如何找到一组独特的最接近的点对? [英] How to find a unique set of closest pairs of points?

问题描述

A 和 B 分别是 m 和 n 点的集合，其中 m< =n .我想从 B 中找到一组名为 C 的 m unique 个点，其中两个点之间的距离之和所有 [A(i)，C(i)] 对都是最小的.

A and B are sets of m and n points respectively, where m<=n. I want to find a set of m unique points from B, named C, where the sum of distances between all [A(i), C(i)] pairs is the minimal.

要解决此问题而没有 uniqueness 约束，我可以找到从 B 到 A 中每个点的最近点:

To solve this without uniqueness constraint I can just find closest points from B to each point in A:

m = 5; n = 8; dim = 2;
A = rand(m, dim);
B = rand(n, dim);
D = pdist2(A, B);
[~, I] = min(D, [], 2);
C2 = B(I, :);

在 C 中可能存在重复的 B 元素的地方.现在第一个解决方案是蛮力搜索:

Where there may be repeated elements of B present in C. Now the first solution is brute-force search:

minSumD = inf;
allCombs = nchoosek(1:n, m);
for i = 1:size(allCombs, 1)
    allPerms = perms(allCombs(i, :));
    for j = 1:size(allPerms, 1)
        ind = sub2ind([m n], 1:m, allPerms(j, :));
        sumD = sum(D(ind));
        if sumD<minSumD
            minSumD = sumD;
            I = allPerms(j, :);
        end
    end
end
C = B(I, :);

我认为 C2 (到每个 A(i)的最接近点的集合)与 C 几乎相同，除了它的重复点.那么如何减少计算时间呢?

I think C2 (set of closest points to each A(i)) is pretty much alike C except for its repeated points. So how can I decrease the computation time?

推荐答案

使用匈牙利算法的变体，用于计算最小/最大重量完美匹配.为未使用的B点创建n-m个虚拟点以与之匹配(或者，如果您愿意付出更多的努力，请使匈牙利算法适用于非平方矩阵).

Use a variant of the Hungarian algorithm, which computes a minimum/maximum weight perfect matching. Create n-m dummy points for the unused B points to match with (or, if you're willing to put in more effort, adapt the Hungarian algorithm machinery to non square matrices).

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