The "
Cadillac Quadratic Solver" by Palmer O. Hanson, Jr. for the HP-33s seems like a straight-forward port to the HP-32Sii. Simply replace as follows:
HP33S: B0015 SGN with HP-32Sii: B0015 x<0?; B0016 -1; and B0017 1.
The program works well on the 33S (haven't tried it on the 35S yet) but does not perform as well on the 32Sii when comparing the test case results with those posted at the bottom of the article. Of the eight test cases in the article, I get two incorrect results: test cases '1' and '6' only have one correct root, and there is a minor variation on test case '7'. I can't figure out the problem.
Isn't the replacement code for SGN above applicable? I have double-checked the code and that is the only difference between the 32Sii and 33S programs.
Thanks,
Jeff K
(09-15-2014 11:22 PM)Jeff_Kearns Wrote: [ -> ]HP33S: B0015 SGN with HP-32Sii: B0015 x<0?; B0016 -1; and B0017 1.
These two do different things.
This sequence might work better:
Although the stack action is somewhat different.
Pauli
(09-15-2014 11:32 PM)Paul Dale Wrote: [ -> ]This sequence might work better:
Pauli
Same behaviour...
Interestingly, it solves rather difficult equations correctly like: a = 4,877,361,379, b = -9,754,525,226, c = 4,877,163,849, that give Re = 0.999979750 ; Im = 2.8995463E-10 as answers, but not a simple one like 5x² + 6x + 1 = 0. Instead of x = -0.2 or x = -1, I get x = 0.2 or x = -1.66666667.
--- Jeff
(09-15-2014 11:55 PM)Jeff_Kearns Wrote: [ -> ] (09-15-2014 11:32 PM)Paul Dale Wrote: [ -> ]This sequence might work better:
Pauli
Same behaviour... --- Jeff
I would suggest
ABS
LASTx
x=0?
x!
/
Assuming SGN(0) = 0 on the hp 35s, which I haven't checked.
Gerson
(09-16-2014 12:06 AM)Jeff_Kearns Wrote: [ -> ] (09-16-2014 12:03 AM)Gerson W. Barbosa Wrote: [ -> ]I would suggest
ABS
LASTx
x=0?
x!
/
Gerson
Same result... -- Jeff
Notice your code keeps the argument on the stack. Perhaps you should get rid of it ( x<>y Rv ). Depending on how the stack is being used, this might not help either.
(09-16-2014 12:22 AM)Gerson W. Barbosa Wrote: [ -> ]Notice your code keeps the argument on the stack. Perhaps you should get rid of it ( x<>y Rv ). Depending on how the stack is being used, this might not help either.
This is like the problem. The code snippet from the original article is:
Code:
B0015 SGN
B0016 X
B0017 +
B0018 STO J
B0019 x<>y
B0020 /
Leaving rubbish in Y isn't good at step B0020.
- Pauli
(09-16-2014 12:31 AM)Paul Dale Wrote: [ -> ] (09-16-2014 12:22 AM)Gerson W. Barbosa Wrote: [ -> ]Notice your code keeps the argument on the stack. Perhaps you should get rid of it ( x<>y Rv ). Depending on how the stack is being used, this might not help either.
This is like the problem. The code snippet from the original article is:
Code:
B0015 SGN
B0016 X
B0017 +
B0018 STO J
B0019 x<>y
B0020 /
Leaving rubbish in Y isn't good at step B0020.
- Pauli
Found the following archived
discussion but the suggested workarounds starting at message #8 do not work either... What gives?
Jeff
(09-16-2014 12:43 AM)Jeff_Kearns Wrote: [ -> ] (09-16-2014 12:31 AM)Paul Dale Wrote: [ -> ]This is like the problem. The code snippet from the original article is:
Code:
B0015 SGN
B0016 X
B0017 +
B0018 STO J
B0019 x<>y
B0020 /
Leaving rubbish in Y isn't good at step B0020.
- Pauli
Found the following archived discussion but the suggested workarounds starting at message #8 do not work either... What gives?
Jeff
The ENTER instruction (LSTx in my suggestion) makes the content of the stack register T to be lost, unlike the original SGN function. Any SGN replacement you choose should preserve the stack.
(09-16-2014 12:43 AM)Jeff_Kearns Wrote: [ -> ]What gives?
You'll probably have to track the code through and examine the stack usage at each step.
- Pauli
(09-16-2014 01:02 AM)Paul Dale Wrote: [ -> ]You'll probably have to track the code through and examine the stack usage at each step.
