Im using Google Map on Xamarin Android and iOS
Im trying to create a close polygon from random tap coordinates.
Android like:
var options = new PolygonOptions();
options.InvokeFillColor(int.Parse("80000000", System.Globalization.NumberStyles.HexNumber));
options.InvokeStrokeColor(int.Parse("000000", System.Globalization.NumberStyles.HexNumber));
options.InvokeStrokeWidth(2);
options.AddAll(listOfPosition); //list of tapped coordinates
googleMap.AddPolygon(options);
iOS like:
var polygon = new Polygon();
polygon.Path = path; // this is my list of tapped coordinates
polygon.StrokeWidth = 2;
polygon.StrokeColor = UIColor.Black;
polygon.FillColor = UIColor.FromRGBA(0, 0, 0, 0.3f);
polygon.Map = mapView;
Let say I have tapped coordinates randomly like this :
Debug|LOCATION:: 37.9115841998119 : -122.566957734525
Debug|LOCATION:: 37.9117754470967 : -122.561504803598
Debug|LOCATION:: 37.9085008969633 : -122.566276118159
Debug|LOCATION:: 37.9086561762004 : -122.562640719116
Debug|LOCATION:: 37.9102216477146 : -122.561550065875
Debug|LOCATION:: 37.9098513127501 : -122.567805983126
Debug|LOCATION:: 37.9123967989511 : -122.564518935978
How can I make it close polygon without intersecting lines inside even may tap coordinates reached 100 position data.? Any Idea or Computation? Thanks advance..
You don't have to compute locally, Use Shapes API of Google maps
https://developer.xamarin.com/guides/android/platform_features/maps_and_location/maps/part_2_-_maps_api/#Drawing_on_the_Map
https://developers.google.com/maps/documentation/android-api/shapes
To sort the points locally for closed polygon
https://stackoverflow.com/a/7369943/2700586
Reference:
http://geomalgorithms.com/a10-_hull-1.html
Here's the solution :)
You need to get the most Left and most Right on Lat and Lon.
get the min and max latitude.
Draw a line and seperate first the lower points and the higher points
Check points if its on higher or lower by equation of line.
public static bool CheckIsUpperCoordinate(PositionData a, PositionData b, PositionData c)
{
var position = Math.Sign((b.Latitude - a.Latitude) * (c.Longitude - a.Longitude) - (b.Longitude - a.Longitude) * (c.Latitude - a.Latitude));
return position > 0 ? true : false; //true if upper , false if lower position
}
Now sort lowerlist and upperlist by its longitude.
var lowerPosition = lowerPositionList.OrderBy(x => x.Longitude).ToList();
var higherPosition = higherPositionList.OrderByDescending(x => x.Longitude).ToList();
And connect all points from this sequence left most -> lower points list -> right most -> upper points list
path.AddLatLon(leftMost.Latitude, leftMost.Longitude);
foreach (PositionData lowerData in lowerPosition)
{
path.AddLatLon(lowerData.Latitude, lowerData.Longitude);
}
path.AddLatLon(rightMost.Latitude, rightMost.Longitude);
foreach (PositionData higherData in higherPosition)
{
path.AddLatLon(higherData.Latitude, higherData.Longitude);
}
path.AddLatLon(leftMost.Latitude, leftMost.Longitude);
wrap it on polygon
var polygon = new Polygon();
polygon.Path = path;
polygon.StrokeWidth = 2;
polygon.StrokeColor = UIColor.Black;
polygon.FillColor = UIColor.FromRGBA(0, 0, 0, 0.3f);
polygon.Map = mapView;
Hope it may help someone :)
I appreciated my Math Teacher after this task. lol xd
From the man page for XFillPolygon:
If shape is Complex, the path may self-intersect. Note that contiguous coincident points in the path are not treated as self-intersection.
If shape is Convex, for every pair of points inside the polygon, the line segment connecting them does not intersect the path. If known by the client, specifying Convex can improve performance. If you specify Convex for a path that is not convex, the graphics results are undefined.
