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<think>
Okay, I need to create a self-contained HTML file that animates a recursive backtracking maze generation on a canvas. Let me start by recalling how recursive backtracking works for maze generation. The algorithm starts at a cell, randomly selects an unvisited neighbor, carves a passage to it, pushes the current cell to the stack, and moves to the new cell. When there are no more unvisited neighbors, it backtracks using the stack until it finds a cell with unvisited neighbors again. The process continues until all cells are visited.
First, I'll structure the HTML with a canvas element. Since no external libraries are allowed, I'll use the canvas API directly. The animation should use requestAnimationFrame for smoothness. I need to handle the canvas setup, the maze grid, and the backtracking stack.
I'll start by defining variables: the canvas context, cell size, grid dimensions. The grid size should be adjustable, but the user didn't specify, so maybe a fixed size like 30x30? But I need to keep within the character limit. Let me think, maybe 20x20 to keep the grid manageable.
The maze cells will need to track visited status and walls. Each cell can have walls on all four sides (top, right, bottom, left). Alternatively, I can represent walls between cells. Maybe using an array of cells where each cell has visited status and walls. Wait, perhaps a 2D array of cells, each cell has visited and a list of walls.
Alternatively, for simplicity, each cell can have visited, and when moving between cells, we remove walls between them. Hmm.
The main steps are:
1. Initialize the grid and set up all walls.
2. Start at a starting cell (maybe 0,0).
3. Use a stack to track the path.
4. In each step:
a. Check current cell's unvisited neighbors.
b. If any, pick one randomly, carve a passage (remove walls between), mark as visited, push current to stack, move to new cell.
c. If none, backtrack by popping from stack (if stack not empty), move to that cell.
5. Repeat until all cells visited.
6. Display the maze as it's built, highlighting the current cell and the stack path.
The animation loop needs to perform a step each frame, then render the grid. But since animation is frame-based, each frame can perform one step of the algorithm.
To handle the animation, in the requestAnimationFrame loop, each frame does:
- If the maze is done, reset and restart.
- Else, perform one step of the algorithm (select neighbor, carve, etc.)
- Redraw the maze, highlighting current cell and stack path.
The highlighting: current cell can be colored, and the stack cells (path taken) can be a different color.
Now, structuring the code:
In the HTML file:
- Canvas element with id="canvas".
- JavaScript inside script tags.
Variables needed:
- ctx (canvas context)
- width and height of the canvas, based on cell size and grid size. Let's say each cell is 10px, grid 20x20, so canvas size 200x200.
- cellSize: 10, grid rows and cols: 20 each.
- grid: 2D array of objects with visited and walls. Or maybe an array of visited flags and separate walls. Hmm, maybe better to track visited and have a way to represent walls between cells.
Alternatively, for each cell, track visited and a list of walls. Each cell can have a visited flag, and walls can be represented as booleans for top, right, bottom, left. Wait, but when carving between two cells, need to remove the wall for both. Maybe better to represent walls between cells. Like, for each cell, the right and bottom walls are part of the cell's own walls, so when moving to the right neighbor, you remove the current cell's right wall and the neighbor's left wall.
Alternatively, perhaps each cell has four walls (top, right, bottom, left), and when moving from a cell to a neighbor, you remove the relevant walls. For example, moving to the north would remove the current cell's top wall and the neighbor's bottom wall (since the neighbor is north, so its bottom wall is adjacent to current's top).
This might get a bit complex. Maybe a better way is to have a grid where each cell has visited status and an array of walls (four directions). But when moving between cells, we can handle removing walls between them.
Alternatively, here's a possible structure for the grid:
grid is a 2D array of objects with:
{
visited: boolean,
walls: [top, right, bottom, left] (each boolean)
}
Wait, but the walls between cells need to be shared. Hmm. Alternatively, the cells can have walls in their own coordinates, and when moving between cells, we can manage both cells' walls.
Alternatively, perhaps the walls are stored per cell. For example, each cell has a top wall (which is actually the wall above the cell) and a bottom wall (below it), etc. When carving between two cells, say cell (x,y) and (x+1,y), you remove the right wall of (x,y) and the left wall of (x+1,y).
