A "walk" is a sequence of alternating vertices and edges, starting with a vertex and ending with a vertex with any number of revisiting vertices and retracing of edges. If a walk has the restriction of no repetition of vertices and no edge is retraced it is called a "path". If there is a walk to every vertex from any other vertex of the graph then it is called a "connected" graph.
dijkstra's algorithm (note* there are different kinds of dijkstra's implementation) and growth graph algorithm
Dijkstra's algorithm is used by the OSPF and the IS-IS routing protocols. The last three letters in OSPF (SPF) mean "shortest path first", which is an alternative name for Dijkstra's algorithm.
A tree is a connected graph in which only 1 path exist between any two vertices of the graph i.e. if the graph has no cycles. A spanning tree of a connected graph G is a tree which includes all the vertices of the graph G.There can be more than one spanning tree for a connected graph G.
Prove that the maximum vertex connectivity one can achieve with a graph G on n. 01. Define a bipartite graph. Prove that a graph is bipartite if and only if it contains no circuit of odd lengths. Define a cut-vertex. Prove that every connected graph with three or more vertices has at least two vertices that are not cut vertices. Prove that a connected planar graph with n vertices and e edges has e - n + 2 regions. 02. 03. 04. Define Euler graph. Prove that a connected graph G is an Euler graph if and only if all vertices of G are of even degree. Prove that every tree with two or more vertices is 2-chromatic. 05. 06. 07. Draw the two Kuratowski's graphs and state the properties common to these graphs. Define a Tree and prove that there is a unique path between every pair of vertices in a tree. If B is a circuit matrix of a connected graph G with e edge arid n vertices, prove that rank of B=e-n+1. 08. 09.
- a problem in NP means that it can be solved in polynomial time with a non-deterministic turing machine - a problem that is NP-hard means that all problems in NP are "easier" than this problem - a problem that is NP-complete means that it is in NP and it is NP-hard example - Hamiltonian path in a graph: The problem is: given a graph as input, an algorithm must say whether there is a hamiltonian path in it or not. in NP: here is an algorithm that works in polynomial time on a non-deterministic turing machine: guess a path in the graph. Check that it is really a hamiltonian path. NP-hard: we use reduction from a problem that is NP-comlete (SAT for example). Given an input for the other problem we construct a graph for the hamiltonian-path problem. The graph should have a path iff the original problem should return "true". Therefore, if there is an algorithm that executes in polynomial time, we solve all the problems in NP in polynomial time.j
A practical application is in certain routing protocols, like OSPF. The problem it solves is to search for the "shortest" path to each destination - "shortest" meaning the one that has the lowest "distance" or "metric" according to the criteria used. Dijkstra's algorithm is easy to use and is a good graph search algorithm to use when it is hard to calculate the heuristics.
Which routing protocol depends on the DUAL algorithm to calculate the shortest path to a destination
yes
Path Oriented Decision Making.
This distance-vector algorithm works by computing the shortest path , and considers weights. The algorithm was distributed widely in the RIP protocol.
Answer: shortest path routing
Yes. An example: _____A---------B________ A connected directly to B and D by one path. _____|_______/|\________ B connected directly to A and E by one path, and to C by two paths. _____|______/_|_\_______ _____|_____/___\_|______ _____|__E/_____\|______ E connected directly to B and D by one path. _____|____\_____C______ C connected directly to B and D by two paths. _____|_____\____|\_____ _____|______\___|__\___ _____|_______\__|__/___ _____|________\_|_/____ _____|_________\|/_____ _____-------------D_____ D connected directly to A and E by one path, and to C by two paths. There is an Euler circuit: ABCDEBCDA But a Hamiltonian circuit is impossible: as part of a circuit A can only be reached by the path BAD, but once BAD has been traversed it is impossible to get to both C and E without returning to B or D first. However there is a Hamiltonian Path: ABCDE.