Question

This is a follow up to How to set up and solve simultaneous equations in python but I feel deserves its own reputation points for any answer.

For a fixed integer n, I have a set of 2(n-1) simultaneous equations as follows.

M(p) = 1+((n-p-1)/n)*M(n-1) + (2/n)*N(p-1) + ((p-1)/n)*M(p-1)

N(p) = 1+((n-p-1)/n)*M(n-1) + (p/n)*N(p-1)

M(1) = 1+((n-2)/n)*M(n-1) + (2/n)*N(0)

N(0) = 1+((n-1)/n)*M(n-1)

M(p) is defined for 1 <= p <= n-1. N(p) is defined for 0 <= p <= n-2. Notice also that p is just a constant integer in every equation so the whole system is linear.

Some very nice answers were given for how to set up a system of equations in python. However, the system is sparse and I would like to solve it for large n. How can I use scipy’s sparse matrix representation and http://docs.scipy.org/doc/scipy/reference/sparse.linalg.html for example instead?

Answer 1

I wouldn’t normally keep beating a dead horse, but it happens that my non-vectorized approach to solving your other question, has some merit when things get big. Because I was actually filling the coefficient matrix one item at a time, it is very easy to translate into COO sparse matrix format, which can efficiently be transformed to CSC and solved. The following does it:

import scipy.sparse

def sps_solve(n) :
    # Solution vector is [N[0], N[1], ..., N[n - 2], M[1], M[2], ..., M[n - 1]]
    n_pos = lambda p : p
    m_pos = lambda p : p + n - 2
    data = []
    row = []
    col = []
    # p = 0
    # n * N[0] + (1 - n) * M[n-1] = n
    row += [n_pos(0), n_pos(0)]
    col += [n_pos(0), m_pos(n - 1)]
    data += [n, 1 - n]
    for p in xrange(1, n - 1) :
        # n * M[p] + (1 + p - n) * M[n - 1] - 2 * N[p - 1] +
        #  (1 - p) * M[p - 1] = n
        row += [m_pos(p)] * (4 if p > 1 else 3)
        col += ([m_pos(p), m_pos(n - 1), n_pos(p - 1)] +
                ([m_pos(p - 1)] if p > 1 else []))
        data += [n, 1 + p - n , -2] + ([1 - p] if p > 1 else [])
        # n * N[p] + (1 + p -n) * M[n - 1] - p * N[p - 1] = n
        row += [n_pos(p)] * 3
        col += [n_pos(p), m_pos(n - 1), n_pos(p - 1)]
        data += [n, 1 + p - n, -p]
    if n > 2 :
        # p = n - 1
        # n * M[n - 1] - 2 * N[n - 2] + (2 - n) * M[n - 2] = n
        row += [m_pos(n-1)] * 3
        col += [m_pos(n - 1), n_pos(n - 2), m_pos(n - 2)]
        data += [n, -2, 2 - n]
    else :
        # p = 1 
        # n * M[1] - 2 * N[0] = n
        row += [m_pos(n - 1)] * 2
        col += [m_pos(n - 1), n_pos(n - 2)]
        data += [n, -2]
    coeff_mat = scipy.sparse.coo_matrix((data, (row, col))).tocsc()
    return scipy.sparse.linalg.spsolve(coeff_mat,
                                       np.ones(2 * (n - 1)) * n)

It is of course much more verbose than building it from vectorized blocks, as TheodorosZelleke does, but an interesting thing happens when you time both approaches:

First, and this is (very) nice, time is scaling linearly in both solutions, as one would expect from using the sparse approach. But the solution I gave in this answer is always faster, more so for larger ns. Just for the fun of it, I also timed TheodorosZelleke’s dense approach from the other question, which gives this nice graph showing the different scaling of both types of solutions, and how very early, somewhere around n = 75, the solution here should be your choice:

I don’t know enough about scipy.sparse to really figure out why the differences between the two sparse approaches, although I suspect heavily of the use of LIL format sparse matrices. There may be some very marginal performance gain, although a lot of compactness in the code, by turning TheodorosZelleke’s answer into COO format. But that is left as an exercise for the OP!