I am trying to make a maze generator, and almost everything is working so far. I have been able to set my position to a random pos, and then I repeat the standard() function. In the function, I add pos to posList, and then I choose a random direction. Next, I check if the cell has been visited by running through all of the posList vectors in reverse. I haven't executed the code that backtracks yet. If visited = false then I move to the square and execute the yet-to-be-made path() function. However, for some reason, the mover just doesn't detect if a cell has been visited or not. I am using p5.js. What am I doing wrong?
var posList = [];
var pos;
var tempo;
var boole = false;
var direc;
var mka = 0;
function setup() {
createCanvas(400, 400);
//Set up position
pos = createVector(floor(random(3)), floor(random(3)));
frameRate(1)
}
//Choose a direction
function direct(dire) {
if(dire === 0) {
return(createVector(0, -1));
} else if(dire === 1) {
return(createVector(1, 0));
} else if(dire === 2) {
return(createVector(0, 1));
} else {
return(createVector(-1, 0));
}
}
/foLo stands fo forLoop
function foLo() {
//If we have checked less than three directions and know there is a possibility for moving
if(mka < 4) {
//tempoRARY, this is what we use to see if the cell has been visited
tempo = createVector(pos.x + direct(direc).x, pos.y + direct(direc).y);
//Go through posList backwards
for(var i = posList.length - 1; i >= 0; i --) {
//If the cell has been visited or the cell is off of the screen
if(tempo === posList[i]) {
//Change the direction
direc ++;
//Roll over direction value
if(direc === 4) {
direc = 0;
}
//Re-execute on next frame
foLo();
//The cell has been visited
boole = false;
//Debugging
console.log(direc)
mka++;
} else if(tempo.x < 0 || tempo.x > 2 || tempo.y < 0 || tempo.y > 2) {
direc ++;
if(direc === 4) {
direc = 0;
}
foLo();
boole = false;
console.log(direc)
mka++;
}
}
//If it wasn't visited (Happens every time for some reason)
if(boole === true) {
//position is now the temporary value
pos = tempo;
console.log("works")
mka = 0;
}
}
}
function standard() {
//Add pos to posList
posList.push(pos);
//Random direction
direc = floor(random(4));
//Convert to vector
direct(direc);
foLo();
//Tracks pos
fill(255, 255, 0);
rect(pos.x*100+50, pos.y*100+50, 50, 50)
}
function draw() {
background(255);
fill(0);
noStroke();
//draw grid
for(var i = 0; i < 4; i ++) {
rect(i*100,0,50,350);
rect(0, i*100, 350, 50);
}
standard();
boole = true;
console.log(pos)
console.log(posList);
}
Your issue is on the line where you compare two vectors if(tempo === posList[i]) {: This will never be true.
You can verify that with the following code (in setup() for example):
const v1 = new p5.Vector(1, 0);
const v2 = new p5.Vector(1, 0);
const v3 = new p5.Vector(1, 1);
console.log(v1 === v2) // false
console.log(v1 === v3) // false
This is because despite having the same value v1 and v2 are referencing two different objects.
What you could do is using the p5.Vector.equals function. The doc has the following example:
let v1 = createVector(10.0, 20.0, 30.0);
let v2 = createVector(10.0, 20.0, 30.0);
let v3 = createVector(0.0, 0.0, 0.0);
print(v1.equals(v2)); // true
print(v1.equals(v3)); // false
This might not give you a working algorithm because I suspect you have other logical errors (but I could be wrong or you will debug them later on) but at least this part of the code will do what you expect.
Another solution is to use a Set instead of your list of positions. The cons of this solution is that you will have to adapt your code to handle the "out of grid" position situation. However when you want to keep track of visited items a Set is usually a great solution because the access time is constant. That this means is that to define is a position has already been visited it will always take the same time (you'll do something like visitedSet.has(positionToCheck), whereas with your solution where you iterate through a list the more cells you have visited to longer it will take to check if the cell is in the list.
The Set solution will require that you transform your vectors before adding them to the set though sine, has I explained before you cannot simply compare vectors. So you could check for their string representation with something like this:
const visitedCells = new Set();
const vectorToString = (v) => `${v.x},{$v.y}` // function to get the vector representation
// ...
visitedCells.add(vectorToString(oneCell)); // Mark the cell as visited
visited = visitedCells.has(vectorToString(anotherCell))
Also has a general advice you should pay attention to your variables and functions name. For example
// foLo stands fo forLoop
function foLo() {
is a big smell: Your function name should be descriptive, when you see your function call foLo(); having to find the comment next to the function declaration makes the code less readable. You could call it generateMaze() and this way you'll know what it's doing without having to look at the function code.
Same for
//tempoRARY, this is what we use to see if the cell has been visited
tempo = createVector(pos.x + direct(direc).x, pos.y + direct(direc).y);
You could simply rename tempo to cellToVisit for example.
Or boole: naming a boolean boole doesn't convey a lot of information.
That could look like some minor details when you just wrote the code but when your code will be several hundred lines or when you read it again after taking several days of break, you'll thank past you for taking care of that.
I'm using KineticJS and trying to get something done that seems simple enough: trying to get a shape to change position with the shape that's currently being dragged.
The idea is that:
• You pick up a shape on the canvas. This triggers a mousedown event listener, which saves the current position of the shape you've picked up.
• While holding the shape, if you trigger the mouseover on another shape, that shape's event gets triggered and swaps its position, based on the current shape's saved position.
