TriangSyst

Triangular Systems for the HP-41

Overview

-This program solves one - or several - linear system(s) whose principal matrix has only zeros above its diagonal.
-It may be used after "triangularizing" a linear system ( Gaussian method, Householder method ... )

B = A.X where A is a lower triangular matrix.

Program Listing

Data Registers: R00 is unused ( Registers R01 thru Rn.m are to be initialized before executing "TRS" )

• R01 = b₁ • Rn+1 = a_1,1 ..................... • Rnm-n+1 = a_1,n with a_i,j = 0 for j > i
• R02 = b₂ • Rn+2 = a_2,1 ..................... • Rnm-n = a_2,n

................ ...............................................................................

• Rnn = b_n • R2n = a_n,1 ..................... • Rnm = a_n,n

>>> At the end x₁ , .......... , x_n are in registers R01 to Rnn.

• This program may be used to solve several linear systems with the same "principal" matrix,
especially if you want to invert the lower triangular matrix itself.

Flags: /
Subroutines: /

01 LBL "TRS"
02 STO N
03 STO O
04 X<>Y
05 ST* N
06 .1
07 %
08 STO M
09 +
10 STO Q
11 ISG M

12 ISG N
13 LBL 01
14 CLX
15 STO P
16 LBL 02
17 RCL IND N
18 ST/ IND M
19 RCL M
20 RCL IND X
21 SIGN
22 ISG M

23 X=0?
24 GTO 04
25 ISG Y
26 ISG N
27 LBL 03
28 CLX
29 RCL IND N
30 LASTX
31 *
32 ST- IND Y
33 ISG N

34 CLX
35 ISG Y
36 GTO 03
37 RCL P
38 ISG X
39 CLX
40 STO P
41 ST+ N
42 GTO 02
43 LBL 04
44 RCL Q

45 FRC
46 ST+ M
47 LASTX
48 INT
49 X^2
50 DSE X
51 ST- N
52 DSE O
53 GTO 01
54 CLA
55 END

( 96 bytes / SIZE 1+n(c+n) )

STACK	INPUTS	OUTPUTS
Y	n	/
X	c	/

where n = number of raws
and c = number of systems ( c + n = number of columns )

Example1: Solve the triangular system:

   4 = 2 x
   5 = 3 x + 4 y
   6 = 5 x + 7 y + 3 z

-Store

   4    2    0    0                     R01   R04   R07   R10
   5    3    4    0        into       R02   R05   R08   R11
   6    5    7    3                     R03   R06   R09   R12

3 ENTER^
1 XEQ "TRS" >>>> ( 8 ) ---Execution time = 3s---

-And we have

   R01 = x =    2
   R02 = y = -1/4
   R03 = z = -3/4

Example2: If you want to invert the above square matrix A, store the identity matrix in R01 thru R09 and A in R10 thru R18

R01 R04 R07 R10 R13 R16      1   0   0   2   0   0
R02 R05 R08 R11 R14 R17 = 0   1   0   3   4   0
R03 R06 R09 R12 R15 R18      0   0   1   5   7   3

3 ENTER^
3 XEQ "TRS" >>>> ( 8 ) ---Execution time = 9s---

-And the inverse is in registers R01 thru R09

R01 R04 R07       1/2       0        0
R02 R05 R08 = -3/8      1/4      0
R03 R06 R09      1/24 -7/12   1/3

Notes:

-The square matrix itself is unchanged
-Since synthetic registers P & Q are used, don't interrupt "TRS",
this could change the contents of these registers.
-The alpha register is cleared at the end.

Variant:

-You can also avoid to store the zero-elements of the matrix ( above the diagonal ).
-In this case,

   add   LASTX   +   ENTER^   SIGN   ST+ X   /     after line 49
   replace line 42 by GTO 01
   delete lines 37 to 41 and lines 14 to 16                          ( 53 lines / 90 bytes )

-In example 2, store

R01 R04 R07 R10 R13 R15      1   0   0   2   4   3
R02 R05 R08 R11 R14          = 0   1   0   3   7
R03 R06 R09 R12                       0   0   1   5

3 ENTER^
3 XEQ "TRS" >>>> ( 5 ) ---Execution time = 8s---

-And the inverse is still in registers R01 thru R09

R01 R04 R07       1/2       0        0
R02 R05 R08 = -3/8      1/4      0
R03 R06 R09      1/24 -7/12   1/3

-Since the SIZE becomes 1 + c n + n(n+1)/2 so you can solve a triangular system of 23 equations in 23 unknowns.

hp41programs

Triangular Systems for the HP-41