There are plenty of examples on how to draw lines on canvas, in js.
But for only educational purposes i want to draw line using algorithm. basically method gets two Vector2 points, from them it finds middle point, then it continues like that recursively until minimum distance of 2 pixels is reached.
I have DrawPoint method to basically draw 1 point on canvas, and DrawLine method that does all the job.
For now I have 2 problems:
1: points are not colored red, as they should be.
2:
It doesnt look like a line.
For Vector2 i used "Victor.js" plugin, and it seems to be working well.
this is code i have:
JS:
var point2 = new Victor(100, 100);
var point3 = new Victor(150, 150);
DrawLine(point2, point3);
function DrawLine(vec0, vec1)
{
var point0 = new Victor(vec0.x, vec0.y);
var point1 = new Victor(vec1.x, vec1.y);
var dist = point1.distance(point0);
if (dist < 2)
return;
//this is how it should look like in c# var middlePoint = point0 + (point1 - point0)/2; But looks like i cant just divide by 2 using victor js because i can only divide vector by vector.
var middlePoint = point0.add(point1.subtract(point0).divide(new Victor(2,2)));
DrawPoint(middlePoint);
DrawLine(point0, middlePoint);
DrawLine(middlePoint, point1);
}
function DrawPoint(point){
var c = document.getElementById("screen");
var ctx = c.getContext("2d");
ctx.fillStyle = "FF0000";
ctx.fillRect(point.x, point.y, 3,1);
}
I really appreciate any help you can provide.
The victor.js documentation shows that most functions of Victors do not return new Victors, but operate on the current instance. In a way, v1.add(v2) is semantically more like v1 += v2 and not v1 + v2.
The problem is with calculating the midpoint. You could use the mix() method, which blends two vectors with a weight. You must clone() the Victor first, otherwise point0will be midofied:
var middlePoint = point0.clone().mix(point1, 0.5);
If you don't change the original Vectors, you don't need to create new instances of Victors from the arguments, you can use the arguments directly:
function DrawLine(point0, point1)
{
var dist = point1.distance(point0);
if (dist < 2) return;
var middlePoint = point0.clone().mix(point1, 0.5);
DrawPoint(middlePoint);
DrawLine(point0, middlePoint);
DrawLine(middlePoint, point1);
}
Finally, as Sven the Surfer has already said in a comment, "FF0000" isn't a valid colour. Use "#FF0000", note the hash mark, or one of the named web colours such as "crimson".
I am using a couple of functions from Snap.SVG, mainly path2curve and the functions around it to build a SVG morph plugin.
I've setup a demo here on Codepen to better illustrate the issue. Basically morphing shapes simple to complex and the other way around is working properly as of Javascript functionality, however, the visual isn't very pleasing.
The first shape morph looks awful, the second looks a little better because I changed/rotated it's points a bit, but the last example is perfect.
So I need either a better path2curve or a function to prepare the path string before the other function builds the curves array. Snap.SVG has a function called getClosest that I think may be useful but it's not documented.
There isn't any documentation available on this topic so I would appreciate any suggestion/input from RaphaelJS / SnapSVG / d3.js / three/js developers.
I've provided a runnable code snippet below that uses Snap.svg and that I believe demonstrates one solution to your problem. With respect to trying to find the best way to morph a starting shape into an ending shape, this algorithm essentially rotates the points of the starting shape one position at a time, sums the squares of the distances between corresponding points on the (rotated) starting shape and the (unchanged) ending shape, and finds the minimum of all those sums. i.e. It's basically a least squares approach. The minimum value identifies the rotation that, as a first guess, will provide the "shortest" morph trajectories. In spite of these coordinate reassignments, however, all 'rotations' should result in visually identical starting shapes, as required.
This is, of course, a "blind" mathematical approach, but it might help provide you with a starting point before doing manual visual analysis. As a bonus, even if you don't like the rotation that the algorithm chose, it also provides the path 'd' attribute strings for all the other rotations, so some of that work has already been done for you.