Here is the relevant code I've written to try to get that working:
Board setup:
Simply just setting up the board and calling the needed functions here. I'm not doing anything with the stage, gameBoard, or ctx yet (part of it was for when I tried to get drawImage to work on multiple canvases, but have abandoned that for now).
class BoardView {
constructor(stage, gameBoard, ctx) {
this.stage = stage;
this.gameBoard = gameBoard;
this.ctx = ctx;
this.orbs = [[], [], [], [], []];
this.setupBoard();
}
Board setup functions:
This is where I go about setting up the board and giving each Kinetic Circle the attributes it needs to render on the Layer. Maybe I'm missing something obvious here?
setupBoard () {
for (let colIdx = 0; colIdx < 5; colIdx++) {
this.addRow(colIdx);
}
this.renderBoard();
}
addRow (colIdx) {
for (let rowIdx = 0; rowIdx < 6; rowIdx++) {
let orbType = Math.round(Math.random() * 5);
let orbColor;
if (orbType === 0) {
orbColor = "#990000";
} else if (orbType === 1) {
orbColor = "#112288";
} else if (orbType === 2) {
orbColor = "#005544";
} else if (orbType === 3) {
orbColor = "#776611";
} else if (orbType === 4) {
orbColor = "#772299";
} else {
orbColor = "#dd2277";
}
let orbject = new Kinetic.Circle({
x: (rowIdx + 0.5) * 100,
y: (colIdx + 0.5) * 100,
width: 100,
height: 100,
fill: orbColor,
draggable: true
});
this.orbs[colIdx].push(orbject);
}
}
Board render function:
This is where I add all the Kinetic Circle objects into new layers, and give those layers all its own attributes to work with when I call the event handlers. I also set up the event handlers here after adding the layers to the stage. Am I perhaps messing this up by adding too many layers?
renderBoard () {
for (let row = 0; row < this.orbs.length; row++) {
for (let orb = 0; orb < this.orbs[row].length; orb++) {
let layer = new Kinetic.Layer();
layer.add(this.orbs[row][orb]);
layer.moving = false;
layer.orbId = `orb${row}${orb}`;
layer.pos = [this.orbs[row][orb].attrs.x, this.orbs[row][orb].attrs.y];
this.stage.add(layer);
layer.on("mousedown", this.handleMouseDown);
layer.on("mouseup", this.handleMouseUp);
layer.on("mouseout", this.handleMouseOut);
layer.on("mousemove", this.handleMouseMove);
}
}
}
Mouse event handlers:
This is where I think I'm having my main problem. How I handle the event of moving the mouse to change orbs, perhaps I'm doing something terribly wrong?
handleMouseDown (e) {
window.currentOrb = this;
console.log(window.currentOrb.orbId);
this.moving = true;
}
//handleMouseUp (e) {
// window.currentOrb = undefined;
// this.moving = false;
//}
//handleMouseOut (e) {
//}
handleMouseMove (e) {
if (window.currentOrb !== undefined && window.currentOrb.orbId != this.orbId) {
this.children[0].attrs.x = window.currentOrb.pos[0];
this.children[0].attrs.y = window.currentOrb.pos[1];
this.children[0].draw();
} else {
}
}
}
module.exports = BoardView;
I've tried looking at the KineticJS docs and many StackOverflow answers as I could in hopes of finding a solution that would work for me, but nothing I've seen and tried so far (including the suggestions that came up as I wrote this question) seemed to be of help, and I'm aware the way I've gone about this so far is probably far from the best way to accomplish what I want, so I'm open to any suggestions, pointers, answered questions, or whatever can point me in the right direction to what I'm missing in order to get this to work.
In case this is helpful, here is also a visualization of how things look when the board is rendered.
The circles are all Kinetic Circles (orbs for the purpose of what I'm going for), and clicking and dragging one to another, the one that isn't being dragged but hovered over should move to the original position of the dragged circles.
Thanks!
EDIT:
I made a few changes to the code since then. First off, I changed adding many layers to the stage, to just one:
renderBoard () {
let layer = new Kinetic.Layer();
for (let row = 0; row < this.orbs.length; row++) {
for (let orb = 0; orb < this.orbs[row].length; orb++) {
layer.add(this.orbs[row][orb]);
this.orbCanvases.push(orbCanvas.id);
}
}
this.stage.add(layer);
}
I instead added listeners to the orb objects instead:
addRow (colIdx) {
for (let rowIdx = 0; rowIdx < 6; rowIdx++) {
//same code as before
let orbject = new Kinetic.Circle({
x: (rowIdx + 0.5) * 100, y: (colIdx + 0.5) * 100,
width: 100, height: 100,
fill: orbColor, draggable: true, pos: [rowIdx, colIdx]
});
orbject.on("mousedown", this.handleMouseDown);
orbject.on("mouseup", this.handleMouseUp);
orbject.on("mouseout", this.handleMouseOut);
orbject.on("mousemove", this.handleMouseMove);
this.orbs[colIdx].push(orbject);
}
}
This has had the benefit of making drag and drop much much faster where before, it was going very slow, but I still can't get my objects to swap position.
To be clear, my main issue is knowing which x, y values I should be changing. At the moment in handleMouseMove, I've been trying to change:
e.target.attrs.x = newX;
e.target.attrs.y = newY;
// newX and newY can be any number
However, no matter what I change it to, it has no effect. So it would help me to know if I'm changing the wrong thing/place, for example, maybe I'm supposed to be changing the Kinetic Circle from the array I've stored? Thanks again.
EDIT 2:
I think I got it! However, I had to take this.orbs and set it in the window at window.orbs, and to test it I did:
window.orbs[0][0].x(450);
window.orbs[0][0].draw();
And this caused the x position to change. But putting it in a window does not seem like good practice?
EDIT 3:
I got the orbs to now swap continuously, except for when it's the same orb being swapped again while mouseover is continuing to fire. At mouseup however, it can be swapped again. I also had to set up multiple layers again to get the mouseover events to work while holding another orb, but the performance seems to have improved a little.