This seems manageable. So for each cell:
walls = { top: true, right: true, bottom: true, left: true }
Initially, all walls are present.
So when moving to a neighbor, we can:
- For the direction chosen (e.g., right):
current cell's right wall becomes false.
the neighbor's left wall becomes false.
Thus, the code can handle that.
So steps for initializing the grid:
- Create a grid array of rows and cols, each cell has visited: false, and walls set to all true.
Now, the starting cell is selected (maybe 0,0). Mark it as visited. The stack starts with this cell.
Then, in each step:
currentCell = stack's top? Or maybe the current cell is tracked separately.
Wait, in recursive backtracking, the stack is used to backtrack. So the algorithm proceeds as:
Initialize stack with starting cell.
currentCell = stack.pop? Wait, no. Let me think:
The algorithm steps:
1. Choose a starting cell, mark it as visited, push to stack.
2. While the stack is not empty:
a. Look for all neighbors of current cell that are unvisited.
b. If any:
i. Choose one at random.
ii. Remove the wall between current and chosen neighbor.
iii. Mark neighbor as visited.
iv. Push current to stack.
v. current becomes the neighbor.
c. Else:
i. current = stack.pop().
But how to track current cell? Maybe the current is the top of the stack, but after choosing a neighbor, the neighbor becomes current, and current is pushed to stack before moving.
Hmm, perhaps the standard implementation is to have the stack hold the path. Let me look up recursive backtracking maze steps.
Wait, recursive backtracking can be done iteratively with a stack. The standard iterative approach is:
1. Start at a random cell.
2. While there are unvisited cells:
a. Mark current cell as visited.
b. Shuffle the neighbors of the current cell.
c. For each neighbor in the shuffled list:
i. If neighbor unvisited:
- Remove wall between current and neighbor.
- Push current to stack.
- Move to neighbor as new current.
- Break to start next iteration.
d. If no unvisited neighbors, pop from stack (backtracking).
Wait, perhaps better to structure it as:
currentCell = starting cell
mark as visited, push to stack.
then in loop:
check neighbors of current cell.
if any unvisited:
pick one, carve passage, push current to stack, move to that neighbor, mark as visited.
else:
if stack not empty, current = stack.pop()
else break.
Wait, perhaps the stack is used to keep track of the path. Let me think of code steps:
Let me outline the algorithm steps in code variables:
var stack = [];
var currentCell = startCell;
currentCell.visited = true;
stack.push(currentCell);
while (!allCellsVisited()) {
var neighbors = getUnvisitedNeighbors(currentCell);
if (neighbors.length > 0) {
var next = random choice from neighbors;
remove walls between current and next;
next.visited = true;
stack.push(currentCell);
currentCell = next;
} else {
if (stack.length > 0) {
currentCell = stack.pop();
} else {
break; // all done?
}
}
}
Hmm, but in this way, the stack is storing the path so that when you backtrack, you pop to the previous cell.
Wait, but in this case, the stack is storing the path so that backtracking is done by popping. So each time you move forward, you push the current, and the next becomes the new current.
I think this is the right way. So this can be managed in code.
Now, the animation will need to perform each step one at a time, so in the animation loop, each frame does one step (either proceed forward or backtrack, if applicable).
So the code will need variables to track the current state: current cell, stack, grid, etc.
Now, structuring the code in JavaScript:
First, setup:
const cellSize = 10;
const cols = 20;
const rows = 20;
const canvas = document.getElementById('canvas');
canvas.width = cols * cellSize;
canvas.height = rows * cellSize;
const ctx = canvas.getContext('2d');
Then the grid:
let grid = [];
for (let y = 0; y < rows; y++) {
grid[y] = [];
for (let x = 0; x < cols; x++) {
grid[y][x] = {
visited: false,
walls: [true, true, true, true] // top, right, bottom, left
};
}
}
The starting cell is, say, [0][0]:
let current = {x:0, y:0};
grid[current.y][current.x].visited = true;
Wait, but we need to track current as a coordinate (x,y). So perhaps:
let currentX = 0;
let currentY = 0;
grid[currentY][currentX].visited = true;
stack = [ {x:0, y:0} ]; // or just track the coordinates.
