I'm working on a program in which i want to store the distance the user walked since pressing a button. I retrieve the distance via geolocator package and display it on screen which works just fine.
I know there are some distanceBetween-Function for locations, but as far as i noticed, they are just calculating the distance between 2 points and not the actual distance the user walked (For example, if the user starts at one point X, walks over to Point Y and back to X would end in comparing start-and endpoint (X to X), which results in distance: 0, but i want the distance X -> Y -> X.
I added following function that calculated the distance based on longitude/latitude.
double distance(Position start, Position current){
return double.parse((acos(sin(start.latitude)*sin(current.latitude)+cos(start.latitude)*cos(current.latitude)*cos(current.longitude-start.longitude))*6371).toStringAsFixed(2));
}
I call it every frame and store the distance between the current and last gps position.
Works slowly but fine, except one Problem:
Somewhen, the double suddenly turns into "NaN", and i can't figure out why.
It's completely random when this occurs - At the beginning, it was always around 0.6, but it also occurred around 4.5 and 0.2, so i think the problem may be somewhere else.
Can anybody help?
Or does anybody knows a built-in-function that can solve the same problem?
I tried parsing the double to only have 2 decimal spaces (Didn't round it before) because i thought the number might just got too many decimal spaces to be displayed, but error still occured.
I have a second task that is happening at the same time each time stamp, so i thought it was hindering retrieving the GPS, so i tried disabling it, but it didn't change anything.
It's possible that you are getting numerical stability issues with the spherical law of cosines since you're calculating the distance on every frame? It is known that the formula has conditioning issues for very small distances (less than one meter).
Note that the domain for
arccosine(x) is given by -1 <= x <= 1. If in your case you were to supply a value greater than 1 (or smaller than -1) you would get a NaN result.
If you are still debugging this you can add a simple print statement:
double distance(Position start, Position current){
double x = sin(start.latitude)*sin(current.latitude)+cos(start.latitude)*cos(current.latitude)*cos(current.longitude-start.longitude);
if (x > 1 || x < -1) {
print("error");
}
return ((acos(sin(start.latitude)*sin(current.latitude)+cos(start.latitude)*cos(current.latitude)*cos(current.longitude-start.longitude))*6371));
}
If this is indeed the case, then you have a few options, use the Haversine formula because it is better conditioned for small distances, or simply set x to 1 if it's above 1. This anyway just means that the distance is zero.
For more information (and the Haversine formula) see also: Great circle distance
I really didn't think about the arccosines domain...
So i updated my code with your proposition to:
double distance(Position start, Position current) {
double x = sin(start.latitude) * sin(current.latitude) + cos(start.latitude) * cos(current.latitude) * cos(current.longitude - start.longitude);
if (x > 1 || x < -1) {
if (kDebugMode) {
print("error");
}
return 0;
}
return double.parse((acos(x) * 6371).toStringAsFixed(2));
}
It works fine, thank you for your help!
Related
I am using Google Maps SDK to allow a user to draw a polygon on the map by tapping. Everything works perfectly is the user is to draw a polygon following a path and continues on that path without crossing over lines. If that happens, this result is produced:
However, if the user is to make an error and cross over or change the direction of their "tapping" path this happens:
I need to either:
A) alert the user that they have created an invalid polygon, and must undo that action, or
B) correct the polygon shape to form a complete polygon.
With the research I have done, option A seems much more feasible and simple since option B would require rearranging the path of the polygon points.
I have done research and found algorithms and formulas to detect line intersection, but I am yet to find any piece of a solution in Swift to recognize if a polygon self-intersects based off of points (in this case, latitude and longitude). I don't need to know the point, just TRUE or FALSE to the question, "Does this Polygon self-intersect?" The polygon will typically have less than 20 sides.
Perhaps there is a solution built in with the GoogleMaps SDK, but I am yet to find it. Also, I understand that there are already algorithms for problems such as these, I am just having trouble implementing them into Swift 2 or 3. Any help is appreciated, thanks!
I'm guessing that you're trying to plot out the quickest way to get from point to point as the crow flies. You'll probably want to consider road direction too, which I won't here.
Both your options are possible. It's easy enough to iterate over every existing line when a new line is added and determine if they've crossed over. But your user would definitely rather not be told that they've screwed up, your app should just fix it for them. This is where it gets fun.