I'm going to try and figure out how to get them to be able to swap continuously on the same mouse hold, but in the meantime, here is the code I wrote to achieve this:
addRow (colIdx) {
for (let rowIdx = 0; rowIdx < 6; rowIdx++) {
// same code as before, changed attr given to Kinetic.Circle
let orbject = new Kinetic.Circle({
x: (rowIdx + 0.5) * 100, y: (colIdx + 0.5) * 100,
width: 100, height: 100,
fill: orbColor, draggable: true, orbId: `orb${colIdx}${rowIdx}`
});
}
}
handleMouseDown (e) {
window.currentOrb = window.orbs[e.target.attrs.orbId];
window.newX = e.target.attrs.x;
window.newY = e.target.attrs.y;
}
Mouse down saving currentOrb by ID and its X and Y
handleMouseUp (e) {
window.currentOrb.x(window.newX);
window.currentOrb.y(window.newY);
window.currentOrb.parent.clear();
window.currentOrb.parent.draw();
window.currentOrb.draw();
window.currentOrb = undefined;
for (let i = 0; i < 5; i++) {
for (let j = 0; j < 6; j++) {
window.orbs[`orb${i}${j}`].draw();
}
}
}
When mouse is released, currently, all orbs are redrawn so they can all be used. I plan to refactor this so only orbs hovered over have this change.
handleMouseMove (e) {
if (window.currentOrb !== undefined && (window.currentOrb.attrs.orbId !== e.target.attrs.orbId)) {
window.orbMove.pause();
window.currentTime = 0;
window.orbMove.play();
let targOrbX = e.target.attrs.x;
let targOrbY = e.target.attrs.y;
// This is the target orb that's being changed's value
// We're storing this in targOrb
e.target.x(window.newX);
e.target.y(window.newY);
e.target.parent.clear();
e.target.parent.draw();
e.target.draw();
// Here we have the previously set current orb's position becoming
// the target orb's position
window.newX = targOrbX;
window.newY = targOrbY;
// Now that the move is made, we can set the newX and Y to be the
// target orb's position once mouseup
}
}
Orb swapping logic which works for passing over orbs once, but not if they are passed over again in the same turn.
When does the "hover" officially take place?
When the mouse event's position enters the second orb? If yes, hit test the mouse vs every non-dragging orb:
// pseudo-code -- make this test for every non-dragging orb
var dx=mouseX-orb[n].x;
var dy=mouseY-orb[n].y;
if(dx*dx+dy*dy<orb[n].radius){
// change orb[n]'s x,y to the dragging orb's x,y (and optionally re-render)
}
When the dragging orb intersects the second orb? If yes, collision test the dragging orb vs every non-dragging orb:
// pseudo-code -- make this test for every non-dragging orb
var dx=orb[draggingIndex].x-orb[n].x;
var dy=orb[draggingIndex].y-orb[n].y;
var rSum=orb[draggingIndex].radius+orb[n].radius;
if(dx*dx+dy*dy<=rSum*rSum){
// change orb[n]'s x,y to the dragging orb's x,y (and optionally re-render)
}
BTW, if you drag an orb over all other orbs, the other orbs will all stack on the dragging orbs original position -- is that what you want?
I have been working on a JavaScript implementation of the early 90's adventure games and specifically plotting a path from the place the hero is standing to the location the player has clicked on. My approach is to first determine if a strait line (without obstructions) can be drawn, if not then to search for a path of clear way-points using Brian Grinstead's excellent javascript-astar. The problem I'm facing however is the path (while optimal will veer into spaces that would seem to the user an unintended. Here is a classic example of what I'm talking about (the green path is the generated path, the red dots are each turns where the direction of the path changes):
Now I know that A* is only guaranteed to return a path that cannot be simpler (in terms of steps), but I'm struggling to implement a heuristic that weights turns. Here is a picture that show two other paths that would also qualify as just as simple (with an equal number of steps)
The Blue path would present the same number of steps and turns while the red path has the same number of steps and fewer turns. In my code I have a simplifyPath() function that removes steps where the direction changes, so if I could get all possible paths from astar then I could select the one with the least turns, but that's not how A* fundamentally works, so I'm looking for a way to incorporate simplicity into the heuristic.
