Given N points in 2-dimensional space, we need to find three points such that triangle made by choosing these points should not contain any other points inside. All given points will not lie on the same line so solution will always exist.
Examples:
In above diagram possible triangle with no point
inside can be formed by choosing these triplets,
[(0, 0), (2, 0), (1, 1)]
[(0, 0), (1, 1), (0, 2)]
[(1, 1), (2, 0), (2, 2)]
[(1, 1), (0, 2), (2, 2)]
So any of the above triplets can be the final answer.
The solution is based on the fact that if there exist triangle(s) with no points inside, then we can form a triangle with any random point among all points.
We can solve this problem by searching all three points one by one. The first point can be chosen randomly. After choosing the first point, we need two points such that their slope should be different and no point should lie inside the triangle of these three points. We can do this by choosing the second point as nearest point to the first and third point as second nearest point with the different slope. For doing this, we first iterate over all points and choose the point which is nearest to the first one and designate that as second point of the required triangle. Then we iterate one more time to find point which has different slope and has the smallest distance, which will be the third point of our triangle.
C++
// C/C++ program to find triangle with no point inside #include <bits/stdc++.h> using namespace std; // method to get square of distance between // (x1, y1) and (x2, y2) int getDistance(int x1, int y1, int x2, int y2) { return (x2 - x1)*(x2 - x1) + (y2 - y1)*(y2 - y1); } // Method prints points which make triangle with no // point inside void triangleWithNoPointInside(int points[][2], int N) { // any point can be chosen as first point of triangle int first = 0; int second, third; int minD = INT_MAX; // choose nearest point as second point of triangle for (int i = 0; i < N; i++) { if (i == first) continue; // Get distance from first point and choose // nearest one int d = getDistance(points[i][0], points[i][1], points[first][0], points[first][1]); if (minD > d) { minD = d; second = i; } } // Pick third point by finding the second closest // point with different slope. minD = INT_MAX; for (int i = 0; i < N; i++) { // if already chosen point then skip them if (i == first || i == second) continue; // get distance from first point int d = getDistance(points[i][0], points[i][1], points[first][0], points[first][1]); /* the slope of the third point with the first point should not be equal to the slope of second point with first point (otherwise they'll be collinear) and among all such points, we choose point with the smallest distance */ // here cross multiplication is compared instead // of division comparison if ((points[i][0] - points[first][0]) * (points[second][1] - points[first][1]) != (points[second][0] - points[first][0]) * (points[i][1] - points[first][1]) && minD > d) { minD = d; third = i; } } cout << points[first][0] << ", " << points[first][1] << endl; cout << points[second][0] << ", " << points[second][1] << endl; cout << points[third][0] << ", " << points[third][1] << endl; } // Driver code to test above methods int main() { int points[][2] = {{0, 0}, {0, 2}, {2, 0}, {2, 2}, {1, 1}}; int N = sizeof(points) / sizeof(points[0]); triangleWithNoPointInside(points, N); return 0; } |
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Java
// Java program to find triangle // with no point inside import java.io.*; class GFG { // method to get square of distance between // (x1, y1) and (x2, y2) static int getDistance(int x1, int y1, int x2, int y2) { return (x2 - x1)*(x2 - x1) + (y2 - y1)*(y2 - y1); } // Method prints points which make triangle with no // point inside static void triangleWithNoPointInside(int points[][], int N) { // any point can be chosen as first point of triangle int first = 0; int second =0; int third =0; int minD = Integer.MAX_VALUE; // choose nearest point as second point of triangle for (int i = 0; i < N; i++) { if (i == first) continue; // Get distance from first point and choose // nearest one int d = getDistance(points[i][0], points[i][1], points[first][0], points[first][1]); if (minD > d) { minD = d; second = i; } } // Pick third point by finding the second closest // point with different slope. minD = Integer.MAX_VALUE; for (int i = 0; i < N; i++) { // if already chosen point then skip them if (i == first || i == second) continue; // get distance from first point int d = getDistance(points[i][0], points[i][1], points[first][0], points[first][1]); /* the slope of the third point with the first point should not be equal to the slope of second point with first point (otherwise they'll be collinear) and among all such points, we choose point with the smallest distance */ // here cross multiplication is compared instead // of division comparison if ((points[i][0] - points[first][0]) * (points[second][1] - points[first][1]) != (points[second][0] - points[first][0]) * (points[i][1] - points[first][1]) && minD > d) { minD = d; third = i; } } System.out.println(points[first][0] + ", " + points[first][1]); System.out.println(points[second][0]+ ", " + points[second][1]) ; System.out.println(points[third][0] +", " + points[third][1]); } // Driver code public static void main (String[] args) { int points[][] = {{0, 0}, {0, 2}, {2, 0}, {2, 2}, {1, 1}}; int N = points.length; triangleWithNoPointInside(points, N); } } // This article is contributed by vt_m. |
