I came up with the following problem that I do not know the solutions to nor can I find the 'lookup' term to investigate further.
Say we have N ordered nodes (n_1,n_2....n_N) each with a fixed distance of 1 between them. So dist(n_1,n_N) = N-1. Now we are are allowed to connect any two nodes, thus effectively reducing their distance to 1. Assume we can have k such connections.
The problem is : how do we choose which nodes to connect to minimize the total distance between any two nodes ?
Is this problem a known variant of some well-studied problem ? Does an efficient solution exist for this (or a variant where we only want to minimize the max distance between any two nodes)
Thanks
You may be interested in "On the sum of all distances in a graph or digraph". That paper refers to your "total distance" as the "transmission" of a graph. Your "max distance" is generally called the "diameter" of a graph. It discusses the two, proves some properties of a graph's transmission, and shows that the transmission and diameter are independent of one another.
Naively, you've got n-choose-k options to try. That's pretty bad if n and k are large. Not too bad if one of them is small.
There is work on doing better than that. This Mathoverflow question asks about reducing the mean distance between vertices, which is proportional to the transmission of the graph. There are two answers, neither of which I can vouch for. It also refers to a paper that directly addresses this question.
Minimizing the diameter of a graph is dealt with in this paper.
You might consider addressing this question to the Math stackexchange.
Given an undirected graph G = G(V, E), how can I find the size of the largest clique in it in polynomial time? Knowing the number of edges, I could put an upper limit on the maximal clique size with
https://cs.stackexchange.com/questions/11360/size-of-maximum-clique-given-a-fixed-amount-of-edges
, and then I could iterate downwards from that upper limit to 1. Since this upper cap is O(sqrt(|E|)), I think I can check for the maximal clique size in O(sqrt(|E|) * sqrt(|E|) * sqrt(|E|)) time.
Is there a more efficient way to solve this NP-complete problem?
Finding the largest clique in a graph is the clique number of the graph and is also known as the maximum clique problem (MCP). This is one of the most deeply studied problems in the graph domain and is known to be NP-Hard so no polynomial time algorithm is expected to be found to solve it in the general case (there are particular graph configurations which do have polynomial time algorithms). Maximum clique is even hard to approximate (i.e. find a number close to the clique number).
If you are interested in exact MCP algorithms there have been a number of important improvements in the past decade, which have increased performance in around two orders of magnitude. The current leading family of algorithms are branch and bound and use approximate coloring to compute bounds. I name the most important ones and the improvement:
Branching on color (MCQ)
Static initial ordering in every subproblem (MCS and BBMC)
Recoloring: MCS
Use of bit strings to encode the graph and the main operations (BBMC)
Reduction to maximum satisfiability to improve bounds (MaxSAT)
Selective coloring (BBMCL)
and others.
It is actually a very active line of research in the scientific community.
The top algorithms are currently BBMC, MCS and I would say MaxSAT. Of these probably BBMC and its variants (which use a bit string encoding) are the current leading general purpose solvers. The library of bitstrings used for BBMC is publicly available.
Well I was thinking a bit about some dynamic programming approach and maybe I figured something out.
First : find nodes with very low degree (can be done in O(n)). Test them, if they are part of any clique and then remove them. With a little "luck" you can crush graph into few separate components and then solve each one independently (which is much much faster).
(To identify component, O(n) time is required).
Second : For each component, you can find if it makes sense to try to find any clique of given size. How? Lets say, you want to find clique of size 19. Then there has to exist at least 19 nodes with at least 19 degree. Otherwise, such clique cannot exist and you dont have to test it.
I have a bunch of points on a 2-dimensional Grid. I want to group the Points into pairs, while minimizing the sum of the euclidean distances between the points of the pairs.
Example:
Given the points:
p1: (1,1)
p2: (5,5)
p3: (1,3)
p4: (6,6)
Best solution:
pair1 = (p1,p3), distance = 2
pair2 = (p2,p4), distance = 1
Minimized total distance: 1+2 = 3
I suspect this problem might be solvable with a variant of the Hungarian Algorithm?!
