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Good morning, all. I am a newb when it comes to this stuff. Thanks in advance for any help.

I need the following code to deliver 3 unique numbers that are greater than 0 and less than 13.


I keep running into problems with returning a 0 or two answers being the same.

What I have tried:

JavaScript
var numb = Math.floor(Math.random()* 13;
var numb2 = Math.floor(Math.random() * 13;
var numb3 = Math.floor(Math.random() * 13;

if (numb != 0) {
    var index =  numb;
} else {
    var index = 1;
}


if (numb2 != 0 && numb2 != numb) {
    var index2 = numb2;
} else  {
    var index2 = numb + 1;
    
}

if (numb3 != 0 && numb3 != numb1 && numb3 != numb2) {
    var index3 = numb3;
} else if (numb2 + 1 == numb) {
    var index3 = numb2 + 2
    else {
   	index3 = numb2 + 1 	
    
    };
}
document.write(index);

document.write(index2);

document.write(index3);
Posted
Updated 9-Mar-16 4:33am
v2
Comments
CHill60 9-Mar-16 9:36am    
Rather than just checking each number once, use a While loop to keep checking - if any of the numbers are the same or 0 just regenerate them all
random_logic 9-Mar-16 9:44am    
Thank you. I am so new at this i'm not sure how to do that.
Sinisa Hajnal 9-Mar-16 9:58am    
Range is too small to reliably have different numbers. You have to check if you already have the number and randomize again.
random_logic 9-Mar-16 10:00am    
I think I understand what you are saying, however I'm not sure how to write the code to make it do that. What would something like this look like?
Sinisa Hajnal 9-Mar-16 9:58am    
As for returning zero, you need to add +1 at the end so you get range of 1 to whatever

Math.random() returns a random number between 0 (inclusive) and 1 (exclusive), with a bit of trick, you can have
Math.random()*12+1

that will return a random number between 1 (inclusive) and 13 (exclusive).
In this way, there is no need to check for the occurrence of 0.
Since you are re-using the same code to generate multiple random numbers, put this code in a function for better code re-use, e.g.
JavaScript
function getRndNumb(){
  return Math.floor(Math.random()*12+1);
}

Lastly, to repeatedly re-generate random numbers when there is duplication, use a loop, in this case do...while is more appropriate, e.g.
do {
   numb1 = getRndNumb();
   numb2 = getRndNumb();
   numb3 = getRndNumb();
} while (numb1 == numb2 || numb1==numb3 || numb2==numb3)

I shall leave it to you to put together these pieces.
 
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v2
Comments
random_logic 9-Mar-16 13:10pm    
Thank you Peter Leow! It worked like a charm! How would something like this work if I wanted to have 13 unique numbers?
Rather than just checking the numbers once, check them in a while loop
JavaScript while Loop[^]

e.g. you could do something like this (Note - untested)
JavaScript
while (numb1 == numb2 || numb2 == numb3 || numb1 == numb3 || numb1 == 0 || numb2 == 0 || numb3 == 0){
    if(numb1 == 0 || numb1 == numb2 || numb1 == numb3){
        numb1 = numb1 + 1;
    if(numb2 == 0 || numb2 == numb3 || numb2 == numb3){
        numb2 = numb2 + 1;
    if(numb3 == 0 || numb3 == numb1 || numb3 == numb2){
        numb3 = numb3 + 1;
}
 
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