jsfiddle here: http://jsfiddle.net/yw0w18m3/2/
I'm using paper.js to make a background image that looks somthing like this:
Basically, I'm creating a couple thousand triangles over a loop and rotating them on every other iteration.
function Tri(x, y, rotate) {
var tri = new Path([
new Point((x - 42), (y - 48)),
new Point((x - 42), y),
new Point(x, (y - 24)),
new Point((x - 42), (y - 48))
]);
tri.fillColor = {
hue: Math.random() * 360,
saturation: 0,
brightness: ( (( Math.random() ) * .95) + .3 )
};
if(rotate) { tri.rotate(180); }
}
for (var i = 0; i < 2000; i++) {
rotate = false;
if( i % 2 ) {
rotate = true;
}
new Tri(x, y, rotate);
x = x + 42;
if( x > (winWidth + 42) ) {
x = 0 ;
y = y + 24;
}
}
There seems to be a brief 1-2 second pause/freeze though while the shapes are being drawn. Is there a more efficient way to draw all the shapes first (or push to an array) then add that to the canvas all at once?
I based my code off of the example here: http://paperjs.org/examples/candy-crash/ (click "source" in the upper right corner).
Any help is much appreciated.
Thanks!
I would end up creating two triangles, one rotated, so they don't have to be built from new points each time. Then choose the correct triangle based on the rotation variable and clone it, as opposed to create points and a triangle from scratch each time. Finally, just change the position of the cloned triangle.
Last, I would correct the maxTri so it doesn't do more than it needs to. The paren should follow the 48, not the 24. You're doing an order of magnitude more triangles than needed.
Here's a link to the sketch.paperjs.org solution I created based on your code. I find sketch easier to use than jsfiddle for paper examples.
proto1 = new Path([
new Point(0, -24),
new Point(0, 24),
new Point(42, 0)
]);
proto1.closed = true;
proto2 = proto1.clone();
proto2.rotate(180);
function putTriangle(pos, rotate) {
var tri = (rotate ? proto2 : proto1).clone();
tri.position = pos;
tri.position = tri.position.subtract([21, 0])
tri.fillColor = {
hue: Math.random() * 360,
saturation: 0,
brightness: Math.random() * 0.5 + 0.5
}
}
var tris = [],
x = 42,
y = 24,
rotate,
winWidth = paper.view.size.width,
winHeight = paper.view.size.height,
rows = (winHeight + 48) / 24,
cols = (winWidth + 42) / 42,
numTri = rows * cols,
numTriOrig = (winWidth + 42) / 42 * (winHeight + 48 / 24);
//console.log(numTri, numTriOrig);
x = 0;
y = 0;
for (var row = 0; row < rows; row++) {
rowrotate = row % 2;
for (var col = 0; col <= cols; col++) {
rotate = rowrotate ^ col % 2;
putTriangle([x,y], rotate);
x += 42;
}
x = 0;
y = y + 24;
}
Two thoughts:
I see you use rotate to transform you triangles into place. This is an expensive operation. You could replace the rotate with a less geometric & more arithmetic calculation of the triangles orientation.
Also, I see is that the fill color is being changed with each triangle and state changes (like fill) are modestly expensive. You could group all the similarly colored triangles and draw them in a single batch.
Related
I'm stumped on what is probably some pretty simple math. I need to get the X and Y coordinates from each tiles referenced ID. The grid below shows the order the ids are generated in. Each tile has a width and height of 32. Number ones x & y would be equal to (0,0). This is for a game I'm starting to make with canvas using a tileset.
1|2|3
4|5|6
7|8|9
So far for X, I've come up with...
(n % 3) * 32 - 32 // 3 is the width of the source image divded by 32
And for Y...
(n / 3) * 32
This is obviously wrong, but It's the closest I've come, and I don't think I'm too far off from the actual formula.
