For a bit of code where the author defines a function like this:

Function pnpoly(npol As Integer, poly() as Point, x as Integer, Y as
Integer) as Boolean,

I am pretty sure the poly() input defined "as Point" is looking for an xy
point value for an edge of a polygon, but I am unsure how it is derived.

Any ideas? My thanks in advance for any comments or suggestions... this is a
function to identify whether an x-y data point falls inside of or outside of
a convex polygon.

Cheers! Brad

Tom, thanks, but no, it has no declaration for "Point".... this is actually
just a bit of code that someone posted in response to a question regarding
determining whether an x-y point on a graph is inside or outside of a convex
polygon outlined on the same x-y graph.

The whole thing looks like this:

Function pnpoly(npol As Integer, poly() As Point, X As Integer, Y As
Integer) As Boolean
Dim inpoly As Boolean
Dim i, j As Integer

i = 0
j = npol - 1
inpoly = False
While (i < npol)
If ((((poly(i).Y <= Y) And (Y < poly(j).Y)) Or _
((poly(j).Y <= Y) And (Y < poly(i).Y)))) Then
If (X < (poly(j).X - poly(i).X) * (Y - poly(i).Y) / (poly(j).Y -
poly(i).Y) + poly(i).X) Then
inpoly = Not (inpoly)
End If
End If
j = i
i = i + 1
Wend
pnpoly = inpoly
End Function

Now, what I was actually trying to do was take Andy Pope's interesting
"bounding areas" chart where he draws bounding lines around two overlapping
convex polygons, and simply count the number of x-y points inside of and
outside of a polygon. So, I hunted around and found this code, referred to
generally as the crossing number algorithm, for just this sort of thing.
Most often what I found was listed as pseudo-code, but in this one case, a
snippet of VBA code. On the psuedo-code, it does have:

typedef struct (int x, y Point;

and I am certain this is most likely what you asked about.

I am still not sure how to apply it in the function above.... thanks, Brad




"Tom Ogilvy" <(E-Mail Removed)> wrote in message
news:0D6CB1C4-B930-4A41-97BE-(E-Mail Removed)...
> Does he have a declaration at the top like:
>
> Private Type POINT
> x As Long
> y As Long
> End Type
>
>
> Then point would be a structure that contains both and x and y coordinate.
> with the declaration at the top of the module, then
> Usage example:
>
> Dim a() as point
> for i = 1 to 10
> a(i).x = int(rnd()*100 + 1)
> a(i).y = int(rnd()*50 + 1)
> Next
>
> --
> Regards,
> Tom Ogilvy
>
>
>
> "Brad" wrote:
>
>> For a bit of code where the author defines a function like this:
>>
>> Function pnpoly(npol As Integer, poly() as Point, x as Integer, Y as
>> Integer) as Boolean,
>>
>> I am pretty sure the poly() input defined "as Point" is looking for an xy
>> point value for an edge of a polygon, but I am unsure how it is derived.
>>
>> Any ideas? My thanks in advance for any comments or suggestions... this
>> is a
>> function to identify whether an x-y data point falls inside of or outside
>> of
>> a convex polygon.
>>
>> Cheers! Brad
>>
>>
>>

I would see it used like this:

Option Explicit
Option Base 0
Private Type POINT
x As Double
y As Double
End Type

Sub Main()
Dim XP, YP, XP1, YP1
Dim poly() As POINT
Dim b As Boolean
Dim x As Double, y As Double
Dim i As Long, j As Long
ReDim poly(0 To 19)

' Polygon
XP = Array(0#, 1#, 1#, 0#, 0#, 1#, _
-0.5, -1#, -1#, -2#, -2.5, _
-2#, -1.5, -0.5, 1#, 1#, 0#, _
-0.5, -1#, -0.5)
YP = Array(0#, 0#, 1#, 1#, 2#, 3#, 2#, _
3#, 0#, -0.5, -1#, -1.5, -2#, _
-2#, -1.5, -1#, -0.5, -1#, _
-1#, -0.5)
' Test Points
XP1 = Array(0.5, 0.5, -0.5, 0.75, 0, _
-0.5, -1, -1.5, -2.25, 0.5, 0.5, -0.5)
YP1 = Array(0.5, 1.5, 1.5, 2.25, 2.01, _
2.5, -0.5, 0.5, -1, -0.25, -1.25, -2.5)

