fixed dependencies
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85
vendor/github.com/nuknal/goNum/LEs_SORIterate.go
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85
vendor/github.com/nuknal/goNum/LEs_SORIterate.go
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// LEs_SORIterate
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/*
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------------------------------------------------------
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作者 : Black Ghost
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日期 : 2018-11-22
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版本 : 0.0.0
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------------------------------------------------------
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解n阶线性方程组的SOR(逐次超松弛, successive over
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relaxation)迭代法
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理论:
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参考 李信真, 车刚明, 欧阳洁, 等. 计算方法. 西北工业大学
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出版社, 2000, pp 68-72.
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收敛的条件:(B为变化后的系数矩阵)
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1. 系数矩阵A严格对角占优,且0 < omega <= 1,或者
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2. 系数矩阵A对称正定,且0 < omega < 2
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------------------------------------------------------
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输入 :
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A 系数矩阵
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b 常数值向量
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tol 最大容许误差
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omega 松弛因子,0 < omega < 2, omega = 1: Siedel,
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omega < 1: 低松弛, omega > 1: 超松弛
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n 最大迭代步数
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输出 :
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sol 解向量
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err 解出标志:false-未解出或达到步数上限;
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true-全部解出
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------------------------------------------------------
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*/
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package goNum
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import "math"
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// LEs_SORIterate 解n阶线性方程组的SOR(逐次超松弛, successive over relaxation)迭代法
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func LEs_SORIterate(A, b, x0 Matrix, tol, omega float64, n int) ([]float64, bool) {
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/*
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解n阶线性方程组的SOR(逐次超松弛, successive over relaxation)迭代法
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输入 :
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A 系数矩阵
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b 常数值向量
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tol 最大容许误差
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omega 松弛因子,0 < omega < 2, omega = 1: Siedel,
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omega < 1: 低松弛, omega > 1: 超松弛
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n 最大迭代步数
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输出 :
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sol 解向量
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err 解出标志:false-未解出或达到步数上限;
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true-全部解出
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*/
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x1 := ZeroMatrix(A.Rows, 1)
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sol := ZeroMatrix(A.Rows, 1)
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var err bool = false
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//求解
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for i := 0; i < n; i++ {
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for i0 := 0; i0 < A.Rows; i0++ {
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sum0 := 0.0
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for j := 0; j < i0; j++ {
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sum0 += A.GetFromMatrix(i0, j) * x1.GetFromMatrix(j, 0)
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}
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sum1 := 0.0
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for j := i0 + 1; j < A.Columns; j++ {
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sum1 += A.GetFromMatrix(i0, j) * x0.GetFromMatrix(j, 0)
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}
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x1.SetMatrix(i0, 0, (1-omega)*x0.GetFromMatrix(i0, 0)+omega*(b.Data[i0]-sum0-sum1)/A.GetFromMatrix(i0, i0))
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}
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//判断收敛
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sol = SubMatrix(x1, x0)
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max, _, _ := Max(sol.Data)
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if math.Abs(max) < tol {
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sol = x1
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err = true
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return sol.Data, err
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}
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//准备下次迭代
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for i0 := 0; i0 < x0.Rows; i0++ {
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x0.Data[i0] = x1.Data[i0]
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}
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}
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return make([]float64, A.Rows), err
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}
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