Network & Matrix Computations

David Gleich

Purdue University

Fall 2011

Course number CS 59000-NMC

Tuesdays and Thursday, 10:30am-11:45am

CIVL 2123

In-class quiz 4

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CS 59000-NMC, 6 September

Please answer the following questions. You may not use any outside references or technology. Justify and explain all answers. This quiz is for my own evaluation, so that I can provide better instruction in the course.

Question

Let $\mA$ be a binary matrix. Suppose this matrix is composed of mostly ones and that the zeros are stored with a compressed-sparse row data structure. Write down an efficient algorithm to compute $\vy = \mA \vx$ give the compressed sparse row data structure for $\mA$’s zeros in the arrays pointer and columns.

Solution

The key insight is that the zero pattern tells us what to exclude from the matrix-vector product, rather than what to include. Consider that if $\mA$ was composed entirely of ones, then:

$$[\mA \vx]_i = \sum x_i$$

Consequently, if we set $\alpha = \sum x_i$, then

$$\mA \vx = \alpha \ve$$

where $$\ve$$ is the vector of all ones.

Once we have this property, let $\mO$ be the matrix of all ones. Now consider $\mA$ from the problem

$$\mA = \mO - \mB$$ where $\mB$ is a *sparse matrix*with the *zero pattern* from $\mA$ and each entry is a one. Consequently, we can just compute $$\mA \vx = \alpha \ve - \mB \vx$$

where $\mB \vx$ is just a standard matrix-vector product.

function sparse_zero_matvec(n,pointers,columns,x)
""" Compute a mat-vec with a mostly one matrix, with a sparse zero pattern.

This function will multiply a binary matrix $A$, which is all
ones except for a sparse pattern of zeros, by a vector $x$.

Here the arrays are zero indexed.

@param n the dimension of the matrix
@param pointers the array of pointers for a CSR pattern of zeros in the matrix
@param columns the array of columns for the CSR pattern of zeros
@param an array with the values of x
@return an array such 
"""

alpha = 0
for xi in x: # assumes x implements an interable interface
  alpha += xi

y = [alpha for _ in xrange(n)] # initialize y
for i in xrange(n):
  change = 0
  for nzi in xrange(pointers[i],pointers[i+1]):
    col = columns[nzi]
    change += x[nzi]
  
  y[i] -= change # adjust the value

return y