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Javascript Snake Game - too much recursion error

so I'm working on this snake game, and I'm basically trying to prevent the food from spawning on top of the snake tail. My setup variables:
let headX = 10; //snake starting position
let headY = 10;
let appleX = 5; //food starting position
let appleY = 5;
This is the function that checks head/food collision
function checkAppleCollision() {
if (appleX === headX && appleY === headY) {
generateApplePosition();
tailLength++;
score++;
}
}
And this is the function that randomizes the apple position after collision, and also returns the "too much recursion" error, after a couple of collisions:
function generateApplePosition() {
let collisionDetect = false;
let newAppleX = Math.floor(Math.random() * tileCount);
let newAppleY = Math.floor(Math.random() * tileCount);
for (let i = 0; i < snakeTail.length; i++) {
let segment = snakeTail[i];
if (newAppleX === segment.x && newAppleY === segment.y) {
collisionDetect = true;
}
}
while (collisionDetect === true) {
generateApplePosition();
}
appleX = newAppleX;
appleY = newAppleY;
}
Please help, I have no idea what to do here. Everything else works as intended.

Using recursions or do while is a bad idea (I'll explain later)
meanwhile, you could simplify your logic by creating:
reusable samePos() and collides() functions
a recursive createApple() function, which will return itself if the randomly generated x,y positions are occupied by the snake body
const world = {w:6, h:1}; // height set to 1 for this demo only
const snake = [{x:0, y:0}, {x:1, y:0}, {x:2, y:0}, {x:3, y:0}];
const apple = {pos: {x:0, y:0}};
// Check if two object's x,y match
const samePos = (a, b) => a.x === b.x && a.y === b.y;
// Check if object x,y is inside an array of objects
const collides = (ob, arr) => arr.some(o => samePos(ob, o));
const createApple = () => {
const randPos = {
x: ~~(Math.random() * world.w),
y: ~~(Math.random() * world.h),
};
if (collides(randPos, snake)) {
console.log(`position ${randPos.x} ${randPos.y} is occupied by snake`);
return createApple(); // Try another position.
}
// Finally a free spot!
apple.pos = randPos;
console.log(`Apple to free position: ${apple.pos.x} ${apple.pos.y}`);
}
createApple();
Run this demo multiple times
The problem
Useless random guesswork!
As you can see from the example above, if you run it multiple times, very often the randomly generated number is the same as the previously generated one:
...
position 2 0 is occupied by snake <<<
position 1 0 is occupied by snake
position 2 0 is occupied by snake <<<
position 2 0 is occupied by snake <<<
position 1 0 is occupied by snake
position 2 0 is occupied by snake <<<
...
therefore, as your snake grows in size, the recursion might go wild — ad absurdum, iterating way too many times, repeating and failing on the same xy positions, until finally hitting a rare free spot...
This is a really bad design.
Solutions
One solution would be to keep track of the already used randomized positions inside an Array - but that implies unnecessarily to go trough such an Array.
A best solution would be to actually treat the game not as a 2D game, but as a 1D game:
Consider this 2D map of size 4x3 as indexes:
0 1 2 3
4 5 6 7
8 9 10 11
now, let's place a snake into this map:
0 ⬛ 2 3
4 ⬛ ⬛ 7
8 9 ⬛ 11
here's the linear map with the Snake as a 1D list:
[ 0 ⬛ 2 3 4 ⬛ ⬛ 7 8 9 ⬛ 11 ]
therefore, instead of using an array of objects {x:n, y:n} for the snake body positions, all you need is:
[1, 5, 6, 10] // Snake body as indexes
Now that you know all the indexes where you're not allowed to place an Apple, all you need to do when creating the new apple is:
Create an Array of 0-N indexes of length: world.w * world.h
Loop the snake body indexes and delete those indexes from the array of indexes to get an Array of free spots indexes
Simply get only once a random key from that array of free spots!
const indexToXY = (index, width) => ({ x: index%width, y: Math.trunc(index/width) });
const world = {w:4, h:3};
const snakeBody = [1, 5, 6, 10];
const createApple = () => {
const arr = [...Array(world.w * world.h).keys()]; // [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
snakeBody.forEach(i => delete arr[i]);
const freeIndexes = arr.filter(k => k !== undefined); // [0, 2, 3, 4, 7, 8, 9, 11]
const appleIndex = freeIndexes[~~(Math.random() * freeIndexes.length)];
const applePos = indexToXY(appleIndex, world.w);
console.log("New apple position: %o", applePos);
};
createApple();
Run this demo multiple times
Having that free spot index simply draw your apple at the XY coordinates using this simple formula
X = index % mapWidth
Y = floor(index / mapWidth)

