the candidate key. Hence, AB, ABC are super keys.

Solution

The given SQL query can be replaced by “Select name, course_id from instructor natural join teaches”

Solution

• Transaction Ti needs to be undone if the log contains the record <Ti start> but does not contain either the record <Ti commit> or the record <Ti abort>.
• Transaction Ti needs to be redone if log contains record <Ti start> and either the record <Ti commit> or the record <Ti abort>.

Solution

Indexing will be an occupation field because the occupation field lexicographically sorted will be 4, 5, 2, 1, 3.

Solution

In a snowflake schema, dimension tables are normalized into multiple related tables, which reduces redundancy.

Solution

The number of 2-attribute combinations is given by the combination formula C(n, k):

C(10, 2) = 10! / (2!(10-2)!) = 10 * 9 / 2 = 45

Solution

Decimal scaling normalization shifts the decimal point based on the number of digits in the maximum value.

Since 1,000 has 4 digits, divide the value by 10^3.

Normalized value = 250 / 1000 = 0.25

Solution

As model complexity increases, bias decreases because the model better fits the training data, but variance increases because the model becomes more sensitive to small data fluctuations, risking overfitting.

Solution

4.8,17.6,32

Solution

The Naive Bayes classifier assumes that features are independent given the class labels. This means each feature contributes independently to the probability of the class label.

Solution

Recall=TP/(FN+TP)=6/(6+3)

Solution

precision=TP/(TP+FP)=6/10

Solution

F1 score=2*precision*recall/(precision+recall)

Solution

In 20-fold cross-validation, the dataset is divided into 20 equal parts. In each fold, one part is used for testing, and the remaining 19 parts are used for training.

Given:

• Total samples = 1000
• Number of folds = 20

To find the training size in each fold:

Training size=1000/20 × 19=50×19=950

Thus, the training size for each fold is 950.

Solution

A.The minimum the Dunn index, the better the clustering i.e. clusters are compact and well separated.

• Incorrect.
  

  The Dunn index is a measure where higher values indicate better clustering because it reflects well-separated and compact clusters. A higher Dunn index suggests better clustering quality, not a minimum value.

B. If A = VΣU^T i.e., SVD Decomposition, then the rank of A == 1 + number of non-zero Rows in Σ

Incorrect.

The rank of a matrix A, based on its Singular Value Decomposition (SVD), is equal to the number of non-zero singular values in Σ, not the number of non-zero rows. It also doesn’t involve "1 +" in the rank calculation.

C. Matrix decomposition is an order of O(n2) algorithm.

• Incorrect.
  

  Most matrix decomposition methods, such as SVD or LU decomposition, have a computational complexity of O(n3) for dense n×n matrices. Sparse matrix methods may have lower complexity, but it is typically not O(n2)

  ---

D. In DBSCAN, the Border Point is a point that has fewer than 2*MinPts within eps but it is in the neighborhood of a core point.

• Incorrect.
  

  In DBSCAN:

  • A Border Point is a point that has fewer than MinPts neighbors (not 2*MinPts) within the radius ε\varepsilonε (eps) but lies in the neighborhood of a Core Point

Solution

Solution

Option B is correct; d-separation rules in Bayesian networks allow us to infer conditional independencies. The underlying causal mechanisms do not need to be known to apply this property.

Option C is also true; conditional independence indeed implies that knowing (C) makes (A) irrelevant to (B) in terms of information.

Option A is incorrect because it is possible to have a direct causal relationship while still preserving conditional independence under certain circumstances.

Option D is false; non-adjacency does not necessarily imply conditional independence unless specific nodes are intervening.

Solution

Tautology

Solution

The Markov blanket contains all variables directly influencing the target variable. Using it allows for efficient computation of the conditional probability of a variable during Gibbs Sampling.

Solution

Solution

Solution

Solution

Solution

i.e. the mean value is 21.919 ohms, correct to 3

significant figures.

Solution

Solution

Solution

A positively skewed distribution is one in which the data is distributed more on one side of the scale with a long tail on the right, and in which the mean, median, and mode are all positive values rather than negative or zero. The mean is typically to the right of the data median in this distribution, which is sometimes referred to as the right-skewed distribution.

A distribution type that is negatively skewed is one in which the mean is smaller than the median and mode, the distribution tail is longer on the left side of the graph, and more values plot on the right side of the graph.

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Since determinate of yy^T is zero hence its inverse does not exist.

• Eigen values for yy^T are 0,0, 3

Hence y^Ty is an eigen value for yyT i.e.3

• Rank of yy^T is 1

Solution

Solution

Solution

Solution

Solution

Solution

Solution

result = product_except_self([1, 2, 3, 4])

The function uses two auxiliary arrays to calculate the product of all elements to the left and right of each index. For [1, 2, 3, 4], the result is [24, 12, 8, 6].

Solution

Lists are mutable and passed by reference in Python. Appending to `y` modifies `x` because they reference the same object in memory.

Solution

The `json.dumps()` method serializes a Python dictionary into a JSON-formatted string.

Solution

Option A uses Depth-First Search (DFS) and appends nodes to the stack after visiting all their neighbors, ensuring the correct topological order. Option B uses Kahn's algorithm, but it is incomplete and may fail for cyclic graphs. Options C and D are incorrect as they don't ensure a valid topological order.

Solution

Option A iteratively generates all possible subsets by combining the current number with all existing subsets. Other options fail to generate all subsets.