LU Decompositions
Posted: Thu Feb 26, 2015 1:59 pm
Here's an early version of an attempted LU decomposition (partial pivot method). It's not overly efficient but I will possibly modify it to also work for matrices where the leading diagonal isn't zero at some point soon, and post below.
Code: Select all
!n=@rows(A)
matrix L = @identity(!n)
matrix P= @identity(!n)
matrix U=A
for !k=1 to !n-1
vector temp = @subextract(U,!k,!k,!n,!k)
!pivot=@max(@abs(temp))
for !j = 1 to @rows(temp)
if @abs(temp(!j))=!pivot then
!ind=!j+!k-1
' !j=!n+1
endif
next
d temp
vector temp_k = @subextract(U,!k,!k,!k,!n)
vector temp_ind = @subextract(U,!ind,!k,!ind,!n)
matplace(U,temp_k,!ind,!k)
matplace(U,temp_ind,!k,!k)
d temp_k temp_ind
vector temp_k = @subextract(L,!k,1,!k,!k-1)
vector temp_ind = @subextract(L,!ind,1,!ind,!k-1)
matplace(L, temp_k,!ind,1)
matplace(L, temp_ind,!k,1)
d temp_k temp_ind
vector temp_k = @rowextract(P,!k)
vector temp_ind = @rowextract(P,!ind)
matplace(P,temp_k,!ind,1)
matplace(P,temp_ind,!k,1)
d temp_k temp_ind
for !j= !k+1 to !n
L(!j,!k)=U(!j,!k)/U(!k,!k)
matplace(U,@subextract(U,!j,!k,!j,!n)-L(!j,!k)*@subextract(U,!k,!k,!k,!n),!j,!k)
next
next