Wait, maybe better to have an array stack of objects with x and y.
Alternatively, store the stack as an array of coordinates.
Wait, this might get a bit tangled. Let me see:
Alternatively, the stack can be an array of objects {x,y}.
Wait, perhaps the stack is an array of cell objects, but to avoid copying, maybe just track coordinates.
Alternatively, the stack holds the previous positions.
Wait, perhaps:
let stack = [];
let currentX = 0;
let currentY = 0;
grid[currentY][currentX].visited = true;
stack.push( {x: currentX, y: currentY } );
Wait, but in the first step, do we push the current cell to stack? Maybe need to re-examine the algorithm.
Alternatively, maybe the stack starts empty, and when moving to a neighbor, push the previous cell onto the stack.
Hmm, perhaps an alternative approach is better. Let me think of code variables:
stack = [];
currentX = 0, currentY =0;
grid[currentY][currentX].visited = true;
stack.push( {x: currentX, y: currentY} ); // initial push? Or not.
Wait, perhaps the standard approach is:
Initialize stack with the starting cell. Then:
while stack is not empty:
currentCell = stack.pop()
process currentCell (mark as visited?)
add adjacent unvisited neighbors to stack (random order?)
Wait, but that would be a depth-first search, but with random order. That's the recursive backtracking.
Alternatively, here's another way to structure the loop:
The stack holds the path. Each step, check if current has unvisited neighbors.
Hmm, perhaps it's better to structure the code as follows:
The animation loop (via requestAnimationFrame) will take a step each frame. Each step can be either:
- Choosing a neighbor and moving forward.
- Backtracking.
So the step function may do:
function step() {
let neighbors = getUnvisitedNeighbors(currentX, currentY);
if (neighbors.length > 0) {
let next = pick random neighbor;
// remove walls between current and next
if (next.x === currentX + 1) { // right
grid[currentY][currentX].walls[1] = false; // right wall
grid[next.y][next.x].walls[3] = false; // left wall
} else if (next.x === currentX -1) { // left
grid[currentY][currentX].walls[3] = false;
grid[next.y][next.x].walls[1] = false;
} else if (next.y === currentY +1) { // bottom
grid[currentY][currentX].walls[2] = false;
grid[next.y][next.x].walls[0] = false;
} else if (next.y === currentY -1) { // top
grid[currentY][currentX].walls[0] = false;
grid[next.y][next.x].walls[2] = false;
}
stack.push( {x: currentX, y: currentY} );
currentX = next.x;
currentY = next.y;
grid[currentY][currentX].visited = true;
cellsCarved++;
} else {
// backtrack
if (stack.length > 0) {
let back = stack.pop();
currentX = back.x;
currentY = back.y;
} else {
// finished
reset();
}
}
}
Wait, but how to track the neighbors correctly. The getUnvisitedNeighbors function must find all adjacent cells (up, down, left, right) within grid bounds, and check if they are unvisited.
So, the getUnvisitedNeighbors function can be written as:
function getUnvisitedNeighbors(x, y) {
let neighbors = [];
// check all four directions
if (y > 0 && !grid[y-1][x].visited) {
neighbors.push( {x: x, y: y-1} );
}
if (y < rows-1 && !grid[y+1][x].visited) {
neighbors.push( {x: x, y: y+1} );
}
if (x > 0 && !grid[y][x-1].visited) {
neighbors.push( {x: x-1, y: y} );
}
if (x < cols-1 && !grid[y][x+1].visited) {
neighbors.push( {x: x+1, y: y} );
}
return neighbors;
}
Wait, but in this code, the order of checking is up, down, left, right. To randomize, we need to shuffle the neighbors.
Alternatively, in step function:
let neighbors = getUnvisitedNeighbors(currentX, currentY);
if (neighbors.length) {
let randomIndex = Math.floor( Math.random() * neighbors.length );
let next = neighbors[randomIndex];
... process next...
}
That way, a random neighbor is selected each time.
So that's good.
