I guess choosing n equally spaced points withing the given interval will be
the desired solution.
2009/10/20 Mithun Kumar Singh <mithunsi...@gmail.com>
> Hi,
> This problem is same as a Travelling salesman problem........In
> travelling salesman we need to cover points in Min distance...here we need
> to just opposite..
>
> PS: Answer may be misleading ...if so Pls correct me... :)
>
> -Mithun
>
>
>
>
> On Tue, Oct 20, 2009 at 6:16 PM, monty 1987 <1986mo...@gmail.com> wrote:
>
>> Hi,
>> Still waiting for solution...........
>>
>> On Wed, Oct 7, 2009 at 3:18 PM, monty 1987 <1986mo...@gmail.com> wrote:
>>
>>> The important thing is all the points do not lie in same range i.e.
>>> x1 ,x2 ,x3 each of them have their own range.
>>>
>>>
>>> On Wed, Oct 7, 2009 at 3:15 PM, monty 1987 <1986mo...@gmail.com> wrote:
>>>
>>>> The min. distance between two points i.e. the euclidean distance between
>>>> two points.
>>>>
>>>>
>>>> On Tue, Oct 6, 2009 at 5:52 PM, MrM <maleki...@gmail.com> wrote:
>>>>
>>>>>
>>>>> you can arrange them with equal distances !
>>>>> if n=1 then, it does not matter where you put the point !
>>>>> if n>1 then, put them with distances = (r2i-r1i) / (n-1) !
>>>>> it means ou put the first point on r1i and the last point on r2i, the
>>>>> remaining point are distributed with equal distances !
>>>>>
>>>>> On Oct 5, 5:22 pm, monty 1987 <1986mo...@gmail.com> wrote:
>>>>> > We have to locate n points on the x-axis
>>>>> > For each point xi
>>>>> > the x co-ordinate of it lies between a
>>>>> range
>>>>> > [r1i,r2i]
>>>>> > Now we have to decide the location of points such that
>>>>> > minimum { distance between any two points } is maximum.
>>>>> >
>>>>> > Any answer is welcomed.
>>>>>
>>>>>
>>>>>
>>>>
>>>
>>
>>
>>
>
> >
>
--
"Reduce, Reuse and Recycle"
Regards,
Vivek.S
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