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).

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。

真的很喜欢Nirg发布的解决方案，由bobobobo编辑。我只是让它javascript友好，更容易读懂我的使用:

function insidePoly(poly, pointx, pointy) {
    var i, j;
    var inside = false;
    for (i = 0, j = poly.length - 1; i < poly.length; j = i++) {
        if(((poly[i].y > pointy) != (poly[j].y > pointy)) && (pointx < (poly[j].x-poly[i].x) * (pointy-poly[i].y) / (poly[j].y-poly[i].y) + poly[i].x) ) inside = !inside;
    }
    return inside;
}

这似乎在R中工作(为丑陋道歉，希望看到更好的版本!)。

pnpoly <- function(nvert,vertx,verty,testx,testy){
          c <- FALSE
          j <- nvert 
          for (i in 1:nvert){
              if( ((verty[i]>testy) != (verty[j]>testy)) && 
   (testx < (vertx[j]-vertx[i])*(testy-verty[i])/(verty[j]-verty[i])+vertx[i]))
            {c <- !c}
             j <- i}
   return(c)}

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");
    }
}