Now, the initial setup:
When the page loads, we need to initialize the grid, stack, current position, and start the animation.
Variables needed:
- currentX, currentY (starting at 0,0)
- stack (empty initially?)
Wait, in the code above's step function, the stack is being pushed with the current before moving to next. Let me see:
Initially, when starting:
currentX = 0, currentY =0, grid[0][0].visited = true.
Then, in step:
neighbors = getUnvisited... which would be all except current.
Wait, first step: neighbors would be cells adjacent to (0,0). For example, (0,1) (down), (1,0) (right). Assuming rows and cols are at least 2.
So first, neighbors would have those cells.
Then, a random is chosen. Let's say (0,1). Then, the walls between (0,0) and (0,1) are removed: (0,0)'s bottom wall (wall index 2), and (0,1)'s top wall (index 0).
Then, stack is pushed with {0,0}, current becomes (0,1), which is marked as visited.
Next step, check neighbors of (0,1), etc.
This seems okay.
Now, the stack is used for backtracking.
Now, the reset function: when all cells are visited (the maze is done), then we need to restart with a fresh grid. So reset() would re-initialize grid, current, stack, etc., and set an animation flag to restart.
Now, the animation loop:
let animationFrame = null;
let isRunning = true;
function animate() {
if (!isRunning) {
// Check if maze is done and restart
if (allVisited()) {
reset();
isRunning = true; // restart
}
return;
}
step();
draw();
animationFrame = requestAnimationFrame(animate);
}
function allVisited() {
for (let y = 0; y < rows; y++) {
for (let x = 0; x < cols; x++) {
if (!grid[y][x].visited) return false;
}
}
return true;
}
The reset function would re-initialize everything:
function reset() {
// Reset grid
grid = [];
for (let y = 0; y < rows; y++) {
grid[y] = [];
for (let x = 0; x < cols; x++) {
grid[y][x] = { visited: false, walls: [true, true, true, true] };
}
}
// Reset current and stack
currentX = 0;
currentY = 0;
grid[currentY][currentX].visited = true;
stack = [];
cellsCarved = 1;
isRunning = false; // to allow next iteration to start
// Maybe clear the canvas here? Or let draw() handle.
}
Wait, but in the initial setup, the first step after reset is to push to stack?
Wait, in the code above, during step(), stack is pushed with the current (before moving). So after reset:
currentX =0, currentY=0; grid[0][0].visited is true.
So first step:
neighbors = getUnvisitedNeighbors(0,0) which includes cells around.
Thus, it should proceed.
Now, the drawing function: draw() needs to display the grid.
function draw() {
ctx.clearRect(0,0, canvas.width, canvas.height);
for (let y = 0; y < rows; y++) {
for (let x = 0; x < cols; x++) {
const cell = grid[y][x];
const px = x * cellSize;
const py = y * cellSize;
// draw walls
if (cell.walls[0]) { // top
ctx.fillRect(px, py, cellSize, 1);
}
if (cell.walls[1]) { // right
ctx.fillRect(px + cellSize -1, py, 1, cellSize);
}
if (cell.walls[2]) { // bottom
ctx.fillRect(px, py + cellSize -1, cellSize, 1);
}
if (cell.walls[3]) { // left
ctx.fillRect(px, py, 1, cellSize);
}
// highlight current cell
if (y === currentY && x === currentX) {
ctx.fillStyle = 'rgba(0,255,0,0.3)';
ctx.fillRect(px+1, py+1, cellSize-2, cellSize-2);
}
}
}
// draw stack path? Not sure
// Also show cellsCarved count?
ctx.fillStyle = 'black';
ctx.fillText(`Cells carved: ${cellsCarved}`, 10, 20);
}
Hmm. The cellsCarved variable is incremented each time a new cell is visited (when moving to next). Since initially current cell is visited, that should be 1. Each step that moves to a new cell adds 1. So when all cells are visited, it should be rows*cols.
Now, putting all together, the variables:
let stack = [];
let currentX = 0, currentY = 0;
let cellsCarved = 1; // starts at 1, because initial cell is visited.
