fork(1) download
  1. #include <stdio.h>
  2. #include <math.h>
  3.  
  4. #define N 4
  5. #define MAX_ITER 150
  6. #define EPS 1e-7
  7.  
  8. void print_system(double A[N][N], double b[N]);
  9. void jacobi_solver(double A[N][N], double b[N]);
  10. void gauss_seidel_solver(double A[N][N], double b[N]);
  11. void verify_solution(double A[N][N], double b[N], double x[N]);
  12.  
  13. int main() {
  14. double A[N][N] = {
  15. { 5.0, -1.0, -2.0, 0.0},
  16. { 0.0, 8.0, 1.0, 7.0},
  17. {-2.0, -7.0, -4.0, 0.0},
  18. { 4.0, 0.0, -3.0, -25.0}
  19. };
  20. double b[N] = {-1.32, 14.41, 0.70, -0.03};
  21.  
  22. print_system(A, b);
  23.  
  24. printf("\n======================================================\n");
  25. printf(" 1. ヤコビの反復法\n");
  26. printf("======================================================\n");
  27. jacobi_solver(A, b);
  28.  
  29. printf("\n======================================================\n");
  30. printf(" 2. ガウス・ザイデル法\n");
  31. printf("======================================================\n");
  32. gauss_seidel_solver(A, b);
  33.  
  34. return 0;
  35. }
  36.  
  37. void print_system(double A[N][N], double b[N]) {
  38. printf("対角優位化後の方程式:\n");
  39. for (int i = 0; i < N; i++) {
  40. printf("[ ");
  41. for (int j = 0; j < N; j++) {
  42. printf("%6.1f ", A[i][j]);
  43. }
  44. printf("] [x%d] = [ %6.2f ]\n", i + 1, b[i]);
  45. }
  46. }
  47.  
  48. void jacobi_solver(double A[N][N], double b[N]) {
  49. double x[N] = {0.0, 0.0, 0.0, 0.0};
  50. double x_new[N];
  51. int converged = 0;
  52.  
  53. printf("計算回数k\t残差2乗和\t\tx1\t\tx2\t\tx3\t\tx4\n");
  54. printf("------------------------------------------------------------------------\n");
  55.  
  56. for (int k = 1; k <= MAX_ITER; k++) {
  57. for (int i = 0; i < N; i++) {
  58. double sum = 0.0;
  59. for (int j = 0; j < N; j++) {
  60. if (j != i) {
  61. sum += A[i][j] * x[j];
  62. }
  63. }
  64. x_new[i] = (b[i] - sum) / A[i][i];
  65. }
  66.  
  67. double rss = 0.0;
  68. for (int i = 0; i < N; i++) {
  69. rss += pow(x_new[i] - x[i], 2);
  70. }
  71.  
  72. printf("%d\t\t%.2e\t\t%.5f\t%.5f\t%.5f\t%.5f\n", k, rss, x_new[0], x_new[1], x_new[2], x_new[3]);
  73.  
  74. for (int i = 0; i < N; i++) {
  75. x[i] = x_new[i];
  76. }
  77.  
  78. if (rss < EPS) {
  79. printf("\n>> 収束判定条件を満足しました(計算回数: %d回)\n", k);
  80. converged = 1;
  81. break;
  82. }
  83. }
  84.  
  85. if (!converged) {
  86. printf("\n警告: 指定された計算回数以内に収束しませんでした。\n");
  87. }
  88.  
  89. printf("\n=== 連立方程式の最終解 (ヤコビの反復法) ===\n");
  90. for (int i = 0; i < N; i++) {
  91. printf("x[%d] = %10.6f\n", i + 1, x[i]);
  92. }
  93.  
  94. verify_solution(A, b, x);
  95. }
  96.  
  97. void gauss_seidel_solver(double A[N][N], double b[N]) {
  98. double x[N] = {0.0, 0.0, 0.0, 0.0};
  99. double x_old[N];
  100. int converged = 0;
  101.  
  102. printf("計算回数k\t残差2乗和\t\tx1\t\tx2\t\tx3\t\tx4\n");
  103. printf("------------------------------------------------------------------------\n");
  104.  
  105. for (int k = 1; k <= MAX_ITER; k++) {
  106. for (int i = 0; i < N; i++) {
  107. x_old[i] = x[i];
  108. }
  109.  
  110. for (int i = 0; i < N; i++) {
  111. double sum = 0.0;
  112. for (int j = 0; j < N; j++) {
  113. if (j != i) {
  114. sum += A[i][j] * x[j];
  115. }
  116. }
  117. x[i] = (b[i] - sum) / A[i][i];
  118. }
  119.  
  120. double rss = 0.0;
  121. for (int i = 0; i < N; i++) {
  122. rss += pow(x[i] - x_old[i], 2);
  123. }
  124.  
  125. printf("%d\t\t%.2e\t\t%.5f\t%.5f\t%.5f\t%.5f\n", k, rss, x[0], x[1], x[2], x[3]);
  126.  
  127. if (rss < EPS) {
  128. printf("\n>> 収束判定条件を満足しました(計算回数: %d回)\n", k);
  129. converged = 1;
  130. break;
  131. }
  132. }
  133.  
  134. if (!converged) {
  135. printf("\n警告: 指定された計算回数以内に収束しませんでした。\n");
  136. }
  137.  
  138. printf("\n=== 連立方程式の最終解 (ガウス・ザイデル法) ===\n");
  139. for (int i = 0; i < N; i++) {
  140. printf("x[%d] = %10.6f\n", i + 1, x[i]);
  141. }
  142.  
  143. verify_solution(A, b, x);
  144. }
  145.  
  146. void verify_solution(double A[N][N], double b[N], double x[N]) {
  147. printf("\n--- 求まったものが解である確認(コード内での検算: Ax == b) ---\n");
  148. for (int i = 0; i < N; i++) {
  149. double lhs_calc = 0.0;
  150. for (int j = 0; j < N; j++) {
  151. lhs_calc += A[i][j] * x[j];
  152. }
  153. double error = lhs_calc - b[i];
  154. printf("%d行目: 計算値 Ax = %10.5f | 設定値 b = %10.5f | 誤差 = %e\n",
  155. i + 1, lhs_calc, b[i], error);
  156. }
  157. }
Success #stdin #stdout 0s 5324KB
stdin
Standard input is empty
stdout
対角優位化後の方程式:
[    5.0   -1.0   -2.0    0.0 ] [x1] = [  -1.32 ]
[    0.0    8.0    1.0    7.0 ] [x2] = [  14.41 ]
[   -2.0   -7.0   -4.0    0.0 ] [x3] = [   0.70 ]
[    4.0    0.0   -3.0  -25.0 ] [x4] = [  -0.03 ]

