C# - Writing numbers 1 through 10 in random order

Question

Part 1: All I am wanting to achieve is to write the numbers 1, 2, 3 ... 8, 9, 10 to the console window in random order. So all the numbers will need to be written to console window, but the order of them must be random.

Part 2: In my actual project I plan to write all of the elements in an array, to the console window in random order. I am assuming that if I can get the answer to part 1, I should easily be able to implement this with an array.

Here is the Fischer Yates solution, which is O(n) http://stackoverflow.com/questions/1287567/is-using-random-and-orderby-a-good-shuffle-algorithm — Andras Sebo, Jun 20 '13 at 09:47

score 0 · Answer 1 · answered Jun 20 '13 at 09:44

0

Enumerable.Range(1, 10).OrderBy(i => Guid.NewGuid()) works nicely.

answered Jun 20 '13 at 09:44

Rob Lyndon

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OK, so this is very picky. Note that this is O (N Log N) whereas something like the Fisher Yates shuffle is O(N) – Jeff Foster Jun 20 '13 at 09:46
Sure, but if N is small enough that you're writing them to the console, that's probably not too much of a problem in this case. – Rob Lyndon Jun 20 '13 at 09:47
Apart from that you should at least add `ToArray` since OP wants an array as result. – Tim Schmelter Jun 20 '13 at 10:42
Not correct. The OP wants to read from an existing array, randomise it, and then write the results to the console. – Rob Lyndon Jun 20 '13 at 11:11

score 0 · Answer 2 · answered Jun 20 '13 at 09:50

using System;
using System.Collections;

namespace ConsoleApplication
{
    class Numbers
    {
        public ArrayList RandomNumbers(int max)
        {
            // Create an ArrayList object that will hold the numbers
            ArrayList lstNumbers = new ArrayList();
            // The Random class will be used to generate numbers
            Random rndNumber = new Random();

            // Generate a random number between 1 and the Max
            int number = rndNumber.Next(1, max + 1);
            // Add this first random number to the list
            lstNumbers.Add(number);
            // Set a count of numbers to 0 to start
            int count = 0;

            do // Repeatedly...
            {
                // ... generate a random number between 1 and the Max
                number = rndNumber.Next(1, max + 1);

                // If the newly generated number in not yet in the list...
                if (!lstNumbers.Contains(number))
                {
                    // ... add it
                    lstNumbers.Add(number);
                }

                // Increase the count
                count++;
            } while (count <= 10 * max); // Do that again

            // Once the list is built, return it
            return lstNumbers;
        }
    }

Main

  class Program
    {
        static int Main()
        {
            Numbers nbs = new Numbers();
            const int Total = 10;
            ArrayList lstNumbers = nbs.RandomNumbers(Total);

            for (int i = 0; i < lstNumbers.Count; i++)
                Console.WriteLine("{0}", lstNumbers[i].ToString());

            return 0;
        }
    }
}

score 0 · Accepted Answer · answered Jun 20 '13 at 09:57

0

/// <summary>
/// Returns all numbers, between min and max inclusive, once in a random sequence.
/// </summary>
IEnumerable<int> UniqueRandom(int minInclusive, int maxInclusive)
{
    List<int> candidates = new List<int>();
    for (int i = minInclusive; i <= maxInclusive; i++)
    {
        candidates.Add(i);
    }
    Random rnd = new Random();
    while (candidates.Count > 0)
    {
        int index = rnd.Next(candidates.Count);
        yield return candidates[index];
        candidates.RemoveAt(index);
    }
}

In your program

Console.WriteLine("All numbers between 0 and 10 in random order:");
foreach (int i in UniqueRandom(0, 10)) {
    Console.WriteLine(i);
}

answered Jun 20 '13 at 09:57

Matheno

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Nice! I consolidated the code slightly for my own needs. Other than that thanks for answer, will be a good starting point. – SomeRandomProgrammer Jun 20 '13 at 10:10
Can you mark this post as the answer when this was helpfull to you? – Matheno Jun 20 '13 at 10:11
9

This works, but it's good form to credit the original source when copy/pasting answers: http://stackoverflow.com/a/1011408/106159 – Matthew Watson Jun 20 '13 at 10:21

Matheno · Answer 4 · 2013-06-20T09:52:35.737

-2

int[] ints = new int[11];
Random rand = new Random();

Random is a class built into .NET, and allows us to create random integers really, really easily. Basically all we have to do is call a method inside our rand object to get that random number, which is nice. So, inside our loop, we just set each element to the results of that method:

for (int i = 0; i < ints.Length; i++)
{
  ints[i] = rand.Next(11);
}

We are essentially filling our entire array with random numbers here, all between 0 and 10. At this point all we have to do is display the contents for the user, which can be done with a foreach loop:

foreach (int i in ints)
{
  Console.WriteLine(i.ToString());
}

edited Jun 20 '13 at 09:52

answered Jun 20 '13 at 09:46

Matheno

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OP doesn't want duplicate numbers, so this is wrong. Also this gives numbers in the range 0..9 and the OP wants 1..10 – Matthew Watson Jun 20 '13 at 09:49
This was just a rough example, OP can change the range of max numbers himself I hope. ;) – Matheno Jun 20 '13 at 09:53
Yes but the more fundamental problem is that your code can produce duplicate numbers, and miss other numbers out altogether. The OP wants the numbers 1..10 arranged in a random order, not a set of random numbers each of which is in the range 1..10. (And note that your recent edit now produces random numbers in the range 0..10, not 1..10 anyway) – Matthew Watson Jun 20 '13 at 09:56
I added a new solution, which works for me. – Matheno Jun 20 '13 at 09:59
1

Aye, that new answer will work. – Matthew Watson Jun 20 '13 at 10:20

C# - Writing numbers 1 through 10 in random order

4 Answers4