HP 25 Fibonacci sequence in one program step
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10-14-2021, 10:03 AM
(This post was last modified: 10-14-2021 10:05 AM by C.Ret.)
Post: #21
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RE: HP 25 Fibonacci sequence in one program step
(10-13-2021 04:42 PM)Gerson W. Barbosa Wrote: RPN: Y, Z, T <- k1; X <-k2 Then * * * … Thank to remind me this old matrix arithmetic multiplication on RPL. I remember a few decade when I try to use power exponentiation of [[1 1][1 0]] to the n to directly get Fn value. Hard time discovering sadly that the matrix exponentiation on the HP-28C/S doesn't work with ^ instruction. (10-13-2021 06:26 PM)Dave Britten Wrote: I think you guys have figured out the gist of it. OK, now I can publish my completed trace print. [attachment=9914] (10-13-2021 07:35 PM)David Hayden Wrote:(10-13-2021 07:11 PM)Gerson W. Barbosa Wrote: 0 1 + Then repeat RCL + ST LThat's it! I can't applied this trick neither on my HP-41C that miss memory recall arithmetic, nor on my HP-15C who have recall arithmetic but no stack register manipulation. The amazing HP 42S (or succedanea DM 42S or Free 42) is really a blend of the two anthology systems. At least, like the HP-25, the HP-41 and HP-15C need a two steps naked program « LASTx + », close to a simple RPL code leaving the Fibonacci sequence reversed in the stack initiate by 1 DUP and repeating « OVER OVER + » over and over... |
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10-14-2021, 02:13 PM
(This post was last modified: 10-14-2021 04:10 PM by Albert Chan.)
Post: #22
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RE: HP 25 Fibonacci sequence in one program step
(10-14-2021 10:03 AM)C.Ret Wrote: We can use this to get fib(n+m) = fib(m)*fib(n+1) + fib(m-1)*fib(n) If n=m, we have fib(2n) = fib(n) * (fib(n+1) + fib(n-1)) Using this, we can can 1/sqrt(5) another way φ^n ≈ fib(2n) / fib(n) = fib(n+1) + fib(n-1) ≈ fib(n) * (φ+1/φ) = fib(n) * √5 fib(n) = round(φ^n / √5) |
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10-14-2021, 07:04 PM
(This post was last modified: 10-17-2021 03:41 PM by rprosperi.)
Post: #23
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RE: HP 25 Fibonacci sequence in one program step
(10-14-2021 02:13 PM)Albert Chan Wrote:(10-14-2021 10:03 AM)C.Ret Wrote: Looks like more than 1 program step to me.... Also, where are those functions on an HP-25? I've looked at mine again today, and just can't see them anywhere... --Bob Prosperi |
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10-16-2021, 02:55 PM
(This post was last modified: 10-16-2021 03:00 PM by Gerson W. Barbosa.)
Post: #24
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RE: HP 25 Fibonacci sequence in one program step
(10-13-2021 08:34 PM)rprosperi Wrote: Many other models support recall arithmetic (45, 27, 15C, 32S, 32SII, others?) but only the 42S includes recall arithmetic AND stack register access. Since we’re at it, I decided to write an HP-42S program to display the Fibonacci sequence trying to use the least number of steps (Not sure whether I’ve succeeded or not – another one I tried has exactly the same number of steps and bytes). Interestingly an equivalent HP-15C program, which lacks recall stack arithmetic but does have recall arithmetic, is possible using the same number of steps (8). Code:
XEQ “Fib” → 0 R/S → 1 R/S → 1 R/S → 2 R/S → 3 R/S → 5 R/S → 8 … |
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10-17-2021, 02:17 PM
(This post was last modified: 10-17-2021 02:45 PM by Gerson W. Barbosa.)
Post: #25
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RE: HP 25 Fibonacci sequence in one program step
(10-16-2021 02:55 PM)Gerson W. Barbosa Wrote: (Not sure whether I’ve succeeded or not – another one I tried has exactly the same number of steps and bytes) Surely I hadn’t. It is possible to do it in 6 steps and 17 steps. Thanks Mike (Stgt) for his comment “Many think a program must start with a global label, but no, it needs not” and examples which easily lead to Code:
Edited for a minor grammar issue |
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