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Create local repository for matrix toolboxes. Step #0 done.
author Antonio Pino Robles <data.script93@gmail.com>
date Wed, 06 May 2015 14:56:53 +0200
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8f23314345f4 Create local repository for matrix toolboxes. Step #0 done.
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1 function [L, U, P, Q, rho, ncomp] = gep(A, piv)
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2 %GEP Gaussian elimination with pivoting: none, complete, partial or rook.
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3 % [L, U, P, Q, RHO] = GEP(A, piv) computes the factorization P*A*Q = L*U
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4 % of the m-by-n matrix A, where m >= n,
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5 % where L is m-by-n unit lower triangular, U is n-by-n upper triangular,
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6 % and P and Q are permutation matrices. RHO is the growth factor.
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7 % PIV controls the pivoting strategy:
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8 % PIV = 'c': complete pivoting,
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9 % PIV = 'p': partial pivoting,
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10 % PIV = 'r': rook pivoting.
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11 % The default is no pivoting (PIV = '').
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12 % For PIV = 'r' only, NCOMP is the total number of comparisons.
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13 %
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14 % By itself, GEP(A) returns the final reduced matrix from the
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15 % elimination containing both L and U.
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16
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17 % Reference:
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18 % N. J. Higham, Accuracy and Stability of Numerical Algorithms,
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19 % Second edition, Society for Industrial and Applied Mathematics,
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20 % Philadelphia, PA, 2002; chap. 9.
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21
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22 [m, n] = size(A);
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23 if m < n, error('Matrix must be m-by-n with m >= n.'), end
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24 if nargin < 2, piv = ''; end
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25 pp = 1:m; qq = 1:n;
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26 if nargout >= 5
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27 maxA = norm(A(:), inf);
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28 rho = maxA;
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29 end
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30 ncomp = 0;
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31
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32 for k = 1:min(m-1,n)
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33
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34 if findstr(piv, 'cpr')
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35 if strcmp(piv, 'c')
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36
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37 % Find largest element in remaining square submatrix.
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38 % Note: when tie for max, no guarantee which element is chosen.
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39 [colmaxima, rowindices] = max( abs(A(k:m, k:n)) );
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40 [biggest, colindex] = max(colmaxima);
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41 row = rowindices(colindex)+k-1; col = colindex+k-1;
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42
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43 elseif strcmp(piv, 'p')
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44
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45 % Find largest element in k'th column.
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46 [colmaxima, rowindices] = max( abs(A(k:m, k)) );
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47 row = rowindices(1)+k-1; col = k;
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48
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49 elseif strcmp(piv, 'r')
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50
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51 % Find element that is largest in its row and its column.
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52 col_last = k;
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53 for it = 1:inf
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54 [colmaxima, rowindices] = max( abs(A(k:m, col_last)) );
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55 ncomp = ncomp + m-k;
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56 row = rowindices(1)+k-1;
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57 new_abs = abs(A(row,col_last));
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58 if it > 1
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59 if new_abs == last_abs
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60 row = row_last;
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61 break
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62 end
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63 end
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64 last_abs = new_abs;
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65 row_last = row;
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66 [rowmaxima, colindices] = max( abs(A(row, k:n)) );
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67 ncomp = ncomp + n-k;
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68 col = colindices(1)+k-1;
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69 new_abs = abs(A(row,col));
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70 if new_abs == last_abs
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71 col = col_last;
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72 break
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73 end
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74 last_abs = new_abs;
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75 col_last = col;
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76 end
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77
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78 end
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79
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80 % Permute largest element into pivot position.
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81 A( [k, row], : ) = A( [row, k], : );
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82 A( :, [k, col] ) = A( :, [col, k] );
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83 pp( [k, row] ) = pp( [row, k] ); qq( [k, col] ) = qq( [col, k] );
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84 end
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85
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86 if A(k,k) == 0
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87 if findstr(piv, 'c')
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88 break
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89 elseif strcmp(piv, '') % Zero pivot is problem only for no pivoting.
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90 error('Elimination breaks down with zero pivot. Quitting...')
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91 end
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92 end
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93
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94 i = k+1:m;
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95 if A(k,k) ~= 0 % Simplest way to handle zero pivot for partial and rook.
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96 A(i,k) = A(i,k)/A(k,k); % Multipliers.
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97 end
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98
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99 if k+1 <= n
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100 % Elimination
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101 j = k+1:n;
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102 A(i,j) = A(i,j) - A(i,k) * A(k,j);
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103 if nargout >= 5, rho = max( rho, max(max(abs(A(i,j)))) ); end
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104 end
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105
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106 end
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107
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108 if nargout <= 1
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109 L = A;
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110 return
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111 end
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112
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113 L = tril(A,-1) + eye(m,n);
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114 U = triu(A);
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115 U = U(1:n,:);
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116
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117 if nargout >= 3, P = eye(m); P = P(pp,:); end
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118 if nargout >= 4, Q = eye(n); Q = Q(:,qq); end
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119 if nargout >= 5, rho = rho/maxA; end