You can modify the snippet to provide any starting and ending shapes you want. The limitations are as follows:
Each shape should have the same number of points (although the point types, e.g. 'lineto', 'cubic bezier curve', 'horizontal lineto', etc., can completely vary)
Each shape should be closed, i.e. end with "Z"
The morph desired should involve only translation. If scaling or rotation is desired, those should be applied after calculating the morph based only on translation.
By the way, in response to some of your comments, while I find Snap.svg intriguing, I also find its documentation to be somewhat lacking.
Update: The code snippet below works in Firefox (Mac or Windows) and Safari. However, Chrome seems to have trouble accessing the Snap.svg library from its external web site as written (<script...github...>). Opera and Internet Explorer also have problems. So, try the snippet in the working browsers, or try copying the snippet code as well as the Snap library code to your own computer. (Is this an issue of accessing third party libraries from within the code snippet? And why browser differences? Insightful comments would be appreciated.)
var
s = Snap(),
colors = ["red", "blue", "green", "orange"], // colour list can be any length
staPath = s.path("M25,35 l-15,-25 C35,20 25,0 40,0 L80,40Z"), // create the "start" shape
endPath = s.path("M10,110 h30 l30,20 C30,120 35,135 25,135Z"), // create the "end" shape
staSegs = getSegs(staPath), // convert the paths to absolute values, using only cubic bezier
endSegs = getSegs(endPath), // segments, & extract the pt coordinates & segment strings
numSegs = staSegs.length, // note: the # of pts is one less than the # of path segments
numPts = numSegs - 1, // b/c the path's initial 'moveto' pt is also the 'close' pt
linePaths = [],
minSumLensSqrd = Infinity,
rotNumOfMin,
rotNum = 0;
document.querySelector('button').addEventListener('click', function() {
if (rotNum < numPts) {
linePaths.forEach(function(linePath) {linePath.remove();}); // erase any previous coloured lines
var sumLensSqrd = 0;
for (var ptNum = 0; ptNum < numPts; ptNum += 1) { // draw new lines, point-to-point
var linePt1 = staSegs[(rotNum + ptNum) % numPts]; // the new line begins on the 'start' shape
var linePt2 = endSegs[ ptNum % numPts]; // and finished on the 'end' shape
var linePathStr = "M" + linePt1.x + "," + linePt1.y + "L" + linePt2.x + "," + linePt2.y;
var linePath = s.path(linePathStr).attr({stroke: colors[ptNum % colors.length]}); // draw it
var lineLen = Snap.path.getTotalLength(linePath); // calculate its length
sumLensSqrd += lineLen * lineLen; // square the length, and add it to the accumulating total
linePaths[ptNum] = linePath; // remember the path to facilitate erasing it later
}
if (sumLensSqrd < minSumLensSqrd) { // keep track of which rotation has the lowest value
minSumLensSqrd = sumLensSqrd; // of the sum of lengths squared (the 'lsq sum')
rotNumOfMin = rotNum; // as well as the corresponding rotation number
}
show("ROTATION OF POINTS #" + rotNum + ":"); // display info about this rotation
var rotInfo = getRotInfo(rotNum);
show(" point coordinates: " + rotInfo.ptsStr); // show point coordinates
show(" path 'd' string: " + rotInfo.dStr); // show 'd' string needed to draw it
show(" sum of (coloured line lengths squared) = " + sumLensSqrd); // the 'lsq sum'
rotNum += 1; // analyze the next rotation of points
} else { // once all the rotations have been analyzed individually...
