I am unable to understand that does simple path in graph always contains first and last vertices as same?Can simple path be simple cycle ? Can simple cycle be simple path in graph data structure?Please help me out!.
A path is a path(sequences of vertices where each vertex is adjacent to vertex next to it), simple path does not repeat vertices. So, a simple path is not a cycle. simple path does not contain same vertex as ending and starting.
A simple cycle is a cycle in which only(no other vertices are repeated) the starting and ending vertices are repeated. so a simple cycle is a path but not simple path.
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I was thinking on how to find a longest possible path in a complete, directed graph for every single vertex.
Example of such a graph
So for every single vertex I want to find the maximum possible amount of vertices that one can travel through (not going through any vertex more than once) and the specific path that has that specific length.
For example in the given graph, for starting vertex nr.1, the maximum length is 4 and the path:
1,4,2,3 ,or 1,2,3,4 (I just need to get one of them, not all).
What kind of algoritm could handle that?
In case it matters I use C++.
Algorithms like the Bellman-Ford algorithm and Dijkstra's algorithm exist to find the shortest path from a single starting vertex on a graph to every other vertex. However, in the program I'm writing, the starting vertex changes a lot more often than the destination vertex does. What algorithm is there that does the reverse--that is, given a single destination vertex, to find the shortest path from every starting vertex?
Just reverse all the edges, and treated destination as start node. Problem solved.
If this is an undirected graph: I think inverting the problem would give you advantages. View the actual destination as the start and find the shortest path from that to all of the other nodes in the graph. By doing this you can use the traditional pathing algorithms.
The problem is testing whether a graph G contains a Hamiltonian path or not with the one use of hamiltonian cycle Hcycle(V,E) function which gives output true of false whether the G contains Hamiltonian cycle.
I must write a program with polynomial time complexity, which has to decide whether the unoriented graph G contains at least one Hamiltonian path with the use of one Hamiltonian Cycle function which has to give output to this problem.
Also I need to write a program with the opposite problem. (use of Hpath function to find out whether the graph contains Hemiltonian Cycle).
I can't find a solution to this problem.
I can use both Hcycle and Hpath only once.
We assume that the function Hcycle and Hpath run in linear time complexity.
Path by Hcycle(V,E): Call Hcycle() on a graph created by adding one vertex that is connected to all other vertices. If new graph has a cycle than removing new node from that cycle we get path on original graph.
Cycle by Hpath(V,E): Call Hpath() on a graph created by adding one vertex and connecting it to same neighbours as one existing vertex. That means these 2 vertices will have same naighbours. If new graph has a path than at least one path end is on these two vertices. If other vertex is not end, than it is on path third position and by reordering path we can set both vertices to be path end points. Merging these two vertices (since they have same neighbours) we get a cycle in original graph.
Path by Hcycle(V,E): If graph has cycle than it has path also. If graph doesn't have cycle than for each unconnected pair of vertices (v1, v2) add edge between them and check is there a cycle with Hcycle(V,E+(v1,v2)). If there is a cycle than there is a path between v1 and v2 in original graph. Hcycle is called max |V|^2 times.
Cycle by Hpath(V,E): Idea is to force a path to has end vertices we know about. That can be done by constructing graph where two vertices have degree one. Let N(v) be neighbours of v. For an edge (v1,v2) and n1 from N(v1)-v2 and n2 from N(v2)-v1 construct a graph by removing all edges from v1 except to n1, and removing all edges from v2 except to n2. If that graph has a path, than it's ends are on v1 and v2, and our original graph has a circle. Hpath is called max |E|*|V|^2 times.
Every hamiltonian cycle is a hamiltonian path, just break the cycle somewhere.
The other way around doesn't work quite easily. The brute-force solution is:
for all hamiltonian paths p, check whether there is edge between start and end of p, if there is, make a cycle. However if HPath returns just some path, there is no way to make a cycle out of it (I guess).
Check Wikipedia
One of the classical Travelling Salesman Problem (TSP) definitions is:
Given a weighted complete undirected graph where triangle inequality holds return an Hamiltonian path of minimal total weight.
In my case I do not want an Hamiltonian path, I need a path between two well known vertexes. So the formulation would be:
Given a weighted complete undirected graph where triangle inequality holds and two special vertexes called source and destination return a minimal weighted path that visits all nodes exactly once and starts from the source and ends to the destination.
I recall that an Hamiltonian path is a path in an undirected graph that visits each vertex exactly once.
