When I call this statement 4 times in a loop, it gives me duplicate random numbers.
var a = (parseInt(Math.random() * 4))+1
For example, sometimes it gives me 1,3,2,4 (fine!!), but sometimes it produes like 1, 3, 1,4 etc.
How I make sure that when my loop runs 4 times, I get unique set everytime
Shuffling a pre-filled array would be a good solution:
// shuffle function taken from here
// http://stackoverflow.com/a/2450976/603003
alert( shuffle( [1,2,3,4] ) );
Another apporach consists in the following function:
function gen4Numbers() {
var numbers = [];
while (numbers.length < 4) {
var newNr = (parseInt(Math.random() * 4))+1;
if (numbers.indexOf(newNr) == -1) {
numbers.push(newNr);
}
}
return numbers;
}
alert(gen4Numbers());
You should be warned that there is a very probability that this code will loop forever.
[1,2,3,4] and shuffle it. This will remove the risk of an endless loop.
You need to buffer your answers, but a much better way is to pre-compute them.
So for example you want random integers from 1 to 4 and unique, this is the same thing as randomly sorting an array of values from 1 to 4
for the general case:
// numbers of values
var n = 10;
// make initially sorted array
var arr = [];
for (var i=1; i<=n; ++i) { arr.push(i); }
// randomly sort the array
arr.sort(function () { return Math.random() - 0.5; });
now arr will have all the numbers from 1 to 10 in a random order, so if you want 4 unique random numbers in that range they are in arr[0], arr[1] arr[2], arr[3]
make n = 4 and you have your problem solved
i am using time
var d = new Date();
var uniqId = d.getTime();
Store the numbers in an array, and each time check if the number was generated before.
If you aren't generating a lot of numbers, this is pretty fast since you don't have to
loop a lot.
============= Edit 1 =============
However, if you have to generate a lot of random numbers, I suggest putting an array of
all the numbers, shuffling it, and then taking the first n numbers that you need.
Array shuffling can be found here:
http://stackoverflow.com/questions/2450954/how-to-randomize-shuffle-a-javascript-array
In case you want a huge set of numbers, you should store every distinct number, then pick one up every iteration of the loop.
var maxnB = 1000; // or 4
var list = [];
for (var i = 1; i <= maxNb; ++i) // store all number in a list [1, 2, 3, ... , maxNb]
list.push(i);
while (list.length > 0)
{
var pos = (parseInt(Math.random() * list.length))
var myNumber = list[pos]; // get your unique random number
list[pos] = list[list.length - 1]; // replace the number used with last number
list.pop(); //remove last number, decrease list.length by 1
// do something with your new number
}
Doing this, you have a 0 probability of looping forever, and if the pop() method and [] operator has a o(1) time complexity access, it will improve the time processing, instead of looking for your number in the buffer list every time you got a new number to check.
do this.
for( i = 0; i< 4; i++)
{
num = Math.floor((Math.random() * 3) + 1);
if(randomArray.indexOf(num)<0)
{
randomArray[i]=num;
}//endif
else
{
i=i-1;
}//endElse
}//endForLoop
basically, if a number is repeated -lets say at position 3 (randomArray[3] has same value as randomArray[atAnotherIndex], your counter goes back to 3 and do math.random again until a unique value is attained.
