@Sankalp: I wanted a point far enough outside the region that a line
from any point in the region could not contain any other point in the
region. There are several implications: 1) if the point is to the left
of the region, it can't have an integer y coordinate between 0 and B.
That is where the 0.5 comes from. Second, it seemed reasonable, but
not necessary, to put Y near the middle of the range [0, B]. Then I
just solved for an X that guaranteed that the slope of the line from
any point in the region to the point (X, Y) had slope m satisfying -1/
A < m < 1/A. Then a line through any point (x1,y) could not intersect
any point (x2,y-1) or (x3,y+1) within the region.
Regarding your algorithm: What does it do with A = B = 5 and points
{(1,1), (2,2), (3,3), (4,4), (1,4)}. This example shows that you can't
always use a line through (0,0).
Dave
On Jan 24, 10:15 pm, sankalp srivastava <[email protected]>
wrote:
> @
> Dave
> How did you come up with this solution?
> Also Y=floor(B)/2+.5 , X=-A*(B-Y) or X=-AB +AY or Y=X/A+B . this is
> the equation of a line with slope 1/A and an intercept of B on Y
> axis .I don't quite get this.!!Please elaborate .
> Meanwhile , this is my approach .
>
> The slope of the line wil be between the maximum's of the two points
> i.e in the case of (10,0)..(10,10)...It will be between 0 and 90
> degrees as all the points lie between them . Now we can just binary
> search over this slope checking for the slope values . The slope and
> set cardinality is also a monotonic function , so I guess binary
> search approach will work , but the time taken will be nlog n . Please
> correct me if I'm wrong .
> #include<iostream>
> struct point
> {
> int x ;
> int y ;};
>
> using namespace std ;
> // the given points
> point p[100];
> int main()
> {
> int n;
> cin>>n;//number of points
> for(int i=0;i<n;i++)
> {
> cin>>p[i].x>>p[i].y;
> }
> int high=90;
> int low=0;
> int mid;
> //the line is supposed to start from 0,0
> while(low<=high)
> {
> mid=(high+low)/2;
> if(haspoints(mid)<0)
> //upper has less points than below
> low=mid+1;
> else if(haspoints(mid)>0)
> //lower has more points than upper
> high=mid-1;
> else
> {//we found our answer
> cout<<mid<<endl;
> return 0;
> }
> }
> return 0;
>
> }
>
> On Jan 24, 9:30 am, Dave <[email protected]> wrote:
>
>
>
> > Generalizing the problem, let there be n points (x_i, y_i), where x_i
> > and y_i are integers with 1 <= x_i <= A and 1 <= y_i <= B. A line
> > separating the points into two sets of equal cardinality can be found
> > as follows: Let Y = floor(B/2) + 0.5, and let X = -A * (B - Y). Find
> > the slopes of the lines from the point (X, Y) to each (x_i, y_i). The
> > point (X, Y) is constructed so that each of these slopes will be
> > distinct. Find the median M of these slopes. Then the line
>
> > y = M(x - X) + Y
>
> > will separate the points as desired. It will pass through exactly one
> > of the points if n is odd, and will miss all of the points if n is
> > even. This is O(n) in time and O(n) in space, where n is the number of
> > points.
>
> > Dave
>
> > On Jan 21, 11:45 pm, Divya Jain <[email protected]> wrote:
>
> > > assume the region to be (0,0) , (10,0), (0,10), (10,10)
>
> > > On 22 January 2011 08:33, Dave <[email protected]> wrote:
>
> > > > @Divya: The coordinates of the points are between 0 and 1 and are
> > > > integers. That can't be right.
>
> > > > Dave
>
> > > > On Jan 21, 1:46 pm, divya <[email protected]> wrote:
> > > > > Within a 2D space, there is a batch of points(no duplicate) in the
> > > > > region (0,0),(0,1),(1,0),(1,1), try to find a line which can divide
> > > > > the region to 2 parts with half points in each .the input will be an
> > > > > array of points and the length of the array.
> > > > > struct point{
> > > > > int x;
> > > > > int y;};
>
> > > > > input : struct point * points, int length
>
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