Question $4($k$-$means, $40/100)$ Using the same terminology adopted in our course, we shall

refer to the $"$k$-$means" algorithm as the algorithm that initializes the centroids randomly,

followed by a "refinement phase" where the clusters are improved further. We shall refer to

$"$k$-$means$++"$ as the algorithm that only selects the centroids, so as to provide some

theoretical guarantees. Consider the following points in the $2D$ Euclidean space: $p1=(0,0),$

$p2=(0,1),p3=(0,2),p4=(2,0),p5=(3,0),p6=(4,1),p7=(5,0),p8=(7,0),p9=(8,0),p10=(8,1).$ Let

$k=3.$

a$)(10/100)$ Run the k$-$means$++$ algorithm to select the initial centroids, assuming that $1)$

p$4$ is selected as first centroid $2)$ the remaining two centroids are chosen assuming

that at each step the point with $3$rd largest probability is selected by k$-$means$++$

$($breaking ties arbitrarily$).$ In other words, let $q1,q2,q3,..,qn$ be the input points sorted

non$-$increasingly according to their probability of being selected at a given step of $k-$

means$++.$ Then, the point q$3$ is going to be selected at that step. Which centroids

have been selected at the end of this initialization step?

b$)(10/100)$ What is the probability that p$5$ is selected as centroid at step $3$ of $k-$

means$++?$

c$)(10/100)$ Run the refinement phase of the k$-$means algorithm until it terminates while

using the centroids selected in a$).$ What are the final clusters?

d$)(10/100)$ Consider the variant of the k$-$means algorithm, where $1)$ any given point $p$

can be moved from cluster $C$ with centroid $c$ to a cluster $C\text{'}$ with centroid c$\text{'}$ even if

$\{$:$d(p,c)=d(p,c^{\text{'}}),2)$ we stop as soon as we obtain the same clustering in two

consecutive iterations. Show that this variant of k$-$means might never terminate by

providing an example with at most $5$ points in the $1-$dimensional Euclidean space:Question $4($k$-$means, $40/100)$ Using the same terminology adopted in our course, we shall

refer to the $"$k$-$means" algorithm as the algorithm that initializes the centroids randomly,

followed by a "refinement phase" where the clusters are improved further. We shall refer to

$"$k$-$means$++"$ as the algorithm that only selects the centroids, so as to provide some

theoretical guarantees. Consider the following points in the $2D$ Euclidean space: $p1=(0,0),$

$p2=(0,1),p3=(0,2),p4=(2,0),p5=(3,0),p6=(4,1),p7=(5,0),p8=(7,0),p9=(8,0),p10=(8,1).$ Let

$k=3.$

a$)(10/100)$ Run the k$-$means$++$ algorithm to select the initial centroids, assuming that $1)$

p$4$ is selected as first centroid $2)$ the remaining two centroids are chosen assuming

that at each step the point with $3$rd largest probability is selected by k$-$means$++$

$($breaking ties arbitrarily$).$ In other words, let $q1,q2,q3,..,qn$ be the input points sorted

non$-$increasingly according to their probability of being selected at a given step of $k-$

means$++.$ Then, the point q$3$ is going to be selected at that step. Which centroids

have been selected at the end of this initialization step?

b$)(10/100)$ What is the probability that p$5$ is selected as centroid at step $3$ of $k-$

means$++?$

c$)(10/100)$ Run the refinement phase of the k$-$means algorithm until it terminates while

using the centroids selected in a$).$ What are the final clusters?

d$)(10/100)$ Consider the variant of the k$-$means algorithm, where $1)$ any given point $p$

can be moved from cluster $C$ with centroid $c$ to a cluster $C\text{'}$ with centroid c$\text{'}$ even if

$\{$:$d(p,c)=d(p,c^{\text{'}}),2)$ we stop as soon as we obtain the same clustering in two

consecutive iterations. Show that this variant of k$-$means might never terminate by

providing an example with at most $5$ points in the $1-$dimensional Euclidean space: