What Was The Main Conclusion Reached From Piaget's Updated Study On Retribution Versus Restitution?

Introduction

The thought of creating machines which larn by themselves has been driving humans for decades now. For fulfilling that dream, unsupervised learning and clustering is the key. Unsupervised learning provides more flexibility, but is more challenging as well.

Clustering plays an important role to draw insights from unlabeled data. It classifies the data in similar groups which improves various business decisions by providing a meta understanding.

In this skill test, we tested our community on clustering techniques. A total of 1566 people registered in this skill test. If you missed taking the test, here is your opportunity for you to detect out how many questions you lot could have answered correctly.

If you lot are just getting started with Unsupervised Learning, here are some comprehensive resource to assist y'all in your journey:

Machine Learning Certification Form for Beginners
The Nearly Comprehensive Guide to Thou-Means Clustering You'll E'er Demand
Certified AI & ML Blackbelt+ Program

Overall Results

Below is the distribution of scores, this will help you lot evaluate your performance:

Yous tin can access your performance here. More than 390 people participated in the skill exam and the highest score was 33. Here are a few statistics about the distribution.

Overall distribution

Mean Score: 15.11

Median Score: fifteen

Mode Score: 16

Helpful Resources

An Introduction to Clustering and different methods of clustering

Getting your clustering right (Function I)

Getting your clustering right (Part II)

Questions & Answers

Q1. Flick Recommendation systems are an example of:

Classification
Clustering
Reinforcement Learning
Regression

Options:

B. A. two Only

C. 1 and ii

D. ane and 3

E. ii and 3

F. i, 2 and iii

H. 1, ii, 3 and iv

Solution: (E)

More often than not, movie recommendation systems cluster the users in a finite number of similar groups based on their previous activities and contour. Then, at a fundamental level, people in the same cluster are made like recommendations.

In some scenarios, this tin can too exist approached every bit a classification problem for assigning the most appropriate movie course to the user of a specific grouping of users. Also, a movie recommendation system can be viewed as a reinforcement learning problem where information technology learns by its previous recommendations and improves the future recommendations.

Q2. Sentiment Analysis is an example of:

Regression
Nomenclature
Clustering
Reinforcement Learning

Options:

A. 1 But

B. 1 and 2

C. one and three

D. 1, ii and 3

E. 1, 2 and iv

F. i, ii, 3 and 4

Solution: (E)

Sentiment analysis at the fundamental level is the task of classifying the sentiments represented in an paradigm, text or speech into a fix of defined sentiment classes similar happy, lamentable, excited, positive, negative, etc. It tin can also be viewed as a regression problem for assigning a sentiment score of say 1 to 10 for a corresponding image, text or speech.

Another manner of looking at sentiment analysis is to consider it using a reinforcement learning perspective where the algorithm constantly learns from the accurateness of past sentiment analysis performed to improve the future performance.

Q3. Can decision trees be used for performing clustering?

A. True

B. Imitation

Q4. Which of the following is the most appropriate strategy for data cleaning before performing clustering analysis, given less than desirable number of data points:

Capping and flouring of variables
Removal of outliers

Options:

A. 1 but

B. two only

C. one and ii

D. None of the above

Q5. What is the minimum no. of variables/ features required to perform clustering?

A. 0

B. one

C. 2

D. 3

Q6. For two runs of One thousand-Mean clustering is information technology expected to go same clustering results?

A. Yes

B. No

Solution: (B)

K-Means clustering algorithm instead converses on local minima which might also correspond to the global minima in some cases but not always. Therefore, it'due south advised to run the K-Means algorithm multiple times before drawing inferences nearly the clusters.

Nevertheless, note that it'south possible to receive aforementioned clustering results from K-means past setting the aforementioned seed value for each run. Just that is washed by just making the algorithm choose the gear up of same random no. for each run.

Q7. Is it possible that Consignment of observations to clusters does not modify between successive iterations in G-Means

A. Yes

B. No

C. Tin't say

D. None of these

Solution: (A)

When the Yard-Means algorithm has reached the local or global minima, information technology will not alter the consignment of data points to clusters for 2 successive iterations.

