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按方法整理/矩阵-平方根法.py

@@ -0,0 +1,49 @@
+import math
+
+# 获取下三角矩阵的索引
+# 列 行
+def getIndexFromDownMatrix(col, row):
+    if row > col:
+        row, col = col, row
+    return (1+col)*col//2+row
+
+
+# 平方根法求解
+def SqrtSolve(A,b):
+    n = len(b)
+    L = [0]*(n*(n+1)//2)
+    for j in range(n):
+        for i in range(j,n):
+            if i == j:
+                L[getIndexFromDownMatrix(i,j)] = A[getIndexFromDownMatrix(i,j)]
+                for k in range(j):
+                    L[getIndexFromDownMatrix(i,j)] -= L[getIndexFromDownMatrix(j,k)]**2
+                L[getIndexFromDownMatrix(i,j)] = L[getIndexFromDownMatrix(i,j)]**0.5
+            else:
+                L[getIndexFromDownMatrix(i,j)] = A[getIndexFromDownMatrix(i,j)]
+                for k in range(j):
+                    L[getIndexFromDownMatrix(i,j)] -= L[getIndexFromDownMatrix(i,k)]*L[getIndexFromDownMatrix(j,k)]
+                L[getIndexFromDownMatrix(i,j)] /= L[getIndexFromDownMatrix(j,j)]
+    print(L)
+    for i in range(n):
+        for k in range(i):
+            b[i] -= L[getIndexFromDownMatrix(i,k)]*b[k]
+        b[i] /= L[getIndexFromDownMatrix(i,i)]
+
+    for i in range(n-1,-1,-1):
+        for k in range(i+1,n):
+            b[i] -= L[getIndexFromDownMatrix(k,i)]*b[k]
+        b[i] /= L[getIndexFromDownMatrix(i,i)]
+    return b
+
+#把A,b换成题干的数值###########################################
+if __name__ == "__main__":
+    ##########################################################################
+    # 储存下三角矩阵 a11, a21, a22, a31, a32, a33 ...，按照这个格式写
+    A = [4,2,2,-2,-3,14]
+
+
+    b = [10,5,4]    # b向量，常数项
+
+    print("x:")
+    print(SqrtSolve(A,b))

按方法整理/矩阵-模范数-点积.py

@@ -0,0 +1,35 @@
+#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
+# 计算矩阵的点积
+def Dot(A,B):
+    if len(A[0]) != len(B):
+        return None
+    return [[sum([A[i][j] * B[j][k] for j in range(len(A[0]))]) for k in range(len(B[0]))] for i in range(len(A))]
+        
+#改成题干的矩阵##########################
+if __name__ == "__main__":
+    A = [[1, 3], [-2, 4]]      # 注意储存形式
+    x = [[1], [-1]]            # 注意储存形式，单独一列的相量，每个数字都要中括号
+
+    print("Norm of x with v=1:", Norm(x, 1))
+    print("Norm of x with v=inf:", Norm(x, float("inf")))
+    print("Norm of x with v=2:", Norm(x, 2))
+    print("Dot product of A and x:", Dot(A, x))
+    print("Norm of Ax with v=2:", Norm(Dot(A, x), 2))
+    print("Norm of A with v=inf:", Norm(A, float("inf")))
+    print("Norm of A with v=1:", Norm(A, 1))
+    
+    

按方法整理/矩阵-追赶法.py

@@ -0,0 +1,42 @@
+import math
+
+# 追赶法求解
+def ZGsolve(A,b):
+    n = len(b)
+    beta = [0]*n
+    for i in range(n):
+        if i == 0:
+            beta[i] = A[i][2] / A[i][1]
+        else:
+            beta[i] = A[i][2] / (A[i][1] - A[i][0]*beta[i-1])
+    print("beta:")
+    print(beta[:-1])
+    for i in range(n):
+        if i == 0:
+            b[i] = b[i] / A[i][1]
+        else:
+            b[i] = (b[i] - A[i][0]*b[i-1]) / (A[i][1] - A[i][0]*beta[i-1])
+    print("y:")
+    print(b)
+    for i in range(n-2,-1,-1):
+        b[i] = b[i] - beta[i]*b[i+1]
+
+    return b
+
+
+if __name__ == "__main__":
+    ##########################################################################
+    #把A,b换成题干的数值###########################################
+    # 储存追赶法A矩阵
+    A = [
+        [0,4,-1],
+        [-1,4,-1],
+        [-1,4,-1],
+        [-1,4,-1],
+        [-1,4,0]
+    ]
+    # A第一行前面有个0，最后一行后面有个0
+
+    b = [100,200,200,200,100]
+    print("x:")
+    print(ZGsolve(A,b))

按方法整理/非线性方程-抛物线法.py

@@ -50,6 +50,7 @@ def MullerSolve(fx,x0,x1,x2,err1,err2,N):
 
 
 if __name__ == "__main__":
+    ##############################################################################################################
     err1 = 1e-5           # 精度要求    P152
     err2 = 1e-5           # 精度要求    P152
     N = 100               # 最大迭代次数