How to generate a unique random number in a range without repeating a number in C#

Asked Apr 10 '18 at 13:38

Active Apr 10 '18 at 13:57

Viewed 111 times

I have a requirement to produce a quiz and I need to randomize all numbers but only once between 0 and 24 (25 questions).

For example:

Next 'Question Button' is pressed and it will iterate randomly all 25 numbers until all numbers are used only once. Once it gets to the 25th number, it should go to the 'submit for results' on a last page. I am using C# Multiview to produce the quiz wizard pages but don't quite know how to achieve this efficiently.

So far, I have something like this to generate the numbers, but I don't know the best way to both generate a number and remember that number as the person goes through each page of the quiz to the final number:

 Random rnd = new Random();

        Console.WriteLine("\n25 random integers from 0 to 24:");
        for (int ctr = 1; ctr <= 25; ctr++)
        {
            Console.Write("{0,6}", rnd.Next(0, 24));
            if (ctr % 5 == 0) Console.WriteLine();
        }

edited Apr 10 '18 at 13:57

asked Apr 10 '18 at 13:38

Jason

3

Generate all numbers, shuffle them, take as many as you need. – Corak Apr 10 '18 at 13:40
you can this number to a list and check if the number is in the list not the most efficient way, but it will work – styx Apr 10 '18 at 13:42
2

basically: `Enumerable.Range(0, 25).OrderBy(i => rand.Next()).Take(25)` -- although there are "better" shuffle algorithms (see fisher-yates for example). – Corak Apr 10 '18 at 13:45
@Corak why would `.Take(25)` even be necessary? – Rafalon Apr 10 '18 at 13:47
Yes, I need all 25 unique numbers one time only – Jason Apr 10 '18 at 13:48
@Rafalon - It's not if one is really interested in *all* the numbers. I kept it in to reflect the previous statement: generate all, shuffle, take. – Corak Apr 10 '18 at 13:48
The numbers represent the page number to go to which is why I need all numbers – Jason Apr 10 '18 at 13:49
1

@Jason - so `Enumerable.Range(0, 25).OrderBy(i => rand.Next())` would suffice, as it produces all the numbers. – Corak Apr 10 '18 at 13:50
Was a typo. Sorry -- I fixed it. – Jason Apr 10 '18 at 13:51

How to generate a unique random number in a range without repeating a number in C#

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