8Puzzle game

[Abdulrahman Mostafa]

I have been trying to create a puzzle that can solve itself using BFS algorithm. Currently, the search algorithm and the main but I don't know how to integrate both of them together. my BFS:

#ifndef BFS_h
#define BFS_h
// Program to print BFS traversal from a given source vertex. BFS(int s)
// traverses vertices reachable from s.
#include<iostream>
#include <list>

using namespace std;

// This class represents a directed graph using adjacency list representation
class Graph
{
int V; // No. of vertices
list<int> *adj; // Pointer to an array containing adjacency lists
public:
Graph(int V); // Constructor
void addEdge(int v, int w); // function to add an edge to graph
void BFS(int s); // prints BFS traversal from a given source s
};

Graph::Graph(int V)
{
this->V = V;
adj = new list<int>[V];
}

void Graph::addEdge(int v, int w)
{
adj[v].push_back(w); // Add w to v’s list.
}

void Graph::BFS(int s)
{
// Mark all the vertices as not visited
bool *visited = new bool[V];
for(int i = 0; i < V; i++)
visited[i] = false;

// Create a queue for BFS
list<int> queue;

// Mark the current node as visited and enqueue it
visited\[s\] = true;
queue.push\_back(s);

// 'i' will be used to get all adjacent vertices of a vertex
list<int>::iterator i;

while(!queue.empty())
{
    // Dequeue a vertex from queue and print it
    s = queue.front();
    cout << s << " ";
    queue.pop\_front();
    
    // Get all adjacent vertices of the dequeued vertex s
    // If a adjacent has not been visited, then mark it visited
    // and enqueue it
    for(i = adj\[s\].begin(); i != adj\[s\].end(); ++i)
    {
        if(!visited\[\*i\])
        {
            visited\[\*i\] = true;
            queue.push\_back(\*i);
        }
    }
}

}

and my main:

#include "queue"
#include <iostream>
#include <iomanip>
#include "BFS.h"
#include "A*search.h"

using namespace std;

int main()
{
int s;
int box1=3,box2=6,box3=9,box4=1,box5=s,box6=7,box7=5,box8=8,box9=2;

cout << " -----------" << endl;
cout << " | " << box1 << " | " << box2 << " | " << box3 <<

[Lost User]

You need to create a Graph object, with one of its constructors, and then call the various methods to construct and solve the puzzle:

int main()
{
int s;
int box1=3,box2=6,box3=9,box4=1,box5=s,box6=7,box7=5,box8=8,box9=2;

cout << " -----------" << endl;
cout << " | " << box1 << " | " << box2 << " | " << box3 << " | " << endl;
cout << " -----------" << endl;
cout << " | " << box4 << " | " << box5 << " | " << box6 << " | " << endl;
cout << " -----------" << endl;
cout << " | " << box7 << " | " << box8 << " | " << box9 << " | " << endl;
cout << " -----------" << endl;

// void BFS(int s);
Graph graph = new Graph(V); // where does the value of V come from?
graph.addEdge(v, w); // what are these values?
graph.BFS(int s); // where does the value of s come from?

return 0;

}

[David Crow]

So did someone give you these two pieces of code?

"One man's wage rise is another man's price increase." - Harold Wilson

"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons

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