如果两个线段重叠或相交，则将它们合并在同一圆上 [英] Combine two segments on the same circle if they overlap or intersect

问题描述

如果两个部分重叠或相交，我会尝试合并.我的问题类似于此一个>.但是，我要结合两个部分.

I am try to combine two segments if they overlap or intersect.My question is similar to this and this. However, what I want is to combine two segments.

public class Segment
{
    private readonly double _from;
    private readonly double _to;
    public Segment(double from, double to)
    {
        _from = Normalize(from); // 0 <= x < 360
        _to = Normalize(to);
    }

    public bool Inside(double target)
    {
        if (_from < _to)
            return _from <= target && target <= _to;
        return _from <= target || target <= _to;
    }
}

我正在尝试写一个TryCombine().

public bool TryCombine(Segment other, out Segment result)
{
    // if not intersect
    // return false

    // else if overlap
    // return true, out larger one

    // else if intersect
    // return true, out a new one with merged bound
}

预期结果

var seg1 = new Segment(0, 100);
var seg2 = new Segment(50, 150);
var seg3 = new Segment(110, -100);
var seg4 = new Segment(10, 50);
Segment result;

seg1.TryCombine(seg2, result); // True, result = Segment(0, 150)
seg1.TryCombine(seg3, result); // False
seg1.TryCombine(seg4, result); // True, result = Segment(0, 100)
seg2.TryCombine(seg3, result); // True, result = Segment(260, 150)

推荐答案

您可以在第二个链接中使用我的答案中所述的方法.

You can use approach described in my answer in you second link.

ma = (a2 + a1)/ 2  
mb = (b2 + b1)/ 2  
cda = Cos(da)
cdb = Cos(db)

要检查交叉点是否存在以及发生哪种类型的交叉点，请找到4个布尔值

To check whether intersection exists and what kind o intersection takes place, find 4 boolean values

 BStartInsideA = (Cos(ma - b1) >= cda)
 BEndInsideA  =  (Cos(ma - b2) >= cda)
 AStartInsideB = (Cos(mb - a1) >= cdb)
 AEndInsideB =   (Cos(mb - a2) >= cdb)

这些组合可能形成16种可能的结果(并非所有结果都是可靠的).我将这些结果组合为4位值的位，并在case语句中对其进行处理.

These combinations could form 16 possible results (not all are reliable). I'd combine these results as bits of 4-bit value and treat them in the case statement.

例如，如果第一个和最后一个值均为true(值0b1001 = 9)，则您具有类似于seg1-seg2情况的简单交集-因此获得AStart ans起点，BEnd作为终点并将其标准化(添加360如果小于AStart，则返回BEnd).

For example, if the first and the last values are true (value 0b1001 = 9), you have simple intersection like your seg1-seg2 case - so get AStart ans starting point, BEnd as ending point and normalize them (add 360 to BEnd if it is less than AStart).

预规范化步骤应提供BEnd> = BStart和AEnd> = AStart(例如，将(3,1)弧转换为具有中点182和半角179的(3，361)

Pre-normalization step should provide BEnd>=BStart and AEnd>=AStart (for example, transform (3,1) arc into (3, 361) with middle point 182 and semi-angle 179)

可能的结果(两个额外的案例，4个简单的最终组合，4个单端一致的案例):

Possible results (two extra cases, 4 simple end combinations, 4 one-end coinciding cases):

 0000: no intersection
 1111: full circle

 0011: AStart-AEnd
 1001: AStart-BEnd
 0110: BStart-AEnd
 1100: BStart-BEnd

 0111: AStart-AEnd
 1011: AStart-AEnd
 1110: BStart-BEnd
 1101: BStart-BEnd

一位组合和1010，0101看起来不可能

One-bit combinations and 1010, 0101 look impossible

使用通配符，如作者建议的那样:

with wildcards, as author proposed:

 At first check for 
   0000: no intersection
   1111: full circle
 then
   **11: AStart-AEnd
   1001: AStart-BEnd
   0110: BStart-AEnd
   11**: BStart-BEnd

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