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How can I make so that my Values dont repeat with Random.Range()? [duplicate]

Basically I'm creating a program to randomly generate 6 unique lottery numbers so there is no duplicates in the same line, here is the code I have so far...
//Generate 6 random numbers using the randomiser object
int randomNumber1 = random.Next(1, 49);
int randomNumber2 = random.Next(1, 49);
int randomNumber3 = random.Next(1, 49);
int randomNumber4 = random.Next(1, 49);
int randomNumber5 = random.Next(1, 49);
int randomNumber6 = random.Next(1, 49);
textBox1.Text = randomNumber1.ToString();
textBox2.Text = randomNumber2.ToString();
textBox3.Text = randomNumber3.ToString();
textBox4.Text = randomNumber4.ToString();
textBox5.Text = randomNumber5.ToString();
textBox6.Text = randomNumber6.ToString();
}
I'm getting random numbers but sometimes there is the same number on the same line, how do I make each number unique????
Thanks in advance

You need to store them in a collection and each time you pick a new number you need to make sure it's not present already, otherwise you need to generate a new number until you find a unique number.
Instead of this, I would generate a sequence between 1 and 49, shuffle them and pick 6 number out of the sequence, for example:
var rnd = new Random();
var randomNumbers = Enumerable.Range(1,49).OrderBy(x => rnd.Next()).Take(6).ToList();

You can't. You've only specified that each number be a random number from 1 to 49, not that it shouldn't match any duplicates.
Since you've got a relatively small set of numbers, your best bet is probably to draw the random numbers, put them into a HashSet, then if you need more, pull more. Something like this:
HashSet<int> numbers = new HashSet<int>();
while (numbers.Count < 6) {
numbers.Add(random.Next(1, 49));
}
Here you're taking advantage of the HashSet's elimination of duplicates. This won't work with a List or other collection.

Returning repeat values is a necessity in order for a generator to satisfy a necessary statistical property of randomness: the probability of drawing a number is not dependent on the previous numbers drawn.
You could shuffle the integers in the range 1 to 49 and return the first 6 elements. See http://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle for more details on such a shuffler.
However, I think you get a slight statistical bias by doing this.
The best way is probably to use random.Next(1, 49); and reject any repeat. That will be free from statistical bias and the fact that you're only wanting 6 from 49 possibilities, the number of collisions will not slow the algorithm appreciably.

Using this extension method for reservoir sampling:
public static IList<T> TakeRandom<T>(
this IEnumerable<T> source, int count, Random random)
{
var list = new List<T>(count);
int n = 1;
foreach (var item in source)
{
if (list.Count < count)
{
list.Add(item);
}
else
{
int j = random.Next(n);
if (j < count)
{
list[j] = item;
}
}
n++;
}
return list;
}
You can sample your collection like this:
var random = new Random();
var numbers = Enumerable.Range(1, 49).TakeRandom(6, random);
numbers.Shuffle(random);
Note the returned numbers will be uniformly sampled out of all (49 choose 6) possibilities for a set of 6 numbers out of {1, 2, ..., 49}, but they will neither remain in order nor be uniformly shuffled. If you want to have the order randomized as well, you can easily do a standard Fisher-Yates shuffle afterwards.
public static void Shuffle<T>(this IList<T> list, Random random)
{
for (int i = 0; i < list.Count; i++)
{
int j = random.Next(i, list.Count);
T temp = list[j];
list[j] = list[i];
list[i] = temp;
}
}
Note a more heavily optimized version of Fisher-Yates shuffle can be found in this answer: Randomize a List<T>

List<int> aux = new List<int>();
while(aux.Count < 6)
{
int rnd = random.Next(1,49);
if(!aux.Contains(rnd))aux.add(rnd);
}
if you put all Texbox in the same panel you can do that
int j = 0;
foreach(Control x in MyPanel.Controls)
{
if(x is TexBox)
{
x.Text = aux[j].toString();
j++;
}
}

It's my solution: generate array of number
/// <summary>
/// auto generate a array with number element and max value is max
/// </summary>
/// <param name="number">number element of array</param>
/// <param name="max">max value of array</param>
/// <returns>array of number</returns>
public static int[] createRandomArray(int number, int max)
{
List<int> ValueNumber = new List<int>();
for (int i = 0; i < max; i++)
ValueNumber.Add(i);
int[] arr = new int[number];
int count = 0;
while (count < number)
{
Random rd = new Random();
int index = rd.Next(0,ValueNumber.Count -1);
int auto = ValueNumber[index];
arr[count] = auto;
ValueNumber.RemoveAt(index);
count += 1;
}
return arr;
}

