Let be an array of objects [a, b, c, d, ...] and a function distance(x, y) that gives a numeric value showing how 'different' are objects x and y.
I'm looking for an algorithm to find the subset of the array of length n that maximizes the minimum difference between that subset element.
Of course, I can't simply sort the array by the minimum of the differences with other elements and take the n highest entries, since removing an element can very well change the distances. For instance, if a=b, then removing a means the minimum distance of b with another element will change dramatically.
So far, the only solution I could find was wether to iteratively remove the element with the lowest minimum distance and re-calculate the distance at each iteration, until only n elements are left, or, vice-versa, to iteratively pick new elements, recalculate the distances, add the new pick or replace an existing one based on the distance minimums.
Does anybody know how I could get the same results without those iterations?
PS: here is an example, the matrix shows the 'distance' between each element...
a b c d
a - 1 3 2
b 1 - 4 2
c 3 4 - 5
d 2 2 5 -
If we'd keep only 2 elements, that would be c and d; if we'd keep 3, that would be a or b, c and d.
This problem is NP-hard, so no-one knows an efficient (polynomial time) algorithm for solving it.
Here's a quick sketch that it is NP-hard, by reduction from CLIQUE.
Suppose we have an instance of CLIQUE in the form of a graph G and a number n and we want to know whether there is a clique of size n in G. Construct a distance matrix d such that d(i, j) = 1 if vertices i and j are connected in G, or 0 if they are not. Then find a subset of the vertices of G of size n that maximizes the minimum distance between elements (your problem). If the minimum distance between vertices in this subset is 1, then G has a clique of size n; otherwise it does not.
As Gareth said this is an NP-hard problem, however there has been a lot of research into solving these kind of problems and as such better methods than brute force have been found. Unfortunately this is such a large area that you could spend forever looking at the possible implementations of a solutions.
However if you are interested in a heuristic way of solving this I would suggest looking into Ant Colony Optimization (ACO) which has proven fairly effective at finding optimum paths within graphs.
Related
I want to compute a one-to-one assignment, from A to B where A and B have n elements. Each element of A will be assigned to one element in B. There is an associated cost between each element of A and B. The goal is to minimize the maximum cost of the assignment.
How to compute such an assignment using the cost matrix? I need to select an element from each row/column such that minimax strategy is achieved.
I was able to achieve this by listing all possible assignments and calculating their maximum costs and then selecting the one with the least maximum cost, but I wonder if there's a better approach.
The easiest is probably as a MIP (Mixed-Integer Programming) model:
min z
z >= c[i,j]*x[i,j] ∀ i,j
sum(i, x[i,j]) = 1 ∀ j
sum(j, x[i,j]) = 1 ∀ i
x[i,j] ∈ {0,1}
This can be solved with any MIP solver.
A brute force polynomial cost algorithm based on repeated application of max matching in bipartite graph:
Order the pairs by assignment cost. Take n cheapest pairs. This determines a bipartite graph with A and B as vertices (pairs are edges). Try to find a matching in this graph. If found, it determines the optimal one-to-one assignment.
If not found, add the next pair. Repeat until a matching is found.
I suspect this can be optimized by not running the matching algorithm from the start every time you add an edge but rather keeping the maximum matching and checking if it can be augmented with the newly added edge.
There is an grid consisting of h * w (h, w <= 200) pixels, every pixel is represented by a value, we want to find the largest continuous region.
A continuous region is defined in this way:
Given a point P(x, y), The connected region must include this point.
There exists reference point R(x, y) of value v, any point in the connected region must be connected to this point. Also, there is a value g_critical(g_critical <= 100000). Let the value of a point in the connected region be v, the difference
of u and v must be smaller or equal than g_critical.
The question is to find the size of the largest connected region.
For example the grid. h = 5, w = 5, g_critical = 3, P(x, y) = (2, 4)
1 3 7 9 2
2 5 6 6 8
3 5 9 3 6
2 7 3 2 9
In this case, the bold region is the largest connected region. Notice that R(x, y) is chosen at (2, 3) or (2, 2) in this case. The size of the region is 14.
I have rephrased the question a bit so it is shorter. So if there is any ambiguity, please point it out in the comment. This question is also in our private judge so I am unable to share the problem source here.
My attempt
I have tried to loop through every cell, consider it as the R point and use bfs to find the connected region attached to it. Then, check if P is contained in the region.
The complexity is O(h * h * w * w), which is too large. So any way to optimize it?
I am guessing that maybe starting with p will help, but I am not sure how I should do it. Maybe there is some kind of flood fill algorithms that allow me to do it?
Thanks in advance.