Q8. Which of the post-obit tin can act as possible termination atmospheric condition in K-Means?

For a fixed number of iterations.
Assignment of observations to clusters does not change between iterations. Except for cases with a bad local minimum.
Centroids do non change betwixt successive iterations.
Terminate when RSS falls below a threshold.

Options:

A. 1, 3 and 4

B. 1, 2 and iii

C. i, 2 and four

D. All of the above

Solution: (D)

All iv conditions can be used as possible termination condition in G-Means clustering:

This condition limits the runtime of the clustering algorithm, but in some cases the quality of the clustering will be poor because of an insufficient number of iterations.
Except for cases with a bad local minimum, this produces a good clustering, but runtimes may exist unacceptably long.
This also ensures that the algorithm has converged at the minima.
Terminate when RSS falls beneath a threshold. This criterion ensures that the clustering is of a desired quality subsequently termination. Practically, it's a good practice to combine it with a bound on the number of iterations to guarantee termination.

Q9. Which of the following clustering algorithms suffers from the problem of convergence at local optima?

G- Means clustering algorithm
Agglomerative clustering algorithm
Expectation-Maximization clustering algorithm
Various clustering algorithm

Options:

A. ane only

B. 2 and 3

C. 2 and iv

D. ane and 3

E. 1,2 and 4

F. All of the above

Solution: (D)

Out of the options given, only Grand-Means clustering algorithm and EM clustering algorithm has the drawback of converging at local minima.

Q10. Which of the post-obit algorithm is most sensitive to outliers?

A. G-means clustering algorithm

B. K-medians clustering algorithm

C. K-modes clustering algorithm

D. K-medoids clustering algorithm

Solution: (A)

Out of all the options, Thou-Means clustering algorithm is virtually sensitive to outliers as it uses the mean of cluster information points to notice the cluster center.

Q11. After performing Yard-Means Clustering analysis on a dataset, you observed the following dendrogram. Which of the following conclusion can be fatigued from the dendrogram?

A. At that place were 28 information points in clustering analysis

B. The all-time no. of clusters for the analyzed information points is four

C. The proximity part used is Average-link clustering

D. The above dendrogram interpretation is not possible for K-Means clustering analysis

Solution: (D)

A dendrogram is not possible for K-Means clustering analysis. However, one tin create a cluster gram based on Chiliad-Means clustering assay.

Q12. How can Clustering (Unsupervised Learning) exist used to improve the accurateness of Linear Regression model (Supervised Learning):

Creating different models for different cluster groups.
Creating an input characteristic for cluster ids as an ordinal variable.
Creating an input feature for cluster centroids as a continuous variable.
Creating an input feature for cluster size as a continuous variable.

Options:

A. ane only

B. 1 and 2

C. 1 and 4

D. 3 only

E. 2 and four

F. All of the above

Solution: (F)

Creating an input feature for cluster ids as ordinal variable or creating an input feature for cluster centroids every bit a continuous variable might not convey whatsoever relevant information to the regression model for multidimensional data. Merely for clustering in a unmarried dimension, all of the given methods are expected to convey meaningful data to the regression model. For instance, to cluster people in ii groups based on their pilus length, storing clustering ID equally ordinal variable and cluster centroids as continuous variables will convey meaningful data.

Q13. What could be the possible reason(due south) for producing 2 different dendrograms using agglomerative clustering algorithm for the aforementioned dataset?

A. Proximity function used

B. of data points used

C. of variables used

D. B and c merely

Eastward. All of the higher up

Solution: (E)

Change in either of Proximity function, no. of data points or no. of variables will lead to different clustering results and hence different dendrograms.

Q14. In the figure below, if y'all draw a horizontal line on y-axis for y=two. What will be the number of clusters formed?

A. ane

B. ii

C. 3

D. four

Solution: (B)

Since the number of vertical lines intersecting the red horizontal line at y=two in the dendrogram are 2, therefore, ii clusters will be formed.