It's too late but I use a Method named M_Randomizer created by me. It may look as too much work, but it's technique is different from traditional which is based on generating a random number and checking the previously generated list for uniqueness. This code while generating a new random number, never looks for the previously generated. And if we talk about touching all combinations, I have tested this method till 9 factorial, maybe little bias for some but it touches all.
using System;
class Randomizer
{
public int[] M_Randomizer(int x)
{
bool b = false;
if (x < -1)
{
b = true;
x = -1 * x;
}
if(x == -1)
x = 0;
if (x < 2)
return new int[x];
int[] site;
int k = new Random(Guid.NewGuid().GetHashCode()).Next() % 2;
if (x == 2)
{
site = new int[2];
site[0] = k;
site[1] = 1 - site[0];
return site;
}
else if (x == 3)
{
site = new int[3];
site[0] = new Random(Guid.NewGuid().GetHashCode()).Next(0, 3);
site[1] = (site[0] + k + 1) % 3;
site[2] = 3 - (site[0] + site[1]);
return site;
}
site = new int[x];
int a = 0, m = 0, n = 0, tmp = 0;
int[] p = M_Randomizer(3);
int[] q;
if (x % 3 == 0)
q = M_Randomizer(x / 3);
else
q = M_Randomizer((x / 3) + 1);
if (k == 0)
{
for (m = 0; m < q.Length; m++)
{
for (n = 0; n < p.Length && a < x; n++)
{
tmp = (q[m] * 3) + p[n];
if (tmp < x)
{
site[a] = tmp;
a++;
}
}
}
}
else
{
while (n < p.Length)
{
while (a < x)
{
tmp = (q[m] * 3) + p[n];
if (tmp < x)
{
site[a] = tmp;
a++;
}
m = m + k;
if (m >= q.Length)
break;
}
m = m % q.Length;
n++;
}
}
a = (new Random(Guid.NewGuid().GetHashCode()).Next() % 2) + 1;
k = new Random(Guid.NewGuid().GetHashCode()).Next() % 10;
if (k > 5)
for (int i = a; i < k; i++)
while (a < site.Length)
{
if (k % (a + 1) == 0)
{
tmp = site[a - 1];
site[a - 1] = site[a];
site[a] = tmp;
}
a = a + 2;
}
k = new Random(Guid.NewGuid().GetHashCode()).Next() % 10;
if (k > 5)
{
n = x / 2;
k = 0;
if (x % 2 != 0)
k = (new Random(Guid.NewGuid().GetHashCode()).Next() % 2);
p = new int[n + k];
m = (x - n) - k;
for (a = 0; m < x; a++, m++)
p[a] = site[m];
m = n + k;
for (a = (x - m) - 1; a >= 0; a--, m++)
site[m] = site[a];
for (a = 0; a < p.Length; a++)
site[a] = p[a];
}
int[] site2;
int[] site3 = new int[x];
if (b)
return site;
else
site2 = M_Randomizer(-1 * x);
for (a = 0; a < site.Length; a++)
site3[site2[a]] = site[a];
return site3;
}
public int[] M_Randomizer(int x, int start)
{
int[] dm = M_Randomizer(x);
for(int a = 0; a < x; a++)
dm[a] = dm[a] + start;
return dm;
}
}

Look at using an array to hold your 6 numbers.
Each time you generate one, loop through the array to make sure it is not already there. If it is, then generate another & loop again until you have a non-match.