======================================================
 1. ヤコビの反復法
======================================================
計算回数k	残差2乗和		x1		x2		x3		x4
------------------------------------------------------------------------
1		3.34e+00		-0.26400	1.80125	-0.17500	0.00120
2		9.21e+00		0.02625	1.82208	-3.19519	-0.02004
3		1.81e+00		-1.17766	2.21818	-3.37676	0.38882
4		1.50e-01		-1.17107	1.88312	-3.46799	0.21799
5		3.77e-01		-1.27457	2.04401	-2.88494	0.22999
6		1.38e-01		-1.00917	1.96063	-3.11474	0.14346
7		2.78e-02		-1.11777	2.06506	-3.10151	0.21350
8		2.15e-02		-1.09159	2.00213	-3.22998	0.19454
9		1.50e-02		-1.15557	2.03478	-3.13292	0.21414
10		4.03e-03		-1.11021	2.00549	-3.15808	0.19226
11		1.67e-03		-1.12613	2.02778	-3.12950	0.20253
12		1.41e-03		-1.11024	2.01522	-3.16055	0.19656
13		5.42e-04		-1.12518	2.02433	-3.14651	0.20283
14		1.96e-04		-1.11774	2.01709	-3.15499	0.19875
15		1.30e-04		-1.12258	2.02171	-3.14604	0.20096
16		6.52e-05		-1.11807	2.01866	-3.15171	0.19911
17		2.52e-05		-1.12095	2.02099	-3.14863	0.20051
18		1.31e-05		-1.11925	2.01938	-3.15126	0.19968
19		7.24e-06		-1.12063	2.02043	-3.14929	0.20027
20		3.13e-06		-1.11963	2.01967	-3.15045	0.19981
21		1.46e-06		-1.12024	2.02022	-3.14962	0.20011
22		7.84e-07		-1.11980	2.01985	-3.15026	0.19991
23		3.72e-07		-1.12013	2.02011	-3.14984	0.20006
24		1.70e-07		-1.11992	2.01993	-3.15012	0.19996
25		8.60e-08		-1.12006	2.02005	-3.14991	0.20003

>> 収束判定条件を満足しました(計算回数: 25回)

=== 連立方程式の最終解 (ヤコビの反復法) ===
x[1] =  -1.120063
x[2] =   2.020050
x[3] =  -3.149911
x[4] =   0.200028

--- 求まったものが解である確認(コード内での検算: Ax == b) ---
1行目: 計算値 Ax =   -1.32054 | 設定値 b =   -1.32000 | 誤差 = -5.434164e-04
2行目: 計算値 Ax =   14.41069 | 設定値 b =   14.41000 | 誤差 = 6.884127e-04
3行目: 計算値 Ax =    0.69942 | 設定値 b =    0.70000 | 誤差 = -5.808020e-04
4行目: 計算値 Ax =   -0.03122 | 設定値 b =   -0.03000 | 誤差 = -1.219528e-03

======================================================
 2. ガウス・ザイデル法
======================================================
計算回数k	残差2乗和		x1		x2		x3		x4
------------------------------------------------------------------------
1		1.36e+01		-0.26400	1.80125	-3.19519	0.34238
2		9.66e-01		-1.18182	1.90106	-2.91095	0.16142
3		1.15e-01		-1.04817	2.02387	-3.19270	0.21662
4		1.29e-02		-1.13630	2.01080	-3.12574	0.19448
5		1.74e-03		-1.11214	2.02180	-3.15708	0.20211
6		2.22e-04		-1.12247	2.01904	-3.14709	0.19925
7		2.97e-05		-1.11903	2.02029	-3.15099	0.20027
8		3.88e-06		-1.12034	2.01988	-3.14963	0.19990
9		5.15e-07		-1.11987	2.02004	-3.15013	0.20004
10		6.77e-08		-1.12005	2.01999	-3.14995	0.19999

>> 収束判定条件を満足しました(計算回数: 10回)

=== 連立方程式の最終解 (ガウス・ザイデル法) ===
x[1] =  -1.120045
x[2] =   2.019985
x[3] =  -3.149951
x[4] =   0.199987

--- 求まったものが解である確認(コード内での検算: Ax == b) ---
1行目: 計算値 Ax =   -1.32031 | 設定値 b =   -1.32000 | 誤差 = -3.084483e-04
2行目: 計算値 Ax =   14.40984 | 設定値 b =   14.41000 | 誤差 = -1.626182e-04
3行目: 計算値 Ax =    0.70000 | 設定値 b =    0.70000 | 誤差 = -6.661338e-16
4行目: 計算値 Ax =   -0.03000 | 設定値 b =   -0.03000 | 誤差 = -2.498002e-16