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Anamica
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Quote Anamica Replybullet Topic: Gate Study Material Graph theoretic algorithms
    Posted: 04Apr2007 at 10:41pm

7.1 Single-source shortest path:

Graphs can be used to represent the highway structure of a state or country with vertices representing cities and edges representing sections of highway. The edges can then be assigned weights which may be either the distance between the two cities connected by the edge or the average time to drive along that section of highway. A motorist wishing to drive from city A to B would be interested in answers to the following questions:
    1. Is there a path from A to B?
    2. If there is more than one path from A to B? Which is the shortest path?
The problems defined by these questions are special case of the path problem we study in this section. The length of a path is now defined to be the sum of the weights of the edges on that path. The starting vertex of the path is referred to as the source and the last vertex the destination. The graphs are digraphs representing streets. Consider a digraph G=(V,E), with the distance to be traveled as weights on the edges. The problem is to determine the shortest path from v0 to all the remaining vertices of G. It is assumed that all the weights associated with the edges are positive. The shortest path between v0 and some other node v is an ordering among a subset of the edges. Hence this problem fits the ordering paradigm.     Example: Consider the digraph of fig 7-1. Let the numbers on the edges be the costs of travelling along that route. If a person is interested travel from v1 to v2, then he encounters many paths. Some of them are
    1. v1à v2 = 50 units
    2. v1à v3à v4à v2 = 10+15+20=45 units
    3. v1à v5à v4à v2 = 45+30+20= 95 units
    4. v1à v3à v4à v5à v4à v2 = 10+15+35+30+20=110 units
The cheapest path among these is the path along v1à v3à v4à v2. The cost of the path is 10+15+20 = 45 units. Even though there are three edges on this path, it is cheaper than travelling along the path connecting v1 and v2 directly i.e., the path v1à v2 that costs 50 units. One can also notice that, it is not possible to travel to v6 from any other node. To formulate a greedy based algorithm to generate the cheapest paths, we must conceive a multistage solution to the problem and also of an optimization measure. One possibility is to build the shortest paths one by one. As an optimization measure we can use the sum of the lengths of all paths so far generated. For this measure to be minimized, each individual path must be of minimum length. If we have already constructed i shortest paths, then using this optimization measure, the next path to be constructed should be the next shortest minimum length path. The greedy way to generate these paths in non-decreasing order of path length. First, a shortest path to the nearest vertex is generated. Then a shortest path to the second nearest vertex is generated, and so on. A much simpler method would be to solve it using matrix representation. The steps that should be followed is as follows, Step 1: find the adjacency matrix for the given graph. The adjacency matrix for fig 7.1 is given below

V1 V2 V3 V4 V5 V6
V1 - 50 10 Inf 45 Inf
V2 Inf - 15 Inf 10 Inf
V3 20 Inf - 15 inf Inf
V4 Inf 20 Inf - 35 Inf
V5 Inf Inf Inf 30 - Inf
V6 Inf Inf Inf 3 Inf -
Step 2: consider v1 to be the source and choose the minimum entry in the row v1. In the above table the minimum in row v1 is 10. Step 3: find out the column in which the minimum is present, for the above example it is column v3. Hence, this is the node that has to be next visited. Step 4: compute a matrix by eliminating v1 and v3 columns. Initially retain only row v1. The second row is computed by adding 10 to all values of row v3. The resulting matrix is
  V2 V4 V5 V6
V1à Vw 50 Inf 45 Inf
V1à V3à Vw 10+inf 10+15 10+inf 10+inf
Minimum 50 25 45 inf
Step 5: find the minimum in each column. Now select the minimum from the resulting row. In the above example the minimum is 25. Repeat step 3 followed by step 4 till all vertices are covered or single column is left. The solution for the fig 7.1 can be continued as follows

  V2 V5 V6
V1à Vw 50 45 Inf
V1à V3à V4à Vw 25+20 25+35 25+inf
Minimum 45 45 inf

 

  V5 V6
V1à Vw 45 Inf
V1à V3à V4à V2à Vw 45+10 45+inf
Minimum 45 inf

  V6
V1à Vw Inf
V1à V3à V4à V2à V5à Vw 45+inf
Minimum inf

Finally the cheapest path from v1 to all other vertices is given by V1à V3à V4à V2à V5.



