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Having issues trying to solve N Rook problem . Always get n*n solution and not N factorial

I'm trying to get N ways of solves a N rook problem. The issue I am having is currently, I seem to get n*n solutions while it needs to be N! . Below is my code, I have written it in simple loops and functions, so it's quite long. Any help would be greatly appreciated
Note: Please ignore case for n = 2. I get some duplicates which I thought I would handle via JSON.stringify
var createMatrix = function (n) {
var newMatrix = new Array(n);
// build matrix
for (var i = 0; i < n; i++) {
newMatrix[i] = new Array(n);
}
for (var i = 0; i < n; i++) {
for (var j = 0; j < n; j++) {
newMatrix[i][j] = 0;
}
}
return newMatrix;
};
var newMatrix = createMatrix(n);
// based on rook position, greying out function
var collision = function (i, j) {
var col = i;
var row = j;
while (col < n) {
// set the row (i) to all 'a'
col++;
if (col < n) {
if (newMatrix[col][j] !== 1) {
newMatrix[col][j] = 'x';
}
}
}
while (row < n) {
// set columns (j) to all 'a'
row++;
if (row < n) {
if (newMatrix[i][row] !== 1) {
newMatrix[i][row] = 'x';
}
}
}
if (i > 0) {
col = i;
while (col !== 0) {
col--;
if (newMatrix[col][j] !== 1) {
newMatrix[col][j] = 'x';
}
}
}
if (j > 0) {
row = j;
while (row !== 0) {
row--;
if (newMatrix[i][row] !== 1) {
newMatrix[i][row] = 'x';
}
}
}
};
// checks position with 0 and sets it with Rook
var emptyPositionChecker = function (matrix) {
for (var i = 0; i < matrix.length; i++) {
for (var j = 0; j < matrix.length; j++) {
if (matrix[i][j] === 0) {
matrix[i][j] = 1;
collision(i, j);
return true;
}
}
}
return false;
};
// loop for every position on the board
loop1:
for (var i = 0; i < newMatrix.length; i++) {
var row = newMatrix[i];
for (var j = 0; j < newMatrix.length; j++) {
// pick a position for rook
newMatrix[i][j] = 1;
// grey out collison zones due to the above position
collision(i, j);
var hasEmpty = true;
while (hasEmpty) {
//call empty position checker
if (emptyPositionChecker(newMatrix)) {
continue;
} else {
//else we found a complete matrix, break
hasEmpty = false;
solutionCount++;
// reinitiaze new array to start all over
newMatrix = createMatrix(n);
break;
}
}
}
}

There seem to be two underlying problems.
The first is that several copies of the same position are being found.
If we consider the case of N=3 and we visualise the positions by making the first rook placed red, the second placed green and the third to be placed blue, we get these three boards:
They are identical positions but will count as 3 separate ones in the given Javascript.
For a 3x3 board there are also 2 other positions which have duplicates. The gets the count of unique positions to 9 - 2 - 1 -1 = 5. But we are expecting N! = 6 positions.
This brings us to the second problem which is that some positions are missed. In the case of N=3 this occurs once when i===j==1 - ie the mid point of the board.
This position is reached:
This position is not reached:
So now we have the number of positions that should be found as 9 - 2 - 1 - 1 +1;
There appears to be nothing wrong with the actual Javascript in as much as it is implementing the given algorithm. What is wrong is the algorithm which is both finding and counting duplicates and is missing some positions.
A common way of solving the N Rooks problem is to use a recursive method rather than an iterative one, and indeed iteration might very soon get totally out of hand if it's trying to evaluate every single position on a board of any size.
This question is probably best taken up on one of the other stackexchange sites where algorithms are discussed.

How to extract elements of a matrix in a particular pattern using javascript?