linePaths.forEach(function(linePath) {linePath.remove();}); // erase any coloured lines
show(" ");
show("BEST ROTATION, i.e. rotation with lowest sum of (lengths squared): #" + rotNumOfMin);
// show which rotation to use
show("Use the shape based on this rotation of points for morphing");
$("button").off("click");
}
});
function getSegs(path) {
var absCubDStr = Snap.path.toCubic(Snap.path.toAbsolute(path.attr("d")));
return Snap.parsePathString(absCubDStr).map(function(seg, segNum) {
return {x: seg[segNum ? 5 : 1], y: seg[segNum ? 6 : 2], seg: seg.toString()};
});
}
function getRotInfo(rotNum) {
var ptsStr = "";
for (var segNum = 0; segNum < numSegs; segNum += 1) {
var oldSegNum = rotNum + segNum;
if (segNum === 0) {
var dStr = "M" + staSegs[oldSegNum].x + "," + staSegs[oldSegNum].y;
} else {
if (oldSegNum >= numSegs) oldSegNum -= numPts;
dStr += staSegs[oldSegNum].seg;
}
if (segNum !== (numSegs - 1)) {
ptsStr += "(" + staSegs[oldSegNum].x + "," + staSegs[oldSegNum].y + "), ";
}
}
ptsStr = ptsStr.slice(0, ptsStr.length - 2);
return {ptsStr: ptsStr, dStr: dStr};
}
function show(msg) {
var m = document.createElement('pre');
m.innerHTML = msg;
document.body.appendChild(m);
}
pre {
margin: 0;
padding: 0;
}
<script src="//cdn.jsdelivr.net/snap.svg/0.4.1/snap.svg-min.js"></script>
<p>Best viewed on full page</p>
<p>Coloured lines show morph trajectories for the points for that particular rotation of points. The algorithm seeks to optimize those trajectories, essentially trying to find the "shortest" cumulative routes.</p>
<p>The order of points can be seen by following the colour of the lines: red, blue, green, orange (at least when this was originally written), repeating if there are more than 4 points.</p>
<p><button>Click to show rotation of points on top shape</button></p>
Struggeling translating the position of the mouse to the location of the tiles in my grid. When it's all flat, the math looks like this:
this.position.x = Math.floor(((pos.y - 240) / 24) + ((pos.x - 320) / 48));
this.position.y = Math.floor(((pos.y - 240) / 24) - ((pos.x - 320) / 48));
where pos.x and pos.y are the position of the mouse, 240 and 320 are the offset, 24 and 48 the size of the tile. Position then contains the grid coordinate of the tile I'm hovering over. This works reasonably well on a flat surface.
Now I'm adding height, which the math does not take into account.
This grid is a 2D grid containing noise, that's being translated to height and tile type. Height is really just an adjustment to the 'Y' position of the tile, so it's possible for two tiles to be drawn in the same spot.
I don't know how to determine which tile I'm hovering over.
edit:
Made some headway... Before, I was depending on the mouseover event to calculate grid position. I just changed this to do the calculation in the draw loop itself, and check if the coordinates are within the limits of the tile currently being drawn. creates some overhead tho, not sure if I'm super happy with it but I'll confirm if it works.
edit 2018:
I have no answer, but since this ha[sd] an open bounty, help yourself to some code and a demo
The grid itself is, simplified;
let grid = [[10,15],[12,23]];
which leads to a drawing like:
for (var i = 0; i < grid.length; i++) {
for (var j = 0; j < grid[0].length; j++) {
let x = (j - i) * resourceWidth;
let y = ((i + j) * resourceHeight) + (grid[i][j] * -resourceHeight);
// the "+" bit is the adjustment for height according to perlin noise values
}
}
edit post-bounty:
See GIF. The accepted answer works. The delay is my fault, the screen doesn't update on mousemove (yet) and the frame rate is low-ish. It's clearly bringing back the right tile.
Source
Intresting task.
Lets try to simplify it - lets resolve this concrete case
Solution
Working version is here: https://github.com/amuzalevskiy/perlin-landscape (changes https://github.com/jorgt/perlin-landscape/pull/1 )
Explanation
First what came into mind is:
Just two steps:
find an vertical column, which matches some set of tiles
iterate tiles in set from bottom to top, checking if cursor is placed lower than top line
Step 1
We need two functions here:
Detects column:
function getColumn(mouseX, firstTileXShiftAtScreen, columnWidth) {
return (mouseX - firstTileXShiftAtScreen) / columnWidth;
}
Function which extracts an array of tiles which correspond to this column.
