Javascript matrix

Question

I am trying to make a matrix which has an X of X's. The following code produces a diagonal of X's from the top left to the bottom right but not from the top right to the bottom left. I am unsure of where to even begin. Should another for loop be created with a new variable? Or is there something as simple as adding an else if statement for variable j? Any help will be greatly appreciated.

var nMatrix =  "";
var n = prompt ("enter a number");
n = parseInt(n);
for (var i = 1; i <= n; i++) {
    var row = "| ";
    for (var j = 1; j <= n; j++) {
        if (i == j)
            row += "x "; //top left to bottom right diagonally      
        else 
            row += Math.floor (9*Math.random()+1)+" ";
    }               
    row += "|\n";
    nMatrix += row;
}
document.getElementById ("matrix").innerText = nMatrix;

i == n - j + 1 PS: try used to starting iteration from 0, since almost everything around you in programming is 0-based — zerkms
– zerkms, Commented Jun 4, 2015 at 5:48

Connor Peet · Accepted Answer · 2015-06-04 05:56:18Z

1

Maybe something along the lines of:

function grid (size) {
    var out = '';

    for (var row = 0; row < size; row++) {
        out += '| ';
        for (var col = 0; col < size; col++) {
            out += Math.random() < 0.9 ? '  ' : 'x ';
        }
        out += '|\n';
    }

    return out;
}

Output of grid(12):

 |     x                 x |
 |       x x     x         |
 | x                     x |
 |                         |
 |           x             |
 |                         |
 |     x                   |
 |               x         |
 |                       x |
 |                         |
 |                     x   |
 | x                       |

answered Jun 4, 2015 at 5:56

Connor Peet

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Comments

Jeong Jinmyeong · Accepted Answer · 2015-06-04 06:18:30Z

Assume you have (n*n) matrix looks like this:

   1 2 3 ... n    j
1 |           |
2 |           |
3 |           |
: |           |
n |           |

i

You understand (i == j) is the key to pick following x's, so you coded that successfully. (by the way, i prefer j == i).

   1 2 3 ... n    j
1 |x          |
2 |  x        |
3 |    x      |
: |           |
n |          x|

i

So you need the find out the rule that pick following x's:

   1 2 3 ... n    j
1 |          x|
2 |        x  |
3 |           |
: |  x        |
n |x          |

i

How do you see the rule of x's? Isn't that for every x, i+j = n+1 holds?

so what you need to check is if (j == n + 1 - i). If that holds, you need to draw x.

One thing you maybe want to take consideration into is when (j == i) and (j == n + 1 - i) both holds: this is possible when n is odd. If you're using 'else if' correctly, no problem will occur. However it will be good for you to thinking about this.

Greg Prisament · Accepted Answer · 2015-06-04 06:12:05Z

0

How about:

function grid(size) {
    var out = "";
    for (var i = 0; i < size; i++) {
        for (var j = 0; j < size; j++) {
            out += (i == j || i == (size - j - 1)) ? "X" : " ";
        }
        out += "\n";
    }
    return out;
}

Output of grid(9):

Here's a jsfiddle: https://jsfiddle.net/6w36s1kb/

edited Jun 4, 2015 at 6:12

answered Jun 4, 2015 at 6:04

Greg Prisament

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