Threaded Mode | Linear Mode

Gerson W. Barbosa · (This post was last modified: 05-17-2017 12:37 PM by Gerson W. Barbosa.)

Code:

00 { 25-Byte Prgm }                        

01 LBL "Q"                     

02 X<> ST Z                                        

03 +/-

04 STO/ ST Z

05 STO+ ST X 

06 /                 

07 STO ST Z        

08 X^2                

09 +             

10 SQRT      

11 RCL+ ST Y                     

12 X<>Y               

13 RCL- ST L 

14 END

Examples:

1) Given x² - 5x + 6 = 0, compute 123456*x₁ and 123456*x₂

.................... 123456 ENTER 1 ENTER 5 +/- ENTER 6 XEQ Q -> Y: 3 ; x₂ = 3
................................................................ X: 2 ; x₁ = 2
........................................................ Rv * -> X: 370,368 ; 123456*x₂
........................................................ Rv * -> X: 246,912 ; 123456*x₁

2) 3x² - 12x + 87 = 0

............................... 3 ENTER 12 +/- ENTER 87 XEQ Q -> Y: 2 .i5 ; x₂ = 2 + 5i
................................................................ X: 2 -i5 ; x₁ = 2 - 5i

3) x² - (4 + 6i)x + (-5 + 10i) = 0

.. 1 ENTER 4 ENTER 6 COMPLEX +/- 5 +/- ENTER 10 COMPLEX XEQ Q -> Y: 3 i4 ; x₂ = 3 + 4i
................................................................ X: 1 i2 ; x₁ = 1 + 2i

Edited to change code. Now it is one step and one byte shorter. This is the older code:

Code:

00 { 26-Byte Prgm }                        

01 LBL "Q"                     

02 X<> ST Z                                        

03 +/-

04 STO/ ST Z

05 /                       

06 2                  

07 /                 

08 STO ST Z        

09 X^2                

10 +             

11 SQRT      

12 RCL+ ST Y                     

13 X<>Y               

14 RCL- ST L 

15 END

Edited again to change subject line.

Jeff_Kearns · (This post was last modified: 05-26-2014 11:47 AM by Jeff_Kearns.)

It was interesting to read the evolution of this routine in your 2010 thread (see Gerson's link below in post #3).

Jeff K
- edited to address an invalid link

Gerson W. Barbosa · (This post was last modified: 05-25-2014 09:56 PM by Gerson W. Barbosa.)

(05-25-2014 03:30 PM)Jeff_Kearns Wrote: It was interesting to read the evolution of this routine in your 2010 thread here.

Jeff,

That's a Google Drive link and I cannot follow it. I think you mean this old forum thread:

Short Quadratic Solver (HP-42S)

The influence and inspiration that led me to this even shorter version are described in this thread from 2012:

Fast Quadratic Formula for the HP-41C

Thank you for your interest!

Gerson.

Jeff_Kearns · (This post was last modified: 05-28-2014 10:21 PM by Jeff_Kearns.)

(05-25-2014 09:55 PM)Gerson W. Barbosa Wrote: Thank you for your interest!

Gerson,

I am interested! The following routine for the HP-32sii and HP-33s (slightly modified from Eddie Shore's original to display the roots as x + iy) is the one I use but it is 39 steps long!

Q0001 LBL Q ' Set label Q
Q0002 INPUT A
Q0003 INPUT B
Q0004 INPUT C
Q0005 SQ(B)-4xAxC ' Equation (enter by RS [right shift] EQN)
Q0006 STO D
Q0007 RCL D ' Is the discriminant negative?
Q0008 x<0?
Q0009 SF 1 ' Flag 1 - indicate that the roots are complex
Q0010 RCL D
Q0011 FS? 1
Q0012 +/- ' The HP 33 cannot take square roots of x<0
Q0013 square root
Q0014 STO E ' E = abs(sqrt(D))
Q0015 -B/(2xA) ' enter this as an equation
Q0016 STO F
Q0017 E/(2xA) ' enter this an an equation
Q0018 STO G
Q0019 RCL G
Q0020 0
Q0021 FS? 1
Q0022 x<>y
Q0023 RCL F
Q0024 0
Q0025 CMPLX+
Q0026 x<>y
Q0027 STOP ' press the R/S key: display first root
Q0028 RCL G
Q0029 +/-
Q0030 0
Q0031 FS? 1
Q0032 x<>y
Q0033 RCL F
Q0034 0
Q0035 CMPLX+
Q0036 x<>y
Q0037 STOP
Q0038 CF 1 ' clean up command
Q0039 RTN

Can you think of a way of dramatically shortening it, in line with your 15-liner for the HP-42s, so that it still gives real and complex roots? The memory in the HP-32Sii is so limited and shortening this program might allow me to squeeze one more program in there...