As others have said, this doesn't need to be recursive, and you should also take into account the (however unlikely) possibility where there are no more tiles to spawn on which would result in an infinite loop.
function generateApplePosition() {
// Count how many tiles are left for spawning in
const tilesLeft = (tileCount * tileCount) - snakeTail.length;
let collisionDetect;
if (tilesLeft > 0) {
do {
const newAppleX = Math.floor(Math.random() * tileCount);
const newAppleY = Math.floor(Math.random() * tileCount);
collisionDetect = false;
for (let i = 0; i < snakeTail.length; i++) {
const { x, y } = snakeTail[i];
if (newAppleX === x && newAppleY === y) {
collisionDetect = true; // Collision
break;
}
}
if (!collisionDetect) {
// Found spawn point
appleX = newAppleX;
appleY = newAppleY;
}
} while (collisionDetect);
}
}

Getting two points on the edges of a rectangle

I have a rectangle and would like to:
Get a random point on one (any) of the sides.
Get a random point on one (except for the previously picked) side.
My initial approach is to create arrays for each possible side.
var arr:Array = [[{x:0,y:0}, // Top
{x:width,y:0}], //
[{x:width,y:0}, // Right
{x:width,y:height}], //
[{x:width,y:height}, // Bottom
{x:0,y:height}], //
[{x:0,y:height}, // Left
{x:0,y:0}]]; //
Then, I get the sides.
rand is an instance of Rand and has the methods:
.next() which provides a random number between 0 and 1
.between(x,y) which returns a random number between x and y.
var firstSide:Array = arr[rand.next() * arr.length];
var secondSide:Array;
do {
secondSide = arr[rand.next() * arr.length];
} while(secondSide.equals(firstSide));
Finally, I calculate my points.
var pointOnFirstSide:Object = {x:rand.between(firstSide[0].x, firstSide[1].x),
y:rand.between(firstSide[0].y, firstSide[1].y};
var pointOnSecondSide:Object = {x:rand.between(secondSide[0].x, secondSide[1].x),
y:rand.between(secondSide[0].y, secondSide[1].y};
I don't think this is the most efficient way to solve this.
How would you do it?

Assuming we have the following interfaces and types:
interface Rand {
next(): number;
between(x: number, y: number): number;
}
interface Point {
x: number;
y: number;
}
type PointPair = readonly [Point, Point];
and taking you at your word in the comment that the procedure is: first randomly pick two sides, and then pick random points on those sides... first let's see what's involved in picking two sides at random:
const s1 = Math.floor(rand.between(0, arr.length));
const s2 = (Math.floor(rand.between(1, arr.length)) + s1) % arr.length;
s1 and s2 represent the indices of arr that we are choosing. The first one chooses a whole number between 0 and one less than the length of the array. We do this by picking a real number (okay, floating point number, whatever) between 0 and the length of the array, and then taking the floor of that real number. Since the length is 4, what we are doing is picking a real number uniformly between 0 and 4. One quarter of those numbers are between 0 and 1, another quarter between 1 and 2, another quarter between 2 and 3, and the last quarter are between 3 and 4. That means you have a 25% chance of choosing each of 0, 1, 2 and 3. (The chance of choosing 4 is essentially 0, or perhaps exactly 0 if rand is implemented in the normal way which excludes the upper bound).
For s2 we now pick a number uniformly between 1 and the length of the array. In this case, we are picking 1, 2, or 3 with a 33% chance each. We add that number to s1 and then take the remainder when dividing by 4. Think of what we are doing as starting on the first side s1, and then moving either 1, 2, or 3 sides (say) clockwise to pick the next side. This completely eliminates the possibility of choosing the same side twice.
Now let's see what's involved in randomly picking a point on a line segment (which can be defined as a PointPair, corresponding to the two ends p1 and p2 of the line segment) given a Rand instance:
function randomPointOnSide([p1, p2]: PointPair, rand: Rand): Point {
const frac = rand.next(); // between 0 and 1
return { x: (p2.x - p1.x) * frac + p1.x, y: (p2.y - p1.y) * frac + p1.y };
}
Here what we do is pick a single random number frac, representing how far along the way from p1 to p2 we want to go. If frac is 0, we pick p1. If frac is 1, we pick p2. If frac is 0.5, we pick halfway between p1 and p2. The general formula for this is a linear interpolation between p1 and p2 given frac.
Hopefully between the two of those, you can implement the algorithm you're looking for. Good luck!
Link to code