There's an O(h w √(g_critical) α(h w))-time algorithm (where α is the inverse Ackermann function, constant for practical purposes) that uses a disjoint set data structure with an "undo" operation and a variant of Mo's trick. The idea is, decompose the interval [v − g_critical, v] into about √g_critical subintervals of length about √g_critical. For each subinterval [a, b], prepare a disjoint set data structure representing the components of the matrix where the allowed values are [b, a + 2 g_critical]. Then for each c in [a, b], extend the disjoint set with points whose values lie in [c, b) and (a + 2 g_critical, c + 2 g_critical] and report the number of nodes in the component of P(x,y), then undo these operations (keep a stack of the writes made to the structure, with original values; then pop each one, writing the original values).
There's also an O(h w log(h w))-time algorithm that you're not going to like because it uses dynamic trees. (Sleator–Tarjan's 1985 construction based on splay trees is the simplest and works OK here.) Posting it mainly in case it inspires a more practical approach.
The high-level idea is a kinetic algorithm that "slides" the interval of allowed values over the at most g_critical + 1 possibilities, repeatedly reporting and taking the max over the size of the connected component containing P.
To do this, we need to maintain the maximum spanning forest on a derived graph. Given a node-weighted undirected graph, construct a new, edge-weighted graph by subdividing each edge and setting the weight of each new edge to the weight of the old node to which it is incident. Deleting the lightest nodes in the graph is straightforward – all paths avoid these nodes if possible, so the new maximum spanning forest won't have any more edges. To add the lightest nodes not in the graph, try to link their incident edges one at a time. If a cycle would form, evert one endpoint, query the path minimum from the other endpoint, and unlink that minimum.
To report the size of the component containing P we need another decoration that captures the size of the concrete subtree (as opposed to the represented subtree) of each node. The details get a bit gnarly.
Here's some heuristics which might help:
First some pre-processing in O(h*w*log(h*w)):
Store matrix values in an array, sort it
Now every value in array is a possible candidate for point R
Also maximum component will be of values in range [R-critical, R+critical]
So we can estimate size of component (in best case) with simple binary search
Now some heuristics:
Sort array again this time by estimated component size in descending order
Now try BFS in this order if estimated size is bigger than currently found best size
I have a set of N points (in particular this point are binary string) and for each of them I have a discrete metric (the Hamming distance) such that given two points, i and j, Dij is the distance between the i-th and the j-th point.
I want to find a subset of k elements (with k < N of course) such that the distance between this k points is the maximum as possibile.
In other words what I want is to find a sort of "border points" that cover the maximum area in the space of the points.
If k = 2 the answer is trivial because I can try to search the two most distant element in the matrix of distances and these are the two points, but how I can generalize this question when k>2?
Any suggest? It's a NP-hard problem?
Thanks for the answer
One generalisation would be "find k points such that the minimum distance between any two of these k points is as large as possible".
Unfortunately, I think this is hard, because I think if you could do this efficiently you could find cliques efficiently. Suppose somebody gives you a matrix of distances and asks you to find a k-clique. Create another matrix with entries 1 where the original matrix had infinity, and entries 1000000 where the original matrix had any finite distance. Now a set of k points in the new matrix where the minimum distance between any two points in that set is 1000000 corresponds to a set of k points in the original matrix which were all connected to each other - a clique.
This construction does not take account of the fact that the points correspond to bit-vectors and the distance between them is the Hamming distance, but I think it can be extended to cope with this. To show that a program capable of solving the original problem can be used to find cliques I need to show that, given an adjacency matrix, I can construct a bit-vector for each point so that pairs of points connected in the graph, and so with 1 in the adjacency matrix, are at distance roughly A from each other, and pairs of points not connected in the graph are at distance B from each other, where A > B. Note that A could be quite close to B. In fact, the triangle inequality will force this to be the case. Once I have shown this, k points all at distance A from each other (and so with minimum distance A, and a sum of distances of k(k-1)A/2) will correspond to a clique, so a program finding such points will find cliques.
To do this I will use bit-vectors of length kn(n-1)/2, where k will grow with n, so the length of the bit-vectors could be as much as O(n^3). I can get away with this because this is still only polynomial in n. I will divide each bit-vector into n(n-1)/2 fields each of length k, where each field is responsible for representing the connection or lack of connection between two points. I claim that there is a set of bit-vectors of length k so that all of the distances between these k-long bit-vectors are roughly the same, except that two of them are closer together than the others. I also claim that there is a set of bit-vectors of length k so that all of the distances between them are roughly the same, except that two of them are further apart than the others. By choosing between these two different sets, and by allocating the nearer or further pair to the two points owning the current bit-field of the n(n-1)/2 fields within the bit-vector I can create a set of bit-vectors with the required pattern of distances.
