Leetcode: Convex Polygon

Posted 2020-08-24 neverlandly

Given a list of points that form a polygon when joined sequentially, find if this polygon is convex (Convex polygon definition).

Note:

There are at least 3 and at most 10,000 points.
Coordinates are in the range -10,000 to 10,000.
You may assume the polygon formed by given points is always a simple polygon (Simple polygon definition). In other words, we ensure that exactly two edges intersect at each vertex, and that edges otherwise don‘t intersect each other.
Example 1:

[[0,0],[0,1],[1,1],[1,0]]

Answer: True

Explanation:

Example 2: [[0,0],[0,10],[10,10],[10,0],[5,5]] 

Answer: False 

Explanation:

https://discuss.leetcode.com/topic/70706/beyond-my-knowledge-java-solution-with-in-line-explanation

https://discuss.leetcode.com/topic/70664/c-7-line-o-n-solution-to-check-convexity-with-cross-product-of-adajcent-vectors-detailed-explanation

The key observation for convexity is that vector pi+1-pi always turns to the same direction to pi+2-pi formed by any 3 sequentially adjacent vertices, i.e., cross product (pi+1-pi) x (pi+2-pi) does not change sign when traversing sequentially along polygon vertices.

Note that for any 2D vectors v1, v2,

• v1 x v2 = det([v1, v2])

which is the determinant of 2x2 matrix [v1, v2]. And the sign of det([v1, v2]) represents the positive z-direction of right-hand system from v1 to v2. So det([v1, v2]) ≥ 0 if and only if v1 turns at most 180 degrees counterclockwise to v2.

 1 public class Solution {
 2     public boolean isConvex(List<List<Integer>> points) {
 3         // For each set of three adjacent points A, B, C, find the cross product AB · BC. If the sign of
 4         // all the cross products is the same, the angles are all positive or negative (depending on the
 5         // order in which we visit them) so the polygon is convex.
 6         boolean gotNegative = false;
 7         boolean gotPositive = false;
 8         int numPoints = points.size();
 9         int B, C;
10         for (int A = 0; A < numPoints; A++) {
11             // Trick to calc the last 3 points: n - 1, 0 and 1.
12             B = (A + 1) % numPoints;
13             C = (B + 1) % numPoints;
14     
15             int crossProduct =
16                 crossProductLength(
17                     points.get(A).get(0), points.get(A).get(1),
18                     points.get(B).get(0), points.get(B).get(1),
19                     points.get(C).get(0), points.get(C).get(1));
20             if (crossProduct < 0) {
21                 gotNegative = true;
22             }
23             else if (crossProduct > 0) {
24                 gotPositive = true;
25             }
26             if (gotNegative && gotPositive) return false;
27         }
28     
29         // If we got this far, the polygon is convex.
30         return true;
31     }
32     
33     // Return the cross product AB x BC.
34     // The cross product is a vector perpendicular to AB and BC having length |AB| * |BC| * Sin(theta) and
35     // with direction given by the right-hand rule. For two vectors in the X-Y plane, the result is a
36     // vector with X and Y components 0 so the Z component gives the vector‘s length and direction.
37     private int crossProductLength(int Ax, int Ay, int Bx, int By, int Cx, int Cy)
38     {
39         // Get the vectors‘ coordinates.
40         int ABx = Bx - Ax;
41         int ABy = By - Ay;
42         int BCx = Cx - Bx;
43         int BCy = Cy - By;
44     
45         // Calculate the Z coordinate of the cross product.
46         return (ABx * BCy - ABy * BCx);
47     }
48 }