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72 lines
2.5 KiB
Mathematica
72 lines
2.5 KiB
Mathematica
## Based on Algorithm 3 from P. Maponi,
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## p. 283, doi:10.1016/j.laa.2006.07.007
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clc ## Clear the screen
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## Define the matrix to be inverted. This is example 8 from the paper
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## In the future this matrix needs to be read from the function call arguments
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#A=[1,1,-1; ...
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# 1,1,0; ...
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# -1,0,-1];
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#A0=diag(diag(A)); ## The diagonal part of A
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### The modified example that gives all singular updates at some point
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#A=[1,1,1; ...
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# 1,1,0; ...
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# -1,0,-1];
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#A0=diag(diag(A)); ## The diagonal part of A
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## A uniform distributed random square (5x5) integer matrix with entries in [-1,1]
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do
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A=randi([-1,1],5,5);
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#A=rand(5);
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A0=diag(diag(A)); ## The diagonal part of A
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until (det(A)!=0 && det(A0)!=0) ## We need both matrices to be simultaniously non-singular
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A
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printf("Determinant of A is: %d\n",det(A))
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printf("Determinant of A0 is: %d\n\n",det(A0))
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Ar=A-A0; ## The remainder of A
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nCols=columns(Ar); ## The number of coluns of A (M in accompanying PDF)
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Id=eye(nCols); ## Identity matrix, used for the Cartesian basis vectors
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Ainv=zeros(nCols,nCols,nCols+1); ## 3d matrix to store intermediate inverse of A
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U=zeros(nCols,nCols,nCols); ## 3d matrix to store the nCols rank-1 updates
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a=zeros(1,nCols); ## Vector containing the break-down values
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used=zeros(1,nCols); ## Vector to keep track of which updates have been used
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## Loop to calculate A0_inv and the rank-1 updates and store in U
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Ainv(:,:,1)=eye(nCols);
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for i=1:nCols
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Ainv(i,i,1)=1/A0(i,i);
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U(:,:,i)=Ar(:,i)*transpose(Id(:,i));
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endfor
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## Here starts the calculation of the inverse of A
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for i=1:nCols ## Outer loop iterated over the intermediates A_l^-1, M times in total
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## Here the break-down values are calculated for each intermediate inverse
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for j=1:nCols
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a(j)=1 + transpose(Id(:,j)) * Ainv(:,:,i) * Ar(:,j); ## First time all elmts 1 because A0_inv is diagonal
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printf("a(%d) = %f\n",j,a(j))
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## Select next update ONLY if break-down value != 0 AND not yet used
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if (a(j)!=0 && used(j)!=1);
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k=j;
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break;
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endif
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endfor
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## Do the actual S-M update
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Ainv(:,:,i+1) = Ainv(:,:,i) - Ainv(:,:,i) * U(:,:,k) * Ainv(:,:,i) / a(k);
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used(k)=1; ## Mark this update as used
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endfor
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## Test if the inverse found is really an inverse (does not work if values are floats)
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if (A*Ainv(:,:,nCols+1)==eye(nCols))
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printf("\n");
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printf("Inverse found. A^{-1}:\n");
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Ainv(:,:,nCols+1)
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else
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printf("\n");
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printf("Inverse NOT found! A*A^{-1}:\n");
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A*Ainv(:,:,nCols+1)
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endif
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