This question actually rephrases that one. The code jam problem is the following:
You are given a complete undirected graph with N nodes and K "forbidden" edges. N <= 300, K <= 15. Find the number of Hamiltonian cycles in the graph that do not use any of the K "forbidden" edges.
The straightforward DP approach of O(2^N*N^2) does not work for such N. It looks like that the winning solutions are O(2^K). Does anybody know how to solve this problem ?
Let's find out for each subset S of forbidden edges, how many Hamiltonian cycles there exist, that use all edges of S. If we solve this subtask then we'll solve the problem by inclusion-exclusion formula.
Now how do we solve the subtask? Let's draw all edges of S. If there exist a vertex of degree more than 2, then obviously we cannot complete the cycle and the answer is 0. Otherwise the graph is divided into connected components. Each component is a sole vertex, a cycle or a simple path.
If there exists a cycle, then it must pass through all vertices, otherwise we won't be able to complete the Hamiltonian cycle. If this is the case, the answer is 2. (The cycle can be traversed in 2 directions.) Otherwise the answer is 0.
The remaining case is when there are c paths and k sole vertices. To complete the Hamiltonian cycle we must choose the direction of each path (2^c ways) and then choose the order of components. We've got c+k components, so they can be rearranged in (c+k)! ways. But we are interested in cycles, so we don't distinguish the orderings which turn into one another by cyclic shifts. (So (1,2,3), (2,3,1) and (3,1,2) are the same.) It means that we must divide the answer by the number of shifts, c+k. So the answer (to the subtask) is 2^c (c+k-1)!.
This idea is implemented in bmerry's solution (very clean code, btw).
Hamiltonian cycle problem is a special case of travelling salesman problem (obtained by setting the distance between two cities to a finite constant if they are adjacent and infinity otherwise.)
These are NP Complete problems which in simple words means no fast solution to them is known.
A trivial heuristic algorithm for locating Hamiltonian paths is to construct a path abc... and extend it until no longer possible; when the path abc... xyz cannot be extended any longer because all neighbours of z already lie in the path, one goes back one step, removing the edge yz and extending the path with a different neighbour of y; if no choice produces a Hamiltonian path, then one takes a further step back, removing the edge xy and extending the path with a different neighbour of x, and so on. This algorithm will certainly find an Hamiltonian path (if any) but it runs in exponential time.
For more check NP Complete problem chapter of "Introduction to Algos" by Cormen
Related
I'm trying to find something on the literature to help me to solve this problem:
Given a graph with non negative edge weights, find a simple s-t positive path of any size, i.e., a path that goes from s to t with length greater than 0. I was trying to use dijkstra algorithm to find a positive shortest path by avoiding relax the edges that have cost zero, but this is wrong. I don't want to believe that this problem is NP-Complete =/. Sometimes it seems to be NP-Complete, because there may be a case where the only positive path is the longest path. But this is such a specific case. I think that on this kind of instance the longest path problem is polynomially solvable.
Can someone help me identifying this problem or showing that it is NP-Complete ?
Observations: The only requirement is to be some positive path (not necessarily the lowest or longest). In case of multiple positive paths it can be anyone. In case of non existence of such path, the algorithm should flag that the graph has no positive path.
Dijkstra's algorithm produces an answer in polynomial time (so in P rather than NP) and provides a shortest path between any 2 points on the graph.
Please refer to Dijkstra's Algorithm Wiki for further details.
You don't really need any more proof.
I'm not quite sure how "relax the edges that have cost 0" has any impact on this question at all.
The problem is NP-complete even with only one single edge having value 1, and all the others having value 0. Reduce from Two Directed Paths:
Two Directed Paths
Input: A directed graph G and two pairs of vertices (s1, t1) and (s2, t2)
Output: Two vertex-disjoint paths, one from s1 to t1 and from from s2 to t2.
Create a new instance with all edges having weight 0, and create an edge from t1 to s2 with weight 1.
Bellman-Ford algorithm is famously known to solve the single source shortest path problem (SSSPP) for any arbitrary connected graph G(V,E) with additive edge weights, whenever one exists.
