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Why Andrew's Monotone Algorithm takes O(N LogN) times?

Andrew's monotone chain convex hull algorithm constructs the convex hull of a set of 2-dimensional points in O(n\log n) time.
I followed the steps of the algorithm and found out that it has O(n Logn) time complexity. Sorting is just finding the lowest X and then in case of equal, find the lower Y. I am not using heap or other sorts initially. On which line, does it has Log operation?
More information can be found from the link provided below.
Link: https://en.wikibooks.org/wiki/Algorithm_Implementation/Geometry/Convex_hull/Monotone_chain
points.Sort();
if (points.Count <= 3)
{
return new List<PlaceOfInterest>(points);
}
List<PlaceOfInterest> upperHull = new List<PlaceOfInterest>();
foreach (PlaceOfInterest point in points)
{
PlaceOfInterest p2 = point;
while (upperHull.Count >= 2)
{
PlaceOfInterest pivot = upperHull[upperHull.Count - 2];
PlaceOfInterest p1 = upperHull[upperHull.Count - 1];
if (Calculation.SignedArea(pivot, p1, p2) <= 0)
{
upperHull.RemoveAt(upperHull.Count - 1);
}
else
{
break;
}
}
upperHull.Add(p2);
}
upperHull.RemoveAt(upperHull.Count - 1);
List<PlaceOfInterest> lowerHull = new List<PlaceOfInterest>();
for (int i = points.Count - 1; i >= 0; i--)
{
PlaceOfInterest p2 = points[i];
while (lowerHull.Count >= 2)
{
PlaceOfInterest pivot = lowerHull[lowerHull.Count - 2];
PlaceOfInterest p1 = lowerHull[lowerHull.Count - 1];
if (Calculation.SignedArea(pivot, p1, p2) <= 0)
{
lowerHull.RemoveAt(lowerHull.Count - 1);
}
else
{
break;
}
}
lowerHull.Add(p2);
}
lowerHull.RemoveAt(lowerHull.Count - 1);
if (!(Enumerable.SequenceEqual(upperHull, lowerHull)))
{
upperHull.AddRange(lowerHull);
}
return upperHull;

points.Sort() takes O(N log N) time.
The rest takes O(N) time. (If you did it right -- I didn't check)

grahams scan to implement polygon partition in c#

I am trying to solve polygon partition problem using graham scan
below is the problem statement and input and expected output
I have implemented graham scan logic but i need to display output
according to the sets formed and display all the points with indexes;
The first line will contain a single integer ().
The next lines will contain two integers (), denoting a point with coordinates . Note that points are not guaranteed to be distinct.
SAMPLE INPUT
3
0 0
0 1
1 0
On the first line, output a single integer (), the number of sets in your partition.
On the next lines, print the elements of the partition. On the -th of these lines, first print the integer (), and then indices of points forming the -th set. Points are numbered starting from . Each index from to must appear in exactly one set.
SAMPLE OUTPUT
1
3 1 2 3
below is the code implemented according to the output mentioned above,
how to print for more sets?
private static void convexHull(Point[] points, int N, int set)
{
int index = 0;
List<int> result = new List<int>();
// There must be at least 3 points
if (N < 3) return;
// Initialize Result
MyStack<Point> hull = new MyStack<Point>(N);
// Find the leftmost point
int l = 0;
for (int i = 1; i < N; i++)
if (points[i].x < points[l].x)
l = i;
// Start from leftmost point, keep moving
// counterclockwise until reach the start point
// again. This loop runs O(h) times where h is
// number of points in result or output.
int p = l, q;
do
{
// Add current point to result
hull.push(points[p]);
// Search for a point 'q' such that
// orientation(p, x, q) is counterclockwise
// for all points 'x'. The idea is to keep
// track of last visited most counterclock-
// wise point in q. If any point 'i' is more
// counterclock-wise than q, then update q.
q = (p + 1) % N;
for (int i = 0; i < N; i++)
{
// If i is more counterclockwise than
// current q, then update q
if (orientation(points[p], points[i], points[q])
== 2)
q = i;
}
// Now q is the most counterclockwise with
// respect to p. Set p as q for next iteration,
// so that q is added to result 'hull'
p = q;
} while (p != l); // While we don't come to first
set += 1;
var ele = hull.GetAllStackElements();
foreach (Point pt in ele)
{
index += 1;
}
Console.WriteLine(set);
Console.Write(string.Format("{0} ", index));
for (int s = 1; s <= index; s++)
{
Console.Write(string.Format("{0} ", s));
}
Console.Write(Environment.NewLine);
}

