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Dijkstra's algorithm to find all the shortest paths possible

Asked 2010-05-12T13:45:15.613
23

I'm working on Dijkstra's algorithm, and I really need to find all the possible shortest paths, not just one. I'm using an adjacency matrix and I applied Dijkstra's algorithm, and I can find the shortest path. But I need to find all the paths with that minimum cost, I mean all the possible solutions, if they exist.

This is how my algorithm works, for a single solution:

public void dijkstra( int graph[][] )
{
    int d[] = new int[ graph.length ];
    int dC[] = new int[ graph.length ];
    int p[] = new int[ graph.length ];

    for( int i = 0; i < graph.length; i++ ){
        d[ i ] = 100; dC[ i ] = 100; p[ i ] = -1;
    }
    d[ 0 ] = 0; dC[ 0 ] = 0;

    int i = 0, min = 200, pos = 0; //You can change the min to 1000 to make it the largest number
    while( i < graph.length ){
        //extract minimum
        for( int j = 0; j < dC.length; j++ ){
            if( min > d[ j ] && dC[ j ] != -1 ){
                min = d[ j ]; pos = j;
            }
        }
        dC[ pos ] = -1;

        //relax
        for( int j = 0; j < graph.length; j++ ){
            if( d[ j ] > graph[ pos ][ j ] + d[ pos ] ){
                d[ j ] = graph[ pos ][ j ] + d[ pos ];
                p[ j ] = pos;
            }
        }
        i++; min = 200;
    }

    for( int j = 0; j < p.length; j++ ){
        System.out.print( p[ j ] + " " );
    }
    System.out.print( "\n" );
    for( int j = 0; j < d.length; j++ ){
        System.out.print( d[ j ] + " " );
    }
    System.out.print( "\n" );
}
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1 Answer

4

If your graphs allows edges with weight = 0 and also allows cycles, bear in mind that there are infinitely many shortest paths, and you cannot hope to output them all!

If there are either no zero-weight edges, or your graph is a DAG, then your solution is still problematic: it either doesn't produce all shortest paths as required, or it does so withO(2^N) space and time complexity. But then, perhaps there might be as many shortest paths, so you can not hope to do better in the general case.

This is the best source for a slightly more general problem: Finding the k Shortest Paths (Eppstein). You could apply this by iterating the shortest paths, stopping at the first path that is longer than the previous.

For the restricted problem of finding only the (equivalent) shortest paths, Anderson Imes' hint above (is this homework? Otherwise someone should spell it out) is good and much simpler than the above.

answered 2010-05-25T16:11:22.670

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