The basic implementation version of the algorithm for e.g.: Bellman-Ford-Wiki-page and its proof of correctness, when used parallel relaxation of all edges, as per my understanding, implies an interesting by-product, which I call as "intermediate optimality property", (which might be very helpful for some applications like this question) is stated as below:
After k iterations, we have every node identified with its shortest path from the same source, under the constraint of #edges in the path is <= k
This, under the assumption of simple shortest-path existence, will ensure producing the shortest path solution to SSSPP, for every destination node, after at most |V|-1 iterations.
Is the above property correct?
According to some people (for e.g. comments just below this question), it is not correct and I fail to understand why!!
(UPDATED: I am using a parallel update on all the vertices.)
Suppose we have a simple graph A->B->C->D with all weights equal to 1.
If we visit the vertices in the order A,B,C,D then during the first iteration we will relax all of the following:
A->B, finds shortest path to B is 1
B->C, finds shortest path to C is 2
C->D, finds shortest path to D is 3.
So in the first iteration we have found the shortest path to D despite this path needing 3 edges.
However, if the vertices were visited in the order D,C,B,A it would take more iterations to find the shortest path to D.
In other words, after k iterations we will have certainly found any shortest paths with #edges <= k, however we may also have found a better route that uses more edges.
I am trying to do c++ program.I am trying to do problem in which i have numbers of points. Now i need to find the path that goes through all the points. This is not actually TSP because as per my knowledge in TSP it is possible to travel from all points to every other points. But in my case the path network between the points is fixed and i just need to find the suitable path that goes through all the points provided that all points may not have connection to every other point..so what algorithm am i supposed to follow.
It seems you are looking for a way to traverse a graph? If so have you tried Breadth first search http://en.wikipedia.org/wiki/Breadth-first_search or Depth first search http://en.wikipedia.org/wiki/Depth-first_search to traverse your graph.
You want to find a Hamiltonian path for a graph.
In the mathematical field of graph theory, a Hamiltonian path (or
traceable path) is a path in an undirected graph that visits each
vertex exactly once. A Hamiltonian cycle (or Hamiltonian circuit) is a
Hamiltonian path that is a cycle. Determining whether such paths and
cycles exist in graphs is the Hamiltonian path problem, which is
NP-complete.
Some techniques that exist :
There are n! different sequences of vertices that might be Hamiltonian
paths in a given n-vertex graph (and are, in a complete graph), so a
brute force search algorithm that tests all possible sequences would
be very slow. There are several faster approaches. A search procedure
by Frank Rubin divides the edges of the graph into three classes:
those that must be in the path, those that cannot be in the path, and
undecided. As the search proceeds, a set of decision rules classifies
the undecided edges, and determines whether to halt or continue the
search. The algorithm divides the graph into components that can be
solved separately. Also, a dynamic programming algorithm of Bellman,
Held, and Karp can be used to solve the problem in time O(n2 2n). In
this method, one determines, for each set S of vertices and each
vertex v in S, whether there is a path that covers exactly the
vertices in S and ends at v. For each choice of S and v, a path exists
for (S,v) if and only if v has a neighbor w such that a path exists
for (S − v,w), which can be looked up from already-computed
information in the dynamic program.
Andreas Björklund provided an alternative approach using the
inclusion–exclusion principle to reduce the problem of counting the
number of Hamiltonian cycles to a simpler counting problem, of
counting cycle covers, which can be solved by computing certain matrix
determinants. Using this method, he showed how to solve the
Hamiltonian cycle problem in arbitrary n-vertex graphs by a Monte
Carlo algorithm in time O(1.657n); for bipartite graphs this algorithm
can be further improved to time O(1.414n).
For graphs of maximum degree three, a careful backtracking search can
find a Hamiltonian cycle (if one exists) in time O(1.251n).
I believe that the Hamiltonian cycle problem can be summed up as the following:
Given an undirected graph G = (V, E), a
Hamiltonian circuit is a tour in G passing through
every vertex of G once and only once.