How to extract elements of a matrix in a particular pattern using javascript?
following is the code for generating a 2d matrix of size any
var matrix = [];
for(var i=0; I<size; i++) {
matrix[i] = [];
for(var j=0; j<size; j++) {
matrix[i][j] = undefined;
}
}
so the first one is the original matrix. second one is the matrix made out by removing some elements/ cells.
// don't worry about splicing... this is just a demo.
[[3,5], [4,5], [4,2] ... ].forEach(([i,j], _, arr) => delete arr[i][j])
***** This is where you begin *****
(from the above splices matrix the following has to be done)
so what I want to achieve is that I want to separate or group cells from the above spliced matrix in the following manner. (like concentric circles/squares fashion by keeping in mind the fact that some cells are removed)
first you start at the central element [i,j], then move to the next level 3 x 3 & assign a value to all concentric cells (or separate cell address out)... then move to the next level 4 x 4 assign a value to all concentric cells (or separate cell address out) and so on...
Note that the spliced matrix is the starting point
travel in this fashion outwards (like concentric squares):
the idea is that:\
generate a matrix of size n * n.
remove some elements (that will be any)
... from here the algorithm starts...
separate cells from the above splices matrix in a concentric square fashion. --> this is where I want help (step 3 only)
In simple terms I want to separate out the cells from the matrix in the following order (one level at a time 3 x 3 first then 4 x 4 then 5 x 5 and so on.... ( from the above spliced matrix, which means that some cells will be already removed.)
keep in mind the fact that some cells are already removed (so you'll have to skip them -)

First, figure out the centroid of the square. The center is going to be at {length / 2, width / 2}. We're dealing in integer units, so floor. Using the convention that the top left of the figure is {0, 0} , then ie a 7x7 square has its center at
{floor(7/2), floor(7/2)} === {3,3}
Then, figure out a distance formula. In the pictures, the unit squares in purple are all the squares where either the x coordinate or y coordinate are N units away from the center. In other words
max(x - centerX, y-centerY) === N
I agree with #audzzy, you don't want to delete anything since that actually changes the length of an individual row, which we don't want. Instead just set it to another value.
So the idea is to:
find the center
iterate over each unit square
clear any unit squares which are N distance from the center
const clearCircle = (mat, N) => {
// find the center
const maxI = mat.length, maxJ = mat[0].length,
center = {
i: Math.floor(mat.length / 2),
j: Math.floor(mat[0].length / 2)
};
// iterate over all units
for (let i = 0; i < maxI; i++) {
for (let j = 0; j < maxJ; j++) {
// check if its N units from the center
if (Math.max(Math.abs(center.i - i), Math.abs(center.j - j)) === N) {
mat[i][j] = " ";
}
}
}
return mat;
}
Full example:
/*jshint esnext: true */
const generateMat = size => {
const mat = [];
for (let i = 0; i < size; i++) {
mat[i] = [];
for (let j = 0; j < size; j++) {
mat[i][j] = "[ ]";
}
}
return mat;
}
const mat = generateMat(9);
const logMat = mat => console.log("\n" + mat.map(row => (row).join(" ")).join("\n") + "\n");
const clearCircle = (mat, N) => {
// find the center
const maxI = mat.length, maxJ = mat[0].length,
center = {
i: Math.floor(mat.length / 2),
j: Math.floor(mat[0].length / 2)
};
// iterate over all units
for (let i = 0; i < maxI; i++) {
for (let j = 0; j < maxJ; j++) {
// check if its N units from the center
if (Math.max(Math.abs(center.i - i), Math.abs(center.j - j)) === N) {
mat[i][j] = " ";
}
}
}
return mat;
}
logMat(mat);
logMat(clearCircle(mat, 1));
logMat(clearCircle(mat, 2));
logMat(clearCircle(mat, 3));

Well... you just need exclude squares outside boundaries, after that you can just select elements which are in the most external layer of the matrix!
Something like that:
function extractElements(layer, matrix) {
const elements = []
const n = matrix.length; // matrix NxN
const frontIndex = layer -1;
const backIndex = n - layer;
if(layer <= 0 || layer > Math.ceil(n/2)) {
console.log("invalid layer");
return [];
}
for(let i = 0; i < n; ++i){
if(i < frontIndex || i > backIndex)
continue;
for(let j = 0; j < n; ++j){
if(j < frontIndex || j > backIndex)
continue;
if(i === frontIndex || i === backIndex || j === frontIndex || j === backIndex) {
if(matrix[i][j] !== undefined)
elements.push(matrix[i][j]);
}
}
}
return elements;
}
So, if you want the most external layer of a matrix 5x5, you can call extractElements(1, matrix)
Good code!

