another question is ,which algorithm used in ZDPointInPolygon?

[iknow34languages]

@BertoldVdb

Can you show me some docs to help me to read your code?

Thanks.

[iknow34languages]

@BertoldVdb i Think it's Ray casting algorithm ;

[BertoldVdb]

Hello,

It is the winding algorithm. Please check the point in polygon Wikipedia page.

Sincerely,
Bertold

[iknow34languages]

@BertoldVdb

Hello.

i saw that the winding algorithm is simple :

// wn_PnPoly(): winding number test for a point in a polygon
// Input: P = a point,
// V[] = vertex points of a polygon V[n+1] with V[n]=V[0]
// Return: wn = the winding number (=0 only when P is outside)
int
wn_PnPoly( Point P, Point* V, int n )
{
int wn = 0; // the winding number counter

// loop through all edges of the polygon
for (int i=0; i<n; i++) {   // edge from V[i] to  V[i+1]
    if (V[i].y <= P.y) {          // start y <= P.y
        if (V[i+1].y  > P.y)      // an upward crossing
             if (isLeft( V[i], V[i+1], P) > 0)  // P left of  edge
                 ++wn;            // have  a valid up intersect
    }
    else {                        // start y > P.y (no test needed)
        if (V[i+1].y  <= P.y)     // a downward crossing
             if (isLeft( V[i], V[i+1], P) < 0)  // P right of  edge
                 --wn;            // have  a valid down intersect
    }
}
return wn;

}
isLeft();
// isLeft(): test if a point is Left|On|Right of an infinite 2D line.
// Input: three points P0, P1, and P2
// Return: >0 for P2 left of the line through P0 to P1
// =0 for P2 on the line
// <0 for P2 right of the line
inline int
isLeft( Point P0, Point P1, Point P2 )
{
return ( (P1.x - P0.x) * (P2.y - P0.y)
- (P2.x - P0.x) * (P1.y - P0.y) );
}

your code seems much more difficult ;

Why?

[BertoldVdb]

Hello, The code is not really more difficult, just written in less concise way but the logic is the same. I also calculate the distance to the nearest edge to know when to do a new lookup. Sincerely, Bertold Van den Bergh

…

On Thu, Jul 26, 2018 at 1:34 PM, iknow34languages ***@***.***> wrote: @BertoldVdb <https://github.com/BertoldVdb> Hello. i saw that the winding algorithm is simple : // wn_PnPoly(): winding number test for a point in a polygon // Input: P = a point, // V[] = vertex points of a polygon V[n+1] with V[n]=V[0] // Return: wn = the winding number (=0 only when P is outside) int wn_PnPoly( Point P, Point* V, int n ) { int wn = 0; // the winding number counter // loop through all edges of the polygon for (int i=0; i<n; i++) { // edge from V[i] to V[i+1] if (V[i].y <= P.y) { // start y <= P.y if (V[i+1].y > P.y) // an upward crossing if (isLeft( V[i], V[i+1], P) > 0) // P left of edge ++wn; // have a valid up intersect } else { // start y > P.y (no test needed) if (V[i+1].y <= P.y) // a downward crossing if (isLeft( V[i], V[i+1], P) < 0) // P right of edge --wn; // have a valid down intersect } } return wn; } isLeft(); // isLeft(): test if a point is Left|On|Right of an infinite 2D line. // Input: three points P0, P1, and P2 // Return: >0 for P2 left of the line through P0 to P1 // =0 for P2 on the line // <0 for P2 right of the line inline int isLeft( Point P0, Point P1, Point P2 ) { return ( (P1.x - P0.x) * (P2.y - P0.y) - (P2.x - P0.x) * (P1.y - P0.y) ); } your code seems much more difficult ; Why? — You are receiving this because you were mentioned. Reply to this email directly, view it on GitHub <#5 (comment)>, or mute the thread <https://github.com/notifications/unsubscribe-auth/AGCFmZjcCHcFK9K0LFCfCe6IJMgEP7nKks5uKakvgaJpZM4VgP-S> .

[iknow34languages]

@BertoldVdb

ok! i ll learn your code carefully , you are my hero.~
👍 👍 👍 👍 👍

[BertoldVdb]

:)