创建权重邻接矩阵

问题描述

我需要根据以下论文为图的边缘分配权重:

I need to assign weights to edges of a graph, from the following papers:

L.Xiao和S. Boyd的用于分布平均的快速线性迭代"S. Boyd的图拉普拉斯特征值的凸优化"

"Fast linear iterations for distributed averaging" by L. Xiao and S. Boyd"Convex Optimization of Graph Laplacian Eigenvalues" by S. Boyd

我的图形有一个邻接矩阵(一个50 x 50矩阵)，具有512个非零值.

I have the adjacency matrix for my graph (a 50 by 50 matrix), with 512 non-zero values.

我还有一个权重为256 x 1的矢量.

I also have a 256 by 1 vector with the optimal weights.

对于我使用的软件，我需要一个50 x 50的矩阵，其边缘(i，j)的权重在邻接矩阵的相关位置(并且边缘(j，i)的符号相反) .

For the software I'm using, I need a 50 by 50 matrix with the weight of edge (i,j) in the relevant position of the adjacency matrix (and with the opposite sign for edge (j,i)).

我的尝试在下面，但我无法使其正常工作.

My attempt is below, but I can't get it working.

function weights = construct_weight_mtx(weight_list, Adj)

weights = zeros(size(Adj));
positions = find(Adj);

for i=1:length(positions)/2
    if Adj(i) == 1
        weights(i) = weight_list(i);
    end
end

weights = weights - weights';

find(Adj) == find(weights);

end

推荐答案

您正在原始邻接矩阵中找到非零位置，但在其中找到了全部.为了解决这个问题，您只需要占据这些职位的前一半即可.

You're finding the nonzero positions in the original adjacency matrix, but you're finding all of them. To get around this, you then take only the first half of those positions.

for i=1:length(positions)/2 ...

不幸的是，这会从完整的列中获取索引，而不仅仅是对角线以下的位置.因此，如果您的矩阵全为1，则可以采用:

Unfortunately, this takes the indices from complete columns rather than just the positions below the diagonal. So if your matrix was all 1's, you'd be taking:

1 1 1 0 0 ...
1 1 1 0 0 ...
1 1 1 0 0 ...
...

代替:

1 0 0 0 0 ...
1 1 0 0 0 ...
1 1 1 0 0 ...
...

要获取正确的值，我们只需获取Adj的下三角部分，然后找到该位置的非零位置:

To take the correct values, we just take the lower triangular portion of Adj and then find the nonzero positions of that:

positions = find(tril(Adj));

现在，我们只有对角线以下的256个位置，并且可以遍历所有位置.接下来，我们需要在循环中修复分配:

Now we have only the 256 positions below the diagonal and we can loop over all of the positions. Next, we need to fix the assignment in the loop:

for i=1:length(positions)
    if Adj(i) == 1   %// we already know Adj(i) == 1 for all indices in positions
        weights(i) = weight_list(i);   %// we need to update weights(positions(i))
    end
end

所以它变成:

for i=1:length(positions)
    weights(positions(i)) = weight_list(i);
end

但是，如果我们要做的就是将256个值分配给256个位置，那么我们可以在没有for循环的情况下做到这一点:

But if all we're doing is assigning 256 values to 256 positions, we can do that without a for loop:

weights(position) = weight_list;

请注意，weight_list的元素必须以正确的顺序排列，下三角部分的非零元素按列排序.

Note that the elements of weight_list must be in the proper order with the nonzero elements of the lower-triangular portion ordered by columns.

完整代码:

function weights = construct_weight_mtx(weight_list, Adj)

weights = zeros(size(Adj));
positions = find(tril(Adj));

weights(positions) = weight_list;

weights = weights - weights.';   %// ' is complex conjugate; not a big deal here, but something to know

find(Adj) == find(weights);   %// Not sure what this is meant to do; maybe an assert?

end

这篇关于创建权重邻接矩阵的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！