' writing to the worksheet is not required
' but supports graphing the data to verify

Cells(2,1) = "XP" : Cells(3, 1) = "YP"
Cells(4,1) = "XP1" : Cells(5,1) = "YP1"
Cells(2, 2).Resize(1, 20) = XP
Cells(3, 2).Resize(1, 20) = YP
Cells(4, 2).Resize(1, 12) = XP1
Cells(5, 2).Resize(1, 12) = YP1
j = 0
For i = LBound(XP) To UBound(XP)
poly(j).x = XP(i)
poly(j).y = YP(i)
j = j + 1
Next

j = 1
For i = LBound(XP1) To UBound(XP1)
x = XP1(i): y = YP1(i)
b = pnpoly(20, poly, x, y)
Cells(6, j).Offset(0, 1) = b
j = j + 1
Next

End Sub

Function pnpoly(npol As Integer, poly() As POINT, _
x As Double, y As Double) As Boolean
Dim inpoly As Boolean
Dim i As Integer, j As Integer

i = 0
j = npol - 1
inpoly = False
While (i < npol)
If ((poly(i).y <= y) And (y < poly(j).y)) Or _
((poly(j).y <= y) And (y < poly(i).y)) Then
If x < (poly(j).x - poly(i).x) * (y - poly(i).y) / _
(poly(j).y - poly(i).y) + poly(i).x Then
inpoly = Not (inpoly)
End If
End If
j = i
i = i + 1
Wend
pnpoly = inpoly
End Function

Note that I have changed the function a little to support the data used.
(integer=>double)


See description at:
http://tinyurl.com/y93rhr

or "spelled out"

http://209.85.165.104/search?q=cache...s&ct=clnk&cd=5

--
Regards,
Tom Ogilvy


"Brad" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> Tom, thanks, but no, it has no declaration for "Point".... this is
> actually just a bit of code that someone posted in response to a question
> regarding determining whether an x-y point on a graph is inside or outside
> of a convex polygon outlined on the same x-y graph.
>
> The whole thing looks like this:
>
> Function pnpoly(npol As Integer, poly() As Point, X As Integer, Y As
> Integer) As Boolean
> Dim inpoly As Boolean
> Dim i, j As Integer
>
> i = 0
> j = npol - 1
> inpoly = False
> While (i < npol)
> If ((((poly(i).Y <= Y) And (Y < poly(j).Y)) Or _
> ((poly(j).Y <= Y) And (Y < poly(i).Y)))) Then
> If (X < (poly(j).X - poly(i).X) * (Y - poly(i).Y) /
> (poly(j).Y - poly(i).Y) + poly(i).X) Then
> inpoly = Not (inpoly)
> End If
> End If
> j = i
> i = i + 1
> Wend
> pnpoly = inpoly
> End Function
>
> Now, what I was actually trying to do was take Andy Pope's interesting
> "bounding areas" chart where he draws bounding lines around two
> overlapping convex polygons, and simply count the number of x-y points
> inside of and outside of a polygon. So, I hunted around and found this
> code, referred to generally as the crossing number algorithm, for just
> this sort of thing. Most often what I found was listed as pseudo-code, but
> in this one case, a snippet of VBA code. On the psuedo-code, it does have:
>
> typedef struct (int x, y Point;
>
> and I am certain this is most likely what you asked about.
>
> I am still not sure how to apply it in the function above.... thanks,
> Brad
>
>
>
>
> "Tom Ogilvy" <(E-Mail Removed)> wrote in message
> news:0D6CB1C4-B930-4A41-97BE-(E-Mail Removed)...
>> Does he have a declaration at the top like:
>>
>> Private Type POINT
>> x As Long
>> y As Long
>> End Type
>>
>>
>> Then point would be a structure that contains both and x and y
>> coordinate.
>> with the declaration at the top of the module, then
>> Usage example:
>>
>> Dim a() as point
>> for i = 1 to 10
>> a(i).x = int(rnd()*100 + 1)
>> a(i).y = int(rnd()*50 + 1)
>> Next
>>
>> --
>> Regards,
>> Tom Ogilvy
>>
>>
>>
>> "Brad" wrote:
>>
>>> For a bit of code where the author defines a function like this:
>>>
>>> Function pnpoly(npol As Integer, poly() as Point, x as Integer, Y as
>>> Integer) as Boolean,
>>>
>>> I am pretty sure the poly() input defined "as Point" is looking for an
>>> xy
>>> point value for an edge of a polygon, but I am unsure how it is derived.
>>>
>>> Any ideas? My thanks in advance for any comments or suggestions... this
>>> is a
>>> function to identify whether an x-y data point falls inside of or
>>> outside of
>>> a convex polygon.
>>>
>>> Cheers! Brad
>>>
>>>
>>>
>
>

