On 4/20/11 Wed  Apr 20, 2011  3:17 PM, "shawn wilson" <ag4ve...@gmail.com>
scribbled:

> On Apr 20, 2011 1:05 AM, "ai nguyen" <aichuab...@gmail.com> wrote:
>> 
>> 
>> A population of 20 cows, each one has age and weight (known).
> 
> Irrelevant

I don't think so. The fact that there are 20 cows and 2 attributes for each
is central to the problem.

> 
>> Device this population into 2 group, each group has 10 cows.
>> 
> two data sets to compare

Not quite. Two sets to create out of the given initial set.

> 
>> Questions:
>> How to pick a cow on each group so that a distribution of AGE and
>> WEIGHT on each group is similar.
> 
> Which property should have more weight? If neither, than just add them up
> and use that result to compare. The sort and loop for each set and store the
> index in a value if set1 - set2 is less than the old result.

Both properties should probably have equal weight.

You must first define what you mean by "distribution". There are several
measures of distribution: average, standard deviation, mean, median, etc.
There are probably more complex measures that only students of statistics
will know about (not me).

Having selected a measurement method for comparing distribution, you must
then think of a partitioning algorithm that will create two sets with the
closest distribution values. I bet a random selection won't do too bad.

You might then want to try picking a pair of cows, swapping them, and seeing
if it improves the distribution comparison.

> 
> Show your strategy or/and implement
>> it in PERL.
> 
> Irrelevant.

Not totally irrelevant to the problem of implementing the solution in Perl.

> 
>> 
>> Thanks
>> 
> Yeah. But when you ask questions in the future, think about the task, try to
> do something (hell, try to fail if you must) and then post. When everyone
> knows its homework that probably means that you've failed at failing and
> should sleep off your hangover and try again.

I canb agree on that point.



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