here's some basic code to do what you want (here I colored the cells, you could add them to a result array, or do whateve else you want)
notice it has very little iterations- only goes over the "wanted" cells every time- and colors the relevant lines and columns,
(for visual reasons -
'-' is a regular cell,
' ' is deleted cell,
'a' is a "purple" cell)
setMatrix is basically the interesting part that gets the interesting cells of every level..
let setMatrix = (matrix, level, value) => {
let size = matrix.length;
for(let x=Math.floor(size/2)-level;x<=Math.floor(size/2)+level;x++){
matrix[x][Math.floor(size/2)-level] = matrix[x][Math.floor(size/2)-level] == ' ' ? ' ' : value;
matrix[x][Math.floor(size/2)+level] = matrix[x][Math.floor(size/2)+level] == ' ' ? ' ' : value;
if(x!=Math.floor(size/2)-level && x!=Math.floor(size/2)+level){
matrix[Math.floor(size/2)-level][x] = matrix[Math.floor(size/2)-level][x] == ' ' ? ' ' : value;
matrix[Math.floor(size/2)+level][x] = matrix[Math.floor(size/2)+level][x] == ' ' ? ' ' : value;
}
}
}
and here's some code to call it for each level:
function doWork(){
// init
let size = 7;
var matrix = [];
for(var i=0; i<size; i++) {
matrix[i] = [];
for(var j=0; j<size; j++) {
matrix[i][j] = '-';
}
}
// delete
[[1,1], [2,3], [2,5], [5,1], [6,4]].forEach(([i,j])=> matrix[i][j] = ' ');
// go over each level
for(let i=0;i<=Math.floor(size/2);i++){
// set
setMatrix(matrix, i, 'a');
// print
for(let x=0;x<size;x++){
let line='';
for(let y=0;y<size;y++){
line+=matrix[x][y];
}
console.log(line);
}
console.log();
//reset
setMatrix(matrix, i, '-');
}
}
doWork();

