A game has a large number of unit objects held in an array.
These units have x/y coordinates locating them on a 2d map.
let units = [
{id: 1, x=3450, y = 1456},
{id: 2, x=5560, y = 2423},
{id: 3, x=1321, y = 3451}
]
Approx 50 times a second the game requires each unit to generate a list of other units within a given distance (to interact with them by fighting/avoiding etc).
As the unit count grows into the thousands the current process where each unit checks distance vs each other unit slows down dramatically as you start getting exponentially more tests required.
Looking into similar problems posted online we started grouping the units into row/column cell collections then only performing the distance tests on those that 'might' be close enough to be relevant. However we found that constructing this grouping actually takes longer to build and maintain than the gains it provided.
A testable version of the current code is below - on my fairly typical browser this takes about a second to complete and it needs to improve this substantially - suggestions for optimisations welcome.
//create the world
let mapWidth = 5000;
let mapHeight = 2000;
let releventDistance = 200;
let unitCount = 5000;
//new unit function creates a unit in a random position on the map
function newUnit(id){
let newUnit = {};
newUnit.id = id;
newUnit.x = Math.floor(Math.random()*mapWidth);
newUnit.y = Math.floor(Math.random()*mapHeight);
//this array of 'relevent' neigbours is the collection of other units close enough to interact with
newUnit.neighbours = [];
return newUnit;
}
//simple distance test
function distance (unit1, unit2){
let dx = unit1.x - unit2.x;
let dy = unit1.y - unit2.y;
return Math.sqrt(dx * dx + dy * dy);
}
//collection of units
var myUnits = [];
//populate the units
for (let i =0; i<unitCount; i++){
myUnits.push(newUnit(i));
}
console.log(unitCount + " units created");
//complete a full-scan with a nested foreach
let timeStamp1 = new Date();
myUnits.forEach(unit => {
myUnits.forEach(unit2 => {
//don't test a unit against itself
if(unit.id != unit2.id){
let unitDist = distance(unit, unit2);
if (unitDist <= releventDistance){
unit.neighbours.push({unit : unit2, distance : unitDist});
}
}
})
})
//print results
console.log((new Date() - timeStamp1) + "ms: to complete bruteforce fullscan");
//print average number of neigbours
let totalNeighbourCount = 0;
myUnits.forEach(myUnit => {totalNeighbourCount += myUnit.neighbours.length});
console.log(Math.floor(totalNeighbourCount/myUnits.length) + ": average number of neighbours");
You could iterate only from the index plus one for the inner loop and avoid visiting already visited pairs.
This approach requires to add the pair to each neighbor.
//create the world
let mapWidth = 5000;
let mapHeight = 2000;
let releventDistance = 200;
let unitCount = 5000;
//new unit function creates a unit in a random position on the map
function newUnit(id){
let newUnit = {};
newUnit.id = id;
newUnit.x = Math.floor(Math.random()*mapWidth);
newUnit.y = Math.floor(Math.random()*mapHeight);
//this array of 'relevent' neigbours is the collection of other units close enough to interact with
newUnit.neighbours = [];
return newUnit;
}
//simple distance test
function distance (unit1, unit2){
let dx = unit1.x - unit2.x;
let dy = unit1.y - unit2.y;
return Math.sqrt(dx * dx + dy * dy);
}
//collection of units
var myUnits = [];
//populate the units
for (let i =0; i<unitCount; i++){
myUnits.push(newUnit(i));
}
console.log(unitCount + " units created");
let timeStamp1 = new Date();
for (let i = 0, l1 = myUnits.length - 1; i < l1; i++) {
const unit = myUnits[i];
for (let j = i + 1, l2 = myUnits.length; j < l2; j++) {
const unit2 = myUnits[j];
let unitDist = distance(unit, unit2);
if (unitDist <= releventDistance) {
unit2.neighbours.push({ unit: unit, distance: unitDist });
unit.neighbours.push({ unit: unit2, distance: unitDist });
}
}
}
//print results
console.log((new Date() - timeStamp1) + "ms: to complete bruteforce fullscan");
//print average number of neigbours
let totalNeighbourCount = 0;
myUnits.forEach(myUnit => {totalNeighbourCount += myUnit.neighbours.length});
console.log(Math.floor(totalNeighbourCount/myUnits.length) + ": average number of neighbours");
I am woking on a project of weighted interpolation. Each station has a coordinate point on the map as shown below.
