Union and Intersection of two Graphs
Given two graphs G1 and G2, the task is to find the union and intersection of the two given graphs, i.e. (G1 ∪ G2) and (G1 ∩ G2).
Examples:
Input: G1 = { (“e1”, 1, 2), (“e2”, 1, 3), (“e3”, 3, 4), (“e4”, 2, 4) }, G2 = = { (“e4”, 2, 4), (“e5”, 2, 5), (“e6”, 4, 5) }
Output:
G1 union G2 is
e1 1 2
e2 1 3
e3 3 4
e4 2 4
e5 2 5
e6 4 5G1 intersection G2 is
e4 2 4
Explanation:
Union of the graphs G1 and G2:Intersection of the graphs G1 and G2:
Approach: Follow the steps below to solve the problem:
- Define a function, say Union(G1, G2), to find the union of the G1 and G2:
- Initialize a map, say added, that stores if an edge is already been added or not.
- Iterate over the edges of the graph G1 and push all the edges in a graph, say G, and mark all the edges visited in added.
- Now, again traverse over the edges of the graph G2 and push the edge in the G if the edge is not already been added, and then mark the edge added in the map added.
- Define a function say Intersection(G1, G2) to find the Intersection of the G1 and G2:
- Initialize a map, say added , that stores if an edge is already been added or not.
- Traverse over the edges of the graph G1 and marked all the edges visited in the map added.
- Now, again traverse over the edges of the graph G2 and push the edge in the graph G, if the edge is already been added. Then, mark the edge added in the map.
- Now, print the graphs obtained after the function call of Union(G1, G2) and Intersection(G1, G2).
Below is the implementation of the above approach:
C++14
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find union of two graphs void find_union( vector<tuple<string, int , int > > G1, vector<tuple<string, int , int > > G2) { // Stores an edge of the graph G1 map<string, pair< int , int > > added; // Stores the unioun graph G1 vector<tuple<string, int , int > > G; // Iterate over the edges // of the graph G1 for ( auto p : G1) { string a = get<0>(p); // Get the edges int b = get<1>(p); int c = get<2>(p); // Insert the current // edges into graph G G.push_back( make_tuple(a, b, c)); added[a] = { b, c }; } // Iterate over the edges // of the graph G1 for ( auto p : G2) { string a = get<0>(p); int b = get<1>(p); int c = get<2>(p); pair< int , int > x = { b, c }; pair< int , int > y = { c, b }; // If either edge x or // y is already added if (added[a] == x || added[a] == y) continue ; // Otherwise G.push_back(make_tuple(a, b, c)); } // Print the unioun cout << "G1 union G2 is\n" ; for ( auto p : G) { string a = get<0>(p); int b = get<1>(p); int c = get<2>(p); cout << a << " " << b << " " << c << endl; } } // Function to find intersection of two graphs void find_intersection( vector<tuple<string, int , int > > G1, vector<tuple<string, int , int > > G2) { // Stores an edge map<string, pair< int , int > > added; // Stores the graph of intersection vector<tuple<string, int , int > > G; // Iterate over edges of graph G1 for ( auto p : G1) { string a = get<0>(p); int b = get<1>(p); int c = get<2>(p); added[a] = { b, c }; } // Iterate over edges of graph G2 for ( auto p : G2) { string a = get<0>(p); int b = get<1>(p); int c = get<2>(p); pair< int , int > x = { b, c }; pair< int , int > y = { c, b }; // If either edge x or // y is already added if (added[a] == x || added[a] == y) G.push_back(make_tuple(a, b, c)); } // Print the graph G cout << "G1 intersection G2 is\n" ; for ( auto p : G) { string a = get<0>(p); int b = get<1>(p); int c = get<2>(p); cout << a << " " << b << " " << c << endl; } } // Driver Code int main() { vector<tuple<string, int , int > > G1 = { make_tuple( "e1" , 1, 2), make_tuple( "e2" , 1, 3), make_tuple( "e3" , 3, 4), make_tuple( "e4" , 2, 4) }; vector<tuple<string, int , int > > G2 = { make_tuple( "e4" , 2, 4), make_tuple( "e5" , 2, 5), make_tuple( "e6" , 4, 5) }; // Function call for finding the // Union of the given graph find_union(G1, G2); // Function call for finding the // Intersection of the given graph find_intersection(G1, G2); return 0; } |
G1 union G2 is e1 1 2 e2 1 3 e3 3 4 e4 2 4 e5 2 5 e6 4 5 G1 intersection G2 is e4 2 4
Time Complexity: O(N * log(N))
Auxiliary Space: O(N)