p5.js Prim's Algorithm Maze Generation: Stuck in Infinite Loop

Note: I also posted this question on p5.js' forum: https://discourse.processing.org/t/p5-js-prims-algorithm-maze-generation-stuck-in-infinite-loop/8277
I'm trying to implement a randomized Prim's algorithm to generate a maze. However, the program keeps getting stuck in an infinite loop as the length of the list of walls (wallList) is always in the thousands. Currently I am using an if statement that stops the maze generation after 11500 iterations to prevent an infinite loop.
My pseudocode is based on Wikipedia's description of the algorithm:
Start with a grid full of walls.
Pick a cell, mark it as part of the
maze. Add the walls of the cell to the wall list.
While there are
walls in the list:
Pick a random wall from the list. If only one of
the two cells that the wall divides is visited, then:
Make the wall a
passage and mark the unvisited cell as part of the maze.
Add the
neighboring walls of the cell to the wall list.
Remove the wall from
the list.
HTML:
<!doctype html>
<html>
<head>
<meta charset="utf-8" />
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<script src="https://ajax.aspnetcdn.com/ajax/jQuery/jquery-3.3.1.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/p5.js/0.6.0/p5.js"></script>
</script>
</head>
<body>
<div id="game-wrapper">
<div id="canvas-wrapper">
</div>
</div>
</div>
<script type="text/javascript" src="maze.js"></script>
</body>
</html>
JavaScript (maze.js):
// Note to self: JavaScript variables have function scope, not block scope
var numIterations = 0; // Just for debugging purposes (stop at 1000 iterations otherwise the program goes into
// an infinite loop
// The directions and vectors arrays correspond to each other
// For example, the first element of directions is "N" and the first element of vectors also represents a
// north vector
var directions = ["N", "E", "S", "W"]
var vectors = [
[-1, 0],
[0, 1],
[1, 0],
[0, -1]
];
var wallList = {}; // Structure of a wall [rol (num), col (num), direction (string)]
function Maze(numRows, numColumns) {
/*
Defines a maze given the number of rows and the number of columns in the maze
*/
this.numColumns = numColumns;
this.numRows = numRows;
this.numCells = numRows * numColumns;
this.cellGraph = [];
for (var i = 0; i < numRows; i++) { // For every single row
this.cellGraph.push([]); // Start out with an empty row
}
}
Maze.prototype.computeFrontierWalls = function (cellRow, cellColumn) {
/*
The frontier walls of a cell is defined as all the walls of the adjacent cells
*/
/*
Coordinates of adjacent cells:
Up [cellRow - 1, cellColumn]
Down [cellRow + 1, cellColumn]
Right [cellRow, cellColumn + 1]
Left [cellRow, cellColumn - 1]
*/
var coordinates = [
[cellRow - 1, cellColumn],
[cellRow + 1, cellColumn],
[cellRow, cellColumn + 1],
[cellRow, cellColumn - 1]
];
var computedFrontier = []; // List of frontier cells
var originalCell = this.cellGraph[cellRow][cellColumn]; // We want to calculate the frontier of the original cell
for (var i = 0; i < coordinates.length; i++) {
// Get the coordinates of the adjacent cell
var coordinate = coordinates[i];
var row = coordinate[0];
var col = coordinate[1];
// See if a cell exists at that area
// If there is a cell that exists, add all of the walls of the cell to the computedFrontier array
if (row >= 0 && row < this.cellGraph.length && col >= 0 && col < this.cellGraph[0].length) {
var cell = this.cellGraph[parseInt(row)][parseInt(col)];
for (var j = 0; j < directions.length; j++) {
computedFrontier.push([cell.row, cell.column, directions[j]]);
}
}
}
return computedFrontier;
}
function Cell(cellSize, row, column) {
this.cellSize = cellSize; // The width and height of the cell
this.column = column;
this.row = row;
this.xPos = column * cellSize;
this.yPos = row * cellSize;
this.walls = [true, true, true, true]; // 0 = top, 1 = right, 2 = bottom, 3 = left
this.visited = false; // Whether the cell has been traversed or not
}
function getRandomPos(widthCells, heightCells) {
var row = Math.floor(Math.random() * heightCells); // Generate a random row
var column = Math.floor(Math.random() * widthCells); // Generate a random column
return [row, column];
}
var mazeIntro = function (p) {
var maze = new Maze(20, 35); // Generate a new maze with 20 rows and 35 columns
Maze.prototype.createMaze = function () { // Build an empty maze
for (var i = 0; i < this.numRows; i++) { // Iterate through every row
for (var j = 0; j < this.numColumns; j++) { // Iterate through every column
var cell = new Cell(20, i, j); // Create a new size at row i and column j with size 20
maze.cellGraph[i].push(cell); // Add the cell to the row
}
}
}
maze.createMaze(); // Build the maze
p.setup = function () {
var canvas = p.createCanvas(700, 400);
p.background(255, 255, 255);
p.smooth();
p.noLoop();
// Pick a cell, mark it as part of the maze. Add the walls of the cell to the wall list
var pos = getRandomPos(maze.cellGraph[0].length, maze.cellGraph.length);
;
var row = pos[0];
var column = pos[1];
maze.cellGraph[row][column].visited = true;
for (var k = 0; k < directions.length; k++) {
var key = row.toString() + column.toString() + directions[k].toString();
if (!wallList[key]) {
wallList[key] = [row, column, directions[k]];
}
}
}
Cell.prototype.display = function () {
/*
For each wall:
1. Check if it is on the border of the maze:
2. If it is on the border: don't draw the wall
3. If it isn't on the border: draw the wall
*/
p.stroke(0, 0, 0);
if (this.walls[0] && this.row != 0) { // Top
p.line(this.xPos, this.yPos, this.xPos + this.cellSize, this.yPos);
}
if (this.walls[1] && this.column != maze.widthCells - 1) { // Right
p.line(this.xPos + this.cellSize, this.yPos, this.xPos + this.cellSize, this.yPos + this.cellSize);
}
if (this.walls[2] && this.row != maze.heightCells - 1) { // Bottom
p.line(this.xPos + this.cellSize, this.yPos + this.cellSize, this.xPos, this.yPos + this.cellSize);
}