I am certain algorithms exist for finding the minimal polygon containing all points, but I didn't look them up, because where's the fun in that.
Here's how I would do it. In pseudocode:
if (line has intersected existing line)
find mean point (sum x sum y / n)
find nearest point to centre by:
taking min of: points.map(sqrt((x - centrex)^2 + (y-centrey)^2))
from the line between centre and nearest point, determine angle to every other line.
points.remove(nearest)
angles = points.map(cosine law(nearest to centre, centre, this point))
<- make sure to check if it crossed pi, at which point you must add pi.
sort angles so minimum is first.
starting at nearest point, add line to next point in the array of minimal angle points
I'm sorry I haven't put this into swift. I will update tomorrow with proper Swift 3.
This seems to be working pretty well for what I need. Adopted from Rob's answer here
func intersectionBetweenSegmentsCL(p0: CLLocationCoordinate2D, _ p1: CLLocationCoordinate2D, _ p2: CLLocationCoordinate2D, _ p3: CLLocationCoordinate2D) -> CLLocationCoordinate2D? {
var denominator = (p3.longitude - p2.longitude) * (p1.latitude - p0.latitude) - (p3.latitude - p2.latitude) * (p1.longitude - p0.longitude)
var ua = (p3.latitude - p2.latitude) * (p0.longitude - p2.longitude) - (p3.longitude - p2.longitude) * (p0.latitude - p2.latitude)
var ub = (p1.latitude - p0.latitude) * (p0.longitude - p2.longitude) - (p1.longitude - p0.longitude) * (p0.latitude - p2.latitude)
if (denominator < 0) {
ua = -ua; ub = -ub; denominator = -denominator
}
if ua >= 0.0 && ua <= denominator && ub >= 0.0 && ub <= denominator && denominator != 0 {
print("INTERSECT")
return CLLocationCoordinate2D(latitude: p0.latitude + ua / denominator * (p1.latitude - p0.latitude), longitude: p0.longitude + ua / denominator * (p1.longitude - p0.longitude))
}
return nil
}
I then implemented like this:
if coordArray.count > 2 {
let n = coordArray.count - 1
for i in 1 ..< n {
for j in 0 ..< i-1 {
if let intersection = intersectionBetweenSegmentsCL(coordArray[i], coordArray[i+1], coordArray[j], coordArray[j+1]) {
// do whatever you want with `intersection`
print("Error: Intersection # \(intersection)")
}
}
}
}
Check the following gif: https://i.gyazo.com/72998b8e2e3174193a6a2956de2ed008.gif
I want the cylinder to instantly change location to the nearest empty space on the plane as soon as I put a cube on the cylinder. The cubes and the cylinder have box colliders attached.
At the moment the cylinder just gets stuck when I put a cube on it, and I have to click in some direction to make it start "swimming" through the cubes.
Is there any easy solution or do I have to create some sort of grid with empty gameobjects that have a tag which tells me if there's an object on them or not?
This is a common problem in RTS-like video games, and I am solving it myself. This requires a breadth-first search algorithm, which means that you're checking the closest neighbors first. You're fortunate to only have to solve this problem in a gridded-environment.
Usually what programmers will do is create a queue and add each node (space) in the entire game to that queue until an empty space is found. It will start with e.g. the above, below, and adjacent spaces to the starting space, and then recursively move out, calling the same function inside of itself and using the queue to keep track of which spaces still need to be checked. It will also need to have a way to know whether a space has already been checked and avoid those spaces.
Another solution I'm conceiving of would be to generate a (conceptual) Archimedean spiral from the starting point and somehow check each space along that spiral. The tricky part would be generating the right spiral and checking it at just the right points in order to hit each space once.
Here's my quick-and-dirty solution for the Archimedean spiral approach in c++:
float x, z, max = 150.0f;
vector<pair<float, float>> spiral;
//Generate the spiral vector (run this code once and store the spiral).
for (float n = 0.0f; n < max; n += (max + 1.0f - n) * 0.0001f)
{
x = cos(n) * n * 0.05f;
z = sin(n) * n * 0.05f;
//Change 1.0f to 0.5f for half-sized spaces.
