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  1. #include <stdio.h>
  2. #include <stdlib.h>
  3. #include <math.h>
  4.  
  5. #define SIZE 4
  6. #define EPSILON_POWER 1e-9  
  7. #define EPSILON_NEWTON 1e-7  
  8. #define MAX_ITER 1000  
  9.  
  10.  
  11. double vector_norm(double *v, int n) {
  12.     double sum = 0.0;
  13.     for (int i = 0; i < n; i++) {
  14.         sum += v[i] * v[i];
  15.     }
  16.     return sqrt(sum);
  17. }
  18.  
  19. void mat_vec_mul(double A[SIZE][SIZE], double *x, double *y, int n) {
  20.     for (int i = 0; i < n; i++) {
  21.         y[i] = 0.0;
  22.         for (int j = 0; j < n; j++) {
  23.             y[i] += A[i][j] * x[j];
  24.         }
  25.     }
  26. }
  27.  
  28. double f(double lambda) {
  29.     return pow(lambda, 4) + pow(lambda, 3) - 54.0 * pow(lambda, 2) - 224.0 * lambda - 345.0;
  30. }
  31.  
  32. double df(double lambda) {
  33.     return 4.0 * pow(lambda, 3) + 3.0 * pow(lambda, 2) - 108.0 * lambda - 224.0;
  34. }
  35.  
  36.  
  37. void solve_newton(int attempt, double lambda0) {
  38.     double lambda = lambda0;
  39.     int iter = 0;
  40.  
  41.     while (iter <= MAX_ITER) {
  42.         double f_val = f(lambda);
  43.         double df_val = df(lambda);
  44.  
  45.         if (fabs(df_val) < 1e-12) return; 
  46.  
  47.         double delta = f_val / df_val;
  48.         lambda -= delta;
  49.         iter++;
  50.  
  51.         if (fabs(delta) < EPSILON_NEWTON) {
  52.             printf("試行 #%d (初期値 %5.1f) -> 収束成功 (%2d回) | 固有値解 λ = %.10f\n", 
  53.                    attempt, lambda0, iter, lambda);
  54.             return;
  55.         }
  56.     }
  57. }
  58.  
  59.  
  60. int main() {
  61.     double A[SIZE][SIZE] = {
  62.         {1.0,  2.0, 3.0,  5.0},
  63.         {3.0, -5.0, 1.0,  4.0},
  64.         {5.0,  9.0, 2.0,  -6.0},
  65.         {1.0,  7.0, 4.0,  1.0}
  66.     };
  67.  
  68.     double x[SIZE], y[SIZE], Ax[SIZE];
  69.     double lambda_old = 0.0, lambda_new = 0.0;
  70. 	 printf("                べき乗法              \n");
  71.     for (int i = 0; i < SIZE; i++) {
  72.         x[i] = 1.0;
  73.     }
  74.  
  75.     printf("初期ベクトル x(0): [%.1f, %.1f, %.1f, %.1f]\n\n", x[0], x[1], x[2], x[3]);
  76.     printf("%-5s | %-12s | %-12s | %-30s\n", "Iter", "固有値(新)", "差分値", "固有ベクトルx(k)");
  77.  
  78.     int actual_iters = 0;
  79.     for (int iter = 1; iter <= MAX_ITER; iter++) {
  80.         actual_iters = iter;
  81.         mat_vec_mul(A, x, y, SIZE);
  82.  
  83.         double norm_y = vector_norm(y, SIZE);
  84.         if (norm_y < 1e-12) break;
  85.         for (int i = 0; i < SIZE; i++) {
  86.             y[i] /= norm_y;
  87.         }
  88.  
  89.         mat_vec_mul(A, y, Ax, SIZE);
  90.         double num = 0.0, den = 0.0;
  91.         for (int i = 0; i < SIZE; i++) {
  92.             num += y[i] * Ax[i];
  93.             den += y[i] * y[i];
  94.         }
  95.         lambda_new = num / den;
  96.         double diff = fabs(lambda_new - lambda_old);
  97.  
  98.         if (iter <= 5 || iter % 5 == 0) {
  99.             printf("%-5d | %-12.8f | %-12.4e | [%.5f, %.5f, %.5f, %.5f]\n", 
  100.                    iter, lambda_new, diff, y[0], y[1], y[2], y[3]);
  101.         }
  102.  
  103.         if (diff < EPSILON_POWER) {
  104.             if (iter % 5 != 0 && iter > 5) {
  105.                 printf("%-5d | %-12.8f | %-12.4e | [%.5f, %.5f, %.5f, %.5f]\n", 
  106.                        iter, lambda_new, diff, y[0], y[1], y[2], y[3]);
  107.             }
  108.             break;
  109.         }
  110.  
  111.         for (int i = 0; i < SIZE; i++) {
  112.             x[i] = y[i];
  113.         }
  114.         lambda_old = lambda_new;
  115.     }
  116.  
  117.     printf("反復回数 %d 回で収束しました。\n", actual_iters);
  118.     printf("\n最大固有値 (λ1)          : %.10f\n", lambda_new);
  119.     printf("対応する固有ベクトル (x1) : [");
  120.     for (int i = 0; i < SIZE; i++) {
  121.         printf("%.10f%s", y[i], (i == SIZE - 1) ? "]\n" : ", ");
  122.     }
  123.  
  124.     printf("\n    Ax と λx の比較   \n");
  125.     mat_vec_mul(A, y, Ax, SIZE);
  126.     printf("Ax    : [");
  127.     for (int i = 0; i < SIZE; i++) {
  128.         printf("%.6f%s", Ax[i], (i == SIZE - 1) ? "]\n" : ", ");
  129.     }
  130.     printf("λ * x : [");
  131.     for (int i = 0; i < SIZE; i++) {
  132.         printf("%.6f%s", lambda_new * y[i], (i == SIZE - 1) ? "]\n" : ", ");
  133.     }
  134.  
  135.     printf("\n\n");
  136.     printf("   固有値が最大固有値であることの確認:ニュートン・ラフソン法     \n");
  137.     printf("対象の特性方程式: f(λ) = λ^4 - 2.0λ^3 - 43.0λ^2 - 36.0λ + 130.0 = 0\n\n");
  138.  
  139.     double initial_lambda[] = {10.0, 1.0, -3.5, -6.0};
  140.     for (int i = 0; i < 4; i++) {
  141.         solve_newton(i + 1, initial_lambda[i]);
  142.     }
  143.     return 0;}
  144.  
  145.  
Success #stdin #stdout 0s 5324KB
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                べき乗法              
初期ベクトル x(0): [1.0, 1.0, 1.0, 1.0]

