数值计算方法（高斯消元以及LU分解）
基础方法没啥好叙述的，书上随处可见，给出了具体的代码实现以及注释，错误之处请留言。谢谢
Mathematical expression of gaussian elimination
elimination-step and get a upper triangular matrix
for k=0 to n-1
mki=a(k)ika(k)kk(i=k+1,…,n−1)
ak+1ij=aij(k)−mk∗a(k)kj(i,j=k+1,…,n−1)
bk+1i=bi(k)−mik∗b(k)k(i=k+1,…,n−1)
back-substitution-step
xn−1=b(n−1)n−1an−1(n−1)(n−1)
for i=n-2 to 0
xi=b(n)i−∑n−1j=i+1a(n)ijxjanii
Mathematical expression of LU decomposition
1.implement A=LU
u1i=a1i(i=1,2,…,n)
li1=ai1u11(i=1,2,…,n)
for k=1 to n-1
uki=aki−∑j=0k−1lkjuji(i=k,k+1,…,n−1)
lik=aik−∑k−1j=0lijujkukk(i=k+1,…,n−1,k!=n)
2.计算LY=b
y1=b1
for k=1 to n-1
yk=bk−∑j=0k−1lkjyj(k=2,3,…,n−1)
计算UX=Y
xn−1=bn−1u(n−1)(n−1)
for k= n-2 to 0
xk=yk−∑n−1j=k+1ukjxjukk
import numpy as np #a package to calculate
from scipy.sparse import identity# to get a identity matrix
import time#calculate time of diffient method create a random matrix M with dimension 100*100 ,generate A by adding an identify matrix
np.random.seed(1) #set seed to make s unchanged
s=np.random.rand(100,100)
A=s+identity(100).toarray()
#add a identity to make sure has a inverse matrix
print A
"""[[ 1.41702200e+00 7.20324493e-01 1.14374817e-04 ..., 5.73679487e-01
2.87032703e-03 6.17144914e-01]
[ 3.26644902e-01 1.52705810e+00 8.85942099e-01 ..., 2.34362086e-01
6.16778357e-01 9.49016321e-01]
[ 9.50176119e-01 5.56653188e-01 1.91560635e+00 ..., 7.96260777e-02
9.82817114e-01 1.81612851e-01]
...,
[ 2.47870273e-01 1.11841381e-01 5.13540776e-01 ..., 1.83527618e+00
2.39285522e-01 7.30797255e-02]
[ 9.52966404e-01 1.12326974e-01 8.02396496e-01 ..., 4.95854311e-01
1.50837092e+00 8.47333803e-02]
[ 4.37268153e-01 8.63201246e-01 6.80236396e-01 ..., 5.95336949e-02
1.08043656e-01 1.76279378e+00]]"""gengerate a vector x=(1,2,3,+⋯+100)T
x=np.array(range(1,101)).T generate a vector b as b=AX
b=np.dot(A,x)
#get result by calling function ,also you can calculate by yourself
#for i in range(100):
# b[i]=0
# for j in range(100):
# b[i]+=A[i][j]*x[j][0]calculate x
#Gauss-Jordan elimination
def Gauss(n,A,b):
#elimination-step and get a upper triangular matrix
for k in range(0,n-1):
for i in range(k+1,n):
m=A[i][k]/A[k][k]#calculate multiplier
for j in range(k,n):
A[i][j]=A[i][j]-m*A[k][j] #elimination of ith row
b[i]=b[i]-m*b[k]
"""back-substitution-step,easy to implement according
to arithmetic expression"""
newx=np.zeros((n,1))
newx[n-1]=(b[n-1]/A[n-1][n-1])
i=n-2
while(i>=0):
sum2=0
for j in range(i+1,n):
sum2+=A[i][j]*newx[j]
newx[i]=(b[i]-sum2)/A[i][i]
i-=1
return newx
"""this method is based on Gauss,just exchange kth row with maxkth
row before calculate m to avoid some mistakes caused by float"""
def Gauss_exchange(n,A,b):
for k in range(0,n-1):
maxk=k
maxA=A[k][k]
for t in range(k+1,n):
if(A[t][k]>maxA):
maxk=t
maxA=A[t][k]
for j in range(k,n):
tmp=A[k][j]
A[k][j]=A[maxk][j]
A[maxk][j]=tmp
tmp=b[k]
b[k]=b[maxk]
b[maxk]=tmp #exchange maxk with k
for i in range(k+1,n):
m=A[i][k]/A[k][k]
for j in range(k,n):
A[i][j]=A[i][j]-m*A[k][j]
b[i]=b[i]-m*b[k]
#same as Gauss
newx=np.zeros((n,1))
newx[n-1]=(b[n-1]/A[n-1][n-1])
i=n-2
while(i>=0):
sum1=0
for j in range(i+1,n):
sum1+=A[i][j]*newx[j]
newx[i]=(b[i]-sum1)/A[i][i]
i-=1
return newx
def dolittle(n,A,b):
L=np.zeros(A.shape,dtype="float")+identity(100).toarray()
U=np.zeros(A.shape,dtype="float")
#implement A=LU
for i in range(0,n):
U[0][i]=A[0][i]
for i in range(0,n):
L[i][0]=A[i][0]/U[0][0]
for k in range(1,n):
for i in range(k,n):
sum3=0
for j in range(0,k):
sum3+=L[k][j]*U[j][i]
U[k][i]=A[k][i]-sum3
for i in range(k+1,n):
sum3=0
for j in range(0,k):
sum3=sum3+L[i][j]*U[j][k]
L[i][k]=(A[i][k]-sum3)/U[k][k]
#calculate LY=b
Y=np.zeros((n,1))
Y[0]=b[0]
for k in range(1,n):
sum3=0
for j in range(0,k):
sum3+=L[k][j]*Y[j]
Y[k]=b[k]-sum3
#calculate UX=Y
newx=np.zeros((n,1))
newx[n-1]=(Y[n-1]/U[n-1][n-1])
i=n-2
while(i>=0):
sum3=0
for j in range(i+1,n):
sum3+=U[i][j]*newx[j]
newx[i]=(Y[i]-sum3)/U[i][i]
i-=1
return newx
"""copy() function is to make sure do not change A
and pass correct parameters in follow functions"""
t0=time.time() #start time
newx2=Gauss(100,A.copy(),b.copy())
print "the time of Gaussian elimination is ",time.time()-t0
t0=time.time()
newx1=Gauss_exchange(100,A.copy(),b.copy())
print "the time of column principal element elimination is ",time.time()-t0
t0=time.time()
newx3=dolittle(100,A.copy(),b.copy())
print "the time of LU decomposition method is ",time.time()-t0
print "first ",newx1.T, " second",newx2.T,"\n third",newx3.T
the time of Gaussian elimination is 0.805999994278
the time of column principal element elimination is 0.884999990463
the time of LU decomposition method is 0.444000005722
""" first [[ 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.
15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28.
29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42.
43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56.
57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70.
71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82. 83. 84.
85. 86. 87. 88. 89. 90. 91. 92. 93. 94. 95. 96. 97. 98.
99. 100.]]
second [[ 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.
15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28.
29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42.
43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56.
57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70.
71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82. 83. 84.
85. 86. 87. 88. 89. 90. 91. 92. 93. 94. 95. 96. 97. 98.
99. 100.]]
third [[ 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.
15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28.
29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42.
43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56.
57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70.
71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82. 83. 84.
85. 86. 87. 88. 89. 90. 91. 92. 93. 94. 95. 96. 97. 98.
99. 100.]]conclusion:the data is small ,It’s hard to compare performance
author: cbf
17.10.3