B-spline algorithm

I have code for drawing Bezier curves. Is it posible to modify this code for drawing B-Spline curves?
Here is my code using DeCasteljau algorithm:
private Point getPoint(int r, int i, double t)
{
if (r == 0) return points[i];
Point p1 = getPoint(r - 1, i, t);
Point p2 = getPoint(r - 1, i + 1, t);
return new Point((int)((1 - t) * p1.X + t * p2.X), (int)((1 - t) * p1.Y + t * p2.Y));
}
I found this code for B-Spline curves. It looks similar to my code, but I have XY points and there are only numbers. I don't know how to modify my code. I tried something but it doesn't work.
private double BasisFunction(int k, int i, ParameterCollection u, double t){
if(k==0)
{
if((u[i]<=t) && (t<=u[i+1]))
return 1;
else
return 0;
}
else
{
double memb1, memb2;
if(u[i+k]==u[i])
memb1 = 0;
else
memb1 = ((t-u[i])/(u[i+k]-u[i]))*BasisFunction(k-1, i, u, t);
if(u[i+k+1]==u[i+1])
memb2 = 0;
else
memb2 = ((u[i+k+1]-t)/(u[i+k+1]-u[i+1]))*BasisFunction(k-1, i+1, u, t);
return memb1+memb2;
}
}
Please help.

The function BasisFunction() is for computing the value of B-spline basis function N(n,i)(t), where n is degree and i ranges from 0 to (m-1) with m is the number of control points. So, to use this function, you need to define the following for your B-spline:
degree.
m control points, denoting them as P[i][2] with i=0~(m-1)
knot sequence. This is the input "ParameterCollection" to the BasisFunction. You need to have (m+degree+1) knots in the knot sequence and the knot values need to be monotonically non-decreasing. An example of knot sequence for degree 3 B-spline with 5 control points is [0,0,0,0,u0,1,1,1,1], where u0 is any value between [0,1].
Then you can evaluate any point on the B-spline curve at parameter t by something like:
double point[2]={0.0}; // point on the B-spline curve
for (int ii=0; ii < m; ii++) // loop thru all control points
{
double basisVal = BasisFunction(degree, ii, knotSequence, t);
point[0] += P[ii][0]*basisVal;
point[1] += P[ii][1]*basisVal;
}

Find minimum distance between 2 points in 2 arrays

I have 2 lists containing points (x,y and z) and would like to find the closest points.
I assume I need to do something like this:
for (int i = 0; i < myarray1.Count; i++)
{
for (int j = 0; j < myarray2.Count; j++)
{
// Calculate the quadratic distance between 2 points (point in index i and j)
// Then store the minimum distance I guess?
}
}

another option is to use Kd-tree
using the Nearest neighbour search will give you a O(log n) complexity to find the nearest point to a given set of points, and your code will have O( n log n), instead of O (n^2).
see here for implementation and example of how to use it.

double min_dist = DOUBLE_MAX;
for (i = 0; i < myarray1.Count; i++)
{
for (j = 0; j < myarray2.Count; j++)
{
curr_dist = dist(a[i],a[j]);
if( min_dist > curr_dist)
{
min_dist = curr_dist;
}
}
}
where
double dist(Point a, Point b) {
return sqrt(pow(a.x-b.x,2)+pow(a.y-b.y,2)+pow(a.z-b.z,2);
}

