I have a variable that has a number between 1-3.
I need to randomly generate a new number between 1-3 but it must not be the same as the last one.
It happens in a loop hundreds of times.
What is the most efficient way of doing this?
7
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1is it between 1-3, are we talking whole numbers? so options are 1, 2 or 3?filype– filype2012-04-17 10:26:04 +00:00Commented Apr 17, 2012 at 10:26
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2You realise that generating a number from 1-3 without repeating is a) not random and b) functionally equivalent to selecting one of two numbers, i.e. probability 0.5?Widor– Widor2012-04-17 10:26:41 +00:00Commented Apr 17, 2012 at 10:26
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1ok i understand b) but why a)?Moshe Shaham– Moshe Shaham2012-04-17 10:28:15 +00:00Commented Apr 17, 2012 at 10:28
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What have you tried so far? The efficiency of the loop depends on many things: Are you using this random in the loop only, should it be passed forward as an argument etc.Teemu– Teemu2012-04-17 10:39:20 +00:00Commented Apr 17, 2012 at 10:39
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@Moshe because for it to be truly random, each number would have to be independent of the last. As you have the condition that there must be no repeats, the future numbers are affected by the past.Widor– Widor2012-04-17 10:41:35 +00:00Commented Apr 17, 2012 at 10:41
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6 Answers
May the powers of modular arithmetic help you!!
This function does what you want using the modulo operator:
/**
* generate(1) will produce 2 or 3 with probablity .5
* generate(2) will produce 1 or 3 with probablity .5
* ... you get the idea.
*/
function generate(nb) {
rnd = Math.round(Math.random())
return 1 + (nb + rnd) % 3
}
if you want to avoid a function call, you can inline the code.
1 Comment
Bergi
Nice, I did exactly the same inline :)
Here is a jsFiddle that solves your problem : http://jsfiddle.net/AsMWG/
I've created an array containing 1,2,3 and first I select any number and swap it with the last element. Then I only pick elements from position 0 and 1, and swap them with last element.
Comments
var x = 1; // or 2 or 3
// this generates a new x out of [1,2,3] which is != x
x = (Math.floor(2*Math.random())+x) % 3 + 1;
Comments
You can randomly generate numbers with the random number generator built in to javascript. You need to use Math.random().
If you're push()-ing into an array, you can always check if the previously inserted one is the same number, thus you regenerate the number. Here is an example:
var randomArr = [];
var count = 100;
var max = 3;
var min = 1;
while (randomArr.length < count) {
var r = Math.floor(Math.random() * (max - min) + min);
if (randomArr.length == 0) {
// start condition
randomArr.push(r);
} else if (randomArr[randomArr.length-1] !== r) {
// if the previous value is not the same
// then push that value into the array
randomArr.push(r);
}
}
Comments
As Widor commented generating such a number is equivalent to generating a number with probability 0.5. So you can try something like this (not tested):
var x; /* your starting number: 1,2 or 3 */
var y = Math.round(Math.random()); /* generates 0 or 1 */
var i = 0;
var res = i+1;
while (i < y) {
res = i+1;
i++;
if (i+1 == x) i++;
}
Comments
The code is tested and it does for what you are after.
var RandomNumber = {
lastSelected: 0,
generate: function() {
var random = Math.floor(Math.random()*3)+1;
if(random == this.lastSelected) {
generateNumber();
}
else {
this.lastSelected = random;
return random;
}
}
}
RandomNumber.generate();
2 Comments
Florian Margaine
Why was this downvoted? It's pretty good and not using array like all the others.
Agent Zebra
This did not work for me, it did not produce a new random number every time.