Multi-dimensional Array - Check For Diagonal Consecutive Values

Writing a check for an array to determine if there are any potential consecutive values within it, be it horizontal, vertical, or either way diagonal. The example below is a sample

Solution 1:

Have you tried:

(b[i][j]==="X" && i===j) ? dl++ : 0;
(b[j][i]==="X" && (i+j)===(len-1)) ? dr++ : 0;

?

dl or the left diagonal would have i and j equal(so (0,0) (1,1) and (2,2))

dr or the right diagonal would have the sum of i and j equal to side-length minus 1(so (0,2) (1,1) (2,0))

This would work when the length of the diagonal is the same as the length of the matrix, full-body diagonal in a tic-tac-toe game, for example. For partial diagonals you can modify the code slightly to something like:

var diff = 0;
var sum = len - 1;  
(b[i][j]==="X") ? diff = (i-j) : diff = 0;  
(b[i][j]==="X") ? sum= (i+j) : sum = len;  
(b[i][j]==="X" && (i-j)===diff) ? dl++ : 0;
(b[j][i]==="X" && (i+j)===sum) ? dr++ : 0;

Solution 2:

Final answer (due to @tewathia's answer and some further research)

function testWin() {
    var win=3, len=b.length, r=0, c=0, dr=0, dl=0;
    for(var i=0;i<len;i++){
        for(var j=0;j<len;j++){
            (b[j][i]==="X") ? c++ : c=0;
            (b[i][j]==="X") ? r++ : r=0;
            if(b[i][j]==="X" && i<len-win+1){ dr=0; dl=0;
                for(var z=0;z<win;z++){ 
                    (b[i+z][j+z]==="X") ? dr++ : dr=0;
                    (b[i+z][j-z]==="X") ? dl++ : dl=0;
                }
            }
            if(c===win || r===win || dr===win || dl===win){ alert("YOU WIN!"); return true;}
        } r=0;
    }
}

Fiddle Here: http://jsfiddle.net/atk3V/1/