7.2 Minimum-Cost spanning trees

Let G=(V,E) be an undirected connected graph. A sub-graph t = (V,E1) of G is a spanning tree of G if and only if t is a tree.     Above figure shows the complete graph on four nodes together with three of its spanning tree. Spanning trees have many applications. For example, they can be used to obtain an independent set of circuit equations for an electric network. First, a spanning tree for the electric network is obtained. Let B be the set of network edges not in the spanning tree. Adding an edge from B to the spanning tree creates a cycle. Kirchoff’s second law is used on each cycle to obtain a circuit equation. Another application of spanning trees arises from the property that a spanning tree is a minimal sub-graph G’ of G such that V(G’) = V(G) and G’ is connected. A minimal sub-graph with n vertices must have at least n-1 edges and all connected graphs with n-1 edges are trees. If the nodes of G represent cities and the edges represent possible communication links connecting two cities, then the minimum number of links needed to connect the n cities is n-1. the spanning trees of G represent all feasible choice. In practical situations, the edges have weights assigned to them. Thse weights may represent the cost of construction, the length of the link, and so on. Given such a weighted graph, one would then wish to select cities to have minimum total cost or minimum total length. In either case the links selected have to form a tree. If this is not so, then the selection of links contains a cycle. Removal of any one of the links on this cycle results in a link selection of less const connecting all cities. We are therefore interested in finding a spanning tree of G. with minimum cost since the identification of a minimum-cost spanning tree involves the selection of a subset of the edges, this problem fits the subset paradigm. 7.2.1 Prim’s Algorithm A greedy method to obtain a minimum-cost spanning tree builds this tree edge by edge. The next edge to include is chosen according to some optimization criterion. The simplest such criterion is to choose an edge that results in a minimum increase in the sum of the costs of the edges so far included. There are two possible ways to interpret this criterion. In the first, the set of edges so far selected form a tree. Thus, if A is the set of edges selected so far, then A forms a tree. The next edge(u,v) to be included in A is a minimum-cost edge not in A with the property that A U {(u,v)} is also a tree. The corresponding algorithm is known as prim’s algorithm. For Prim’s algorithm draw n isolated vertices and label them v1, v2, v3,…vn. Tabulate the given weights of the edges of g in an n by n table. Set the non existent edges as very large. Start from vertex v1 and connect it to its nearest neighbor (i.e., to the vertex, which has the smallest entry in row1 of table) say Vk. Now consider v1 and vk as one subgraph and connect this subgraph to its closest neighbor. Let this new vertex be vi. Next regard the tree with v1 vk and vi as one subgraph and continue the process until all n vertices have been connected by n-1 edges. Consider the graph shown in fig 7.3. There are 6 vertices and 12 edges. The weights are tabulated in table given below.
  V1 V2 V3 V4 V5 V6
V1 - 10 16 11 10 17
V2 10 - 9.5 Inf Inf 19.5
V3 16 9.5 - 7 Inf 12
V4 11 Inf 7 - 8 7
V5 10 Inf Inf 8 - 9
V6 17 19.5 12 7 9 -

Start with v1 and pick the smallest entry in row1, which is either (v1,v2) or (v1,v5). Let us pick (v1, v5). The closest neighbor of the subgraph (v1,v5) is v4 as it is the smallest in the rows v1 and v5. The three remaining edges selected following the above procedure turn out to be (v4,v6) (v4,v3) and (v3, v2) in that sequence. The resulting shortest spanning tree is shown in fig 7.4. The weight of this tree is 41.5.

  7.2.3 Kruskal’s Algorithm: There is a second possible interpretation of the optimization criteria mentioned earlier in which the edges of the graph are considered in non-decreasing order of cost. This interpretation is that the set t of edges so far selected for the spanning tree be such that it is possible to complete t into a tree. Thus t may not be a tree at all stages in the algorithm. In fact, it will generally only be a forest since the set of edges t can be completed into a tree if and only if there are no cycles in t. this method is due to kruskal. The Kruskal algorithm can be illustrated as folows, list out all edges of graph G in order of non-decreasing weight. Next select a smallest edge that makes no circuit with previously selected edges. Continue this process until (n-1) edges have been selected and these edges will constitute the desired shortest spanning tree. For fig 7.3 kruskal solution is as follows, V1 to v2 =10 V1 to v3 = 16 V1 to v4 = 11 V1 to v5 = 10 V1 to v6 = 17 V2 to v3 = 9.5 V2 to v6 = 19.5 V3 to v4 = 7 V3 to v6 =12 V4 to v5 = 8 V4 to v6 = 7 V5 to v6 = 9 The above path in ascending order is V3 to v4 = 7 V4 to v6 = 7 V4 to v5 = 8 V5 to v6 = 9 V2 to v3 = 9.5 V1 to v5 = 10 V1 to v2 =10 V1 to v4 = 11 V3 to v6 =12 V1 to v3 = 16 V1 to v6 = 17 V2 to v6 = 19.5 Select the minimum, i.e., v3 to v4 connect them, now select v4 to v6 and then v4 to v5, now if we select v5 to v6 then it forms a circuit so drop it and go for the next. Connect v2 and v3 and finally connect v1 and v5. Thus, we have a minimum spanning tree, which is similar to the figure 7.4.


7.3 Techniques for graphs:

A fundamental problem concerning graphs is the reachability problem. In its simplest form it requires us to determine whether there exists a path in the given graph G=(V,E) such that this path starts at vertex v and ends at vertex u. A more general form is to determine for a given starting Vertex v belonging to V all vertices u such that there is a path from v to u. This latter problem can be solved by starting at vertex v and systematically searching the graph G for vertices that can be reached from v. The 2 search methods for this are :
    1. Breadth first search.
    2. Depth first search.
7.3.1 Breadth first search: In Breadth first search we start at vertex v and mark it as having been reached. The vertex v at this time is said to be unexplored. A vertex is said to have been explored by an algorithm when the algorithm has visited all vertices adjacent from it. All unvisited vertices adjacent from v are visited next. There are new unexplored vertices. Vertex v has now been explored. The newly visited vertices have not been explored and are put onto the end of the list of unexplored vertices. The first vertex on this list is the next to be explored. Exploration continues until no unexplored vertex is left. The list of unexplored vertices acts as a queue and can be represented using any of the standard queue representations. 7.3.2 Depth first search: A depth first search of a graph differs from a breadth first search in that the exploration of a vertex v is suspended as soon as a new vertex is reached. At this time the exploration of the new vertex u begins. When this new vertex has been explored, the exploration of u continues. The search terminates when all reached vertices have been fully explored. This search process is best-described recursively. Algorithm DFS(v) { visited[v]=1 for each vertex w adjacent from v do { If (visited[w]=0)then DFS(w); } }




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