What is the fastest way to solve the problem?
(Little Remark: I always should have less than 12 points.)
The problem you are trying to solve is similar to the shortest path through a fully connected (mesh) network, where you are not allowed to visit each vertex/node more than once, and you don't care about connecting the minimal pairs.
This problem is approachable when using techniques from graph theory, metric spaces, and other results from computational geometry.
This problem is similar the wiki article on the Closest pair of points problem, and the article offers some useful insights regarding Voroni diagrams and Delaunay triangulation, as well as using Recursive Divide and Conquer algorithms to solve the problem.
Note that solving the closest pair of points is not the solution, as you could have four points (A,B,C,D) in a line, where d(B,C) is least, but then you would also have d(A,D), and the sum would be larger than d(A,B) and d(C,D).
This stackoverflow question explains how to find the shortest distance between two points, and has a useful hint to skip computing the square root while comparing distances. Answers suggest using a divide and conquer approach (linear), but observe that splitting both X and Y coordinates might partition more appropriately.
This math stackexchange question addresses a similar problem, and suggests using Prim's algorithm, Kruskal's algorithm, or notes that this is a special case of the Travelling Salesman problem, which is NP-hard.
My approach would be to solve your problem (pairing the closest points) using a greedy algorithm to compute a minimal spanning tree, and then remove from the spanning tree 1/2 the edges (leaving disconnected pairs). Likely using a second (variant) of a greedy algorithm.
There are so few pairings possible for 12 or less points (about 10000 or less as pointed out in a comment), you can check all pairings by brute force and even with this solution you can solve about 10000 problems per second with 12 or less points on a modern personal computer. If you want a faster solution, you can enumerate nearest neighbors in order for each point and then just check pairings that are minimal with respect to which nearest neighbors are used for each point. In the worst-case I don't think this gives a speed-up, but for example if your 12 points come in 6 pairs of very close points (where unpaired points are far away) then you'd find the solution very quickly because the minimal pairing with respect to nearest neighbors would match together each point with its first nearest neighbor.
Problem:
Given a set of 2D points in a plane, find a set of edges E that minimizes both average trip time between any two points and the size of E: ie, by associating a cost r with each unit of trip time and a cost e per edge in the set.
I'm sure that there's a set of algorithms that deal with this problem, but I can't seem to find the right search term. I've considered starting with a complete graph and pruning, but I can't think of an efficient way to calculate the damage done by removing an edge. Any suggestions? Approximate ('good-enough') solutions welcome.
Let me know if my statement of the problem can be improved or clarified.
There is some work in the literature on spanners, which is related to what you describe (the main difference is that spanners control the maximum stretch, while you're concerned with the average). Chew's construction ("There is a planar graph almost as good as the complete graph", SoCG '86) gives an O(1)-approximation for your problem, since the triangulation has less than three times as many edges as a spanning tree (lower bound on the optimum, since the graph must be connected) and adjusts each Euclidean distance by a factor of at most sqrt(10) (hence the sqrt(10) times the average for the optimum).
Im looking for an algorithm to be used in a racing game Im making. The map/level/track is randomly generated so I need to find two locations, start and goal, that makes use of the most of the map.
The algorithm is to work inside a two dimensional space
From each point, one can only traverse to the next point in four directions; up, down, left, right
Points can only be either blocked or nonblocked, only nonblocked points can be traversed
Regarding the calculation of distance, it should not be the "bird path" for a lack of a better word. The path between A and B should be longer if there is a wall (or other blocking area) between them.
Im unsure on where to start, comments are very welcome and proposed solutions are preferred in pseudo code.
Edit: Right. After looking through gs's code I gave it another shot. Instead of python, I this time wrote it in C++. But still, even after reading up on Dijkstras algorithm, the floodfill and Hosam Alys solution, I fail to spot any crucial difference. My code still works, but not as fast as you seem to be getting yours to run. Full source is on pastie. The only interesting lines (I guess) is the Dijkstra variant itself on lines 78-118.
