Numerical analysis of the Beihang Operation II

ImportCmath fromMathImport*SGN=LambdaX:1ifX > 0Else-1#the problem of the Father's assignment! Note Subscript starting from 1definit (): A= [[0 forIinchRange (10)] forIinchRange (10)]    forIinchRange (10):       forJinchRange (10): II, JJ= i + 1, j + 1A[i][j]= sin (0.5 * II + 0.2 * JJ)ifII! = JJElse1.52 * COS (ii + 1.2 *JJ)returnA#Matrix TransposedefTransposition (A): ans= [[0 forIinchRange (len (A))] forJinchRange (len (a[0))] forIinchRange (len (A)): forJinchRange (len (a[i)): Ans[j][i]=A[i][j]returnans#matrix and matrix multiplicationdefMatrix_mul_matrix (A, B):assertLen (a[0]) = =Len (B) ans= [[0] * len (b[0]) forIinchRange (len (A))] forIinchRange (len (ans)): forJinchRange (len (ans[i)): Ans[i][j]= SUM ([a[i][k] * B[k][j] forKinchRange (len (B))]) returnans#multiplication of matrices and vectorsdefMatrix_mul_vector (A, b):assertLen (a[0]) = =Len (b)return[Sum ([a[i][j] * B[j] forJinchRange (len (b))]) forIinchRange (len (A))]#vector and vector multiplicationdefVector_mul_vector (A, B):assertLen (A) = =Len (B)returnSUM ([a[i] * B[i] forIinchRange (len (A))])#vector divided by constant kdefDiv (V, k):return[V[i]/k forIinchRange (len (v))]#on the proposed triangulation of matricesdefquasi_upper_triangular (A): forIinchRange (0, Len (A)-2):#i = number of columns      if  notAny ([a[j][i] forJinchRange (i + 2, Len (A))]):ContinueD= sqrt (sum ([a[j][i] * * 2 forJinchRange (i + 1, Len (A))]) C=-SGN (A[i + 1][i]) *D H= c * (c-a[i + 1][i]) u= [0] * (i + 1) + [A[i + 1][i]-c] + [a[j][i] forJinchRange (i + 2, Len (A))] P=Div (matrix_mul_vector (transposition (A), u), h) Q=Div (Matrix_mul_vector (A, u), h) T= Vector_mul_vector (P, u)/h W= [Q[i]-T * u[i] forIinchRange (len (q))] forIinchRange (len (A)): forJinchRange (len (a[i)): A[i][j]= A[i][j]-w[i] * U[j]-u[i] *P[j]#Print MatrixdefPrintmatrix (a,s):Print(s) forIinchRange (len (A)): forJinchRange (len (a[i)):Print('%.3f'% a[i][j],' ', end=',')      Print()   Print("=========")defQR (B, C): forIinchRange (len (B)-1):      if  notAny ([b[j][i] forJinchRange (i + 1, len (B))]):ContinueD= sqrt (sum ([b[j][i] * * 2 forJinchRange (i, Len (B)))) C=-SGN (B[i][i]) *D H= c * (C-b[i][i]) u= [0] * (i) + [b[i][i]-c] + [b[j][i] forJinchRange (i + 1, Len (B))] V=Div (matrix_mul_vector (transposition (B), u), h) forIinchRange (len (B)): forJinchRange (len (b[i)): B[i][j]-= u[i] *V[j] P=Div (matrix_mul_vector (transposition (C), u), h) Q=Div (Matrix_mul_vector (C, u), h) T= Vector_mul_vector (P, u)/h W= [Q[i]-T * u[i] forIinchRange (len (q))] forIinchRange (len (C)): forJinchRange (len (c[i)): C[i][j]= C[i][j]-w[i] * U[j]-u[i] *P[j]#QR decomposition with two-step displacement to find the characteristic rootdefQr_with_double_shift (A): Root= [0] *Len (A) I= Len (A)-1k= 0#Number of steps    whileI >=0:ifi = =0:root[0]=A[0][0] Break      elifABS (A[i][i-1]) <Epsilon:root[i]=A[i][i] I-= 1ContinueD1, D2, D3, D4= A[i-1][i-1], a[i-1][i], a[i][i-1], A[i][i] Delta= Cmath.sqrt ((d1 + d4) * * 2-4 * (D1 * D4-D2 *D3)) ifi = = 1orABS (A[i-1][i-2]) <Epsilon: (Root[i], root[i-1]) = (((D1 + d4) + Delta)/2, ((D1 + D4)-Delta)/2) I-= 2Continue      ifK = =L:Print("didn ' t get all of the root after {} steps". Format (L)) BreakS, t= D1 + d4, D1 * D4-D2 *D3 M=Matrix_mul_matrix (A, a) forIinchRange (len (M)): forJinchRange (len (m[0)): M[i][j]-= S *A[i][j] M[i][i]+=T-qr (M, A) k+ = 1returnRoot"""Gaussian elimination method applicable scope: Each order master type is greater than 0 column main element Gaussian elimination method applies the scope: the determinant value is greater than 0 the whole main element Gaussian elimination method applies the scope: solves the arbitrary equation, may obtain a solution space to in this question, according to the characteristic value to obtain the characteristic vector to have two methods: 1. The 0-point translation inverse power method iterates the nearest Lamda eigenvalue, and can also find the eigenvector, but it can only find a eigenvector 2. All-principal-element Gaussian elimination method, which is the most perfect"""#A feature vector of the root of a square matrix A with all-principal-element Gaussian elimination methoddefSolve (A, root): N=Len (A) forIinchrange (N): A[i][i]-=Root IND= [I forIinchrange (n)] rank=N forIinchrange (N): x, y=I, I forRowinchRange (I, N): forColinchRange (I, N):ifABS (A[x][y]) <ABS (A[row][col]): x, y=Row, colifABS (A[x][y]) < epsilon:#The maximum value is also 0, starting backRank =I Break      #swap the largest row with the current rowA[X], a[i] =A[i], a[x]#Column y and column I       forRowinchrange (N): A[row][y], A[row][i]=A[row][i], A[row][y] ind[i], Ind[y]=Ind[y], Ind[i]#A single line of unitst=A[i][i] forJinchRange (I, N): A[i][j]/=T forRowinchRange (i + 1, N):#Line J-a[j][i] Timest =A[row][i]ifT==0:Continue          forColinchRange (I,n): A[row][col]-=t*A[i][col]#The process of generation back    forRowinchRange (Rank-1,-1,-1):       forJinchRange (row + 1, N):#Row J of line I-a[i][j] times         #It is important to note that A[i][j must be saved with T, otherwise a[i][j] becomes 0.t =A[row][j] forKinchRange (j, N): A[row][k]-= T *A[j][k]#construct eigenvector space, be sure to pay attention to the various columns to change back!ans = [[0] * Len (A) forIinchRange (N-rank)]  forIinchRange (rank, n): forJinchrange (N): Ans[i-RANK][IND[J]] =-A[j][i] Ans[i-Rank][ind[i]] = 1returnAnsepsilon= 1e-12L= 0xFFFFFFA=init () Printmatrix (A,'the original matrix') Quasi_upper_triangular (a) Printmatrix (a),'a matrix after a quasi-diagonalization') Roots=Qr_with_double_shift (A)Print('each feature root', Roots) Printmatrix (A,'The matrix after the QR two-step displacement method ends') forIinchRoots:ifType (i) = =float:a=init () x=Solve (A, i)Print('feature Root'I'the corresponding eigenvector is \ n', x)

Numerical analysis of the Beihang Operation II