(b) Source CodeWrite solution in Phython Code should be well written and commented. Problem 1 Is Graph Bipartite?There is an undirected graph with n nodes, where each node is numbered between 0 and n- 1.You are given a 2D array graph, where graphlu] is an array of nodes that node u is adjacent to.More formally, for each v in graphlu), there is an undirected edge between node u and node v.The graph has the following properties:• There are no self-edges (graphlu] does not contain u).There are no parallel edges (graphlu] does not contain duplicate values).• If v is in graph[u), then u is in graph[u} (the graph is undirected).• The graph may not be connected, meaning there may be two nodes u and v such that thereis no path between them.A graph is bipartite if the nodes can be partitioned into two disjoint sets A and B such thatevery edge in the graph connects a node in set A and a node in set B.There are other equivalent definitions of a bipartite graph. A graph is bipartite if and only if it hasno odd length cycle, i.e., no cycle in the graph has an odd mumber of edges. Note that if a graph has nocycles it is bipartite (hence trees are bipartite graphs).It is an interesting exercise for you to show that the above two definitions are equivalent (highlyrecommended if you want to get a taste of graph theory). But that is not the goal of this exercise.Using the odd cycle definition one can use DFS or BFS to elegantly check whether a graph is bipartitein linear time, ie., O(m + n) time. Such solutions are discussed in leetcode (note that some of these maynot be fully correct). This is also not the goal of this exercise.accepted and will receive O points.The goal of this exercise is to explore yet another approach to check bipartiteness and is based on thefollowing ides. You have to argue that the idea is correct (as mentioned below) and implement the ideaas an algorithm that checks for bipartiteness of a graph.Let G = (V,E) be the input graph that you want to check for bipartiteness. We first transform Ginto a new graph G' = (V', E') as follows. For each esch node v e V, construct two nodes v1, v2 € V'and for each edge (u, v) € E, create two edges (u1, 12) and (u2, v1) in E'. This completes the descriptionof constructing G' from G. See Figure 1 for an illustration.The idea of this construction is that we reduce the problem of checking bipartiteness of G to checkingthe number of connected components of G as stated in the following theorem.you give such a solution, this will not beTheorem 1Let C be the mumber of connected components in G. Then the mumber of connected componentsin G' is 20 if and only if the graph G is bipartite.Your first goal is to give a proof of the above theorem (soe the submissions instruetions below).To get an idea of why the above theorem might be true, see the two examples given in Figures 1and 2. The graph in Example 1 is bipartite (no cycle) and the number of connected components is 1; it is2 in G'. The graph in Example 2 is not bipartite (odd cycle of length 3) and the number of connectedcomponents is 1; it is 1 also in G'.Your second goal is to use the theorem above to give an efficient algorithm for checking bipartiteness. bdadiFigure 1: Example 1b2daFigure 2: Example 2

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(b) Source CodeWrite solution in Phython Code should be well written and commented. Problem 1 Is Graph Bipartite?There is an undirected graph with n nodes, where each node is numbered between 0 and n- 1.You are given a 2D array graph, where graphlu] is an array of nodes that node u is adjacent to.More formally, for each v in graphlu), there is an undirected edge between node u and node v.The graph has the following properties:• There are no self-edges (graphlu] does not contain u).There are no parallel edges (graphlu] does not contain duplicate values).• If v is in graph[u), then u is in graph[u} (the graph is undirected).• The graph may not be connected, meaning there may be two nodes u and v such that thereis no path between them.A graph is bipartite if the nodes can be partitioned into two disjoint sets A and B such thatevery edge in the graph connects a node in set A and a node in set B.There are other equivalent definitions of a bipartite graph. A graph is bipartite if and only if it hasno odd length cycle, i.e., no cycle in the graph has an odd mumber of edges. Note that if a graph has nocycles it is bipartite (hence trees are bipartite graphs).It is an interesting exercise for you to show that the above two definitions are equivalent (highlyrecommended if you want to get a taste of graph theory). But that is not the goal of this exercise.Using the odd cycle definition one can use DFS or BFS to elegantly check whether a graph is bipartitein linear time, ie., O(m + n) time. Such solutions are discussed in leetcode (note that some of these maynot be fully correct). This is also not the goal of this exercise.accepted and will receive O points.The goal of this exercise is to explore yet another approach to check bipartiteness and is based on thefollowing ides. You have to argue that the idea is correct (as mentioned below) and implement the ideaas an algorithm that checks for bipartiteness of a graph.Let G = (V,E) be the input graph that you want to check for bipartiteness. We first transform Ginto a new graph G' = (V', E') as follows. For each esch node v e V, construct two nodes v1, v2 € V'and for each edge (u, v) € E, create two edges (u1, 12) and (u2, v1) in E'. This completes the descriptionof constructing G' from G. See Figure 1 for an illustration.The idea of this construction is that we reduce the problem of checking bipartiteness of G to checkingthe number of connected components of G as stated in the following theorem.you give such a solution, this will not beTheorem 1Let C be the mumber of connected components in G. Then the mumber of connected componentsin G' is 20 if and only if the graph G is bipartite.Your first goal is to give a proof of the above theorem (soe the submissions instruetions below).To get an idea of why the above theorem might be true, see the two examples given in Figures 1and 2. The graph in Example 1 is bipartite (no cycle) and the number of connected components is 1; it is2 in G'. The graph in Example 2 is not bipartite (odd cycle of length 3) and the number of connectedcomponents is 1; it is 1 also in G'.Your second goal is to use the theorem above to give an efficient algorithm for checking bipartiteness. bdadiFigure 1: Example 1b2daFigure 2: Example 2

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