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How To Iterate Through Every Latitude/Longitude in Java

I'm accessing this API that gives me global weather:
https://callforcode.weather.com/doc/v3-global-weather-notification-headlines/
However, it takes lat/lng as an input parameter, and I need the data for the entire world.
I figure I can loop through every latitude longitude, every 2 latitudes and 2 longitude degrees, giving me a point on the world, every ~120 miles across and roughly 100 degrees north/south, which should give me all the data in 16,200 API calls ((360/2) * (180/2)).
How can I do this effectively in Java?
I'd conceived something like this; but is there a better way of doing this?
for(int i = 0; i < 360; i+2){
var la = i;
for(int x = 0 x < 180; x+2) {
var ln = x;
//call api with lat = i, lng = x;
}
}

It's somewhat of a paradigm shift, but I would NOT use a nested for-loop for this problem. In many situations where you are looking at iterating over an entire result set, it is often possible to trim the coverage dramatically without losing much or any effectiveness. Caching, trimming, prioritizing... these are the things you need: not a for-loop.
Cut sections entirely - maybe you can ignore ocean, maybe you can ignore Antartica and the North Pole (since people there have better ways of checking weather anyway)
Change your search frequency based on population density. Maybe northern Canada doesn't need to be checked as thoroughly as Los Angeles or Chicago.
Rely on caching in low-usage areas - presumably you can track what areas are actually being used and can then more frequently refresh those sections.
So what you end up with is some sort of weighted caching system that takes into account population density, usage patterns, and other priorities to determine what latitude/longitude coordinates to check and how frequently.
High-level code might look something like this:
void executeUpdateSweep(List<CoordinateCacheItem> cacheItems)
{
for(CoordinateCacheItem item : cacheItems)
{
if(shouldRefreshCache(item))
{
//call api with lat = item.y , lng = item.x
}
}
}
boolean shouldRefreshCache(item)
{
long ageWeight = calculateAgeWeight(item);//how long since last update?
long basePopulationWeight = item.getBasePopulationWeight();//how many people (users and non-users) live here?
long usageWeight = calculateUsageWeight(item);//how much is this item requested?
return ageWeight + basePopulationWeight + usageWeight > someArbitraryThreshold;
}

Java HashMap of Vectors

i'm in a situation in which i'd like to have an hash map with an Integer value as key and an array of Double as values (of which i know the length).
So i want to have something like
HashMap<Integer, Double[]> hash = new HashMap<Integer, Double[]>();
Next i scan a ResultSet finding key,value for the first attribute.
At the end of this first phase i will have an hash map with some keys and for each key i will have a Double value representing a particular score.
Next i want to scan another different ResultSet with different key,values and i want to populate my hash with these values.
The problem is that here i can find element for which i don't have an entry and documents for which i already have an entry.
I'd like to arrive to a situation in which for a particular key i can access all different scores.
How can i add iteratively values to those arrays? because if i use the usual hash.put(key,value) i have to use a Double[] as value but i want to add to the hash map different score iteratively.
I think that using a Vector can bring me some problems due to the fact that some keys can have some empty values for which they don't will be populate.

I’m uncertain how well I understand your requirements. I think you may do something sort of:
double[] arrayInMap = hash.get(key);
if (arrayInMap == null) {
hash.put(key, valuesToAdd);
} else {
if (arrayInMap.length != valuesToAdd.length) {
throw new IllegalStateException("Key "+ key + ": cannot add " + valuesToAdd.length
+ " values to array of length " + arrayInMap.length);
}
for (int ix = 0; ix < arrayInMap.length; ix++) {
arrayInMap[ix] += valuesToAdd[ix];
}
}
I hope you’ll at least be able to use it as inspiration.

How to efficiently remove duplicate collision pairs in spatial hash grid?

I'm working on a 2D game for android so performance is a real issue and a must. In this game there might occur a lot of collisions between any objects and I don't want to check in bruteforce o(n^2) whether any gameobject collides with another one. In order to reduce the possible amount of collision checks I decided to use spatial hashing as broadphase algorithm becouse it seems quite simple and efficient - dividing the scene on rows and columns and checking collisions between objects residing only in the same grid element.
Here's the basic concept I quickly scratched:
public class SpatialHashGridElement
{
HashSet<GameObject> gameObjects = new HashSet<GameObject>();
}
static final int SPATIAL_HASH_GRID_ROWS = 4;
static final int SPATIAL_HASH_GRID_COLUMNS = 5;
static SpatialHashGridElement[] spatialHashGrid = new SpatialHashGridElement[SPATIAL_HASH_GRID_ROWS * SPATIAL_HASH_GRID_COLUMNS];
void updateGrid()
{
float spatialHashGridElementWidth = screenWidth / SPATIAL_HASH_GRID_COLUMNS;
float spatialHashGridElementHeight = screenHeight / SPATIAL_HASH_GRID_ROWS;
for(SpatialHashGridElement e : spatialHashGrid)
e.gameObjects.clear();
for(GameObject go : displayList)
{
for(int i = 0; i < go.vertices.length/3; i++)
{
int row = (int) Math.abs(((go.vertices[i*3 + 1] / spatialHashGridElementHeight) % SPATIAL_HASH_GRID_ROWS));
int col = (int) Math.abs(((go.vertices[i*3 + 0] / spatialHashGridElementWidth) % SPATIAL_HASH_GRID_COLUMNS));
if(!spatialHashGrid[row * SPATIAL_HASH_GRID_COLUMNS + col].gameObjects.contains(go))
spatialHashGrid[row * SPATIAL_HASH_GRID_COLUMNS + col].gameObjects.add(go);
}
}
}
The code isn't probably of the highest quality so if you spot anything to improve please don't hesitate to tell me but the most worrying problem that arises currently is that in 2 grid cells there might be same collision pairs checked. Worst case example (assuming none of the objects spans more than 2 cells):
Here we have 2 gameObjects colliding(red and blue). Each of them resides in 4 cells => therefore in each cell there will be the same pair to check.
I can't come up with some efficient approach to remove the possibility of duplicate pairs without a need to filter the grid after creating it in updateGrid(). Is there some brilliant way to detect that some collision pair has been already inserted even during the updateGrid function? I will be very grateful for any tips!

I'm trying to explain my idea using some pseudo-code (C# language elements):
public partial class GameObject {
// ...
Set<GameObject> collidedSinceLastTick = new HashSet<GameObject>();
public boolean collidesWith(GameObject other) {
if (collidedSinceLastTick.contains(other)) {
return true; // or even false, see below
}
boolean collided = false;
// TODO: your costly logic here
if (collided) {
collidedSinceLastTick.add(other);
// maybe return false if other actions depend on a GameObject just colliding once per tick
}
return collided;
}
// ...
}
HashSet and .hashCode() both can be tuned in some cases. Maybe you could even remove displayList and "hold" everything in spatialHashGrid to reduce the memory foot-print a little bit. Of course do that only if you don't need special access to displayList - in XML's DocumentObjectModel objects can be accessed by a path throught the tree, and "hot spots" can be accessed by ID where the ID has to be assigned explicitely. For serializing (saving game state or whatever) it should not be an issue to iterate through spatialHashGrid performance-wise (it's a bit slower than serializing the gameObject set because you may have to suppress duplicates - using Java serialization it even does not save the same object twice using the default settings, saving just a reference after the first occurence of an object).

Algorithm to detect and combine overlapping / colliding circles

I'm trying to write a time efficient algorithm that can detect a group of overlapping circles and make a single circle in the "middle" of the group that will represent that group. The practical application of this is representing GPS locations over a map, put the conversion in to Cartesian co-ordinates is already handled so that's not relevant, the desired effect is that at different zoom levels clusters of close together points just appear as a single circle (that will have the number of points printed in the centre in the final version)
In this example the circles just have a radius of 15 so the distance calculation (Pythagoras) is not being square rooted and compared to 225 for the collision detection. I was trying anything to shave off time, but the problem is this really needs to happen very quickly becasue it's a user facing bit of code that needs to be snappy and good looking.
I've given this a go and I it works with small data sets pretty well. 2 big problems, it takes too long and it can run out of memory if all the points are on top of one another.
The route I've taken is to calculate distance between each point in a first pass, and then take the shortest distance first and start to combine from there, anything that's been combined becomes ineligible for combination on that pass, and the whole list is passed back around to the distance calculations again until nothing changes.
To be honest I think it needs a radical shift in approach and I think it's a little beyond me. I've re factored my code in to one class for ease of posting and generated random points to give an example.
package mergepoints;
import java.awt.Point;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
public class Merger {
public static void main(String[] args) {
Merger m = new Merger();
m.subProcess(m.createRandomList());
}
private List<Plottable> createRandomList() {
List<Plottable> points = new ArrayList<>();
for (int i = 0; i < 50000; i++) {
Plottable p = new Plottable();
p.location = new Point((int) Math.floor(Math.random() * 1000),
(int) Math.floor(Math.random() * 1000));
points.add(p);
}
return points;
}
private List<Plottable> subProcess(List<Plottable> visible) {
List<PlottableTuple> tuples = new ArrayList<PlottableTuple>();
// create a tuple to store distance and matching objects together,
for (Plottable p : visible) {
PlottableTuple tuple = new PlottableTuple();
tuple.a = p;
tuples.add(tuple);
}
// work out each Plottable relative distance from
// one another and order them by shortest first.
// We may need to do this multiple times for one set so going in own
// method.
// this is the bit that takes ages
setDistances(tuples);
// Sort so that smallest distances are at the top.
// parse the set and combine any pair less than the smallest distance in
// to a combined pin.
// any plottable thats been combine is no longer eligable for combining
// so ignore on this parse.
List<PlottableTuple> sorted = new ArrayList<>(tuples);
Collections.sort(sorted);
Set<Plottable> done = new HashSet<>();
Set<Plottable> mergedSet = new HashSet<>();
for (PlottableTuple pt : sorted) {
if (!done.contains(pt.a) && pt.distance <= 225) {
Plottable merged = combine(pt, done);
done.add(pt.a);
for (PlottableTuple tup : pt.others) {
done.add(tup.a);
}
mergedSet.add(merged);
}
}
// if we haven't processed anything we are done just return visible
// list.
if (done.size() == 0) {
return visible;
} else {
// change the list to represent the new combined plottables and
// repeat the process.
visible.removeAll(done);
visible.addAll(mergedSet);
return subProcess(visible);
}
}
private Plottable combine(PlottableTuple pt, Set<Plottable> done) {
List<Plottable> plottables = new ArrayList<>();
plottables.addAll(pt.a.containingPlottables);
for (PlottableTuple otherTuple : pt.others) {
if (!done.contains(otherTuple.a)) {
plottables.addAll(otherTuple.a.containingPlottables);
}
}
int x = 0;
int y = 0;
for (Plottable p : plottables) {
Point position = p.location;
x += position.x;
y += position.y;
}
x = x / plottables.size();
y = y / plottables.size();
Plottable merged = new Plottable();
merged.containingPlottables.addAll(plottables);
merged.location = new Point(x, y);
return merged;
}
private void setDistances(List<PlottableTuple> tuples) {
System.out.println("pins: " + tuples.size());
int loops = 0;
// Start from the first item and loop through, then repeat but starting
// with the next item.
for (int startIndex = 0; startIndex < tuples.size() - 1; startIndex++) {
// Get the data for the start Plottable
PlottableTuple startTuple = tuples.get(startIndex);
Point startLocation = startTuple.a.location;
for (int i = startIndex + 1; i < tuples.size(); i++) {
loops++;
PlottableTuple compareTuple = tuples.get(i);
double distance = distance(startLocation, compareTuple.a.location);
setDistance(startTuple, compareTuple, distance);
setDistance(compareTuple, startTuple, distance);
}
}
System.out.println("loops " + loops);
}
private void setDistance(PlottableTuple from, PlottableTuple to,
double distance) {
if (distance < from.distance || from.others == null) {
from.distance = distance;
from.others = new HashSet<>();
from.others.add(to);
} else if (distance == from.distance) {
from.others.add(to);
}
}
private double distance(Point a, Point b) {
if (a.equals(b)) {
return 0.0;
}
double result = (((double) a.x - (double) b.x) * ((double) a.x - (double) b.x))
+ (((double) a.y - (double) b.y) * ((double) a.y - (double) b.y));
return result;
}
class PlottableTuple implements Comparable<PlottableTuple> {
public Plottable a;
public Set<PlottableTuple> others;
public double distance;
#Override
public int compareTo(PlottableTuple other) {
return (new Double(distance)).compareTo(other.distance);
}
}
class Plottable {
public Point location;
private Set<Plottable> containingPlottables;
public Plottable(Set<Plottable> plots) {
this.containingPlottables = plots;
}
public Plottable() {
this.containingPlottables = new HashSet<>();
this.containingPlottables.add(this);
}
public Set<Plottable> getContainingPlottables() {
return containingPlottables;
}
}
}

