Calculator overflow?

Just a guess but if they calculated the answer as
(343*342304)/(40*391)
the numerator would exceed 1e100 and give you overflow problems.

Ian Smith

This is heading into the science of numerical methods, not my forte. I'll throw
this one to the panel.

Unfortunately manufacturer-specific details are sparse, mainly because the
manufacturers regard the details of their numerical methods as valuable
intellectual property.

I have no specifics on the Casio, and nothing obvious pops up with Google, but
you can get an idea of the sort of thing they use by studying Numerical Methods
in general, eg here is a page for Mathematica:
#

A sample:

Combinatorial functions

Most combinatorial functions use sparse caching and recursion.

Factorial, Binomial and related functions use a divide-and-conquer algorithm to
balance the number of digits in subproducts.

Fibonacci[n] uses an iterative method based on the binary digit sequence of n.

PartitionsP[n] uses Euler's pentagonal formula for small n, and the
non-recursive Hardy-Ramanujan-Rademacher method for larger n.

ClebschGordan and related functions use generalized hypergeometric series.


Of course all of Mathematica's methods are designed in such a way as to avoid
the overflow problems you found, and they have plenty more CPU and RAM at their
disposal.
--
Patrick Hamlyn posting from Perth, Western Australia
Windsurfing capital of the Southern Hemisphere
Moderator: polyforms group ()

It's more likely that it uses some sort of heavily optimised numerical
approximation based on a fast-converging series.
--
Patrick Hamlyn posting from Perth, Western Australia
Windsurfing capital of the Southern Hemisphere
Moderator: polyforms group ()

...

I suppose my question really is, What algorithm does this calculator use
that allows it to compute C(343,39) = 343!/(39!304!) but makes it
unable to compute C(343,40) = 343!/(40!303!) ?
We know that it does not first calculate 343! and then divide,
because calculating 343! will cause overflow. My guess is that
it calculates C(343,39) by (343/39)*(342/38)(306/2)*(305/1),
thus avoiding overflow. So a similar calculation of
C(343,40) by (343/40)*(342/39)(305/2)*(304/1)
should similarly avoid overflow, but doesn't. What explains this?
-- Mark Spahn (West Seneca, NY, USA)

There isn't actually a little person in there doing the calculation, and making
tricky algorithmic decisions on the fly. Very early calculator models that did
this were impractical and only made an appearance in the circus, on stage etc.

What they do nowadays is pre-define an algorithm suitably simple for the tiny
electronic brain of the calculator, which unfortunately is not terribly
creative.
--
Patrick Hamlyn posting from Perth, Western Australia
Windsurfing capital of the Southern Hemisphere
Moderator: polyforms group ()