JavaScript for Random Numbers with Recursion

Question

I'm trying to create a javascript function that accepts 2 parameters min and max and generates a random number between the two integers. That part is easy.

Where things get rough is that I need to create a conditional that says

function generateNum (min, max) {
  var randNumber = Math.ceil(Math.random()*(max - min)+min);

  if (randNumber === max) {
    // store the result (probably in an array) and generate a new 
    // number with the same behavior as randNumber (e.g. it is also
    // stores it's total in the array and recursively "re-generates 
    // a new number until max is not hit)
  }
}

The idea is to recursive-ise this so that a running total of the number of max hits is stored, combined, and then returned.

For example: The script receives min / max parameters of 5/10 generateNum(5,10){}. If the value generated by randNumber were 5, 6, 7, 8, or 9 then there would be no recursion and the function would return that value. If the value generated by randNumber is 10, then the value 10 is stored in an array and the function now "re-tries" recursively (meaning that as many times as 10 is generated, then that value is stored as an additional object in the array and the function re-tries). When the process stops (which could be infinite but has a parabolically decreasing probability of repeating with each recursion). The final number (5, 6, 7, 8, 9) would be added to the total of generated max values and the result would be returned.

Quite an unusual mathematic scenario, let me know how I can clarify if that doesn't make sense.

So what is the "stop" condition and what is the "recurse" condition? Just write that in the code as described. — user166390
– user166390, Commented Jun 5, 2012 at 19:35
Don't understand why you need recursion. Just use a while loop and a counter which you increment whenever you get max. Stop once you get a different number. The result is then counter * max + randNumber (or if you want to return an array that would be easy to create as well, but recursion is not needed either way). — Felix Kling
– Felix Kling, Commented Jun 5, 2012 at 19:37
so what exactly is the question, it seems as though you have all of the logic there? Just store the number in an array that is at a higher scope level than your function and call generateNum(min,max); right afterwards. — Mike McMahon
– Mike McMahon, Commented Jun 5, 2012 at 19:37
Just thought you should know that your function will return 'min' only on extremely rare occasions. There's a proper function for random numbers in the answers, though. — panda-34
– panda-34, Commented Jun 5, 2012 at 19:44

Guffa · Accepted Answer · 2012-06-05 20:39:16Z

3

That part is easy.

Not as easy as you think... The algorithm that you have is broken; it will almost never give you the minimum value. Use the Math.floor method instead, and add one to the range:

var randNumber = Math.floor(Math.random() * (max - min + 1)) + min;

To do this recursively is simple, just call the method from itself:

function generateNum (min, max) {
  var randNumber = Math.floor(Math.random()*(max - min + 1)) + min;
  if (randNumber == max) {
    randNumber += generateNum(min, max);
  }
  return randNumber;
}

You can also solve this without recursion:

function generateNum (min, max) {
  var randNumber = 0;
  do {
    var num = Math.floor(Math.random()*(max - min + 1)) + min;
    randNumber += num;
  } while (num == max);
  return randNumber;
}

There is no need to use an array in either case, as you don't need the seprate values in the end, you only need the sum of the values.

edited Jun 5, 2012 at 20:39

answered Jun 5, 2012 at 19:43

Guffa

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3 Comments

Brian Over a year ago

Great answer @Guffa! I noticed the broken behavior in rarely generating the min value. Also, big +1 for pointing out that the array is unnecessary, that's a relief.

panda-34 Over a year ago

@Guffa, You'd better change =+ to += in your first variant. Otherwise it'll look like you don't test your code.

Brian Over a year ago

Just to re-iterate since other answers are bubbling to the top. This is definitely the best method.

Paul · Accepted Answer · 2012-06-05 20:00:28Z

2

I assume that you don't really need a recursive solution since you tagged this for-loop. This will return the number of times the max number was picked:

function generateNum (min, max) {
    var diff = max - min;
    if(diff <= 0)
        return;

    for(var i = 0; diff == Math.floor(Math.random()*(diff + 1)); i++);

    return i;
}

Example outputs:

generateNum(1,2)  // 3
generateNum(1,2)  // 1
generateNum(1,2)  // 0
generateNum(5,10) // 0
generateNum(5,10) // 1

edited Jun 5, 2012 at 20:00

answered Jun 5, 2012 at 19:36

Paul

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6 Comments

Brian Over a year ago

What is the if(min == max) \ return; for in this code? The min paramater is the floor of the range which is 5 - 10 for generateNum(5,10);... what would be an example of meeting the if(min == max) \ return; condition?

Paul Over a year ago

@Brian That is a precation to stop it from looping infintely. If min == max then max will always be generated and it will continue generating random numbers forever.

Brian Over a year ago

Awesome! Just one small point of clarification: I probably shouldn't be using an array, simply because it's unnecessary. In the end I just want to return the total, which will require another loop to add together the items in results.

Paul Over a year ago

@Brian Okay, I made the function return an array, because that is what you asked for. (Also you would not need a loop, you could use array.length), but it is much better to return the number of occurrences if that is all you need. I will update my answer.

Brian Over a year ago

No worries, I realize that is what I asked for. Apologies that @Guffa was able to see a mistake in what I was asking for, I've got to give him the official nod so that someone else doesn't make my mistake. Leave yours as is though, since it presents a different solution.

|

Benjamin Gruenbaum · Accepted Answer · 2012-06-05 19:38:50Z

1

Two things:

1) the probability to roll 10 stays (theoretically the same on each roll (re-try)), the low probability is of hitting n times 10 in a row

2) I don't see why recursion is needed, what about a while loop?

 var randNumber;
 var arr = [];
 while ((randNumber = Math.ceil(Math.random()*(max - min)+min)) === max) {
   arr.push(
 }

answered Jun 5, 2012 at 19:38

Benjamin Gruenbaum

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Comments

panda-34 · Accepted Answer · 2012-06-05 20:34:58Z

1

I'd consider an idea that you don't need to use not only recursion and arrays but not even a for loop. I think you need a simple expression like this one (separated into three for clarity):

function generateNum (min, max)
{
    var randTail = Math.floor(Math.random()*(max - min)+min);
    var randRepeatMax = Math.floor(Math.log(Math.random()) / Math.log(1/(max-min+1)));
    return randRepeatMax*max + randTail;
}

Assuming one random number is as good as another, this should give you the same distribution of values as the straightforward loop.

edited Jun 5, 2012 at 20:34

answered Jun 5, 2012 at 20:22

panda-34

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Saebekassebil · Accepted Answer · 2012-06-05 20:41:20Z

1

Recursive method:

function generateNum (min, max) {
  var res = Math.floor(Math.random() * (max - min + 1)) + min;

  return (res === max) ? [res].concat(generateNum(min, max)) : res;
}

edited Jun 5, 2012 at 20:41

answered Jun 5, 2012 at 19:36

Saebekassebil

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