On Oct 2, 2013, at 9:51 AM, Tamara Temple <tamouse.li...@gmail.com> wrote:

> 
> On Oct 1, 2013, at 1:51 PM, Floyd Resler <fres...@adex-intl.com> wrote:
> 
>> Here's my task: A group of kids is going to be staying with different host 
>> families throughout the next 8 months.  The number of kids staying with a 
>> host family can range from 2 to 10.  When deciding which kids should stay 
>> together at a host family, the idea is for the system to put together kids 
>> who have stayed with each other the least on past weekends.  So, if a host 
>> family can keep 5 kids, then the group of 5 kids who have stayed together 
>> the least will be chosen.
>> 
>> I can't think of an easy, quick way to accomplish this.  I've tried various 
>> approaches that have resulted in a lot of coding and being very slow.  My 
>> idea was to give each group of kids a score and the lowest score is the 
>> group that is selected.  However, this approach wound of iterating through 
>> several arrays several times which was really slow.  Does anyone have any 
>> ideas on this puzzle?
>> 
>> Thanks!
>> Floyd
>> 
>> 
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> 
> While definitely a tempting coding exercise, I just want to say that if this 
> is urgent in any way, shuffling cards with the kids' names on them by hand 
> might just be faster and less frustrating :)
> 
> OTOH, if this is something you're going to have to figure out week after 
> week, then a software solution might be handy.
> 
> This is also not an *easy* problem to solve; there isn't a simple approach to 
> optimizing this sort of thing because you're building a net between all the 
> various kids based on past stays, in addition to the constraints of host 
> family  capacity. Thus your previous code attempts might in fact be the end 
> result.
> 
> Obviously, structuring the data is the key here.
> 
> I'm thinking of 3 primary models: Kids, Hosts, and Stays.
> 
> Kids and Hosts seem pretty obvious. Stays is the interesing model, and needs 
> to have joining tables with Kids and Hosts.
> 
> A Stay will have one Host, and have many Kids and a date.
> 
> The algorithm then needs to make the graph where it can pull out the number 
> of times any particular kid has stayed with another, looking something like 
> this:
> 
> Amy:
>   Ben: 10
>   Jill: 3
>   Carlos: 7
>   Chen: 2
> Ben:
>   Amy: 10
>   Jill: 5
>   Carlos: 8
>   Chen: 3
> Jill:
>   … and so on
> 
> Then you be able to pull through that graph and find the smallest number of 
> stays for each kid.
> 
> Not simple, but I hope this helps.
> 
> 

That's the only approach I could think of.  I may have to tell the director it 
may be a bit slow but at least she won't have to do it by hand!

Thanks!
Floyd


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