Iter  | 固有値(新) | 差分値    | 固有ベクトルx(k)        
1     | 7.06015038   | 7.0602e+00   | [0.55069, 0.15019, 0.50063, 0.65081]
2     | 7.56272200   | 5.0257e-01   | [0.68485, 0.48918, 0.14675, 0.51975]
3     | 7.79665790   | 2.3394e-01   | [0.53345, 0.20811, 0.56744, 0.59172]
4     | 8.36088120   | 5.6422e-01   | [0.66241, 0.41253, 0.25086, 0.57281]
5     | 8.52180634   | 1.6093e-01   | [0.58882, 0.28456, 0.47179, 0.59139]
10    | 8.69758623   | 9.4221e-03   | [0.62524, 0.34192, 0.38099, 0.58908]
15    | 8.69591999   | 7.0277e-04   | [0.62205, 0.33644, 0.39006, 0.58967]
20    | 8.69624282   | 7.2845e-05   | [0.62237, 0.33699, 0.38916, 0.58962]
25    | 8.69621204   | 7.2757e-06   | [0.62234, 0.33693, 0.38925, 0.58962]
30    | 8.69621514   | 7.2934e-07   | [0.62235, 0.33694, 0.38924, 0.58962]
35    | 8.69621483   | 7.3086e-08   | [0.62234, 0.33694, 0.38924, 0.58962]
40    | 8.69621486   | 7.3240e-09   | [0.62234, 0.33694, 0.38924, 0.58962]
45    | 8.69621486   | 7.3395e-10   | [0.62234, 0.33694, 0.38924, 0.58962]
反復回数 45 回で収束しました。

最大固有値 (λ1)          : 8.6962148610
対応する固有ベクトル (x1) : [0.6223447229, 0.3369369049, 0.3892384556, 0.5896219066]

    Ax と λx の比較   
Ax    : [5.412043, 2.930076, 3.384901, 5.127479]
λ * x : [5.412043, 2.930076, 3.384901, 5.127479]


   固有値が最大固有値であることの確認:ニュートン・ラフソン法     
対象の特性方程式: f(λ) = λ^4 - 2.0λ^3 - 43.0λ^2 - 36.0λ + 130.0 = 0

試行 #1 (初期値  10.0) -> 収束成功 ( 5回) | 固有値解 λ = 8.6962148612
試行 #2 (初期値   1.0) -> 収束成功 (15回) | 固有値解 λ = -5.4892537611
試行 #3 (初期値  -3.5) -> 収束成功 (11回) | 固有値解 λ = -5.4892537611
試行 #4 (初期値  -6.0) -> 収束成功 ( 5回) | 固有値解 λ = -5.4892537611
https://ideone.com/GsbLsF
language:
C (gcc 8.3)
created:
1 day ago
visibility:
public

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