Here is my actual code so far:
function startGame() {
const canvas = document.getElementById("rpg");
const ctx = canvas.getContext("2d");
const tileSet = new Image();
tileSet.src = "dungeon_tiles.png";
let map = {
cols: 10,
rows: 10,
tsize: 32,
getTileX: function(counter, tiles) {
return ((tiles[counter] - 1) % 64) * 32;
},
getTileY: function(counter, tiles) {
return ((tiles[counter] - 1) / 64) * 32;
}
};
let counter = 0;
tileSet.onload = function() {
for (let c = 0; c < map.cols; c++) {
for (let r = 0; r < map.rows; r++) {
let x = map.getTileX(counter, mapObj.layers[0].data); // mapObj.layers[0].data is the array of values
let y = map.getTileY(counter, mapObj.layers[0].data);
counter += 1;
ctx.drawImage(
tileSet, // image
x, // source x
y, // source y
map.tsize, // source width
map.tsize, // source height
r * map.tsize, // target x
c * map.tsize, // target y
map.tsize, // target width
map.tsize // target height
);
}
}
};
}
If 1 is (0,0) and each tile is 32*32, then finding your horizontal position is a simple 32*(t-1) where t is your tile number. t-1 because your tiles start from 1 instead of 0. Now, you have 3 tiles per row so you want to reset every 3, so the final formula for your x is 32*((t-1)%3).
For the vertical position it's almost the same, but you want to increase your position by 32 only once every 3 tiles, so this is your y: 32*floor((t-1)/3).
floor((t-1)/3) is simply integer division since the numbers are always positive.
If I understand this correctly, you want to get the 1|2|3 values based on x, y correct? You can do something like this:
((y * total # of rows) + x) + 1
This would convert the 2D x, y index to a single index which is, as you stated, 1|2|3. This formula is based on your example where count starts at 1 and not 0. If you want to convert it to 0 base, just remove the + 1.
If you have the width and height, or probably location of input/character, you can have a GetX(int posX) and GetY(int posY) to get the x and y based on the position. Once you have converted the position to x, y values, use the formula above.
int GetX(int posX)
{
return (posX / 32);
}
int GetY(int posY)
{
return (posY / 32);
}
int GetIndex(int posX, int posY)
{
return ((GetY(posY) / totalRows) + GetX(posX)) + 1;
}
Let's say I need to put a text in the middle of the area of a triangle.
I can calculate the coordinates of the triangle's center using getBBox():
var triangle = "M0,0 L100,0 100,50 z";
var BBox = paper.path(triangle).getBBox();
var middle;
middle.x = BBox.x + BBox.width/2;
middle.y = BBox.y + BBox.height/2;
This results in the coordinates (50, 25) which are always on the long side of the triangle.
How can I make sure the calculated "middle" is inside the triangle?
The correct coordinates should be approximately: (75, 25).
The code should of course be independent of this particular example, it should work for any kind of shape.
I've done some more research in the topic, and following an advice from another list I got here:
https://en.wikipedia.org/wiki/Centroid
There is an algorithm there to calculate the centroid of an irregular polygon, which I have translated into this code:
function getCentroid(path) {
var x = new Array(11);
var y = new Array(11);
var asum = 0, cxsum = 0, cysum = 0;
var totlength = path.getTotalLength();
for (var i = 0; i < 11; i++) {
var location = path.getPointAtLength(i*totlength/10);
x[i] = location.x;
y[i] = location.y;
if (i > 0) {
asum += x[i - 1]*y[i] - x[i]*y[i - 1];
cxsum += (x[i - 1] + x[i])*(x[i - 1]*y[i] - x[i]*y[i - 1]);
cysum += (y[i - 1] + y[i])*(x[i - 1]*y[i] - x[i]*y[i - 1]);
}
}
return({x: (1/(3*asum))*cxsum, y: (1/(3*asum))*cysum});
}
It's basically an approximation of any path by 10 points (the 11th is equal to the starting point), and the function returns, for that triangle, the coordinates:
Object {x: 65.32077336966377, y: 16.33111549955705}
I've tested it with many other shapes, and it works pretty good.
Hope it helps somebody.
