@Algoose, in worst case, this is still O(n^2), ain't it?

On Wed, Dec 21, 2011 at 12:50 PM, Algoose chase <harishp...@gmail.com>wrote:

> Find the distance between each of the points and the origin(0,0) and sort
> the points based on this distance.
> Also, divide the points based on which quadrant they belong.  If the
> difference between their distance(from origin) between 2 points is less and
> they belong to the same quadrant, then they are likely to be close. So,
> instead of comparing each point with every other point as in the O(N^2)
> solution. We can compare a given point only with a subset of points that
> appear to be close.
>
> On Wed, Dec 21, 2011 at 1:00 AM, SAMMM <somnath.nit...@gmail.com> wrote:
>
>> You are given a list of points in the plane, write a program that
>> outputs each point along with the three other points that are closest
>> to it. These three points ordered by distance.
>> The order is less then O(n^2) .
>>
>> For example, given a set of points where each line is of the form: ID
>> x-coordinate y-coordinate
>>
>>
>> 1  0.0      0.0
>> 2  10.1     -10.1
>> 3  -12.2    12.2
>> 4  38.3     38.3
>> 5 79.99 179.99
>>
>>
>>
>> Your program should output:
>>
>>
>> 1 2,3,4
>> 2 1,3,4
>> 3 1,2,4
>> 4 1,2,3
>> 5 4,3,1
>>
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