If shape is Nonconvex, the path does not self-intersect, but the shape is not wholly convex. If known by the client, specifying Nonconvex instead of Complex may improve performance. If you specify Nonconvex for a self-intersecting path, the graphics results are undefined.
I am having performance problems with fill XFillPolygon and, as the man page suggests, the first step I want to take is to specify the correct shape of the polygon. I am currently using Complex to be on the safe side.
Is there an efficient algorithm to determine if a polygon (defined by a series of coordinates) is convex, non-convex or complex?
You can make things a lot easier than the Gift-Wrapping Algorithm... that's a good answer when you have a set of points w/o any particular boundary and need to find the convex hull.
In contrast, consider the case where the polygon is not self-intersecting, and it consists of a set of points in a list where the consecutive points form the boundary. In this case it is much easier to figure out whether a polygon is convex or not (and you don't have to calculate any angles, either):
For each consecutive pair of edges of the polygon (each triplet of points), compute the z-component of the cross product of the vectors defined by the edges pointing towards the points in increasing order. Take the cross product of these vectors:
given p[k], p[k+1], p[k+2] each with coordinates x, y:
dx1 = x[k+1]-x[k]
dy1 = y[k+1]-y[k]
dx2 = x[k+2]-x[k+1]
dy2 = y[k+2]-y[k+1]
zcrossproduct = dx1*dy2 - dy1*dx2
The polygon is convex if the z-components of the cross products are either all positive or all negative. Otherwise the polygon is nonconvex.
If there are N points, make sure you calculate N cross products, e.g. be sure to use the triplets (p[N-2],p[N-1],p[0]) and (p[N-1],p[0],p[1]).
If the polygon is self-intersecting, then it fails the technical definition of convexity even if its directed angles are all in the same direction, in which case the above approach would not produce the correct result.
This question is now the first item in either Bing or Google when you search for "determine convex polygon." However, none of the answers are good enough.
The (now deleted) answer by #EugeneYokota works by checking whether an unordered set of points can be made into a convex polygon, but that's not what the OP asked for. He asked for a method to check whether a given polygon is convex or not. (A "polygon" in computer science is usually defined [as in the XFillPolygon documentation] as an ordered array of 2D points, with consecutive points joined with a side as well as the last point to the first.) Also, the gift wrapping algorithm in this case would have the time-complexity of O(n^2) for n points - which is much larger than actually needed to solve this problem, while the question asks for an efficient algorithm.
#JasonS's answer, along with the other answers that follow his idea, accepts star polygons such as a pentagram or the one in #zenna's comment, but star polygons are not considered to be convex. As
#plasmacel notes in a comment, this is a good approach to use if you have prior knowledge that the polygon is not self-intersecting, but it can fail if you do not have that knowledge.
#Sekhat's answer is correct but it also has the time-complexity of O(n^2) and thus is inefficient.
#LorenPechtel's added answer after her edit is the best one here but it is vague.
A correct algorithm with optimal complexity
The algorithm I present here has the time-complexity of O(n), correctly tests whether a polygon is convex or not, and passes all the tests I have thrown at it. The idea is to traverse the sides of the polygon, noting the direction of each side and the signed change of direction between consecutive sides. "Signed" here means left-ward is positive and right-ward is negative (or the reverse) and straight-ahead is zero. Those angles are normalized to be between minus-pi (exclusive) and pi (inclusive). Summing all these direction-change angles (a.k.a the deflection angles) together will result in plus-or-minus one turn (i.e. 360 degrees) for a convex polygon, while a star-like polygon (or a self-intersecting loop) will have a different sum ( n * 360 degrees, for n turns overall, for polygons where all the deflection angles are of the same sign). So we must check that the sum of the direction-change angles is plus-or-minus one turn. We also check that the direction-change angles are all positive or all negative and not reverses (pi radians), all points are actual 2D points, and that no consecutive vertices are identical. (That last point is debatable--you may want to allow repeated vertices but I prefer to prohibit them.) The combination of those checks catches all convex and non-convex polygons.
Here is code for Python 3 that implements the algorithm and includes some minor efficiencies. The code looks longer than it really is due to the the comment lines and the bookkeeping involved in avoiding repeated point accesses.
