基于顶点与半径向量的平行四边形内点判定及简化方法问询

Alright, let's tackle this problem clearly—since you're working with a parallelogram defined by a single vertex and two direction vectors (your "radius vectors"), there's a streamlined way to check if a point lies inside it, leveraging cross products (and yes, we can keep it tied to the single cross product used for the parallelogram's area).

First, Define Our Parallelogram #

Let's formalize the setup to avoid confusion:

• Let P₀ be the starting vertex of the parallelogram.
• Let v₁ and v₂ be the two "radius vectors" (the edges extending from P₀ to the adjacent vertices).
• The full parallelogram has vertices: P₀, P₀ + v₁, P₀ + v₂, P₀ + v₁ + v₂.
• We want to check if a point Q lies inside (or on the boundary of) this shape.

Simplified Cross-Product Method (Tied to Area) #

This method avoids complex area summations and leans on the core property of cross products: their magnitude equals the area of the parallelogram formed by two vectors, and their sign tells us relative orientation.

Here's the step-by-step process:

1. Translate to local coordinates: Compute the vector from P₀ to Q:

   u = Q - P₀
   

   This shifts our problem to a coordinate system where P₀ is the origin, making calculations easier.

2. Compute key cross products:

   • First, calculate the area-related cross product for the entire parallelogram:

     cross_v1_v2 = v₁.x * v₂.y - v₁.y * v₂.x  # 2D cross product; extends to 3D with standard cross product
     

     The absolute value of this is exactly the area of the parallelogram, as you noted.
   • Next, compute two cross products involving u:

     cross_u_v2 = u.x * v₂.y - u.y * v₂.x
     cross_v1_u = v₁.x * u.y - v₁.y * u.x
     

3. Check the orientation and bounds:

   • If cross_v1_v2 > 0 (v₁ and v₂ form a counter-clockwise orientation):
     The point Q is inside (or on the boundary) if both are true:
     0 ≤ cross_u_v2 ≤ cross_v1_v2 and 0 ≤ cross_v1_u ≤ cross_v1_v2
   • If cross_v1_v2 < 0 (clockwise orientation):
     The point Q is inside (or on the boundary) if both are true:
     cross_v1_v2 ≤ cross_u_v2 ≤ 0 and cross_v1_v2 ≤ cross_v1_u ≤ 0
   • If cross_v1_v2 = 0:
     The "parallelogram" is actually a line segment or a single point—just check if Q lies on that line/point.

Why This Works #

This method boils down to checking if u (the vector from P₀ to Q) can be written as a linear combination u = s*v₁ + t*v₂ where s ∈ [0,1] and t ∈ [0,1]. The cross products let us calculate s and t without solving linear equations:

• s = cross_u_v2 / cross_v1_v2
• t = cross_v1_u / cross_v1_v2

By checking the bounds of the cross products (instead of computing the division), we avoid floating-point precision issues and keep the calculation fast.

Area-Based Verification (If You Want to Confirm) #

If you want to tie this explicitly to area sums (as you mentioned in your question), here's how it connects:
The total area of the parallelogram is |cross_v1_v2|. For Q to be inside, the sum of the areas of the four triangles formed by Q and each edge of the parallelogram should equal the total area. But this is redundant—our cross product method already encodes this check in a much more efficient way, so there's no need to compute all those triangle areas.

Edge Case Notes #

• If you want to exclude points on the parallelogram's boundary, just replace the ≤ signs with < in the checks.
• This works for 2D and 3D parallelograms (in 3D, use the standard 3D cross product, and check that the resulting vectors have the same direction and magnitude bounds).

内容的提问来源于stack exchange，提问作者Tomilov Anatoliy