修改龙格库塔bug

279-3.py

@@ -7,7 +7,7 @@ def ClassicRK(x0,y0,h,xk,fxy):
         k2 = fxy(x0+h/2,y0+h*k1/2)
         k3 = fxy(x0+h/2,y0+h*k2/2)
         k4 = fxy(x0+h,y0+h*k3)
-        y0 += h*(k1+4*k2+k3)/6
+        y0 += h*(k1+2*k2+2*k3+k4)/6
         x0 += h
         result.append((x0,y0))
     return result

按方法整理/常微分方程-经典龙格-库塔格式.py

@@ -0,0 +1,29 @@
+import math
+
+#经典R-K，龙格-库塔法
+def ClassicRK(x0,y0,h,xk,fxy):
+    k1=k2=k3=k4=0
+    result = [(x0,y0)]
+    while x0<=xk:
+        k1 = fxy(x0,y0)
+        k2 = fxy(x0+h/2,y0+h*k1/2)
+        k3 = fxy(x0+h/2,y0+h*k2/2)
+        k4 = fxy(x0+h,y0+h*k3)
+        y0 += h*(k1+2*k2+2*k3+k4)/6
+        x0 += h
+        result.append((x0,y0))
+    return result
+
+
+if __name__=="__main__":
+    ##########################################################################
+    x0 = 0   # x的初始值（左边界）换成题干里面的#################
+    y0 = -1  # y的初始值换成题干里面的#################
+    fxy = lambda x,y: x + y    #f(x,y)换成题干里面的#################
+    real_fx = lambda x: -x-1     #真实函数换成题干里面的#################
+    h = 0.1         #步长换成题干里面的#################
+    xk = 2           #x的右边界换成题干里面的#################
+    result = ClassicRK(x0, y0, h, xk, fxy)
+    print("x\ty\t\treal_y\t\t\t误差")
+    for x, y in result:
+        print(f"{x:.2f}\t{y:.10f}\t\t{real_fx(x):.10f}\t\t\t{abs(y - real_fx(x)):.5f}")

按方法整理/矩阵-SOR逐次超松弛迭代法.py

@@ -0,0 +1,62 @@
+import math
+
+#模 范数
+def Norm(x,v):
+    if len(x[0]) == 1:
+        if v == 1:
+            return sum([abs(i[0]) for i in x])
+        elif v == 2:
+            return (sum([i[0]**2 for i in x]))**0.5
+        elif v == float("inf"):
+            return max([abs(i[0]) for i in x])
+    else:
+        if v == 1:
+            return max([sum([abs(x[i][j]) for i in range(len(x))]) for j in range(len(x[0]))])
+        elif v == float("inf"):
+            return max([sum([abs(i) for i in x[j]]) for j in range(len(x))])
+    return None
+
+#SOR方法 逐次超松弛迭代
+def SOR(A,b,x,w,err,N):
+    count = 0
+    n = len(A)
+    while True:
+        count += 1
+        for i in range(n):
+            sum1 = sum(A[i][j] * x[j] for j in range(i))
+            sum2 = sum(A[i][j] * x[j] for j in range(i, n))
+            x[i] += w*(b[i] - sum1 - sum2) / A[i][i]
+        r = [[b[i] - sum(A[i][j] * x[j] for j in range(len(A[0])))] for i in range(n)]
+        err_now = Norm(r, float("inf"))
+        x_t = [round(i,5) for i in x]
+        print(f"第{count}次迭代, 误差 = {err_now:.5}, x = {x_t}")
+        if err_now < err:
+            return x, count, 1
+        
+        if count > N:
+            return None,count, 0
+        
+
+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 = [
+        [4,-1,0,-1,0,0],
+        [-1,4,-1,0,-1,0],
+        [0,-1,4,0,0,-1],
+        [-1,0,0,4,-1,0],
+        [0,-1,0,-1,4,-1],
+        [0,0,-1,0,-1,4]
+    ]
+    b = [2,3,2,2,1,2]
+
+    x = [0,0, 0, 0, 0, 0]   # 初始解
+    err = 1e-5              # 精度要求
+    w = 1                   # 松弛因子,题干要求    P201
+    x1,k,sta = SOR(A, b, x, w, err, 100)
+    print(f"w = {w}, 解为: {x1}, 迭代次数: {k}, 状态: {'收敛' if sta == 1 else '未收敛'}")
+
+    w = 1.1
+    x = [0, 0, 0, 0, 0, 0]
+    x2,k,sta = SOR(A, b, x, w, err, 100)
+    print(f"w = {w}, 解为: {x2}, 迭代次数: {k}, 状态: {'收敛' if sta == 1 else '未收敛'}")