This snippet will calculate the center of any polygon by averaging the vertices.
var paper = Raphael(0,0, 320, 200);
var triangle = "M0,0 L100,0 100,50 z";
var tri = paper.path(triangle);
tri.attr('fill', 'blue');
var center = raphaelPathCenter( tri );
var circle = paper.circle( center.x, center.y, 5);
circle.attr("fill", "#f00");
circle.attr("stroke", "#fff");
function raphaelPathCenter( path ) {
path.getBBox(); // forces path to be traced so realPath is not null.
var vertices = parseSVGVertices( path.realPath );
var center = vertices.reduce( function(prev,cur) {
return { x: prev.x + cur.x, y: prev.y + cur.y }
}, {x:0, y:0} );
center.x /= vertices.length;
center.y /= vertices.length;
return center;
}
function parseSVGVertices( svgPath )
{
var vertices = [];
for ( var i = 0; i < svgPath.length; i ++ )
{
var vertex = svgPath[i];
if ( "ML".indexOf( vertex[0] ) > -1 ) // check SVG command
vertices.push( { x: vertex[1], y: vertex[2] } );
}
return vertices;
}
<script src="https://raw.githubusercontent.com/DmitryBaranovskiy/raphael/master/raphael-min.js"></script>
<canvas id='canvas'></canvas>
<pre id='output'></pre>
However there are a few more triangle centers to choose from.
I have an object that sits at point 0,0. This object cannot share space with any other object of its type that may appear on top of it, next to it, above it, etc.. There may be more than a few of these objects present overlapping each other and i have no knowledge of where the other ones are placed until i try the collision detection method.
My thinking is that i'll use a collision detection along side a grid search. Along the lines of the photo below.
The object will first try its default best case location. If that doesn't work then it tries to the left, left-above, left-below, etc, until it has searched all the #1 positions. Then it moves onto the #2 positions and so on until it finds a place to drop the element where it won't be overlapping another.
this is the code i'm playing around with right now but it is choosing some very, very random locations for things. I'm pretty sure it isn't following the algorithm i described above.
for (let i = 0; i < 5 && this._hasCollisions(this._tagWrapper); i++) {
/**
* This algorithm explores positions inside nested boxes.
* The move algorithm behaves the following way. It goes,
* down, up, left, down, up, right * 2, repeat.
*
* For example this is how it works given the height of 5 and a width of 7
* numbers are expressed in the offset generated
* 1: 5,0 4: 5,-7 7: 5,7 10: 10,-14
* 2: -5,0 5: -5,-7 8: -5,7 11: -10,-14
* 3: 0,-7 6: 0,7 9: 0,-14
*/
// Calculate which box the collision detector is working in
// This happens every 9 iterations
let multiplier = (i / 9) + 1;
/**
* Get the x offset
*/
if (i % 3 === 0) {
// Clear the height offset on multiples of 3
xOffset = 0;
} else {
// Set the height to the multiplier
xOffset = this._tagWrapper.offsetWidth * multiplier;
}
if (i % 3 === 2) {
// Get the sequence 2, 5, 8, 11, 14, etc..
