Jacobi method: Difference between revisions
An optimization to the python algorithm (it was calculating in each iteration D_inv) |
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{{Short description|Iterative method used to solve a linear system of equations}} |
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{{Distinguish|Jacobi eigenvalue algorithm}} |
{{Distinguish|Jacobi eigenvalue algorithm}} |
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In [[numerical linear algebra]], the '''Jacobi method''' is an iterative algorithm for determining the solutions of a [[Diagonally dominant matrix|strictly diagonally dominant]] [[system of linear equations]]. Each diagonal element is solved for, and an approximate value is plugged in. The process is then iterated until it converges. This algorithm is a stripped-down version of the [[Jacobi eigenvalue algorithm|Jacobi transformation method of matrix diagonalization]]. The method is named after [[Carl Gustav Jacob Jacobi]]. |
In [[numerical linear algebra]], the '''Jacobi method''' (a.k.a. the '''Jacobi iteration method''') is an iterative algorithm for determining the solutions of a [[Diagonally dominant matrix|strictly diagonally dominant]] [[system of linear equations]]. Each diagonal element is solved for, and an approximate value is plugged in. The process is then iterated until it converges. This algorithm is a stripped-down version of the [[Jacobi eigenvalue algorithm|Jacobi transformation method of matrix diagonalization]]. The method is named after [[Carl Gustav Jacob Jacobi]]. |
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== Description == |
== Description == |
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Let <math>A\mathbf x = \mathbf b</math> be a square system of ''n'' linear equations, where:<math display="block">A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix}, \qquad \mathbf{x} = \begin{bmatrix} x_{1} \\ x_2 \\ \vdots \\ x_n \end{bmatrix} , \qquad \mathbf{b} = \begin{bmatrix} b_{1} \\ b_2 \\ \vdots \\ b_n \end{bmatrix}.</math> |
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Let |
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When <math>A</math> and <math>\mathbf b</math> are known, and <math>\mathbf x</math> is unknown, we can use the Jacobi method to approximate <math>\mathbf x</math>. The vector <math>\mathbf x^{(0)}</math> denotes our initial guess for <math>\mathbf x</math> (often <math>\mathbf x^{(0)}_i=0</math> for <math>i=1,2,...,n</math>). We denote <math>\mathbf{x}^{(k)}</math> as the ''k-''th approximation or iteration of <math>\mathbf{x}</math>, and <math>\mathbf{x}^{(k+1)}</math> is the next (or ''k''+1) iteration of <math>\mathbf{x}</math>. |
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:<math>A\mathbf x = \mathbf b</math> |
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=== Matrix-based formula === |
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be a square system of ''n'' linear equations, where: |
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<math>A=\begin{bmatrix} a_{11} & |
Then ''A'' can be decomposed into a [[diagonal matrix|diagonal]] component ''D'', a lower triangular part ''L'' and an upper triangular part ''U'':<math display="block">A=D+L+U \qquad \text{where} \qquad D = \begin{bmatrix} a_{11} & 0 & \cdots & 0 \\ 0 & a_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\0 & 0 & \cdots & a_{nn} \end{bmatrix} \text{ and } L+U = \begin{bmatrix} 0 & a_{12} & \cdots & a_{1n} \\ a_{21} & 0 & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & 0 \end{bmatrix}. </math>The solution is then obtained iteratively via |
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:<math> \mathbf{x}^{(k+1)} = D^{-1} (\mathbf{b} - (L+U) \mathbf{x}^{(k)}). </math> |
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=== Element-based formula === |
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Then ''A'' can be decomposed into a [[diagonal matrix|diagonal]] component ''D'', a lower triangular part ''L'' and an upper triangular part ''U'': |
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The element-based formula for each row <math>i</math> is thus:<math display="block"> x^{(k+1)}_i = \frac{1}{a_{ii}} \left(b_i -\sum_{j\ne i}a_{ij}x^{(k)}_j\right),\quad i=1,2,\ldots,n. </math>The computation of <math>x_i^{(k+1)}</math> requires each element in <math>\mathbf{x}^{(k)}</math> except itself. Unlike the [[Gauss–Seidel method]], we can't overwrite <math>x_i^{(k)}</math> with <math>x_i^{(k+1)}</math>, as that value will be needed by the rest of the computation. The minimum amount of storage is two vectors of size ''n''. |
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:<math>A=D+L+U \qquad \text{where} \qquad D = \begin{bmatrix} a_{11} & 0 & \cdots & 0 \\ 0 & a_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\0 & 0 & \cdots & a_{nn} \end{bmatrix} \text{ and } L+U = \begin{bmatrix} 0 & a_{12} & \cdots & a_{1n} \\ a_{21} & 0 & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & 0 \end{bmatrix}. </math> |
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The solution is then obtained iteratively via |
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:<math> \mathbf{x}^{(k+1)} = D^{-1} (\mathbf{b} - (L+U) \mathbf{x}^{(k)}), </math> |
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where <math>\mathbf{x}^{(k)}</math> is the ''k''th approximation or iteration of <math>\mathbf{x}</math> and <math>\mathbf{x}^{(k+1)}</math> is the next or ''k'' + 1 iteration of <math>\mathbf{x}</math>. The element-based formula is thus: |
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:<math> x^{(k+1)}_i = \frac{1}{a_{ii}} \left(b_i -\sum_{j\ne i}a_{ij}x^{(k)}_j\right),\quad i=1,2,\ldots,n. </math> |
