Hello,

I went to college from '80-'84, and lived and died by my HP 41CV. I wrote dozens of programs for it as I progressed through my courses.

I recently picked up a new HP 35s, and decided to convert one of my oldest, most-used HP41 programs to the 35 as a learning exercise mainly.

The program just does simple linear interpolation. Suppose we have a table of data such as

...

X1    Y1

X2    Y2

...

Given any value X, it simply estimates Y using linear interpolation. The equation to do that is brutally simple, it is just such a frequent need that it is a great candidate for a quickie program.

As a further requirement however, I decided that I wanted to make it "low footprint," i.e. such that it only used the absolute minimum of the 26 A..Z named storage registers, and relied instead on the 800 indirect-addressed unnamed storage registers for any temporary storage requirements. I'm not sure how strong a need that is, frankly, but I thought it might be, to preserve the named registers for use as variables in equations, and allow those values to remain unmolested and untrampled-upon. That design decision did complicate things a bit, resulting in some muddy-looking code as the program plays games with the stack, but it's not too bad.

Program Name: "L" (it lived for years as "LIP" in my HP41 calculators)

Named storage registers used: I

Un-named indirect addressed storage registers used: "n" through "n+3", where the value of "n" is set early in the program and can be easily changed if needed. Default = 10.

L001	LBL L	

L002	10	Base register 'N' for storage

L003	STO I	N stored in I

L004	SF 10	

L005	X2,X1,X	Prompt for X2 enter X1 enter X

L006	CF 10	

L007	STO(I)	X stored in N

L008	1	

L009	STO+ I	N+1 stored in I

L010	RDN	

L011	RDN	

L012	STO(I)	X1 stored in N+1

L013	1	

L014	STO+ I	N+2 stored in I

L015	RDN	

L016	RDN	

L017	STO(I)	X2 stored in N+2

L018	1	

L019	STO+ I	N+3 stored in I

L020	SF 10	

L021	Y2,Y1	Prompt for Y2 enter Y1

L022	CF 10	

L024	STO(I)	Y1 stored in N+3

L025	-	(Y2-Y1)

L026	1	

L027	STO- I	N+2 stored in I

L028	RDN	

L029	RCL(I)	X2

L030	1	

L031	STO- I	N+1 stored in I

L032	RDN	

L033	RCL(I)	X1

L034	-	(X2-X1)

L035	/	(Y2-Y1)/(X2-X1)

L036	RCL(I)	X1

L037	1	

L038	STO- I	N stored in I

L039	RDN	

L040	RCL(I)	X

L041	X<>Y	

L042	-	(X-X1)

L043	*	(X-X1)*(Y2-Y1)/(X2-X1)

L044	3	

L045	STO+ I	N+3 stored in I

L046	RDN	

L047	RCL(I)	Y1

L048	+	(X-X1)*(Y2-Y1)/(X2-X1) + Y1

L049	RTN	Y is left in X-register

LN=166	CK=FC9A	

As mentioned, the linear interpolation equation is trivial and straightforward and not really worthy of much discussion. The only thing possibly novel in this program is my attempt to make it "low footprint", and use the un-named storage registers as much as possible for temporary storage needs, rather than the named A..Z registers. That decision did complicate the code a bit, as I had to manipulate I in the middle of the "real math", which complicates the listing somewhat.

Best,
Elliott

thanks for the program.

step 23 is missing.

Oops, thanks for noticing. It's actually not missing, my line numbers are just messed up. The LN and CK numbers are correct, and the program should end on line L048.