Here is my current code:
var img,
field = document.getElementById('field'),
EngineBuilder = function(field, size) {
var context = field.getContext("2d"),
graphSettings = { size: size, mid: Math.ceil(size/2)},
engine = {
getPosition: function(event) {
var bounds = field.getBoundingClientRect(),
x = Math.floor(((event.clientX - bounds.left)/field.clientWidth)*field.width),
y = Math.floor(((event.clientY - bounds.top)/field.clientHeight)*field.height),
node = graph.grid[Math.floor(y/graphSettings.size)][Math.floor(x/graphSettings.size)];
return {
x: x,
y: y,
node: node
}
},
drawObstructions: function() {
context.clearRect (0, 0, 320, 200);
if(img) {
context.drawImage(img, 0, 0);
} else {
context.fillStyle = 'rgb(0, 0, 0)';
context.fillRect(200, 100, 50, 50);
context.fillRect(0, 100, 50, 50);
context.fillRect(100, 100, 50, 50);
context.fillRect(0, 50, 150, 50);
}
},
simplifyPath: function(start, complexPath, end) {
var previous = complexPath[1], simplePath = [start, {x:(previous.y*graphSettings.size)+graphSettings.mid, y:(previous.x*graphSettings.size)+graphSettings.mid}], i, classification, previousClassification;
for(i = 1; i < (complexPath.length - 1); i++) {
classification = (complexPath[i].x-previous.x).toString()+':'+(complexPath[i].y-previous.y).toString();
if(classification !== previousClassification) {
simplePath.push({x:(complexPath[i].y*graphSettings.size)+graphSettings.mid, y:(complexPath[i].x*graphSettings.size)+graphSettings.mid});
} else {
simplePath[simplePath.length-1]={x:(complexPath[i].y*graphSettings.size)+graphSettings.mid, y:(complexPath[i].x*graphSettings.size)+graphSettings.mid};
}
previous = complexPath[i];
previousClassification = classification;
}
simplePath.push(end);
return simplePath;
},
drawPath: function(start, end) {
var path, step, next;
if(this.isPathClear(start, end)) {
this.drawLine(start, end);
} else {
path = this.simplifyPath(start, astar.search(graph, start.node, end.node), end);
if(path.length > 1) {
step = path[0];
for(next = 1; next < path.length; next++) {
this.drawLine(step, path[next]);
step = path[next];
}
}
}
},
drawLine: function(start, end) {
var x = start.x,
y = start.y,
dx = Math.abs(end.x - start.x),
sx = start.x<end.x ? 1 : -1,
dy = -1 * Math.abs(end.y - start.y),
sy = start.y<end.y ? 1 : -1,
err = dx+dy,
e2, pixel;
for(;;) {
pixel = context.getImageData(x, y, 1, 1).data[3];
if(pixel === 255) {
context.fillStyle = 'rgb(255, 0, 0)';
} else {
context.fillStyle = 'rgb(0, 255, 0)';
}
context.fillRect(x, y, 1, 1);
if(x === end.x && y === end.y) {
break;
} else {
e2 = 2 * err;
if(e2 >= dy) {
err += dy;
x += sx;
}
if(e2 <= dx) {
err += dx;
y += sy;
}
}
}
},
isPathClear: function(start, end) {
var x = start.x,
y = start.y,
dx = Math.abs(end.x - start.x),
sx = start.x<end.x ? 1 : -1,
dy = -1 * Math.abs(end.y - start.y),
sy = start.y<end.y ? 1 : -1,
err = dx+dy,
e2, pixel;
for(;;) {
pixel = context.getImageData(x, y, 1, 1).data[3];
if(pixel === 255) {
return false;
}
if(x === end.x && y === end.y) {
return true;
} else {
e2 = 2 * err;
if(e2 >= dy) {
err += dy;
x += sx;
}
if(e2 <= dx) {
err += dx;
y += sy;
}
}
}
}
}, graph;
engine.drawObstructions();
graph = (function() {
var x, y, rows = [], cols, js = '[';
for(y = 0; y < 200; y += graphSettings.size) {
cols = [];
for(x = 0; x < 320; x += graphSettings.size) {
cols.push(context.getImageData(x+graphSettings.mid, y+graphSettings.mid, 1, 1).data[3] === 255 ? 0 : 1);
}
js += '['+cols+'],\n';
rows.push(cols);
}
js = js.substring(0, js.length - 2);
js += ']';
document.getElementById('Graph').value=js;
return new Graph(rows, { diagonal: true });
})();
return engine;
}, start, end, engine = EngineBuilder(field, 10);
field.addEventListener('click', function(event) {
var position = engine.getPosition(event);
if(!start) {
start = position;
} else {
end = position;
}
if(start && end) {
engine.drawObstructions();
engine.drawPath(start, end);
start = end;
}
}, false);
#field {
border: thin black solid;
width: 98%;
background: #FFFFC7;
}
#Graph {
width: 98%;
height: 300px;
overflow-y: scroll;
}
<script src="http://jason.sperske.com/adventure/astar.js"></script>
<code>Click on any 2 points on white spaces and a path will be drawn</code>
<canvas id='field' height
='200' width='320'></canvas>
<textarea id='Graph' wrap='off'></textarea>
After digging into Michail Michailidis' excellent answer I've added the following code to my simplifyPath() function) (demo):
simplifyPath: function(start, complexPath, end) {
var previous = complexPath[1],
simplePath = [start, {x:(previous.y*graphSettings.size)+graphSettings.mid, y:(previous.x*graphSettings.size)+graphSettings.mid}],
i,
finalPath = [simplePath[0]],
classification,
previousClassification;
for(i = 1; i < (complexPath.length - 1); i++) {
classification = (complexPath[i].x-previous.x).toString()+':'+(complexPath[i].y-previous.y).toString();
if(classification !== previousClassification) {
simplePath.push({x:(complexPath[i].y*graphSettings.size)+graphSettings.mid, y:(complexPath[i].x*graphSettings.size)+graphSettings.mid});
} else {
simplePath[simplePath.length-1]={x:(complexPath[i].y*graphSettings.size)+graphSettings.mid, y:(complexPath[i].x*graphSettings.size)+graphSettings.mid};
}
previous = complexPath[i];
previousClassification = classification;
}
simplePath.push(end);
previous = simplePath[0];
for(i = 2; i < simplePath.length; i++) {
if(!this.isPathClear(previous, simplePath[i])) {
finalPath.push(simplePath[i-1]);
previous = simplePath[i-1];
}
}
finalPath.push(end);
return finalPath;
}
Basically after it reduces redundant steps in the same direction, it tries to smooth out the path by looking ahead to see if it can eliminate any steps.
Very very intresting problem! Thanks for this question! So...some observations first:
Not allowing diagonal movement fixes this problem but since you are interested in diagonal movement I had to search more.
I had a look at path simplifying algorithms like:
Ramer Douglas Peuker
(http://en.wikipedia.org/wiki/Ramer%E2%80%93Douglas%E2%80%93Peucker_algorithm)
and an implementation: https://gist.github.com/rhyolight/2846020.
I added an implementation to your code without success. This algorithm doesn't take into account obstacles so it was difficult to adapt it.
I wonder what would the behavior be (for diagonal movements) if you had used Dijkstra instead of A* or if you used an 'all shortest paths between a pair of nodes' algorithm and then you sorted them by increasing changes in direction.
