Problem Description
Given two line segments, implement an algorithm to determine whether these two segments intersect.
Segment A is given by points A1(x1, y1) and A2(x2, y2), while segment B is given by points B1(x3, y3) and B2(x4, y4).
The intersection status is determined considering cases where the segments intersect, the endpoints of the segments are located inside the other segment, or the segments lie on a straight line.
Example Input
A1: (1, 1)
A2: (4, 4)
B1: (1, 4)
B2: (4, 1)
Example Output
Intersect
Problem Solving Process
To solve this problem, we will utilize several geometric concepts and mathematical calculations.
To determine whether the segments intersect, we first identify the direction of both segments and then use that to decide the intersection status.
Step 1: Direction Calculation
Using the endpoints of the given two segments, calculate the direction in which each segment lies.
When we have two segments A(A1, A2) and B(B1, B2), we can calculate the direction as follows:
def direction(a1, a2, b1, b2):
# Function to determine direction
diff = (a2[0] - a1[0]) * (b1[1] - a1[1]) - (a2[1] - a1[1]) * (b1[0] - a1[0])
if diff == 0:
return 0 # Same direction
return 1 if diff > 0 else -1 # Left (+) or Right (-)
Step 2: Intersection Conditions
Whether the two segments intersect can be determined by the following conditions:
- The direction of segment A does not match at points B1 and B2, and the direction of segment B does not match at points A1 and A2, then the two segments intersect.
- If the points of segment A exist on the extension of segment B, meaning one of the endpoints of the line is included inside the other line, it is also considered an intersection.
- If all points lie on the same straight line and the endpoints overlap, that is considered an intersection.
Step 3: Implementation
Now, let’s implement this condition in Java.
public class LineIntersection {
static class Point {
int x, y;
Point(int x, int y) { this.x = x; this.y = y; }
}
static int direction(Point a1, Point a2, Point b) {
int diff = (a2.x - a1.x) * (b.y - a1.y) - (a2.y - a1.y) * (b.x - a1.x);
return (diff == 0) ? 0 : (diff > 0) ? 1 : -1;
}
static boolean isIntersect(Point a1, Point a2, Point b1, Point b2) {
int d1 = direction(a1, a2, b1);
int d2 = direction(a1, a2, b2);
int d3 = direction(b1, b2, a1);
int d4 = direction(b1, b2, a2);
// Cross case
if(d1 != d2 && d3 != d4) return true;
// Collinear case
if(d1 == 0 && onSegment(a1, a2, b1)) return true;
if(d2 == 0 && onSegment(a1, a2, b2)) return true;
if(d3 == 0 && onSegment(b1, b2, a1)) return true;
if(d4 == 0 && onSegment(b1, b2, a2)) return true;
return false;
}
static boolean onSegment(Point a, Point b, Point p) {
return (p.x <= Math.max(a.x, b.x) && p.x >= Math.min(a.x, b.x) &&
p.y <= Math.max(a.y, b.y) && p.y >= Math.min(a.y, b.y));
}
public static void main(String[] args) {
Point A1 = new Point(1, 1);
Point A2 = new Point(4, 4);
Point B1 = new Point(1, 4);
Point B2 = new Point(4, 1);
if (isIntersect(A1, A2, B1, B2)) {
System.out.println("Intersect");
} else {
System.out.println("Do not intersect");
}
}
}
Running the code above will allow you to determine whether the two segments intersect.
Step 4: Testing
Try various test cases to check whether the code functions correctly.
// Test Case 1
A1: (1, 1)
A2: (4, 4)
B1: (1, 4)
B2: (4, 1) // Output: Intersect
// Test Case 2
A1: (1, 1)
A2: (1, 3)
B1: (1, 2)
B2: (1, 4) // Output: Intersect
// Test Case 3
A1: (1, 1)
A2: (2, 2)
B1: (3, 3)
B2: (4, 4) // Output: Do not intersect
This way, you can verify whether the algorithm works correctly through various cases.
Conclusion
Through this course, we learned how to implement an algorithm for determining the intersection of line segments.
It was a process of understanding and applying intersection conditions through geometric thinking and code implementation.
This knowledge can be used in various applications, and it will greatly assist in improving understanding of algorithms and data structures.
I hope this course helps all of you to solve various problems and further solidify your fundamentals.