Javascript generating variations with exclusion from only one array

Question

There is an array with IDs.

var ingredientsIDs = new Array();
    ingredientsIDs[0]  = 1;
    ingredientsIDs[1]  = 2;
    ingredientsIDs[2]  = 3;
    ingredientsIDs[3]  = 4;

I tried to find out how to convert these IDs to another array what looks like this.

Array [ ".1", ".1.2", ".1.2.3", ".1.2.3.4", ".2", ".2.3", ".2.3.4", ".3", ".3.4", ".4", ".1.3", ".1.3.4" , "1.4" .... etc]

I would like to get all "permutations" but it is not really that. I wasted the whole day.

The first part is ready but I don't known how to continue.

What I wrote so far

var allVariation = new Array();
var index        = 0;
var n            = 0;
var l            = 0;
var i            = 0;
var j            = 0;

var ingredientsIDs = new Array();
ingredientsIDs[0]  = 1;
ingredientsIDs[1]  = 2;
ingredientsIDs[2]  = 3;
ingredientsIDs[3]  = 4;

for(i = 0; i <= ingredientsIDs.length; i++){
  for(j = 0; j < i; j++){
    if(allVariation[n] == undefined){allVariation[n] = "";}
    allVariation[n] += "." + ingredientsIDs[j];
  }
  if( i == ingredientsIDs.length  && l <= ingredientsIDs.length){
    l++;
    index = ingredientsIDs.indexOf(l);
    ingredientsIDs.splice(index, 1);
    i = 0;
  }
  n++;
}

console.log(allVariation);

So I would be happy if somebody can show me some way, how to continue.

You might be looking for all 'subsets'? See e.g. http://stackoverflow.com/questions/5752002/find-all-possible-subset-combos-in-an-array — le_m, Jun 24 '16 at 21:07
I will try the subsets, maybe it is a good point for me, ty le_m — Michael, Jun 24 '16 at 21:14
yaaaay, tank you so much le_m! that was what I am looking for! Sorry for duplicated question I didn't know how to find it because the "subsets" keyword new for me. — Michael, Jun 25 '16 at 07:27
but it is stoped by browser because my array has 51 items.... QQ — Michael, Jun 25 '16 at 09:25

Javascript generating variations with exclusion from only one array

0 Answers0