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The computation of <math>x_i^{(k+1)}</math> requires each element in '''x'''<sup>(''k'')</sup> except itself. Unlike the [[Gauss–Seidel method]], we can't overwrite <math>x_i^{(k)}</math> with <math>x_i^{(k+1)}</math>, as that value will be needed by the rest of the computation. The minimum amount of storage is two vectors of size ''n''. |
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== Algorithm == |
== Algorithm == |
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'''Input:''' {{nowrap|initial guess < |
'''Input:''' {{nowrap|initial guess ''x''<sup>(0)</sup> to the solution}}, (diagonal dominant) matrix ''A'', right-hand side vector ''b'', convergence criterion |
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'''Output:''' {{nowrap|solution when convergence is reached}} |
'''Output:''' {{nowrap|solution when convergence is reached}} |
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'''Comments:''' |
'''Comments:''' pseudocode based on the element-based formula above |
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{{nowrap| |
{{nowrap|1=''k'' = 0}} |
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'''while''' convergence not reached '''do''' |
'''while''' convergence not reached '''do''' |
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'''for''' i := 1 '''step until''' n '''do''' |
'''for''' ''i'' := 1 '''step until''' n '''do''' |
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{{nowrap| |
{{nowrap|1=''σ'' = 0}} |
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'''for''' j := 1 '''step until''' n '''do''' |
'''for''' ''j'' := 1 '''step until''' n '''do''' |
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'''if''' j ≠ i '''then''' |
'''if''' ''j'' ≠ ''i'' '''then''' |
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{{nowrap| |
{{nowrap|1=''σ'' = ''σ'' + ''a''<sub>''ij''</sub> ''x''<sub>''j''</sub><sup>(''k'')</sup>}} |
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'''end''' |
'''end''' |
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'''end''' |
'''end''' |
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{{nowrap|< |
{{nowrap|1=''x''<sub>''i''</sub><sup>(''k''+1)</sup> = (''b''<sub>''i''</sub> − ''σ'') / ''a''<sub>''ii''</sub>}} |
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'''end''' |
'''end''' |
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increment ''k'' |
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{{nowrap|<math>k = k + 1</math>}} |
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'''end''' |
'''end''' |
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The Jacobi method sometimes converges even if these conditions are not satisfied. |
The Jacobi method sometimes converges even if these conditions are not satisfied. |
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Note that the Jacobi method does not converge for every symmetric [[positive-definite matrix]]. |
Note that the Jacobi method does not converge for every symmetric [[positive-definite matrix]]. For example, |
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<math display="block"> |
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For example |
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:<math> |
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A = |
A = |
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\begin{pmatrix} |
\begin{pmatrix} |
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</math> |
</math> |
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==Examples== |
== Examples == |
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===Example |
=== Example question === |
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A linear system of the form <math>Ax=b</math> with initial estimate <math>x^{(0)}</math> is given by |
A linear system of the form <math>Ax=b</math> with initial estimate <math>x^{(0)}</math> is given by |
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\end{bmatrix} .</math> |
\end{bmatrix} .</math> |
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We use the equation <math> x^{(k+1)}=D^{-1}(b - (L+U)x^{(k)})</math>, described above, to estimate <math>x</math>. First, we rewrite the equation in a more convenient form <math>D^{-1}(b - (L+U)x^{(k)}) = Tx^{(k)} + C</math>, where <math>T=-D^{-1}(L+U)</math> and <math>C = D^{-1}b</math>. From the known values |
We use the equation <math> x^{(k+1)}=D^{-1}(b - (L+U)x^{(k)})</math>, described above, to estimate <math>x</math>. First, we rewrite the equation in a more convenient form <math>D^{-1}(b - (L+U)x^{(k)}) = Tx^{(k)} + C</math>, where <math>T=-D^{-1}(L+U)</math> and <math>C = D^{-1}b</math>. From the known values |
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<math display=block> D^{-1}= |
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:<math> D^{-1}= |
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\begin{bmatrix} |
\begin{bmatrix} |
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1/2 & 0 \\ |
1/2 & 0 \\ |
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\end{bmatrix} .</math> |
\end{bmatrix} .</math> |
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we determine <math> T=-D^{-1}(L+U) </math> as |
we determine <math> T=-D^{-1}(L+U) </math> as |
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<math display=block> T= |
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\begin{bmatrix} |
\begin{bmatrix} |
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1/2 & 0 \\ |
1/2 & 0 \\ |
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\end{bmatrix} .</math> |
\end{bmatrix} .</math> |
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Further, <math>C</math> is found as |
Further, <math>C</math> is found as |
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<math display=block> C = |
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:<math> C = |