I think these exist because I think there is a construction that creates such patterns with high probability. Create n random bit-vectors of length k. Any two such bit-vectors have an expected Hamming distance of k/2 with a variance of k/4 so a standard deviation of sqrt(k)/2. For large k we expect the different distances to be reasonably similar. To create within this set two points that are very close together, make one a copy of the other. To create two points that are very far apart, make one the not of the other (0s in one where the other has 1s and vice versa).
Given any two points their expected distance from each other will be (n(n-1)/2 - 1)k/2 + k (if they are supposed to be far apart) and (n(n-1)/2 -1)k/2 (if they are supposed to be close together) and I claim without proof that by making k large enough the expected difference will triumph over the random variability and I will get distances that are pretty much A and pretty much B as I require.
#mcdowella, I think that probably I don't explain very well my problem.
In my problem I have binary string and for each of them I can compute the distance to the other using the Hamming distance
In this way I have a distance matrix D that has a finite value in each element D(i,j).
I can see this distance matrix like a graph: infact, each row is a vertex in the graph and in the column I have the weight of the arc that connect the vertex Vi to the vertex Vj.
This graph, for the reason that I explain, is complete and it's a clique of itself.
For this reason, if i pick at random k vertex from the original graph I obtain a subgraph that is also complete.
From all the possible subgraph with order k I want to choose the best one.
What is the best one? Is a graph such that the distance between the vertex as much large but also much uniform as possible.
Suppose that I have two vertex v1 and v2 in my subgraph and that their distance is 25, and I have three other vertex v3, v4, v5, such that
d(v1, v3) = 24, d(v1, v4) = 7, d(v2, v3) = 5, d(v2, v4) = 22, d(v1, v5) = 14, d(v1, v5) = 14
With these distance I have that v3 is too far from v1 but is very near to v2, and the opposite situation for v4 that is too far from v2 but is near to v1.
Instead I prefer to add the vertex v5 to my subgraph because it is distant to the other two in a more uniform way.
I hope that now my problem is clear.
You think that your formulation is already correct?
I have claimed that the problem of finding k points such that the minimum distance between these points, or the sum of the distances between these points, is as large as possible is NP-complete, so there is no polynomial time exact answer. This suggests that we should look for some sort of heuristic solution, so here is one, based on an idea for clustering. I will describe it for maximising the total distance. I think it can be made to work for maximising the minimum distance as well, and perhaps for other goals.
Pick k arbitrary points and note down, for each point, the sum of the distances to the other points. For each other point in the data, look at the sum of the distances to the k chosen points and see if replacing any of the chosen points with that point would increase the sum. If so, replace whichever point increases the sum most and continue. Keep trying until none of the points can be used to increase the sum. This is only a local optimum, so repeat with another set of k arbitrary/random points in the hope of finding a better one until you get fed up.
This inherits from its clustering forebear the following property, which might at least be useful for testing: if the points can be divided into k classes such that the distance between any two points in the same class is always less than the distance between any two points in different classes then, when you have found k points where no local improvement is possible, these k points should all be from different classes (because if not, swapping out one of a pair of points from the same class would increase the sum of distances between them).
This problem is known as the MaxMin Diversity Problem (MMDP). It is known to be NP-hard. However, there are algorithms for giving good approximate solutions in reasonable time, such as this one.
I'm answering this question years after it was asked because I was looking for algorithms to solve the same problem, and had trouble even finding out what to call it.
I have two sets, A and B, with N and M points in R^n respectively. I know that N < M always.
The distance between two points, P and Q, is denoted by d( P,Q ). As the problem is generic, this distance could be any function (e.g. Euclidean distance).
I want to find the closest subset of B to A. Mathematically I would say, I want to find the subset C of B with size N with the minimal global distance to A. The global distance between A and C is given by
D(A,C) = min([sum(d(P_i,Q_i),i=1,N) with P_i in A and Q_i in C* for C* in Permutations of C])
I've been thinking about this problem and I did an algorithm that get a local optimum, but not necessarily the optimal:
Step 1) Find the nearest point of each point of A in B. If there are no points repeated, I found the optimal subset and finish the algorithm. However, if there are points repeated, go to step 2.
Step 2) Compare their distances (of course I compare the distance between points with the same closest point). The point with the minimal distance keeps the point previously found and the others change their desired point for the "next" closest point that have not been selected for another point yet.
Step 3) Check if all the points are different. If they are, finish. If not, go back to step 2.
Any idea? Trying all the combinations is not a good one (I should calculate M!/(M-N)! global distances)
If M = N, this problem could be formulated as minimum-weight perfect matching in a bipartite graph or, in other words, an assignment problem. A well-known method for solving an assignment problem is the Hungarian algorithm.
To make Hungarian algorithm applicable in the case N < M, you could extend set A with (M-N) additional elements (each having zero distance to all elements of B).
