1. Give pseudo-code to determine if a point t lies inside a triangle with vertic
ID: 3665927 • Letter: 1
Question
1. Give pseudo-code to determine if a point t lies inside a triangle with vertices p, q, and r (in clockwise order, say). A value of true should be returned if t lies on the boundary or inside the triangle, and false otherwise. See Figure 1.
2. Assume that the vertices p1, p2, . . . , pn of an n-sided convex polygon P are given in clockwise order. Give an algorithm that tests if a query point t lies inside the given convex polygon (i.e., it should return true if t lies inside the polygon or on the boundary, and false otherwise). Your algorithm should run in O(log n) time. Hint: The pseudo-code from part 1 might come in useful here.
Please give mathematically precise answer. Give an algorithm for a problem, presenting the following: the algorithm, a justification of its correctness, and a derivation of its running time.
Explanation / Answer
#include <iostream>
using namespace std;
// Define Infinite (Using INT_MAX caused overflow problems)
#define INF 10000
struct Point
{
int x;
int y;
};
// Given three colinear points p, q, r, the function checks if
// point q lies on line segment 'pr'
bool onSegment(Point p, Point q, Point r)
{
if (q.x <= max(p.x, r.x) && q.x >= min(p.x, r.x) &&
q.y <= max(p.y, r.y) && q.y >= min(p.y, r.y))
return true;
return false;
}
// To find orientation of ordered triplet (p, q, r).
// The function returns following values
// 0 --> p, q and r are colinear
// 1 --> Clockwise
// 2 --> Counterclockwise
int orientation(Point p, Point q, Point r)
{
int val = (q.y - p.y) * (r.x - q.x) -
(q.x - p.x) * (r.y - q.y);
if (val == 0) return 0; // colinear
return (val > 0)? 1: 2; // clock or counterclock wise
}
// The function that returns true if line segment 'p1q1'
// and 'p2q2' intersect.
bool doIntersect(Point p1, Point q1, Point p2, Point q2)
{
// Find the four orientations needed for general and
// special cases
int o1 = orientation(p1, q1, p2);
int o2 = orientation(p1, q1, q2);
int o3 = orientation(p2, q2, p1);
int o4 = orientation(p2, q2, q1);
// General case
if (o1 != o2 && o3 != o4)
return true;
// Special Cases
// p1, q1 and p2 are colinear and p2 lies on segment p1q1
if (o1 == 0 && onSegment(p1, p2, q1)) return true;
// p1, q1 and p2 are colinear and q2 lies on segment p1q1
if (o2 == 0 && onSegment(p1, q2, q1)) return true;
// p2, q2 and p1 are colinear and p1 lies on segment p2q2
if (o3 == 0 && onSegment(p2, p1, q2)) return true;
// p2, q2 and q1 are colinear and q1 lies on segment p2q2
if (o4 == 0 && onSegment(p2, q1, q2)) return true;
return false; // Doesn't fall in any of the above cases
}
// Returns true if the point p lies inside the polygon[] with n vertices
bool isInside(Point polygon[], int n, Point p)
{
// There must be at least 3 vertices in polygon[]
if (n < 3) return false;
// Create a point for line segment from p to infinite
Point extreme = {INF, p.y};
// Count intersections of the above line with sides of polygon
int count = 0, i = 0;
do
{
int next = (i+1)%n;
// Check if the line segment from 'p' to 'extreme' intersects
// with the line segment from 'polygon[i]' to 'polygon[next]'
if (doIntersect(polygon[i], polygon[next], p, extreme))
{
// If the point 'p' is colinear with line segment 'i-next',
// then check if it lies on segment. If it lies, return true,
// otherwise false
if (orientation(polygon[i], p, polygon[next]) == 0)
return onSegment(polygon[i], p, polygon[next]);
count++;
}
i = next;
} while (i != 0);
// Return true if count is odd, false otherwise
return count&1; // Same as (count%2 == 1)
}
// Driver program to test above functions
int main()
{
Point polygon1[] = {{0, 0}, {10, 0}, {10, 10}, {0, 10}};
int n = sizeof(polygon1)/sizeof(polygon1[0]);
Point p = {20, 20};
isInside(polygon1, n, p)? cout << "Yes ": cout << "No ";
p = {5, 5};
isInside(polygon1, n, p)? cout << "Yes ": cout << "No ";
Point polygon2[] = {{0, 0}, {5, 5}, {5, 0}};
p = {3, 3};
n = sizeof(polygon2)/sizeof(polygon2[0]);
isInside(polygon2, n, p)? cout << "Yes ": cout << "No ";
p = {5, 1};
isInside(polygon2, n, p)? cout << "Yes ": cout << "No ";
p = {8, 1};
isInside(polygon2, n, p)? cout << "Yes ": cout << "No ";
Point polygon3[] = {{0, 0}, {10, 0}, {10, 10}, {0, 10}};
p = {-1,10};
n = sizeof(polygon3)/sizeof(polygon3[0]);
isInside(polygon3, n, p)? cout << "Yes ": cout << "No ";
return 0;
}
pseducode for triangle
where
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