没有什么比归纳定义问题更美好的了。为了完整起见，你在序言中有一个版本，它可能也澄清了光线投射背后的思想:

基于仿真的简化算法在http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html

一些helper谓词:

exor(A,B):- \+A,B;A,\+B.
in_range(Coordinate,CA,CB) :- exor((CA>Coordinate),(CB>Coordinate)).

inside(false).
inside(_,[_|[]]).
inside(X:Y, [X1:Y1,X2:Y2|R]) :- in_range(Y,Y1,Y2), X > ( ((X2-X1)*(Y-Y1))/(Y2-Y1) +      X1),toggle_ray, inside(X:Y, [X2:Y2|R]); inside(X:Y, [X2:Y2|R]).

get_line(_,_,[]).
get_line([XA:YA,XB:YB],[X1:Y1,X2:Y2|R]):- [XA:YA,XB:YB]=[X1:Y1,X2:Y2]; get_line([XA:YA,XB:YB],[X2:Y2|R]).

给定两点a和B的直线(直线(a,B))方程为:

                    (YB-YA)
           Y - YA = ------- * (X - XA) 
                    (XB-YB) 

It is important that the direction of rotation for the line is setted to clock-wise for boundaries and anti-clock-wise for holes. We are going to check whether the point (X,Y), i.e the tested point is at the left half-plane of our line (it is a matter of taste, it could also be the right side, but also the direction of boundaries lines has to be changed in that case), this is to project the ray from the point to the right (or left) and acknowledge the intersection with the line. We have chosen to project the ray in the horizontal direction (again it is a matter of taste, it could also be done in vertical with similar restrictions), so we have:

               (XB-XA)
           X < ------- * (Y - YA) + XA
               (YB-YA) 

Now we need to know if the point is at the left (or right) side of the line segment only, not the entire plane, so we need to restrict the search only to this segment, but this is easy since to be inside the segment only one point in the line can be higher than Y in the vertical axis. As this is a stronger restriction it needs to be the first to check, so we take first only those lines meeting this requirement and then check its possition. By the Jordan Curve theorem any ray projected to a polygon must intersect at an even number of lines. So we are done, we will throw the ray to the right and then everytime it intersects a line, toggle its state. However in our implementation we are goint to check the lenght of the bag of solutions meeting the given restrictions and decide the innership upon it. for each line in the polygon this have to be done.

is_left_half_plane(_,[],[],_).
is_left_half_plane(X:Y,[XA:YA,XB:YB], [[X1:Y1,X2:Y2]|R], Test) :- [XA:YA, XB:YB] =  [X1:Y1, X2:Y2], call(Test, X , (((XB - XA) * (Y - YA)) / (YB - YA) + XA)); 
                                                        is_left_half_plane(X:Y, [XA:YA, XB:YB], R, Test).

in_y_range_at_poly(Y,[XA:YA,XB:YB],Polygon) :- get_line([XA:YA,XB:YB],Polygon),  in_range(Y,YA,YB).
all_in_range(Coordinate,Polygon,Lines) :- aggregate(bag(Line),    in_y_range_at_poly(Coordinate,Line,Polygon), Lines).

traverses_ray(X:Y, Lines, Count) :- aggregate(bag(Line), is_left_half_plane(X:Y, Line, Lines, <), IntersectingLines), length(IntersectingLines, Count).

% This is the entry point predicate
inside_poly(X:Y,Polygon,Answer) :- all_in_range(Y,Polygon,Lines), traverses_ray(X:Y, Lines, Count), (1 is mod(Count,2)->Answer=inside;Answer=outside).

我认为下面这段代码是最好的解决方案(从这里开始):

int pnpoly(int nvert, float *vertx, float *verty, float testx, float testy)
{
  int i, j, c = 0;
  for (i = 0, j = nvert-1; i < nvert; j = i++) {
    if ( ((verty[i]>testy) != (verty[j]>testy)) &&
     (testx < (vertx[j]-vertx[i]) * (testy-verty[i]) / (verty[j]-verty[i]) + vertx[i]) )
       c = !c;
  }
  return c;
}

参数

nvert:多边形中的顶点数。是否在末端重复第一个顶点在上面的文章中已经讨论过了。 vertx, verty:包含多边形顶点的x坐标和y坐标的数组。 testx, testy:测试点的X坐标和y坐标。

它既简短又高效，适用于凸多边形和凹多边形。如前所述，您应该首先检查边界矩形，并单独处理多边形孔。

这背后的想法很简单。作者描述如下:

我从测试点水平运行一条半无限射线(增加x，固定y)，并计算它穿过多少条边。在每个十字路口，光线在内部和外部之间切换。这叫做乔丹曲线定理。

当水平射线穿过任意一条边时，变量c从0变为1，从1变为0。基本上它记录了交叉边的数量是偶数还是奇数。0表示偶数，1表示奇数。

对于检测多边形上的命中，我们需要测试两件事:

如果点在多边形区域内。(可通过Ray-Casting算法实现) 如果点在多边形边界上(可以用与在折线(线)上检测点相同的算法来完成)。

在大多数情况下，这是一个比其他算法都快的算法。

它又新又雅致。我们花费O(n * log(n))时间构建一个表，允许我们在O(log(n) + k)时间内测试多边形中的点。

与光线跟踪或角度不同，使用扫描光束表可以更快地对同一多边形进行多次检查。我们必须预先构建一个扫描束活动边表，这是大多数代码正在做的事情。

We calculate the scanbeam and the active edges for that position in the y-direction. We make a list of points sorted by their y-component and we iterate through this list, for two events. Start-Y and End-Y, we track the active edges as we process the list. We process the events in order and for each scanbeam we record the y-value of the event and the active edges at each event (events being start-y and end-y) but we only record these when our event-y is different than last time (so everything at the event point is processed before we mark it in our table).

我们得到我们的表格:

[] p6p5、p6p7 p6p5, p6p7, p2p3, p2p1 p6p7, p5p4, p2p3, p3p1 p7p8, p5p4, p2p3, p2p1 p7p8, p5p4, p3p4, p2p1 p7p8 p2p1、 p7p8、p1p0 p8p0、p1p0 []

在构建该表之后，实际执行工作的代码只有几行。

注意:这里的代码使用复数值作为点。所以。real是。x。imag是。y。

def point_in_scantable(actives_table, events, xi, point):
    beam = bisect(events, point.imag) - 1  # Binary search in sorted array.
    actives_at_y = actives_table[beam]
    total = sum([point.real > xi(e, point.imag) for e in actives_at_y])
    return bool(total % 2)

我们对事件进行二进制搜索，以找到特定值的actives_at_y。对于在y点的所有活动，我们计算在我们点的特定y点的x段值。每次x截距大于点的x分量时加1。然后对总数乘以2。(这是偶数-奇数填充规则，你可以很容易地适应任何其他填充规则)。

完整的代码:

from bisect import bisect

def build_edge_list(polygon):
    edge_list = []
    for i in range(1, len(polygon)):
        if (polygon[i].imag, polygon[i].real) < (polygon[i - 1].imag, polygon[i - 1].real):
            edge_list.append((polygon[i], i))
            edge_list.append((polygon[i - 1], ~i))
        else:
            edge_list.append((polygon[i], ~i))
            edge_list.append((polygon[i - 1], i))

    def sort_key(e):
        return e[0].imag, e[0].real, ~e[1]

    edge_list.sort(key=sort_key)
    return edge_list


def build_scanbeam(edge_list):
    actives_table = []
    events = []
    y = -float("inf")
    actives = []
    for pt, index in edge_list:
        if y != pt.imag:
            actives_table.append(list(actives))
            events.append(y)
        if index >= 0:
            actives.append(index)
        else:
            actives.remove(~index)
        y = pt.imag
    return actives_table, events

def point_in_polygon(polygon, point):
    def x_intercept(e, y):
        pt0 = polygon[e-1]
        pt1 = polygon[e]
        if pt1.real - pt0.real == 0:
            return pt0.real
        m = (pt1.imag - pt0.imag) / (pt1.real - pt0.real)
        b = pt0.imag - (m * pt0.real)
        return (y - b) / m

    edge_list = build_edge_list(polygon)
    actives_table, events = build_scanbeam(edge_list)
    try:
        if len(point):
            return [point_in_scantable(actives_table, events, x_intercept, p) for p in point]
    except TypeError:
        return point_in_scantable(actives_table, events, x_intercept, point)

def point_in_scantable(actives_table, events, xi, point):
    beam = bisect(events, point.imag) - 1  # Binary search in sorted array.
    actives_at_y = actives_table[beam]
    total = sum([point.real > xi(e, point.imag) for e in actives_at_y])
    return bool(total % 2)

如果忽略，则扫描表的构建时间为O(n*log(n))。我们实际上是在O(log(n) + k)时间内查到的。其中n是多边形中段数的大小，k是该多边形中典型的活动边数。其他的光线追踪方法实际上需要O(n)时间。每次我们检查一个点，它迭代整个多边形。所以即使有这个明显的次优实现，它也轻而易举地打败了其他所有的。

There's a few performance tricks that could be done, for example, we can lower the time complexity to O(log(n) + log(k)) time. To do this we would implement Bentley-Ottmann into the sweep line, and rather than processing the intersections as different events, we split the lines at the intersections. We then also sort the active edges by their x-intercepts. We then know that no intersections occur during a scanbeam and since we sorted our segments (taking care to order them correctly within the scanbeam even if they start at the same initial point (you need to look at the slopes, or just compare midpoints of the segments). We then have a sorted intersection-less actives lists scanbeam table which means we can binary search into active edge list as well. Though that sounds like a lot of work for a value of k which is going to be typically 2 or maybe 4.

此外，由于这基本上变成了一个查找表和一些x截距的最小计算，它更能用GPU完成。你不再需要在多边形上循环了。所以你可以用numpy这样的东西来大量计算这些点，这样你就可以一次做所有的计算，从而提高性能。

Like David Segonds' answer suggests I use an approach of angle summation derived from my concave polygon drawing algorithm. It relies of adding up the approximate angles of subtriangles around the point to obtain a weight. A weight around 1.0 means the point is inside the triangle, a weight around 0.0 means outside, a weight around -1.0 is what happens when inside the polygon but in reverse order (like with one of the halves of a bowtie-shaped tetragon) and a weight of NAN if exactly on an edge. The reason it's not slow is that angles don't need to be estimated accurately at all. Holes can be handled by treating them as separate polygons and subtracting the weights.

typedef struct { double x, y; } xy_t;

xy_t sub_xy(xy_t a, xy_t b)
{
    a.x -= b.x;
    a.y -= b.y;
    return a;
}

double calc_sharp_subtriangle_pixel_weight(xy_t p0, xy_t p1)
{
    xy_t rot, r0, r1;
    double weight;

    // Rotate points (unnormalised)
    rot = sub_xy(p1, p0);
    r0.x = rot.x*p0.y - rot.y*p0.x;
    r0.y = rot.x*p0.x + rot.y*p0.y;
    r1.y = rot.x*p1.x + rot.y*p1.y;

    // Calc weight
    weight = subtriangle_angle_approx(r1.y, r0.x) - subtriangle_angle_approx(r0.y, r0.x);

    return weight;
}

double calc_sharp_polygon_pixel_weight(xy_t p, xy_t *corner, int corner_count)
{
    int i;
    xy_t p0, p1;
    double weight = 0.;

    p0 = sub_xy(corner[corner_count-1], p);
    for (i=0; i < corner_count; i++)
    {
        // Transform corner coordinates
        p1 = sub_xy(corner[i], p);

        // Calculate weight for each subtriangle
        weight += calc_sharp_subtriangle_pixel_weight(p0, p1);
        p0 = p1;
    }

    return weight;
}

因此，对于多边形的每一段，都形成一个子三角形，并计算点，然后旋转每个子三角形以计算其近似角度并添加到权重。

调用subtriangle_angle_approx(y, x)可以替换为atan2(y, x) / (2.*pi)，但是一个非常粗略的近似值就足够精确了:

double subtriangle_angle_approx(double y, double x)
{
    double angle, d;
    int obtuse;

    if (x == 0.)
        return NAN;

    obtuse = fabs(y) > fabs(x);
    if (obtuse)
        swap_double(&y, &x);

    // Core of the approximation, a very loosely approximate atan(y/x) / (2.*pi) over ]-1 , 1[
    d = y / x;
    angle = 0.13185 * d;

    if (obtuse)
        angle = sign(d)*0.25 - angle;

    return angle;
}

下面是nirg给出的答案的c#版本，它来自RPI教授。请注意，使用来自RPI源代码的代码需要归属。

在顶部添加了一个边界框复选。然而，正如James Brown所指出的，主代码几乎和边界框检查本身一样快，所以边界框检查实际上会减慢整体操作，因为您正在检查的大多数点都在边界框内。所以你可以让边界框签出，或者另一种选择是预先计算多边形的边界框，如果它们不经常改变形状的话。

public bool IsPointInPolygon( Point p, Point[] polygon )
{
    double minX = polygon[ 0 ].X;
    double maxX = polygon[ 0 ].X;
    double minY = polygon[ 0 ].Y;
    double maxY = polygon[ 0 ].Y;
    for ( int i = 1 ; i < polygon.Length ; i++ )
    {
        Point q = polygon[ i ];
        minX = Math.Min( q.X, minX );
        maxX = Math.Max( q.X, maxX );
        minY = Math.Min( q.Y, minY );
        maxY = Math.Max( q.Y, maxY );
    }

    if ( p.X < minX || p.X > maxX || p.Y < minY || p.Y > maxY )
    {
        return false;
    }

    // https://wrf.ecse.rpi.edu/Research/Short_Notes/pnpoly.html
    bool inside = false;
    for ( int i = 0, j = polygon.Length - 1 ; i < polygon.Length ; j = i++ )
    {
        if ( ( polygon[ i ].Y > p.Y ) != ( polygon[ j ].Y > p.Y ) &&
             p.X < ( polygon[ j ].X - polygon[ i ].X ) * ( p.Y - polygon[ i ].Y ) / ( polygon[ j ].Y - polygon[ i ].Y ) + polygon[ i ].X )
        {
            inside = !inside;
        }
    }

    return inside;
}