凸多边形间固定斜率最长约束线段及扩展问题算法咨询

Alright, let's break this down step by step—this is a classic convex geometry problem with some neat extensions, so I'll walk through each part clearly, leaning on the properties of convex shapes that make these problems solvable efficiently.

Given convex polygon A containing convex polygon B entirely, we need the longest segment with a fixed slope, with one endpoint on B's boundary and the other on A's boundary.

Solution Approach #

Convex polygons play nice with projections and coordinate transformations, so here's the straightforward method:

1. Rotate the Coordinate System
   Convert the problem to a horizontal line problem by rotating all vertices of A and B so that the target slope aligns with the x-axis. The rotation matrix for an angle θ (where θ = arctan(target_slope)) is:

   [[cosθ, sinθ], [-sinθ, cosθ]]
   

   Rotation preserves distances, so we don't have to worry about distorting segment lengths.
2. Compute Axis Projections
   For the rotated polygons, calculate their projection intervals on the x-axis:
   • A's projection: [A_min_x, A_max_x] (the leftmost and rightmost x-values of A's vertices/edges)
   • B's projection: [B_min_x, B_max_x]
3. Find the Longest Valid Segment
   Since both shapes are convex, the longest fixed-slope segment will always connect an extreme point of B to an extreme point of A in the rotated direction:
   • From B's leftmost boundary point to A's rightmost boundary point
   • From B's rightmost boundary point to A's leftmost boundary point
     Calculate both segment lengths and take the larger one—this is your answer.

Why This Works #

Convex polygons have the property that their projections onto any axis are intervals whose endpoints map directly to boundary points (vertices or edge endpoints). The longest fixed-slope segment must span the maximum possible distance between these extreme projections, as any intermediate point would result in a shorter segment.

We now have two segments with distinct fixed slopes, sharing a common endpoint P (which lies inside or on B's boundary), with the other endpoints on A's boundary. We need to maximize the sum of their lengths.

Solution Approach #

This relies on the behavior of distance functions over convex sets:

1. Distance Functions Are Concave
   For any fixed slope (direction), the maximum distance from a point P to A's boundary along that direction is a concave function of P. The sum of two concave functions is also concave, and concave functions attain their maximum on a convex set (like B) at the set's vertices.
2. Check All Vertices of B
   For each vertex Pᵢ of B:
   • Draw a line through Pᵢ with the first slope, find its intersection Q₁ with A's boundary, compute the length |PᵢQ₁|
   • Draw a line through Pᵢ with the second slope, find its intersection Q₂ with A's boundary, compute the length |PᵢQ₂|
   • Calculate the sum Sᵢ = |PᵢQ₁| + |PᵢQ₂|
3. Take the Maximum Sum
   The largest Sᵢ across all B's vertices is your optimal solution. You don't need to check points on B's edges—since the sum function is concave, edge points can't yield a larger sum than the vertices.

Absolutely—this problem generalizes cleanly to higher dimensions, thanks to the same convexity properties that make the 2D case manageable.

Core Problem in High Dimensions #

Instead of a fixed slope (2D direction), we have a fixed direction vector v. We need the longest segment from B's boundary to A's boundary, where the segment aligns with v.

• Use Support Functions
  For a convex set, the support function in direction v gives the maximum dot product of any point in the set with v (this corresponds to the "extreme" point in direction v). Compute:
  • h_A(v): Support function of A in direction v
  • h_B(-v): Support function of B in direction -v (the extreme point opposite to v)
  • h_A(-v): Support function of A in direction -v
  • h_B(v): Support function of B in direction v
    The longest segment length is the maximum of:

  (h_A(**v**) - h_B(-**v**)) / ||**v**|| , (h_A(-**v**) - h_B(**v**)) / ||**v**||
  

  The division by ||v|| converts the projection difference to an actual segment length.

Extension 1 in High Dimensions #

For two fixed direction vectors v₁ and v₂, sharing an endpoint P ∈ B, maximize the sum of segment lengths from P to A's boundary along each direction.

• Just like in 2D, the distance functions for each direction are concave, so their sum is concave. The maximum will be at one of B's vertices.
• For each vertex Pᵢ of B, compute the maximum distance to A's boundary along v₁ and v₂, sum them, and take the largest total.

内容的提问来源于stack exchange，提问作者Sharan