TWO_PI = 2 * pi
def is_convex_polygon(polygon):
"""Return True if the polynomial defined by the sequence of 2D
points is 'strictly convex': points are valid, side lengths non-
zero, interior angles are strictly between zero and a straight
angle, and the polygon does not intersect itself.
NOTES: 1. Algorithm: the signed changes of the direction angles
from one side to the next side must be all positive or
all negative, and their sum must equal plus-or-minus
one full turn (2 pi radians). Also check for too few,
invalid, or repeated points.
2. No check is explicitly done for zero internal angles
(180 degree direction-change angle) as this is covered
in other ways, including the `n < 3` check.
"""
try: # needed for any bad points or direction changes
# Check for too few points
if len(polygon) < 3:
return False
# Get starting information
old_x, old_y = polygon[-2]
new_x, new_y = polygon[-1]
new_direction = atan2(new_y - old_y, new_x - old_x)
angle_sum = 0.0
# Check each point (the side ending there, its angle) and accum. angles
for ndx, newpoint in enumerate(polygon):
# Update point coordinates and side directions, check side length
old_x, old_y, old_direction = new_x, new_y, new_direction
new_x, new_y = newpoint
new_direction = atan2(new_y - old_y, new_x - old_x)
if old_x == new_x and old_y == new_y:
return False # repeated consecutive points
# Calculate & check the normalized direction-change angle
angle = new_direction - old_direction
if angle <= -pi:
angle += TWO_PI # make it in half-open interval (-Pi, Pi]
elif angle > pi:
angle -= TWO_PI
if ndx == 0: # if first time through loop, initialize orientation
if angle == 0.0:
return False
orientation = 1.0 if angle > 0.0 else -1.0
else: # if other time through loop, check orientation is stable
if orientation * angle <= 0.0: # not both pos. or both neg.
return False
# Accumulate the direction-change angle
angle_sum += angle
# Check that the total number of full turns is plus-or-minus 1
return abs(round(angle_sum / TWO_PI)) == 1
except (ArithmeticError, TypeError, ValueError):
return False # any exception means not a proper convex polygon
The following Java function/method is an implementation of the algorithm described in this answer.
public boolean isConvex()
{
if (_vertices.size() < 4)
return true;
boolean sign = false;
int n = _vertices.size();
for(int i = 0; i < n; i++)
{
double dx1 = _vertices.get((i + 2) % n).X - _vertices.get((i + 1) % n).X;
double dy1 = _vertices.get((i + 2) % n).Y - _vertices.get((i + 1) % n).Y;
double dx2 = _vertices.get(i).X - _vertices.get((i + 1) % n).X;
double dy2 = _vertices.get(i).Y - _vertices.get((i + 1) % n).Y;
double zcrossproduct = dx1 * dy2 - dy1 * dx2;
if (i == 0)
sign = zcrossproduct > 0;
else if (sign != (zcrossproduct > 0))
return false;
}
return true;
}
The algorithm is guaranteed to work as long as the vertices are ordered (either clockwise or counter-clockwise), and you don't have self-intersecting edges (i.e. it only works for simple polygons).
Here's a test to check if a polygon is convex.
Consider each set of three points along the polygon--a vertex, the vertex before, the vertex after. If every angle is 180 degrees or less you have a convex polygon. When you figure out each angle, also keep a running total of (180 - angle). For a convex polygon, this will total 360.
This test runs in O(n) time.
Note, also, that in most cases this calculation is something you can do once and save — most of the time you have a set of polygons to work with that don't go changing all the time.
To test if a polygon is convex, every point of the polygon should be level with or behind each line.
Here's an example picture:
The answer by #RoryDaulton
seems the best to me, but what if one of the angles is exactly 0?
Some may want such an edge case to return True, in which case, change "<=" to "<" in the line :
if orientation * angle < 0.0: # not both pos. or both neg.
