Check if there is a cycle of length 3 in the graph that satisfies the given condition

Checks the graph for a loop of length 3 that satisfies the given criteria, preparing to repeatedly traverse each vertex and look at its neighboring vertices. If a vertex has two neighbors that are too connected, a cycle of length 3 exists. This condition guarantees that there is an edge between two neighbors, thus forming a triangle. By filtering all vertices and their neighboring vertices, we will identify whether such a cycle exists. If we find that a vertex has two related neighbors, we can conclude that the graph shows a cycle of length 3 that satisfies the given condition.

usage instructions

• Adjacency matrix method

• Adjacency list method

Adjacency method

To check if there is a cycle of length 3 in the graph that satisfies a given condition, we can make use of the contagious method. In this approach, we iterate for each vertex in the graph and check its neighboring vertices. For each vertex, we check whether any two of its neighboring vertices are too closely related. If such a match is found, we check whether the conditions for that match are met. If the condition is met, it indicates a loop of length 3 that is close to satisfying the given condition. By looking at all the vertices in the graph, we can determine if such a cycle exists.

algorithm

• Initialize the Boolean variable named "cycleExists" to false.

• Iterate over each vertex in the graph:

  • For each vertex, repeat its adjacent vertices.

  • For each adjacent vertex, emphasize its adjacent vertices (except the current vertex).

  • If any two adjacent vertices are related, continue to the next step.

• Check whether the combination of associated vertices found in step 2c satisfies the condition.

  • If the condition is met, set "cycleExists" to true and break out of the loop.

• After completing the cycle, check the value of "cycleExists".

  • If "cycleExists" is true, then there is a cycle of length 3 in the graph that satisfies the given condition.

  • If "cycleExists" is wrong, no such cycle exists.

• Output results.

• This calculation repeats the vertices of the graph, analyzes their adjacent vertices, and checks whether any matches of adjacent vertices form a cycle of length 3 that satisfies the given condition.

  李>

Example

#include <iostream>
#include <vector>

using namespace std;

bool checkCycle(vector<vector<int>>& graph, int v, vector<bool>& visited, int parent, int condition) {
    visited[v] = true;

    for (int u : graph[v]) {
        if (!visited[u]) {
            visited[u] = true;

            for (int w : graph[u]) {
                if (visited[w] && w != parent && condition == graph[v][u] + graph[u][w]) {
                    return true;
                }
            }

            visited[u] = false;
        }
    }

    return false;
}

bool hasCycleOfLength3(vector<vector<int>>& graph, int condition) {
    int numVertices = graph.size();
    vector<bool> visited(numVertices, false);

    for (int v = 0; v < numVertices; v++) {
        visited[v] = true;

        for (int u : graph[v]) {
            if (checkCycle(graph, u, visited, v, condition)) {
                return true;
            }
        }

        visited[v] = false;
    }

    return false;
}

int main() {
    int numVertices, numEdges;
    cout << "Enter the number of vertices and edges: ";
    cin >> numVertices >> numEdges;

    vector<vector<int>> graph(numVertices);

    cout << "Enter the connections between vertices (u, v) and their corresponding weights: " << endl;
    for (int i = 0; i < numEdges; i++) {
        int u, v, weight;
        cin >> u >> v >> weight;

        graph[u].push_back(v);
        graph[v].push_back(u);

        // Store the weight/condition between u and v
        graph[u][v] = weight;
        graph[v][u] = weight;
    }

    int condition;
    cout << "Enter the condition to be satisfied: ";
    cin >> condition;

    if (hasCycleOfLength3(graph, condition)) {
        cout << "Cycle of length 3 satisfying the condition exists." << endl;
    } else {
        cout << "Cycle of length 3 satisfying the condition does not exist." << endl;
    }

    return 0;
}

Output

Enter the number of vertices and edges:  

Adjacency list method

Adjacent list methods can be information structures used to talk to the diagram. In this approach, each vertex of the graph is associated with a list containing all its adjacent vertices. To check if there is a cycle of length 3 in the graph that satisfies the given condition, we will iterate over each vertex and its neighboring vertices. For each adjacent vertex, we check if it contains an adjacent vertex in common with the current vertex. If such a common vertex exists, then a ring of length 3 is found. This approach guarantees efficient investigation of the graph by storing essential data about almost all vertices in the infectious list and their associations.

algorithm

• Make an infectious list that talks to the graph, where each vertex contains a list of its neighboring vertices.

• Iterate over each vertex in the graph.

• For each vertex, repeat its adjacent vertices.

• For each adjacent vertex, emphasize its adjacent vertices (except the current vertex).

• Check whether there is a common vertex between the current vertex and the adjacent vertex of the adjacent vertex.

• If a common vertex is found, a cycle of length 3 exists. Return true.

• If no ring of length 3 is found, return false.

Example

#include <iostream>
#include <vector>
#include <unordered_set>

using namespace std;

bool hasCycleOfLength3(vector<vector<int>>& graph) {
    int n = graph.size();
    
    for (int u = 0; u < n; ++u) {
        unordered_set<int> adjSet(graph[u].begin(), graph[u].end());

        for (int v : graph[u]) {
            for (int w : graph[v]) {
                if (w != u && adjSet.count(w) > 0) {
                    return true; // Cycle of length 3 found
                }
            }
        }
    }

    return false; // No cycle of length 3 found
}

int main() {
    // Create the graph as an adjacency list
    vector<vector<int>> graph = {
        {1, 2},
        {0, 2},
        {0, 1, 3},
        {2, 4},
        {3}
    };

    // Check if a cycle of length 3 exists
    bool cycleExists = hasCycleOfLength3(graph);

    // Print the result
    if (cycleExists) {
        cout << "A cycle of length 3 exists in the graph." << endl;
    } else {
        cout << "No cycle of length 3 exists in the graph." << endl;
    }

    return 0;
}

Output

A cycle of length 3 exists in the graph.

in conclusion

This article examines methods for checking whether there is a loop of length 3 in a graph that satisfies a given condition. It illustrates two approaches, specifically the contagious frame approach and the contagious list approach. This article traces the calculation process and gives bits of C code for both methods. The contagious network approach involves emphasizing each vertex and its neighboring vertices to identify cycles of length 3 that satisfy the condition. The contagious list method exploits information structures that talk to the graph and examines common vertices between adjacent vertices to determine the proximity of cycles.

The above is the detailed content of Check if there is a cycle of length 3 in the graph that satisfies the given condition. For more information, please follow other related articles on the PHP Chinese website!