Now, what I would like to do is reduce my problem to this. My problem is:
Given a weighted undirected graph G, integer k, and vertices u, v
both in G, is there a simple path in G from u to v
with total weight of AT LEAST k?
So knowing that the Hamiltonian cycle problem is NP-complete, by reducing this problem to the Hamiltonian, this problem is also proved NP-complete. My issue is the function reducing it to Hamiltonian.
The big issue is that the Hamiltonian problem does not deal with edge weights, so I must convert my graph to one that doesn't have any weights.
On top of that, this problem has a designated start and finish (u and v), whereas the Hamiltonian finds a cycle, so any start is the same as the finish.
For (1), I am thinking along the lines of passing a graph with all simple paths of total weight LESS THAN k taken out.
For (2), I am thinking that this is not really an issue, because if there is a Hamiltonian cycle, then the simple path from u to v can be sliced out of it.
So, my real questions are:
Is my solution going to give me the right answer?
If yes, then how can I take out the edges that will produce simple paths of total weight less than k WITHOUT affecting the possibility that one of those edges may be required for the actual solution? Because if an edge e is taken out because it produces a simple path of weight < k for a subset of E, it can still be used in a simple path with a different combination of edges to produce a path of weight >= k.
Thanks!
Your reduction is in the wrong direction. To show that your problem is NP-complete, you need to reduce Hamilton Circuit to it, and that is very easy. I.e. show that every Hamilton Circuit problem can be expressed in terms of your problem variant.
More of a hint than an answer:
A unweighted graph can be interpreted as a weighted graph where each edge has weight 1. What would the cost of a Hamiltonian cycle be in a graph like that?
The part about the mismatch between cycles and paths is correct and is a problem you need to solve. Basically you need to prove that the Hamiltonian Path problem is also NP complete (a relatively straightfoward reduction to the Cycle problem)
As for the other part, you are doing things in the wrong direction. You need to show that your more complicated problem can solve the Hamiltonian Path problem (and is thus NP-hard). To do this you just basically have to use the special case of unit-cost edges.
So, I understand the problem of finding the longest simple path in a graph is NP-hard, since you could then easily solve the Hamiltonian circuit problem by setting edge weights to 1 and seeing if the length of the longest simple path equals the number of edges.
My question is: What kind of path would you get if you took a graph, found the maximum edge weight, m, replaced each edge weight w with m - w, and ran a standard shortest path algorithm on that? It's clearly not the longest simple path, since if it were, then NP = P, and I think the proof for something like that would be a bit more complicated =P.
If you could solve shortest path problems with negative weights you would find a longest path, the two are equivalent, this could be done by putting a weight of -w instead of w
The standard algorithm for negative weights is the Bellman-Ford algorithm. However the algorithm will not work if there is a cycle in the graph such that the sum of edges is negative. In the graph that you create, all such cycles have negative sum weights and so the algorithm won't work. Unless of course you have no cycles, in which case you have a tree (or a forest) and the problem is solvable via dynamic programming.
If we replace a weight of w by m-w, which guarantees that all weights will be positive, then the shortest path can be found via standard algorithms. If the shortest path P in this graph has k edges then the length is k*m-w(P) where w(P) is the length of the path with the original weights. This path is not necessarily the longest one, however, of all paths with k edges, P is the longest one.
alt text http://dl.getdropbox.com/u/317805/path2.jpg
The graph above is transformed to below using your algorithm.
The Longest path is the red line in the above graph.And depending on how ties are broken and algorithm you use, the shortest path in the transformed graph could be the blue line or the red line. So transforming graph edge weights using the constant that you mentioned yields no significant results. This is why you cannot find the longest path using the shortest path algorithms no matter how clever you are. A simpler transformation could be to negate all the edge weights and run the algorithm. I dont know if I have answered your question but as far as the path property goes the transformed graph doesnt have any useful information regarding the distance.
However this particular transformation is useful in other areas. For example you could force the algorithm to select a particular edge weight in bipatrite matching if you have more than one constraint by adding a huge constant.
Edit: I have been told to add this statement: The above graph is not just about the physical distance. They need not hold the triangle inequality. Thanks.