var stationCoor = [[408,352],[525,348],[535,495],[420,400],[272,145],[175,195],[197,335]];
I am taking points that are located in the lake and I am using those to create weighted averages for inputs from those stations. Here is my function for determining the weighted numbers.
function findWeightSpeed(xPos, yPos){
var totalHypt = 0;
var arrHpyt = [];
var arrWeight = [];
for(var l=0;l<7;l++){
var xDis = Math.abs(xPos-stationCoor[l][0]);
var yDis = Math.abs(yPos-stationCoor[l][1]);
var hptSq = Math.pow(xDis,2)+Math.pow(yDis,2);
var hypt = Math.sqrt(hptSq);
totalHypt = totalHypt+hypt;
arrHpyt.push(hypt);
}
for(var j=0;j<7;j++){
arrWeight.push(arrHpyt[j]/totalHypt)
}
return arrWeight;
}
This finds the hypotenuse between the point (xPos,yPos) and the stations. It then adds the data up and divides each station by the total yielding the weighted numbers.
I need to use these points to weight wind direction from these stations. I was using the funciotn below to calculate an average of points.
function averageAngles(){
var x = 0;
var y = 0;
var pi = 22/7;
var angle = [2.7925,2.8797,2.9670,3.0543, 0.0872]; // 310,320,330,340,10
for(var i = 0; i < angle.length; i++) {
x += Math.cos(angle[i]);
y += Math.sin(angle[i]);
}
var average_angle = Math.atan2(y, x);
console.log((average_angle/pi)*360);
}
This gave me accurate information for a weighted average of .20 for all points. However, the weighted average points for the 7 stations (as seen below on the map) is similar to [0.1076839005418769, 0.08051796093187284, 0.003987308213631277, 0.08458358029618485, 0.2463427297217639, 0.26463834002675196, 0.21224618026791833].
How would I go about making a function that takes the weighted average numbers from the findWeightSpeed() and using that to weight the circular quantities in averageAngles()?
I used this How do you calculate the average of a set of circular data? to make the function for averaging angles.
Many thanks for any suggestions given.
Here is a link I found online that explains the entire procedure.
Computing Weighted Averages for Wind Speed and Direction
The code is similar to this.
function weightAllData(xPos,yPos,windData){
var uVecSum = 0;
var vVecSum = 0;
var arrayWeightSpeed = findWeightSpeed(xPos, yPos); //using weighted interpolation based on distance
var arrayWindSpeed = [WSData];
var arrayWindDirection = [WDData];
for(var m=0;m<7;m++){
uVecSum = uVecSum + (arrayWeightSpeed[m] * getUVector(arrayWindSpeed[m],(arrayWindDirection[m]/180)*Math.PI));
vVecSum = vVecSum + (arrayWeightSpeed[m] * getVVector(arrayWindSpeed[m],(arrayWindDirection[m]/180)*Math.PI));
}
var weightWS = Math.sqrt(Math.pow(uVecSum,2)+Math.pow(vVecSum,2));
if(vVecSum!=0){
weightWDRad = Math.atan(uVecSum/vVecSum);
}
if(vVecSum==0){
weightWDRad = Math.atan(uVecSum/(0.0001+vVecSum));
}
if(weightWDRad<0){
weightWDRad = weightWDRad + Math.PI
}
weightWD = (weightWDRad * (180/Math.PI));
}
Let me know if you want an explanation
Is it possible to get a random number between 1-100 and keep the results mainly within the 40-60 range? I mean, it will go out of that range rarely, but I want it to be mainly within that range... Is it possible with JavaScript/jQuery?
Right now I'm just using the basic Math.random() * 100 + 1.
The simplest way would be to generate two random numbers from 0-50 and add them together.
This gives a distribution biased towards 50, in the same way rolling two dice biases towards 7.
