Jeff Aa wrote at Thu, 19 Sep 2002 19:08:44 +0200:
>> I liked your algorithm as it is quick and easy.
>> But on the other hand the results aren't as good as possible.
>
> Actually, I do believe that the results are as good as possible, for any
> distribution of numbers. 8-) consider that the firs
> -Original Message-
> From: Janek Schleicher [mailto:[EMAIL PROTECTED]]
> Sent: 19 September 2002 16:47
> To: [EMAIL PROTECTED]
> Subject: RE: lots of numbers...
>
> I liked your algorithm as it is quick and easy.
> But on the other hand the results ar
Jeff wrote at Thu, 19 Sep 2002 01:27:27 +0200:
>> But could it be, we missed the simplest, but acceptable algorithm:
>>
>> [snipped my algorithm for shortness]
>
> mmm - looks like a more complex approach to me?
Hrm, yep.
I liked your algorithm as it is quick and easy.
But on the other hand the
-Original Message-
From: Janek Schleicher [mailto:[EMAIL PROTECTED]]
Sent: 18 September 2002 18:32
To: [EMAIL PROTECTED]
Subject: RE: lots of numbers...
> You meant my $iterations = int( scalar @numbers / 3 );
> (However, the scalar cast isn't necessary)
>
> my
Jeff Aa wrote at Wed, 18 Sep 2002 18:56:52 +0200:
> Your analysis is interesting. I am no mathemagician either, but it
> occurs to me that it would be easy to code your analysis more
> efficiently than using the sum rank, something like this: (not
> debuggered)
>
> use strict;
>
> # Note that t
> -Original Message-
> From: Janek Schleicher [mailto:[EMAIL PROTECTED]]
> Sent: 18 September 2002 16:34
> To: [EMAIL PROTECTED]
> Subject: Re: lots of numbers...
>
>
> I think, it's better to weight the sorted numbers
> with their sum rank, in the abo
Angerstein wrote at Wed, 18 Sep 2002 17:20:29 +0200:
> if you have a list of numbers:
> ...
> What you really want are groups each with 3 numbers.
>
> The summe of the numbers in a group should be near as possible on the
> average value.
>
> My "simple" idea:
> sort the numbers.
> put the firs