After reading a bit about A* here http://buildnewgames.com/astar/ I thought that the implementation of A-star you are using is the problem or the heuristics. I tried all the heuristics on the a-star of your code including euclidean that I coded myself and tried also all the heuristics in the http://buildnewgames.com/astar code Unfortunately all of the diagonal allowing heuristics were having the same issue you are describing.
I started working with their code because it is a one-to-one grid and yours was giving me issues drawing. Your simplifyPath that I tried to remove was also causing additional problems. You have to keep in mind that since
you are doing a remapping this could be an issue based on that
On a square grid that allows 4 directions of movement, use Manhattan distance (L1).
On a square grid that allows 8 directions of movement, use Diagonal distance (L∞).
On a square grid that allows any direction of movement, you might or might not want Euclidean distance (L2). If A* is finding paths on the grid but you are allowing movement not on the grid, you may want to consider other representations of the map. (from http://theory.stanford.edu/~amitp/GameProgramming/Heuristics.html)
What is my pseudocode algorithm:
var path = A-star();
for each node in path {
check all following nodes till some lookahead limit
if you find two nodes in the same row but not column or in the same column but not row {
var nodesToBeStraightened = push all nodes to be "straightened"
break the loop;
}
skip loop index to the next node after zig-zag
}
if nodesToBeStraightened length is at least 3 AND
nodesToBeStraightened nodes don't form a line AND
the resulting Straight line after simplification doesn't hit an obstruction
var straightenedPath = straighten by getting the first and last elements of nodesToBeStraightened and using their coordinates accordingly
return straightenedPath;
Here is the visual explanation of what is being compared in the algorithm:
Visual Explanation:
How this code will be used with yours (I did most of the changes - I tried my best but there are so many problems like with how you do drawing and because of the rounding of the grid etc - you have to use a grid and keep the scale of the paths accurate - please see also assumptions below):
var img,
field = document.getElementById('field'),
EngineBuilder = function(field, size) {
var context = field.getContext("2d"),
graphSettings = { size: size, mid: Math.ceil(size/2)},
engine = {
//[...] missing code
removeZigZag: function(currentPath,lookahead){
//for each of the squares on the path - see lookahead more squares and check if it is in the path
for (var i=0; i<currentPath.length; i++){
var toBeStraightened = [];
for (var j=i; j<lookahead+i+1 && j<currentPath.length; j++){
var startIndexToStraighten = i;
var endIndexToStraighten = i+1;
//check if the one from lookahead has the same x xor the same y with one later node in the path
//and they are not on the same line
if(
(currentPath[i].x == currentPath[j].x && currentPath[i].y != currentPath[j].y) ||
(currentPath[i].x == currentPath[j].y && currentPath[i].x != currentPath[j].x)
) {
endIndexToStraighten = j;
//now that we found something between i and j push it to be straightened
for (var k = startIndexToStraighten; k<=endIndexToStraighten; k++){
toBeStraightened.push(currentPath[k]);
}
//skip the loop forward
i = endIndexToStraighten-1;
break;
}
}
if (toBeStraightened.length>=3
&& !this.formsALine(toBeStraightened)
&& !this.lineWillGoThroughObstructions(currentPath[startIndexToStraighten], currentPath[endIndexToStraighten],this.graph?????)
){
//straightening:
this.straightenLine(currentPath, startIndexToStraighten, endIndexToStraighten);
}
}
return currentPath;
},
straightenLine: function(currentPath,fromIndex,toIndex){
for (var l=fromIndex; l<=toIndex; l++){
if (currentPath[fromIndex].x == currentPath[toIndex].x){
currentPath[l].x = currentPath[fromIndex].x;
}
else if (currentPath[fromIndex].y == currentPath[toIndex].y){
currentPath[l].y = currentPath[fromIndex].y;
}
}
},
lineWillGoThroughObstructions: function(point1, point2, graph){
var minX = Math.min(point1.x,point2.x);
var maxX = Math.max(point1.x,point2.x);
var minY = Math.min(point1.y,point2.y);
var maxY = Math.max(point1.y,point2.y);
//same row
if (point1.y == point2.y){
for (var i=minX; i<=maxX && i<graph.length; i++){
if (graph[i][point1.y] == 1){ //obstacle
return true;
}
}
}
//same column
if (point1.x == point2.x){
for (var i=minY; i<=maxY && i<graph[0].length; i++){
if (graph[point1.x][i] == 1){ //obstacle
return true;
}
}
}
return false;
},
formsALine: function(pointsArray){
//only horizontal or vertical
if (!pointsArray || (pointsArray && pointsArray.length<1)){
return false;
}
var firstY = pointsArray[0].y;
var lastY = pointsArray[pointsArray.length-1].y;
var firstX = pointsArray[0].x;
var lastX = pointsArray[pointsArray.length-1].x;
//vertical line
if (firstY == lastY){
for (var i=0; i<pointsArray.length; i++){
if (pointsArray[i].y!=firstY){
return false;
}
}
return true;
}
//horizontal line
else if (firstX == lastX){
for (var i=0; i<pointsArray.length; i++){
if (pointsArray[i].x!=firstX){
return false;
}
}
return true;
}
return false;
}
//[...] missing code
}
//[...] missing code
}
Assumptions and Incompatibilities of the above code:
obstacle is 1 and not 0
the orginal code I have in the demo is using array instead of {x: number, y:number}
in case you use the other a-star implementation, the point1 location is on the column 1 and row 2.
converted to your {x: number, y:number} but haven't checked all the parts:
I couldn't access the graph object to get the obstacles - see ?????
You have to call my removeZigZag with a lookahead e.g 7 (steps away) in the place where you were
doing your path simplification
admittedly their code is not that good compared to the a-star implementation from Stanford but for our purposes it should be irelevant
possibly the code has bugs that I don't know of and could be improved. Also I did my checks only in this specific world configuration
I believe the code has complexity O(N x lookahead)~O(N) where N is the number of steps in the input A* shortest path.