jcalz already gave an excellent answer. Here is an alternate version for the variant I asked about in the comments: When you want your points uniformly chosen over two sides of the perimeter, so that if your w : h ratio was 4 : 1, the first point is four times as likely to lie on a horizontal side as a vertical one. (This means that the chance of hitting two opposite long sides is 24/45; two opposite short side, 1/45; and one of each, 20/45 -- by a simple but slightly tedious calculation.)
const rand = {
next: () => Math. random (),
between: (lo, hi) => lo + (hi - lo) * Math .random (),
}
const vertices = (w, h) => [ {x: 0, y: h}, {x: w, y: h}, {x: w, y: 0}, {x: 0, y: 0} ]
const edges = ([v1, v2, v3, v4]) => [ [v1, v2], [v2, v3], [v3, v4], [v4, v1] ]
const randomPoint = ([v1, v2], rand) => ({
x: v1 .x + rand .next () * (v2 .x - v1 .x),
y: v1 .y + rand .next () * (v2 .y - v1 .y),
})
const getIndex = (w, h, x) => x < w ? 0 : x < w + h ? 1 : x < w + h + w ? 2 : 3
const twoPoints = (w, h, rand) => {
const es = edges (vertices (w, h) )
const perimeter = 2 * w + 2 * h
const r1 = rand .between (0, perimeter)
const idx1 = getIndex (w, h, r1)
const r2 = (
rand. between (0, perimeter - (idx1 % 2 == 0 ? w : h)) +
Math .ceil ((idx1 + 1) / 2) * w + Math .floor ((idx1 + 1) / 2) * h
) % perimeter
const idx2 = getIndex (w, h, r2)
return {p1: randomPoint (es [idx1], rand), p2: randomPoint (es [idx2], rand)}
}
console .log (
// Ten random pairs on rectangle with width 5 and height 2
Array (10) .fill () .map (() => twoPoints (5, 2, rand))
)
The only complicated bit in there is the calculation of r2. We calculate a random number between 0 and the total length of the remaining three sides, by adding all four sides together and subtracting off the length of the current side, width if idx is even, height if it's odd. Then we add it to the total length of the sides up to and including the index (where the ceil and floor calls simply count the number of horizontal and vertical sides, these values multiplied by the width and height, respectively, and added together) and finally take a floating-point modulus of the result with the perimeter. This is the same technique as in jcalz's answer, but made more complex by dealing with side lengths rather than simple counts.
I didn't make rand an instance of any class or interface, and in fact didn't do any Typescript here, but you can add that yourself easily enough.