Rotate image 45 deg in mind. The red numbers are columnNo. 3 column is highlighted. X axis is horizontal
function tileExists(x, y, width, height) {
return x >= 0 & y >= 0 & x < width & y < height;
}
function getTilesInColumn(columnNo, width, height) {
let startTileX = 0, startTileY = 0;
let xShift = true;
for (let i = 0; i < columnNo; i++) {
if (tileExists(startTileX + 1, startTileY, width, height)) {
startTileX++;
} else {
if (xShift) {
xShift = false;
} else {
startTileY++;
}
}
}
let tilesInColumn = [];
while(tileExists(startTileX, startTileY, width, height)) {
tilesInColumn.push({x: startTileX, y: startTileY, isLeft: xShift});
if (xShift) {
startTileX--;
} else {
startTileY++;
}
xShift = !xShift;
}
return tilesInColumn;
}
Step 2
A list of tiles to check is ready. Now for each tile we need to find a top line. Also we have two types of tiles: left and right. We already stored this info during building matching tiles set.
function getTileYIncrementByTileZ(tileZ) {
// implement here
return 0;
}
function findExactTile(mouseX, mouseY, tilesInColumn, tiles2d,
firstTileXShiftAtScreen, firstTileYShiftAtScreenAt0Height,
tileWidth, tileHeight) {
// we built a set of tiles where bottom ones come first
// iterate tiles from bottom to top
for(var i = 0; i < tilesInColumn; i++) {
let tileInfo = tilesInColumn[i];
let lineAB = findABForTopLineOfTile(tileInfo.x, tileInfo.y, tiles2d[tileInfo.x][tileInfo.y],
tileInfo.isLeft, tileWidth, tileHeight);
if ((mouseY - firstTileYShiftAtScreenAt0Height) >
(mouseX - firstTileXShiftAtScreen)*lineAB.a + lineAB.b) {
// WOHOO !!!
return tileInfo;
}
}
}
function findABForTopLineOfTile(tileX, tileY, tileZ, isLeftTopLine, tileWidth, tileHeight) {
// find a top line ~~~ a,b
// y = a * x + b;
let a = tileWidth / tileHeight;
if (isLeftTopLine) {
a = -a;
}
let b = isLeftTopLine ?
tileY * 2 * tileHeight :
- (tileX + 1) * 2 * tileHeight;
b -= getTileYIncrementByTileZ(tileZ);
return {a: a, b: b};
}
Please don't judge me as I am not posting any code. I am just suggesting an algorithm that can solve it without high memory usage.
The Algorithm:
Actually to determine which tile is on mouse hover we don't need to check all the tiles. At first we think the surface is 2D and find which tile the mouse pointer goes over with the formula OP posted. This is the farthest probable tile mouse cursor can point at this cursor position.
This tile can receive mouse pointer if it's at 0 height, by checking it's current height we can verify if this is really at the height to receive pointer, we mark it and move forward.
Then we find the next probable tile which is closer to the screen by incrementing or decrementing x,y grid values depending on the cursor position.
Then we keep on moving forward in a zigzag fashion until we reach a tile which cannot receive pointer even if it is at it's maximum height.
When we reach this point the last tile found that were at a height to receive pointer is the tile that we are looking for.
In this case we only checked 8 tiles to determine which tile is currently receiving pointer. This is very memory efficient in comparison to checking all the tiles present in the grid and yields faster result.
One way to solve this would be to follow the ray that goes from the clicked pixel on the screen into the map. For that, just determine the camera position in relation to the map and the direction it is looking at:
const camPos = {x: -5, y: -5, z: -5}
const camDirection = { x: 1, y:1, z:1}
The next step is to get the touch Position in the 3D world. In this certain perspective that is quite simple:
const touchPos = {
x: camPos.x + touch.x / Math.sqrt(2),
y: camPos.y - touch.x / Math.sqrt(2),
z: camPos.z - touch.y / Math.sqrt(2)
};
Now you just need to follow the ray into the layer (scale the directions so that they are smaller than one of your tiles dimensions):
for(let delta = 0; delta < 100; delta++){
const x = touchPos.x + camDirection.x * delta;
const y = touchPos.y + camDirection.y * delta;
const z = touchPos.z + camDirection.z * delta;
Now just take the tile at xz and check if y is smaller than its height;
const absX = ~~( x / 24 );
const absZ = ~~( z / 24 );
if(tiles[absX][absZ].height >= y){
// hanfle the over event
}
I had same situation on a game. first I tried with mathematics, but when I found that the clients wants to change the map type every day, I changed the solution with some graphical solution and pass it to the designer of the team. I captured the mouse position by listening the SVG elements click.
