Gaussian elimination 高斯消元

Gaussian elimination 高斯消元

It can be used to solve linear equation systems or to invert a matrix.
高斯消元法用于解决线性代数求多元方程组的解，或者用于求可逆矩阵的逆
嘿嘿，python代码现成可用：

def gauss_jordan(m ,  eps  =   1.0 / ( 10 ** 10 )) :
   """ Puts given matrix (2D array) into the Reduced Row Echelon Form.
     Returns True if successful, False if 'm' is singular.
     NOTE: make sure all the matrix items support fractions! Int matrix will NOT work!
     Written by J. Elonen in April 2005, released into Public Domain """
  (h ,  w)  =  (len(m) ,  len(m[ 0 ]))
   for  y in  range ( 0 , h) :
    maxrow  =  y
     for  y2 in  range (y + 1 ,  h) :      #  Find max pivot
       if   abs (m[y2][y])  >   abs (m[maxrow][y]) :
        maxrow  =  y2
    (m[y] ,  m[maxrow])  =  (m[maxrow] ,  m[y])
     if   abs (m[y][y])  <=  eps :       #  Singular?
       return   False
     for  y2 in  range (y + 1 ,  h) :      #  Eliminate column y
      c  =  m[y2][y]  /  m[y][y]
       for  x in  range (y ,  w) :
        m[y2][x]  -=  m[y][x]  *  c
   for  y in  range (h - 1 ,   0 - 1 ,   - 1 ) :   #  Backsubstitute
    c   =  m[y][y]
     for  y2 in  range ( 0 , y) :
       for  x in  range (w - 1 ,  y - 1 ,   - 1 ) :
        m[y2][x]  -=   m[y][x]  *  m[y2][y]  /  c
    m[y][y]  /=  c
     for  x in  range (h ,  w) :         #  Normalize row y
      m[y][x]  /=  c
   return   True

使用方法 :

If your matrix is of form [A:x] (as is usual when solving systems), items of A and x both have to be divisible by items of A but not the other way around. Thus, you could, for example, use floats for A and vectors for x. Example:

mtx = [[1.0, 1.0, 1.0, Vec3(0.0,  4.0, 2.0), 2.0],

       [2.0, 1.0, 1.0, Vec3(1.0,  7.0, 3.0), 3.0],

       [1.0, 2.0, 1.0, Vec3(15.0, 2.0, 4.0), 4.0]]

if gauss_jordan(mtx):

  print mtx

else:

  print "Singular!"

# Prints out (approximately):

#

# [[1.0, 0.0, 0.0, (  1.0,  3.0,  1.0),  1.0],

#  [0.0, 1.0, 0.0, ( 15.0, -2.0,  2.0),  2.0],

#  [0.0, 0.0, 1.0, (-16.0,  3.0, -1.0), -1.0]]

Auxiliary functions contributed by Eric Atienza (also released in Public Domain):

def solve(M, b):

  """

  solves M*x = b

  return vector x so that M*x = b

  :param M: a matrix in the form of a list of list

  :param b: a vector in the form of a simple list of scalars

  """

  m2 = [row[:]+[right] for row,right in zip(M,b) ]

  return [row[-1] for row in m2] if gauss_jordan(m2) else None

def inv(M):

  """

  return the inv of the matrix M

  """

  #clone the matrix and append the identity matrix

  # [int(i==j) for j in range_M] is nothing but the i(th row of the identity matrix

  m2 = [row[:]+[int(i==j) for j in range(len(M) )] for i,row in enumerate(M) ]

  # extract the appended matrix (kind of m2[m:,...]

  return [row[len(M[0]):] for row in m2] if gauss_jordan(m2) else None

def zeros( s , zero=0):

    """

    return a matrix of size `size`

    :param size: a tuple containing dimensions of the matrix

    :param zero: the value to use to fill the matrix (by default it's zero )

    """

    return [zeros(s[1:] ) for i in range(s[0] ) ] if not len(s) else zero

算法伪代码:
i := 1

j : =   1
while  (i ≤ m and j ≤ n)  do
  Find pivot  in  column j, starting  in  row i:
  maxi : =  i
   for  k : =  i + 1  to m  do
     if  abs(A[k,j])  >  abs(A[maxi,j]) then
      maxi : =  k
    end  if
  end  for
   if  A[maxi,j] ≠  0  then
    swap rows i and maxi, but  do  not change the value of i
    Now A[i,j] will contain the old value of A[maxi,j].
    divide each entry  in  row i by A[i,j]
    Now A[i,j] will have the value  1 .
     for  u : =  i + 1  to m  do
      subtract A[u,j]  *  row i from row u
      Now A[u,j] will be  0 , since A[u,j]  -  A[i,j]  *  A[u,j]  =  A[u,j]  -   1   *  A[u,j]  =   0 .
    end  for
    i : =  i  +   1
  end  if
  j : =  j  +   1
end  while