Tom, thanks so much! Brad


"Tom Ogilvy" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
>I would see it used like this:
>
> Option Explicit
> Option Base 0
> Private Type POINT
> x As Double
> y As Double
> End Type
>
> Sub Main()
> Dim XP, YP, XP1, YP1
> Dim poly() As POINT
> Dim b As Boolean
> Dim x As Double, y As Double
> Dim i As Long, j As Long
> ReDim poly(0 To 19)
>
> ' Polygon
> XP = Array(0#, 1#, 1#, 0#, 0#, 1#, _
> -0.5, -1#, -1#, -2#, -2.5, _
> -2#, -1.5, -0.5, 1#, 1#, 0#, _
> -0.5, -1#, -0.5)
> YP = Array(0#, 0#, 1#, 1#, 2#, 3#, 2#, _
> 3#, 0#, -0.5, -1#, -1.5, -2#, _
> -2#, -1.5, -1#, -0.5, -1#, _
> -1#, -0.5)
> ' Test Points
> XP1 = Array(0.5, 0.5, -0.5, 0.75, 0, _
> -0.5, -1, -1.5, -2.25, 0.5, 0.5, -0.5)
> YP1 = Array(0.5, 1.5, 1.5, 2.25, 2.01, _
> 2.5, -0.5, 0.5, -1, -0.25, -1.25, -2.5)
>
> ' writing to the worksheet is not required
> ' but supports graphing the data to verify
>
> Cells(2,1) = "XP" : Cells(3, 1) = "YP"
> Cells(4,1) = "XP1" : Cells(5,1) = "YP1"
> Cells(2, 2).Resize(1, 20) = XP
> Cells(3, 2).Resize(1, 20) = YP
> Cells(4, 2).Resize(1, 12) = XP1
> Cells(5, 2).Resize(1, 12) = YP1
> j = 0
> For i = LBound(XP) To UBound(XP)
> poly(j).x = XP(i)
> poly(j).y = YP(i)
> j = j + 1
> Next
>
> j = 1
> For i = LBound(XP1) To UBound(XP1)
> x = XP1(i): y = YP1(i)
> b = pnpoly(20, poly, x, y)
> Cells(6, j).Offset(0, 1) = b
> j = j + 1
> Next
>
> End Sub
>
> Function pnpoly(npol As Integer, poly() As POINT, _
> x As Double, y As Double) As Boolean
> Dim inpoly As Boolean
> Dim i As Integer, j As Integer
>
> i = 0
> j = npol - 1
> inpoly = False
> While (i < npol)
> If ((poly(i).y <= y) And (y < poly(j).y)) Or _
> ((poly(j).y <= y) And (y < poly(i).y)) Then
> If x < (poly(j).x - poly(i).x) * (y - poly(i).y) / _
> (poly(j).y - poly(i).y) + poly(i).x Then
> inpoly = Not (inpoly)
> End If
> End If
> j = i
> i = i + 1
> Wend
> pnpoly = inpoly
> End Function
>
> Note that I have changed the function a little to support the data used.
> (integer=>double)
>
>
> See description at:
> http://tinyurl.com/y93rhr
>
> or "spelled out"
>
> http://209.85.165.104/search?q=cache...s&ct=clnk&cd=5
>
> --
> Regards,
> Tom Ogilvy
>
>
> "Brad" <(E-Mail Removed)> wrote in message
> news:(E-Mail Removed)...
>> Tom, thanks, but no, it has no declaration for "Point".... this is
>> actually just a bit of code that someone posted in response to a question
>> regarding determining whether an x-y point on a graph is inside or
