>>> import gndemo
>>> from scipy import *
>>> import numpy as N
>>> n = 5;
>>> n = 4;
>>> e = gndemo.path(n)  # the edge list of a path graph
>>> print e
[[0, 1], [1, 2], [2, 3]]
>>> a = gndemo.edges2adjmat(e)  # generate an adjacency matrix
>>> print a
  (0, 1)	1
  (1, 0)	1
  (1, 2)	1
  (2, 1)	1
  (2, 3)	1
  (3, 2)	1
>>> l = gndemo.adjmat2lap(a)   # generate the laplacian
>>> print l
  (0, 0)	1.0
  (1, 0)	-1.0
  (0, 1)	-1.0
  (1, 1)	2.0
  (2, 1)	-1.0
  (1, 2)	-1.0
  (2, 2)	2.0
  (3, 2)	-1.0
  (2, 3)	-1.0
  (3, 3)	1.0
>>> b = N.zeros(n)     # create a right-hand side to solve
>>> b[0] = 1
>>> b[-1] = -1
>>> print b
[ 1  0  0 -1]
>>> ans = linalg.iterative.cg(l,b)  # solve the system
>>> x = ans[0]
>>> print x
[ 1.5  0.5 -0.5 -1.5]
>>> print l*x
[ 1.  0.  0. -1.]
>>> print linalg.norm(l*x - b)   # check that we got a good answer
0.0
>>> n = 10000                  # now, let's do a much larger path graph
>>> e = gndemo.path(n)
>>> a = gndemo.edges2adjmat(e)
l =
>>>  
>>> l = gndemo.adjmat2lap(a)
>>> b = zeros(n)
>>> b[0] = 1
>>> b[-1] = -1
>>> ans = linalg.iterative.cg(l,b)  # this only took a few seconds
>>> x = ans[0]
>>> print linalg.norm(l*x - b)
0.0

# note: path graphs are about the worst case for this algorithm
# random graphs, which we try next, are the best case
# also note that my code for generating the matrices is painfully slow

>>> e = gndemo.randGraph(n, 40*n)
>>> a = gndemo.edges2adjmat(e)
>>> l = gndemo.adjmat2lap(a)
>>> b = zeros(n)
>>> b[0] = 1
>>> b[-1] = -1
>>> ans = linalg.iterative.cg(l,b)
>>> x = ans[0]
>>> print linalg.norm(l*x - b)
9.90255546596e-006

# note: that was still pretty small error


# here is a demo of solving for the flow given potentials at particular
# vertices.  In this exmple, I create a path graph on 10 vertices (
# numbered from 0 to 9), set the potential of vertex 0 to -1, the potential
# of vertex 9 to 1, and then get all the induced potentials:
#
# note that this demo was done using ipython

In [1]: from numpy import *

In [2]: import gndemo

In [3]: e = gndemo.path(10)

In [4]: verts = array([0, 9])

In [5]: pots = array([-1, 1])

In [6]: x = gndemo.solvePotentials(e, verts, pots)

In [7]: print x
[-1.         -0.77777778 -0.55555556 -0.33333333 -0.11111111  0.11111111
  0.33333333  0.55555556  0.77777778  1.        ]