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\begin{bmatrix} |
\begin{bmatrix} |
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1/2 & 0 \\ |
1/2 & 0 \\ |
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\end{bmatrix}. </math> |
\end{bmatrix}. </math> |
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With <math>T</math> and <math>C</math> calculated, we estimate <math>x</math> as <math> x^{(1)}= Tx^{(0)}+C </math>: |
With <math>T</math> and <math>C</math> calculated, we estimate <math>x</math> as <math> x^{(1)}= Tx^{(0)}+C </math>: |
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<math display=block> x^{(1)}= |
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\begin{bmatrix} |
\begin{bmatrix} |
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0 & -1/2 \\ |
0 & -1/2 \\ |
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\end{bmatrix} .</math> |
\end{bmatrix} .</math> |
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The next iteration yields |
The next iteration yields |
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<math display=block> x^{(2)}= |
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\begin{bmatrix} |
\begin{bmatrix} |
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0 & -1/2 \\ |
0 & -1/2 \\ |
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.</math> |
.</math> |
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===Example 2=== |
===Example question 2=== |
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Suppose we are given the following linear system: |
Suppose we are given the following linear system: |
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If we choose {{math|(0, 0, 0, 0)}} as the initial approximation, then the first approximate solution is given by |
If we choose {{math|(0, 0, 0, 0)}} as the initial approximation, then the first approximate solution is given by |
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<math display=block> |
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:<math> |
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\begin{align} |
\begin{align} |
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x_1 & = (6 + 0 - (2 * 0)) / 10 = 0.6, \\ |
x_1 & = (6 + 0 - (2 * 0)) / 10 = 0.6, \\ |
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\end{align} |
\end{align} |
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</math> |
</math> |
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Using the approximations obtained, the iterative procedure is repeated until the desired accuracy has been reached. The following are the approximated solutions after five iterations. |
Using the approximations obtained, the iterative procedure is repeated until the desired accuracy has been reached. The following are the approximated solutions after five iterations. |
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The exact solution of the system is {{math|(1, 2, −1, 1)}}. |
The exact solution of the system is {{math|(1, 2, −1, 1)}}. |
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=== |
=== Python example === |
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<syntaxhighlight lang="numpy" line="1"> |
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The following numerical procedure simply iterates to produce the solution vector. |
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import numpy as np |
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ITERATION_LIMIT = 1000 |
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<syntaxhighlight lang="python"> |
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def jacobi(A, b, x_init, epsilon=1e-10, max_iterations=500): |
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D = np.diag(np.diag(A)) |
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LU = A - D |
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x = x_init |
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D_inv = np.diag(1 / np.diag(D)) |
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for i in range(max_iterations): |
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x_new = np.dot(D_inv, b - np.dot(LU, x)) |
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if np.linalg.norm(x_new - x) < epsilon: |
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return x_new |
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x = x_new |
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return x |
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# initialize the matrix |
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# problem data |
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A = np.array([ |
A = np.array([[10., -1., 2., 0.], |
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[ |
[-1., 11., -1., 3.], |
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[2, |
[2., -1., 10., -1.], |
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[ |
[0.0, 3., -1., 8.]]) |
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# initialize the RHS vector |
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[1, 1, 2, 8] |
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b = np.array([6., 25., -11., 15.]) |
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]) |
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b = np.array([29, 31, 26, 19]) |
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# prints the system |
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# you can choose any starting vector |
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print("System:") |
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x_init = np.zeros(len(b)) |
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for i in range(A.shape[0]): |
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x = jacobi(A, b, x_init) |
