![]() ![]() To apply GAE to the large-scale graph data is another research topic, so we don't report the result in the "Tencent" dataset. Bipartite graph code#N/A (*): For the GAE model, the code of the original GAE paper can not simply applied to the large-scale bipartite graph due to the memory constrain. Thereby return a false indicating the given graph is not bipartite otherwise, keep on returning true.Ĭonsider the following graph and its adjacency list.Ĭonsider the following illustration to understand the colouring of the nodes using DFS traversal.Only Linux (*): For the Node2Vec model, its binary file is only ELF 64-bit LSB executable, x86-64, for GNU/Linux. If at any moment, we get an adjacent node from the adjacency list which is already coloured and has the same colour as the current node, we can say it is not possible to colour it, hence it cannot be bipartite.For every uncoloured node, initialise it with the opposite colour to that of the current node. In DFS traversal, we travel in-depth to all its uncoloured neighbours using the adjacency list. ![]() We will start with the colour 0, you can start with 1 as well, just make sure for the adjacent node, it should be opposite of what the current node has. We will try to colour with 0 and 1, but you can choose other colours as well. In the DFS function call, make sure to pass the value of the assigned colour, and assign the same in the colour array.For DFS traversal, we need a start node and a visited array but in this case, instead of a visited array, we will take a colour array where all the nodes are initialised to -1 indicating they are not coloured yet.We will be defining the DFS traversal below, but this check has to be done for every component, for that we can use the simple for loop concept that we have learnt, to call the traversals for unvisited nodes. DFS goes in-depth, i.e., traverses all nodes by going ahead, and when there are no further nodes to traverse in the current path, then it backtracks on the same path and traverses other unvisited nodes. In this article, we will be solving it using DFS traversal.ĭFS is a traversal technique which involves the idea of recursion and backtracking. We can follow either of the traversal techniques. If at any moment of traversal, we find the adjacent nodes to have the same colour, it means that there is an odd cycle, or it cannot be a bipartite graph. The intuition is the brute force of filling colours using any traversal technique, just make sure no two adjacent nodes have the same colour. So, any graph with an odd cycle length can never be a bipartite graph. With a cycle, any graph with an even cycle length can also be a bipartite graph. Any linear graph with no cycle is always a bipartite graph. Intuition.Ī bipartite graph is a graph which can be coloured using 2 colours such that no adjacent nodes have the same colour. If we are able to colour a graph with two colours such that no adjacent nodes have the same colour, it is called a bipartite graph.ĭisclaimer: Don’t jump directly to the solution, try it out yourself first. Check whether the graph is bipartite or not. Problem Statement: Given an adjacency list of a graph adj of V no. ![]()
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