written by Jarno Elonen <email@example.com>, april 2005, released into the Public Domain
The following ultra-compact Python function performs in-place Gaussian elimination for given matrix, putting it into the Reduced Row Echelon Form. It can be used to solve linear equation systems or to invert a matrix.
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 Jarno Elonen in April 2005, released into Public Domain""" (h, w) = (len(m), len(m)) 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
Warning! Integers will not work.
Make sure your matrix items support fractions. For instance, an int matrix will give you wrong results because int / int = int, i.e. they will be truncated.
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):] 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 ) ] if not len(s) else zero