//fmod is float modulus (remainder).
x = x - fmod(x, 1.0f);
z = z - fmod(z, 1.0f);
pair<float, float> currentPoint = make_pair(x, z);
//Make sure this pair isn't at (0.0f, 0.0f) and that it's not already in the spiral.
if ((x != 0.0f || z != 0.0f) && find(spiral.begin(), spiral.end(), currentPoint) == spiral.end())
{
spiral.push_back(currentPoint);
}
}
//Loop through the results (run this code per usage of the spiral).
for (unsigned int n = 0U; n < spiral.size(); ++n)
{
//Draw or test the spiral.
}
It generates a vector of unique points (float pairs) that can be iterated through in order, which will allow you to draw or test every space around the starting space in a nice, outward (breadth-first), gridded spiral. With 1.0f-sized spaces, it generates a circle of 174 test points, and with 0.5f-sized spaces, it generates a circle of 676 test points. You only have to generate this spiral once and then store it for usage numerous times throughout the rest of the program.
Note:
This spiral samples differently as it grows further and further out from the center (in the for loop: n += (max + 1.0f - n) * 0.0001f).
If you use the wrong numbers, you could very easily break this code or cause an infinite loop! Use at your own risk.
Though more memory intensive, it is probably much more time-efficient than the traditional queue-based solutions due to iterating through each space exactly once.
It is not a 100% accurate solution to the problem, however, because it is a gridded spiral; in some cases it may favor the diagonal over the lateral. This is probably negligible in most cases though.
I used this solution for a game I'm working on. More on that here. Here are some pictures (the orange lines in the first are drawn by me in Paint for illustration, and the second picture is just to demonstrate what the spiral looks like if expanded):
I would like to have a function where I can input a radius value and have said function spit out the area for that size circle. The catch is I want it to do so for integer based coordinates only.
I was told elsewhere to look at Gauss's circle problem, which looks to be exactly what I'm interested in, but I don't really understand the math behind it (assuming it is actually accurate in calculating what I'm wanting).
As a side note, I currently use a modified circle drawing algorithm which does indeed produce the results I desire, but it just seems so incredibly inefficient (both the algorithm and the way in which I'm using it to get the area).
So, possible answers for this to me would be actual code or pseudocode for such a function if such a thing exists or something like a thorough explanation of Gauss's circle problem and why it is/isn't what I'm looking for.
The results I would hope the function would produce:
Input: Output
0: 1
1: 5
2: 13
3: 29
4: 49
5: 81
6: 113
7: 149
8: 197
9: 253
I too had to solve this problem recently and my initial approach was that of Numeron's - iterate on x axis from the center outwards and count the points within the upper right quarter, then quadruple them.
I then improved the algorithm around 3.4 times.
What I do now is just calculating how many points there are within an inscribed square inside that circle, and what's between that square and the edge of the circle (actually in the opposite order).
This way I actually count one-eighth of the points between the edge of the circle, the x axis and the right edge of the square.
Here's the code:
public static int gaussCircleProblem(int radius) {
int allPoints=0; //holds the sum of points
double y=0; //will hold the precise y coordinate of a point on the circle edge for a given x coordinate.
long inscribedSquare=(long) Math.sqrt(radius*radius/2); //the length of the side of an inscribed square in the upper right quarter of the circle
int x=(int)inscribedSquare; //will hold x coordinate - starts on the edge of the inscribed square
while(x<=radius){
allPoints+=(long) y; //returns floor of y, which is initially 0
x++; //because we need to start behind the inscribed square and move outwards from there
y=Math.sqrt(radius*radius-x*x); // Pythagorean equation - returns how many points there are vertically between the X axis and the edge of the circle for given x
}
allPoints*=8; //because we were counting points in the right half of the upper right corner of that circle, so we had just one-eightth
allPoints+=(4*inscribedSquare*inscribedSquare); //how many points there are in the inscribed square
allPoints+=(4*radius+1); //the loop and the inscribed square calculations did not touch the points on the axis and in the center
return allPoints;
}
Here's a picture to illustrate that:
Round down the length of the side of an inscribed square (pink) in the upper right quarter of the circle.
Go to next x coordinate behind the inscribed square and start counting orange points until you reach the edge.
Multiply the orange points by eight. This will give you the yellow
ones.
Square the pink points. This will give you the dark-blue ones. Then
multiply by four, this will get you the green ones.