Thanks,

Jeff K

Thomas Klemm · 05-27-2014, 06:57 PM

(05-27-2014 05:30 PM)Jeff_Kearns Wrote: Can you think of a way of dramatically shortening it, (...) so that it still gives real and complex roots?

Copied most of it from here: Short quadratic solver (HP-15C)

Code:

Q01 LBL Q

Q02 ENTER

Q03 R^

Q04 /

Q05 R^

Q06 LASTx

Q07 /

Q08 2

Q09 +/-

Q10 /

Q11 ENTER

Q12 ENTER

Q13 x^2

Q14 R^

Q15 -

Q16 SQRT

Q17 ENTER

Q18 +/-

Q19 CMPLX-

Q20 RTN

Doesn't give you the complex solution but that should be easy to add. There are probably shorter solutions possible by using a register.

Cheers
Thomas

Gerson W. Barbosa · 05-28-2014, 04:18 AM

(05-27-2014 06:57 PM)Thomas Klemm Wrote:
(05-27-2014 05:30 PM)Jeff_Kearns Wrote: Can you think of a way of dramatically shortening it, (...) so that it still gives real and complex roots?

Copied most of it from here: Short quadratic solver (HP-15C)

Code:

Q01 LBL Q Q02 ENTER Q03 R^ Q04 / Q05 R^ Q06 LASTx Q07 / Q08 2 Q09 +/- Q10 / Q11 ENTER Q12 ENTER Q13 x^2 Q14 R^ Q15 - Q16 SQRT Q17 ENTER Q18 +/- Q19 CMPLX- Q20 RTN

Doesn't give you the complex solution but that should be easy to add. There are probably shorter solutions possible by using a register.

Cheers
Thomas

Nice use of CMPLX- for saving a step or two!

This might do while we don't think of something better:

Code:

Q01 LBL Q

Q02 CF 1

Q03 ENTER

Q04 R^

Q05 /

Q06 R^

Q07 LASTx

Q08 /

Q09 -2

Q10 /

Q11 ENTER

Q12 ENTER

Q13 x^2

Q14 R^

Q15 -

Q16 x<0?

Q17 SF 1

Q18 ABS

Q19 SQRT

Q20 FS? 1

Q21 RTN

Q22 ENTER

Q23 +/-

Q24 CMPLX-

Q25 RTN

If the roots are complex, the flag 1 announciator will be lit and the real and complex parts will be in registers Y and X, respectively.

Cheers,

Gerson[/code]

Jeff_Kearns · (This post was last modified: 05-28-2014 11:12 PM by Jeff_Kearns.)

(05-28-2014 04:18 AM)Gerson W. Barbosa Wrote: If the roots are complex, the flag 1 annunciator will be lit and the real and complex parts will be in registers Y and X, respectively.

Cheers,

Gerson

Gerson -

The code (without INPUT prompts) works OK for real roots but not for complex roots (IMHO)...

I prefer to see the results (if complex) displayed as x + iy, as per the 39 step routine based on Eddie Shore's program to which I added a couple of x<>y statements to display 'correctly'. There are two issues with this particular routine that I can't quite figure out:

1) even when I insert a x<>y before the final RTN, I still get the imaginary part in x, i.e. ix + y. With or without a x<>y before the final RTN, the displayed result is the same.

2) Only one complex root is solved correctly. As an example, if you try solving 5x² + 2x + 1 = 0, you get the following: Flag 1 is set with the imaginary part 0.4 in x, and the real part -0.2 in y. Upon R/S the second root (x = -0.2 - 0.4i) is not shown. Granted, all complex roots are of the form x = D ± Ei, so one can deduce the second root and it doesn't really matter I guess BUT...

Can you suggest a better solution?

Regards,

Jeff K

Gerson W. Barbosa · (This post was last modified: 05-29-2014 03:53 PM by Gerson W. Barbosa.)

(05-28-2014 10:17 PM)Jeff_Kearns Wrote:
(05-28-2014 04:18 AM)Gerson W. Barbosa Wrote: If the roots are complex, the flag 1 annunciator will be lit and the real and complex parts will be in registers Y and X, respectively.

Cheers,

Gerson

Gerson -

The code (without INPUT prompts) works OK for real roots but not for complex roots (IMHO)...