the main graphic directly used to capture and translate the mouse position to my required pixel.
https://blog.lavrton.com/hit-region-detection-for-html5-canvas-and-how-to-listen-to-click-events-on-canvas-shapes-815034d7e9f8
https://code.sololearn.com/Wq2bwzSxSnjl/#html
Here is the grid input I would define for the sake of this discussion. The output should be some tile (coordinate_1, coordinate_2) based on visibility on the users screen of the mouse:
I can offer two solutions from different perspectives, but you will need to convert this back into your problem domain. The first methodology is based on coloring tiles and can be more useful if the map is changing dynamically. The second solution is based on drawing coordinate bounding boxes based on the fact that tiles closer to the viewer like (0, 0) can never be occluded by tiles behind it (1,1).
Approach 1: Transparently Colored Tiles
The first approach is based on drawing and elaborated on here. I must give the credit to #haldagan for a particularly beautiful solution. In summary it relies on drawing a perfectly opaque layer on top of the original canvas and coloring every tile with a different color. This top layer should be subject to the same height transformations as the underlying layer. When the mouse hovers over a particular layer you can detect the color through canvas and thus the tile itself. This is the solution I would probably go with and this seems to be a not so rare issue in computer visualization and graphics (finding positions in a 3d isometric world).
Approach 2: Finding the Bounding Tile
This is based on the conjecture that the "front" row can never be occluded by "back" rows behind it. Furthermore, "closer to the screen" tiles cannot be occluded by tiles "farther from the screen". To make precise the meaning of "front", "back", "closer to the screen" and "farther from the screen", take a look at the following:
.
Based on this principle the approach is to build a set of polygons for each tile. So firstly we determine the coordinates on the canvas of just box (0, 0) after height scaling. Note that the height scale operation is simply a trapezoid stretched vertically based on height.
Then we determine the coordinates on the canvas of boxes (1, 0), (0, 1), (1, 1) after height scaling (we would need to subtract anything from those polygons which overlap with the polygon (0, 0)).
Proceed to build each boxes bounding coordinates by subtracting any occlusions from polygons closer to the screen, to eventually get coordinates of polygons for all boxes.
With these coordinates and some care you can ultimately determine which tile is pointed to by a binary search style through overlapping polygons by searching through bottom rows up.
It also matters what else is on the screen. Maths attempts work if your tiles are pretty much uniform. However if you are displaying various objects and want the user to pick them, it is far easier to have a canvas-sized map of identifiers.
