(09-16-2014 12:33 PM)Dieter 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.
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
In order to address the case F=0, which seems to be a consequence of B=0, I had suggested later in the thread the insertion of x=0? x! between ABS and /. Your suggestion is one step shorter and more efficient, so I would go for it. That's what I was able to come up with while watching an interview on TV. When memory is not an issue (sometimes it is on the 32S-II), I'll definitely use the SGN subroutine replacement, since no stack analysis is necessary.
Regards,
Gerson.
You don't really need the
SIGN function. You can exchange the
SQRT with register
F and use register arithmetic (lines B12 - B17). But then you don't push
F onto the stack. That's why we need to recall
F later in line E07.
Code:
B01 LBL B
B02 FIX 9
B03 RCL H
B04 x^2
B05 STO(i)
B06 RCL D
B07 RCL I
B08 SUM -
B09 SUM xy
B10 x<0?
B11 GTO E
B12 SQRT
B13 x<> F
B14 x<0?
B15 RCL- F
B16 x>=0?
B17 RCL+ F
B18 STO J
B19 x<>y
B20 /
B21 RCL J
B22 x#0?
B23 GTO F
B24 CLx
B25 STOP
F01 LBL F
F02 RCL G
F03 x<>y
F04 /
F05 STOP
E01 LBL E
E02 SF 0
E03 +/-
E04 SQRT
E05 RCL D
E06 /
E07 RCL F
E08 R UP
E09 /
E10 STOP
Currently I have troubles with my HP-32II and therefore can't test these changes. But as an inspiration you can have a look at the original code for the HP-15C (lines 057-061 of program B).
HTH
Thomas
PS: Sorry for being late to the party.
The suggestions in my previous post are flawed. Meanwhile I could fix it. This program works fine with my HP-32SII:
Code:
H01 LBL H
H02 INPUT A
H03 STO D
H04 INPUT B
H05 -2
H06 ÷
H07 STO F
H08 STO H
H09 INPUT C
H10 STO G
H11 STO I
H12 CF 0
H13 SCI 2
A01 LBL A
A02 CLΣ
A03 28
A04 STO i
A05 4
A06 STO(i)
A07 33
A08 STO i
A09 RCL H
A10 STO(i)
A11 RCL÷ D
A12 RND
A13 RCL D
A14 Σ-
A15 RCL I
A16 Σxy
A17 x<>y
A18 STO(i)
A19 R↓
A20 x<>y
A21 RCL H
A22 Σ-
A23 R↓
A24 Σ-
A25 Σxy
A26 ABS
A27 RCL I
A28 ABS
A29 x≤y?
A30 GTO B
A31 ENTER
A32 R↑
A33 STO H
A34 Σxy
A35 STO I
A36 ABS
A37 1E24
A38 ×
A39 RCL G
A40 ABS
A41 x≤y?
A42 GTO A
B01 LBL B
B02 FIX 9
B03 RCL H
B04 x²
B05 STO(i)
B06 RCL D
B07 RCL I
B08 Σ-
B09 Σxy
B10 x<0?
B11 GTO E
B12 SQRT
B13 x<> F
B14 x<0?
B15 RCL- F
B16 x≥0?
B17 RCL+ F
B18 x<> G
B19 x≠0?
B20 RCL÷ G
B21 RCL G
B22 RCL÷ D
B23 STOP
E01 LBL E
E02 SF 0
E03 +/-
E04 SQRT
E05 RCL÷ D
E06 x<>y
E07 RCL F
E08 x<>y
E09 ÷
E10 STOP
This post made me try to understand the algorithm that is used. I posted my findings in an
article.
Cheers
Thomas
Dear Members, I have a question about Palmer Hanson program for Hp 35s. I don´t know how to insert steps Q038, 040 and 065= SUM - , from the program listing of Cadillac Quadratic Solver.
Can you please assit me about the order of the keys for that instruction
Thank you in advance, Pedro
(02-17-2015 07:36 PM)PedroLeiva Wrote: [ -> ]Can you please assit me about the order of the keys for that instruction
I see this is your first post here, so welcome to the forum. ;-)
In the listing, "Sum" refers to the greek Sigma, so "Sum–" means Σ– (yellow-shifted function of the Σ+ key in the bottom key row). Equally, "Sum xy" means Σxy which is accessed via the SUMS menu (blue-shifted Minus-key).
Dieter
(02-17-2015 08:26 PM)Dieter Wrote: [ -> ] (02-17-2015 07:36 PM)PedroLeiva Wrote: [ -> ]Can you please assit me about the order of the keys for that instruction
I see this is your first post here, so welcome to the forum. ;-)
In the listing, "Sum" refers to the greek Sigma, so "Sum–" means Σ– (yellow-shifted function of the Σ+ key in the bottom key row). Equally, "Sum xy" means Σxy which is accessed via the SUMS menu (blue-shifted Minus-key).
Dieter
How I did not realize it! Thank you very much for your assistance. Pedro