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).

Scala版本的解决方案由nirg(假设边界矩形预检查是单独完成的):

def inside(p: Point, polygon: Array[Point], bounds: Bounds): Boolean = {

  val length = polygon.length

  @tailrec
  def oddIntersections(i: Int, j: Int, tracker: Boolean): Boolean = {
    if (i == length)
      tracker
    else {
      val intersects = (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
      oddIntersections(i + 1, i, if (intersects) !tracker else tracker)
    }
  }

  oddIntersections(0, length - 1, tracker = false)
}

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。

nirg的c#版本的答案在这里:我只分享代码。这可能会节省一些时间。

public static bool IsPointInPolygon(IList<Point> polygon, Point testPoint) {
            bool result = false;
            int j = polygon.Count() - 1;
            for (int i = 0; i < polygon.Count(); i++) {
                if (polygon[i].Y < testPoint.Y && polygon[j].Y >= testPoint.Y || polygon[j].Y < testPoint.Y && polygon[i].Y >= testPoint.Y) {
                    if (polygon[i].X + (testPoint.Y - polygon[i].Y) / (polygon[j].Y - polygon[i].Y) * (polygon[j].X - polygon[i].X) < testPoint.X) {
                        result = !result;
                    }
                }
                j = i;
            }
            return result;
        }

下面是golang版本的@nirg答案(灵感来自于@@m-katz的c#代码)

func isPointInPolygon(polygon []point, testp point) bool {
    minX := polygon[0].X
    maxX := polygon[0].X
    minY := polygon[0].Y
    maxY := polygon[0].Y

    for _, p := range polygon {
        minX = min(p.X, minX)
        maxX = max(p.X, maxX)
        minY = min(p.Y, minY)
        maxY = max(p.Y, maxY)
    }

    if testp.X < minX || testp.X > maxX || testp.Y < minY || testp.Y > maxY {
        return false
    }

    inside := false
    j := len(polygon) - 1
    for i := 0; i < len(polygon); i++ {
        if (polygon[i].Y > testp.Y) != (polygon[j].Y > testp.Y) && testp.X < (polygon[j].X-polygon[i].X)*(testp.Y-polygon[i].Y)/(polygon[j].Y-polygon[i].Y)+polygon[i].X {
            inside = !inside
        }
        j = i
    }

    return inside
}

VBA版本:

注意:请记住，如果你的多边形是地图中的一个区域，纬度/经度是Y/X值，而不是X/Y(纬度= Y，经度= X)，因为从我的理解来看，这是历史含义，因为经度不是一个测量值。

类模块:CPoint

Private pXValue As Double
Private pYValue As Double

'''''X Value Property'''''

Public Property Get X() As Double
    X = pXValue
End Property

Public Property Let X(Value As Double)
    pXValue = Value
End Property

'''''Y Value Property'''''

Public Property Get Y() As Double
    Y = pYValue
End Property

Public Property Let Y(Value As Double)
    pYValue = Value
End Property

模块:

Public Function isPointInPolygon(p As CPoint, polygon() As CPoint) As Boolean

    Dim i As Integer
    Dim j As Integer
    Dim q As Object
    Dim minX As Double
    Dim maxX As Double
    Dim minY As Double
    Dim maxY As Double
    minX = polygon(0).X
    maxX = polygon(0).X
    minY = polygon(0).Y
    maxY = polygon(0).Y

    For i = 1 To UBound(polygon)
        Set q = polygon(i)
        minX = vbMin(q.X, minX)
        maxX = vbMax(q.X, maxX)
        minY = vbMin(q.Y, minY)
        maxY = vbMax(q.Y, maxY)
    Next i

    If p.X < minX Or p.X > maxX Or p.Y < minY Or p.Y > maxY Then
        isPointInPolygon = False
        Exit Function
    End If


    ' SOURCE: http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html

    isPointInPolygon = False
    i = 0
    j = UBound(polygon)

    Do While i < UBound(polygon) + 1
        If (polygon(i).Y > p.Y) Then
            If (polygon(j).Y < p.Y) Then
                If p.X < (polygon(j).X - polygon(i).X) * (p.Y - polygon(i).Y) / (polygon(j).Y - polygon(i).Y) + polygon(i).X Then
                    isPointInPolygon = True
                    Exit Function
                End If
            End If
        ElseIf (polygon(i).Y < p.Y) Then
            If (polygon(j).Y > p.Y) Then
                If p.X < (polygon(j).X - polygon(i).X) * (p.Y - polygon(i).Y) / (polygon(j).Y - polygon(i).Y) + polygon(i).X Then
                    isPointInPolygon = True
                    Exit Function
                End If
            End If
        End If
        j = i
        i = i + 1
    Loop   
End Function

Function vbMax(n1, n2) As Double
    vbMax = IIf(n1 > n2, n1, n2)
End Function

Function vbMin(n1, n2) As Double
    vbMin = IIf(n1 > n2, n2, n1)
End Function


Sub TestPointInPolygon()

    Dim i As Integer
    Dim InPolygon As Boolean

'   MARKER Object
    Dim p As CPoint
    Set p = New CPoint
    p.X = <ENTER X VALUE HERE>
    p.Y = <ENTER Y VALUE HERE>

'   POLYGON OBJECT
    Dim polygon() As CPoint
    ReDim polygon(<ENTER VALUE HERE>) 'Amount of vertices in polygon - 1
    For i = 0 To <ENTER VALUE HERE> 'Same value as above
       Set polygon(i) = New CPoint
       polygon(i).X = <ASSIGN X VALUE HERE> 'Source a list of values that can be looped through
       polgyon(i).Y = <ASSIGN Y VALUE HERE> 'Source a list of values that can be looped through
    Next i

    InPolygon = isPointInPolygon(p, polygon)
    MsgBox InPolygon

End Sub