Here are my test cases which highlight the issue :
# A square
assert is_convex_polygon( ((0,0), (1,0), (1,1), (0,1)) )
# This LOOKS like a square, but it has an extra point on one of the edges.
assert is_convex_polygon( ((0,0), (0.5,0), (1,0), (1,1), (0,1)) )
The 2nd assert fails in the original answer. Should it?
For my use case, I would prefer it didn't.
This method would work on simple polygons (no self intersecting edges) assuming that the vertices are ordered (either clockwise or counter)
For an array of vertices:
vertices = [(0,0),(1,0),(1,1),(0,1)]
The following python implementation checks whether the z component of all the cross products have the same sign
def zCrossProduct(a,b,c):
return (a[0]-b[0])*(b[1]-c[1])-(a[1]-b[1])*(b[0]-c[0])
def isConvex(vertices):
if len(vertices)<4:
return True
signs= [zCrossProduct(a,b,c)>0 for a,b,c in zip(vertices[2:],vertices[1:],vertices)]
return all(signs) or not any(signs)
I implemented both algorithms: the one posted by #UriGoren (with a small improvement - only integer math) and the one from #RoryDaulton, in Java. I had some problems because my polygon is closed, so both algorithms were considering the second as concave, when it was convex. So i changed it to prevent such situation. My methods also uses a base index (which can be or not 0).
These are my test vertices:
// concave
int []x = {0,100,200,200,100,0,0};
int []y = {50,0,50,200,50,200,50};
// convex
int []x = {0,100,200,100,0,0};
int []y = {50,0,50,200,200,50};
And now the algorithms:
private boolean isConvex1(int[] x, int[] y, int base, int n) // Rory Daulton
{
final double TWO_PI = 2 * Math.PI;
// points is 'strictly convex': points are valid, side lengths non-zero, interior angles are strictly between zero and a straight
// angle, and the polygon does not intersect itself.
// NOTES: 1. Algorithm: the signed changes of the direction angles from one side to the next side must be all positive or
// all negative, and their sum must equal plus-or-minus one full turn (2 pi radians). Also check for too few,
// invalid, or repeated points.
// 2. No check is explicitly done for zero internal angles(180 degree direction-change angle) as this is covered
// in other ways, including the `n < 3` check.
// needed for any bad points or direction changes
// Check for too few points
if (n <= 3) return true;
if (x[base] == x[n-1] && y[base] == y[n-1]) // if its a closed polygon, ignore last vertex
n--;
// Get starting information
int old_x = x[n-2], old_y = y[n-2];
int new_x = x[n-1], new_y = y[n-1];
double new_direction = Math.atan2(new_y - old_y, new_x - old_x), old_direction;
double angle_sum = 0.0, orientation=0;
// Check each point (the side ending there, its angle) and accum. angles for ndx, newpoint in enumerate(polygon):
for (int i = 0; i < n; i++)
{
// Update point coordinates and side directions, check side length
old_x = new_x; old_y = new_y; old_direction = new_direction;
int p = base++;
new_x = x[p]; new_y = y[p];
new_direction = Math.atan2(new_y - old_y, new_x - old_x);
if (old_x == new_x && old_y == new_y)
return false; // repeated consecutive points
// Calculate & check the normalized direction-change angle
double angle = new_direction - old_direction;
if (angle <= -Math.PI)
angle += TWO_PI; // make it in half-open interval (-Pi, Pi]
else if (angle > Math.PI)
angle -= TWO_PI;
if (i == 0) // if first time through loop, initialize orientation
{
if (angle == 0.0) return false;
orientation = angle > 0 ? 1 : -1;
}
else // if other time through loop, check orientation is stable
if (orientation * angle <= 0) // not both pos. or both neg.
