How to generate a sparse matrix for a given array?

Dear All,

I want to generate a sparse matrix B for a given array A. A contains two columns of indecis. For example,

A = [1 3;
     1 4;
     2 3;
     2 5;
     3 5;
     4 5]

The sparse matrix B should be as follows:

B = [ 2  0 -1 -1  0;
      0  2 -1  0 -1;
     -1 -1  2  0  0;
     -1  0  0  2 -1;
      0 -1 -1 -1  3]

The characteristics of B:

1. Sum of each row is zero.
2. If we consider matrix A gives the information of edges of a graph, B(i,i) = sum of number of edges, B(i,j) = -1 if there is an edge between i and j.

Thanks a lot.

Benson

Answer 1

[Matt J]

B=laplacian(  graph(A(:,1),A(:,2))  );

[Benson Gou]

Hi, Matt,

I tried your code, but I got the following error:

"Graph has multiple edges."

Thanks a lot.

Benson

[Matt J]

Works fine for me:

A = [1 3;
     1 4;
     2 3;
     2 5;
     3 5;
     4 5];

B=laplacian(  graph(A(:,1),A(:,2))  );;

full(B)

ans = 5×5

2 0 -1 -1 0 0 2 -1 0 -1 -1 -1 3 0 -1 -1 0 0 2 -1 0 -1 -1 -1 3

[Matt J]

You probably need to do,

A=unique(sort(A,2),'rows')

[Benson Gou]

yes, it works well. Thanks a lot.

Benson

[Matt J]

You're welcome, but please Accept-click the answer that you deem best resolves your question.

Answer 2

[the cyclist]

Here is one way:

A = [1 3;
     1 4;
     2 3;
     2 5;
     3 5;
     4 5];

d = max(A(:));

B = sparse(A(:,1),A(:,2),-1,d,d) + sparse(A(:,2),A(:,1),-1,d,d);

for ii = 1:d
    B(ii,ii) = -sum(B(ii,:));
end

disp(B)

   (1,1)        2
   (3,1)       -1
   (4,1)       -1
   (2,2)        2
   (3,2)       -1
   (5,2)       -1
   (1,3)       -1
   (2,3)       -1
   (3,3)        3
   (5,3)       -1
   (1,4)       -1
   (4,4)        2
   (5,4)       -1
   (2,5)       -1
   (3,5)       -1
   (4,5)       -1
   (5,5)        3

[Benson Gou]

Hi, the Cyclist,

Thanks for your answer. It looks pretty good. But I am wondering if it is possible to avoid using the for iterations, because matrix A has 5000 rows.

Thanks a lot again.

Benson

[the cyclist]

A = [1 3;
     1 4;
     2 3;
     2 5;
     3 5;
     4 5];

d = max(A(:));

B = sparse(A(:,1),A(:,2),-1,d,d) + sparse(A(:,2),A(:,1),-1,d,d);

% for ii = 1:d
%     B(ii,ii) = -sum(B(ii,:));
% end

B = B - diag(sum(B,2));

disp(B)

   (1,1)        2
   (3,1)       -1
   (4,1)       -1
   (2,2)        2
   (3,2)       -1
   (5,2)       -1
   (1,3)       -1
   (2,3)       -1
   (3,3)        3
   (5,3)       -1
   (1,4)       -1
   (4,4)        2
   (5,4)       -1
   (2,5)       -1
   (3,5)       -1
   (4,5)       -1
   (5,5)        3

[Benson Gou]

Hi, Cyclist,

Thanks a lot for your prompt reply.

I tried your code on my case with 5000 rows in A. But I found there are some different results between your code and my original code which uses for iterations.

For example, row 3 in B, yours is

(1,5) 8

(1,7) -1

(1,28) -3

(1,33) -1

(1,34) -1

(1,397) -1

(1,434) -1

My result is

(1,5) 6

(1,7) -1

(1,28) -1

(1,33) -1

(1,34) -1

(1,397) -1

(1,434) -1

The locations of non-zero elemnts are correct, but the values are different. The differency may be caused by the double edges between two nodes. There are 2 edges between 5 and 28.

Thanks

Benson

[Bruno Luong]

(1,28)      -3

"There are 2 edges between 5 and 28."

Make it three.

[Benson Gou]

Hi, Bruno,

Instead of making it 3, it should be -1, as my code did. It means I would like to ignore the mulitple edges between two nodes.

Thanks.

Benson

[the cyclist]

I assume that doing what @Matt J suggested in his answer:

A=unique(sort(A,2),'rows')

is what you would need to do here as well.

If you posted a small example that exhibits the problem, it would help. But I'm guessing you have your answer.

[Benson Gou]

Hi, the Cyclist,

Thanks a lot for your great help. You have a good day!

Benson