如果两个线段重叠或相交,则将它们合并在同一个圆上 [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,我会尝试将它们组合起来.我的问题类似于 this 和 this一>.但是,我想要的是组合两个部分.
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;
}
}
我正在尝试编写一个 TryCombine()
.
I am trying to write a 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
您可以使用我在第二个链接中的回答中描述的方法.
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)
要检查是否存在交集以及发生什么样的交集,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)
这些组合可以形成 16 种可能的结果(并非所有结果都可靠).我会将这些结果组合为 4 位值的位,并在 case
语句中处理它们.
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都为真(值 0b1001 = 9
),你有简单的交集,就像你的 seg1-seg2 案例 - 所以得到 AStart ans 起点,BEnd 作为结束指向并标准化它们(如果 BEnd 小于 AStart,则将 360 添加到 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).
预归一化步骤应提供 BEnd>=BStart 和 AEnd>=AStart(例如,将 (3,1) 弧变换为中点 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 个简单的结束组合,4 个单端重合的案例):
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
带有通配符,正如作者所建议的:
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
这篇关于如果两个线段重叠或相交,则将它们合并在同一个圆上的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!