def problem18Verify(A2, B2):
ex2 = AxClass(A2)
x = B2.copy()
x, i = conjGrad(ex2,x,B2)
print("\nVerification for problem 18:")
print("\nNumber of iterations =", i)
print("\nThe solution is:\n", x)
def problem19Verify(A3, B3):
ex2 = AxClass(A3)
x = B3.copy()
x, i = conjGrad(ex2,x,B3)
print("\nVerification for problem 19:")
print("\nNumber of iterations =", i)
print("\nThe solution is:\n", x)
def problem19():
b = np.array([0.0, 0.0, 100.0, 0.0, 0.0, 100.0, 200.0, 200.0, 300.0],dtype=float)
n = 9
x = np.zeros(n)
x, numIter = conjGrad(Ax2, x, b)
print("\nNumber of iterations =",numIter)
print("\nThe solution for 19 is:\n",x)
input("\nPress return to exit")
def problem18():
n = eval(input("Number of equations ==> "))
b = np.zeros(n)
b[n-1] = 100.00
x = np.zeros(n)
x, numIter = conjGrad(Ax,x,b)
print("\nNumber of iterations =", numIter)
print("\nThe solution for 18 is:\n", x)
input("\nPress return to exit")
# fill in the other 8 rows of Ax (2 points per row of Ax for
# a total of 18 points)
Ax[1] = v[0] - 4.0*v[1] + v[2] + v[4]
Ax[2] = v[1] - 4.0*v[2] + v[5]
Ax[3] = v[0] - 4.0*v[3] + v[4] + v[6]
Ax[4] = v[1] + v[3] - 4.0*v[4] + v[5] + v[7]
Ax[5] = v[2] + v[4] - 4.0*v[5] + v[8]
Ax[6] = v[3] - 4.0*v[6] + v[7]
Ax[7] = v[4] + v[6] - 4.0*v[7] + v[8]
Ax[8] = v[5] + v[7] - 4.0*v[8]
return Ax
b = array([0,0,100,0,0,100,200,200,300])*(-1.0)
x = zeros((9))*1.0
tol = 1e-06
s1,numIter = conjGrad(Ax, x, b, tol) # 5 points for correct call here
print("\nThe solution is:\n",s1)
print("\nNumber of iterations =",numIter, "using Tol: ", 1e-06)
print("\n CG Convergence Test")
print("Iterations Tolerance")
#
# Use the loop below to call conjGrad with various tolerances,
# compare the number of iterations.
#
for tol in [1e-02, 1e-04, 1e-06, 1e-08, 1e-10, 1e-12, 1e-14, 1e-16]:
x = zeros((9),dtype=float)
s2,numIter = conjGrad(Ax, x, b, tol) # 5 points for correct call here
print('%6d %8.1e'%(numIter,tol))
print("\nError between vectors obtained with tol=1e-06 and tol=1e-16: ", linalg.norm(s1-s2), "\n")
[0., 1., 0., 1., -4., 1., 0., 1., 0.],
[0., 0., 1., 0., 1., -4., 0., 0., 1.],
[0., 0., 0., 1., 0., 0., -4., 1., 0.],
[0., 0., 0., 0., 1., 0., 1., -4., 1.],
[0., 0., 0., 0., 0., 1., 0., 1., -4.]]
Ax = array(Ax)
Ax = dot(Ax, v)
return Ax
# b and x are the right-hand side and solution vectors from problem 19
b = array([0, 0, 100, 0, 0, 100, 200, 200, 300]) * (-1.0)
x = zeros((9)) * 1.0
tol = 1e-06
# Solves the equation using a tolerance of 1e-6
s1, numIter = conjGrad(Ax, x, b, tol=tol)
print("\nThe solution is:\n", s1)
print("\nNumber of iterations =", numIter, "using Tol: ", 1e-06)
print("\n CG Convergence Test")
print("Iterations Tolerance")
#
# Use the loop below to call conjGrad with various tolerances,
# compare the number of iterations.
#
for tol in [1e-02, 1e-04, 1e-06, 1e-08, 1e-10, 1e-12, 1e-14, 1e-16]:
x = zeros((9), dtype=float)
s2, numIter = conjGrad(Ax, x, b, tol=tol)
print('%6d %8.1e' % (numIter, tol))
# Prints the error measure between the solutions with tolerances of
#!/usr/bin/python
## example2_18
import numpy
from conjGrad import *
def Ax(v):
n = len(v)
Ax = numpy.zeros((n), dtype=float)
Ax[0] = 2.0 * v[0] - v[1] + v[n - 1]
Ax[1:n - 1] = -v[0:n - 2] + 2.0 * v[1:n - 1] - v[2:n]
Ax[n - 1] = -v[n - 2] + 2.0 * v[n - 1] + v[0]
return Ax
n = eval(raw_input("Number of equations ==> "))
b = numpy.zeros((n), dtype=float)
b[n - 1] = 1.0
x = numpy.zeros((n), dtype=float)
x, numIter = conjGrad(Ax, x, b)
