GATE CSE 2002 | Question: 12

Question

Fill in the blanks in the following template of an algorithm to compute all pairs shortest path lengths in a directed graph $G$ with $n*n$ adjacency matrix $A$. $A[i,j]$ equals $1$ if there is an edge in $G$ from $i$ to $j$, and $0$ otherwise. Your aim in filling in the blanks is to ensure that the algorithm is correct.

INITIALIZATION: For i = 1 ... n
    {For j = 1 ... n
        { if a[i,j] = 0 then P[i,j] =_______ else P[i,j] =_______;}
    }
    
ALGORITHM: For i = 1 ... n
    {For j = 1 ... n
        {For k = 1 ... n
            {P[__,__] = min{_______,______}; }
        }
    }    

1. Copy the complete line containing the blanks in the Initialization step and fill in the blanks.
2. Copy the complete line containing the blanks in the Algorithm step and fill in the blanks.
3. Fill in the blank: The running time of the Algorithm is $O$(___).

Comments

here we are not given any information about the weights between the edges, then how can we solve it. The only way I think it can be solved is when A[i,j]= Weight[i,j] if there is an edge in G from i to j.

PLEASE HELP !!!

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@Bikram

sir need to  edit answer 

P[  j , k ] = min(  p[ j , k ] ,  p[ j , i ] +   p[  i , k ] ) }

Answer 1

INITIALIZATION: 
For i = 1 ... n {
   For j = 1 ... n { 
      if a[i,j] = 0 then P[i,j] = infinite  
        // i.e. if there is no direct path then put path length as infinite
       else P[i,j] =a[i,j];
   }
}

ALGORITHM: 
For i = 1 ... n { //i loops over the intermediate vertices
   For j = 1 ... n {
      For k = 1 ... n {
               P[j, k] = min( p[j,k] , p[j,i] + p[i,k]);
      }
   }
}              

Time complexity $O(n^3)$

This algorithm is for weighted graph but it will work for unweighted graph too because if $p[i,j]=1$, $p[i,k]=1$ and $p[k,j]=1$  then according to the algorithm $p[i,j] = \min(p[i,j] ,p[i,k] + p[k,j])  = \min(1,2) =1$

And all the other cases are also satisfied. $($like if $p[i,j]$ was $0$ in last iteration and there exist a path via $k)$

Comments

yes @Himanshu1 is correct here. First create an edge with inner most two loops and then checking an external point with outer most loop. discuss

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good explanation

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Some extra points:-

1. In order to convert an undirected graph to a directed graph, just draw the arrows on both sides, i.e, put $a[i][j]$ as well as $a[j][i]$ the same value. 
2. Inorder to detect negative cycle, we can put the condition as:

for(i=0; i<n; i++){
    if(P[i][i] < 0){
        printf("Negative Cycle detected"); // Because, P[i][i] 
         // must be zero, but due to the presence of negative cycle, 
         //the value became negative.
    }
}

Answer 2

I'll try to answer in order to provide maximum clarity.

INITIALIZATION: For i = 1 ... n
    {For j = 1 ... n
        { if a[i,j] = 0 then P[i,j] =_______ else P[i,j] =_______;}
    }

We, intialise our final distance matrix initially with the Adjacency matrix of the graph. If we have a path from vertex i to j, then we define it's cost, otherwise we set it to infinite.

so

if a[i,j] = 0 (There is no direct path from vertex i to j)  then P[i,j] =0 else P[i,j]=a[i][j] (Path cost to reach vertex j from i);

Now comes the second part

ALGORITHM: For i = 1 ... n //Consider each vertex to be an intermediate vertex.
    {For j = 1 ... n
        {For k = 1 ... n
            {
          for Each pair of vertices (j,k), find the minimum path to reach k from j and this minimum path has 2 ways
   (1)Either directly go from j to k or
   (2)Using i as an intermediate vertex, go to vertex i from j and then go to k from i.
  Find minimum of above two.
   The ith loop considers for each path, the intermediate vertex that can give minimum cost path to reach k from j.
        P[j,k] = min{P[j][k],p[j][i]+p[i][k]}; }
        }
    }    

Running time of our algorithm is $O(V^3)$ where V is the number of the vertices in the graph.

Comments

@Ayush Upadhyaya

As it is given 0 or 1 we donot consider the edge weights here so finally it boils down to a boolean  variable having 0 or 1

so  can this

P[j,k] = min{P[j][k],p[j][i]+p[i][k];

be not written as:

P[j,k] = min{P[j][k],p[j][i] & p[i][k]);

&=logical 'and'.

Answer 3

It is a Floyd Warshall algorithm(Dynamic Programming approach).

for i = 1 to N     
    for j = 1 to N        
        if there is an edge from i to j          
            dist[0][i][j] = the length of the edge from i to j        
        else           dist[0][i][j] = INFINITY      
for k = 1 to N     
    for i = 1 to N        
        for j = 1 to N           
            dist[k][i][j] = min(dist[k-1][i][j], dist[k-1][i][k] + dist[k-1][k][j])

Time Complexity: $O(n^3)$

 

References:

http://www.algorithmist.com/index.php/Floyd-Warshall's_Algorithm

http://www.geeksforgeeks.org/dynamic-programming-set-16-floyd-warshall-algorithm/