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| courses:cs211:winter2018:journals:holmesr:section_3.1 [2018/02/06 03:41] – holmesr | courses:cs211:winter2018:journals:holmesr:section_3.1 [2018/02/06 04:53] (current) – holmesr | ||
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| The one thing I didn't understand about this chapter is the how a BFS could be implemented using a queue. The chapter promised that this was true but didn't really explain how, at least in my eyes. | The one thing I didn't understand about this chapter is the how a BFS could be implemented using a queue. The chapter promised that this was true but didn't really explain how, at least in my eyes. | ||
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| + | ===== Section 3.4 Testing Bipartiteness: | ||
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| + | A bipartite graph is one that can be separated into two sets such that every edge has one edge in X and the other in Y. This can be represented by a BFS tree by labelling all the odd layers as the layers which contain nodes in the X partition and the even layers contain nodes in the Y partition. Then it is easy to see that there can not be an edge that begins and ends in the same layer, and by the definition of a BFS, there can not be an edge that spans more than one layer difference from the layer in which it has one end. It follows from this then that a bipartite graph can not contain any odd cycles, since an odd cycle must contain an edge that begins and ends in the same layer of the BFS tree. | ||
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| + | This lays down the basis for a very easy addition to be made to the BSF algorithm that will help determine whether a graph is bipartite or not. Simply add an array of size n called Partition and assign all the nodes //n// in an odd layer to be Partition[n] = X and all the nodes in even layers to be Partition[n] = Y. If any edge has both ends being Partition[n] = X or Partition[n] = Y, then the graph is bipartite. Since this all occurs in constant time, it does not bump up the O(n+m) running time of BFS. | ||
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| + | This section was fascinating in that it demonstrated that determining bipartiteness of a graph is reliant on the same algorithmic skeleton as the BFS algorithm. | ||
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| + | ===== Section 3.5 Connectivity in Directed Graphs ===== | ||
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| + | First, directed graphs require two adjacency lists in order to be properly represented since an edge may go from node //u// to node //v// but not return from //v// to //u//. One lists represents the edges that leave a node and the nodes that those edges connect to, and the other list represents the edges which lead to a node and from which node they come. They are called the //from list// and the //to list//, respectively. | ||
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| + | The graph search algorithms BFS and DFS perform almost the same with directed graphs as they do with undirected graphs. The major difference being that they use a node's //from list// to traverse through the nodes which the current node has an edge from itself to them. It is important to note that in a directed graph, a node //n// can have a path to node //r//, but node //r// is not required to have a path back to node //n//. The graph traversals are discovering nodes to which //n// has a path; these paths are not necessarily reciprocated. | ||
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| + | It is interesting that these traversals are finding all the nodes to which node //n// has a path, but to find all the nodes with paths to //n//, the same traversal can be run on an identical graph with the directions of the edges reversed. This leads into a discussion of strong connectivity and mutual reachability. | ||
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| + | A graph is strongly connected if all nodes u and v which have paths (u,v) also have paths (v,u). In other words, if all nodes are mutually reachable to one another. An interesting property of mutual reachability is that if two nodes are mutually reachable and a third node is mutually reachable to one of those two, then it is mutually reachable to the other also. | ||
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| + | The strong component is an analog to the connected component in an undirected graph. The strong component in a graph G is the set of nodes //n// which are mutually reachable from the nodes //s//. The a set of nodes is a strong component if that set is returned by a BFS started at node //n// on graph G and also returned by a BFS started at node //n// on a graph G< | ||
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| + | ===== Section 3.6 DAGS and Topological Orderings | ||
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| + | A Directed Acyclic Graph (DAG) is just what it sounds like, a directed graph containing no cycles. These are useful for representing precedence relations and dependencies because they don't contain cycles. A cycle in a dependency relation would mean that one process could not start until another had taken place, which would never happen because none could start first. | ||
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| + | These types of graphs have can also be represented by a structure known as a topological ordering, which the book defines as "an ordering of the nodes v< | ||
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| + | It is also true that every DAG has a topological ordering, since there must be a node that has no incoming edges due to the fact that there are no cycles in a DAG. Once this node has been removed and removing a node cannot create a cycle, then there is a DAG remaining with all the nodes that were in the previous DAG, except the removed node. This truth can be used to devise an algorithm that runs in O(n< | ||
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| + | 1.find the node with no incoming edges | ||
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| + | 2.delete this node | ||
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| + | 3.recursively compute the topological ordering of the DAG without the removed node and add this after the deleted node. | ||
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| + | There is yet a faster, linear-time method to computing the topological ordering of the DAG, but I could not quite grasp what this was. It dealt with active nodes and edges coming from those nodes into the current node. | ||
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