First problem was that i couldn't find an algorithm that,given an directed graph as input, gives as output a list of all cycles present in the graph. (This problem should be NP-complete).
After thinking about the problem for a while I realized that what probably I really needed was to find the circuit (it can have duplicate vertex but not duplicate edges) with max weight (sum of the weights of the edges).
It should be a NP-complete problem too, and a way to proceed could be to list all circuits present in the graph and then to sort them by sum of edge weights.
Do you know some algorithm that gives as output a list of all circuits present in a directed graph? Or one that find the circuit with max weight ?
I have found this, but it's not exactly what i need.
http://epubs.siam.org/doi/abs/10.1137/0205007
However do you confirm the computational complexity of these problems ?
You could do sth. like here: Finding all cycles in a directed graph.
You do this search for every node and parallelize that so as to reduce runtime. Afterwards you apply an efficient sort-algorithm to your list of cycle where each cycle is a list of nodes. Sorting algorithms may me Mergesort or Quicksort for instance, but choose which ever u prefer..
I hope that brings u forward.
I have a undirected graph with about 100 nodes and about 200 edges. One node is labelled 'start', one is 'end', and there's about a dozen labelled 'mustpass'.
I need to find the shortest path through this graph that starts at 'start', ends at 'end', and passes through all of the 'mustpass' nodes (in any order).
( http://3e.org/local/maize-graph.png / http://3e.org/local/maize-graph.dot.txt is the graph in question - it represents a corn maze in Lancaster, PA)
Everyone else comparing this to the Travelling Salesman Problem probably hasn't read your question carefully. In TSP, the objective is to find the shortest cycle that visits all the vertices (a Hamiltonian cycle) -- it corresponds to having every node labelled 'mustpass'.
In your case, given that you have only about a dozen labelled 'mustpass', and given that 12! is rather small (479001600), you can simply try all permutations of only the 'mustpass' nodes, and look at the shortest path from 'start' to 'end' that visits the 'mustpass' nodes in that order -- it will simply be the concatenation of the shortest paths between every two consecutive nodes in that list.
In other words, first find the shortest distance between each pair of vertices (you can use Dijkstra's algorithm or others, but with those small numbers (100 nodes), even the simplest-to-code Floyd-Warshall algorithm will run in time). Then, once you have this in a table, try all permutations of your 'mustpass' nodes, and the rest.
Something like this:
//Precomputation: Find all pairs shortest paths, e.g. using Floyd-Warshall
n = number of nodes
for i=1 to n: for j=1 to n: d[i][j]=INF
for k=1 to n:
for i=1 to n:
for j=1 to n:
d[i][j] = min(d[i][j], d[i][k] + d[k][j])
//That *really* gives the shortest distance between every pair of nodes! :-)
//Now try all permutations
shortest = INF
for each permutation a[1],a[2],...a[k] of the 'mustpass' nodes:
shortest = min(shortest, d['start'][a[1]]+d[a[1]][a[2]]+...+d[a[k]]['end'])
print shortest
(Of course that's not real code, and if you want the actual path you'll have to keep track of which permutation gives the shortest distance, and also what the all-pairs shortest paths are, but you get the idea.)
It will run in at most a few seconds on any reasonable language :)
[If you have n nodes and k 'mustpass' nodes, its running time is O(n3) for the Floyd-Warshall part, and O(k!n) for the all permutations part, and 100^3+(12!)(100) is practically peanuts unless you have some really restrictive constraints.]
run Djikstra's Algorithm to find the shortest paths between all of the critical nodes (start, end, and must-pass), then a depth-first traversal should tell you the shortest path through the resulting subgraph that touches all of the nodes start ... mustpasses ... end
This is two problems... Steven Lowe pointed this out, but didn't give enough respect to the second half of the problem.
You should first discover the shortest paths between all of your critical nodes (start, end, mustpass). Once these paths are discovered, you can construct a simplified graph, where each edge in the new graph is a path from one critical node to another in the original graph. There are many pathfinding algorithms that you can use to find the shortest path here.
Once you have this new graph, though, you have exactly the Traveling Salesperson problem (well, almost... No need to return to your starting point). Any of the posts concerning this, mentioned above, will apply.
Actually, the problem you posted is similar to the traveling salesman, but I think closer to a simple pathfinding problem. Rather than needing to visit each and every node, you simply need to visit a particular set of nodes in the shortest time (distance) possible.
The reason for this is that, unlike the traveling salesman problem, a corn maze will not allow you to travel directly from any one point to any other point on the map without needing to pass through other nodes to get there.
I would actually recommend A* pathfinding as a technique to consider. You set this up by deciding which nodes have access to which other nodes directly, and what the "cost" of each hop from a particular node is. In this case, it looks like each "hop" could be of equal cost, since your nodes seem relatively closely spaced. A* can use this information to find the lowest cost path between any two points. Since you need to get from point A to point B and visit about 12 inbetween, even a brute force approach using pathfinding wouldn't hurt at all.
