BFS Shortest Path: Edge weight either 1 or 2

Question

I am trying to implement a shortest path algorithm using BFS. That is I am trying to find the shortest path from a specified vertex to every other vertex. However, its a special case where all edge weights are either 1 or 2. I know it could be done with Dijkstra's algorithm but I must use Breadth First Search.

So far I have a working version of BFS that searches first for a vertex connected with an edge of weight 1. If it cannot find it, then returns a vertex connected with an edge of weight 2. After thinking about it, this is not the correct way to find the shortest path. The problem is I cannot think of any reasoning why BFS would work with weights 1 or 2, as opposed to any weight.

Here is the code:

public void addEdge(int start, int end, int weight)
  {
  adjMat[start][end] = 1;
  adjMat[end][start] = 1;
  edge_weight[start][end] = weight; 
  edge_weight[end][start] = weight; 
  }

// -------------------------------------------------------------
public void bfs()                   // breadth-first search
  {                                // begin at vertex 0
  vertexList[0].wasVisited = true; // mark it
  displayVertex(0);                // display it
  theQueue.insert(0);              // insert at tail
  int v2;

  while( !theQueue.isEmpty() )     // until queue empty,
     {
     int v1 = theQueue.remove();   // remove vertex at head
     // until it has no unvisited neighbors
     while( (v2=getAdjUnvisitedVertex(v1)) != -1 ){// get one,
        vertexList[v2].wasVisited = true;  // mark it
        displayVertex(v2);                 // display it
        theQueue.insert(v2);               // insert it
        }
     }  // end while(queue not empty)

  // queue is empty, so we're done
  for(int j=0; j<nVerts; j++)             // reset flags
     vertexList[j].wasVisited = false;
  }  // end bfs()
// -------------------------------------------------------------
// returns an unvisited vertex adj to v -- ****WITH WEIGHT 1****
public int getAdjUnvisitedVertex(int v) {
    for (int j = 0; j < nVerts; j++)
        if (adjMat[v][j] == 1 && vertexList[j].wasVisited == false && edge_weight[v][j] == 1){
            //System.out.println("Vertex found with 1:"+ vertexList[j].label);
            return j;
        }
    for (int k = 0; k < nVerts; k++)
        if (adjMat[v][k] == 1 && vertexList[k].wasVisited == false && edge_weight[v][k] == 2){
            //System.out.println("Vertex found with 2:"+vertexList[k].label);
            return k;
        }
    return -1;
}  // end getAdjUnvisitedVertex()
   // -------------------------------------------------------------
}  
////////////////////////////////////////////////////////////////
public class BFS{
public static void main(String[] args)
  {
  Graph theGraph = new Graph();
  theGraph.addVertex('A');    // 0  (start for bfs)
  theGraph.addVertex('B');    // 1
  theGraph.addVertex('C');    // 2

  theGraph.addEdge(0, 1,2);     // AB
  theGraph.addEdge(1, 2,1);     // BC
  theGraph.addEdge(2, 0,1);     // AD


  System.out.print("Visits: ");
  theGraph.bfs();             // breadth-first search
  System.out.println();
  }  // end main()
   }

The problem then is, that I don't know why BFS can work for the shortest path problem with edges of weight 1 or 2 as opposed to any edges of any weight.

Any help is appreciated. Thanks!

Edit: Here is the problem: You wish to find the shortest path from s to the rest of the vertices. This could be done using Dijkstra's algorithm, but you need to use a breadth-first search strategy. You should be able to do this because any edge weight is either 1 or 2. Describe the changes to BFS that will solve your problem

Answer 1

The significance of the edge weight limit to BFS is that it ensures that if there is an edge from A to B then there is no shorter path from A to B. The maximum weight of the AB edge is 2. The minimum total weight of a path from A to C to B is also 2.

Answer 2

I am trying to implement a shortest path algorithm using BFS. That is I am trying to find the shortest path from a specified vertex to every other vertex. However, its a special case where all edge weights are either 1 or 2. I know it could be done with Dijkstra's algorithm but I must use Breadth First Search.

So far I have a working version of BFS that searches first for a vertex connected with an edge of weight 1. If it cannot find it, then returns a vertex connected with an edge of weight 2. After thinking about it, this is not the correct way to find the shortest path. The problem is I cannot think of any reasoning why BFS would work with weights 1 or 2, as opposed to any weight.

Here is the code:

public void addEdge(int start, int end, int weight)
  {
  adjMat[start][end] = 1;
  adjMat[end][start] = 1;
  edge_weight[start][end] = weight; 
  edge_weight[end][start] = weight; 
  }

// -------------------------------------------------------------
public void bfs()                   // breadth-first search
  {                                // begin at vertex 0
  vertexList[0].wasVisited = true; // mark it
  displayVertex(0);                // display it
  theQueue.insert(0);              // insert at tail
  int v2;

  while( !theQueue.isEmpty() )     // until queue empty,
     {
     int v1 = theQueue.remove();   // remove vertex at head
     // until it has no unvisited neighbors
     while( (v2=getAdjUnvisitedVertex(v1)) != -1 ){// get one,
        vertexList[v2].wasVisited = true;  // mark it
        displayVertex(v2);                 // display it
        theQueue.insert(v2);               // insert it
        }
     }  // end while(queue not empty)

  // queue is empty, so we're done
  for(int j=0; j<nVerts; j++)             // reset flags
     vertexList[j].wasVisited = false;
  }  // end bfs()
// -------------------------------------------------------------
// returns an unvisited vertex adj to v -- ****WITH WEIGHT 1****
public int getAdjUnvisitedVertex(int v) {
    for (int j = 0; j < nVerts; j++)
        if (adjMat[v][j] == 1 && vertexList[j].wasVisited == false && edge_weight[v][j] == 1){
            //System.out.println("Vertex found with 1:"+ vertexList[j].label);
            return j;
        }
    for (int k = 0; k < nVerts; k++)
        if (adjMat[v][k] == 1 && vertexList[k].wasVisited == false && edge_weight[v][k] == 2){
            //System.out.println("Vertex found with 2:"+vertexList[k].label);
            return k;
        }
    return -1;
}  // end getAdjUnvisitedVertex()
   // -------------------------------------------------------------
}  
////////////////////////////////////////////////////////////////
public class BFS{
public static void main(String[] args)
  {
  Graph theGraph = new Graph();
  theGraph.addVertex('A');    // 0  (start for bfs)
  theGraph.addVertex('B');    // 1
  theGraph.addVertex('C');    // 2

  theGraph.addEdge(0, 1,2);     // AB
  theGraph.addEdge(1, 2,1);     // BC
  theGraph.addEdge(2, 0,1);     // AD


  System.out.print("Visits: ");
  theGraph.bfs();             // breadth-first search
  System.out.println();
  }  // end main()
   }

The problem then is, that I don't know why BFS can work for the shortest path problem with edges of weight 1 or 2 as opposed to any edges of any weight.

Any help is appreciated. Thanks!

Edit: Here is the problem: You wish to find the shortest path from s to the rest of the vertices. This could be done using Dijkstra's algorithm, but you need to use a breadth-first search strategy. You should be able to do this because any edge weight is either 1 or 2. Describe the changes to BFS that will solve your problem