Q15. What is the most appropriate no. of clusters for the data points represented past the following dendrogram:

A. 2

B. 4

C. half-dozen

D. 8

Solution: (B)

The decision of the no. of clusters that can all-time depict different groups can be chosen by observing the dendrogram. The all-time selection of the no. of clusters is the no. of vertical lines in the dendrogram cut by a horizontal line that tin transverse the maximum altitude vertically without intersecting a cluster.

In the above example, the all-time choice of no. of clusters will exist four every bit the crimson horizontal line in the dendrogram below covers maximum vertical distance AB.

Q16. In which of the following cases will K-Means clustering neglect to give good results?

Data points with outliers
Data points with different densities
Data points with circular shapes
Information points with non-convex shapes

Options:

A. i and 2

B. 2 and 3

C. 2 and 4

D. 1, two and 4

E. one, 2, 3 and 4

Solution: (D)

K-Ways clustering algorithm fails to requite skilful results when the data contains outliers, the density spread of information points beyond the information space is different and the data points follow non-convex shapes.

Q17. Which of the following metrics, exercise we take for finding dissimilarity betwixt two clusters in hierarchical clustering?

Single-link
Consummate-link
Average-link

Options:

A. 1 and 2

B. i and three

C. two and 3

D. ane, ii and 3

Solution: (D)

All of the iii methods i.e. single link, complete link and average link can be used for finding dissimilarity between two clusters in hierarchical clustering.

Q18. Which of the following are true?

Clustering analysis is negatively afflicted by multicollinearity of features
Clustering analysis is negatively afflicted by heteroscedasticity

Options:

A. 1 only

B. 2 only

C. 1 and 2

D. None of them

Solution: (A)

Clustering assay is not negatively affected by heteroscedasticity only the results are negatively impacted past multicollinearity of features/ variables used in clustering equally the correlated feature/ variable will carry extra weight on the distance calculation than desired.

Q19. Given, six points with the following attributes:

Which of the following clustering representations and dendrogram depicts the use of MIN or Single link proximity role in hierarchical clustering:

Solution: (A)

For the single link or MIN version of hierarchical clustering, the proximity of ii clusters is defined to be the minimum of the distance between any two points in the different clusters. For instance, from the tabular array, nosotros come across that the altitude between points iii and 6 is 0.11, and that is the height at which they are joined into i cluster in the dendrogram. As another example, the distance betwixt clusters {3, half-dozen} and {2, 5} is given by dist({3, 6}, {2, 5}) = min(dist(three, ii), dist(6, 2), dist(iii, five), dist(half-dozen, five)) = min(0.1483, 0.2540, 0.2843, 0.3921) = 0.1483.

Q20 Given, half-dozen points with the post-obit attributes:

Which of the following clustering representations and dendrogram depicts the use of MAX or Complete link proximity function in hierarchical clustering:

Solution: (B)

For the single link or MAX version of hierarchical clustering, the proximity of ii clusters is defined to be the maximum of the distance betwixt any 2 points in the different clusters. Similarly, here points iii and 6 are merged first. Nevertheless, {three, 6} is merged with {four}, instead of {ii, five}. This is because the dist({3, 6}, {four}) = max(dist(3, iv), dist(6, iv)) = max(0.1513, 0.2216) = 0.2216, which is smaller than dist({iii, 6}, {ii, five}) = max(dist(three, 2), dist(vi, 2), dist(3, 5), dist(6, five)) = max(0.1483, 0.2540, 0.2843, 0.3921) = 0.3921 and dist({3, 6}, {i}) = max(dist(iii, 1), dist(6, 1)) = max(0.2218, 0.2347) = 0.2347.