Just an alternative to consider. It does look remarkably like the traveling salesman problem, and those are good papers to read up on, but look closer and you'll see that its only overcomplicating things. ^_^ This coming from the mind of a video game programmer who's dealt with these kinds of things before.
This is not a TSP problem and not NP-hard because the original question does not require that must-pass nodes are visited only once. This makes the answer much, much simpler to just brute-force after compiling a list of shortest paths between all must-pass nodes via Dijkstra's algorithm. There may be a better way to go but a simple one would be to simply work a binary tree backwards. Imagine a list of nodes [start,a,b,c,end]. Sum the simple distances [start->a->b->c->end] this is your new target distance to beat. Now try [start->a->c->b->end] and if that's better set that as the target (and remember that it came from that pattern of nodes). Work backwards over the permutations:
[start->a->b->c->end]
[start->a->c->b->end]
[start->b->a->c->end]
[start->b->c->a->end]
[start->c->a->b->end]
[start->c->b->a->end]
One of those will be shortest.
(where are the 'visited multiple times' nodes, if any? They're just hidden in the shortest-path initialization step. The shortest path between a and b may contain c or even the end point. You don't need to care)
Andrew Top has the right idea:
1) Djikstra's Algorithm
2) Some TSP heuristic.
I recommend the Lin-Kernighan heuristic: it's one of the best known for any NP Complete problem. The only other thing to remember is that after you expanded out the graph again after step 2, you may have loops in your expanded path, so you should go around short-circuiting those (look at the degree of vertices along your path).
I'm actually not sure how good this solution will be relative to the optimum. There are probably some pathological cases to do with short circuiting. After all, this problem looks a LOT like Steiner Tree: http://en.wikipedia.org/wiki/Steiner_tree and you definitely can't approximate Steiner Tree by just contracting your graph and running Kruskal's for example.
Considering the amount of nodes and edges is relatively finite, you can probably calculate every possible path and take the shortest one.
Generally this known as the travelling salesman problem, and has a non-deterministic polynomial runtime, no matter what the algorithm you use.
http://en.wikipedia.org/wiki/Traveling_salesman_problem
The question talks about must-pass in ANY order. I have been trying to search for a solution about the defined order of must-pass nodes. I found my answer but since no question on StackOverflow had a similar question I'm posting here to let maximum people benefit from it.
If the order or must-pass is defined then you could run dijkstra's algorithm multiple times. For instance let's assume you have to start from s pass through k1, k2 and k3 (in respective order) and stop at e. Then what you could do is run dijkstra's algorithm between each consecutive pair of nodes. The cost and path would be given by:
dijkstras(s, k1) + dijkstras(k1, k2) + dijkstras(k2, k3) + dijkstras(k3, 3)
How about using brute force on the dozen 'must visit' nodes. You can cover all the possible combinations of 12 nodes easily enough, and this leaves you with an optimal circuit you can follow to cover them.
Now your problem is simplified to one of finding optimal routes from the start node to the circuit, which you then follow around until you've covered them, and then find the route from that to the end.
Final path is composed of :
start -> path to circuit* -> circuit of must visit nodes -> path to end* -> end
You find the paths I marked with * like this
Do an A* search from the start node to every point on the circuit
for each of these do an A* search from the next and previous node on the circuit to the end (because you can follow the circuit round in either direction)
What you end up with is a lot of search paths, and you can choose the one with the lowest cost.
There's lots of room for optimization by caching the searches, but I think this will generate good solutions.
It doesn't go anywhere near looking for an optimal solution though, because that could involve leaving the must visit circuit within the search.
One thing that is not mentioned anywhere, is whether it is ok for the same vertex to be visited more than once in the path. Most of the answers here assume that it's ok to visit the same edge multiple times, but my take given the question (a path should not visit the same vertex more than once!) is that it is not ok to visit the same vertex twice.
So a brute force approach would still apply, but you'd have to remove vertices already used when you attempt to calculate each subset of the path.
I'm trying to solve the standard bipartization problem, i.e., find a subset of the edges such that the output graph is a bipartite graph.
My additional constraints are:
The number of vertices on each side must be equal.
Each vertex has exactly 1 edge.
In fact, it would suffice to know whether such a subset exists at all - I don't really need the construction itself.
Optimally, the algorithm should be fast as I need to run it for O(400) nodes repeatedly.
If each vertex is to be incident on exactly one edge, it seems what you want is a matching. If so, Edmonds's blossom algorithm will do the job. I haven't used an implementation of the algorithm to recommend. You might check out http://www.algorithmic-solutions.com/leda/ledak/index.htm
I was thinking about an extension to the Shortest Hamiltonian Path (SHP) problem, and I couldn't find a way of solving it. I know it is NP-complete, but I figured I'd ask here for ideas, since I do not want to simply brute force the problem.