Sort Objects in 3d

So I'm developing a kind of 3d game with a fixed camera position. I have many entitys, each of which gets these propertys:
Position (coordinateX, coordinateY, coordinateZ)
Hitbox (sizeX, sizeY, sizeZ)
For example, an entity could be structured in this way:
Position (3, 4, 9)
Hitbox (2, 3, 1)
Now I have on my 3d Map many Entitys, which I want to draw. Since I want to draw the objects in the foreground in front of the objects in the background, I have to sort the drawing of the objects in the canvas. That means objects that are behind other objects have to be drawn first and so on...
I have been trying for many hours to figure out how to sort the objects. With this Sort function I came to the end, because unfortunately the calculation doesn't work as it should. Tbh I do not really know how to fix this calculation at all but I thought adding my thoughts to the question is not that bad for all of us.
let objects = {
entity1: {
position: {
x: 0,
y: 10,
z: 5
},
hitbox: {
sizeX: 4,
sizeY: 1
}
},
entity2: {
position: {
x: 4,
y: 2,
z: 4
},
hitbox: {
sizeX: 3,
sizeY: 3
}
}
}
let layeredObjects = Object.values(objects).sort((itemA, itemB) =>  {
if ((itemA.position.x - itemA.hitbox.sizeX < itemB.position.x + itemB.hitbox.sizeX) &&
(itemA.position.y + itemA.hitbox.sizeY < itemB.position.y - itemB.hitbox.sizeY) ||
(itemA.position.x + itemA.hitbox.sizeX < itemB.position.x - itemB.hitbox.sizeX)) return -1;
return 1;
})
console.log(layeredObjects)
So the best solution for my problem would be a method that is able to return the enities sorted (layers) in the way I can easily draw them into my canvas.
I would be very happy about your help.

Algorithm to convert a 2d voxel map to line sometimes gets stuck

I have a 2D voxel map for a game, which is a 2D array where 1 means ground and 0 means sky.
Example: all 1's in the array (ground) are green boxes
The algorithm starts at the leftmost ground voxel that touches the sky (red box in picture).
It will explore 8 neighbours of the current position to check if one of them is a ground voxel and also touches a sky voxel. This means it should be added to the groundline.
Example of the algorithm working (it's able to go in 'caves' too)
On this map it figured it out and returned a line across the ground.
In some situations it suddenly stops though, like on this map:
After about 10 loops it stopped creating the line.
Here's the code, with some explanatory comments in there:
voxelToLine() {
let voxels = this.voxels.length,//this.voxels is the 2d array
lineGround = [],
checkedVoxels = [],
nowChecking,
toCheck = [],
otherPaths = [],
done = false;
for (let y = 1; y < voxels - 1; y++)//sets first coordinate for line
if (this.voxels[0][y] && (!this.voxels[0][y - 1] || !this.voxels[1][y] || !this.voxels[0][y + 1])) {
lineGround[0] = [0, y / voxels];
nowChecking = [1, y];//search starts from this point
}
let looped = 0;
while (!done) {//continues search untill right side is located, or it got stuk (max 10*voxelmap width loops)
toCheck = nowChecking.neighbours(8, (n) => n[0] > 0 && n[0] < voxels - 1);//gets 8 neighbour points around current point, neighbours between 1 and (voxelwidth -1) get returned
let foundNew = false;
for (let i = 0; i < toCheck.length; i++) {//check every neighbour
let x = toCheck[i][0],
y = toCheck[i][1],
index = y * voxels + x;
if (!checkedVoxels.includes(index)) {
if (this.voxels[x][y] && (!this.voxels[x][y - 1] || !this.voxels[x + 1][y] || !this.voxels[x - 1][y] || !this.voxels[x][y + 1])) {
//if the neighbour is a floor voxel, and touches a skyvoxel this neighbour is added to the line
checkedVoxels.push(index);
if (foundNew) {//if a valid neighbour is already found, this means there are 2 possible paths from the current point
otherPaths.push([x, y]);
} else {
lineGround.push([x / voxels, y / voxels]);
nowChecking = [x, y];
//valid point gets added to the line and currently explored point get updated
foundNew = true;
}
if (x >= voxels) done = true;
}
} else if (i == toCheck.length - 1 && !foundNew) {
if (otherPaths.length > 0) {
nowChecking = otherPaths.pop();
//if none of the neighbours are correct an alternative path gets explored
foundNew = true;
}
}
}
if (!foundNew || looped++ > voxels * 10) {
//if it never found a valid neighbour, or it's looped too often break from the whileloop
console.log('loops: ', looped);
break;
}
}
if (lineGround[0][0] !== 0) lineGround.splice(0, 0, [0, lineGround[0][1]]);
if (lineGround[lineGround.length - 1][0] !== 1) lineGround.push([1, lineGround[lineGround.length - 1][1]]);
//x=0 and x=1 have to exist, so if they don't exist yet, add them
return lineGround;
}
You can also test it here: game. If you click you remove (set to 0) a few voxels within a radius of where you clicked. Also the line gets recalculated.
I'm stuck on this, because I have no idea why the line stops in some situations.
All code is here. The relevant file is js/Level.js