Q21 Given, six points with the following attributes:

Which of the following clustering representations and dendrogram depicts the use of Group average proximity function in hierarchical clustering:

B.
C.

Solution: (C)

For the group average version of hierarchical clustering, the proximity of 2 clusters is defined to be the average of the pairwise proximities betwixt all pairs of points in the different clusters. This is an intermediate approach between MIN and MAX. This is expressed by the post-obit equation:

Here, the altitude betwixt some clusters. dist({iii, vi, 4}, {1}) = (0.2218 + 0.3688 + 0.2347)/(3 ∗ 1) = 0.2751. dist({2, five}, {1}) = (0.2357 + 0.3421)/(2 ∗ ane) = 0.2889. dist({3, six, 4}, {2, 5}) = (0.1483 + 0.2843 + 0.2540 + 0.3921 + 0.2042 + 0.2932)/(half-dozen∗1) = 0.2637. Because dist({3, six, 4}, {ii, 5}) is smaller than dist({3, 6, 4}, {1}) and dist({2, 5}, {1}), these two clusters are merged at the 4th stage

Q22. Given, six points with the following attributes:

Which of the following clustering representations and dendrogram depicts the use of Ward's method proximity function in hierarchical clustering:

Solution: (D)

Ward method is a centroid method. Centroid method calculates the proximity between two clusters by calculating the distance between the centroids of clusters. For Ward'south method, the proximity betwixt two clusters is divers as the increase in the squared mistake that results when ii clusters are merged. The results of applying Ward'southward method to the sample data set of six points. The resulting clustering is somewhat unlike from those produced by MIN, MAX, and grouping boilerplate.

Q23. What should be the all-time choice of no. of clusters based on the following results:

A. 1

B. 2

C. 3

D. four

Solution: (C)

The silhouette coefficient is a measure of how like an object is to its ain cluster compared to other clusters. Number of clusters for which silhouette coefficient is highest represents the best choice of the number of clusters.

Q24. Which of the post-obit is/are valid iterative strategy for treating missing values before clustering analysis?

A. Imputation with mean

B. Nearest Neighbor assignment

C. Imputation with Expectation Maximization algorithm

D. All of the above

Solution: (C)

All of the mentioned techniques are valid for treating missing values before clustering assay but only imputation with EM algorithm is iterative in its functioning.

Q25. K-Mean algorithm has some limitations. One of the limitation it has is, information technology makes hard assignments(A point either completely belongs to a cluster or not belongs at all) of points to clusters.

Notation: Soft assignment can exist consider every bit the probability of beingness assigned to each cluster: say Yard = 3 and for some point xn, p1 = 0.seven, p2 = 0.ii, p3 = 0.ane)

Which of the following algorithm(s) allows soft assignments?

Gaussian mixture models
Fuzzy One thousand-means

Options:

A. ane only

B. ii only

C. ane and 2

D. None of these

Solution: (C)

Both, Gaussian mixture models and Fuzzy K-means allows soft assignments.

Q26. Presume, y'all desire to cluster 7 observations into 3 clusters using Yard-Means clustering algorithm. Afterward offset iteration clusters, C1, C2, C3 has following observations:

C1: {(2,2), (4,iv), (6,half-dozen)}

C2: {(0,4), (iv,0)}

C3: {(5,5), (9,9)}

What will be the cluster centroids if you want to proceed for second iteration?

A. C1: (4,iv), C2: (2,2), C3: (vii,vii)

B. C1: (6,half dozen), C2: (iv,four), C3: (9,9)

C. C1: (2,2), C2: (0,0), C3: (five,5)

D. None of these

Solution: (A)

Finding centroid for information points in cluster C1 = ((2+iv+6)/three, (2+four+half dozen)/3) = (4, 4)

Finding centroid for data points in cluster C2 = ((0+4)/2, (4+0)/2) = (2, 2)

Finding centroid for data points in cluster C3 = ((5+ix)/2, (5+9)/2) = (7, 7)

Hence, C1: (4,four), C2: (ii,2), C3: (7,vii)

Q27. Assume, you want to cluster 7 observations into 3 clusters using G-Means clustering algorithm. Afterwards first iteration clusters, C1, C2, C3 has following observations:

C1: {(two,ii), (4,4), (6,6)}

C2: {(0,4), (4,0)}

C3: {(5,5), (nine,9)}

What will exist the Manhattan distance for observation (9, ix) from cluster centroid C1. In second iteration.

