Generating random numbers without repeating.C#

Question

Hi everyone I am trying to generate 6 different numbers on the same line in c# but the problem that i face is some of the numbers are repeating on the same line.Here is my code to

var rand = new Random();
List<int> listNumbers = new List<int>();
int numbers = rand.Next(1,49);
for (int i= 0 ; i < 6 ;i++)
    {
         listNumbers.Add(numbers);
         numbers = rand.Next(1,49);
    }

somewhere my output is

17 23 23 31 33 48

Answer 1

Check each number that you generate against the previous numbers:

List<int> listNumbers = new List<int>();
int number;
for (int i = 0; i < 6; i++)
{
  do {
     number = rand.Next(1, 49);
  } while (listNumbers.Contains(number));
  listNumbers.Add(number);
}

Another approach is to create a list of possible numbers, and remove numbers that you pick from the list:

List<int> possible = Enumerable.Range(1, 48).ToList();
List<int> listNumbers = new List<int>();
for (int i = 0; i < 6; i++)
{
  int index = rand.Next(0, possible.Count);
  listNumbers.Add(possible[index]);
  possible.RemoveAt(index);
}

Answer 2

listNumbers.AddRange(Enumerable.Range(1, 48)
                               .OrderBy(i => rand.Next())
                               .Take(6))

Answer 3

Create a HashSet and generate a unique random numbers

public List<int> GetRandomNumber(int from,int to,int numberOfElement)
{
    var random = new Random();
    HashSet<int> numbers = new HashSet<int>();
    while (numbers.Count < numberOfElement)
    {
        numbers.Add(random.Next(from, to));
    }
    return numbers.ToList();
}

Answer 4

Make it a while loop and add the integers to a hashset. Stop the loop when you have six integers.

Answer 5

I've switched your for loop with a do...while loop and set the stopping condition on the list count being smaller then 6. This might not be the best solution but it's the closest to your original code.

List<int> listNumbers = new List<int>();
do
    {
        int numbers = rand.Next(1,49);
        if(!listNumbers.Contains(number)) {
            listNumbers.Add(numbers);
        }
    } while (listNumbers.Count < 6)

Answer 6

Instead of using a List, you should use an HashSet. The HashSet<> prohibites multiple identical values. And the Add method returns a bool that indicates if the element was added to the list, Please find the example code below.

public static IEnumerable<int> GetRandomNumbers(int count)
{
HashSet<int> randomNumbers = new HashSet<int>();

for (int i = 0; i < count; i++) 
    while (!randomNumbers.Add(random.Next()));

return randomNumbers;
}

Answer 7

The best approach (CPU time-wise) for such tasks is creating an array of all possible numbers and taking 6 items from it while removing the item you just took from the array. Example:

const int min = 1, max = 49;
List<int> listNumbers = new List<int>();
int[] numbers = new int[max - min + 1];
int i, len = max - min + 1, number;

for (i = min; i < max; i++) numbers[i - min] = i;
for (i = 0; i < 6; i++) {
    number = rand.Next(0, len - 1);
    listNumbers.Add(numbers[number]);
    if (number != (len - 1)) numbers[number] = numbers[len - 1];
    len--;
}

Answer 8

If you are not worried about the min, max, and range then you can use this.

 var nexnumber = Guid.NewGuid().GetHashCode();
        if (nexnumber < 0)
        {
            nexnumber *= -1;
        }

Answer 9

What you do is to generate a random number each time in the loop. There is a chance of course that the next random number may be the same as the previous one. Just add one check that the current random number is not present in the sequence. You can use a while loop like: while (currentRandom not in listNumbers): generateNewRandomNumber

Answer 10

Paste the below in the class as a new method

    public int randomNumber()
    {
        var random = new Random();
        int randomNumber = random.Next(10000, 99999);
        return randomNumber;
    }

And use the below anywhere in the tests wherever required

var RandNum = randomNumber();

driver.FindElement(By.CssSelector("[class='test']")).SendKeys(**RandNum**);

Answer 11

   public partial class Form1 : Form
{
    public Form1()
    {
        InitializeComponent();
    }

    private void button1_Click(object sender, EventArgs e)
    {
        int[] que = new int[6];
        int x, y, z;
        Random ran = new Random();            
        for ( x = 0; x < 6; x++)
        {
            que[x] = ran.Next(1,49);
            for (y = x; y >= 0; y--)
            {
                if (x == y)
                {
                    continue;
                }
                if (que[x] == que[y])
                {
                    que[x] = ran.Next(1,49);
                    y = x;
                }


            }
        }
        listBox1.Items.Clear();
        for (z = 0; z < 6; z++)
        {
            listBox1.Items.Add(que[z].ToString());
        }
    }
}