return false;
// Accumulate the direction-change angle
angle_sum += angle;
// Check that the total number of full turns is plus-or-minus 1
}
return Math.abs(Math.round(angle_sum / TWO_PI)) == 1;
}
And now from Uri Goren
private boolean isConvex2(int[] x, int[] y, int base, int n)
{
if (n < 4)
return true;
boolean sign = false;
if (x[base] == x[n-1] && y[base] == y[n-1]) // if its a closed polygon, ignore last vertex
n--;
for(int p=0; p < n; p++)
{
int i = base++;
int i1 = i+1; if (i1 >= n) i1 = base + i1-n;
int i2 = i+2; if (i2 >= n) i2 = base + i2-n;
int dx1 = x[i1] - x[i];
int dy1 = y[i1] - y[i];
int dx2 = x[i2] - x[i1];
int dy2 = y[i2] - y[i1];
int crossproduct = dx1*dy2 - dy1*dx2;
if (i == base)
sign = crossproduct > 0;
else
if (sign != (crossproduct > 0))
return false;
}
return true;
}
For a non complex (intersecting) polygon to be convex, vector frames obtained from any two connected linearly independent lines a,b must be point-convex otherwise the polygon is concave.
For example the lines a,b are convex to the point p and concave to it below for each case i.e. above: p exists inside a,b and below: p exists outside a,b
Similarly for each polygon below, if each line pair making up a sharp edge is point-convex to the centroid c then the polygon is convex otherwise it’s concave.
blunt edges (wronged green) are to be ignored
N.B
This approach would require you compute the centroid of your polygon beforehand since it doesn’t employ angles but vector algebra/transformations
Adapted Uri's code into matlab. Hope this may help.
Be aware that Uri's algorithm only works for simple polygons! So, be sure to test if the polygon is simple first!
% M [ x1 x2 x3 ...
% y1 y2 y3 ...]
% test if a polygon is convex
function ret = isConvex(M)
N = size(M,2);
if (N<4)
ret = 1;
return;
end
x0 = M(1, 1:end);
x1 = [x0(2:end), x0(1)];
x2 = [x0(3:end), x0(1:2)];
y0 = M(2, 1:end);
y1 = [y0(2:end), y0(1)];
y2 = [y0(3:end), y0(1:2)];
dx1 = x2 - x1;
dy1 = y2 - y1;
dx2 = x0 - x1;
dy2 = y0 - y1;
zcrossproduct = dx1 .* dy2 - dy1 .* dx2;
% equality allows two consecutive edges to be parallel
t1 = sum(zcrossproduct >= 0);
t2 = sum(zcrossproduct <= 0);
ret = t1 == N || t2 == N;
end
I have a trouble when I try locate the sprite more near of my player. I have a little map and side scroll. only have a grow and a player on middle. He can looks and walk left or right, and if your player looks left and you press "shot", I search on my linkedlist of mobs a the most near Mob of my player and I do this mob receibe the damage.
I have this code:
- mobs is my linkedlist of mobs. (Extends of AnimatedSprite)
This is my first game, and I dont know if there are a better method of do this, this dont search the more near, only the first element of my list, any idea? :)
public void shot(){
float playerx = player.getX();
Mob target = mobs.element();
if(player.getDireccion()==Entidad.DIR_IZQUIERDA){//If direction if left
for(Mob z:mobs){
if(z.getX()<playerx &&
z.getX()>target.getX())
target= z;
}
}else if(player.getDireccion()==Entidad.DIR_DERECHA){//If direction is right
for(Mob z:mobs){
if(z.getX()>playerx && z.getX()<target.getX())
target= z;
}
}
target.recibeDaño();//receibe damaget (loss life basically)
if(objetivo.getVida()<=0){ //These delete body and sprite of the game
final Mob eliminar = target;
eliminarZombie(eliminar,this);
mobs.remove(target);
System.gc();
}
}
Sorry my english.
loop over all enemies and calculate the distance which is
distance = x2 - x1
where x2 is the x attribute of the enemy and x1 is the x attribute of the player
only take into account the positive distances if you are facing right , and only take the negative distances if you are facing left
then chose the smallest absolute value of the distance
so it's something like this
float shortest = 1000; //just put a large number here
for(Mob z:mobs){
distance = z.x - player.x;
if((player.getDirection == Direction.RIGHT) && distance > 0 && distance < shortest){
//set the target to the current mob and change the value of the shortest
}
if((player.getDirection == Direction.LEFT) && distance < 0 && Math.abs(distance) < shortest){
//same as above
}
}
//damage the target here and remove it
just notice that I didn't take care of the case where the target is right on top of the enemy and the distance == 0
your code should work too , but instead of looping over the list twice , it's better to loop once
other notes on your code are:
don't call System.gc(), not a good idea
if you want to remove an item from a linked list , remove it on updateThread using this.runOnUpdateThread() and give it a runnable [if you don't do it this way your game will crash randomly]
you might also use an iterator to go over the items , then use .remove() to remove the latest item you reached , it's thread safe
I am trying to solve a problem with drawing a path from huge (100k+) set of GeoPoints to a MapView on Android.