But speed is not the main issue here. I would really appreciate the help if someone would be kind enough to point out the differences in the algorithms.
In Hosam Alys algorithm, is the only difference that he scans from the borders instead of every node?
In Dijkstras you keep track and overwrite the distance walked, but not in floodfill, but thats about it?
Assuming the map is rectangular, you can loop over all border points, and start a flood fill to find the most distant point from the starting point:
bestSolution = { start: (0,0), end: (0,0), distance: 0 };
for each point p on the border
flood-fill all points in the map to find the most distant point
if newDistance > bestSolution.distance
bestSolution = { p, distantP, newDistance }
end if
end loop
I guess this would be in O(n^2). If I am not mistaken, it's (L+W) * 2 * (L*W) * 4, where L is the length and W is the width of the map, (L+W) * 2 represents the number of border points over the perimeter, (L*W) is the number of points, and 4 is the assumption that flood-fill would access a point a maximum of 4 times (from all directions). Since n is equivalent to the number of points, this is equivalent to (L + W) * 8 * n, which should be better than O(n2). (If the map is square, the order would be O(16n1.5).)
Update: as per the comments, since the map is more of a maze (than one with simple obstacles as I was thinking initially), you could make the same logic above, but checking all points in the map (as opposed to points on the border only). This should be in order of O(4n2), which is still better than both F-W and Dijkstra's.
Note: Flood filling is more suitable for this problem, since all vertices are directly connected through only 4 borders. A breadth first traversal of the map can yield results relatively quickly (in just O(n)). I am assuming that each point may be checked in the flood fill from each of its 4 neighbors, thus the coefficient in the formulas above.
Update 2: I am thankful for all the positive feedback I have received regarding this algorithm. Special thanks to #Georg for his review.
P.S. Any comments or corrections are welcome.
Follow up to the question about Floyd-Warshall or the simple algorithm of Hosam Aly:
I created a test program which can use both methods. Those are the files:
maze creator
find longest distance
In all test cases Floyd-Warshall was by a great magnitude slower, probably this is because of the very limited amount of edges that help this algorithm to achieve this.
These were the times, each time the field was quadruplet and 3 out of 10 fields were an obstacle.
Size Hosam Aly Floyd-Warshall
(10x10) 0m0.002s 0m0.007s
(20x20) 0m0.009s 0m0.307s
(40x40) 0m0.166s 0m22.052s
(80x80) 0m2.753s -
(160x160) 0m48.028s -
The time of Hosam Aly seems to be quadratic, therefore I'd recommend using that algorithm.
Also the memory consumption by Floyd-Warshall is n2, clearly more than needed.
If you have any idea why Floyd-Warshall is so slow, please leave a comment or edit this post.
PS: I haven't written C or C++ in a long time, I hope I haven't made too many mistakes.
It sounds like what you want is the end points separated by the graph diameter. A fairly good and easy to compute approximation is to pick a random point, find the farthest point from that, and then find the farthest point from there. These last two points should be close to maximally separated.
For a rectangular maze, this means that two flood fills should get you a pretty good pair of starting and ending points.
I deleted my original post recommending the Floyd-Warshall algorithm. :(
gs did a realistic benchmark and guess what, F-W is substantially slower than Hosam Aly's "flood fill" algorithm for typical map sizes! So even though F-W is a cool algorithm and much faster than Dijkstra's for dense graphs, I can't recommend it anymore for the OP's problem, which involves very sparse graphs (each vertex has only 4 edges).
For the record:
An efficient implementation of Dijkstra's algorithm takes O(Elog V) time for a graph with E edges and V vertices.
Hosam Aly's "flood fill" is a breadth first search, which is O(V). This can be thought of as a special case of Dijkstra's algorithm in which no vertex can have its distance estimate revised.
The Floyd-Warshall algorithm takes O(V^3) time, is very easy to code, and is still the fastest for dense graphs (those graphs where vertices are typically connected to many other vertices). But it's not the right choice for the OP's task, which involves very sparse graphs.