A. ten

B. 5*sqrt(two)

C. 13*sqrt(2)

D. None of these

Solution: (A)

Manhattan altitude between centroid C1 i.e. (four, 4) and (nine, 9) = (9-4) + (9-four) = 10

Q28. If ii variables V1 and V2, are used for clustering. Which of the following are true for K means clustering with k =iii?

If V1 and V2 has a correlation of 1, the cluster centroids will exist in a straight line
If V1 and V2 has a correlation of 0, the cluster centroids will be in straight line

Options:

A. ane only

B. two just

C. 1 and 2

D. None of the above

Solution: (A)

If the correlation between the variables V1 and V2 is 1, then all the data points volition exist in a direct line. Hence, all the three cluster centroids will class a directly line as well.

Q29. Characteristic scaling is an important step before applying One thousand-Mean algorithm. What is reason behind this?

A. In distance calculation it will give the same weights for all features

B. You always become the aforementioned clusters. If you use or don't utilise feature scaling

C. In Manhattan distance it is an of import step merely in Euclidian it is not

D. None of these

Solution; (A)

Feature scaling ensures that all the features get same weight in the clustering assay. Consider a scenario of clustering people based on their weights (in KG) with range 55-110 and height (in inches) with range 5.6 to 6.4. In this example, the clusters produced without scaling can be very misleading as the range of weight is much higher than that of height. Therefore, its necessary to bring them to aforementioned scale then that they have equal weightage on the clustering result.

Q30. Which of the following method is used for finding optimal of cluster in K-Mean algorithm?

A. Elbow method

B. Manhattan method

C. Ecludian mehthod

D. All of the higher up

E. None of these

Solution: (A)

Out of the given options, only elbow method is used for finding the optimal number of clusters. The elbow method looks at the percentage of variance explained every bit a office of the number of clusters: 1 should choose a number of clusters and then that adding another cluster doesn't give much better modeling of the data.

Q31. What is true almost 1000-Hateful Clustering?

Thousand-means is extremely sensitive to cluster center initializations
Bad initialization tin can lead to Poor convergence speed
Bad initialization tin can lead to bad overall clustering

Options:

A. 1 and 3

B. ane and two

C. 2 and iii

D. 1, 2 and three

Solution: (D)

All 3 of the given statements are true. Chiliad-means is extremely sensitive to cluster center initialization. Also, bad initialization tin can pb to Poor convergence speed every bit well every bit bad overall clustering.

Q32. Which of the following tin be applied to get expert results for Thou-means algorithm corresponding to global minima?

Try to run algorithm for dissimilar centroid initialization
Adjust number of iterations
Notice out the optimal number of clusters

Options:

A. ii and iii

B. i and 3

C. 1 and ii

D. All of above

Solution: (D)

All of these are standard practices that are used in gild to obtain good clustering results.

Q33. What should exist the best choice for number of clusters based on the following results:

A. 5

B. 6

C. 14

D. Greater than 14

Solution: (B)

Based on the higher up results, the best selection of number of clusters using elbow method is six.

Q34. What should be the best choice for number of clusters based on the following results:

A. two

B. 4

C. half-dozen

D. eight

Solution: (C)

Generally, a college average silhouette coefficient indicates ameliorate clustering quality. In this plot, the optimal clustering number of grid cells in the study surface area should be 2, at which the value of the boilerplate silhouette coefficient is highest. However, the SSE of this clustering solution (m = two) is also big. At k = vi, the SSE is much lower. In add-on, the value of the average silhouette coefficient at k = 6 is besides very high, which is simply lower than k = 2. Thus, the best option is m = 6.

Q35. Which of the following sequences is correct for a One thousand-Means algorithm using Forgy method of initialization?