In fact, by using a larger number of "dice" (as #Falco suggests), you can make a closer approximation to a bell-curve:
function weightedRandom(max, numDice) {
let num = 0;
for (let i = 0; i < numDice; i++) {
num += Math.random() * (max/numDice);
}
return num;
}
JSFiddle: http://jsfiddle.net/797qhcza/1/
You have some good answers here that give specific solutions; let me describe for you the general solution. The problem is:
I have a source of more-or-less uniformly distributed random numbers between 0 and 1.
I wish to produce a sequence of random numbers that follow a different distribution.
The general solution to this problem is to work out the quantile function of your desired distribution, and then apply the quantile function to the output of your uniform source.
The quantile function is the inverse of the integral of your desired distribution function. The distribution function is the function where the area under a portion of the curve is equal to the probability that the randomly-chosen item will be in that portion.
I give an example of how to do so here:
http://ericlippert.com/2012/02/21/generating-random-non-uniform-data/
The code in there is in C#, but the principles apply to any language; it should be straightforward to adapt the solution to JavaScript.
Taking arrays of numbers, etc. isn't efficient. You should take a mapping which takes a random number between 0 to 100 and maps to the distribution you need. So in your case, you could take f(x)=-(1/25)x2+4x to get a distribution with the most values in the middle of your range.
I might do something like setup a "chance" for the number to be allowed to go "out of bounds". In this example, a 20% chance the number will be 1-100, otherwise, 40-60:
$(function () {
$('button').click(function () {
var outOfBoundsChance = .2;
var num = 0;
if (Math.random() <= outOfBoundsChance) {
num = getRandomInt(1, 100);
} else {
num = getRandomInt(40, 60);
}
$('#out').text(num);
});
function getRandomInt(min, max) {
return Math.floor(Math.random() * (max - min + 1)) + min;
}
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<button>Generate</button>
<div id="out"></div>
fiddle: http://jsfiddle.net/kbv39s9w/
I needed to solve this problem a few years ago and my solution was easier than any of the other answers.
I generated 3 randoms between the bounds and averaged them. This pulls the result towards the centre but leaves it completely possible to reach the extremities.
It looks stupid but you can use rand twice:
var choice = Math.random() * 3;
var result;
if (choice < 2){
result = Math.random() * 20 + 40; //you have 2/3 chance to go there
}
else {
result = Math.random() * 100 + 1;
}
Sure it is possible. Make a random 1-100. If the number is <30 then generate number in range 1-100 if not generate in range 40-60.
There is a lot of different ways to generate such random numbers. One way to do it is to compute the sum of multiple uniformly random numbers. How many random numbers you sum and what their range is will determine how the final distribution will look.
The more numbers you sum up, the more it will be biased towards the center. Using the sum of 1 random number was already proposed in your question, but as you notice is not biased towards the center of the range. Other answers have propose using the sum of 2 random numbers or the sum of 3 random numbers.
You can get even more bias towards the center of the range by taking the sum of more random numbers. At the extreme you could take the sum of 99 random numbers which each were either 0 or 1. That would be a binomial distribution. (Binomial distributions can in some sense be seen as the discrete version of normal distributions). This can still in theory cover the full range, but it has so much bias towards the center that you should never expect to see it reach the endpoints.
This approach means you can tweak just how much bias you want.
What about using something like this:
var loops = 10;
var tries = 10;
var div = $("#results").html(random());
function random() {
var values = "";
for(var i=0; i < loops; i++) {
var numTries = tries;
do {
var num = Math.floor((Math.random() * 100) + 1);
numTries--;
}
while((num < 40 || num >60) && numTries > 1)
values += num + "<br/>";
}
return values;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<div id="results"></div>
The way I've coded it allows you to set a couple of variables:
loops = number of results
tries = number of times the function will try to get a number between 40-60 before it stops running through the while loop
Added bonus: It uses do while!!! Awesomeness at its best
You can write a function that maps random values between [0, 1) to [1, 100] according to weight. Consider this example:
Here, the value 0.95 maps to value between [61, 100].
In fact we have .05 / .1 = 0.5, which, when mapped to [61, 100], yields 81.