There are more problems than the one you raised. I played a bit on your site and there are many patterns where things go wrong.
I tried to follow the logic of your code, but got lost in details. So I rewrote most of the code. The main idea is that you should keep record of which direction (slope) you are travelling along the ground in order to know in which order you should look among the neighbours for one that is part of the ground.
Let's say the neighbours are numbered as follows, from 0 to 7:
+---+---+---+
| 7 | 0 | 1 |
+---+---+---+
| 6 | * | 2 |
+---+---+---+
| 5 | 4 | 3 |
+---+---+---+
The cell marked with * is the last cell you found to be on ground level. Now let's say the previous one found was at 6, then the search among the neighbours should start at 7, then 0, 1, 2, ... 5. The first one that is found to be solid, should be the next cell added to ground level.
Another example: if the previous one found was at 4 (we're going upward), then the neighbours should be searched starting at 5, then 6, 7, 0, 1, 2 and 3.
The first neighbour that is found to be solid (ground) is the one you want to add to your ground line. This way you will follow every curve, into "caves", upward or downward, left or right.
Of course, things can still go weird if you start on an island. But I did not attempt to solve that particular case.
I've implemented the above idea in the following version of your method:
voxelToLine() {
let voxels = this.voxels.length, x, y, i;
// neighbors' relative coordinates listed in clockwise order
const neighbor = [ [0,-1], [1,-1], [1,0], [1,1], [0,1], [-1,1], [-1,0], [-1,-1] ];
for (y = 0; y < voxels; y++) //sets first coordinate for line.
if (this.voxels[0][y]) break; // found ground, don't look further down
let lineGround = [[0, y / voxels]];
let [curX, curY] = [0, y]; //search starts here
let direction = 0; // upward
let looped = 0;
do {// Continues search until right side is located,
// or it got stuk (max 10*voxelmap width loops)
for (i = 0; i < 8; i++) {//check every neighbour, starting at `direction`
[x, y] = [curX + neighbor[direction][0], curY + neighbor[direction][1]];
// if we found ground, then pick that cell as the next one on the line
if (x>=0 && x<voxels && y>=0 && y<voxels && this.voxels[x][y]) break;
direction = (direction + 1) % 8; // turn clockwise to get next neighbour
}
//if it never found a valid neighbour
if (i === 8) break;
lineGround.push([x / voxels, y / voxels]);
// prepare for next round
[curX, curY] = [x, y];
direction = (direction + 5) % 8;
} while (looped++ <= voxels*10 && curX < voxels - 1);
//x=0 and x=1 have to exist, so if they don't exist yet, add them
if (lineGround[0][0] !== 0) lineGround.splice(0, 0, [0, lineGround[0][1]]);
if (lineGround[lineGround.length - 1][0] !== 1)
lineGround.push([1, lineGround[lineGround.length - 1][1]]);
return lineGround;
}

Looks like it's skipping over the voxel right below the last legitimate ground voxel because it's already been "checked" (added to the checkedVoxels array).
Interestingly, this would prevent your ground path to ever turn 90 degrees (you'll notice your example picture doesn't have such a voxel pattern).