Add the points on the axis and the one in the center. This gives you
the light-blue ones and the red one.
This is an old question but I was recently working on the same thing. What you are trying to do is as you said, Gauss's circle problem, which is sort of described here
While I too have difficulty understaning the serious maths behind it all, what it more or less pans out to when not using wierd alien symbols is this:
1 + 4 * sum(i=0, r^2/4, r^2/(4*i+1) - r^2/(4*i+3))
which in java at least is:
int sum = 0;
for(int i = 0; i <= (radius*radius)/4; i++)
sum += (radius*radius)/(4*i+1) - (radius*radius)/(4*i+3);
sum = sum * 4 + 1;
I have no idea why or how this works and to be honest Im a bit bummed I have to use a loop to get this out rather than a single line, as it means the performance is O(r^2/4) rather than O(1).
Since the math wizards can't seem to do better than a loop, I decided to see whether I could get it down to O(r + 1) performance, which I did. So don't use the above, use the below. O(r^2/4) is terrible and will be slower even despite mine using square roots.
int sum = 0;
for(int x = 0; x <= radius; x++)
sum += Math.sqrt(radius * radius - x * x);
sum = sum * 4 + 1;
What this code does is loop from centre out to the edge along an orthogonal line, and at each point adding the distance from line to edge in a perpendicualr direction. At the end it will have the number of points in a quater, so it quadruples the result and adds one because there is also central point. I feel like the wolfram equation does something similar, since it also multiplies by 4 and adds one, but IDK why it loops r^2/4.
Honestly these aren't great solution, but it seems to be the best there is. If you are calling a function which does this regularly then as new radii come up save the results in a look-up table rather than doing a full calc each time.
Its not a part of your question, but it may be relevant to someone maybe so I'll add it in anyway. I was personally working on finding all the points within a circle with cells defined by:
(centreX - cellX)^2 + (centreY - cellY)^2 <= radius^2 + radius
Which puts the whole thing out of whack because the extra +radius makes this not exactly the pythagorean theorem. That extra bit makes the circles look a whole lot more visually appealing on a grid though, as they don't have those little pimples on the orthogonal edges. It turns out that, yes my shape is still a circle, but its using sqrt(r^2+r) as radius instead of r, which apparently works but dont ask me how. Anyway that means that for me, my code is slightly different and looks more like this:
int sum = 0;
int compactR = ((radius * radius) + radius) //Small performance boost I suppose
for(int j = 0; j <= compactR / 4; j++)
sum += compactR / (4 * j + 1) - compactR / (4 * j + 3);
sum = sum * 4 + 1;
I got a line graph with Y axis having value and X axis having time. The X axis have 5 minute resolution. I'm looking for some kind of an algorithm to help me teach the iPhone to understand where the line is going. I've never taken an algorithms class, so any help would be appreciated. What I need to know is if the line has been rising for a certain number of segments continuously.
Right now I'm implementing the following:
If the current data point has Y value greater than the previous one, increment the slope counter by one. If it is equal, increment slope counter by 0. If the value is less, decrement the slope counter.
if(current>previous)
{
counter++;
}
else if(current<previous)
{
counter--;
}
This produces a sawtooth like graphs, which is easier to analyze. However due to the problems with the window size, the graph may "bounce". This is where I expect my logic to have problems.
I hope there's some kind of CS algorithm to help me with this task, as I don't even know what kinds of keywords to type into google for this problem.
If all you need to know is if the line has been rising for a certain number of segments continuously, why not just have a counter that increments until it hits that certain number of segments, or resets if the line goes down, like:
int counter = 0;
for (int i = 1; i < datasize; i++) {
if (data[i] > data[i - 1]) {
++counter;
if (counter == THRESHOLD) {
println("trending up at %d.", i);
}
} else if (data[i] < data[i - 1]) {
counter = 0;
}
}
If you're just looking to see if the line is trending up or down overall, could you just do this:
if (data[datasize - 1] > data[0]) {
println("Overall trend is up.");
} else if (data[datasize - 1] < data[0]) {
println("Overall trend is down.");
} else {
print("Overall trend is flat.");
}
If you want better prediction -- like, here's the line up to this point in time, here's a guess at what it's going to look like in the future, there are two avenues to explore. The first is "regression analysis" or "regression lines". This will work best for data which is generally increasing or decreasing as time goes on, and will get you the rate of those increases or decreases (the average slope of the line).
