Solution

To merge 5 individual points into a single cluster, you need (n-1) merges, where (n) is the number of points. So, (5-1 = 4) merges are required.

Solution

All the Above

Solution

A larger k-value means less bias towards overestimating the truly expected error (as training folds will be closer to the total dataset) and higher running time (as you are getting closer to the limit case: Leave-One-Out CV). We also need to consider the variance between the k folds accuracy while selecting the k.

Solution

All three nearest point are of +class so this point will be classified as +class.

Solution

Now this point will be classified as – class because there are 4 – class and 3 +class point are in nearest circle.

Solution

The boundary becomes smoother with increasing value of K.

Solution

If C increases, ||W|| would decreases, that means the margin( 2/||W|| ) increases.

Similarly, if K increases, hinge loss should decrease, which means errors are low which implies that margin is decreasing.( It is given wrong in the options).

And the number of support vectors may decrease as the margin decreases.

If K=0, that means there is no loss term. So it is hard margin svm.

Solution

Both I and II are incorrect

Solution

Solution

Points 6 and 7 are not support vectors and SVM margin only gets affected because of support vectors hence A is incorrect and B is correct. As point 4 is support vector its absence may or may not affect the margin and decision boundary hence, D is incorrect.

When we remove point 5 only the margin passing through that point and the decision boundary will get affected. The other margin passing through points 1,2,3,4 will remain intact.

Solution

Solution

Solution

Solution

Solution

Error will be same for both a and b.

Solution

PCA ⇒ Dimensional Reduction Technique (unsupervised learning)

Naive bayes ⇒ Generative Technique (model of a joint probability function of input and labels)

Logistic Regression ⇒ Discriminative Technique (works on , y P x (x → input/features) y → label)

Solution

if (1, 1) and (–1, 1) are part of same circle than line joining them will also be part of same circle and (0, 1) is on line connecting these points.

Solution

y^​=3+2x

For each x:

• When x=1, y^=3+2(1)=5
• When x=2, y^=3+2(2)=7
• When x=3, y^=3+2(3)=9
• When x=4 y^=3+2(4)=11
• When x=5, y^=3+2(5)=13

Error will be 0

Solution

• Since the support vectors are symmetric around the origin, the decision boundary is a line equidistant from both classes.
• The line that equidistantly separates these points is x1+x2=0

Solution

1. Calculate Eigenvalues:
   

   Det(Σ−λI)=(4−λ)(3−λ)−4=λ2−7λ+10=0
   

   Solving gives eigenvalues λ1=5, λ2=2.5

2. Calculate the Percentage of Variance:

   • The total variance is 5+2.5=7.5
   • The variance explained by the first component is 5/7.5=0.6667×100=66.67%

Solution

The nearest neighbors are (3,3),(2,3),(4,5) corresponding to classes 1, 0, and 1.

Solution

1. Pre-pruning:

   • Stops tree growth based on criteria like depth, minimum samples per leaf, or information gain threshold.

2. Post-pruning:

   • Fully grow the tree and then remove branches or nodes that don't improve model performance, usually evaluated on a validation set.

Solution

1. Symmetry Breaking:

   • Initializing all weights to zero would prevent the network from learning anything meaningful due to symmetry.
   • Small random values are used to break symmetry and allow different neurons to learn different features.

Solution

Calculate the probabilities:

P(A)=20/50=0.4,P(B)=30/50=0.6

Calculate the Gini Index:

Gini=1−(P(A)2+P(B)2)=1−(0.4^2+0.6^2)=1−(0.16+0.36)=0.49

Solution

Calculate the probabilities:

P(positive)=30/40=0.75,P(negative)=10/40=0.25

Calculate the entropy:

Entropy=−(0.75⋅log⁡2(0.75)+0.25⋅log⁡2(0.25))≈0.8

Solution

ReLU is preferred because it helps mitigate the vanishing gradient problem, which is common with sigmoid and tanh functions. ReLU activation allows for faster training and can handle deeper networks more effectively.

Solution

The kernel trick allows SVM to operate in a high-dimensional space without explicitly calculating the coordinates of the data in that space. This enables the SVM to find a linear separator in the transformed space, which may correspond to a non-linear separator in the original space.

Solution

Principal Component Analysis (PCA) is a technique used for dimensionality reduction while preserving as much variance as possible in the data. It transforms the original data into a new set of dimensions (principal components) that are linear combinations of the original dimensions.

The principal components (the new dimensions) are orthogonal to each other, meaning they are independent or uncorrelated. In other words, their dot product should be zero.

let's analyze the options:

• Option A: [1,2,3], [5,-2,2]

  • Dot product: 1∗5+2∗(−2)+3∗2=5−4+6=7
  • These vectors are not orthogonal because the dot product is not zero.

• Option B: [4,5,6], [-2,5,-4]

  • Dot product: 4∗(−2)+5∗5+6∗(−4)=−8+25−24=−7
  • These vectors are not orthogonal because the dot product is not zero.

• Option C: [1,2,3], [-7,2,1]

  • Dot product: 1∗(−7)+2∗2+3∗1=−7+4+3=0
  • These vectors are orthogonal because the dot product is zero. This suggests that these two vectors could represent principal components.

• Option D: [6,7,8], [2,4,5]

  • Dot product: 6∗2+7∗4+8∗5=12+28+40=80
  • These vectors are not orthogonal because the dot product is not zero.

So option C is the correct one.

Solution

The accuracy metric is not a good idea for imbalanced class problems.

Precision and recall metrics are good for imbalanced class problems.

Solution

Time taken = Iterations × Time per iteration =20×0.5=10

Solution

Single linkage, also known as the nearest-neighbor method, tends to create elongated clusters because it links clusters based on the shortest distance between their nearest points.

Solution

The first cluster will include the points 1 and 3, so the new centroid is (1+3)/2=2

Solution

Naive Bayes assumes that features are conditionally independent given the class label.