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row = [f"{A[i, j]}*x{j + 1}" for j in range(A.shape[1])] |
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print(f'{" + ".join(row)} = {b[i]}') |
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print() |
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x = np.zeros_like(b) |
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print("x:", x) |
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for it_count in range(ITERATION_LIMIT): |
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print("computed b:", np.dot(A, x)) |
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if it_count != 0: |
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print("real b:", b) |
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print(f"Iteration {it_count}: {x}") |
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</syntaxhighlight> |
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x_new = np.zeros_like(x) |
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for i in range(A.shape[0]): |
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Produces the output: |
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s1 = np.dot(A[i, :i], x[:i]) |
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s2 = np.dot(A[i, i + 1:], x[i + 1:]) |
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x_new[i] = (b[i] - s1 - s2) / A[i, i] |
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if x_new[i] == x_new[i-1]: |
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break |
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if np.allclose(x, x_new, atol=1e-10, rtol=0.): |
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<pre> |
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break |
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x: [3.99275362 2.95410628 2.16183575 0.96618357] |
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computed b: [29. 31. 26. 19.] |
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x = x_new |
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real b: [29 31 26 19] |
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</pre> |
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print("Solution: ") |
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print(x) |
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error = np.dot(A, x) - b |
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print("Error:") |
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print(error) |
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</syntaxhighlight> |
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== Weighted Jacobi method == |
== Weighted Jacobi method == |
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=== Convergence in the symmetric positive definite case === |
=== Convergence in the symmetric positive definite case === |
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In case that the system matrix <math> A </math> is of symmetric [[Positive-definite matrix|positive-definite]] type one can show convergence. |
In case that the system matrix <math> A </math> is of symmetric [[Positive-definite matrix|positive-definite]] type one can show convergence. |
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Let <math> C=C_\omega = I-\omega D^{-1}A </math> be the iteration matrix. |
Let <math> C=C_\omega = I-\omega D^{-1}A </math> be the iteration matrix. |
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The spectral radius can be minimized for a particular choice of <math> \omega = \omega_\text{opt} </math> as follows |
The spectral radius can be minimized for a particular choice of <math> \omega = \omega_\text{opt} </math> as follows |
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<math display="block"> |
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\min_\omega \rho (C_\omega) = \rho (C_{\omega_\text{opt}}) = 1-\frac{2}{\kappa(D^{-1}A)+1} |
\min_\omega \rho (C_\omega) = \rho (C_{\omega_\text{opt}}) = 1-\frac{2}{\kappa(D^{-1}A)+1} |
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\quad \text{for} \quad |
\quad \text{for} \quad |
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\omega_\text{opt} := \frac{2}{\lambda_\text{min}(D^{-1}A)+\lambda_\text{max}(D^{-1}A)} \,, |
\omega_\text{opt} := \frac{2}{\lambda_\text{min}(D^{-1}A)+\lambda_\text{max}(D^{-1}A)} \,, |
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</math> |
</math> |
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where <math> \kappa </math> is the [[ |
where <math> \kappa </math> is the [[Condition number#Matrices|matrix condition number]]. |
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==See also== |
==See also== |
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{{Numerical linear algebra}} |
{{Numerical linear algebra}} |
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{{Authority control}} |
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[[Category:Numerical linear algebra]] |
[[Category:Numerical linear algebra]] |
Latest revision as of 13:26, 18 April 2024
In numerical linear algebra, the Jacobi method (a.k.a. the Jacobi iteration method) is an iterative algorithm for determining the solutions of a strictly diagonally dominant system of linear equations. Each diagonal element is solved for, and an approximate value is plugged in. The process is then iterated until it converges. This algorithm is a stripped-down version of the Jacobi transformation method of matrix diagonalization. The method is named after Carl Gustav Jacob Jacobi.
Description
[edit]Let be a square system of n linear equations, where:
When and are known, and is unknown, we can use the Jacobi method to approximate . The vector denotes our initial guess for (often for ). We denote as the k-th approximation or iteration of , and is the next (or k+1) iteration of .
