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k= j generate, create pdf417 none in .net projects iPad j = 2, 3, . . . , N Using these equations, we can nd a shortest path to a node as follows. First, nd a node k with edge (k, j) such that u j = u k + ak j . Then nd an arc (l, k) such that u k = u l + alk , and continue in this fashion.

Eventually, we would reach node 1. Unfortunately, Bellman s equations do not lead to a solution directly..

A4.2 Shortest path in an acyclic network A4.2 Shortest path in an acyclic network In an acyclic network, as shown in Fig. A4.

3, it is easy to use Bellman s equations to nd a shortest path. The nodes in such a network can be renumbered in such a fashion that an edge (i, j) exists if and only if i < j. In this case we can rewrite Bellman s equations as u1 = 0 and u j = min{u k + ak j }.

k< j j = 2, 3, . . . , N These equations can th PDF 417 for .NET en be solved as u 1 is known, u 2 only depends on u 1 , u 3 only depends on u 1 and u 2 , and so on. The complexity of this problem is O(N 2 ).

. A4.2.1 Dijkstra method For non-cyclic graphs, we need another method given by Dijkstra [73].

This method is applicable to a graph for which edge weights are positive. This algorithm starts with labeling nodes in stages. At each stage of computation, some labels are designated permanent and others remain tentative.

A permanent label on a node represents the true length of the shortest path from that node. After including the new labeled nodes, distances to all other nodes are computed again. Let di j denote the distance from node i to node j.

Let i be the source node. Then dii is set to zero and di j , i = j is set to a large value if j is not a neighbor of i. Otherwise, it is set equal to the weight of the direct link, ai j .

Next, the algorithm nds a node j with minimum di j and labels it permanent. It then uses it to improve distances to other nodes by computing dik min(dik , di j + a jk ) At each stage in the process, the value of dik represents the best known shortest distance from i to k. Using these labels of the nodes, the algorithm then marks another unlabeled node with a minimum value of dik as permanently labeled.

The same computation is carried out again. Since all edge weights are positive, in the next iteration, none of the marked nodes can have any smaller value. An example of the execution of the algorithm is shown in Fig.

A4.1. Node A is the source node.

A dark node is a permanently labeled node. At each step, one node is marked labeled and the value associated with a node is its shortest distance from the source thus far with L being a large value. The algorithm terminates in N 1 steps.

. Graph algorithms Fig. A4.1. Example execution of Dijkstra s shortest-path algorithm. A4.3 Shortest paths be .NET pdf417 tween all pairs of nodes Now, suppose we want to compute the shortest path between all pairs of nodes.

This may be necessary as communication may occur between any pair of nodes. It is desirable to use the shortest path as this reduces the requirements for network resources. Sometimes this may cause congestion, as has been shown by many researchers.

For example, suppose that the network graph is such that it can be partitioned into two parts, A and B, and the two parts are only connected by two links, one with a low-weight link and the other with a high-weight link, as shown in Fig. A4.2.

All communication between the two halves will use the low-weight link and the other link remains unused. The second link should not have been included in the design, but if it exists then its use will reduce the congestion on the lowweight link. The shortest-path routing algorithm does not utilize the second link at all.

Coming back to the all-to-all communication problem, we can compute paths from every node to every other node. Thus, we need to solve the problem N times. Alternately, we may use an integrated procedure developed separately, that may be more advantageous.

We investigate the latter approach next. Let u i j denote the length of the shortest path from node i to node j and let u im be the shortest path j such that the path contains no more than m edges. It is clear that u iN will be u i j , the j.

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