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0

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javascript to return values on decrement

My code is following

var points = 4;
var yModifier = [];
for (var i = 0; i <= points; i++) {
    yModifier.push([]);
};
yModifier.forEach(
    (a, j) => {
        var start = j; 
        var end = start + points; 
        for (var i = start, k = 0; i <= end; i++, k++) {
            yModifier[j].push([k, i])
        };
    }
);
console.log(yModifier);

which produces the following

0: [[0,0],[1,1],[2,2],[3,3],[4,4]]
1: [[0,1],[1,2],[2,3],[3,4],[4,5]]
2: [[0,2],[1,3],[2,4],[3,5],[4,6]] 
3: [[0,3],[1,4],[2,5],[3,6],[4,7]]  
4: [[0,4],[1,5],[2,6],[3,7],[4,8]] 

How can I modify the forEach to give me the following

0: [[0,0],[1,1],[2,2],[3,3],[4,4]]
1: [[0,-1],[1,0],[2,1],[3,2],[4,3]]
2: [[0,-2],[1,-1],[2,0],[3,1],[4,2]] 
3: [[0,-3],[1,-2],[2,-1],[3,0],[4,1]]  
4: [[0,-4],[1,-3],[2,-2],[3,-1],[4,0]] 
about 4 years ago · Juan Pablo Isaza
2 answers
Answer question

0

var points = 4;
var yModifier = [];
for (var i = 0; i <= points; i++) {
    yModifier.push([]);
};
yModifier.forEach(
    (a, j) => {
        var start = j; //a =0, start=0
        var end = start + points; //a =1, end=360
        for (var i = start, k = 0; i <= end; i++, k++) {
            yModifier[j].push([k, k - j])
        };
    }
);
console.log(yModifier);

about 4 years ago · Juan Pablo Isaza Report

0

Looks like the pattern for values is [col, col - row]:

var points = 4;
var yModifier = Array.from({ length: points + 1 }, function(_, row) {
  return Array.from({ length: points + 1 }, function(_, col) {
    return [col, col - row];
  });
});
console.log(yModifier);

about 4 years ago · Juan Pablo Isaza Report
Answer question
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