Matrix-based formula
[edit]Then A can be decomposed into a diagonal component D, a lower triangular part L and an upper triangular part U:The solution is then obtained iteratively via
Element-based formula
[edit]The element-based formula for each row is thus:The computation of requires each element in except itself. Unlike the Gauss–Seidel method, we can't overwrite with , as that value will be needed by the rest of the computation. The minimum amount of storage is two vectors of size n.
Algorithm
[edit]Input: initial guess x(0) to the solution, (diagonal dominant) matrix A, right-hand side vector b, convergence criterion Output: solution when convergence is reached Comments: pseudocode based on the element-based formula above k = 0 while convergence not reached do for i := 1 step until n do σ = 0 for j := 1 step until n do if j ≠ i then σ = σ + aij xj(k) end end xi(k+1) = (bi − σ) / aii end increment k end
Convergence
[edit]The standard convergence condition (for any iterative method) is when the spectral radius of the iteration matrix is less than 1:
A sufficient (but not necessary) condition for the method to converge is that the matrix A is strictly or irreducibly diagonally dominant. Strict row diagonal dominance means that for each row, the absolute value of the diagonal term is greater than the sum of absolute values of other terms:
The Jacobi method sometimes converges even if these conditions are not satisfied.
Note that the Jacobi method does not converge for every symmetric positive-definite matrix. For example,
Examples
[edit]Example question
[edit]A linear system of the form with initial estimate is given by
We use the equation , described above, to estimate . First, we rewrite the equation in a more convenient form , where and . From the known values we determine as Further, is found as With and calculated, we estimate as : The next iteration yields This process is repeated until convergence (i.e., until is small). The solution after 25 iterations is
Example question 2
[edit]Suppose we are given the following linear system:
If we choose (0, 0, 0, 0) as the initial approximation, then the first approximate solution is given by Using the approximations obtained, the iterative procedure is repeated until the desired accuracy has been reached. The following are the approximated solutions after five iterations.
0.6 | 2.27272 | -1.1 | 1.875 |
1.04727 | 1.7159 | -0.80522 | 0.88522 |
0.93263 | 2.05330 | -1.0493 | 1.13088 |
1.01519 | 1.95369 | -0.9681 | 0.97384 |
0.98899 | 2.0114 | -1.0102 | 1.02135 |
The exact solution of the system is (1, 2, −1, 1).
Python example
[edit]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 = [f"{A[i, j]}*x{j + 1}" for j in range(A.shape[1])]
print(f'{" + ".join(row)} = {b[i]}')
print()
x = np.zeros_like(b)
for it_count in range(ITERATION_LIMIT):
if it_count != 0:
print(f"Iteration {it_count}: {x}")
x_new = np.zeros_like(x)
for i in range(A.shape[0]):
s1 = np.dot(A[i, :i], x[:i])
s2 = np.dot(A[i, i + 1:], x[i + 1:])
x_new[i] = (b[i] - s1 - s2) / A[i, i]
if x_new[i] == x_new[i-1]:
break
if np.allclose(x, x_new, atol=1e-10, rtol=0.):
break
x = x_new
print("Solution: ")
print(x)
error = np.dot(A, x) - b
print("Error:")
print(error)
Weighted Jacobi method
[edit]The weighted Jacobi iteration uses a parameter to compute the iteration as
with being the usual choice.[1] From the relation , this may also be expressed as
- .
Convergence in the symmetric positive definite case
[edit]In case that the system matrix is of symmetric positive-definite type one can show convergence.
Let be the iteration matrix. Then, convergence is guaranteed for
where is the maximal eigenvalue.
The spectral radius can be minimized for a particular choice of as follows where is the matrix condition number.
See also
[edit]- Gauss–Seidel method
- Successive over-relaxation
- Iterative method § Linear systems
- Gaussian Belief Propagation
- Matrix splitting
References
[edit]- ^ Saad, Yousef (2003). Iterative Methods for Sparse Linear Systems (2nd ed.). SIAM. p. 414. ISBN 0898715342.
External links
[edit]- This article incorporates text from the article Jacobi_method on CFD-Wiki that is under the GFDL license.
- Black, Noel; Moore, Shirley & Weisstein, Eric W. "Jacobi method". MathWorld.
- Jacobi Method from www.math-linux.com