print "\nThe solution is:\n", x
print "\nNumber of iterations =", numIter
#!/usr/bin/python3
## example2_18 on p. 97 of Kiusalaas, 3rd Ed.
from numpy import zeros, float64, sqrt
from conjGrad import *
def Ax(v): # define Matrix-vector multiplication function
n = len(v)
Ax = zeros((n), dtype=float64)
Ax[0] = 2.0 * v[0] - v[1] + v[n - 1]
Ax[1:n - 1] = -v[0:n - 2] + 2.0 * v[1:n - 1] - v[2:n]
Ax[n - 1] = -v[n - 2] + 2.0 * v[n - 1] + v[0]
return Ax
n = eval(input("Number of equations ==> "))
b = zeros((n), dtype=float64)
b[n - 1] = 1.0
x = zeros((n), dtype=float64)
x, numIter = conjGrad(Ax, x, b, 1.0e-2)
print("\nThe solution is:\n", x)
print("\nNumber of iterations =", numIter)
input("\nPress return to exit")
[1,0,0,-4,1,0,1,0,0],
[0,1,0,1,-4,1,0,1,0],
[0,0,1,0,1,-4,0,0,1],
[0,0,0,1,0,0,-4,1,0],
[0,0,0,0,1,0,1,-4,1],
[0,0,0,0,0,1,0,1,-4]])
Ax = dot(A, v) #matrix mult of A*v
return Ax
b = array([0,0,100,0,0,100,200,200,300])*(-1.0) #solution vector
x = zeros((9))*1.0
tol = 1e-06 #set the tolerance
s1,numIter = conjGrad(Ax, x, b, tol)
print("\nThe solution is:\n",s1)
print("\nNumber of iterations =",numIter, "using Tol: ", 1e-06)
print("\n CG Convergence Test")
print("Iterations Tolerance")
#
# Use the loop below to call conjGrad with various tolerances,
# compare the number of iterations.
#
for tol in [1e-02, 1e-04, 1e-06, 1e-08, 1e-10, 1e-12, 1e-14, 1e-16]:
x = zeros((9),dtype=float)
s2,numIter = conjGrad(Ax, x, b, tol)
print('%6d %8.1e'%(numIter,tol))
print("\nError between vectors obtained with tol=1e-06 and tol=1e-16: ", linalg.norm(s1-s2), "\n")
Ax[2] = -4.0 * v[2] + 1.0 * (v[1] + v[5])
Ax[3] = -4.0 * v[3] + 1.0 * (v[0] + v[4] + v[6])
Ax[4] = -4.0 * v[4] + 1.0 * (v[1] + v[3] + v[5] + v[7])
Ax[5] = -4.0 * v[5] + 1.0 * (v[2] + v[4] + v[8])
Ax[6] = -4.0 * v[6] + 1.0 * (v[3] + v[7])
Ax[7] = -4.0 * v[7] + 1.0 * (v[4] + v[6] + v[8])
Ax[8] = -4.0 * v[8] + 1.0 * (v[5] + v[7])
return Ax
b = array([0,0,100,0,0,100,200,200,300])*(-1.0)
x = zeros((9))*1.0
tol = 1e-06
#Conjugate Gradient Method being called with the parameters Ax as the function
#x being the temperature variables and b being the - solution vector, and a 1e-06 tolerance.
s1,numIter = conjGrad(Ax, x, b, 1e-06)
print("\nThe solution is:\n",s1)
print("\nNumber of iterations =",numIter, "using Tol: ", 1e-06)
print("\n CG Convergence Test")
print("Iterations Tolerance")
#
# Use the loop below to call conjGrad with various tolerances,
# compare the number of iterations.
#
for tol in [1e-02, 1e-04, 1e-06, 1e-08, 1e-10, 1e-12, 1e-14, 1e-16]:
x = zeros((9),dtype=float)
#calling the conjGrad function with a varying tolerance and printing out the new solutions.
s2,numIter = conjGrad(Ax, x, b, tol)
print('%6d %8.1e'%(numIter,tol))
#!/usr/bin/python
## example2_18
from numarray import zeros,Float64,sqrt
from conjGrad import *
def Ax(v):
n = len(v)
Ax = zeros((n),type=Float64)
Ax[0] = 2.0*v[0] - v[1]+v[n-1]
Ax[1:n-1] = -v[0:n-2] + 2.0*v[1:n-1] -v [2:n]
Ax[n-1] = -v[n-2] + 2.0*v[n-1] + v[0]
return Ax
n = eval(raw_input("Number of equations ==> "))
b = zeros((n),type=Float64)
b[n-1] = 1.0
x = zeros((n),type=Float64)
x,numIter = conjGrad(Ax,x,b)
print "\nThe solution is:\n",x
print "\nNumber of iterations =",numIter
raw_input("\nPress return to exit")