Smoothing out values of an array

If I had an array of numbers such as [3, 5, 0, 8, 4, 2, 6], is there a way to “smooth out” the values so they’re closer to each other and display less variance?
I’ve looked into windowing the data using something called the Gaussian function for a 1-dimensional case, which is my array, but am having trouble implementing it. This thread seems to solve exactly what I need but I don’t understand how user naschilling (second post) came up with the Gaussian matrix values.
Context: I’m working on a music waveform generator (borrowing from SoundCloud’s design) that maps the amplitude of the song at time t to a corresponding bar height. Unfortunately there’s a lot of noise, and it looks particularly ugly when the program maps a tiny amplitude which results in a sudden decrease in height. I basically want to smooth out the bar heights so they aren’t so varied.
The language I'm using is Javascript.
EDIT: Sorry, let me be more specific about "smoothing out" the values. According to the thread linked above, a user took an array
[10.00, 13.00, 7.00, 11.00, 12.00, 9.00, 6.00, 5.00]
and used a Gaussian function to map it to
[ 8.35, 9.35, 8.59, 8.98, 9.63, 7.94, 5.78, 7.32]
Notice how the numbers are much closer to each other.
EDIT 2: It worked! Thanks to user Awal Garg's algorithm, here are the results:
No smoothing
Some smoothing
Maximum smoothing
EDIT 3: Here's my final code in JS. I tweaked it so that the first and last elements of the array were able to find its neighbors by wrapping around the array, rather than calling itself.
var array = [10, 13, 7, 11, 12, 9, 6, 5];
function smooth(values, alpha) {
var weighted = average(values) * alpha;
var smoothed = [];
for (var i in values) {
var curr = values[i];
var prev = smoothed[i - 1] || values[values.length - 1];
var next = curr || values[0];
var improved = Number(this.average([weighted, prev, curr, next]).toFixed(2));
smoothed.push(improved);
}
return smoothed;
}
function average(data) {
var sum = data.reduce(function(sum, value) {
return sum + value;
}, 0);
var avg = sum / data.length;
return avg;
}
smooth(array, 0.85);

Interesting question!
The algorithm to smooth out the values obviously could vary a lot, but here is my take:
"use strict";
var array = [10, 13, 7, 11, 12, 9, 6, 5];
function avg (v) {
return v.reduce((a,b) => a+b, 0)/v.length;
}
function smoothOut (vector, variance) {
var t_avg = avg(vector)*variance;
var ret = Array(vector.length);
for (var i = 0; i < vector.length; i++) {
(function () {
var prev = i>0 ? ret[i-1] : vector[i];
var next = i<vector.length ? vector[i] : vector[i-1];
ret[i] = avg([t_avg, avg([prev, vector[i], next])]);
})();
}
return ret;
}
function display (x, y) {
console.clear();
console.assert(x.length === y.length);
x.forEach((el, i) => console.log(`${el}\t\t${y[i]}`));
}
display(array, smoothOut(array, 0.85));
NOTE: It uses some ES6 features like fat-arrow functions and template strings. Firefox 35+ and Chrome 45+ should work fine. Please use the babel repl otherwise.
My method basically computes the average of all the elements in the array in advance, and uses that as a major factor to compute the new value along with the current element value, the one prior to it, and the one after it. I am also using the prior value as the one newly computed and not the one from the original array. Feel free to experiment and modify according to your needs. You can also pass in a "variance" parameter to control the difference between the elements. Lowering it will bring the elements much closer to each other since it decreases the value of the average.
A slight variation to loosen out the smoothing would be this:
"use strict";
var array = [10, 13, 7, 11, 12, 9, 6, 5];
function avg (v) {
return v.reduce((a,b) => a+b, 0)/v.length;
}
function smoothOut (vector, variance) {
var t_avg = avg(vector)*variance;
var ret = Array(vector.length);
for (var i = 0; i < vector.length; i++) {
(function () {
var prev = i>0 ? ret[i-1] : vector[i];
var next = i<vector.length ? vector[i] : vector[i-1];
ret[i] = avg([t_avg, prev, vector[i], next]);
})();
}
return ret;
}
function display (x, y) {
console.clear();
console.assert(x.length === y.length);
x.forEach((el, i) => console.log(`${el}\t\t${y[i]}`));
}
display(array, smoothOut(array, 0.85));
which doesn't take the averaged value as a major factor.
Feel free to experiment, hope that helps!

The technique you describe sounds like a 1D version of a Gaussian blur. Multiply the values of the 1D Gaussian array times the given window within the array and sum the result. For example
Assuming a Gaussian array {.242, .399, .242}
To calculate the new value at position n of the input array - multiply the values at n-1, n, and n+1 of the input array by those in (1) and sum the result. eg for [3, 5, 0, 8, 4, 2, 6], n = 1:
n1 = 0.242 * 3 + 0.399 * 5 + 0.242 * 0 = 2.721
You can alter the variance of the Gaussian to increase or reduce the affect of the blur.

