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let's say I have a grid i.e. 2d array
const grid = [
[0, 0, A, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 0, B, 0],
[D, E, 0, C, F],
[0, 0, 0, 0, 0],
]
if some cell in the grid can visit all adjacent cells 4-diredtionally, for example, C is at [3, 3] so it can visit [3, 3 + 1], [3 - 1, 3],[3 +1, 3]``[3, 3 - 1], so normally I would have to hard code this like
// 👇 hard-coded directions
const dirs = [
[1, 0],
[-1, 0],
[0, 1],
[0, -1],
]
const possibleMoves = []
for (const [dx, dy] of dirs) {
possibleMoves.push([dx + x, dy +y])
}
then if it can move 8-directionally then you have to hard code more directions
const dirs = [[1, 0], [-1, 0] , [0,1], [0,-1], [1,1], [-1,1], [-1,-1], [1,-1]]
Is there a smarter way to generate the dirs array for the next moves?
Yes!
First: any time you're doing grid-logic, start by checking what Amit Patel has to say.
Honestly, that link has everything you could ever need.
The short version is: if you know the grid width and cell layout, you can easily calculate coordinate offsets of any cell neighbor for any definition of "neighbor."
That logic can be implemented as a pure function that requires both the grid dimensions and the coordinates of the cell whose neighbors you want (aka the "target" cell):
let myNeighbors = getCellNeighbors(
{ x: 2, y: 2 }, // coords of target cell
{ width: 10, height: 10 } // grid dimensions, in cells
)
Or you can create a stateful thing that takes the grid dimensions at creation and calculates the offsets once, to be re-used for all getNeighbors calls:
let myGrid = new Grid(10, 10)
let myNeighbors = myGrid.getNeighbors(2, 5)
let myBiggerGrid = new Grid(25, 25)
let otherNeighbors = myBiggerGrid(2, 5)
I have a table of cells that need to be manipulated in a few specific spots accords to data I get
let arr = Array(3).fill(Array(3).fill(0));
[{x: 0, y: 0, value: 1}, {x: 1, y: 0, value: 2},{x: 2, y: 0, value: 3}].map(pos =>
arr[pos.x][pos.y] = pos.value
)
console.log(arr)
I expected the code to give [[1,0,0],[2,0,0],[3,0,0]] but instead it give [[3,0,0],[3,0,0],[3,0,0]], in other words it draws all as last y (value 3) and ignore the [pos.x] for some reason, don't sure why.
I'd liked to get some help with a possible workaround as explanation why this code not working as I expecting
thanks in advance!
the problem is when you do the outer fill , that fill function is executed only once, so basically you are having reference to same array in the nested arrays, so when you map you keep updating that same reference and hence all are same in the end
you might wanna do
let arr = [];
for(let i = 0; i < 3; i++) { arr.push(Array(3).fill(0));}
Try this instead:
var arr = Array.from({length:3},()=>Array(3).fill(0));
[{x: 0, y: 0, value: 1}, {x: 1, y: 0, value: 2},{x: 2, y: 0, value: 3}].map(pos =>
arr[pos.x][pos.y] = pos.value
)
console.log(arr);
The Array.from() method creates a new, shallow-copied Array instance from an array-like or iterable object.
I'm writing a board game in java script, and what i'm trying to accomplish is: layout the board(chess/checkers format) Then add pieces to the board based on position. So for example i want to be able to write code for piece a to be moved onto tile 10.
So far in my code i have a loop to create the board but don't a method to properly name the tiles, so that the piece can correctly be placed on the tile.
for (i=0; i<64; i++){
var tile = cc.Sprite.create(res.myTile_png);
this.addChild(tile,0);
x = centerpos.x + ((i % 8) - 3.5) * tile.getBoundingBox().width;
y = centerpos.y + (Math.floor(i / 8) - 3.5) * tile.getBoundingBox().height;
tile.setPosition(x,y);
}
One way to go about doing this would be to assign a unique integer identifier to every distinct piece in the game, and then maintain a matrix of dimensions equal to the # of rows x # of columns on the board, with values of the piece identifiers in the correct address in the matrix that would correspond to their position on the board.
