Matrix Sparsification

Problem Statement Reduce the number of 1s in a matrix

Measuring Sparsity The way I measured sparsity was by adding up the total number of 1s in a matrix and dividing by the total number of elements This gives you a number between 0 and 1 that tells you what percentage of the matrix is filled with ones

Experiment Setup First I generated H, which is a sparse 300x582 matrix with a column weight of 3, sparsity=.01 Then I multiplied it by a random invertible matrix. D is the resulting dense matrix, sparsity≈.5 Then I tried to make D as sparse as the original matrix

Comparing rowspaces From the sparse H matrix, generate a G matrix so that mod(H*G,2)=0 Then test D and any sparsified versions of D to ensure that mod2 of multiplication by G still results in a 0 matrix

GF2 difficulties Matrix sparsification is difficult in GF2 because it requires a different set of math rules For example:  With real numbers, the vectors [0 1 1],[1 0 1], and [1 1 0] are independent  In GF2, any one of those vectors can be made from the other two

Row Echelon Form The dense matrix starts with sparsity≈.5 In an [m x n] matrix, row reduction will give m columns with only one 1 in them. The rest of the columns should be approximately half 1s Now sparsity≈(m+.5*(n-m)*m)/(n*m) The more square a matrix is, the more this step helps.

Null Space Row Echelon Form tries to make the pivot columns the earliest possible columns By computing the null space of the original matrix, and then computing the null space of that null space, you get back the original rowspace Now the pivot columns are in different locations

SP2 function Find the row which reduces the number of 1s the most. A(row,:)*A T is a vector that gives the number of matching 1s in each row (i.e. the number of 1s that will be eliminated if the rows are added) (1-A(row,:))*A T = a vector giving you the number of 1s that will be introduced if the rows are added.

SP2 function If B= A(row,:)*A T - (1-A(row,:))*A T, then finding the maximum of B will reduce the sparsity the most per row addition. But first, you have to set B(row)= -n, because otherwise B(row) will always be the maximum, and you can’t add a row to itself.

SP2 function If max(B) is positive, then adding rows helps sparsify the matrix If max(B) is 0, then adding rows keeps the sparsity the same, but changes the location of the 1s When max(B) is negative, it makes the matrix more dense, but it can be helpful in overall sparsification because it helps get you out of local minimums

SP3 function First I make a matrix of all possible combinations of two different rows and also the original rows that I’m testing on This new matrix has (m-1)Choose2 + (m-1) rows Then I follow the same process as with the SP2 function, but using this new matrix, instead of one made up of only the original rows

Why not add 3? The matrix size grows too quickly. With 300 rows, the matrix for adding 1 or 2 rows has rows, and the matrix for adding 1 to 3 rows would have rows.

Why not add 3? You could break the matrix down into multiple smaller matrices, then save the best row from each matrix and add in the best overall row One reason for the significant improvement when adding two rows instead of one is that in GF2 1+1=0.

Orthogonal Projection The easiest way to span a space is with orthogonal vectors If the projection of vector a onto b is greater than.5, then b is a significant component of a, so removing it will make the vectors closer to orthogonal, and the matrix more sparse i.e. if((a·b)/(|b| 2 )>.5) rowa=mod2(rowa+rowb)

LDPC Decoding results In LDPC codes, every 1 in a matrix represents a connection between check nodes and variable nodes. Reducing the sparsity of a matrix makes LDPC decoding faster, and more reliable

SNR Vs. Probability of Error