Specify the number of clusters
Assign cluster centroids randomly
Assign each information indicate to the nearest cluster centroid
Re-assign each point to nearest cluster centroids
Re-compute cluster centroids

Options:

A. 1, two, 3, 5, 4

B. 1, iii, two, 4, 5

C. two, i, 3, 4, five

D. None of these

Solution: (A)

The methods used for initialization in K means are Forgy and Random Partition. The Forgy method randomly chooses k observations from the information ready and uses these as the initial means. The Random Sectionalization method showtime randomly assigns a cluster to each observation and so proceeds to the update stride, thus calculating the initial hateful to be the centroid of the cluster'southward randomly assigned points.

Q36. If yous are using Multinomial mixture models with the expectation-maximization algorithm for clustering a set of data points into two clusters, which of the assumptions are of import:

A. All the information points follow two Gaussian distribution

B. All the information points follow northward Gaussian distribution (n >ii)

C. All the data points follow two multinomial distribution

D. All the data points follow northward multinomial distribution (n >2)

Solution: (C)

In EM algorithm for clustering its essential to choose the same no. of clusters to classify the data points into as the no. of different distributions they are expected to be generated from and too the distributions must be of the same blazon.

Q37. Which of the following is/are non true almost Centroid based M-Means clustering algorithm and Distribution based expectation-maximization clustering algorithm:

Both starts with random initializations
Both are iterative algorithms
Both have strong assumptions that the data points must fulfill
Both are sensitive to outliers
Expectation maximization algorithm is a special case of Grand-Means
Both requires prior noesis of the no. of desired clusters
The results produced by both are non-reproducible.

Options:

A. 1 only

B. five only

C. ane and 3

D. 6 and 7

E. iv, vi and vii

F. None of the above

Solution: (B)

All of the above statements are truthful except the 5^th as instead K-Ways is a special case of EM algorithm in which just the centroids of the cluster distributions are calculated at each iteration.

Q38. Which of the following is/are non true about DBSCAN clustering algorithm:

For information points to be in a cluster, they must exist in a distance threshold to a core indicate
It has strong assumptions for the distribution of information points in dataspace
It has substantially high time complexity of order O(n³)
It does non require prior knowledge of the no. of desired clusters
It is robust to outliers

Options:

A. 1 only

B. ii only

C. 4 only

D. 2 and 3

E. 1 and 5

F. 1, 3 and 5

Solution: (D)

DBSCAN can form a cluster of any arbitrary shape and does not have strong assumptions for the distribution of data points in the dataspace.
DBSCAN has a low fourth dimension complexity of club O(n log northward) only.

Q39. Which of the following are the high and low premises for the existence of F-Score?

A. [0,1]

B. (0,1)

C. [-1,1]

D. None of the higher up

Solution: (A)

The lowest and highest possible values of F score are 0 and i with one representing that every data point is assigned to the correct cluster and 0 representing that the precession and/ or recollect of the clustering analysis are both 0. In clustering analysis, high value of F score is desired.

Q40. Post-obit are the results observed for clustering 6000 data points into three clusters: A, B and C:

What is the F₁-Score with respect to cluster B?

A. three

B. 4

C. 5

D. 6

Solution: (D)

Here,

True Positive, TP = 1200

True Negative, TN = 600 + 1600 = 2200

False Positive, FP = 1000 + 200 = 1200

False Negative, FN = 400 + 400 = 800

Therefore,

Precision = TP / (TP + FP) = 0.5

Recollect = TP / (TP + FN) = 0.6

Hence,

F_ane= 2 * (Precision * Recall)/ (Precision + recall) = 0.54 ~ 0.5

Terminate Notes

I hope yous enjoyed taking the test and institute the solutions helpful. The test focused on conceptual as well as practical noesis of clustering fundamentals and its various techniques.

I tried to clear all your doubts through this commodity, but if we have missed out on something so let us know in comments beneath. Besides, If you lot have whatsoever suggestions or improvements you think we should brand in the next skilltest, you lot tin can let the states know past dropping your feedback in the comments department.