For instance, the starting arrangement of pieces in checkers can be represented by:
[
[ 0, -1, 0, -1, 0, -1, 0, -1 ],
[ -1, 0, -1, 0, -1, 0, -1, 0 ],
[ 0, -1, 0, -1, 0, -1, 0, -1 ],
[ 0, 0, 0, 0, 0, 0, 0, 0 ],
[ 0, 0, 0, 0, 0, 0, 0, 0 ],
[ 1, 0, 1, 0, 1, 0, 1, 0 ],
[ 0, 1, 0, 1, 0, 1, 0, 1 ],
[ 1, 0, 1, 0, 1, 0, 1, 0 ]
]
with, say, -1 representing white, and 1 representing red pieces on the board.
The tile elements of the board can also be kept in a matrix, so that the two matrices can be iterated over together to place the pieces in corresponding locations.
The unique ids can also then be used as CSS class names, or image file names to be attached to the element representing the piece.
I am working on my first full program with two weeks of programming under my belt, and have run into a road block I can't seem to figure out. I am making a connect 4 game, and have started by building the logic in JavaScript before pushing to the DOM. I have started to make it with cell objects made by a constructor, that are then pushed into a game object in the form of a 2D array. I have managed to create a function that makes the play each time, and changes the value of the cell at the lowest point of that column with a 2 day array. However, I am not sure how to get my check for wins function to operate.
So far my logic is that, for each point in the 2D array, you can check by row, by column, and by diagonals. I understand the logic of how to check for win, but I don't understand how to traverse through the arrays by row and column. In the example below, this.cellsArray is an array of cell objects in the Board Constructor. The array has 7 column arrays, with 6 rows each, as I flipped the typical row column logic to account for Connect Four's column based nature. However I can't access the array like this.cellsArray[col][row], as col and row aren't defined, and I'm not sure how to define an index value? Any help would be appreciated!
Connect 4
Example:
//array location is equal to an instance of this.cellsArray[col][row]
Board.prototype.checkRowRight = function (arrayLocation) {
if ((arrayLocation[i+1][i].value === arrayLocation.value) && (arrayLocation[i+2][i]=== arrayLocation.value) && (arrayLocation[i+3][i].value === arraylocation.value)){
this.winner = this.currentPlayer;
this.winnerFound = true;
console.log('Winner has been found!')
}
};
Referencing back to my logic found here and refactoring out the winning line detection code, this can easily be converted into Javascript as follows:
function chkLine(a,b,c,d) {
// Check first cell non-zero and all cells match
return ((a != 0) && (a ==b) && (a == c) && (a == d));
}
function chkWinner(bd) {
// Check down
for (r = 0; r < 3; r++)
for (c = 0; c < 7; c++)
if (chkLine(bd[r][c], bd[r+1][c], bd[r+2][c], bd[r+3][c]))
return bd[r][c];
// Check right
for (r = 0; r < 6; r++)
for (c = 0; c < 4; c++)
if (chkLine(bd[r][c], bd[r][c+1], bd[r][c+2], bd[r][c+3]))
return bd[r][c];
// Check down-right
for (r = 0; r < 3; r++)
for (c = 0; c < 4; c++)
if (chkLine(bd[r][c], bd[r+1][c+1], bd[r+2][c+2], bd[r+3][c+3]))
return bd[r][c];
// Check down-left
for (r = 3; r < 6; r++)
for (c = 0; c < 4; c++)
if (chkLine(bd[r][c], bd[r-1][c+1], bd[r-2][c+2], bd[r-3][c+3]))
return bd[r][c];
return 0;
}
And a test call:
x =[ [0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 1, 1, 0, 0],
[0, 0, 0, 1, 1, 0, 0],
[0, 0, 1, 2, 2, 2, 0],
[0, 1, 2, 2, 1, 2, 0] ];
alert(chkWinner(x));
The chkWinner function will, when called with the board, return the first (and only, assuming each move changes only one cell and you're checking after every move) winning player.
The idea is to basically limit the checks to those that make sense. For example, when checking cells to the right (see the second loop), you only need to check each row 0-6 starting in each of the leftmost four columns 0-3.
