zoukankan      html  css  js  c++  java
  • a fast algorithm to compute the area of a polygon

    Assume there is a polygon (v1, v2,...vn), where vi, (1<=i<=n) are its vertices. What is the area of this polygon?

    We have learnt cross product, which can be used to calculate the area of a triangle. We can also use this to calculate the area of a polygon by dividing the polygon with n segments into triangles. So the question is how to divide the polygon into triangles. We can choose a vertex A, and connect this vertex, A, to the vertices of the polygon. 

    There are various ways to achieve this which need proving. One way is use one vertex of the polygon as vertex A. Another way is to choose origin (0, 0) as vertex A. We can connect A(0, 0) with vi (1<=i<=n) to form triangles, A-v1-v2, A-v2-v3, A-v3-v4,...A-vn-1-vn, A-vn-v1. We can sum all the areas of these triangle, and the sum is 2 times the size of the area of polygon, labled as S.

    2*S = |S(A-v1-v2) + S(A-v2-v3)+...+S(A-vn-1-vn)+S(A-vn-v1)|

          = |x1y2-x2y1 + x2y3-x3y2 +...+ x(n-1)yn-xny(n-1)+xny1-x1yn|

    which needs 2*n multiplications of double type.

    We can reduce the calling of mulplications of double type to n by an observation, x2y2-x1y1 + x3y3-x2y2 + ... + ynxn - x(n-1)y(n-1) + x1y1 - xnyn = 0.

    So  2*S = |x1y2-x2y1 + x2y3-x3y2 +...+ x(n-1)yn-xny(n-1)+xny1-x1yn     +   x2y2-x1y1 + x3y3-x2y2 + ... + ynxn - x(n-1)y(n-1) + x1y1 - xnyn |

                = |(x1+x2)*(y2-y1) + (x2+x3)(y3-y2) +...+(x(n-1)+xn)*(yn-y(n-1)) + (xn+x1)*(y1-yn)|

    which needs n multiplications of double type. This is the fast algorithm I have seen to compute the area of a polygon.

  • 相关阅读:
    day40_jQuery学习笔记_01
    jQuery选择什么版本 1.x? 2.x? 3.x?
    6个关于dd命令备份Linux系统的例子
    快速掌握grep命令及正则表达式
    Linux下删除乱码或特殊字符文件
    在 Linux 中永久修改 USB 设备权限
    CentOS 7 中 hostnamectl 的使用
    申请红帽企业版Linux开发者订阅
    CentOS6 下rsync服务器配置
    Centos6下DRBD的安装配置
  • 原文地址:https://www.cnblogs.com/Torstan/p/2580569.html
Copyright © 2011-2022 走看看