To compute the distance:
double sqr(double x) {return x*x;}
double distance(MyPoint a, MyPoint b) {
return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y)+sqr(a.z-b.z);
}
Then in the second loop you store the minimum distance found so far:
double d = distance(myarray1[i],myarray2[j]);
if (d<min_d) min_d = d;
where min_d is defined at the beginning:
double min_d = Float.MAX_VALUE;

In C# I would do this using Linq.
First I would define the function that calculates the distance between two points as in Emanuelle Paolini's answer:
public double Distance(Point p1, Point p2)
{
return Math.Sqrt((p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p1.y) * (p1.y - p1.y) + (p1.z - p1.z) * (p1.z - p1.z));
}
Then I would query the two lists as follows:
var distanceQuery = from p1 in myarray1
from p2 in myarray2
select Dist(p1, p2);
And finally I would retrieve the minimum distance:
var minimumDistance = distanceQuery.Min();

Formula is
Math.sqrt(Math.pow(Math.abs(x1-x2),2) + Math.pow(Math.abs(y1-y2),2)+ Math.pow(Math.abs(z1-z2),2))

How can I convert this divide and conquer code to compare one point to a list of points?

I found this code on the website http://rosettacode.org/wiki/Closest-pair_problem and I adopted the C# version of the divide and conquer method of finding the closest pair of points but what I am trying to do is adapt it for use to only find the closest point to one specific point. I have googled quite a bit and searched this website to find examples but none quite like this. I am not entirely sure what to change so that it only checks the list against one point rather than checking the list to find the two closest. I'd like to make my program operate as fast as possible because it could be searching a list of several thousand Points to find the closest to my current coordinate Point.
public class Segment
{
public Segment(PointF p1, PointF p2)
{
P1 = p1;
P2 = p2;
}
public readonly PointF P1;
public readonly PointF P2;
public float Length()
{
return (float)Math.Sqrt(LengthSquared());
}
public float LengthSquared()
{
return (P1.X - P2.X) * (P1.X - P2.X)
+ (P1.Y - P2.Y) * (P1.Y - P2.Y);
}
}
public static Segment Closest_BruteForce(List<PointF> points)
{
int n = points.Count;
var result = Enumerable.Range(0, n - 1)
.SelectMany(i => Enumerable.Range(i + 1, n - (i + 1))
.Select(j => new Segment(points[i], points[j])))
.OrderBy(seg => seg.LengthSquared())
.First();
return result;
}
public static Segment MyClosestDivide(List<PointF> points)
{
return MyClosestRec(points.OrderBy(p => p.X).ToList());
}
private static Segment MyClosestRec(List<PointF> pointsByX)
{
int count = pointsByX.Count;
if (count <= 4)
return Closest_BruteForce(pointsByX);
// left and right lists sorted by X, as order retained from full list
var leftByX = pointsByX.Take(count / 2).ToList();
var leftResult = MyClosestRec(leftByX);
var rightByX = pointsByX.Skip(count / 2).ToList();
var rightResult = MyClosestRec(rightByX);
var result = rightResult.Length() < leftResult.Length() ? rightResult : leftResult;
// There may be a shorter distance that crosses the divider
// Thus, extract all the points within result.Length either side
var midX = leftByX.Last().X;
var bandWidth = result.Length();
var inBandByX = pointsByX.Where(p => Math.Abs(midX - p.X) <= bandWidth);
// Sort by Y, so we can efficiently check for closer pairs
var inBandByY = inBandByX.OrderBy(p => p.Y).ToArray();
int iLast = inBandByY.Length - 1;
for (int i = 0; i < iLast; i++)
{
var pLower = inBandByY[i];
for (int j = i + 1; j <= iLast; j++)
{
var pUpper = inBandByY[j];
// Comparing each point to successivly increasing Y values
// Thus, can terminate as soon as deltaY is greater than best result
if ((pUpper.Y - pLower.Y) >= result.Length())
break;
Segment segment = new Segment(pLower, pUpper);
if (segment.Length() < result.Length())
result = segment;// new Segment(pLower, pUpper);
}
}
return result;
}
I used this code in my program to see the actual difference in speed and divide and conquer easily wins.
var randomizer = new Random(10);
var points = Enumerable.Range(0, 10000).Select(i => new PointF((float)randomizer.NextDouble(), (float)randomizer.NextDouble())).ToList();
Stopwatch sw = Stopwatch.StartNew();
var r1 = Closest_BruteForce(points);
sw.Stop();
//Debugger.Log(1, "", string.Format("Time used (Brute force) (float): {0} ms", sw.Elapsed.TotalMilliseconds));
richTextBox.AppendText(string.Format("Time used (Brute force) (float): {0} ms", sw.Elapsed.TotalMilliseconds));
Stopwatch sw2 = Stopwatch.StartNew();
var result2 = MyClosestDivide(points);
sw2.Stop();
//Debugger.Log(1, "", string.Format("Time used (Divide & Conquer): {0} ms", sw2.Elapsed.TotalMilliseconds));
richTextBox.AppendText(string.Format("Time used (Divide & Conquer): {0} ms", sw2.Elapsed.TotalMilliseconds));
//Assert.Equal(r1.Length(), result2.Length());