The extension is fairly simply: Given an undirected, complete, weighted graph with n vertices, find the shortest hamiltonian path with end vertices v and u.
So, bruteforce would still take O(n!) time, since the remaining n-2 vertices can be visited in (n-2)! ways. I was trying to find a way to maybe solve this slightly faster. My efforts for finding a way to solve this problem in a beneficial manner has so far been fruitless.
Would anyone have an idea how to exploit the knowledge of the end-vertices? Preferably explained alongside some pseudocode. It is required for the solution found to be optimal.
I guess it could be solved by integer programming, since the knowledge of end nodes are fairly limiting, and makes it easy to avoid cycles, but it wouldn't really exploit the composition of the problem.
If you want to find the shortest path to connect all nodes, then you should look at travelling salesman algorithms. I don't exactly see why you approach it as an HSP. The only reason I can think of is that you want to specify your starting cities, but it should be easy to fix that (if you need that I can post it) by changing your graph a bit.
edit: adding how to tweak your graph
Add 1 node (call it E) and only connect it to your starting and ending nodes. A TSP will compute a solution to your problem by connecting all your nodes. As E is only reachable by going from start to E and then to end, the solution (a cycle yes) will contain start - E - end. Then you remove the 2 edges to and from E and you have the optimal solution. Within the TSP, the path from start to end will be optimal.
You can use a metaheuristic algorithm. There are many kinds of them (Local search, constructive search, etc.). One of them could be based on:
For each x belonging to the set of vertices X:
- Find the set of closest vertices to x C.
- Filter the set C in order to include one of them in the path P.
Repeat until all vertices of X have been included in the path P.
End.
Someone somewhere has had to solve this problem. I can find many a great website explaining this problem and how to solve it. While I'm sure they are well written and make sense to math whizzes, that isn't me. And while I might understand in a vague sort of way, I do not understand how to turn that math into a function that I can use.
So I beg of you, if you have a function that can do this, in any language, (sure even fortran or heck 6502 assembler) - please help me out.
prefer an analytical to iterative solution
EDIT: Meant to specify that its a cubic bezier I'm trying to work with.
What you're asking for is the inverse of the arc length function. So, given a curve B, you want a function Linv(len) that returns a t between 0 and 1 such that the arc length of the curve between 0 and t is len.
If you had this function your problem is really easy to solve. Let B(0) be the first point. To find the next point, you'd simply compute B(Linv(w)) where w is the "equal arclength" that you refer to. To get the next point, just evaluate B(Linv(2*w)) and so on, until Linv(n*w) becomes greater than 1.
I've had to deal with this problem recently. I've come up with, or come across a few solutions, none of which are satisfactory to me (but maybe they will be for you).
Now, this is a bit complicated, so let me just give you the link to the source code first:
http://icedtea.classpath.org/~dlila/webrevs/perfWebrev/webrev/raw_files/new/src/share/classes/sun/java2d/pisces/Dasher.java. What you want is in the LengthIterator class. You shouldn't have to look at any other parts of the file. There are a bunch of methods that are defined in another file. To get to them just cut out everything from /raw_files/ to the end of the URL. This is how you use it. Initialize the object on a curve. Then to get the parameter of a point with arc length L from the beginning of the curve just call next(L) (to get the actual point just evaluate your curve at this parameter, using deCasteljau's algorithm, or zneak's suggestion). Every subsequent call of next(x) moves you a distance of x along the curve compared to your last position. next returns a negative number when you run out of curve.
Explanation of code: so, I needed a t value such that B(0) to B(t) would have length LEN (where LEN is known). I simply flattened the curve. So, just subdivide the curve recursively until each curve is close enough to a line (you can test for this by comparing the length of the control polygon to the length of the line joining the end points). You can compute the length of this sub-curve as (controlPolyLength + endPointsSegmentLen)/2. Add all these lengths to an accumulator, and stop the recursion when the accumulator value is >= LEN. Now, call the last subcurve C and let [t0, t1] be its domain. You know that the t you want is t0 <= t < t1, and you know the length from B(0) to B(t0) - call this value L0t0. So, now you need to find a t such that C(0) to C(t) has length LEN-L0t0. This is exactly the problem we started with, but on a smaller scale. We could use recursion, but that would be horribly slow, so instead we just use the fact that C is a very flat curve. We pretend C is a line, and compute the point at t using P=C(0)+((LEN-L0t0)/length(C))*(C(1)-C(0)). This point doesn't actually lie on the curve because it is on the line C(0)->C(1), but it's very close to the point we want. So, we just solve Bx(t)=Px and By(t)=Py. This is just finding cubic roots, which has a closed source solution, but I just used Newton's method. Now we have the t we want, and we can just compute C(t), which is the actual point.
I should mention that a few months ago I skimmed through a paper that had another solution to this that found an approximation to the natural parameterization of the curve. The author has posted a link to it here: Equidistant points across Bezier curves