The second is "Fast Fourier Transform" - this is useful for lines which are like waves, in that they stay between a min and a max bound and have some regular cycle (or a number of regular cycles, which is what the equation will divine).
Have fun. This is an enjoyable problem to solve.
What you might be looking for is Linear Regression , which is estimating a good straight line match to your data (in one least squares sense). The slope of this line might help you tell "where it is going", depending on the behavior of the underlying model.
I'm solving the following problem: I have an object and I know its position now and its position 300ms ago. I assume the object is moving. I have a point to which I want the object to get.
What I need is to get the angle from my current object to the destination point in such a format that I know whether to turn left or right.
The idea is to assume the current angle from the last known position and the current position.
I'm trying to solve this in MATLAB. I've tried using several variations with atan2 but either I get the wrong angle in some situations (like when my object is going in circles) or I get the wrong angle in all situations.
Examples of code that screws up:
a = new - old;
b = dest - new;
alpha = atan2(a(2) - b(2), a(1) - b(1);
where new is the current position (eg. x = 40; y = 60; new = [x y];), old is the 300ms old position and dest is the destination point.
Edit
Here's a picture to demonstrate the problem with a few examples:
In the above image there are a few points plotted and annotated. The black line indicates our estimated current facing of the object.
If the destination point is dest1 I would expect an angle of about 88°.
If the destination point is dest2 I would expect an angle of about 110°.
If the destination point is dest3 I would expect an angle of about -80°.
Firstly, you need to note the scale on the sample graph you show above. The x-axis ticks move in steps of 1, and the y-axis ticks move in steps of 20. The picture with the two axes appropriately scaled (like with the command axis equal) would be a lot narrower than you have, so the angles you expect to get are not right. The expected angles will be close to right angles, just a few degrees off from 90 degrees.
The equation Nathan derives is valid for column vector inputs a and b:
theta = acos(a'*b/(sqrt(a'*a) * sqrt(b'*b)));
If you want to change this equation to work with row vectors, you would have to switch the transpose operator in both the calculation of the dot product as well as the norms, like so:
theta = acos(a*b'/(sqrt(a*a') * sqrt(b*b')));
As an alternative, you could just use the functions DOT and NORM:
theta = acos(dot(a,b)/(norm(a)*norm(b)));
Finally, you have to account for the direction, i.e. whether the angle should be positive (turn clockwise) or negative (turn counter-clockwise). You can do this by computing the sign of the z component for the cross product of b and a. If it's positive, the angle should be positive. If it's negative, the angle should be negative. Using the function SIGN, our new equation becomes:
theta = sign(b(1)*a(2)-b(2)*a(1)) * acos(dot(a,b)/(norm(a)*norm(b)));
For your examples, the above equation gives an angle of 88.85, 92.15, and -88.57 for your three points dest1, dest2, and dest3.
NOTE: One special case you will need to be aware of is if your object is moving directly away from the destination point, i.e. if the angle between a and b is 180 degrees. In such a case you will have to pick an arbitrary turn direction (left or right) and a number of degrees to turn (180 would be ideal ;) ). Here's one way you could account for this condition using the function EPS:
theta = acos(dot(a,b)/(norm(a)*norm(b))); %# Compute theta
if abs(theta-pi) < eps %# Check if theta is within some tolerance of pi
%# Pick your own turn direction and amount here
else
theta = sign(b(1)*a(2)-b(2)*a(1))*theta; %# Find turn direction
end
You can try using the dot-product of the vectors.
Define the vectors 'a' and 'b' as:
a = new - old;
b = dest - new;
and use the fact that the dot product is:
a dot b = norm2(a) * norm2(b) * cos(theta)
where theta is the angle between two vectors, and you get:
cos(theta) = (a dot b)/ (norm2(a) * norm2(b))
The best way to calculate a dot b, assuming they are column vectors, is like this:
a_dot_b = a'*b;
and:
norm2(a) = sqrt(a'*a);
so you get:
cos(theta) = a'*b/(sqrt((a'*a)) * sqrt((b'*b)))
Depending on the sign of the cosine you either go left or right
Essentially you have a line defined by the points old and new and wish to determine if dest is on right or the left of that line? In which case have a look at this previous question.