In your class, many students are friends. Let's assume that two students sharing a friend must be friends themselves; in other words, if students O and 1 are friends and students 1 and 2 are friends, then students 0 and 2 must be friends. Using this rule, we can partition the students into circles of friends. To do this, implement a function networks() that takes two input arguments. The first is the number n of students in the class. We assume students are identified using integers 0 through n 1. The second input argument is a list of tuple objects that define friends. For example, tuple (0; 2) defines students O and 2 as friends. Function networks() should print the partition of students into circles of friends as illustrated: >>> networks(5, [(0, 1), (1, 2), (3, 4)]) Social network O is (0, 1, 2) Social network 1 is (3, 4)

See explaination

Explanation:

#for finding root

def find_root(i,root):

while i!=root[i]:

root[i] = find_root(root[i],root)

i = root[i]

return i

#for joining two disjoint set

def union(i,j,root):

I = find_root(i,root)

J = find_root(j,root)

if I == J: #belongs to same set , so return

return

root[I] = J #make root of one equal to another

def networks(n,friend_list):

root = [i for i in range(n)] #initially all the friend are root and having no other friend

for (i,j) in friend_list: #by relationships union of disjoint sets

union(i,j,root)

social_networks = {} #to gather the friends under all the roots

for i in range(n):

I = find_root(i,root) #root of the ith friend

#add ith student to social group of root I

if I in social_networks:

social_networks[I].append(i)

else:

social_networks[I] = [i]

i = 0

for value in social_networks.values():

print('social network ',i,' is ',value)

i += 1

networks(5,[(0,1),(1,2),(3,4)])

the answer would be d

: )

i guess the correct option is letter a.

use contrasting font sizes, but the same font style