That's because starting anywhere else would run off the right hand side of the board before finding a possible win. In other words, column sets {0,1,2,3}, {1,2,3,4}, {2,3,4,5} and {3,4,5,6} would be valid but {4,5,6,7} would not (the seven valid columns are 0-6).
This is an old thread but i'll throw my solution into the mix since this shows up as a top search result for "how to calculate connect4 win javascript"
I tackled this problem by using matrix addition.
Assume your game board is stored in memory as a 2D array like this:
[ [0, 0, 0, 0, 0, 0, 0],
[0, 0, Y, 0, 0, 0, 0],
[0, 0, Y, 0, 0, 0, 0],
[0, 0, R, 0, 0, 0, 0],
[0, 0, Y, 0, 0, 0, 0],
[0, 0, R, R, R, 0, 0] ];
On each "Coin Drop" you should call a function passing the x/y position of the coin.
THIS is where you calculate weather the user has won the game
let directionsMatrix = {
vertical: { south: [1, 0], north: [-1, 0] },
horizontal: { east: [0, 1], west: [0, -1] },
backward: { southEast: [1, 1], northWest: [-1, -1] },
forward: { southWest: [1, -1], northEast: [-1, 1] },
};
NOTE: "South" in matrix notation is [1,0], meaning "Down 1 cell, Right 0 cells"
Now we can loop through each Axis/Direction to check if there is 4 in a row.
const playerHasWon = (colnum, rowNum, playerColor, newGrid) => {
//For each [North/South, East/West, NorthEast/Northwest, SouthEast/Southwest]
for (let axis in directionsMatrix) {
// We difine this variable here so that "East" and "West" share the same count,
// This allows a coin to be dropped in a middle cell
let numMatches = 1;
// For each [North, South]
for (let direction in directionsMatrix[axis]) {
// Get X/Y co-ordinates of our dropped coin
let cellReference = [rowNum, colnum];
// Add co-ordinates of 1 cell in test direction (eg "North")
let testCell = newGrid[cellReference[0]][cellReference[1]];
// Count how many matching color cells are in that direction
while (testCell == playerColor) {
try {
// Add co-ordinates of 1 cell in test direction (eg "North")
cellReference[0] += directionsMatrix[axis][direction][0];
cellReference[1] += directionsMatrix[axis][direction][1];
testCell = newGrid[cellReference[0]][cellReference[1]];
// Test if cell is matching color
if (testCell == playerColor) {
numMatches += 1;
// If our count reaches 4, the player has won the game
if (numMatches >= 4) {
return true;
}
}
} catch (error) {
// Exceptions are to be expected here.
// We wrap this in a try/catch to ignore the array overflow exceptions
// console.error(error);
break;
}
}
// console.log(`direction: ${direction}, numMatches: ${numMatches}`);
// If our count reaches 4, the player has won the game
if (numMatches >= 4) {
return true;
}
}
}
// If we reach this statement: they have NOT won the game
return false;
};
Here's a link to the github repo if you wish to see the full code.
Here's a link to a live demo
I ve created a array with eleven values . I am trying to slide the values of my array during intervals. What i am trying, every n ms to slide a value to the next position of the array that i ve created. Every inteval i initialize the first value, so i want the slide effect.
var barArray = [0,0,0,0,0,0,0,0,0,0,0];
var interval = 0;
setInterval(function() {
temporal = getNewValue; //getting with a function new value
barArray[0] = temporal;
if(interval == barArray.length)
{
interval = 0;
}
for (var i = 0; barArray.length; i++){
// code missing
}
}, 1000);
I am tried many things without finding a solution.
Output:
1st interval: [76, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
2nd interval: [55, 76, 0, 0, 0, 0, 0, 0, 0, 0, 0]
3rd interval: [32,55, 76, 0, 0, 0, 0, 0, 0, 0, 0]
11th interval: [..., 32, 55, 76]
12th [..., 32,55] ect. `
What you have described here is a queue. You input elements at one end and silently drop them at the other end. JavaScript arrays have functions to add and extract elements at both ends of the array (push/pop and shift/unshift).
In the end, a complete solution would be:
var barArray = [0,0,0,0,0,0,0,0,0,0,0];
setInterval(function() {
barArray.unshift(getNewValue());
barArray.pop();
}, 1000);