travelling salesman local search heuristic
java,debugging,traveling-salesman,hamiltonian-cycle
The problem is that when you swap outgoing[i] and outgoing[j], you are creating two subtours -- two smaller cycles. For example, suppose numcities=6 and your starting tour is 0 1 2 3 4 5. Suppose your if statement is true for i=1, j=3. Your code sets outgoing[1] = 4 and...
Find shortest path from X,Y coordinates (with start ≠ end)
r,traveling-salesman,hamiltonian-cycle
As the comments suggest, your question is related to the Traveling Salesman Problem. It is in fact an example of a Hamiltonian path (a path which visits each node once, but does not end where it started). As you might expect, in R there's already a package for that: TSP....
prove connected graph with degree = 2 has hamiltonian cycle [closed]
algorithm,graph,hamiltonian-cycle
Let the given graph be G. Starting from a vertex v in the graph, let us trace an arbitrary walk(path with repeated vertices allowed), P, by repeatedly picking a vertex adjacent to the last vertex added to P, without repeated any edges. Terminate if you cannot add any more vertices...
Is it possible to use Dijkstra's Shortest Path Algorithm to find the shortest Hamiltonian path? (in Polynomial Time)
algorithm,graph-theory,dijkstra,np-complete,hamiltonian-cycle
No, this is not possible. Your simplified problem is still NP-hard. A reduction from travelling salesman: Given a graph (V, E), find the shortest path that visits each v in V exactly once. Take an arbitrary vertex v in V. Split v into two vertices v_source and v_sink. Use your...
Algorithm for creating most efficient undirected Hamiltonian path
algorithm,graph,cycle,hamiltonian-cycle
Finding an hamiltonian path is a NP-complete problem. You probably won't find a much better solution than trying every permutation of nodes. And if somehow you find it, you'll be instantly rich and famous :) ...