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Check if a Given Graph is Bipartite Using DFS in C++
Suppose we have a connected graph; we have to check whether the graph is bipartite or not. If the graph coloring is possible applying two colors such that nodes in a set are colored with the same color.
So, if the input is like
then the output will be True
To solve this, we will follow these steps −
- Define a function insert_edge(), this will take an edge array adj, u, v,
- insert v at the end of adj[u]
- insert u at the end of adj[v]
- From the main method do the following,
- for each u in adj[v],do
- if visited[u] is same as false, then −
- visited[u] := true
- color[u] := invert of color[v]
- if not is_bipartite_graph(adj, u, visited, color), then −
- return false
- otherwise when color[u] is same as color[v], then −
- return false
- if visited[u] is same as false, then −
- return true
Example (C++)
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h> using namespace std; void insert_edge(vector<int> adj[], int u, int v){ adj[u].push_back(v); adj[v].push_back(u); } bool is_bipartite_graph(vector<int> adj[], int v, vector<bool>& visited, vector<int>& color){ for (int u : adj[v]) { if (visited[u] == false) { visited[u] = true; color[u] = !color[v]; if (!is_bipartite_graph(adj, u, visited, color)) return false; } else if (color[u] == color[v]) return false; } return true; } int main() { int N = 6; vector<int> adj_list[N + 1]; vector<bool> visited(N + 1); vector<int> color(N + 1); insert_edge(adj_list, 1, 2); insert_edge(adj_list, 2, 3); insert_edge(adj_list, 3, 4); insert_edge(adj_list, 4, 5); insert_edge(adj_list, 5, 6); insert_edge(adj_list, 6, 1); visited[1] = true; color[1] = 0; cout << (is_bipartite_graph(adj_list, 1, visited, color)); }
Input
insert_edge(adj_list, 1, 2); insert_edge(adj_list, 2, 3); insert_edge(adj_list, 3, 4); insert_edge(adj_list, 4, 5); insert_edge(adj_list, 5, 6); insert_edge(adj_list, 6, 1);
Output
1
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