if (this.walls[3] && this.column != 0) { // Left
p.line(this.xPos, this.yPos + this.cellSize, this.xPos, this.yPos);
}
p.noStroke();
}
Cell.prototype.toString = function () {
/*
Mainly used for debugging purposes, converts the object into a string containing the row and the column of the cell
*/
return this.row + "|" + this.column;
}
function deleteWall(current, neighbor) {
/*
Deletes two walls separating two cells: current and neighbor
Calculates if neighbor is to the left, right, top, or bottom of cell
Removes the current's wall and the corresponding neighbor's wall
*/
var deltaX = current.column - neighbor.column;
var deltaY = current.row - neighbor.row;
if (deltaX == 1) { // Current is to the right of the neighbor
current.walls[3] = false;
neighbor.walls[1] = false;
}
if (deltaX == -1) { // Current is to the left of the neighbor
current.walls[1] = false;
neighbor.walls[3] = false;
}
if (deltaY == 1) { // Current is to the bottom of the neighbor
current.walls[0] = false;
neighbor.walls[2] = false;
}
if (deltaY == -1) { // Current is to the top of the neighbor
current.walls[2] = false;
neighbor.walls[0] = false;
}
}
function isWall(cellA, cellB) {
// Whether there's a wall or not depends on the orientation of the blocks
// If it's vertical, it has to be false between even numbers
// If it's horizontal, it has to be false between odd numbers
for (var j = 0; j < cellA.walls.length; j++) {
for (var k = 0; k < cellB.walls.length; k++) {
if (Math.abs(j - k) == 2 && !cellA.walls[j] && !cellB.walls[k]) {
var rA = cellA.row;
var cA = cellA.column;
var rB = cellB.row;
var cB = cellB.column
if ((rA - rB) == 1 && j == 0 || (rA - rB) == -1 && j == 2 || (cA - cB) == 1 && j == 3 || (cA - cB) == -1 && j == 1) {
return false;
}
}
}
}
return true;
}
function calculateCellDivision(wall) {
// Calculate the two cells that the wall divides
// For example:
// If the wall is [10, 11, "N"]
// The two cells that the wall divides are (10, 11) and (9, 11)
var row = wall[0];
var col = wall[1];
var cell1 = maze.cellGraph[row][col]; // Get the cell of the wall
// Get the corresponding vector based upon the direction of the wall
var vectorIndex = directions.indexOf(wall[2]);
// Add the vector to the position of cell1
var cell2Row = parseInt(cell1.row) + vectors[vectorIndex][0];
var cell2Column = parseInt(cell1.column) + vectors[vectorIndex][1];
if (cell2Row < 0 || cell2Row >= maze.cellGraph.length || cell2Column < 0 || cell2 >= maze.cellGraph[0].length) {
return -1;
}
var cell2 = maze.cellGraph[cell2Row][cell2Column]; // Get the corresponding cell
var cellsVisited = 0;
var unvisitedCell;
if (cell1.visited) {
cellsVisited += 1;
unvisitedCell = cell2;
}
if (!cell2) { // This means that the wall is a border wall
return -1;
}
if (cell2.visited) {
cellsVisited += 1;
unvisitedCell = cell1;
}
if (cellsVisited == 1) {
return [cell1, cell2, cellsVisited, unvisitedCell];
}
return -1;
}
function getCellWalls(row, col) {
// Gets a cell's walls
var cellWalls = [];
for (var i = 0; i < directions.length; i++) {
cellWalls.push(row + col + directions[i]);
}
return cellWalls;
}
p.draw = function () {
while (Object.keys(wallList).length > 0) { // While there are still walls in the list
console.log("Object.keys(wallList).length = " +
// Pick a random wall of the list
var wallListKeys = $.map(wallList, function (value, key) {
return key;
});
var randomKey = wallListKeys[Math.floor(Math.random() * wallListKeys.length)];
var randomWall = wallList[randomKey];
var components = calculateCellDivision(randomWall);
if (components != -1) {
var numVisited = components[2];
var cell1 = components[0];
var cell2 = components[1];
// If only one of the two cells that the wall divides is visited, then:
// 1. Make the wall a passage and mark the unvisited cell as part of the maze.
// 2. Add the neighboring walls of the cell to the wall list.
// Remove the wall from the list.
if (numVisited == 1) {
deleteWall(cell1, cell2);
var unvisitedCell = maze.cellGraph[components[3].row][components[3].column];
unvisitedCell.visited = true;
var unvisitedString = unvisitedCell.row + "|" + unvisitedCell.column;
// Add the neighboring walls of the cell to the wall list
// Format of the walls (by index):
// 0 = top, 1 = right, 2 = bottom, 3 = left
var computedFrontierWalls = maze.computeFrontierWalls(unvisitedCell.row, unvisitedCell.column);
for (var k = 0; k < computedFrontierWalls.length; k++) {
var computedWall = computedFrontierWalls[k];
var keyString = computedWall[0].toString() + computedWall[1].toString() + computedWall[2];
if (!wallList[keyString]) {
wallList[keyString] = computedWall;
}
}
// Delete randomKey from the list of walls, and then delete the same wall from the corresponding cell
delete wallList[randomKey];
// Calculate the corresponding cell
var direction = randomWall[2];
var directionIndex = directions.indexOf(direction);
var oppositeDirectionIndex = -1;
if (directionIndex == 0) {
oppositeDirectionIndex = 2;
}
if (directionIndex == 2) {
oppositeDirectionIndex = 0;
}
if (directionIndex == 1) {
oppositeDirectionIndex = 3;
}
if (directionIndex == 3) {
oppositeDirectionIndex = 1;
}
var vector = vectors[directionIndex];
var correspondingString = (randomWall[0] + vector[0]).toString() + (randomWall[1] + vector[1]).toString() + directions[oppositeDirectionIndex];
delete wallList[correspondingString];
}
}
numIterations += 1;
if (numIterations == 11500) { // Prevents infinite loop
break;
}
}
p.clear();
// Draw the maze
for (var i = 0; i < maze.cellGraph.length; i++) { // Iterate through row
for (var j = 0; j < maze.cellGraph[i].length; j++) { // Iterate through every column
maze.cellGraph[i][j].display(); // Display the cell
}
}
p.line(0, 400, 400, 400);
}
};
var myp5 = new p5(mazeIntro, "canvas-wrapper"); // Initialize the graphics engine for the canvas
The generation does work, as shown in the screenshot below. But I'm pretty sure my implementation is not correct as the wall list shouldn't contain thousands of walls.

