Arithmetic-Geometric Mean

[Weregoose]

Teaser/Challenge #9

Make a program that will calculate and display the arithmetic-geometric mean
of two numbers that reside in Ans as a list of two elements. The program size
minus the number of characters in its title should end up as exactly 28 bytes.

Note: The result displayed at the end must be a number and cannot be a list.

[Harrierfalcon]

Umm...what exactly is a "arithmetic-geometric mean"? Obviously it's not just the average...

[Weregoose]

[Harrierfalcon]

Sorry.

Not even a quarter of the way there and 36 bytes. Hmm...

[nitacku]

Wow... had to use some calculus on this one.

My finished algorithm is 46 bytes though.
EDIT1: Saved 1 byte. Now it is 45 bytes
EDIT2: Saved 2 bytes. Now it is 43 bytes

Decided to post the code as a screenshot rather than try to type out all the symbols
Spoiler:[attachment=1311:attachment]

[Weregoose]

Well built. :)

[EDIT]

I'm looking more at a bivariate iterative process that depends on a few built-in
instructions in some strategic locations, consisting in total of four lines of code.

[DarkerLine]

I think I got it. If you don't mind the algorithm overflowing when one of the values is too big to be squared, you can cut it down by another byte.

[Weregoose]

The 27 byte version, while slower, definitely wins the contest. The only other
piece that was different from mine was in the last line, where many functions
could easily act as replacements (1–6 in the LIST MATH submenu, mostly). :)

So, who can come up a new challenge?

[nitacku]

<> What about mine!</> :biggrin:
It's 100% accurate according to Gauss's constant

So you mean I didn't have to do all that calculus?

[bfr]

I would hope that finding the arithmetic-geometric mean wouldn't require calculus. :P

[thornahawk]

A late addition, and certainly not the most compact one, but this particular algorithm for the AGM is useful when the product of the two original arguments will overflow:

(highlight below)

1→V:max(A,B→U
√(1−(min(A,B​)/Ans)²→C
While Ans
.5(V+√(V²−Ans²→V
C²/4/Ans→C
End
UV


(the above routine assumes that the two arguments are stored in A and B​).

Appropriate modifications would of course be needed if the assumption of positive A and B is no longer valid.

thornahawk

[DarkerLine]

Wouldn't prod(√(Ans be a valid way to prevent the product from overflowing? Except for the use of stdDev( (which can be avoided), I don't see any way that the routine Goose and I came up with is problematic.

Of course, negative numbers are kind of annoying...

[WikiGuru]

wow, for a second i thought it said calculate the arithmetic and geometric mean of two numbers, i had a program that did it in 23 bytes! hmmm... arithmetic-geometric mean...

[thornahawk]

After more literature-searching, it turns out (to my chagrin! ) that still the best way to do an AGM iteration so that it reliably works for too big or too small numbers (or for that matter, complex numbers), is to use the so-called "homogeneity relation":

AGM(c*a,c*B​) == c*AGM(a,B​)

along with the classical a_i+1 := (a_i+b_i)/2 , b_i+1 := √(a_i*b_i).

So your method, with that modification (whichever of A and B that has the largest magnitude is the appropriate "c"), would still beat the method I had posted. For true robustness, one would also check for the case of either one of A and B being zero, in which case the result should be zero.

On my part, I got interested in this topic again because I have had the need to write programs for computing elliptic integrals and elliptic functions, where AGM-based methods are one of the speediest.

thornahawk

P.S. ...and it turns out, in those particular applications, the actual function needed is AGM(1,√(a))!

[DarkerLine]

What magnitude would you want to reduce the numbers to, though? I can't think of any reason computing AGM(1,2) would be more accurate or faster than computing AGM(1000000,2000000).