修改龙格库塔bug

This commit is contained in:
2025-06-17 00:07:14 +08:00
parent 08d293657d
commit fc0e7cec5f
7 changed files with 357 additions and 1 deletions

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@@ -0,0 +1,102 @@
import math
#列主元高斯消元法
def SovleRowMain(A,b,round_num=15):
ks = 0.00000001
n = len(A)
if len(A[0]) != n:
print("A要为方阵")
return None, None, None, None
if len(b) != n:
print("b与A的行数不匹配")
return None, None, None, None
p = list(range(n))
for i in range(n):
row_max = abs(A[i][i])
row_max_index = i
for j in range(i + 1, n):
if abs(A[j][i]) > row_max:
row_max = abs(A[j][i])
row_max_index = j
A[i], A[row_max_index] = A[row_max_index], A[i]
b[i], b[row_max_index] = b[row_max_index], b[i]
p[i], p[row_max_index] = p[row_max_index], p[i]
if abs(A[i][i]) < ks:
print("A矩阵奇异无法进行高斯消元")
return None, None, None, None
for j in range(i + 1, n):
m = round(A[j][i] / A[i][i],round_num)
A[j][i] = m
for k in range(i + 1, n):
A[j][k] -= round(m * A[i][k],round_num)
b[j] -= round(m * b[i],round_num)
if abs(A[n - 1][n - 1]) < ks:
print("A矩阵奇异无法进行高斯消元")
return None, None, None, None
# 回代求解
b[n - 1] = round(b[n - 1]/A[n - 1][n - 1],round_num)
for i in range(n - 2, -1, -1):
for j in range(i + 1, n):
b[i] -= round(A[i][j] * b[j],round_num)
b[i] /= round(A[i][i])
b = [round(b[i], round_num) for i in range(n)]
# 得到L,U和P矩阵
L = [[0 for i in range(n)] for j in range(n)]
U = [[0 for i in range(n)] for j in range(n)]
P = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
for j in range(n):
if i == j:
L[i][j] = 1
U[i][j] = A[i][j]
elif i < j:
U[i][j] = A[i][j]
else:
L[i][j] = A[i][j]
P[i][p[i]] = 1
return P,L,U,b
#迭代改善法
def IterativeMethod(A, b, err, N, fake_round_num=15):
b_c = [b[i] for i in range(len(b))]
A_c = [[A[i][j] for j in range(len(A[0]))] for i in range(len(A))]
P,L,U,x0 = SovleRowMain(A_c, b_c,fake_round_num)
print(L)
print(U)
print(f"初始解为: {x0}")
count = 0
while count<N:
r1 = [b[i] - sum([A[i][j] * x0[j] for j in range(len(A[0]))]) for i in range(len(A))]
A_c = [[A[i][j] for j in range(len(A[0]))] for i in range(len(A))]
d1 = SovleRowMain(A_c, r1,fake_round_num)[3]
x0 = [x0[i] + d1[i] for i in range(len(x0))]
print(f"{count+1}次迭代, r{count+1} = {r1}, d{count+1} = {d1}, x{count+2} = {x0}")
err_now = max(abs(r1[i]) for i in range(len(r1)))
count += 1
if err_now < err:
break
return x0,count
if __name__ == "__main__":
##########################################################################
#把矩阵改成题干的矩阵b改成题干结果err精度要求修改##########################
A = [
[51,82],
[151/3,81]
]
b = [235,232]
err = 1e-4 # 精度要求
N = 1000 # 迭代次数上限
fake_round_num = 4 # 模拟的四舍五入精度,根据题目情况或者凑过程修改
x = IterativeMethod(A, b, err, N, fake_round_num)[0]
print(f"解为: {x}")
# 先求范数与逆矩阵(条件数)