Situation
Suppose we are given n points on the unit square [0, 1]x[0, 1] and a positive real number r. We define the graph G(point 1, point 2, ..., point n, r) as the graph on vertices {1, 2, ..., n} such that there is an edge connecting two given vertices if and only if the distance between the corresponding points is less than or equal to r. (You can think of the points as transmitters, which can communicate with each other as long as they are within range r.)
Given n points on the unit square [0, 1]x[0, 1], we define the connectivity distance as the smallest possible r for which G(point 1, point 2, ..., point n, r) is connected.
Problem 1) find an algorithm that determines if G(point 1, point 2, ..., point n, r) is connected
Problem 2) find an algorithm that finds the connectivity distance for any n given points
My partial solution
I have an algorithm (Algorithm 1) in mind for problem 1. I haven't implemented it yet, but I'm convinced it works. (Roughly, the idea is to start from vertex 1, and try to reach all other vertices through the edges. I think it would be somewhat similar to this.)
All that remains is problem 2. I also have an algorithm in mind for this one. However, I think it is not efficient time wise. I'll try to explain how it works:
You must first convince yourself that the connectivity distance rmin is necessarily the distance between two of the given points, say p and q. Hence, there are at most *n**(n-1)/2 possible values for rmin.
So, first, my algorithm would measure all *n**(n-1)/2 distances and store them (in an array in C, for instance) in increasing order. Then it would use Algorithm 1 to test each stored value (in increasing order) to see if the graph is connected with such range. The first value that does the job is the answer, rmin.
My question is: is there a better (time wise) algorithm for problem 2?
Remarks: the points will be randomly generated (something like 10000 of them), so that's the type of thing the algorithm is supposed to solve "quickly". Furthermore, I'll implement this in C. (If that makes any difference.)
Here is an algorithm which requires O(n2) time and O(n) space.
It's based on the observation that if you partition the points into two sets, then the connectivity distance cannot be less than the distance of the closest pair of points one from each set in the partition. In other words, if we build up the connected graph by always adding the closest point, then the largest distance we add will be the connectivity distance.
Create two sets, A and B. Put a random point into A and all the remaining points into B.
Initialize r (the connectivity distance) to 0.
Initialize a map M with the distance to every point in B of the point in A.
While there are still points in B:
Select the point b in B whose distance M[b] is the smallest.
If M[b] is greater than r, set r to M[b]
Remove b from B and add it to A.
For each point p in M:
If p is b, remove it from M.
Otherwise, if the distance from b to p is less than M[p], set M[p] to that distance.
When all the points are in A, r will be the connectivity distance.
Each iteration of the while loop takes O(|B|) time, first to find the minimum value in M (whose size is equal to the size of B); second, to update the values in M. Since a point is moved from B to A in each iteration, there will be exactly n iterations, and thus the total execution time is O(n2).
The algorithm presented above is an improvement to a previous algorithm, which used an (unspecified) solution to the bichromatic closest pair (BCP) problem to recompute the closest neighbour to A in every cycle. Since there is an O(n log n) solution to BCP, this implied a solution to the original problem in O(n2 log n). However, maintaining and updating the list of closest points is actually much simpler, and only requires O(n). Thanks to #LajosArpad for a question which triggered this line of thought.
I think your ideas are reasonably good, however, I have an improvement for you.
In fact you build up an array based on measurement and you sort your array. Very nice. At least with not too many points.
The number of n(n-1)/2 is a logical consequence of your pairing requirement. So, for 10000 elements, you will have 49995000 elements. You will need to increase significantly the speed! Also, this number of elements would eat a lot of your memory storage.
How can you achieve greater speed?
First of all, don't build arrays. You already have an array. Secondly, you can easily solve your problem by traversing. Let's suppose you have a function, which determines whether a given distance is enough to connect all the nodes, lets call this function "valid". It is not enough, because you need to find the minimal possible value. So, if you don't have more information about the nodes prior the execution of the algorithm, then my suggestion is this solution:
lowerBound <- 0
upperBound <- infinite
i <- 0
while i < numberOfElements do
j <- i + 1
while j < numberOfElements do
distance <- d(elements[i], elements[j])
if distance < upperBound and distance > lowerBound then
if valid(distance) then
upperBound <- distance
else
lowerBound <- distance
end if
end if
j <- j + 1
end while
i <- i + 1
end while
After traversing all the elements the value of upperBound will hold the smallest distance which still connects the network. You didn't store all the distances, as they were far too many and you have solved your problem in a single cycle. I hope you find my answer helpful.
If some distance makes graph connected, any larger distance would make it connected too. To find minimal connecting distance just sort all distances and use binary search.
Time complexity is O(n^2*log n), space complexity is O(n^2).
You can start with some small distance d then check for connectivity. If the Graph is connected, you're done, if not, increment d by a small distance then check again for connectivity.
You also need a clever algorithm to avoid O(N^2) in case N is big.