如何计算矩阵元素相对位置？如何判断点k在i-j线段右侧？

Alright, let's break down your two questions clearly—they’re closely linked, so we’ll work through them in order.

First, let’s clarify what matrix A represents: each entry A_ij counts how many distinct points k (where k ≠ i and k ≠ j) lie strictly to the right of the line segment connecting point i to point j.

To compute this, you’ll need the coordinates of all your points (let’s assume each point has an (x, y) pair). For every pair of points i(x₁, y₁) and j(x₂, y₂), follow these steps for each other point k(x₃, y₃):

• Calculate the cross product of vectors ij and ik:

  cross = (x₂ - x₁) * (y₃ - y₁) - (y₂ - y₁) * (x₃ - x₁)
  

• If cross < 0, point k is strictly to the right of segment i-j (note: this assumes a standard Cartesian coordinate system where the y-axis points upward; if your system uses a downward-pointing y-axis, the sign flips to cross > 0).
• Count all such k points—this count is exactly the value of A_ij.

Example from Your Question #

You mentioned 3 points (0, 1, 2) where A_02 = 1. Let’s use coordinates that fit this scenario:

• Point 0: (0, 0)
• Point 1: (1, -1)
• Point 2: (2, 0)

Calculating the cross product for segment 0→2 and point 1:

cross = (2-0)*(-1-0) - (0-0)*(1-0) = 2*(-1) - 0 = -2 < 0

This confirms point 1 is on the right of segment 0→2, so A_02 = 1—which matches your example.

This is a direct, individual check using the same cross product logic we used above. For points i(x₁, y₁), j(x₂, y₂), k(x₃, y₃):

• Compute the cross product as shown earlier: cross = (x₂ - x₁)*(y₃ - y₁) - (y₂ - y₁)*(x₃ - x₁)
• Interpret the result:
  • If cross < 0: k is strictly to the right of segment i-j
  • If cross > 0: k is strictly to the left
  • If cross = 0: k is collinear with segment i-j (not on either side)

You don’t actually need matrix A to perform this check—A_ij just aggregates the count of such k points for each i-j pair, but verifying a specific k only requires the cross product calculation.

内容的提问来源于stack exchange，提问作者oldselflearner1959