凸包中的额外点（使用格雷厄姆扫描）错误+ java

Posted 2023-02-17

【中文标题】凸包中的额外点（使用格雷厄姆扫描）错误+ java

【英文标题】：extra point in convex hull (using graham scan) error + java

【发布时间】：2013-05-15 15:10:51

【问题描述】：

我的代码使用格雷厄姆算法找到凸包效果很好（它向我展示了它应该显示的多边形）但我可以看到它向我发送了一个额外的共线点（尽管我正在处理共线点在我的代码中） 这是我的代码：

    public Collection<Coord> territoire()
    
        double checkPoints;
        Collection<Coord> sommets = new ArrayList<Coord>(this);
        Stack<Coord> stackOfConvexHull = new Stack<Coord>();
        ArrayList<Coord> thisToArrList = new ArrayList<Coord>();
        for(Coord c : this)
            thisToArrList.add(c);


        //sorting the array by Y
        ArrayList<Coord> sortedPointsByY = sortedArrayByY(thisToArrList); 

        //sorting the set of points with increasing order of angle
        // they and the base point (basePoint) make with X axis 
        List<Coord> sortedPointsByAngle = new ArrayList<Coord>(sortPointsByAngle(sortedPointsByY));

        if(sortedPointsByAngle.size() < 3)
            System.out.println("the convex hull of less than 3 points is not possible");
        if(collinear(sortedPointsByAngle)) 
            System.out.println("can't make a convex hull of collinear points");


        stackOfConvexHull.push(sortedPointsByAngle.get(0));
        stackOfConvexHull.push(sortedPointsByAngle.get(1));

            for (int i = 2; i < sortedPointsByAngle.size(); i++) 
            

                Coord p1 = sortedPointsByAngle.get(i);
                Coord p2 = stackOfConvexHull.pop();
                Coord p3 = stackOfConvexHull.peek();

                checkPoints = ccw(p3, p2, p1);

                // counter-clockwise
                if(checkPoints > 0) 
                
                    stackOfConvexHull.push(p2);
                    stackOfConvexHull.push(p1);
                

                // collinear
                if(checkPoints == 0)
                    stackOfConvexHull.push(p1);

                // clockwise
                else
                    i--;
            

            // end of the hull
            stackOfConvexHull.push(sortedPointsByAngle.get(0));
            sommets = new ArrayList<Coord>(stackOfConvexHull);

        return sommets;

    

//**********************************Auxiliary méthods****************************************************

                            //***** sorting points by Y and angles *****

    //sorting the points by their y in ascending order
    public ArrayList<Coord> sortedArrayByY(ArrayList<Coord> arrayOfPoints)
    
        Coord temp = null;
        for(int i = 0; i< arrayOfPoints.size(); i++)
         
            for(int j = 0; j< arrayOfPoints.size()-1; j++)
            
                if(arrayOfPoints.get(j+1).y < arrayOfPoints.get(j).y)
                
                    temp = arrayOfPoints.get(j+1);
                    arrayOfPoints.set(j+1, arrayOfPoints.get(j));
                    arrayOfPoints.set(j, temp);  
                
            
        
        return arrayOfPoints;
    


    public Set<Coord> sortPointsByAngle(ArrayList<Coord> points) 
    
        int min = minYIndex(points);
        final Coord basePoint = points.get(min);

        TreeSet<Coord> set = new TreeSet<Coord>(new Comparator<Coord>() 
        
            public int compare(Coord a, Coord b) 

                if(a == b || a.equals(b))
                    return 0;

               double firstAngle = angle(basePoint, a);
               double secondAngle = angle(basePoint, b);

                if(firstAngle < secondAngle) 
                    return -1;

                else if(firstAngle > secondAngle) 
                    return 1;

                else 
                
                    // collinear with the 'basePoint' point, let the point closest to it come first
                    double firstDistance = findDistance(basePoint, a);
                    double secondDistance = findDistance(basePoint, b);

                    if(firstDistance < secondDistance) 
                        return -1;    

                    else 
                        return 1;
                
            
        );

        set.addAll(points);
        return set;
    


    // find the buttom most point (minimum Y)
    // if If the lowest y-coordinate exists in 
    // more than one point in the set, the point with the one with the lowest x-coordinate 
    // will be chosen

    public int minYIndex(ArrayList<Coord> sortedPointsByY)
       
        int min = 0;
        for ( int i = 1; i < sortedPointsByY.size(); i++ ) // O(n) => n number of points
        
            if ( sortedPointsByY.get(i).y == sortedPointsByY.get(min).y) 
            
                if ( sortedPointsByY.get(i).x < sortedPointsByY.get(min).x)
                    min = i;
            
            else if ( sortedPointsByY.get(i).y < sortedPointsByY.get(min).y)
                min = i;
        
        return min;
    

    public double angle(Coord basePoint, Coord a) 
    
        return Math.atan2(a.y - basePoint.y, a.x - basePoint.x);
    


    public double findDistance(Coord basePoint, Coord a)
    
        return Math.sqrt(((basePoint.x - a.x) * (basePoint.x - a.x)) + 
                ((basePoint.y - a.y) * (basePoint.y - a.y)));
    


    //if the result is zero, the point is collinear 
    //if it is positive, the three points constitute left turn (counter clockwise)
    //else the three points constitute right turn (clockwise)    
    public double ccw(Coord p1, Coord p2, Coord p3)
    
        return (p2.x - p1.x)*(p3.y - p1.y) - (p2.y - p1.y)*(p3.x - p1.x);
    

 // check if the points are collinear   
    public boolean collinear(List<Coord> sortedPointsByAngle) 
    

        Coord a, b, c;
        if(sortedPointsByAngle.size() < 2) 
            return true;

        a = sortedPointsByAngle.get(0);
        b = sortedPointsByAngle.get(1);

        for(int i = 2; i < sortedPointsByAngle.size(); i++) 
        

            c = sortedPointsByAngle.get(i);

            if(ccw(a, b, c) != 0) 
                return false;
        

        return true;
    

我正在等待一些提示来帮助我找到我的问题

【问题讨论】：

在这里拍张照片会很好

如何将图片放到本站？

您能提供您的输入示例吗？这样更容易发现问题。

这是我的输入：0 0 0 3.75 1 2.5 1 0.5 1.5 1.5 -4 1.5 -3 1.3 -2 1.1 -1 0.9 0 0.7 -2.9 1.5 -1.8 1.5 -0.7 1.5 0.6 1.5 -3 1.7 -2 1.9 -1 2.1 0 2.3 -3.2 1.2 -2.4 0.9 -1.6 0.6 -0.8 0.3 -3.2 1.95 -2.4 2.4 -1.6 2.85 -0.8 3.3 这是我的输出：0.0 0.0 1.0 0.5 1.5 1.5 1.751 2.5 0.0 0.0 2.85 -4.0 1.5 0.0 0.0

哇，那是相当大，我希望有一个正方形左右；）你能找到一个同样问题的小例子吗？

【参考方案1】：

这可能是一个舍入问题。你计算这个表达式（函数ccw）

(p2.x - p1.x)*(p3.y - p1.y) - (p2.y - p1.y)*(p3.x - p1.x)

双精度。 恰好0 的可能性很小。

我通常用来解决这个问题（虽然这不是一个非常干净的做法）只是测试“几乎 0”：

if (Math.abs(checkPoints) < 0.0000001) // colinear

【讨论】：

你的意思是我把这个条件放在我的 ccw 函数中？

@NavidKoochooloo 不，我的意思是在通话之后（territoire），而不是if(checkPoints == 0)