ଗସ୍ - ସିଡ଼ାଲ୍ ପଦ୍ଧତି

${\displaystyle A\mathbf {x} =\mathbf {b} }$ .

iterationଦ୍ୱାରା ନିମ୍ନ ମତେ ଏହାର ସଂଜ୍ଞା ପ୍ରକରଣ କରାଯାଇପାରେ

ଏବଂ  ${\displaystyle \mathbf {x} ^{(k)}}$  is the kth approximation or iteration of ${\displaystyle \mathbf {x} ,\,\mathbf {x} ^{k+1}}$  is the next or k + 1 iteration of ${\displaystyle \mathbf {x} }$ , and the matrix A is decomposed into a lower triangular component ${\displaystyle L_{*}}$ , and a strictly upper triangular component U: ${\displaystyle A=L_{*}+U}$ .

Inputs: A, A Output: ${\displaystyle \phi }$   Choose an initial guess ${\displaystyle \phi }$  to the solution repeat until convergence     for A from 1 until A do         ${\displaystyle \sigma \leftarrow 0}$          for A from 1 until A do             if A ≠ A then                 ${\displaystyle \sigma \leftarrow \sigma +a_{ij}\phi _{j}}$              end if         end (A-loop)         ${\displaystyle \phi _{i}\leftarrow {\frac {1}{a_{ii}}}(b_{i}-\sigma )}$      end (A-loop)     check if convergence is reached end (repeat) 

ମାଟ୍ରିକ୍ସ ସଂସ୍କରଣ ପାଇଁ ଗୋଟିଏ ଉଦାହରଣ

${\displaystyle A\mathbf {x} =\mathbf {b} }$  ଭାବେ ଗୋଟିଏ linear system ଦର୍ଶଯାଇଛି, ଯଥା:

${\displaystyle A={\begin{bmatrix}16&3\\7&-11\\\end{bmatrix}}}$  ଏବଂ ${\displaystyle b={\begin{bmatrix}11\\13\end{bmatrix}}.}$ 

ଆମେ ନିମ୍ନୋକ୍ତ ସମୀକରଣ

କୁ ନିମ୍ନୋକ୍ତ ରୂପେ ବ୍ୟବହାର କରିବାକୁ ଚାହୁଁ

ଯେଉଁଠାରେ କି:

${\displaystyle T=-L_{*}^{-1}U}$  ଏବଂ ${\displaystyle C=L_{*}^{-1}\mathbf {b} .}$ 

ପାଇଥନ ୩ ଏବଂ NumPy ବ୍ୟବହାର କରାଯାଇଥିବା ଏକ ଉଦାହରଣ

The following numerical procedure simply iterates to produce the solution vector.

import numpy as np  ITERATION_LIMIT = 1000  # initialize the matrix A = np.array([[10., -1., 2., 0.],               [-1., 11., -1., 3.],               [2., -1., 10., -1.],               [0.0, 3., -1., 8.]]) # initialize the RHS vector b = np.array([6., 25., -11., 15.])  # prints the system print("System:") for i in range(A.shape[0]):     row = ["{}*x{}".format(A[i, j], j + 1) for j in range(A.shape[1])]     print(" + ".join(row), "=", b[i]) print()  x = np.zeros_like(b) for it_count in range(ITERATION_LIMIT):     print("Current solution:", x)     x_new = np.zeros_like(x)      for i in range(A.shape[0]):         s1 = np.dot(A[i, :i], x_new[:i])         s2 = np.dot(A[i, i + 1:], x[i + 1:])         x_new[i] = (b[i] - s1 - s2) / A[i, i]      if np.allclose(x, x_new, rtol=1e-8):         break      x = x_new  print("Solution:") print(x) error = np.dot(A, x) - b print("Error:") print(error) 

ଏହାର output ଆସିବ:

System: 10.0*x1 + -1.0*x2 + 2.0*x3 + 0.0*x4 = 6.0 -1.0*x1 + 11.0*x2 + -1.0*x3 + 3.0*x4 = 25.0 2.0*x1 + -1.0*x2 + 10.0*x3 + -1.0*x4 = -11.0 0.0*x1 + 3.0*x2 + -1.0*x3 + 8.0*x4 = 15.0  Current solution: [ 0.  0.  0.  0.] Current solution: [ 0.6         2.32727273 -0.98727273  0.87886364] Current solution: [ 1.03018182  2.03693802 -1.0144562   0.98434122] Current solution: [ 1.00658504  2.00355502 -1.00252738  0.99835095] Current solution: [ 1.00086098  2.00029825 -1.00030728  0.99984975] Current solution: [ 1.00009128  2.00002134 -1.00003115  0.9999881 ] Current solution: [ 1.00000836  2.00000117 -1.00000275  0.99999922] Current solution: [ 1.00000067  2.00000002 -1.00000021  0.99999996] Current solution: [ 1.00000004  1.99999999 -1.00000001  1.        ] Current solution: [ 1.  2. -1.  1.] Solution: [ 1.  2. -1.  1.] Error: [  2.06480930e-08  -1.25551054e-08   3.61417563e-11   0.00000000e+00] 

Program to solve arbitrary no. of equations using Matlab

The following code uses the formula

${\displaystyle x_{i}^{(k+1)}={\frac {1}{a_{ii}}}\left(b_{i}-\sum _{ji}a_{ij}x_{j}^{(k)}\right),\quad i,j=1,2,\ldots ,n}$ 

function [x] = gauss_seidel(A, b, x0, iters)     n = length(A);     x = x0;     for k = 1:iters         for i = 1:n             x(i) = (1/A(i, i))*(b(i) - A(i, 1:n)*x + A(i, i)*x(i));         end     end end 

• Jacobi method
• Successive over-relaxation
• Iterative method. Linear systems
• Gaussian belief propagation
• Matrix splitting
• Richardson iteration