function poly(ctx){var a=arguments;ctx.beginPath();ctx.moveTo(a[1],a[2]);
for(var i=3;i<a.length;i+=2)ctx.lineTo(a[i],a[i+1]);ctx.closePath();ctx.fill();ctx.stroke();}
function circle(ctx,x,y,r){ctx.beginPath();ctx.arc(x,y,r,0,2*Math.PI);ctx.fill();ctx.stroke();}
function Tile(h,c,f){
var cnv=document.createElement("canvas");cnv.width=100;cnv.height=h;
var ctx=cnv.getContext("2d");ctx.lineWidth=3;ctx.lineStyle="black";
ctx.fillStyle=c;poly(ctx,2,h-50,50,h-75,98,h-50,50,h-25);
poly(ctx,50,h-25,2,h-50,2,h-25,50,h-2);
poly(ctx,50,h-25,98,h-50,98,h-25,50,h-2);
f(ctx);return ctx.getImageData(0,0,100,h);
}
function put(x,y,tile,image,id,map){
var iw=image.width,tw=tile.width,th=tile.height,bdat=image.data,fdat=tile.data;
for(var i=0;i<tw;i++)
for(var j=0;j<th;j++){
var ijtw4=(i+j*tw)*4,a=fdat[ijtw4+3];
if(a!==0){
var xiyjiw=x+i+(y+j)*iw;
for(var k=0;k<3;k++)bdat[xiyjiw*4+k]=(bdat[xiyjiw*4+k]*(255-a)+fdat[ijtw4+k]*a)/255;
bdat[xiyjiw*4+3]=255;
map[xiyjiw]=id;
}
}
}
var cleanimage;
var pickmap;
function startup(){
var water=Tile(77,"blue",function(){});
var field=Tile(77,"lime",function(){});
var tree=Tile(200,"lime",function(ctx){
ctx.fillStyle="brown";poly(ctx,50,50,70,150,30,150);
ctx.fillStyle="forestgreen";circle(ctx,60,40,30);circle(ctx,68,70,30);circle(ctx,32,60,30);
});
var sheep=Tile(200,"lime",function(ctx){
ctx.fillStyle="white";poly(ctx,25,155,25,100);poly(ctx,75,155,75,100);
circle(ctx,50,100,45);circle(ctx,50,80,30);
poly(ctx,40,70,35,80);poly(ctx,60,70,65,80);
});
var cnv=document.getElementById("scape");
cnv.width=500;cnv.height=400;
var ctx=cnv.getContext("2d");
cleanimage=ctx.getImageData(0,0,500,400);
pickmap=new Uint8Array(500*400);
var tiles=[water,field,tree,sheep];
var map=[[[0,0],[1,1],[1,1],[1,1],[1,1]],
[[0,0],[1,1],[1,2],[3,2],[1,1]],
[[0,0],[1,1],[2,2],[3,2],[1,1]],
[[0,0],[1,1],[1,1],[1,1],[1,1]],
[[0,0],[0,0],[0,0],[0,0],[0,0]]];
for(var x=0;x<5;x++)
for(var y=0;y<5;y++){
var desc=map[y][x],tile=tiles[desc[0]];
put(200+x*50-y*50,200+x*25+y*25-tile.height-desc[1]*20,
tile,cleanimage,x+1+(y+1)*10,pickmap);
}
ctx.putImageData(cleanimage,0,0);
}
var mx,my,pick;
function mmove(event){
mx=Math.round(event.offsetX);
my=Math.round(event.offsetY);
if(mx>=0 && my>=0 && mx<cleanimage.width && my<cleanimage.height && pick!==pickmap[mx+my*cleanimage.width])
requestAnimationFrame(redraw);
}
function redraw(){
pick=pickmap[mx+my*cleanimage.width];
document.getElementById("pick").innerHTML=pick;
var ctx=document.getElementById("scape").getContext("2d");
ctx.putImageData(cleanimage,0,0);
if(pick!==0){
var temp=ctx.getImageData(0,0,cleanimage.width,cleanimage.height);
for(var i=0;i<pickmap.length;i++)
if(pickmap[i]===pick)
temp.data[i*4]=255;
ctx.putImageData(temp,0,0);
}
}
startup(); // in place of body.onload
<div id="pick">Move around</div>
<canvas id="scape" onmousemove="mmove(event)"></canvas>
Here the "id" is a simple x+1+(y+1)*10 (so it is nice when displayed) and fits into a byte (Uint8Array), which could go up to 15x15 display grid already, and there are wider types available too.
(Tried to draw it small, and it looked ok on the snippet editor screen but apparently it is still too large here)
Computer graphics is fun, right?
This is a special case of the more standard computational geometry "point location problem". You could also express it as a nearest neighbour search.
To make this look like a point location problem you just need to express your tiles as non-overlapping polygons in a 2D plane. If you want to keep your shapes in a 3D space (e.g. with a z buffer) this becomes the related "ray casting problem".
One source of good geometry algorithms is W. Randolf Franklin's website and turf.js contains an implementation of his PNPOLY algorithm.