According to the Wikipedia article, a wall must be removed once is picked randomly and analysed, regardless of the results of said analysis. In your code, the lines delete wallList[randomKey]; and delete wallList[correspondingString]; must be outside of the conditional for the wall to be eliminated from the list.
After deleting said lines, replace this:
numIterations += 1;
if (numIterations == 11500) { // Prevents infinite loop
break;
}
with this:
delete wallList[randomKey];
delete wallList[correspondingString];
and you're good to go. (Disclaimer: I tested it and it worked; but that's a hefty amount of code you have there, so I'm not sure if anything else breaks).
I wanted to try and make it more readable so I could understand better and came up with this. Hope it helps.
// 1. Start with a grid full of walls.
const _WALL = '█';
const _PATH = ' ';
const _COLS = 60;
const _ROWS = 60;
let maze = [];
for(let i = 0; i < _COLS; i++){
maze.push([]);
for(let j = 0; j < _ROWS; j++)
maze[i][j] = _WALL;
}
// 2. Pick a cell, mark it as part of the maze.
let cell = {x:Math.floor(Math.random()*_COLS), y:Math.floor(Math.random()*_ROWS)};
maze[cell.x][cell.y] = _PATH;
// 2.1 Add the walls of the cell to the wall list.
let walls = [];
if(cell.x+1 < _COLS) walls.push({x:cell.x+1, y:cell.y});
if(cell.x-1 >= 0) walls.push({x:cell.x-1, y:cell.y});
if(cell.y+1 < _ROWS) walls.push({x:cell.x, y:cell.y+1});
if(cell.y-1 >= 0) walls.push({x:cell.x, y:cell.y-1});
// 3. While there are walls in the list:
while(walls.length > 0){
// 3.1 Pick a random wall from the list.
let wallIndex = Math.floor(Math.random() * walls.length);
let wall = walls[wallIndex];
// 3.2 If only one of the two cells that the wall divides is visited, then:
let uc = []; // uc will be short for 'unvisited cell'
if(wall.x+1 < _COLS && maze[wall.x+1][wall.y] === _PATH) uc.push({x:wall.x-1, y:wall.y});
if(wall.x-1 >= 0 && maze[wall.x-1][wall.y] === _PATH) uc.push({x:wall.x+1, y:wall.y});
if(wall.y+1 < _ROWS && maze[wall.x][wall.y+1] === _PATH) uc.push({x:wall.x, y:wall.y-1});
if(wall.y-1 >= 0 && maze[wall.x][wall.y-1] === _PATH) uc.push({x:wall.x, y:wall.y+1});
if(uc.length === 1){
// 3.2.1 Make the wall a passage and mark the unvisited cell as part of the maze.
maze[wall.x][wall.y] = _PATH;
if(uc[0].x >=0 && uc[0].x <_COLS && uc[0].y >=0 && uc[0].y<_ROWS){
maze[uc[0].x][uc[0].y] = _PATH;
// 3.2.2 Add the neighboring walls of the cell to the wall list.
if(uc[0].x+1 < _COLS && maze[uc[0].x+1][uc[0].y] === _WALL) walls.push({x:uc[0].x+1, y:uc[0].y});
if(uc[0].x-1 >= 0 && maze[uc[0].x-1][uc[0].y] === _WALL) walls.push({x:uc[0].x-1, y:uc[0].y});
if(uc[0].y+1 < _ROWS && maze[uc[0].x][uc[0].y+1] === _WALL) walls.push({x:uc[0].x, y:uc[0].y+1});
if(uc[0].y-1 >= 0 && maze[uc[0].x][uc[0].y-1] === _WALL) walls.push({x:uc[0].x, y:uc[0].y-1});
}
}
// 3.3 Remove the wall from the list.
walls.splice(wallIndex, 1);
}
console.table(maze);
function setup(){
createCanvas(400, 400);
fill(0);
let widthUnit = width / _COLS;
let heightUnit = height / _ROWS;
for(let i = 0; i < _COLS; i++)
for(let j = 0; j < _ROWS; j++)
if(maze[i][j] === _WALL){
//rect(i*widthUnit, j*heightUnit, widthUnit, heightUnit);
if(i-1 >= 0 && i+1 < _COLS){
if(maze[i-1][j] === _WALL) line((i+0.5)*widthUnit, (j+0.5)*heightUnit, i*widthUnit, (j+0.5)*heightUnit);
if(maze[i+1][j] === _WALL) line((i+0.5)*widthUnit, (j+0.5)*heightUnit, (i+1)*widthUnit, (j+0.5)*heightUnit);
}
if(j-1 >= 0 && j+1 < _ROWS){
if(maze[i][j-1] === _WALL) line((i+0.5)*widthUnit, (j+0.5)*heightUnit, (i+0.5)*widthUnit, j*heightUnit);
if(maze[i][j+1] === _WALL) line((i+0.5)*widthUnit, (j+0.5)*heightUnit, (i+0.5)*widthUnit, (j+1)*heightUnit);
}
}
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/p5.js/0.7.3/p5.min.js"></script>