- Pauli
Agreed. But where's Thomas Klemm when I need him?? He's a master at stack-tracking. -- Jeff
You can try
RCLx F
RCL F
ABS
/
Instead of
RCLF
SGN
x
starting at step B0014, assuming F is never zero.
(09-16-2014 01:20 AM)Gerson W. Barbosa Wrote: [ -> ]You can try
RCLx F
RCL F
ABS
/
Instead of
RCLF
SGN
x
starting at step B0014, assuming F is never zero.
WOW Thanks Gerson! I can't say I readily follow the logic (will have to think about it...) and I end up with two consecutive /'s but it works like a charm now. What, therefore, is the 'general' substitute for SGN on a 32sii? Might make for a separate thread/article.
Jeff K
(09-16-2014 01:33 AM)Jeff_Kearns Wrote: [ -> ] (09-16-2014 01:20 AM)Gerson W. Barbosa Wrote: [ -> ]You can try
RCLx F
RCL F
ABS
/
Instead of
RCLF
SGN
x
starting at step B0014, assuming F is never zero.
WOW Thanks Gerson! I can't say I readily follow the logic (will have to think about it...) and I end up with two consecutive /'s but it works like a charm now. What, therefore, is the 'general' substitute for SGN on a 32sii? Might make for a separate thread/article.
Jeff K
Although it works for all test cases, I fear it won't work when B is zero. In this case the following might be a fix:
RCL*F
RCL F
ABS
x=0?
x!
/
Notice this is not the replacement for SGN, rather for RCL F SGN x.
Gerson.
This subroutine should substitute for SGN, change register Z to something you're happy to lose.
Code:
LBL S
x=0?
x^2 Get Last X correct for a zero input
x=0?
RTN Finished with zero input
R^
STO Z Save T in a register
Rv
ABS |x| / x is SGN(x) for x not zero
Last X
/
R^ Now restore T from the register
CLx Disable stack lift
RCL Z
Rv
RTN Finished
This routine doesn't damage the stack and returns with the correct result and the correct value in Last X. I'm sure someone will be able to improve this.
Then replace SGN with XEQ S in you program.
Pauli
Thanks Pauli and Gerson
Jeff
(09-16-2014 02:49 AM)Paul Dale Wrote: [ -> ]This subroutine should substitute for SGN, change register Z to something you're happy to lose.
Code:
LBL S
x=0?
x^2 Get Last X correct for a zero input
x=0?
RTN Finished with zero input
R^
STO Z Save T in a register
Rv
ABS |x| / x is SGN(x) for x not zero
Last X
/
R^ Now restore T from the register
CLx Disable stack lift
RCL Z
Rv
RTN Finished
A little shorter, using a user flag instead of register Z, same stack & Last X management:
Code:
S01 LBL S
S02 CF 4 Clear flag 4
S03 x<0?
S04 SF 4 Set flag 4 if X < 0
S05 ABS Set lastX
S06 x=0?
S07 RTN If X = 0 stop there, SGN(0)=0
S08 CLx Disable stack lift
S09 1 Prepare answer SGN(X) = 1
S10 FS? 4
S11 +/- If X was negative, SGN(X) = -1
S12 CF 4 Clear flag 4
S13 RTN
(09-16-2014 11:26 AM)Didier Lachieze Wrote: [ -> ]A little shorter, using a user flag instead of register Z, same stack & Last X management:
Nice effort. I knew it could be improved.
Another step forward would be to preserve the flag as well by using an extra label and another subroutine level:
Code:
LBL T
FS? 4
GTO S
GSB S
CF 4
RTN
And change the sign function to always set flag 4 at the end instead of clearing it.
Pauli
(09-16-2014 11:46 AM)Paul Dale Wrote: [ -> ]Another step forward would be to preserve the flag as well by using an extra label and another subroutine level:
[...]
And change the sign function to always set flag 4 at the end instead of clearing it.
Good idea!
Otherwise if you're ready to loose register Z (or another one), you can do even shorter with:
Code:
S01 LBL S
S02 STO Z
S03 ABS
S04 x=0?
S05 RTN
S06 STO/ Z
S07 X<> Z
S08 RTN
(09-16-2014 01:20 AM)Gerson W. Barbosa Wrote: [ -> ]You can try
RCLx F
RCL F
ABS
/
Instead of
RCLF
SGN
x
starting at step B0014, assuming F is never zero.
If I understand correctly, the sequence
RCL F
SGN
x
shall be replaced by something that does not require a sign function and that also does not use more than one stack level. Your suggestion will do so, but the combination of a multiplication and a subsequent division may degrade accuracy, and, more important, it will not work for F = 0.
So how about this one?
RCL F
ABS
X≠0?
RCL/ F
x
Dieter