You can store the points in a better data structure that takes advantage of their position. Something like a quadtree.
The divide and conquer algorithm that you are trying to use doesn't really apply to this problem.

Don't use this algorithm at all, just go through the list one at a time comparing the distance to your reference point and at the end return the point that was the closest. This will be O(n).
You can probably add some extra speed ups but this should be good enough.
I can write some example code if you want.

You're mixing up two different problems. The only reason divide and conquer for the closest pair problem is faster than brute force is that it avoids comparing every point to every other point, so that it gets O(n log n) instead of O(n * n). But finding the closest point to just one point is just O(n). How can you find the closest point in a list of n points, while examining less than n points? What you're trying to do doesn't even make sense.
I can't say why your divide and conquer runs in less time than your brute force; maybe the linq implementation runs slower. But I think you'll find two things: 1) Even if, in absolute terms, your implementation of divide and conquer for 1 point runs in less time than your implementation of brute force for 1 point, they still have the same O(n). 2) If you just try a simple foreach loop and record the lowest distance squared, you'll get even better absolute time than your divide and conquer - and, it will still be O(n).

public static float LengthSquared(PointF P1, PointF P2)
{
return (P1.X - P2.X) * (P1.X - P2.X)
+ (P1.Y - P2.Y) * (P1.Y - P2.Y);
}
If, as your question states, you want to compare 1 (known) point to a list of points to find the closest then use this code.
public static Segment Closest_BruteForce(PointF P1, List<PointF> points)
{
PointF closest = null;
float minDist = float.MaxValue;
foreach(PointF P2 in points)
{
if(P1 != P2)
{
float temp = LengthSquared(P1, P2);
if(temp < minDist)
{
minDist = temp;
closest = P2;
}
}
}
return new Segment(P1, closest);
}
However, if as your example shows, you want to find the closest 2 points from a list of points try the below.
public static Segment Closest_BruteForce(List<PointF> points)
{
PointF closest1;
PointF closest2;
float minDist = float.MaxValue;
for(int x=0; x<points.Count; x++)
{
PointF P1 = points[x];
for(int y = x + 1; y<points.Count; y++)
{
PointF P2 = points[y];
float temp = LengthSquared(P1, P2);
if(temp < minDist)
{
minDist = temp;
closest1 = P1;
closest2 = P2;
}
}
}
return new Segment(closest1, closest2);
}
note the code above was written in the browser and may have some syntax errors.
EDIT Odd... is this an acceptable answer or not? Down-votes without explanation, oh well.