Javascript snake pattern grid

I am trying to draw a grid on screen numbered in a snake pattern in Javascript, I have a working grid but it follows the pattern of
12345
67890
And what I need is
12345
09876
I have seen this done with modulo and have tried to implement but im having trouble getting the right number sequence.
Here is my function
function createGrid(length, height) {
var ledNum = 0;
for (var rows = 0; rows < height; rows++) {
for (var columns = 0; columns < length; columns++) {
var backwards = ledNum + columns;
if (rows % 2 == 0 || rows != 0) {
$("#container").append("<div class='grid' id='" + ledNum + "'>" + //HERE IS MY PROBLEM+"</div>");
}
else if (!rows % 2 == 0) {
$("#container").append("<div class='grid' id='" + ledNum + "'>" + ledNum + "</div>");
}
ledNum++;
};
};
$(".grid").width(960 / length);
$(".grid").height(960 / height);
};
How do I work out the true modulo case to show the numbers correctly in snake pattern?
I am not well versed with 2d arrays but perhaps that might be a better way?

The best way I can think of is to use an object with arrays and exploit its inbuilt functions to ease your job...for example
function createGrid(length,height) {
var lednum = 0;
var grid = [];
for (var row = 0; row < height; row++) {
grid[row] = [];
for (var col = 0; col < length; col++) {
if ((row % 2) === 0) {
grid[row].push(lednum);
} else {
grid[row].unshift(lednum);
}
lednum++;
}
}
return grid;
}
console.log(createGrid(10, 10))
Then you can just print out above grid
Update : How to print above data. You could simply use two for loops.
var length = 10;
var height = 15;
var brNode = document.createElement('br');
var grid = createGrid(length, height));
for (var row = 0; row < height; row++) {
var rowPrint = "";
for (var col = 0; col < length; col++) {
rowPrint += String(grid[row][col]) + " ";
}
var rowNode = document.createTextNode(rowPrint)
$("#container").appendChild(rowNode);
$("#container").appendChild(brNode);
}
Note that this will create rows of textNode broken by <br/> tags. if you want it formatted in some other way..well you have the preformatted data..all you need to do is traverse through it and print it how you want.