>> outside of a convex polygon outlined on the same x-y graph.
>>
>> The whole thing looks like this:
>>
>> Function pnpoly(npol As Integer, poly() As Point, X As Integer, Y As
>> Integer) As Boolean
>> Dim inpoly As Boolean
>> Dim i, j As Integer
>>
>> i = 0
>> j = npol - 1
>> inpoly = False
>> While (i < npol)
>> If ((((poly(i).Y <= Y) And (Y < poly(j).Y)) Or _
>> ((poly(j).Y <= Y) And (Y < poly(i).Y)))) Then
>> If (X < (poly(j).X - poly(i).X) * (Y - poly(i).Y) /
>> (poly(j).Y - poly(i).Y) + poly(i).X) Then
>> inpoly = Not (inpoly)
>> End If
>> End If
>> j = i
>> i = i + 1
>> Wend
>> pnpoly = inpoly
>> End Function
>>
>> Now, what I was actually trying to do was take Andy Pope's interesting
>> "bounding areas" chart where he draws bounding lines around two
>> overlapping convex polygons, and simply count the number of x-y points
>> inside of and outside of a polygon. So, I hunted around and found this
>> code, referred to generally as the crossing number algorithm, for just
>> this sort of thing. Most often what I found was listed as pseudo-code,
>> but in this one case, a snippet of VBA code. On the psuedo-code, it does
>> have:
>>
>> typedef struct (int x, y Point;
>>
>> and I am certain this is most likely what you asked about.
>>
>> I am still not sure how to apply it in the function above.... thanks,
>> Brad
>>
>>
>>
>>
>> "Tom Ogilvy" <(E-Mail Removed)> wrote in message
>> news:0D6CB1C4-B930-4A41-97BE-(E-Mail Removed)...
>>> Does he have a declaration at the top like:
>>>
>>> Private Type POINT
>>> x As Long
>>> y As Long
>>> End Type
>>>
>>>
>>> Then point would be a structure that contains both and x and y
>>> coordinate.
>>> with the declaration at the top of the module, then
>>> Usage example:
>>>
>>> Dim a() as point
>>> for i = 1 to 10
>>> a(i).x = int(rnd()*100 + 1)
>>> a(i).y = int(rnd()*50 + 1)
>>> Next
>>>
>>> --
>>> Regards,
>>> Tom Ogilvy
>>>
>>>
>>>
>>> "Brad" wrote:
>>>
>>>> For a bit of code where the author defines a function like this:
>>>>
>>>> Function pnpoly(npol As Integer, poly() as Point, x as Integer, Y as
>>>> Integer) as Boolean,
>>>>
>>>> I am pretty sure the poly() input defined "as Point" is looking for an
>>>> xy
>>>> point value for an edge of a polygon, but I am unsure how it is
>>>> derived.
>>>>
>>>> Any ideas? My thanks in advance for any comments or suggestions... this
>>>> is a
>>>> function to identify whether an x-y data point falls inside of or
>>>> outside of
>>>> a convex polygon.
>>>>
>>>> Cheers! Brad
>>>>
>>>>
>>>>
>>
>>
>
>