![6.29 In your class, many students are friends. Lets 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)](https://media.cheggcdn.com/media%2F5a5%2F5a539e61-a5b2-41cc-8efa-6ee9dc8a37b1%2Fimage.png)
6.29 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)
![Θ l Chegg.com d File Edit Selection View Go Debug Terminal Help program.py- Visual Studio Code Ccode.cpp Most Visited Getting StartedMost Visited # Well this question is best solved by disjoint set datastructure Chegg. Study 3 # Think of this as there are different sets which dont have any common element and # each set is having some root TEX # Now if two sets are made to have union operation , then just make root of one # equal to root of another. Like you have two disjoint sets a single set 6 tuple objects that define friends. For networks() should print the partition >>> networks(5, [(0, 1), (1, 2), (3,4)) Social network O is (0, 1, 2) Social network 1 is (3, 4) 7 # To find root you can use the root finding with path compression 9 1e 12 13 #for finding root 14 def find_root(i,root): 15 16 17 18 19 2e 21 #for joining two disjoint set 22 def union(i,j,root): 23 24 25 26 27 28 while i!-root[i]: Skip & Get next questio root[i]-find_root (root[i],root) -roo returni Your answer I-find root(i,root) ] - find_root(j,root) if I-J: #belongs to same set , so return return root[1] - ] #make root of one equal to another networks(5, [(e,1), (1,2),(3, PROBLEMS OUTPUT DEBUG CONSOLE TERMINAL 1: cmd (base) C:UserslLucky DesktopICPC>python program.py social network θ is [0, 1, 2] social network 1 is [3, 4] (base) C:UsersLucky Desktop ICPC](https://media.cheggcdn.com/media%2Ff0e%2Ff0e9596a-1b86-4e5a-abd4-1aa25b0ba8d9%2FphpdSt3KJ.png)