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Python 3
# Python3 program to find triangle # with no point inside import sys # method to get square of distance between # (x1, y1) and (x2, y2) def getDistance(x1, y1, x2, y2): return (x2 - x1) * (x2 - x1) + \ (y2 - y1) * (y2 - y1) # Method prints points which make triangle # with no point inside def triangleWithNoPointInside(points, N): # any point can be chosen as # first point of triangle first = 0 second = 0 third = 0 minD = sys.maxsize # choose nearest point as # second point of triangle for i in range(0, N): if i == first: continue # Get distance from first point and choose # nearest one d = getDistance(points[i][0], points[i][1], points[first][0], points[first][1]) if minD > d: minD = d second = i # Pick third point by finding the second closest # point with different slope. minD = sys.maxsize for i in range (0, N): # if already chosen point then skip them if i == first or i == second: continue # get distance from first point d = getDistance(points[i][0], points[i][1], points[first][0], points[first][1]) """ the slope of the third point with the first point should not be equal to the slope of second point with first point (otherwise they'll be collinear) and among all such points, we choose point with the smallest distance """ # here cross multiplication is compared instead # of division comparison if ((points[i][0] - points[first][0]) * (points[second][1] - points[first][1]) != (points[second][0] - points[first][0]) * (points[i][1] - points[first][1]) and minD > d) : minD = d third = i print(points[first][0], ', ', points[first][1]) print(points[second][0], ', ', points[second][1]) print(points[third][0], ', ', points[third][1]) # Driver code points = [[0, 0], [0, 2], [2, 0], [2, 2], [1, 1]] N = len(points) triangleWithNoPointInside(points, N) # This code is contributed by Gowtham Yuvaraj |
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C#
using System; // C# program to find triangle // with no point inside public class GFG { // method to get square of distance between // (x1, y1) and (x2, y2) public static int getDistance(int x1, int y1, int x2, int y2) { return (x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1); } // Method prints points which make triangle with no // point inside public static void triangleWithNoPointInside(int[][] points, int N) { // any point can be chosen as first point of triangle int first = 0; int second = 0; int third = 0; int minD = int.MaxValue; // choose nearest point as second point of triangle for (int i = 0; i < N; i++) { if (i == first) { continue; } // Get distance from first point and choose // nearest one int d = getDistance(points[i][0], points[i][1], points[first][0], points[first][1]); if (minD > d) { minD = d; second = i; } } // Pick third point by finding the second closest // point with different slope. minD = int.MaxValue; for (int i = 0; i < N; i++) { // if already chosen point then skip them if (i == first || i == second) { continue; } // get distance from first point int d = getDistance(points[i][0], points[i][1], points[first][0], points[first][1]); /* the slope of the third point with the first point should not be equal to the slope of second point with first point (otherwise they'll be collinear) and among all such points, we choose point with the smallest distance */ // here cross multiplication is compared instead // of division comparison if ((points[i][0] - points[first][0]) * (points[second][1] - points[first][1]) != (points[second][0] - points[first][0]) * (points[i][1] - points[first][1]) && minD > d) { minD = d; third = i; } } Console.WriteLine(points[first][0] + ", " + points[first][1]); Console.WriteLine(points[second][0] + ", " + points[second][1]); Console.WriteLine(points[third][0] + ", " + points[third][1]); } // Driver code public static void Main(string[] args) { int[][] points = new int[][] { new int[] {0, 0}, new int[] {0, 2}, new int[] {2, 0}, new int[] {2, 2}, new int[] {1, 1} }; int N = points.Length; triangleWithNoPointInside(points, N); } } // This code is contributed by Shrikant13 |
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Output:
0, 0 1, 1 0, 2
This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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