按方法整理/矩阵-特征值-谱半径.py

@@ -0,0 +1,94 @@
+import math
+
+#抛物线法解方程
+def MullerSolve(fx,x0,x1,x2,err1,err2,N):
+    count = 0
+    f0 = fx(x0)
+    f1 = fx(x1)
+    f2 = fx(x2)
+    q = (x2 - x1) / (x1 - x0)
+    p = 0
+    a = 0
+    b = 0
+    c = 0
+    while True:
+        p = (x2 - x0) / (x1 - x0)
+
+        a = q**2 * f0 - q*p*f1 + q*f2
+        b = q**2 *f0 - p**2 *f1 + (p + q)*f2
+        c = p*f2
+        h1 = 0
+        
+        if  b.real < 0:
+            h1 = -2 * c / (b - (b**2 - 4*a*c)**0.5)
+        else:
+            h1 = -2 * c / (b + (b**2 - 4*a*c)**0.5)
+        x3 = x2 + h1 * (x2 - x1)
+        f3 = fx(x3)
+        k = err1 + 1
+        if abs(f3) < 1:
+            k = abs(x3 - x2)
+        else:
+            k = abs(x3 - x2) / abs(f3)
+
+        if abs(f3) < err2 or k < err1:
+            return x3, 1
+        count += 1
+        if count > N:
+            return None, 0
+        x0 = x1
+        x1 = x2
+        x2 = x3
+        f0 = f1
+        f1 = f2
+        f2 = f3
+        q = h1
+#计算矩阵的行列式
+def Det(A):
+    if len(A) == 2:
+        return A[0][0] * A[1][1] - A[0][1] * A[1][0]
+    det = 0
+    for c in range(len(A)):
+        sub_matrix = [row[:c] + row[c+1:] for row in A[1:]]
+        det += ((-1) ** c) * A[0][c] * Det(sub_matrix)
+    return det
+
+
+if __name__ == "__main__":
+    ##########################################################################
+    #把矩阵换成题干的矩阵，通常建议自己再验算一遍，这个能不能算不一定#########################
+    A =[
+        [1,0,1],
+        [2,2,1],
+        [-1,0,0]
+    ] 
+    
+    lam = []
+
+    count = 0
+    k = -100
+    fx = lambda x: Det([[A[i][j] - x * (1 if i == j else 0) for j in range(len(A))] for i in range(len(A))])
+    k = 10
+    while len(lam) < len(A):
+        for i in range(-k,k):
+            re,sta = MullerSolve(fx, i, i + 1, i + 2, 1e-10, 1e-10, 100)
+            if sta == 1:
+                a = round(re.real, 9)
+                b = round(re.imag, 9)
+                re_t = complex(a, b)
+                if re_t not in lam:
+                    if re_t.imag != 0:
+                        lam.append(re_t)
+                        lam.append(re_t.conjugate())
+                    else:
+                        lam.append(a)
+                if len(lam) == len(A):
+                    break
+        k *= 10
+    p = abs(lam[0])
+    for i in range(len(lam)):
+        print(f"λ{i+1} = {lam[i]}")
+        if abs(lam[i]) > p:
+            p = abs(lam[i])
+    
+    print(f"谱半径 = {p:.3f}")

按方法整理/矩阵-迭代改善法.py

@@ -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"第{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）
+

按方法整理/矩阵-逆矩阵-条件数.py

@@ -0,0 +1,69 @@
+import math
+
+#模 范数
+def Norm(x,v):    
+    if len(x[0]) == 1:
+        if v == 1:
+            return sum([abs(i[0]) for i in x])
+        elif v == 2:
+            return (sum([i[0]**2 for i in x]))**0.5
+        elif v == float("inf"):
+            return max([abs(i[0]) for i in x])
+    else:
+        if v == 1:
+            return max([sum([abs(x[i][j]) for i in range(len(x))]) for j in range(len(x[0]))])
+        elif v == float("inf"):
+            return max([sum([abs(i) for i in x[j]]) for j in range(len(x))])
+    return None
+
+# 计算矩阵的行列式
+def Det(A):
+    if len(A) == 2:
+        return A[0][0] * A[1][1] - A[0][1] * A[1][0]
+    det = 0
+    for c in range(len(A)):
+        sub_matrix = [row[:c] + row[c+1:] for row in A[1:]]
+        det += ((-1) ** c) * A[0][c] * Det(sub_matrix)
+    return det
+# 计算矩阵的逆矩阵
+def Inverse(A):
+    n = len(A)
+    # 计算代数余子式矩阵
+    B = [[0 for i in range(n)] for j in range(n)]
+    for i in range(n):
+        for j in range(n):
+            minor = [row[:j] + row[j+1:] for row in (A[:i] + A[i+1:])]
+            B[j][i] = ((-1) ** (i + j)) * sum(minor[k][l] * (-1) ** (k + l) for k in range(n - 1) for l in range(n - 1))
+    det = Det(A)
+    print("det(A):",det)
+    if det == 0:
+        print("矩阵不可逆")
+        return None
+    A_inv = [[B[i][j] / det for j in range(n)] for i in range(n)]
+    return A_inv
+
+# 计算矩阵的条件数
+def Cond(A,v):
+    inv_A = Inverse(A)
+    print(f"inv_A: {inv_A}, Norm(A, v): {Norm(A, v)}, Norm(inv_A, v): {Norm(inv_A, v)}")
+    # print(inv_A,Norm(A, v), Norm(inv_A, v))
+    return Norm(A, v) * Norm(inv_A, v)
+
+
+
+if __name__ == "__main__":
+    ##########################################################################
+    # 把矩阵换成题干的矩阵 #########################
+    A = [
+        [1,2],
+        [1.001,2.001]
+    ]
+    # 把范数的种类数换成题干的要求，inf是无穷范数 #########################
+    print(f"矩阵A的条件数为: {Cond(A, float('inf')):.5f}")     # 1 1范数；2 2范数；float('inf') 无穷范数
+    # 把矩阵换成题干的矩阵 #########################
+    A = [
+        [1,2],
+        [3,4]
+    ]
+    # 把范数的种类数换成题干的要求，inf是无穷范数 ########################
+    print(f"矩阵A的条件数为: {Cond(A, float('inf')):.5f}")     # 1 1范数；2 2范数；float('inf') 无穷范数