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" wrote in message
...
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...ct=clnk& cd=5

--
Regards,
Tom Ogilvy


"Brad" wrote in message
...
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" wrote in message
...
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" wrote in message
...
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...ct=clnk& cd=5

--
Regards,
Tom Ogilvy


"Brad" wrote in message
...
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" wrote in message
...
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