i stumbled upon this post having the same problem with trying to achieve smooth circular waves from fft averages.
i've tried normalizing, smoothing and wildest math to spread the dynamic of an array of averages between 0 and 1. it is of course possible but the sharp increases in averaged values remain a bother that basically makes these values unfeasable for direct display.
instead i use the fft average to increase amplitude, frequency and wavelength of a separately structured clean sine.
imagine a sine curve across the screen that moves right to left at a given speed(frequency) times the current average and has an amplitude of current average times whatever will then be mapped to 0,1 in order to eventually determine 'the wave's' z.
the function for calculating size, color, shift of elements or whatever visualizes 'the wave' will have to be based on distance from center and some array that holds values for each distance, e.g. a certain number of average values.
that very same array can instead be fed with values from a sine - that is influenced by the fft averages - which themselves thus need no smoothing and can remain unaltered.
the effect is pleasingly clean sine waves appearing to be driven by the 'energy' of the sound.
like this - where 'rings' is an array that a distance function uses to read 'z' values of 'the wave's x,y positions.
const wave = {
y: height / 2,
length: 0.02,
amplitude: 30,
frequency: 0.5
}
//var increment = wave.frequency;
var increment = 0;
function sinewave(length,amplitude,frequency) {
ctx.strokeStyle = 'red';
ctx.beginPath();
ctx.moveTo(0, height / 2);
for (let i = 0; i < width; i+=cellSize) {
//ctx.lineTo(i, wave.y + Math.sin(i * wave.length + increment) * wave.amplitude)
ctx.lineTo(i, wave.y + Math.sin(i * length + increment) * amplitude);
rings.push( map( Math.sin(i * length + increment) * amplitude,0,20,0.1,1) );
rings.shift();
}
ctx.stroke();
increment += frequency;
}
the function is called each frame (from draw) with the current average fft value driving the sine function like this - assuming that value is mapped to 0,1:
sinewave(0.006,averg*20,averg*0.3)
allowing fluctuating values to determine wavelength or frequency can have some visually appealing effect. however, the movement of 'the wave' will never seem natural.
i've accomplished a near enough result in my case.
for making the sine appear to be driven by each 'beat' you'd need beat detection to determine the exact tempo of 'the sound' that 'the wave' is supposed to visualize.
continuous averaging of distance between larger peaks in the lower range of fft spectrum might work there with setting a semi fixed frequency - with edm...
i know, the question was about smoothing array values.
forgive me for changing the subject. i just thought that the objective 'sound wave' is an interesting one that could be achieved differently.
and just so this is complete here's a bit that simply draws circles for each fft and assign colour according to volume.
with linewidths relative to total radius and sum of volumes this is quite nice:
//col generator
function getCol(n,m,f){
var a = (PIx5*n)/(3*m) + PIdiv2;
var r = map(sin(a),-1,1,0,255);
var g = map(sin(a - PIx2/3),-1,1,0,255);
var b = map(sin(a - PIx4/3),-1,1,0,255);
return ("rgba(" + r + "," + g + "," + b + "," + f + ")");
}
//draw circles for each fft with linewidth and colour relative to value
function drawCircles(arr){
var nC = 20; //number of elem from array we want to use
var cAv = 0;
var cAvsum = 0;
//get the sum of all values so we can map a single value with regard to this
for(var i = 0; i< nC; i++){
cAvsum += arr[i];
}
cAv = cAvsum/nC;
var lastwidth = 0;
//draw a circle for each elem from array
//compute linewith a fraction of width relative to value of elem vs. sum of elems
for(var i = 0; i< nC; i++){
ctx.beginPath();
var radius = lastwidth;//map(arr[i]*2,0,255,0,i*300);
//use a small col generator to assign col - map value to spectrum
ctx.strokeStyle = getCol(map(arr[i],0,255,0,1280),1280,0.05);
//map elem value as fraction of elem sum to linewidth/total width of outer circle
ctx.lineWidth = map(arr[i],0,cAvsum,0,width);
//draw
ctx.arc(centerX, centerY, radius, 0, Math.PI*2, false);
ctx.stroke();
//add current radius and linewidth to lastwidth
var lastwidth = radius + ctx.lineWidth/2;
}
}
codepen here: https://codepen.io/sumoclub/full/QWBwzaZ
always happy about suggestions.