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.
Related
You are given a weighted connected graph (20 nodes) with all edges having positive weight. We have a robot that starts at point A and it must pass at points B, D and E for example. The idea is to find the shortest path that connects all these 4 points. The robot also has a limited battery, but it can be recharged in some points.
After researching on the internet I have two algorithms in mind: Dijkstra's and TSP. Dijkstra's will find the shortest path between a node and every other node and TSP will find the shortest path that connects all points. Is there any variant of the TSP that only finds the shortest path between a set of nodes? After all, in the TSP all nodes are labeled "must-pass". I'm still not taking in account the battery constraint.
Thanks in advance!
You can reduce your graph to a TSP and then invoke a TSP algorithm on it:
Use Floyd-Warshall algorithm to find the distance u,v for ALL pairs of vertices u and v.
Create a new graph, containing only the "desired" vertices, and set the weight between two such vertices u and v as the distance found by Floyd-Warshall.
Run TSP Solver on the modified graph to get the path in the modified graph, and switch each edge in the modified graph with a shortest path from the original graph.
The above is optimal, because assume there is a shorter path.
D0=u->...D1->...->D2->...->Dk->...->t=D{k+1}
Di->...->D{i+1} has at least the weight of FloydWarshall(Di,D{i+1}) (correctness of Floyd-Warshall), and thus the edges (D0,D1),(D1,D2),...,(Dk,D{k+1) exist in the modified graph with a weight smaller/equal the weight in the given path.
Thus, from correctness of your TSP-Solver, by using D0->D1->...->Dk->D{k+1}, you get a path that is at least as good as the candidate optimal path.
You might also want to look into the generalized traveling salesman problem (GTSP): The nodes are partitioned into subsets, and the problem is to find the minimum-length route that visits exactly one node in each subset. The model is allowed to choose whichever node it wants from each subset. If there are nodes that must be visited, you can put them in a subset all by themselves.
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.
Hello and thanks again for reading this.
I need to know now if the problem of finding the simple path with maximum cost in a weighted undirected graph with the same number of vertex and edges is NP-Complete or not?
Input: Graph G = (V,E) with V (vertex) = E (edges)
Output: The cost of the most expensive path in the graph G.
Could you provide any reference to an article where I can review this.
Thank you very much for your time.
Sincerely,
Alex.
If the graph is not necessarily connected, then any instance of the longest path problem can be reduced to this problem by adding extra isolated vertices to the input graph to make the number of nodes and edges the same. If this isn't thyroid case, and the graph must be connected, then the input graph must have exactly one cycle, since a graph with n-1 edges is a tree. IF you find this cycle with a DFS and contract it to a single node, you then have a tree. It's easy to do longest path computations here; just consider all pairs of edges and get the cost of the unique path between them. If you take this path ans then expand it in the original graph by walking around the cycle where you originally went through the contracted node, I think you get the longest path in polynomial time.
This problem is called the Longest Path Problem, and is NP-complete.
The restriction |V| = |E| doesn't help at all. You can solve an arbitrary graph by adding unconnected vertices until you satisfy the relation.
Is there a graph algorithm that given a start(v) and an end(u) will find a shortest path through the given set of edges, but if u is a disconnected vertex, it will also determine the shortest path to add missing edges until u is no longer disconnected?
I have a pixel matrix where lines are made of 255's(black) and 0's(white). lines(255) can have breaks or spurs and I must get rid of both. I could have a pixel matrix forest with say 7 or so trees of black pixels. I need to find the true end points of each tree, find the single shortest path of each tree, then union all skew trees together to form 1 single line(ie, a single shortest path from the furthest 2 end points in the original matrix). all edge weights could be considered 1.
Thanks
How about running Dijkstra's algorithm and if disconnected, connect v and u? What's your criteria for "best place to add a missing edge?" Do edges have weights (like distance)?
Edit:
For one idea of "the best place," you could try the path that has minimal sum of shortest paths between all connected pairs. Floyd–Warshall algorithm can be used to find shortest paths between all pairs. So, run Floyd-Warshall for each node in v's tree and u.
Your problem isn't well defined for disconnected graphs. I can always add and edge between v and u.
If you meant that given an acyclic undirected disconnected graph, actually known as a forest, and given a subset of edges as a subgraph, can you find the shortest path between vertices, than this problem is trivial since if there is a path in the full graph, there is one path only.
If this is a general graph G, and you are talking about a forest subgraph G', than we need more info. Is this weighted? Is it only positive weights? If it is unweighted, do some variant of Dijksta. Define the diameter of a tree to be the length of the longest path between two leaves (this is well defined in a tree, since ther is only one such path). Let S be the sum of the diameters of all the trees in G', then set the weight all edges out of G' to 2S, then Dijkstra's algorithm will automatically prefer to step through G', stepping outside G' only when there is no choice, since walking through G' would always be cheaper.
Here you have following scenarios:
case:1. (all edges>0)
Negate all edges
find the longest path in graph using this:
https://www.geeksforgeeks.org/longest-path-undirected-tree/
negate the final result
case2: edges might be or might not be negative
Read Floyd–Warshall algorithm for this