For the original problem a good approximation (at worse 3/2 of the best solution) is the Christodes' algorithm, it is possible to modify for my case? Or you know another way?
Add an edge (= road) from your destination node to your source node with cost 0 and you've got a TSP (for which the triangle inequality doesn't hold though).
The book "In pursuit of the Traveling Salesman" briefly mentions this technique.
Why don't you use dijkstra's algorithm with extra book keeping on each node for the path information. i.e., the list of vertices passed in the shortest path to that particular vertex from the source.
And stop when you reach your end vertex. Then your path will be
Path to the starting vertex of the current edge + current edge.
where current edge is the last edge which lead you to your destination.
You can alter a TSP algorithm to make sure you use the edge between start and finish. Then simply remove the edge from the TSP result and you get your path.
In other words, if you add the edge from start to finish to your path, you get a TSP solution, but not necessary the optimal one.
In greedy algorithm, you have a sorted list of all edges L and an empty list I. You keep adding the shortest edge that does not form a cycle until you pass all vertexes. Simply remove the edge from start to finish from L and add it to I before you add the rest of the edges.
In nearest neighbor, you start from one vertex, then add the shortest edge that starts at that vertex and leads to any vertex that has not been visited. Mark your finish as visited, then start from your start vertex and add the edge between the last vertex in the resulted path and the finish and you got your path from start to finish.
There are other TSP algorithms that can be altered to provide this path.
Can some one tell me the difference between hamiltonian path and euler path. They seem similar!
An Euler path is a path that passes through every edge exactly once. If it ends at the initial vertex then it is an Euler cycle.
A Hamiltonian path is a path that passes through every vertex exactly once (NOT every edge). If it ends at the initial vertex then it is a Hamiltonian cycle.
In an Euler path you might pass through a vertex more than once.
In a Hamiltonian path you may not pass through all edges.
Graph Theory Definitions
(In descending order of generality)
Walk: a sequence of edges where the end of one edge marks the beginning of the next edge
Trail: a walk which does not repeat any edges. All trails are walks.
Path: a walk where each vertex is traversed at most once.
(paths used to refer to open walks, the definition has changed now) The property of traversing vertices at most once means that edges are also crossed at most once, hence all paths are trails.
Hamiltonian paths & Eulerian trails
Hamiltonian path: visits every vertex in the graph (exactly once, because it is a path)
Eulerian trail: visits every edge in the graph exactly once (because it is a trail, vertices may well be crossed more than once.)
Eulerian path must visit each edge exactly once, while Hamiltonian path must visit each vertex exactly once.
They are related but are neither dependent nor mutually exclusive. If a graph has an Eurler cycle, it may or may not also have a Hamiltonian cyle and vice versa.
Euler cycles visit every edge in the graph exactly once. If there are vertices in the graph with more than two edges, then by definition, the cycle will pass through those vertices more than once. As a result, vertices can be repeated but edges cannot.
Hamiltonian cycles visit every vertex in the graph exactly once (similar to the travelling salesman problem). As a result, neither edges nor vertices can be repeated.
I'll use a common example in biology; reconstructing a genome by making DNA samples.
De-novo assembly
To construct a genome from short reads, it's necessary to construct a graph of those reads. We do it by breaking the reads into k-mers and assemble them into a graph.
We can reconstruct the genome by visiting each node once as in the diagram. This is known as Hamiltonian path.
Unfortunately, constructing such path is NP-hard. It's not possible to derive an efficient algorithm for solving it. Instead, in bioinformatics we construct a Eulerian cycle where an edge represents an overlap.
A Hamiltonian path visits every node (or vertex) exactly once, and a Eulerian path traverses every edge exactly once.
Euler Path - An Euler path is a path in which each edge is traversed exactly once.
Hamiltonian Path - An Hamiltonian path is path in which each vertex is traversed exactly once.
If you have ever confusion remember E - Euler E - Edge.
An Euler path is a path that uses every edge of a
graph exactly once.and it must have exactly two odd vertices.the path starts and ends at different vertex. A Hamiltonian cycle is a cycle that contains every vertex of the graph hence you may not use all the edges of the graph.
Euler path is a graph using every edge(NOTE) of the graph exactly once. Euler circuit is a euler path that returns to it starting point after covering all edges.
While hamilton path is a graph that covers all vertex(NOTE) exactly once. When this path returns to its starting point than this path is called hamilton circuit.