xOffset *= -1;
}
/**
* Get the y offset
*/
if (i > 2) {
// Set the width to a multiple of the multiplier and assign the existing negativeness
yOffset = this._tagWrapper.offsetHeight * multiplier * (yOffset > 0 ? 1 : -1);
}
if (i % 3 === 0) {
// Flip the sign every 3 numbers
yOffset *= -1;
}
console.log('iteration', i);
this._tagWrapper.style.top = (basePosition.y + yOffset) + 'px';
this._tagWrapper.style.left = (basePosition.x + xOffset) + 'px';
}
What is the best way to go about performing this search? I already hav
Something like this work? (most of the code is just for the visualization)
// just draw a table to visualize
var SIZE = 15;
for (var i = 0; i < SIZE; i++) {
$("#a").append("<tr>");
for (var j = 0; j < SIZE; j++) {
$("#a > tr").last().append("<td>.</td>");
}
}
// where to start searching from
var startX = 8;
var startY = 8;
function loop() {
// tell the world which grid we are on
$("#a > tr:nth-child(" + y + ") > td:nth-child(" + x + ")").css("backgroundColor", "red");
// check if done here!!! - x and y are our positions in the grid
// also do bounds checking here
if (isX) {
x = x + xDirection;
i--;
if (!i) {
// switch dimension
isX = !isX;
i = moveFor;
// switch direction
xDirection *= -1;
}
} else {
y = y + yDirection;
i--;
if (!i) {
// switch dimension
isX = !isX;
// increase the width / height we are spanning
moveFor += 1;
i = moveFor;
// switch direction
yDirection *= -1;
}
}
// jsut so that we have a nice animation
if (x > 0 && y > 0 && x <= SIZE && y <= SIZE) {
setTimeout(loop, 10)
}
}
var x = startX;
var y = startY;
var moveFor = 1;
// our step (down) counter
var i = moveFor;
var xDirection = -1;
var yDirection = -1;
// are we moving along x or y
var isX = true;
loop();
body {
font-family: monospace;
}
td {
height: 20px;
width: 20px;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<table>
<tbody id="a"></tbody>
</table>
I would suggest using distance solution
point1 has x1 and y1
point2 has x2 and y2
var d = Math.sqrt( (x2-=x1)*x2 + (y2-=y1)*y2 );
link here: Get distance between two points in canvas
Here is an implementation of scanning points in a ring around a center point. You define the center point and the distance you want to sample and it returns the list of points in clock-wise order. It is in Python not JavaScript but it is simple enough that you can translate if needed.
I am creating a Tangram puzzle game using Javascript. And I need to detect when a user has drawn a circle (or circle like shape) with their finger. I have been able to gather hundreds (if not thousands) of x and y points with:
var touchX = event.targetTouches[0].pageX - canvas.offsetLeft;
var touchY = event.targetTouches[0].pageY - canvas.offsetTop;
I then push each x and y coordinate into an array:
touchMoveX.push(touchX);
touchMoveY.push(touchY);
I then loop through each array and create two points:
for(var i = 0; i < touchMoveX.length; i++)
{
for(var l=0; l < touchMoveY.length; l++)
{
var xPosition = touchMoveX[i];
var yPosition = touchMoveY[l];
var v1x = touchMoveX[i];
var v2x = touchMoveX[i + 1];
var v1y = touchMoveY[l];
var v2y = touchMoveY[l + 1];
Then using those two points, I use the following formula to figure out the angle between these two points in degrees:
var v1 = {x: v1x, y: v1y}, v2 = {x: v2x, y: v2y},
angleRad = Math.acos( (v1.x * v2.x + v1.y * v2.y) /
(Math.sqrt(v1.x*v1.x + v1.y*v1.y) * Math.sqrt(v2.x*v2.x + v2.y*v2.y) ) ),
angleDeg = angleRad * 180 / Math.PI;
I then sum up all of the angles and see if they are around 360 degrees.
But the above code I have described isn't working very well. Does someone out there have a better way to do this? Thank you very much.
yeah compute the average of all points (giving you a cheaply approximated center) then check if more than a certain percent of points are within a certain threshold. You can tune those values to adjust the precision until it feels right.
edit: Didn't consider that the circle could have multiple sizes, but you could just add another step computing the average of all distances. Adjusted the example for that.
var totalAmount = touchMoveX.length;
// sum up all coordinates and divide them by total length
// the average is a cheap approximation of the center.
var averageX = touchMoveX.reduce( function ( previous, current) {
return previous + current;
} ) / totalAmount ;
var averageY = touchMoveY.reduce( function ( previous, current) {
return previous + current;
} ) / totalAmount ;
// compute distance to approximated center from each point
var distances = touchMoveX.map ( function ( x, index ) {
var y = touchMoveY[index];
return Math.sqrt( Math.pow(x - averageX, 2) + Math.pow(y - averageY, 2) );
} );
// average of those distance is
var averageDistance = distances.reduce ( function ( previous, current ) {
return previous + current;
} ) / distances.length;
var min = averageDistance * 0.8;
var max = averageDistance * 1.2;
// filter out the ones not inside the min and max boundaries
var inRange = distances.filter ( function ( d ) {
return d > min && d < max;
} ).length;
var minPercentInRange = 80;
var percentInRange = inRange.length / totalAmount * 100;
// by the % of points within those boundaries we can guess if it's circle
if( percentInRange > minPercentInRange ) {
//it's probably a circle
}
EDIT: So apparently, PI is finite in JavaScript (which makes sense). But that leaves me with a major problem. What's the next best way to calculate the angles I need?