For this special case we can be even faster than the general algorithms by treating our prior knowledge about the shape of the tiles as a coarse R-tree (a type of spatial index).
I have a canvas on which i'm drawing 4 lines.
I need to then compare the accuracy of those lines drawn against the coordinates of 4 predefined lines.
Can i do this with canvas?
thanks.
As JJPA says, give it a try!
One route you could take would be to store the coordinates of the drawn line and the expected line. Each line forms a vector. You can take the Euclidean distance between the lines using a simple formula. The closer the distance, the more accurate the lines.
I just wrote this.
var expected = {from: {x:10, y:10}, to:{x:20, y:20}}
var identical = {from: {x:10, y:10}, to:{x:20, y:20}}
var close = {from: {x:9, y:9}, to:{x:21, y:21}}
var lessClose = {from: {x:8, y:13}, to:{x:25, y:15}}
var distance = function(a, b) {
return Math.sqrt(Math.pow(a.from.x - b.from.y, 2) +
Math.pow(a.to.x - b.to.y, 2))
}
console.log("identical", distance(expected, identical))
console.log("close", distance(expected, close))
console.log("lessclose", distance(expected, lessClose))
gives
identical 0
close 1.4142135623730951
lessclose 5.830951894845301
If you don't know which way round the lines are, you can swap start and end points and take the minimum distance.
Well, you can get the pixels of your canvas in an rgb-array in javascript:
var pixelarray= canvas.getImageData(x, y, width, height).data;
then pixelarray[0] as integer is the red-value of the first pixel, pixelarray[3] the red-value of the second and so on.
In a canvas of 100px width (pixel-index 0 to 299 in first line), pixelarray[299] would be the blue-value of the last pixel in the first line and pixelarray[300] the red-value of the first in second line.
Then you could get either the functions representing the 4 lines you want to compare to and do something like
function isblack(x,y,width,height){
return pixelarray[(y*width+x)*3]==255&&pixelarray[(y*width+x)*3+1]==255&&pixelarray[(y*width+x)*3+2]==255||x<0||x>width||y<0||y>height;
}
function f(x){return a*x+b;}
var lineblackness=0;
for(var x=0;x<picturewidth;x++)if(isblack(x,f(x),picturewidth,pictureheight))lineblackness++;
if(lineblackness==picturewidth)line_is_drawn();
Where fx(x) would represent the line ...
well, if you use a picture or coordinates for the line, it might get easier:
getcoord(data){
var ar=new array();
for(var i=0;i<data.length/3;i++)if(data[i*3]==255)ar.push(i);
return ar;
}
var comparecoords=getcoord(comparecanvas.getImageData(x, y, width, height).data);
var thisdata=canvas.getImageData(x, y, width, height).data;
var linehits=0;
for(var i=0;i<comparecoords.length;i++){if(thisdata[comparecoords[i]*3]==255)linehits++;}
if(comparecoords.length==linehits)all_coords_are_hit();
Here I only compared the red-value, assuming you have a black&White-image.
If you want to compare lines that aren't exactly fits, you can change the ==255 with a few additional ifs e.g.
function isblack(x,y,width,height,blurr,treshhold){
for(var i=-blurr;i<blurr;i++)if(x-i>0&&x+i<width&&pixelarray[(y*width+(x+i))*3]>=treshhold)return true;
for(var i=-blurr;i<blurr;i++)if(y-i>0&&y+i<height&&pixelarray[((y+i)*width+(x))*3]>=treshhold)return true;
return false;
}
This function uses also only red, but uses a treshhold to let also gray be valid. it also accepts a blurr-value for checking in x-direction before/after the pixel and then again in y-direction.
You could also combine the two loops for checking in an rectangle around the pixel.
Changing the return true into return i will also give you the distance from the pixel you want to check. if you combine the two loops into an i and a j-value, return sqrt(i^2+j^2) and sum all checks up, you can get an indicator how accurate your line is.
To implement sqareroot-error, only comparing the y-distance and summing up the squared distance i^2 should be enough.
Please note that these programs are suggestions, better implementations may exists and my codes could still contain errors.