Generate unique random numbers between 1 and 100

Generate permutation of 100 numbers and then choose serially.

Use Knuth Shuffle(aka the Fisher-Yates shuffle) Algorithm.

JavaScript:

  function fisherYates ( myArray,stop_count ) {
  var i = myArray.length;
  if ( i == 0 ) return false;
  int c = 0;
  while ( --i ) {
     var j = Math.floor( Math.random() * ( i + 1 ) );
     var tempi = myArray[i];
     var tempj = myArray[j];
     myArray[i] = tempj;
     myArray[j] = tempi;

     // Edited thanks to Frerich Raabe
     c++;
     if(c == stop_count)return;

   }
}

CODE COPIED FROM LINK.

EDIT:

Improved code:

function fisherYates(myArray,nb_picks)
{
    for (i = myArray.length-1; i > 1  ; i--)
    {
        var r = Math.floor(Math.random()*i);
        var t = myArray[i];
        myArray[i] = myArray[r];
        myArray[r] = t;
    }

    return myArray.slice(0,nb_picks);
}

Potential problem:

Suppose we have array of 100 numbers {e.g. [1,2,3...100]} and we stop swapping after 8 swaps; then most of the times array will look like {1,2,3,76,5,6,7,8,...numbers here will be shuffled ...10}.

Because every number will be swapped with probability 1/100 so prob. of swapping first 8 numbers is 8/100 whereas prob. of swapping other 92 is 92/100.

But if we run algorithm for full array then we are sure (almost)every entry is swapped.

Otherwise we face a question : which 8 numbers to choose?

var bombout=0;
var checkArr=[];
var arr=[];
while(arr.length < 8 && bombout<100){
  bombout++;
  var randomNumber=Math.ceil(Math.random()*100);
  if(typeof checkArr[randomNumber] == "undefined"){
    checkArr[randomNumber]=1;
    arr.push(randomNumber);
  }
}​

// untested - hence bombout