Wait, in the initial setup after reset:
currentX/Y are set to 0,0, and grid[0][0].visited is set to true, so cellsCarved should be initialized to 1.
Now, in the step function, when moving to next cell (which is unvisited), cellsCarved increments by 1.
Now, the initial animation starts with:
reset();
and then animate();
But need to call animate() once to start.
Wait, in the setup:
window.onload = function() {
reset();
animate();
};
But in HTML, the canvas is in the body, so maybe better to use an init function.
Alternatively, in the JavaScript code, after defining all functions, call reset() and start animate().
Putting all this into code structure.
Now, also, the problem mentions to create a program that draws the maze with the current path, and highlights the current cell. The stack may be used for backtracking.
Now, considering all this, the code should work.
Potential issues:
1. The walls between cells are properly managed. For example, when moving to the right, the current cell's right wall and next cell's left wall are both set to false. That's correct.
2. The current cell's highlight: using fillRect with some transparency.
3. The reset function is called when all cells are visited, so the animation should restart.
Now, putting all together:
The HTML file should have:
<canvas id="canvas"></canvas>
The JS code would be placed in a script tag or external file.
Now, let me check variable names and functions:
- stack is an array of objects {x,y}. Each time we move forward, we push the current position to stack.
Wait in the step function:
before moving to next, we push currentX/Y to stack?
Wait in the code:
stack.push( {x: currentX, y: currentY} );
Then currentX/Y are set to next.x and next.y.
Yes. So that way, when we backtrack, we pop from the stack and return to the previous position.
This should work.
Now, for drawing walls:
The walls array is [top, right, bottom, left].
Wait, in the code:
For a cell at (x,y), the top wall is the top edge of the cell. The right wall is the right edge, etc.
So when two adjacent cells are connected, both walls are removed.
For example, moving from (0,0) to (0,1) (up? Wait, in code above, when moving up (y decreases), like y-1.
Wait need to make sure that direction is correctly handled.
Wait in getUnvisitedNeighbors:
For up, it's y-1, which is the cell above.
Wait, if current cell is (0,0), then up would be y-1 = -1, which is invalid. So the first neighbor is checked:
if (y > 0) so if currentY is 0, then y=0, so up is not allowed.
The direction checks are:
- Up: y-1 (so check y>0)
- Down: y+1 (y < rows-1)
- Left: x-1 (x>0)
- Right: x+1 (x < cols-1)
Therefore, moving down would be to y+1.
Thus, in the step function when moving to next, which direction is it?
For example, if next is to (0,1) (down), then current cell (0,0)'s bottom wall (index 2) is removed, and next cell (0,1)'s top wall (index 0) is removed.
Yes, that's correct.
Another example: moving to the right (next.x = currentX+1):
current.x =0, next.x=1, same y.
The right wall of current cell (index1) and left wall (index3) of next cell.
Yes.
Okay, the walls are managed correctly.
Now, the drawing function: for each cell, it draws the four walls if they exist.
The walls are:
- Top: at the top edge (y*cellSize) to y*cellSize +1.
Wait, the fillRect uses:
For top wall (present):
ctx.fillRect(px, py, cellSize, 1); // top edge from x=px to px+cellSize, y=py, height=1.
Similarly, right wall:
x is px + cellSize -1 (start at right edge), y from py to py + cellSize, width 1.
Similarly for others.
This should correctly draw the walls.
Now, the current cell is highlighted with a green rectangle inside the cell.
Now, the cellsCarved is displayed, which counts how many cells have been visited.
Now, the reset function must reset all variables, including stack, currentX/Y, grid, and cellsCarved.
Now, the animate loop calls step(), then draw(), then requests next frame.
The isRunning flag is set to false when the maze is done, so that when allVisited() is true, it calls reset(), and then sets isRunning back to true? Wait no:
Wait in the animate function:
if (!isRunning) {
// Check if maze is done and restart
if (allVisited()) {
reset();
isRunning = true; // restart
}
return;
}
Wait, if isRunning is false, then during the step where all cells are visited, reset() is called, and isRunning is set to true.
But after reset(), the next time animate is called, isRunning is true, so the step() is executed again.
