Sherman-Morrison/octave-tests/SMMaponiA3.m

122 lines
3.6 KiB
Matlab

## Algorithm 3 from P. Maponi,
## p. 283, doi:10.1016/j.laa.2006.07.007
clc ## Clear the screen
### Define the matrix to be inverted. This is example 8 from the paper
### In the future this matrix needs to be read from the function call arguments
#A=[1,1,-1; ...
# 1,1,0; ...
# -1,0,-1];
#A0=diag(diag(A)); ## The diagonal part of A
### The modified example that gives all singular updates at some point
#A=[1,1,1; ...
# 1,1,0; ...
# -1,0,-1];
#A0=diag(diag(A)); ## The diagonal part of A
### A square uniform distributed random integer matrix with entries in [-1,1]
#do
# A=randi([-1,1],4,4);
# A0=diag(diag(A)); ## The diagonal part of A
#until (det(A)!=0 && det(A0)!=0) ## We need both matrices to be simultaniously non-singular
## Define the matrix to be inverted. This is example 8 from the paper
## In the future this matrix needs to be read from the function call arguments
A=[1,0, 1,-1; ...
0, 1, 1, 0; ...
-1,0,-1, 0; ...
1,1, 1, 1];
A0=diag(diag(A)); ## The diagonal part of A
### A square uniform distributed random float matrix with entries in (0,1)
#do
# A=rand(5);
# A0=diag(diag(A)); ## The diagonal part of A
#until (det(A)!=0 && det(A0)!=0) ## We need both matrices to be simultaniously non-singular
Ar=A-A0; ## The remainder of A
nCols=columns(Ar); ## The number of coluns of A (M in accompanying PDF)
Id=eye(nCols);
A0inv=eye(nCols);
Ainv=zeros(nCols,nCols);
ylk=zeros(nCols,nCols,nCols);
p=zeros(nCols,1);
breakdown=zeros(nCols,1);
cutOff=1e-10
A,A0
printf("Determinant of A is: %d\n",det(A))
printf("Determinant of A0 is: %d\n",det(A0))
## Calculate the inverse of A0 and populate p-vector
for i=1:nCols
A0inv(i,i) = 1 / A0(i,i);
p(i)=i;
endfor
## Calculate all the y0k in M^2 multiplications instead of M^3
for k=1:nCols
for i=1:nCols
#printf("(i,k,1) = (%d,%d,1)\n",i,k);
ylk(i,k,1) = A0inv(i,i) * Ar(i,k);
endfor
endfor
## Calculate all the ylk from the y0k calculated previously
for l=2:nCols
## Calculate break-down conditions and put in a vector
for j=l-1:nCols
breakdown(j) = abs(1+ylk(p(j),p(j),l-1));
#printf("|1 + ylk(%d,%d,%d)|\n", p(j), p(j), l-1);
endfor
[val, lbar] = max(breakdown); ## Find the index of the max value
breakdown=zeros(nCols,1); ## Reset the entries to zero for next l-round
## Swap p(l) and p(lbar)
tmp=p(l-1);
p(l-1)=p(lbar);
p(lbar)=tmp;
for k=l:nCols
for i=1:nCols
ylk(i,p(k),l) = ylk(i,p(k),l-1) - (ylk(p(l-1),p(k),l-1)) / (1+ylk(p(l-1),p(l-1),l-1)) * (ylk(i,p(l-1),l-1));
#printf("ylk(%d,%d,%d) = ylk(%d,%d,%d) - (ylk(%d,%d,%d) / (1+ylk(%d,%d,%d) * (ylk(%d,%d,%d);\n", i,p(k),l,i,p(k),l-1,p(l-1),p(k),l-1,p(l-1),p(l-1),l-1,i,p(l-1),l-1);
endfor
endfor
endfor
## Construct A-inverse from A0-inverse and the ylk
Ainv=A0inv;
for l=1:nCols
Ainv=(Id - ylk(:,p(l),l) * transpose(Id(:,p(l))) / (1 + ylk(p(l),p(l),l))) * Ainv;
#printf("Ainv=(Id - ylk(:,%d,%d) * transpose(Id(:,%d)) / (1 + ylk(%d,%d,%d))) * Ainv\n",p(l),l,p(l),p(l),p(l),l);
endfor
## Test if the inverse found is really an inverse (does not work if values are floats)
IdTest=A*Ainv;
if (IdTest==eye(nCols))
printf("\n");
printf("Inverse of A^{-1} FOUND!\n");
Ainv
else
printf("\n");
printf("Inverse of A^{-1} NOT found yet.\nRunning another test...\n");
for i=1:nCols
for j=1:nCols
if (abs(IdTest(i,j))<cutOff)
IdTest(i,j)=0;
elseif (abs(IdTest(i,j))-1<cutOff)
IdTest(i,j)=1;
endif
endfor
endfor
if (IdTest==eye(nCols))
printf("\n");
printf("Inverse of A^{-1} FOUND!\n");
Ainv
else
printf("\n");
printf("Still not found. Giving up!\n");
IdTest
endif
endif