I prefer to see the results (if complex) displayed as x + iy, as per the 39 step routine based on Eddie Shore's program to which I added a couple of x<>y statements to display 'correctly'. There are two issues with this particular routine that I can't quite figure out:

1) even when I insert a x<>y before the final RTN, I still get the imaginary part in x, i.e. ix + y. With or without a x<>y before the final RTN, the displayed result is the same.

2) Only one complex root is solved correctly. As an example, if you try solving 5x² + 2x + 1 = 0, you get the following: Flag 1 is set with the imaginary part 0.4 in x, and the real part -0.2 in y. Upon R/S the second root (x = -0.2 - 0.4i) is not shown. Granted, all complex roots are of the form x = D ± Ei, so one can deduce the second root and it doesn't really matter I guess BUT...

Can you suggest a better solution?

Regards,

Jeff K

Jeff,

I hope this is better, although not so short anymore:

Code:

Q01 LBL Q

Q02 CF 1

Q03 ENTER

Q04 R↑

Q05 ÷

Q06 R↑

Q07 LASTx

Q08 ÷

Q09 ⁻2

Q10 ÷

Q11 ENTER

Q12 ENTER

Q13 x²

Q14 R↑

Q15 -

Q16 x<0?

Q17 SF 1

Q18 ABS

Q19 SQRT

Q20 ENTER

Q21 +/-

Q22 CMPLX+

Q23 FS? 1

Q24 CMPLX-

Q25 FS? 1

Q26 R↑

Q27 RTN

CK=3BA1 LEN 048.5 (HP-32SII)

Usage examples on the HP-32SII

1 ENTER 5 +/- ENTER 6 XEQ Q --> 2 x<>y --> 3 (Two real roots, because the flag 1 annunciator is off)

1 ENTER 1 +/- ENTER 1 XEQ Q --> 0.5 x<>y 8.66025404E-1
R↓ 0.5 x<>y -0.86602540378 (Two complex roots, 0.5 ± (√3)÷2 i, because the flag 1 annunciator is on)

Alternatively, you can use a slightly modified version of Eddie Shore's program to save a few steps:

Code:

Q0001 LBL Q 

Q0002 CF 1

Q0003 INPUT A

Q0004 INPUT B

Q0005 INPUT C

Q0006 SQ(B)-4×A×C 

Q0007 x<0?

Q0008 SF 1 

Q0009 ABS

Q0010 √x

Q0011 STO E 

Q0012 -B÷2÷A

Q0013 STO F

Q0014 E÷2÷A

Q0015 STO G

Q0016 0

Q0017 FS? 1

Q0018 x<>y

Q0019 RCL F

Q0020 0

Q0021 CMPLX+

Q0022 x<>y

Q0023 STOP 

Q0024 RCL G

Q0025 +/-

Q0026 0

Q0027 FS? 1

Q0028 x<>y

Q0029 RCL F

Q0030 0

Q0031 CMPLX+

Q0032 x<>y

Q0033 RTN

CK=545E

LN=169    (hp 33s)

Regards,

Gerson.

P.S.: Regarding issue #1, that happens because the first RTN instruction (step Q21), not the last, is executed when flag 1 is set.

Thomas Klemm · 05-29-2014, 05:10 AM

(05-29-2014 04:23 AM)Gerson W. Barbosa Wrote:

Code:

Q0007 STO D

You can remove this step as D is never used.

Cheers
Thomas

Gerson W. Barbosa · 05-29-2014, 03:48 PM

(05-29-2014 05:10 AM)Thomas Klemm Wrote:
(05-29-2014 04:23 AM)Gerson W. Barbosa Wrote:

Code:

Q0007 STO D

You can remove this step as D is never used.

Duely removed above. Thank you very much!

Gerson.

Jeff_Kearns · 05-30-2014, 03:56 AM

(05-29-2014 03:48 PM)Gerson W. Barbosa Wrote: Thank you very much!
Gerson.

And I thank you both! --- Jeff K

Gerson W. Barbosa · 05-31-2014, 03:31 AM

(05-30-2014 03:56 AM)Jeff_Kearns Wrote:
(05-29-2014 03:48 PM)Gerson W. Barbosa Wrote: Thank you very much!
Gerson.
And I thank you both! --- Jeff K

You're most welcome!

BTW, the HP-42S code (post #1) has been shortened to 14 steps and 25 bytes, including LBL and END. Perhaps a new world record for the HP-42S :-) (Not being the most accurate, at least that might be the shortest one ever - well, until someone manages to save yet another step or byte)

Regards,

Gerson.