Determining whether or not a polygon (2D) has its vertices ordered clockwise or counterclockwise

本文介绍了一种利用交叉乘积来判断2D多边形顶点顺序是否为顺时针或逆时针的方法,并通过实例展示了如何判断多边形是否为凸形或凹形。适用于计算机图形学和几何处理。

Determining whether or not a polygon (2D) has itsvertices ordered clockwise or counterclockwise

Written by Paul Bourke
March 1998

The following describes a method for determining whether or nota polygon has its vertices ordered clockwise or anticlockwise.As a consequence the test can also be used to determine whether ornot a polygon is concave or convex.A polygon will be assumed to be described by N vertices, ordered

(x 0,y 0), (x 1,y 1), (x 2,y 2), . . . (x n-1,y n-1)

A convenient definition of clockwise is based on considerations ofthe cross product between adjacent edges. If the crossproduct ispositive then it rises above the plane (z axis up out of the plane)and if negative then the cross product is into the plane.

cross product = ((x i - x i-1),(y i - y i-1)) x ((x i+1 - x i),(y i+1 - y i))

=(xi - xi-1) * (yi+1 - yi)- (yi - yi-1) * (xi+1 - xi)

If the polygon is known to be convex then one only has to considerthe cross product between any two adjacent edges. A positive cross product means we have a counterclockwise polygon. There are sometests that may need to be done if the polygons may not be "clean".In particular two vertices must not be coincident and two edges mustnot be colinear.

For the more general case where the polygons may be convex, it isnecessary to consider the sign of the cross product between adjacentedges as one moves around the polygon. If there are more positivecross products then the overall polygon is ordered counterclockwise.There are pathological cases to consider here as well, all the edgescannot be colinear, there must be at least 3 vertices, the polygonmust be simple, that is, it cannot intersect itself or have holes.

Test for concave/convex polygon

A similar argument to the above can be used to determine whether a polygon is concave or convex.For a convex polygon all the cross products of adjacent edges willbe the same sign, a concave polygon will have a mixture of crossproduct signs.

Source Code

Example and test program for testingwhether a polygon is ordered clockwise or counterclockwise.For MICROSOFT WINDOWS, contributed by G. Adam Stanislav.

C function by Paul Bourke

/*
   Return the clockwise status of a curve, clockwise or counterclockwise
   n vertices making up curve p
   return 0 for incomputables eg: colinear points
          CLOCKWISE == 1
          COUNTERCLOCKWISE == -1
   It is assumed that
   - the polygon is closed
   - the last point is not repeated.
   - the polygon is simple (does not intersect itself or have holes)
*/
int ClockWise(XY *p,int n)
{
   int i,j,k;
   int count = 0;
   double z;

   if (n < 3)
      return(0);

   for (i=0;i<n;i++) {
      j = (i + 1) % n;
      k = (i + 2) % n;
      z  = (p[j].x - p[i].x) * (p[k].y - p[j].y);
      z -= (p[j].y - p[i].y) * (p[k].x - p[j].x);
      if (z < 0)
         count--;
      else if (z > 0)
         count++;
   }
   if (count > 0)
      return(COUNTERCLOCKWISE);
   else if (count < 0)
      return(CLOCKWISE);
   else
      return(0);
}

Example and test program for testing whether a polygon is convex or concave.For MICROSOFT WINDOWS, contributed by G. Adam Stanislav.

/*
   Return whether a polygon in 2D is concave or convex
   return 0 for incomputables eg: colinear points
          CONVEX == 1
          CONCAVE == -1
   It is assumed that the polygon is simple
   (does not intersect itself or have holes)
*/
int Convex(XY *p,int n)
{
   int i,j,k;
   int flag = 0;
   double z;

   if (n < 3)
      return(0);

   for (i=0;i<n;i++) {
      j = (i + 1) % n;
      k = (i + 2) % n;
      z  = (p[j].x - p[i].x) * (p[k].y - p[j].y);
      z -= (p[j].y - p[i].y) * (p[k].x - p[j].x);
      if (z < 0)
         flag |= 1;
      else if (z > 0)
         flag |= 2;
      if (flag == 3)
         return(CONCAVE);
   }
   if (flag != 0)
      return(CONVEX);
   else
      return(0);
}


转自:http://debian.fmi.uni-sofia.bg/~sergei/cgsr/docs/clockwise.htm

内容概要:本文详细介绍了“秒杀商城”微服务架构的设计与实战全过程,涵盖系统从需求分析、服务拆分、技术选型到核心功能开发、分布式事务处理、容器化部署及监控链路追踪的完整流程。重点解决了高并发场景下的超卖问题,采用Redis预减库存、消息队列削峰、数据库乐观锁等手段保障数据一致性,并通过Nacos实现服务注册发现与配置管理,利用Seata处理跨服务分布式事务,结合RabbitMQ实现异步下单,提升系统吞吐能力。同时,项目支持Docker Compose快速部署和Kubernetes生产级编排,集成Sleuth+Zipkin链路追踪与Prometheus+Grafana监控体系,构建可观测性强的微服务系统。; 适合人群:具备Java基础和Spring Boot开发经验,熟悉微服务基本概念的中高级研发人员,尤其是希望深入理解高并发系统设计、分布式事务、服务治理等核心技术的开发者;适合工作2-5年、有志于转型微服务或提升架构能力的工程师; 使用场景及目标:①学习如何基于Spring Cloud Alibaba构建完整的微服务项目;②掌握秒杀场景下高并发、超卖控制、异步化、削峰填谷等关键技术方案;③实践分布式事务(Seata)、服务熔断降级、链路追踪、统一配置中心等企业级中间件的应用;④完成从本地开发到容器化部署的全流程落地; 阅读建议:建议按照文档提供的七个阶段循序渐进地动手实践,重点关注秒杀流程设计、服务间通信机制、分布式事务实现和系统性能优化部分,结合代码调试与监控工具深入理解各组件协作原理,真正掌握高并发微服务系统的构建能力。
评论
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
红包 添加红包
表情包 插入表情
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值