Raimund Seidel gives a simple method using matrix multiplication to compute the all-pairs distance matrix on an unweighted, undirected graph (which is exactly what you want) in the first section of his paper On the All-Pairs-Shortest-Path Problem in Unweighted Undirected Graphs
[pdf].
The input is the adjacency matrix and the output is the all-pairs shortest-path distance matrix. The run-time is O(M(n)*log(n)) for n points where M(n) is the run-time of your matrix multiplication algorithm.
The paper also gives the method for computing the actual paths (in the same run-time) if you need this too.
Seidel's algorithm is cool because the run-time is independent of the number of edges, but we actually don't care here because our graph is sparse. However, this may still be a good choice (despite the slightly-worse-than n^2 run-time) if you want the all pairs distance matrix, and this might also be easier to implement and debug than floodfill on a maze.
Here is the pseudocode:
Let A be the nxn (0-1) adjacency matrix of an unweighted, undirected graph, G
All-Pairs-Distances(A)
Z = A * A
Let B be the nxn matrix s.t. b_ij = 1 iff i != j and (a_ij = 1 or z_ij > 0)
if b_ij = 1 for all i != j return 2B - A //base case
T = All-Pairs-Distances(B)
X = T * A
Let D be the nxn matrix s.t. d_ij = 2t_ij if x_ij >= t_ij * degree(j), otherwise d_ij = 2t_ij - 1
return D
To get the pair of points with the greatest distance we just return argmax_ij(d_ij)
Finished a python mockup of the dijkstra solution to the problem.
Code got a bit long so I posted it somewhere else: http://refactormycode.com/codes/717-dijkstra-to-find-two-points-furthest-away-from-each-other
In the size I set, it takes about 1.5 seconds to run the algorithm for one node. Running it for every node takes a few minutes.
Dont seem to work though, it always displays the topleft and bottomright corner as the longest path; 58 tiles. Which of course is true, when you dont have obstacles. But even adding a couple of randomly placed ones, the program still finds that one the longest. Maybe its still true, hard to test without more advanced shapes.
But maybe it can at least show my ambition.
Ok, "Hosam's algorithm" is a breadth first search with a preselection on the nodes.
Dijkstra's algorithm should NOT be applied here, because your edges don't have weights.
The difference is crucial, because if the weights of the edges vary, you need to keep a lot of options (alternate routes) open and check them with every step. This makes the algorithm more complex.
With the breadth first search, you simply explore all edges once in a way that garantuees that you find the shortest path to each node. i.e. by exploring the edges in the order you find them.
So basically the difference is Dijkstra's has to 'backtrack' and look at edges it has explored before to make sure it is following the shortest route, while the breadth first search always knows it is following the shortest route.
Also, in a maze the points on the outer border are not guaranteed to be part of the longest route.
For instance, if you have a maze in the shape of a giant spiral, but with the outer end going back to the middle, you could have two points one at the heart of the spiral and the other in the end of the spiral, both in the middle!
So, a good way to do this is to use a breadth first search from every point, but remove the starting point after a search (you already know all the routes to and from it).
Complexity of breadth first is O(n), where n = |V|+|E|. We do this once for every node in V, so it becomes O(n^2).
Your description sounds to me like a maze routing problem. Check out the Lee Algorithm. Books about place-and-route problems in VLSI design may help you - Sherwani's "Algorithms for VLSI Physical Design Automation" is good, and you may find VLSI Physical Design Automation by Sait and Youssef useful (and cheaper in its Google version...)
If your objects (points) do not move frequently you can perform such a calculation in a much shorter than O(n^3) time.
All you need is to break the space into large grids and pre-calculate the inter-grid distance. Then selecting point pairs that occupy most distant grids is a matter of simple table lookup. In the average case you will need to pair-wise check only a small set of objects.
This solution works if the distance metrics are continuous. Thus if, for example there are many barriers in the map (as in mazes), this method might fail.