Wait, but after reset(), the maze is restarted, so when reset is called, the current state is the initial state (current cell is 0,0, visited, etc.), so the next step() will proceed.
This should work.
Now, testing all variables:
cellsCarved is initialized to 1 in reset().
In the first step, moving to next cell (which is unvisited), so cellsCarved becomes 2.
Each step that moves to a new cell adds 1, so when all cells are visited, it should be 20x20 =400, so cellsCarved will reach 400, then allVisited() returns true, and reset is called.
The animation should then restart every time the maze is complete.
Now, putting all this code together into the script.
Now, possible optimizations:
- Precompute the grid so that the drawing is faster.
But for 20x20 grid, it's manageable.
Now, let's write the full code:
HTML:
<canvas id="canvas"></canvas>
JavaScript:
const canvas = document.getElementById('canvas');
const ctx = canvas.getContext('2d');
let cols = 20, rows = 20;
const cellSize = 10;
let grid;
let stack = [];
let currentX = 0, currentY = 0;
let cellsCarved = 1;
let isRunning = false;
function reset() {
canvas.width = cols * cellSize;
canvas.height = rows * cellSize;
grid = [];
for (let y = 0; y < rows; y++) {
grid[y] = [];
for (let x = 0; x < cols; x++) {
grid[y][x] = {
walls: [true, true, true, true], // top, right, bottom, left
visited: false
};
}
}
grid[currentY][currentX].visited = true;
stack = [];
cellsCarved = 1;
isRunning = true; // start immediately after reset
}
function step() {
if (stack.length === 0 && grid[currentY][currentX].visited) {
const neighbors = getUnvisitedNeighbors();
if (neighbors.length === 0) {
return; // maze complete
}
const next = neighbors[Math.floor(Math.random() * neighbors.length)];
stack.push({x: currentX, y: currentY});
// remove walls between current and next
const dx = next.x - currentX;
const dy = next.y - currentY;
if (dx === 1) { // right
grid[currentY][currentX].walls[1] = false;
grid[next.y][next.x].walls[3] = false;
} else if (dx === -1) { // left
grid[currentY][currentX].walls[3] = false;
grid[next.y][next.x].walls[1] = false;
} else if (dy === 1) { // down
grid[currentY][currentX].walls[2] = false;
grid[next.y][next.x].walls[0] = false;
} else if (dy === -1) { // up
grid[currentY][currentX].walls[0] = false;
grid[next.y][next.x].walls[2] = false;
}
currentX = next.x;
currentY = next.y;
grid[currentY][currentX].visited = true;
cellsCarved++;
} else {
// backtrack
const prev = stack.pop();
currentX = prev.x;
currentY = prev.y;
}
}
function getUnvisitedNeighbors() {
const neighbors = [];
const dirs = [
{dx: 1, dy: 0}, // right
{dx: -1, dy: 0}, // left
{dx: 0, dy: 1}, // down
{dx: 0, dy: -1} // up
];
for (const dir of dirs) {
const nx = currentX + dir.dx;
const ny = currentY + dir.dy;
if (nx >= 0 && nx < cols && ny >=0 && ny < rows) {
if (!grid[ny][nx].visited) {
neighbors.push({x: nx, y: ny});
}
}
}
return neighbors;
}
function draw() {
ctx.clearRect(0,0, canvas.width, canvas.height);
for (let y = 0; y < rows; y++) {
for (let x = 0; x < cols; x++) {
const cell = grid[y][x];
const px = x * cellSize;
const py = y * cellSize;
if (cell.walls[0]) { // top
ctx.fillRect(px, py, cellSize, 1);
}
if (cell.walls[1]) { // right
ctx.fillRect(px + cellSize - 1, py, 1, cellSize);
}
if (cell.walls[2]) { // bottom
ctx.fillRect(px, py + cellSize -1, cellSize, 1);
}
if (cell.walls[3]) { // left
ctx.fillRect(px, py, 1, cellSize);
}
// highlight current cell
if (y === currentY && x === currentX) {
ctx.fillStyle = 'rgba(0,255,0,0.3)';
ctx.fillRect(px + 1, py +1, cellSize -2, cellSize -2);
}
}
}
ctx.fillStyle = 'black';
ctx.fillText(`Cells carved: ${cellsCarved}`, 10, 20);
}
function animate() {
if (isRunning) {
step();
}
draw();
requestAnimationFrame(animate);
}
reset();
animate();
Wait, in this code, the grid is represented with an object for each cell.