This general idea seems to work...
// Input variables
var data = 'abcdefghijklmnopqrstuvwxyz';
var width = 5;
// The actual algorithm.
var rows = Math.ceil(data.length / width);
for (var y = 0; y < rows; y++) {
var rowText = "";
for (var x = 0; x < width; x++) {
// Basically, for every other row (y % 2 == 1),
// we count backwards within the row, as it were, while still
// outputting forward.
var offset = y * width + (y % 2 == 1 ? width - 1 - x : x);
rowText += data[offset] || " ";
}
console.log(rowText);
}
$ node so51356871.js
abcde
jihgf
klmno
tsrqp
uvwxy
z

As I mentioned in comments, there is a lot wrong with the boolean logic in your code:
The first if condition always evaluates to true, except in the first iteration
The second if condition is therefor only evaluated once, and it will be false.
I would split the functionality in two parts:
Create a 2D array with the numbers in "snake" sequence
Create the DOM elements from such a matrix, using some CSS to control the line breaks
function createSnake(width, height) {
const numbers = [...Array(width*height).keys()];
return Array.from({length:height}, (_, row) =>
numbers.splice(0, width)[row % 2 ? "reverse" : "slice"]()); // 2D array
}
function createGrid(matrix) {
$("#grid").empty().append(
[].concat(...matrix.map(row => row.map((val,i) =>
$("<div>").addClass("grid").toggleClass("newline", !i).text(val))))
);
}
// Demo generating a 3 x 3 grid
createGrid(createSnake(3,3));
.grid {
float: left;
padding: 3px;
}
.newline {
clear:left
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="grid"></div>

text string to matrix in Javascript

Beginner here. I've got a text area. I want to break the text up into a matrix and then transpose rows and columns. I've got function that splits the text into groups. another function that puts them into an array and that's as far as I've got. I don't really know how to combine them into one function or if that is needed. i believe i'm supposed convert the array into a matrix by creating sub arrays, then i can finally transpose the text from [ i ][ j ] to [ j ][ i ]. Is this right so far or have i left something out? I'd love to see a working example someplace but can't seem to find one.
function G_Group(size, count)
{
if (size <= 0)
{
size = document.encoder.group_size.value;
if (size <= 0)
{
alert('Invalid group size');
return false;
}
}
if (count <= 0)
{
count = document.encoder.group_count.value;
if (count <= 0)
{
alert('Invalid group count');
return false;
}
}
var t = document.encoder.text.value;
var o = '', groups = 0;
t = Tr(t, " \r\n\t");
while (t.length > 0)
{
if (o.length > 0)
{
o += ' ';
}
if (groups >= count)
{
o += "\n";
groups = 0;
}
groups ++;
o += t.slice(0, size);
t = t.slice(size, t.length);
}
document.encoder.text.value = o;
return false;
}
function toArray()
{
var str = document.encoder.text.value;
var nstr = str.split(" ");
document.encoder.text.value = nstr;
}
//i am having trouble with this part. don't know how to call the document.encoder.text.value into the list and don't know how to call group_size into elementsPerSubArray
function toMatrix(list, elementsPerSubArray) {
var matrix = [], i, k;
for (i = 0, k = -1; i < list.length; i++) {
if (i % elementsPerSubArray === 0) {
k++;
matrix[k] = [];
}
matrix[k].push(list[i]);
}
return matrix;
}
Finally, i don't know how to post result back into the textarea
So far i can group the text into block of how many characters i choose from a selection box (i've got 30 options) and can make line breaks wherever i want. as far as the matrix goes, line breaks will be made after 1 group of (X) characters.
Given this text:
OOMUCHABOUTMYCOUNTRYICAREALOTH
I can break it into groups like this.
OOMUC HABOU TMYCO UNTRY ICARE ALOTH
or like this if i want.
OOMUC
HABOU
TMYCO
UNTRY
ICARE
ALOTH
What i want to do is be able to flip the matrix so it looks like this
AIUTHO
LCNMAO
...ect
HEYOUC