Okay so I need the output to print hollow squares and I'm at a loss right now. I don't want the answer but I would like some hints to help get me on the right track. Thanks!
"use strict"
if (process.argv.length < 3) {
console.log("Not enough command-line arguments given.");
console.log("Usage: node lab13_4.js num");
process.exit();
}
var width = parseInt(process.argv[2]);
function makeLine(width) {
var L = "";
for(var w = 0; w < width; w += 1) { // repeated width many times
L = L + ".";
}
return L;
}
// print the line some number of times.
function printLines(line, howMany) {
// print the right number of lines
for (var i = 0; i < howMany; i += 1) { // repeated height many times
console.log(line);
}
}
for (var x = 0; x <= width; x += 1) {
var line = makeLine(x);
printLines(line, x);
}
If LENGTH be the length of the sides of the square.
You have to make nested loop and check boundary conditions.
for (var i = 0; i < LENGTH; i++) {
for (var j = 0; j < LENGTH; j++) {
if (i == 0 || i == LENGTH - 1 || j == 0 || j == LENGTH - 1) {
process.stdout.write("*")
} else {
process.stdout.write(" ")
}
}
process.stdout.write("\n")
}
We are looking for a solution like this:
*****
* *
* *
* *
*****
The top and bottom sides of the square have the full amount of stars.
All lines in between will only have one star in the beginning of the line and one star at the end of the line.
I use the following function:
function square(input) {
// top line
console.log('*'.repeat(input));
// middle lines
for(let i = 1; i < input - 1; i++) {
console.log('*' + ' '.repeat(input-2) + '*');
}
// bottom line
console.log('*'.repeat(input));
};
square(5);
hope this makes sense :)
Related
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?
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();
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>
This code:
function drawStars(){
for (var i = 1; i < 4; i = i + 1){
console.log('*');
for (var l = 1; l < 6; l = l + 1){
console.log('*');
}
}
}
once run, prints stars into the console.
My question is, how can I make it print a 5 by 2 pattern of starts into the console if I type in drawStars(5,2) into the console?
It should also be able to print other patterns corresponding to what I type into the console.
P.S. Can you also fix the unnecessary stars in between the other stars?
You just need to add parameters in your method and use them in them as the loop end-conditions. Also, the unncessary stars are the one you draw in the first loop only. Here the p parameter is the pattern.
function drawStars(x, y, p){
for (var i = 0; i < x; i++){
var line = ""
for (var j = 0; j < y; j++){
line += p;
}
console.log(line);
}
}
If you call it
drawStars(5, 2, '#')
Then it will output a 5x2 rectangle filled with #:
##
##
##
##
##
I made a chip8 emulator, but ran into some problems with detecting collision when drawing. Chip8 draws onto the screen by XOR'ing individual pixels onto the screen, and setting a flag if a pixel is turned off. My code is as follows:
_DXYN: function(X, Y, N) {
console.log("_DXYN");
for (var i = 0; i < N; i++) {
for (var j = 0; j < 8; j++) {
var bitBefore = graphics[(i + cpu.registers.V[Y]) * 64 + j + cpu.registers.V[X]];
var bitAfter = bitBefore ^ cpu.getBits(memory[cpu.registers.I + i])[j];
graphics[(i + cpu.registers.V[Y]) * 64 + j + cpu.registers.V[X]] = bitAfter;
if ((bitBefore != bitAfter) && bitBefore == 1)
cpu.registers.V[0xF] = 0x1;
else
cpu.registers.V[0xF] = 0x0;
}
}
}
graphics is a 1-dimensional array of ints, each corresponding to a pixel on the screen. A pixel is on if its corresponding integer in the array is a 1, and off it it is a 0. The cpu object contains all the methods, including this one, as well as the registers. X is the opcode parameter for the register which contains the x coordinate of the sprite to draw, and Y is the opcode parameter for the register with the y coordinate. The I register is the location in memory to read from. The cpu.getBits function is as follows:
getBits: function(opcode) {
var bits = [];
for (var i = 0; i < 8; i++) {
bits.push((opcode >> i) & 1)
}
return bits.reverse();
}
The complete code can be found on github here: https://github.com/ichub/Chip8
Look in chip8cpu.js for the implementation.
I found this website : http://www.multigesture.net/articles/how-to-write-an-emulator-chip-8-interpreter which explains in detail how to write a Chip-8 interpreter. From that and the Wikipedia aticle on CHIP-8 I think that your routine should be the following:
_DXYN: function(X, Y, N) {
console.log("_DXYN");
cpu.registers.V[0xF] = 0x0;
for (var i = 0; i < N; i++) { // rows of the sprite
for (var j = 0; j < 8; j++) {
var bitBefore = graphics[(i + cpu.registers.V[Y]) * 64 + j +
cpu.registers.V[X]];
var bitAfter = bitBefore ^ cpu.getBits(memory[cpu.registers.I + i])[j];
graphics[(i + cpu.registers.V[Y]) * 64 + j + cpu.registers.V[X]] = bitAfter;
if ((bitBefore != bitAfter) && bitBefore == 1)
cpu.registers.V[0xF] = 0x1;
}
}
}
In your routine you are VF is cleared at each pixel if there is no collision so its value depends solely on the last pixel drawn. You should clear VF before starting to draw, and set it only when there is a collision.