Firstly I would like to say, I searched through StackOverflow a lot and haven't found an answer.The bottleneck of my code is not actually drawing into canvas, but Projection.toPixels(GeoPoint, Point) or Rect.contains(point.x, point.y) method..I am skipping points not visible on screen and also displaying only every nth point according to current zoom-level. When the map is zoomed-in I want to display as accurate path as possible so I skipping zero (or nearly to zero) points, so that when finding visible points I need to call the projection method for every single point in the collection. And that is what really takes a lot of time (not seconds, but map panning is not fluid and I am not testing it on HTC Wildfire:)). I tried caching calculated points, but since points be recalculated after every map pan/zoom it haven't helped
at all.
I thought about usage of some kind of prune and search algorithm instead of iterate the array, but I figured out the input data is not sorted (I can't throw away any branch stacked between two invisible points). That could I possible solve with simple sort at the beginning, but I am still not sure even the logarithmic count of getProjection() and Rect.contains(point.x, point.y) calls instead of linear would solve the performance problem.
Bellow is my current code. Please help me if you know how to make this better. Thanks a lot!
public void drawPath(MapView mv, Canvas canvas) {
displayed = false;
tmpPath.reset();
int zoomLevel = mapView.getZoomLevel();
int skippedPoints = (int) Math.pow(2, (Math.max((19 - zoomLevel), 0)));
int mPointsSize = mPoints.size();
int mPointsLastIndex = mPointsSize - 1;
int stop = mPointsLastIndex - skippedPoints;
mapView.getDrawingRect(currentMapBoundsRect);
Projection projection = mv.getProjection();
for (int i = 0; i < mPointsSize; i += skippedPoints) {
if (i > stop) {
break;
}
//HERE IS THE PROBLEM I THINK - THIS METHOD AND THE IF CONDITION BELOW
projection.toPixels(mPoints.get(i), point);
if (currentMapBoundsRect.contains(point.x, point.y)) {
if (!displayed) {
Point tmpPoint = new Point();
projection.toPixels(mPoints.get(Math.max(i - 1, 0)),
tmpPoint);
tmpPath.moveTo(tmpPoint.x, tmpPoint.y);
tmpPath.lineTo(point.x, point.y);
displayed = true;
} else {
tmpPath.lineTo(point.x, point.y);
}
} else if (displayed) {
tmpPath.lineTo(point.x, point.y);
displayed = false;
}
}
canvas.drawPath(tmpPath, this.pathPaint);
}
So I figured out how to make it all much faster!
I will post it here, somebody could possibly found it useful in the future.
It has emerged that usage of projection.toPixels() can really harm application performance. So I figured out that way better than take every single GeoPoint, convert it to Point and then check if it is contained in map viewport is, when I count actuall viewport radius of the map as following:
mapView.getGlobalVisibleRect(currentMapBoundsRect);
GeoPoint point1 = projection.fromPixels(currentMapBoundsRect.centerX(), currentMapBoundsRect.centerY());
GeoPoint point2 = projection.fromPixels(currentMapBoundsRect.left, currentMapBoundsRect.top);
float[] results2 = new float[3];
Location.distanceBetween(point1.getLatitudeE6()/1E6, point1.getLongitudeE6()/1E6, point2.getLatitudeE6()/1E6, point2.getLongitudeE6()/1E6, results2);
The radius is in results2[0]..
Then I can take every single GeoPoint and count the distance between it and the center of the map mapView.getMapCenter(). Then I can compare the radius with computed distance and decide whether ot not diplay the point.
So that's it, hope It will be helpful.