<分享>使用GH C#找出最近點連線並檢查是否交集

目標 : 使用GH C#找出最近點連線並檢查是否交集。

檔案 : 

https://dl.dropbox.com/u/29860061/closet%20line%20with%20intersction%20check.gh

流程步驟 : 

1.散佈隨機點在範圍內，設index 0 為起始點。

使用舊點雲(intput x)計算出基準點(basePoint)與新點雲(calNewCloudPts2)。

新點雲中不包含基準點(basePoint)。

public List<Point3d> calNewCloudPts(int basePtIndex, List<Point3d> oldCloudPts){...}

2.使用基準點(basePoint)與新點雲(calNewCloudPts2)搜尋最近點，並儲存當下indexNumber (calNewCloudPts2[closePtIndex])。

public int calculateDist(List < Point3d > calPts, Point3d startPt){...}

3. 使用基準點(basePoint)與最近點(calNewCloudPts2[closePtIndex])相連線(尚未放入LIST中)。

public Line LinkLn(Point3d startPt, Point3d endPt){...}

4.檢查最新連線(myLine)是否與舊有線段(mylnList)相互交集，如果有交集->不連線，如果沒交集->產生新線段，並放入LIST當中。

public bool checkInt(List<Line> lnA, Line lnB){...}

5. 更新基準點(basePoint)與最近點(calNewCloudPts2[closePtIndex])。

initialnum = closePtIndex;

x = calNewCloudPts2;

6.重複1-5步驟，完成最近距離連線

Main Code:

 private void RunScript(List<Point3d> x, Point3d y, ref object A)

  {

    List < Line > mylnList = new List < Line >();

    List<Point3d> calNewCloudPts2 = new List<Point3d>();

    Point3d basePoint = new Point3d();

    //the initial index is 0

    int initialnum = 0;

    int closePtIndex = 0;

    int totaloldPtsLength = x.Count - 1;

    Line myLine;

    bool mytest = false;

    for(int i = 0;i < totaloldPtsLength;i++){

      basePoint = x[initialnum];

      calNewCloudPts2 = calNewCloudPts(initialnum, x);

      closePtIndex = calculateDist(calNewCloudPts2, basePoint);

      //Print("closePtIndex  = {0}", closePtIndex);

      myLine = LinkLn(basePoint, calNewCloudPts2[closePtIndex]);

      mytest = checkInt(mylnList, myLine);

      Print("ff = {0}", mytest);

      if(mytest == false){

        mylnList.Add(myLine);

      }

      initialnum = closePtIndex;

      x = calNewCloudPts2;

      Print("calNewCloudPts2 List = {0}", calNewCloudPts2.Count.ToString());

    }

    //Print("x count = {0}", x.Count.ToString());

    A = mylnList;

  }

結果圖片:

points number - > 100 

 points number - > 500 

 points number - > 1000

心得:

下次將會發布 recursion and timer test