Here is the code in my github repository (you can run the demo)
https://github.com/zifnab87/AstarWithDiagonalsFixedZigZag
based on this A* Javascript implementation downloaded from here: http://buildnewgames.com/astar/
Their clickHandling and world boundary code is broken as when you click on the right side of the map the path calculation is not working sometimes. I didn't have time to find their bug. As a result my code has the same issue
Probably it is because the map I put from your question is not square - but anyway my algorithm should be unaffected. You will see this weird behavior is happening if non of my remove ZigZag code runs. (Edit: It was actually because the map was not square - I updated the map to be square for now)
Feel free to play around by uncommenting this line to see the before-after:
result = removeZigZag(result,7);
I have attached 3 before after image sets so the results can be visualized:
(Keep in mind to match start and goal if you try them - direction DOES matter ;) )
Case 1: Before
Case 1: After
Case 2: Before
Case 2: After
Case 3: Before
Case 3: After
Case 4: Before
Case 4: After
Resources:
My code (A* diagonal movement zig zag fix demo): https://github.com/zifnab87/AstarWithDiagonalsFixedZigZag
Original Javascript A* implementation of my demo can be found above (first commit) or here: - http://buildnewgames.com/astar/
A* explanation: http://buildnewgames.com/astar/
A* explanation from Stanford: http://theory.stanford.edu/~amitp/GameProgramming/Heuristics.html
JavaScript A* implementation used by OP's question (Github):
Ramer Douglas Peuker Algorithm (Wikipedia): http://en.wikipedia.org/wiki/Ramer%E2%80%93Douglas%E2%80%93Peucker_algorithm
Javascript implementation of Douglas Peuker Algorithm: https://gist.github.com/rhyolight/2846020
A* Algorithm (Wikipedia): http://en.wikipedia.org/wiki/A*_search_algorithm
You can use a modified A* algorithm to account for changes in direction. While simplifying the result of the standard A* algorithm may yield good results, it may not be optimal. This modified A* algorithm will return a path of minimal length with the least number of turns.
In the modified A* algorithm, each position corresponds to eight different nodes, each with their own heading. For example, the position (1, 1) corresponds to the eight nodes
(1,1)-up, (1,1)-down, (1,1)-right, (1,1)-left,
(1,1)-up-left, (1,1)-up-right, (1,1)-down-left, and (1,1)-down-right
The heuristic distance from a node to the goal is the the heuristic distance from the corresponding point to the goal. In this case, you probably want to use the following function:
H(point) = max(abs(goal.xcor-point.xcor), abs(goal.ycor-point.ycor))
The nodes that correspond to a particular position are connected to the nodes of the neighboring positions with the proper heading. For example, the nodes corresponding to the position (1,1) are all connected to the following eight nodes
(1,2)-up, (1,0)-down, (2,1)-right, (0,1)-left,
(0,2)-up-left, (2,2)-up-right, (0,0)-down-left, and (2,0)-down-right
The distance between any two connected nodes depends on their heading. If they have the same head, then the distance is 1, otherwise, we have made a turn, so the distance is 1+epsilon. epsilon represents an arbitrarily small value/number.
We know need to have a special case for the both the start and goal. The start and goal are both represented as a single node. At the start, we have no heading, so the distance between the start node and any connected node is 1.
We can now run the standard A* algorithm on the modified graph. We can map the returned path to a path in the original grid, by ignoring the headings. The total length of the returned path will be of the form n+m*epsilon. n is the total length of the corresponding path in the original grid, and m is the number of turns. Because A* returns a path of minimal length, the path in the original grid is a path of minimal length that makes the least turns.
I have come up with somewhat of a fix that is a simple addition to your original code, but it doesn't work in all situations (see image below) because we are limited to what the A* returns us. You can see my jsfiddle here
I added the following code to your simplifyPath function right before the return. What it does is strips out extra steps by seeing if there is a clear path between non-adjacent steps (looking at larger gaps first). It could be optimized, but you should get the gist from what I've got.
do{
shortened = false;
loop:
for(i = 0; i < simplePath.length; i++) {
for(j = (simplePath.length - 1); j > (i + 1); j--) {
if(this.isPathClear(simplePath[i],simplePath[j])) {
simplePath.splice((i + 1),(j - i - 1));
shortened = true;
break loop;
}
}
}
} while(shortened == true);
You can see below that this removes the path that goes in on the left (as in the question) but also that not all the odd turns are removed. This solution only uses the points provided from the A*, not points in between on the path - for example, because the 2nd point does not have a straight unobstructed line to the 4th or 5th points, it cannot optimize point 3 out. It happens a lot less than the original code, but it still does give weird results sometimes.
In edition to nodes having references to their parent nodes. Also store which direction that node came from inside a variable. In my case there was only two possibilities horizontally or vertically. So I created two public static constants for each possibility. And a helper function named "toDirection" which takes two nodes and return which direction should be taken in order to go from one to another:
public class Node {
final static int HORIZONTALLY = 0;
final static int VERTICALLY = 1;
int col, row;
boolean isTravelable;
int fromDirection;
double hCost;
double gCost;
double fCost;
Node parent;
public Node(int col, int row, boolean isTravelable) {
this.col = col;
this.row = row;
this.isTravelable = isTravelable;
}
public static int toDirection(Node from, Node to) {
return (from.col != to.col) ? Node.HORIZONTALLY : Node.VERTICALLY;
}
}
Then you can change your weight calculation function to take turns into account. You can now give a small punishment for turns like:
public double calcGCost(Node current, Node neighbor) {
if(current.fromDirection == Node.toDirection(current, neighbor)) {
return 1;
} else{
return 1.2;
}
}
Full code: https://github.com/tezsezen/AStarAlgorithm
At the risk of potential down voting, I will try my best to suggest an answer. If you weren't using a third party plugin I would suggest a simple pop/push stack object be built however since you are using someone else's plugin it might be best to try and work along side it rather than against it.
