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LU decomposition: Difference between revisions
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[0, 1, 0, 0],
[0, 0, 0, 1]]</pre>
=={{header|J}}==
Taken with slight modification from [http://www.jsoftware.com/jwiki/Essays/LU%20Decomposition]
'''Solution:'''
<lang j> mp=: +/ .*
LU=: 3 : 0
'm n'=. $ A=. y
if. 1=m do.
p ; (=1) ; p{"1 A [ p=. C. (n-1);~.0,(0~:,A)i.1
else.
m2=. >.m%2
'p1 L1 U1'=. LU m2{.A
D=. (/:p1) {"1 m2}.A
F=. m2 {."1 D
E=. m2 {."1 U1
FE1=. F mp %. E
G=. m2}."1 D - FE1 mp U1
'p2 L2 U2'=. LU G
p3=. (i.m2),m2+p2
H=. (/:p3) {"1 U1
(p1{p3) ; (L1,FE1,.L2) ; H,(-n){."1 U2
end.
)
permtomat=. 1 {.~"0 -@>:@:/:
LUdecompose=: (permtomat&.>@{. , }.)@:LU</lang>
'''Example use:'''
<lang j> A=.3 3$1 3 5 2 4 7 1 1 0
LUdecompose A
┌─────┬─────┬───────┐
│1 0 0│1 0 0│1 3 5│
│0 1 0│2 1 0│0 _2 _3│
│0 0 1│1 1 1│0 0 _2│
└─────┴─────┴───────┘
mp/> LUdecompose A
1 3 5
2 4 7
1 1 0</lang>
=={{header|Python}}==
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