The step function now uses a different approach, perhaps a recursive backtracking algorithm.
Wait, the original approach was using a stack-based method similar to recursive backtracking.
The step function here is part of a loop.
Wait, perhaps this code is a different implementation, but the user's initial approach may have some errors.
Alternatively, the problem might require to write the code that uses a stack to build the maze.
Wait, the user's initial code might have a different structure, but let me think.
Wait in the problem, the user asks to create a program that draws the maze with the current path, highlighting the current cell. The stack is used for backtracking.
The code above is following the recursive backtracking approach but implemented iteratively with a stack.
However, in the code I wrote, the step function may not be correctly handling the stack.
Wait step function:
If the stack is empty and the current cell has no unvisited neighbors, the maze is complete.
Else, if there are unvisited neighbors, pick one, push current to stack, modify walls, move there.
Else, backtrack (pop from stack).
Wait, but in the code above, stack is pushed with current before moving to next.
This way, when backtracking, the current is set back to the previous position.
This should work.
But in the step function, when the stack is not empty, it backtracks.
Wait, the condition in step is:
if (stack.length === 0 && ... ) ?
Wait the condition is:
if (stack.length ===0 and current cell is visited),
which isn't the right condition.
Wait perhaps the step function should be:
function step() {
let next;
// check if the current cell has any unvisited neighbors
const neighbors = getUnvisitedNeighbors();
if (neighbors.length > 0) {
// choose a random neighbor
next = neighbors[ Math.floor( Math.random() * neighbors.length ) ];
// knock out walls between current and next
const dx = next.x - currentX;
const dy = next.y - currentY;
if (dx === 1) { // right
grid[currentY][currentX].walls[1] = false;
grid[next.y][next.x].walls[3] = false;
} else if (dx === -1) { // left
grid[currentY][currentX].walls[3] = false;
grid[next.y][next.x].walls[1] = false;
} else if (dy === 1) { // down
grid[currentY][currentX].walls[2] = false;
grid[next.y][next.x].walls[0] = false;
} else if (dy === -1) { // up
grid[currentY][currentX].walls[0] = false;
grid[next.y][next.x].walls[2] = false;
}
// push current to stack, move to next
stack.push( {x: currentX, y: currentY} );
currentX = next.x;
currentY = next.y;
grid[currentY][currentX].visited = true;
cellsCarved++;
} else {
// no neighbors, backtrack
if (stack.length > 0) {
const prev = stack.pop();
currentX = prev.x;
currentY = prev.y;
}
}
}
This is a better approach.
The previous code may have an off-by-one error.
Thus, the step function should first check if there are unvisited neighbors. If yes, pick one, remove walls, push current to stack, move to next. Else, backtrack.
This is the standard iterative backtracking algorithm.
So in this corrected step function, when the current cell has unvisited neighbors, it proceeds; otherwise, backtracks.
The variable isRunning should control whether to proceed.
The animate function should call step every frame, only if isRunning is true.
Wait in the previous code:
function animate() {
if (isRunning) {
step();
}
draw();
requestAnimationFrame(animate);
}
This would call step() every frame, but that may be too fast, making the animation too quick.
Alternatively, we can use a timer, but in the problem, it's okay to animate in real-time.
Thus, the code should work.
Now, setting up the reset function properly.
In reset():
- The grid is initialized with all walls present, and visited set to false.
currentX and Y are set to 0,0.
grid[0][0].visited is set to true.
cellsCarved is 1.
stack is cleared.
isRunning is set to true.
Then, the animate loop calls step() and draw() each frame.
Now, the getUnvisitedNeighbors() function is as before.
The drawing code is okay.
With these corrections, the code should work.
I think this is the correct approach.
Therefore, the final answer is the code as written above, with proper structure and corrections where needed.