That being said I might just do something simple like track my output results and try to logically determine the correct answer. I would make a simple entity type object literal for storage within an array of all possible path's? So the entire object's life span is only to hold position information. Then you could later parse that array of objects looking for the smallest turn count.
Also, since this third party plugin will do most of the work behind the scenes and doesn't seem very accessible to extract, you might need to feed it criteria on your own. For example if its adding more turns then you want, i.e. inside the door looking square, then maybe sending it the coordinates of the start and end arent enouugh. Perhaps its better to stop at each turn and send in the new coordinates to see if a straight line is now possible. If you did this then each turn would have a change to look and see if there is an obstruction stopping a straight line movement.
I feel like this answer is too simple so it must be incorrect but I will try nonetheless...
//Entity Type Object Literal
var pathsFound = function() {
//Path Stats
straightLine: false,
turnCount: 0,
xPos: -1, //Probably should not be instantiated -1 but for now it's fine
yPos: -1,
//Getters
isStraightLine: function() { return this.straightLine; },
getTurnCount: function() { return this.turnCount; },
getXPos: function() { return this.xPos; },
getYPos: function() { return this.yPos; },
//Setters
setStraightLine: function() { this.straightLine = true; },
setCrookedLine: function() { this.straightLine = false; },
setXPos: function(val) { this.xPos = val; },
setYPos: function(val) { this.yPos = val; },
//Class Functionality
incrementTurnCounter: function() { this.turnCount++; },
updateFullPosition: function(xVal, yVal) {
this.xPos = xVal;
this.yPos = yVal.
},
}
This way you could report all the data every step of the way and before you draw to the screen you could iterate through your array of these object literals and find the correct path by the lowest turnCount.
var img,
field = document.getElementById('field'),
EngineBuilder = function(field, size) {
var context = field.getContext("2d"),
graphSettings = { size: size, mid: Math.ceil(size/2)},
engine = {
getPosition: function(event) {
var bounds = field.getBoundingClientRect(),
x = Math.floor(((event.clientX - bounds.left)/field.clientWidth)*field.width),
y = Math.floor(((event.clientY - bounds.top)/field.clientHeight)*field.height),
node = graph.grid[Math.floor(y/graphSettings.size)][Math.floor(x/graphSettings.size)];
return {
x: x,
y: y,
node: node
}
},
drawObstructions: function() {
context.clearRect (0, 0, 320, 200);
if(img) {
context.drawImage(img, 0, 0);
} else {
context.fillStyle = 'rgb(0, 0, 0)';
context.fillRect(200, 100, 50, 50);
context.fillRect(0, 100, 50, 50);
context.fillRect(100, 100, 50, 50);
context.fillRect(0, 50, 150, 50);
}
},
simplifyPath: function(start, complexPath, end) {
var previous = complexPath[1], simplePath = [start, {x:(previous.y*graphSettings.size)+graphSettings.mid, y:(previous.x*graphSettings.size)+graphSettings.mid}], i, classification, previousClassification;
for(i = 1; i < (complexPath.length - 1); i++) {
classification = (complexPath[i].x-previous.x).toString()+':'+(complexPath[i].y-previous.y).toString();
if(classification !== previousClassification) {
simplePath.push({x:(complexPath[i].y*graphSettings.size)+graphSettings.mid, y:(complexPath[i].x*graphSettings.size)+graphSettings.mid});
} else {
simplePath[simplePath.length-1]={x:(complexPath[i].y*graphSettings.size)+graphSettings.mid, y:(complexPath[i].x*graphSettings.size)+graphSettings.mid};
}
previous = complexPath[i];
previousClassification = classification;
}
simplePath.push(end);
return simplePath;
},
drawPath: function(start, end) {
var path, step, next;
if(this.isPathClear(start, end)) {
this.drawLine(start, end);
} else {
path = this.simplifyPath(start, astar.search(graph, start.node, end.node), end);
if(path.length > 1) {
step = path[0];
for(next = 1; next < path.length; next++) {
this.drawLine(step, path[next]);
step = path[next];
}
}
}
},
drawLine: function(start, end) {
var x = start.x,
y = start.y,
dx = Math.abs(end.x - start.x),
sx = start.x<end.x ? 1 : -1,
dy = -1 * Math.abs(end.y - start.y),
sy = start.y<end.y ? 1 : -1,
err = dx+dy,
e2, pixel;
for(;;) {
pixel = context.getImageData(x, y, 1, 1).data[3];
if(pixel === 255) {
context.fillStyle = 'rgb(255, 0, 0)';
} else {
context.fillStyle = 'rgb(0, 255, 0)';
}
context.fillRect(x, y, 1, 1);
if(x === end.x && y === end.y) {
break;
} else {
e2 = 2 * err;
if(e2 >= dy) {
err += dy;
x += sx;
}
if(e2 <= dx) {
err += dx;
y += sy;
}
}
}
},
isPathClear: function(start, end) {
var x = start.x,
y = start.y,
dx = Math.abs(end.x - start.x),
sx = start.x<end.x ? 1 : -1,
dy = -1 * Math.abs(end.y - start.y),
sy = start.y<end.y ? 1 : -1,
err = dx+dy,
e2, pixel;
for(;;) {
pixel = context.getImageData(x, y, 1, 1).data[3];
if(pixel === 255) {
return false;
}
if(x === end.x && y === end.y) {
return true;
} else {
e2 = 2 * err;
if(e2 >= dy) {
err += dy;
x += sx;
}
if(e2 <= dx) {
err += dx;
y += sy;
}
}
}
}
}, graph;
engine.drawObstructions();
graph = (function() {
var x, y, rows = [], cols, js = '[';
for(y = 0; y < 200; y += graphSettings.size) {
cols = [];
for(x = 0; x < 320; x += graphSettings.size) {
cols.push(context.getImageData(x+graphSettings.mid, y+graphSettings.mid, 1, 1).data[3] === 255 ? 0 : 1);
}
js += '['+cols+'],\n';
rows.push(cols);
}
js = js.substring(0, js.length - 2);
js += ']';
document.getElementById('Graph').value=js;
return new Graph(rows, { diagonal: true });
})();
return engine;
}, start, end, engine = EngineBuilder(field, 10);
field.addEventListener('click', function(event) {
var position = engine.getPosition(event);
if(!start) {
start = position;
} else {
end = position;
}
if(start && end) {
engine.drawObstructions();
engine.drawPath(start, end);
start = end;
}
}, false);
#field {
border: thin black solid;
width: 98%;
background: #FFFFC7;
}
#Graph {
width: 98%;
height: 300px;
overflow-y: scroll;
}
<script src="http://jason.sperske.com/adventure/astar.js"></script>
<code>Click on any 2 points on white spaces and a path will be drawn</code>
<canvas id='field' height
='200' width='320'></canvas>
<textarea id='Graph' wrap='off'></textarea>
I am new to JavaScript and HTML and am working on a small game.