Here is the function:
/*
* Function that returns a function that maps random number to value according to map of probability
*/
function createDistributionFunction(data) {
// cache data + some pre-calculations
var cache = [];
var i;
for (i = 0; i < data.length; i++) {
cache[i] = {};
cache[i].valueMin = data[i].values[0];
cache[i].valueMax = data[i].values[1];
cache[i].rangeMin = i === 0 ? 0 : cache[i - 1].rangeMax;
cache[i].rangeMax = cache[i].rangeMin + data[i].weight;
}
return function(random) {
var value;
for (i = 0; i < cache.length; i++) {
// this maps random number to the bracket and the value inside that bracket
if (cache[i].rangeMin <= random && random < cache[i].rangeMax) {
value = (random - cache[i].rangeMin) / (cache[i].rangeMax - cache[i].rangeMin);
value *= cache[i].valueMax - cache[i].valueMin + 1;
value += cache[i].valueMin;
return Math.floor(value);
}
}
};
}
/*
* Example usage
*/
var distributionFunction = createDistributionFunction([
{ weight: 0.1, values: [1, 40] },
{ weight: 0.8, values: [41, 60] },
{ weight: 0.1, values: [61, 100] }
]);
/*
* Test the example and draw results using Google charts API
*/
function testAndDrawResult() {
var counts = [];
var i;
var value;
// run the function in a loop and count the number of occurrences of each value
for (i = 0; i < 10000; i++) {
value = distributionFunction(Math.random());
counts[value] = (counts[value] || 0) + 1;
}
// convert results to datatable and display
var data = new google.visualization.DataTable();
data.addColumn("number", "Value");
data.addColumn("number", "Count");
for (value = 0; value < counts.length; value++) {
if (counts[value] !== undefined) {
data.addRow([value, counts[value]]);
}
}
var chart = new google.visualization.ColumnChart(document.getElementById("chart"));
chart.draw(data);
}
google.load("visualization", "1", { packages: ["corechart"] });
google.setOnLoadCallback(testAndDrawResult);
<script src="https://www.google.com/jsapi"></script>
<div id="chart"></div>
Here's a weighted solution at 3/4 40-60 and 1/4 outside that range.
function weighted() {
var w = 4;
// number 1 to w
var r = Math.floor(Math.random() * w) + 1;
if (r === 1) { // 1/w goes to outside 40-60
var n = Math.floor(Math.random() * 80) + 1;
if (n >= 40 && n <= 60) n += 40;
return n
}
// w-1/w goes to 40-60 range.
return Math.floor(Math.random() * 21) + 40;
}
function test() {
var counts = [];
for (var i = 0; i < 2000; i++) {
var n = weighted();
if (!counts[n]) counts[n] = 0;
counts[n] ++;
}
var output = document.getElementById('output');
var o = "";
for (var i = 1; i <= 100; i++) {
o += i + " - " + (counts[i] | 0) + "\n";
}
output.innerHTML = o;
}
test();
<pre id="output"></pre>
Ok, so I decided to add another answer because I felt like my last answer, as well as most answers here, use some sort of half-statistical way of obtaining a bell-curve type result return. The code I provide below works the same way as when you roll a dice. Therefore, it is hardest to get 1 or 99, but easiest to get 50.
var loops = 10; //Number of numbers generated
var min = 1,
max = 50;
var div = $("#results").html(random());
function random() {
var values = "";
for (var i = 0; i < loops; i++) {
var one = generate();
var two = generate();
var ans = one + two - 1;
var num = values += ans + "<br/>";
}
return values;
}
function generate() {
return Math.floor((Math.random() * (max - min + 1)) + min);
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<div id="results"></div>
I'd recommend using the beta distribution to generate a number between 0-1, then scale it up. It's quite flexible and can create many different shapes of distributions.
Here's a quick and dirty sampler:
rbeta = function(alpha, beta) {
var a = 0
for(var i = 0; i < alpha; i++)
a -= Math.log(Math.random())
var b = 0
for(var i = 0; i < beta; i++)
b -= Math.log(Math.random())
return Math.ceil(100 * a / (a+b))
}
var randNum;
// generate random number from 1-5
var freq = Math.floor(Math.random() * (6 - 1) + 1);
// focus on 40-60 if the number is odd (1,3, or 5)
// this should happen %60 of the time
if (freq % 2){
randNum = Math.floor(Math.random() * (60 - 40) + 40);
}
else {
randNum = Math.floor(Math.random() * (100 - 1) + 1);
}
The best solution targeting this very problem is the one proposed by BlueRaja - Danny Pflughoeft but I think a somewhat faster and more general solution is also worth mentioning.
When I have to generate random numbers (strings, coordinate pairs, etc.) satisfying the two requirements of
The result set is quite small. (not larger than 16K numbers)
The result set is discreet. (like integer numbers only)
I usually start by creating an array of numbers (strings, coordinate pairs, etc.) fulfilling the requirement (In your case: an array of numbers containing the more probable ones multiple times.), then choose a random item of that array. This way, you only have to call the expensive random function once per item.
Distribution
5% for [ 0,39]
90% for [40,59]
5% for [60,99]
Solution
var f = Math.random();
if (f < 0.05) return random(0,39);
else if (f < 0.95) return random(40,59);
else return random(60,99);
Generic Solution
random_choose([series(0,39),series(40,59),series(60,99)],[0.05,0.90,0.05]);
function random_choose (collections,probabilities)
{
var acc = 0.00;
var r1 = Math.random();
var r2 = Math.random();
for (var i = 0; i < probabilities.length; i++)
{
acc += probabilities[i];
if (r1 < acc)
return collections[i][Math.floor(r2*collections[i].length)];
}
return (-1);
}
function series(min,max)
{
var i = min; var s = [];
while (s[s.length-1] < max) s[s.length]=i++;
return s;
}
You can use a helper random number to whether generate random numbers in 40-60 or 1-100:
// 90% of random numbers should be between 40 to 60.
var weight_percentage = 90;
var focuse_on_center = ( (Math.random() * 100) < weight_percentage );
if(focuse_on_center)
{
// generate a random number within the 40-60 range.
alert (40 + Math.random() * 20 + 1);
}
else
{
// generate a random number within the 1-100 range.
alert (Math.random() * 100 + 1);
}
If you can use the gaussian function, use it. This function returns normal number with average 0 and sigma 1.
95% of this number are within average +/- 2*sigma. Your average = 50, and sigma = 5 so
randomNumber = 50 + 5*gaussian()
The best way to do that is generating a random number that is distributed equally in a certain set of numbers, and then apply a projection function to the set between 0 and a 100 where the projection is more likely to hit the numbers you want.
Typically the mathematical way of achieving this is plotting a probability function of the numbers you want. We could use the bell curve, but let's for the sake of easier calculation just work with a flipped parabola.
Let's make a parabola such that its roots are at 0 and 100 without skewing it. We get the following equation:
f(x) = -(x-0)(x-100) = -x * (x-100) = -x^2 + 100x
Now, all the area under the curve between 0 and 100 is representative of our first set where we want the numbers generated. There, the generation is completely random. So, all we need to do is find the bounds of our first set.
The lower bound is, of course, 0. The upper bound is the integral of our function at 100, which is
F(x) = -x^3/3 + 50x^2
F(100) = 500,000/3 = 166,666.66666 (let's just use 166,666, because rounding up would make the target out of bounds)
So we know that we need to generate a number somewhere between 0 and 166,666. Then, we simply need to take that number and project it to our second set, which is between 0 and 100.
We know that the random number we generated is some integral of our parabola with an input x between 0 and 100. That means that we simply have to assume that the random number is the result of F(x), and solve for x.
In this case, F(x) is a cubic equation, and in the form F(x) = ax^3 + bx^2 + cx + d = 0, the following statements are true:
a = -1/3
b = 50
c = 0
d = -1 * (your random number)
Solving this for x yields you the actual random number your are looking for, which is guaranteed to be in the [0, 100] range and a much higher likelihood to be close to the center than the edges.
This answer is really good. But I would like to post implementation instructions (I'm not into JavaScript, so I hope you will understand) for different situation.
Assume you have ranges and weights for every range:
ranges - [1, 20], [21, 40], [41, 60], [61, 100]
weights - {1, 2, 100, 5}
Initial Static Information, could be cached:
Sum of all weights (108 in sample)
Range selection boundaries. It basically this formula: Boundary[n] = Boundary[n - 1] + weigh[n - 1] and Boundary[0] = 0. Sample has Boundary = {0, 1, 3, 103, 108}
Number generation:
Generate random number N from range [0, Sum of all weights).
for (i = 0; i < size(Boundary) && N > Boundary[i + 1]; ++i)
Take ith range and generate random number in that range.
Additional note for performance optimizations. Ranges don't have to be ordered neither ascending nor descending order, so for faster range look-up range that has highest weight should go first and one with lowest weight should go last.
I have created a GPS system using coordinates.
I have found the highest and lowest longitude and latitude coordinates using this function:
var maxLng = 0;
var maxLat = 0;
var minLng = 180;
var minLat = 180;
for(var i=0; i<coordinates.length; i++)
{
//get max coordinates (the +90 is to remove negative numbers)
if (coordinates[i][0]+90 > maxLat)
{
maxLat = coordinates[i][0] + 90;
}
if (coordinates[i][1]+90 > maxLng)
{
maxLng = coordinates[i][1]+ 90;
}
//get min coordinates
if (coordinates[i][0]+90 < minLat)
{
minLat = coordinates[i][0] + 90;
}
if (coordinates[i][1]+90 < minLng)
{
minLng = coordinates[i][1] + 90;
}
}
console.log(maxLat, maxLng,minLat, minLng);
//calculate distance between max and min points
var lngDistance = maxLng - minLng;
var latDistance = maxLat - minLat;
console.log(lngDistance, latDistance);
This outputs the distance between the 2 furthest apart longitude and latitude points, which I then plan to use to create a basic 2d map looking like this:
I need to convert the points, they can be a range of value such as:
0.0009321
19.332
1.9432123
0.0013432423
0.23432423
0.000034324
I want to basically convert all the numbers to 2 significant figures in front of the decimal point and store the result in the array stating how many shifts I have used.
I would want the output to be something like this (original number, converted number, shifts used):
[0.0009321, 93.21, 4]
[19.332, 19.332, 0]
[1.9432123, 19.432123, 1]
[0.0013432423, 13.432423, 3]
...
I am thinking find the first significant figure then count how far away from the decimal point this is. I found a few posts about using REGEX to do this, but I have never used REGEX before.
Can anybody give me some pointers?
Cheers.
That was a fun one figuring out.
Made a basic function for you, and as far as I can see it works.
function change(data){
var ret=new Array();
for(var i=0,len=data.length;i<len;i++){
// make string from number, remove the dot
var st=(''+data[i]).replace('.','');
//remove leading zero's from string
var no_zero=st.replace(/^0+/,'');
//make value
var val=parseInt(no_zero)/(Math.pow(10,no_zero.length-2));
//calculate amount of steps
var steps=Math.round(Math.log(Math.round(val/data[i]))/Math.log(10));
//correction for floating point
if(data[i]<0.01)steps--;
ret.push([data[i],val,steps]);
}
return ret;
}
And a working Fiddle
Say I have array [1,2,5,18,17,8] and I want to turn that into an array of length 40 that follows the same path.
a = [1,2,5,18,17,8];
stepSize = 1 / (40 / a.length);
then i think i could do something like
steps = [];
for( var i = 0; i < 1; i+= stepSize) {
steps.push(d3.interpolate(a[0],a[1])(i));
}
and then repeat that for all the elements. My question is there a better way to do this?
I can only guess what your real problem is but I think you want to plot these values and have a smooth curve. In that case use line.interpolate() https://github.com/mbostock/d3/wiki/SVG-Shapes#line_interpolate
In case you DO know what you need and your solution works for you, take this tip:
Never iterate over stepSize. Calculate it once and multiply it with i in every loop where i goes from 0 to 40. This way you work around precision problems.
Your algorithm cleaned up, tested and working:
var a = [1,5,12,76,1,2];
var steps = 24;
var ss = (a.length-1) / (steps-1);
var result = new Array(steps);
for (var i=0; i<steps; i++) {
var progress = ss * i;
var left = Math.floor(progress);
var right = Math.ceil(progress);
var factor = progress - left;
result[i] = (1 - factor) * a[left] + (factor) * a[right];
// alternative that actually works the same:
//result[i] = d3.interpolateNumber(a[left], a[right], factor);
}
console.log(result);