为了完整性，这里是nirg提供的算法的lua实现，由Mecki讨论:

function pnpoly(area, test)
    local inside = false
    local tx, ty = table.unpack(test)
    local j = #area
    for i=1, #area do
        local vxi, vyi = table.unpack(area[i])
        local vxj, vyj = table.unpack(area[j])
        if (vyi > ty) ~= (vyj > ty)
        and tx < (vxj - vxi)*(ty - vyi)/(vyj - vyi) + vxi
        then
            inside = not inside
        end
        j = i
    end
    return inside
end

变量区域是一个点的表，这些点依次存储为2D表。例子:

> A = {{2, 1}, {1, 2}, {15, 3}, {3, 4}, {5, 3}, {4, 1.5}}
> T = {2, 1.1}
> pnpoly(A, T)
true

GitHub Gist的链接。

当使用qt (qt 4.3+)时，可以使用QPolygon的函数containsPoint

Obj-C版本nirg的答案与样本方法测试点。Nirg的回答对我很有效。

- (BOOL)isPointInPolygon:(NSArray *)vertices point:(CGPoint)test {
    NSUInteger nvert = [vertices count];
    NSInteger i, j, c = 0;
    CGPoint verti, vertj;

    for (i = 0, j = nvert-1; i < nvert; j = i++) {
        verti = [(NSValue *)[vertices objectAtIndex:i] CGPointValue];
        vertj = [(NSValue *)[vertices objectAtIndex:j] CGPointValue];
        if (( (verti.y > test.y) != (vertj.y > test.y) ) &&
        ( test.x < ( vertj.x - verti.x ) * ( test.y - verti.y ) / ( vertj.y - verti.y ) + verti.x) )
            c = !c;
    }

    return (c ? YES : NO);
}

- (void)testPoint {

    NSArray *polygonVertices = [NSArray arrayWithObjects:
        [NSValue valueWithCGPoint:CGPointMake(13.5, 41.5)],
        [NSValue valueWithCGPoint:CGPointMake(42.5, 56.5)],
        [NSValue valueWithCGPoint:CGPointMake(39.5, 69.5)],
        [NSValue valueWithCGPoint:CGPointMake(42.5, 84.5)],
        [NSValue valueWithCGPoint:CGPointMake(13.5, 100.0)],
        [NSValue valueWithCGPoint:CGPointMake(6.0, 70.5)],
        nil
    ];

    CGPoint tappedPoint = CGPointMake(23.0, 70.0);

    if ([self isPointInPolygon:polygonVertices point:tappedPoint]) {
        NSLog(@"YES");
    } else {
        NSLog(@"NO");
    }
}

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).

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。

计算点p与每个多边形顶点之间的有向角和。如果总倾斜角是360度，那么这个点在里面。如果总数为0，则点在外面。

我更喜欢这种方法，因为它更健壮，对数值精度的依赖更小。

计算交集数量的均匀性的方法是有限的，因为你可以在计算交集数量的过程中“击中”一个顶点。

编辑:顺便说一下，这种方法适用于凹凸多边形。

编辑:我最近在维基百科上找到了一篇关于这个话题的完整文章。