I have four 'enemies' whose position on the canvas is determined by the values in the arrays 'enemyX' and 'enemyY'.
Very simply, I want to detect if the enemies have 'collided', i.e have moved within 30px of each other (the enemy image is 30px by 30px).
What I want to do is subtract the value of the i th value in the array with the other values in the same array and see if this value is less than 30. The less than 30 part is an if statement, so how do I go about subtracting all the values from each other without many lines of code?
Here's what I have tried based on the answers below:
var count = 0;
var innercount = 0;
while (count <= 3) {
while (innercount<=3) {
collisionDetect(count, innercount, enemyX[count], enemyY[count], enemyX[innercount], enemyY[innercount])
innercount++
}
count++
}
var i = 0;
while (i < enemyX.length) {
if (collisionX[i] == 1) {
directionX = directionX*-1;
}
if (collisionY[i] == 1) {
direction = directionY*-1;
}
}
}
}
function collisionDetect(count, innercount, x, y, xX, yY ) {
if ((Math.abs(x-xX)) <=30) {
collisionX[count] = 1
collisionX[innercount] = 1
}
if ((Math.abs(y - yY)) <=30) {
collisionY[count] = 1
collisionY[innercount] = 1
}
return collisionX, collisionY;
}
This code gives me a blank canvas.
Detection of an intersection between two objects (assuming rectangular shape) and the position defines the center of the object.
function getRect(x, y, w, h)
{
var left = x - Math.floor(w / 2),
top = y - Math.floor(h / 2);
return {
left: left,
top: top,
right: left + w,
bottom: top + h
};
}
function intersects(A, B)
{
return A.right >= B.left &&
A.left <= B.right &&
A.bottom >= B.top &&
A.top <= B.bottom;
}
alert(intersects(getRect(12, 56, 30, 30), getRect(30, 40, 30, 30))); // true
The getRect() function can be modified to accommodate for a different anchor position, e.g. top left.
You could use a function:
function colliding(x1, y1, x2, y2){
return Math.abs(x1-x2) <= 30 && Math.abs(y1-y2) <= 30;
}
then use the function to test the different combination of enemies: (1,2), (1,3), (1,4), (2,3), (2,4), and (3,4).
So, for example, you would use: colliding(enemyX[2], enemyY[2], enemyX[3], enemyY[3]) to check if enemy 2 and 3 are colliding. Do that with all the combinations above.
EDIT: to make it more readable, you could define an additional function:
function enemiesColliding(e1, e2){
return colliding(enemyX[e1], enemyY[e1], enemyX[e2], enemyY[e2])
}
and then use it:
enemiesColliding(1,2) || enemiesColliding(1,3) || enemiesColliding(1,4) ||
enemiesColliding(2,3) || enemiesColliding(2,4) || enemiesColliding(3,4)
I am going to restate my understanding of you question, just so there is no confusion.
You have two arrays in paralleled , one for x cords, and one for y cords. Each ship has a element in both arrays.
So, for example, ship 12 could be found at xPos[12] and yPos[12], where xPos and yPos are the arrays from above.
It also stands to reason that this a communicative. If ship[a] collides with ship[b] then ship[b] has collided with ship[a]. And I think that hold for 3 + ships.
I would start by writing a distance function.
dist(x1,y1,x2,y2)
{
return Math.sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
Then I would write code run through the arrays.
Every element must be compared to every other element, one time.
var counter = 0;
var indexOfShipThatHaveCollided = [];
while(counter < Xpos.length)
{
var innerCounter = counter;
while(innerCounter < Xpos.length)
{
t = dist(Xpos[counter],Ypos[counter],Xpos[innerCounter],Ypos[innerCounter])
if(t < 30)
{
indexOfShipThatHaveCollided.push(counter)
indexOfShipThatHaveCollided.push(innerCounter)
}
}
}
The code above compares every ship to every other ship ONCE.
It will compare ship[1] to ship[8], but it will not compare ship[8] to ship[1].
I did not test ANY of this code, but I hope it moves you in the right direction.
If you have anymore question just give me a comment.
According my understanding of your question below is my solution:
I think you need to find distance between two points. by using distance you can apply your logic.
To find distance following is the formula:
Given the two points (x1, y1) and (x2, y2),
the distance between these points is given by the formula: