I have two square matrices A
and B
I must convert B
to CSR Format
and determine the product C
A * B_csr = C
I have found a lot of information online regarding CSR Matrix - Vector multiplication. The algorithm is:
for (k = 0; k < N; k = k + 1)
result[i] = 0;
for (i = 0; i < N; i = i + 1)
{
for (k = RowPtr[i]; k < RowPtr[i+1]; k = k + 1)
{
result[i] = result[i] + Val[k]*d[Col[k]];
}
}
However, I require Matrix - Matrix
multiplication.
Further, it seems that most algorithms apply A_csr - vector
multiplication where I require A * B_csr
. My solution is to transpose the two matrices before converting then transpose the final product.
Can someone explain how to compute a Matrix - CSR Matrix
product and/or a CSR Matrix - Matrix
product?
Here is a simple solution in Python for the Dense Matrix X CSR Matrix
. It should be self-explanatory.
def main():
# 4 x 4 csr matrix
# [1, 0, 0, 0],
# [2, 0, 3, 0],
# [0, 0, 0, 0],
# [0, 4, 0, 0],
csr_values = [1, 2, 3, 4]
col_idx = [0, 0, 2, 1]
row_ptr = [0, 1, 3, 3, 4]
csr_matrix = [
csr_values,
col_idx,
row_ptr
]
dense_matrix = [
[1, 3, 3, 4],
[1, 2, 3, 4],
[1, 4, 3, 4],
[1, 2, 3, 5],
]
res = [
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
]
# matrix order, assumes both matrices are square
n = len(dense_matrix)
# res = dense X csr
csr_row = 0 # Current row in CSR matrix
for i in range(n):
start, end = row_ptr[i], row_ptr[i + 1]
for j in range(start, end):
col, csr_value = col_idx[j], csr_values[j]
for k in range(n):
dense_value = dense_matrix[k][csr_row]
res[k][col] += csr_value * dense_value
csr_row += 1
print res
if __name__ == '__main__':
main()
CSR Matrix X Dense Matrix
is really just a sequence of CSR Matrix X Vector
product for each row of the dense matrix right? So it should be really easy to extend the code you show above to do this.
Moving forward, I suggest you don't code these routines yourself. If you are using C++ (based on the tag), then you could have a look at Boost ublas for example, or Eigen. The APIs may seem a bit cryptic at first but it's really worth it in the long term. First, you gain access to a lot more functionality, which you will probably require in the future. Second these implementations will be better optimised.