Map all your circles on a 2D grid first. You then only need to compare the circles in a cell with the other circles in that cell and in it's 9 neighbors (you can reduce that to five by using a brick pattern instead of a regular grid).
If you only need to be really approximate, then you can just group all the circles that fall into a cell together. You will probably also want to merge cells that only have a small number of circles together with there neighbors, but this will be fast.

This problem is going to take a reasonable amount of computation no matter how you do it, the question then is: can you do all the computation up-front so that at run-time it's just doing a look-up? I would build a tree-like structure where each layer is all the points that need to be drawn for a given zoom level. It takes more computation up-front, but at run-time you are simply drawing a list of point, fast.
My idea is to decide what the resolution of each zoom level is (ie at zoom level 1 points closer than 15 get merged; at zoom level 2 points closer than 30 get merged), then go through your points making groups of points that are within the 15 of each other and pick a point to represent group that group at the higher zoom. Now you have a 2 layer tree. Then you pass over the second layer grouping all points that are within 30 of each other, and so on all the way up to your highest zoom level. Now save this tree structure to file, and at run-time you can very quickly change zoom levels by simply drawing all points at the appropriate tree level. If you need to add or remove points, that can be done dynamically by figuring out where to attach them to the tree.
There are two downsides to this method that come to mind: 1) it will take a long time to compute the tree, but you only have to do this once, and 2) you'll have to think really carefully about how you build the tree, based on how you want the groupings to be done at higher levels. For example, in the image below the top level may not be the right grouping that you want. Maybe instead building the tree based off the previous layer, you always want to go back to the original points. That said, some loss of precision always happens when you're trying to trade-off for faster run-time.
EDIT
So you have a problem which requires O(n^2) comparisons, you say it has to be done in real-time, can not be pre-computed, and has to be fast. Good luck with that.
Let's analyze the problem a bit; if you do no pre-computation then in order to decide which points can be merged you have to compare every pair of points, that's O(n^2) comparisons. I suggested building a tree before-hand, O(n^2 log n) once, but then runtime is just a lookup, O(1). You could also do something in between where you do some work before and some at run-time, but that's how these problems always go, you have to do a certain amount of computation, you can play games by doing some of it earlier, but at the end of the day you still have to do the computation.
For example, if you're willing to do some pre-computation, you could try keeping two copies of the list of points, one sorted by x-value and one sorted by y-value, then instead of comparing every pair of points, you can do 4 binary searches to find all the points within, say, a 30 unit box of the current point. More complicated so would be slower for a small number of points (say <100), but would reduce the overall complexity to O(n log n), making it faster for large amounts of data.
EDIT 2
If you're worried about multiple points at the same location, then why don't you do a first pass removing the redundant points, then you'll have a smaller "search list"
list searchList = new list()
for pt1 in points :
boolean clean = true
for pt2 in searchList :
if distance(pt1, pt2) < epsilon :
clean = false
break
if clean :
searchList.add(pt1)
// Now you have a smaller list to act on with only 1 point per cluster
// ... I guess this is actually the same as my first suggestion if you make one of these search lists per zoom level. huh.
EDIT 3: Graph Traversal
A totally new approach would be to build a graph out of the points and do some sort of longest-edge-first graph traversal on them. So pick a point, draw it, and traverse its longest edge, draw that point, etc. Repeat this until you come to a point which doesn't have any untraversed edges longer than your zoom resolution. The number of edges per point gives you an easy way to tradeoff speed for correctness. If the number of edges per point was small and constant, say 4, then with a bit of cleverness you could build the graph in O(n) time and also traverse it to draw points in O(n) time. Fast enough to do it on the fly with no pre-computation.

