The number of horizontal or vertical line segments required to connect 3 points

Suppose that given three different points (or coordinates), you want to find the number of horizontal or vertical line segments that can be formed by connecting these three points. Such line segments are also called polylines. In order to solve this problem, you need the concept of computational geometry. In this article, we will discuss various ways to solve this problem in C.

Input and output scenarios

Assume that c1, c2 and c3 are the coordinates of 3 points on the Cartesian plane. The number of horizontal or vertical line segments connecting these 3 points will be as shown below.

Input: c1 = (-1, -1), c2 = (-2, 3), c3 = (4, 3) Output: 1 Input: c1 = (1, -1), c2 = (1, 3), c3 = (4, 3) Output: 2 Input: c1 = (1, 1), c2 = (2, 6), c3 = (5, 2) Output: 3

Note− Horizontal and vertical line segments must be aligned with the coordinate axes.

Use If statement

We can use if statement to check if there is a horizontal or vertical line between these three points.

• Create a function by combiningc1.xwithc2.x, c1.xwithc3.xandc2.xandc3.x. If either condition is met, it means there is a horizontal line segment and the count is incremented.

• Similarly, this function combinesc1.ywithc2.y, c1.ywithc3.yandc2 .yandc3.y. If any of the conditions is met, the vertical line segment does exist. The count increases again.

Example

#include  using namespace std; struct Coordinate { int x; int y; }; int countLineSegments(Coordinate c1, Coordinate c2, Coordinate c3) { int count = 0; // Check for the horizontal segment if (c1.x == c2.x || c1.x == c3.x || c2.x == c3.x) count++; // Check for the vertical segment if (c1.y == c2.y || c1.y == c3.y || c2.y == c3.y) count++; return count; } int main() { Coordinate c1, c2, c3; c1.x = -1; c1.y = -5; c2.x = -2; c2.y = 3; c3.x = 4; c3.y = 3; int numSegments = countLineSegments(c1, c2, c3); std::cout << "Number of horizontal or vertical line segments: " << numSegments << std::endl; return 0; }

Output

Number of horizontal or vertical line segments: 1

Note−If all three points are on the same axis of the Cartesian plane, that is, the X axis or the Y axis, the number of line segments required to connect them is 1. If the points form an L shape, the result is 2, otherwise the result is 3.

Use auxiliary functions

• Here, we can use auxiliary functions (horizontalLineandverticalLine) to calculate line segments.

• We exploit the fact that in the Cartesian system all points of a horizontal line lie on the same y-coordinate.horizontalLineThe function checks whether two points can form a horizontal line segment by comparing their y coordinates. If the y-coordinates are the same, there is a horizontal line.

• Similarly, all points of a vertical line lie at the same x-coordinate.verticalLineThe function checks whether two points can form a vertical line segment by comparing their x-coordinates. If the x-coordinates are the same, there is a vertical line.

• Next, we have thecountLineSegmentsfunction, which is used to count the number of horizontal and vertical line segments. If there are horizontal or vertical line segments, the count is incremented after each iteration.

Example

#include  using namespace std; struct Coordinate { int x; int y; }; // Helper functions bool horizontalLine(Coordinate c1, Coordinate c2) { return c1.y == c2.y; } bool verticalLine(Coordinate c1, Coordinate c2) { return c1.x == c2.x; } int countLineSegments(Coordinate c1, Coordinate c2, Coordinate c3) { int count = 0; // Check for horizontal segment if (horizontalLine(c1, c2) || horizontalLine(c1, c3) || horizontalLine(c2, c3)) count++; // Check for vertical segment if (verticalLine(c1, c2) || verticalLine(c1, c3) || verticalLine(c2, c3)) count++; return count; } int main() { Coordinate c1, c2, c3; c1.x = -1; c1.y = -5; c2.x = -2; c2.y = 3; c3.x = 4; c3.y = 3; int numSegments = countLineSegments(c1, c2, c3); std::cout << "Number of horizontal or vertical line segments: " << numSegments << std::endl; return 0; }

Output

Number of horizontal or vertical line segments: 1

in conclusion

In this article, we explore various methods using C to find the number of horizontal and vertical lines that can connect 3 different points in the Cartesian plane. We have discussed theifstatement approach to solving this problem. However, due to the large number of iterations, the time complexity also increases. We can effectively solve this problem by usingauxiliary functions, which reduces the number of iterations and thereby reduces the time complexity.

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