Alright, first, my code:
http://jsfiddle.net/joshlalonde/vtfyj/34/
I'm drawing cubes that open up to a 120 degree angle.
So the coordinates are calculated based on (h)eight and theta (120).
On line 46, I have a for loop that contains a nested for loop used for creating rows/columns.
It's somewhat subtle, but I noticed that the lines aren't matching up exactly. The code for figuring out each cubes position is on line 49. One of the things in the first parameter (my x value) for the origin of the cube is off. Can anyone help figure out what it is?
var cube = new Cube(
origin.x + (j * -w * (Math.PI)) +
(i * w * (Math.PI))
, origin.y + j * (h / 2) +
i * (h / 2) +
(-k*h), h);
Sorry if that's confusing. I,j, and k refer to the variable being incremented by the for loops. So basically, a three dimensional for loop.
I think the problem lies with Math.PI.
The width isn't the problem, or so I believe. I originally used 3.2 (which I somehow guessed and it seemed to line up pretty good. But I have no clue what the magical number is). I'm guessing it has to do with the angle being converted to Radians, but I don't understand why Math.PI/180 isn't the solution. I tried multiple things. 60 (in degrees) * Math.PI/180 doesn't work. What is it for?
EDIT: It might just be a JavaScript related math problem. The math is theoretically correct but can't be calculated correctly. I'll accept the imperfection to spare myself from re-writing code in unorthodox manners. I can tell it would take a lot to circumvent using trig math.
There are 2 problems...
Change line 35 to var w=h*Math.sin(30);. The 30 here matches the this.theta / 4 in the Cube getWidthmethod since this.theta equals 120.
Use the following code to generate the position of your new cube. You don't need Math.Pi. You needed to use both the cube width and height in your calculation.
var cube = new Cube(
origin.x+ -j*w - i*h,
origin.y + -j*w/2 + i*h/2,
h);
Alright I found the solution!
It's really simple - I was using degrees instead of radians.
function Cube(x, y, h) {
this.x = x
this.y = y
this.h = h;
this.theta = 120*Math.PI/180;
this.getWidth = function () {
return (this.h * Math.sin(this.theta / 2));
};
this.width = this.getWidth();
this.getCorner = function () {
return (this.h / 2);
};
this.corner = this.getCorner();
}
So apparently Javascript trig functions use Radians, so that's one problem.
Next fix I made was to the offset of each point in the cube. It doesn't need one! (o.O idk why. But whatever it works. I left the old code just in case I discover why later on).
function draw() {
var canvas = document.getElementById("canvas");
var ctx = canvas.getContext("2d");
ctx.fillStyle = "#000";
ctx.fillRect(0, 0, canvas.width, canvas.height); // Draw a black canvas
var h = 32;
var width = Math.sin(60*Math.PI/180);
var w = h*width;
var row = 9; // column and row will always be same (to make cube)
var column = row;
var area = row * column;
var height = 1;
row--;
column--;
height--;
var origin = {
x: canvas.width / 2,
y: (canvas.height / 2) - (h * column/2) + height*h
};
var offset = Math.sqrt(3)/2;
offset = 1;
for (var i = 0; i <= row; i++) {
for (var j = 0; j <= column; j++) {
for (var k = 0; k <= height; k++) {
var cube = new Cube(
origin.x + (j * -w * offset) +
(i * w * offset)
, origin.y + (j * (h / 2) * offset) +
(i * (h / 2) * offset) +
(-k*h*offset), h);
var cubes = {};
cubes[i+j+k] = cube; // Store to array
if (j == column) {
drawCube(2, cube);
}
if (i == row) {
drawCube(1, cube);
}
if (k == height) {
drawCube(0,cube);
}
}
}
}
}
See the full Jsfiddle here: http://jsfiddle.net/joshlalonde/vtfyj/41/