Just a wild guess and something that occurred to me while reading responses from others.
Do a multi-step comparison. Assume your combining distance at the current zoom level is 20 meters. First, subtract (X1 - X2). If This is bigger than 20 meters then you are done, the points are too far. Next, subtract (Y1 - Y2) and do the same thing to reject combining the points.
You could stop here and be happy if you are good with using only horizontal/vertical distances as your metric for combining. Much less math (no squaring or square roots). Pythagoras wouldn't be happy but your users might.
If you really insist on exact answers, do the two subtraction/comparison steps above. If the points are within horizontal and vertical limits, THEN you do the full Pythagoras check with square roots.
Assuming all your points are not highly clustered very close to the combining limit, this should save some CPU cycles.
This is still approximately an O(n^2) technique, but the math should be simpler. If you have the memory, you could store distances between each set of points and then you never have to compute it again. This could take up more memory than you have and also grows at a rate of approximately O(n^2), so be careful.
Also, you could make a linked list or sorted array of all your points, sorted in order of increasing X or increasing Y. (I don't think you need both, just one). Then walk through the list in sorted order. For each point, check the neighbors out until (X1 - X2) is bigger than your combining distance. and then stop. You don't have to compare each set of points for O(N^2), you only have to compare neighbors that are close in one dimension to quickly prune your large list to a small one. As you move through the list, you only have to compare points that have a bigger X than your current candidate, because you already compared and combined with all previous X values. This gets you closer to the O(n) complexity you want. Of course, you would need to check the Y dimension and fully qualify the points to be combined before you actually do it. Don't just use the X distance to make your combining decision.

different for loops java

I'm having some difficulties with the following problem:
I'm making a little game where you're at a specific spot and each spot has each some possible directions.
The available directions are N(ord),E(ast),S,W . I use the function getPosDirections to get the possible directions of that spot. The function returns the directions into an ArrayList<String> e.g. for spot J3: [E,W]
Now the game goes like this: 2 dice will be rolled so you get a number between 2 and 12, this number represents the number of steps you can make.
What I want is an ArrayList of all the possible routes
clarification of all the possible routes:
When I'm at the current position I check what the possibilities are from there. Let's say that's go East and go West. So we get 2 new positions and from there on we need to check for the next possibilities again for both positions (until we took x directions)
(x equals the number thrown by the dice).
e.g.: I throw 3 and I'm currently at spot J3:
[[E,N,E],[E,N,S],[E,S,E],[E,S,S],[W,N,E],[W,N,S],[W,S,E],[W,S,S]]
How would obtain the last mentioned Array(list)?

First, you might wish to think about your approach some more. In the worst case (a 12 is rolled, and all 4 directions are possible at every location), there will be 4^12 ~= 160 million routes. Is it really necessary to iterate over them all? And is it necessary to fill about 1 GB of memory to store that list?
Next, it is probably a good idea to represent directions in a type-safe manner, for instance using an enum.
That being said, recursion is your friend:
private void iteratePaths(Location currentLoc, List<Direction> currentPath, List<List<Direction>> allPaths, int pathLength) {
if (currentPath.size() >= pathLength) {
allPaths.add(new ArrayList<Direction>(currentPath));
return;
}
for (Direction d : currentLoc.getPosDirections()) {
currentPath.add(d);
Location newLoc = currentLoc.walk(d);
iteratePaths(newLoc, currentPath, allPaths, pathLength);
currentPath.remove(currentPath.size() - 1);
}
}
public void List<List<Direction>> getAllPaths(Location loc, int length) {
List<List<Direction>> allPaths = new ArrayList<List<Direction>>();
List<Direction> currentPath = new ArrayList<Direction>();
iteratePaths(loc, currentPath, allPaths, length);
return allPaths;
}

You can assume that your field of spots is a complete graph. Then you need to implement BFS or DFS with saving pathes.
You can implement all logic in any of these algorithms (like getting a list of possible directions from a certain node).