It's so easy with array and OOP (Object Oriented Programming). Before you start you have to add Linq (using System.Linq) library to your project.
Random random = new Random();
int[] array = new int[6];
int number;
for (int i = 0; i < 6; i++)
{
number = random.Next(1, 50);
if (!array.Contains(number)) //If it's not contains, add number to array;
array[i] = number;
else //If it contains, restart random process
i--;
}
for (int i = 1; i < 7; i++)
{
foreach (Control c in this.Controls) //Add random numbers to all Textboxes
{
if (c is TextBox && c.Name.EndsWith(i.ToString()))
{
c.Text = array[i - 1].ToString();
}
}
}

A functional approach could be to generate an infinite sequence of random numbers, filter out non-unique numbers and take the number of unique numbers you need.
For example:
private IEnumerable<int> RandomDigitStream(int seed)
{
Random random = new Random(seed);
while (true)
{
yield return random.Next(DIGIT_MIN, DIGIT_MAX);
}
}
private List<int> GenerateUniqueRandomNumbers(int seed, int count)
{
// Assert that DIGIT_MAX - DIGIT_MIN > count to ensure
// algorithm can finish
return RandomDigitStream(seed)
.Distinct()
.Take(count)
.ToList();
}
The efficiency of this algorithm is mainly dependent on how Distinct is implemented by the .NET team. Its memory usage would grow with the number of digits you require and the range of digits produced by the random function. It also has an unpredictable running time as it depends on the probability distribution of the random function. In fact it is possible for this algorithm to get stuck in an infinite loop if the range of digits produced by the random algorithm is less than the number of digits you require.
Looking at it practically however, it should be fine for a small amount of digits but if you are looking at a large number (100 +) you might want to look at other methods.
It would be more efficient to craft a random algorithm that only produces unique numbers in the first place if that is even possible without using a lookup table.

Here is a small program using recursion to generate number lines, and also uses recursion to randomize and get unique numbers.
using System;
using System.Linq;
using System.Collections.Generic;
public class Program
{
public static Random random;
public static List<int> lottoNumbers = Enumerable.Range(1, 49).ToList();
public static void Main()
{
random = new Random((int)DateTime.Now.Ticks);
var LinesToGenerate = 10;
GenerateNumbers(LinesToGenerate);
}
public static void GenerateNumbers(int LineCount)
{
int[] SelectedNumbers = new int[6];
for (var i = 0; i < 6; i++)
{
var number = GetRandomNumber(lottoNumbers.ToArray());
while (SelectedNumbers.Contains(number))
number = GetRandomNumber(lottoNumbers.ToArray());
SelectedNumbers[i] = number;
}
var numbersOrdered = SelectedNumbers.OrderBy(n => n).Select(n => n.ToString().PadLeft(2, '0'));
Console.WriteLine(string.Join(" ", numbersOrdered));
if (LineCount > 1)
GenerateNumbers(--LineCount);
}
//Recursively and randomly removes numbers from the array until only one is left, and returns it
public static int GetRandomNumber(int[] arr)
{
if (arr.Length > 1)
{
//Remove random number from array
var r = random.Next(0, arr.Length);
var list = arr.ToList();
list.RemoveAt(r);
return GetRandomNumber(list.ToArray());
}
return arr[0];
}
}

Yes. Use array.
Loop how many times you want:
Generate a random number,
Loop through array and compare all with the generated number.
If there's a match then loop again till there's no match.
Then store it.
Done:)

Lottery program that generates 6 random numbers between 1 and 59 C#

I have created a simple program which randomly generates 6 winning numbers. While the program works, I would also like for it to ensure that the same number isn't outputted twice as well as sorting them into numerical order when outputted. How would I go about doing such a thing while sticking to similar techniques already used? My code is down below. Any help is very much appreciated.
int temp;
int[] lotto = new int[6];
Random rand = new Random();
for (int i = 0; i < 6; i++)
{
temp = rand.Next(1, 59);
lotto[i] = temp;
}
Console.Write($"The lotterry winning numbers are: ");
for (int i = 0; i < 6; i++)
{
Console.Write(lotto[i] + " ");
}
Console.ReadKey();

Based on a Fisher-Yates shuffle, but saves some work because we know we don't need all the values (if we only need 6 values out of 10 million potentials, we only need to take the first six iterations of the fisher-yates algorithm).
public IEnumerable<int> DrawNumbers(int count, int MaxNumbers)
{
var r = new Random(); //ideally, make this a static member somewhere
var possibles = Enumerable.Range(1, MaxNumbers).ToList();
for (int i = 0; i < count; i++)
{
var index = r.Next(i, MaxNumbers);
yield return possibles[index];
possibles[index] = possibles[i];
}
}
var lottoNumbers = DrawNumbers(6, 59);
Console.Write("The lotterry winning numbers are: ");
Console.WriteLine(string.Join(" ", lottoNumbers.OrderBy(n => n)));
See it work here:
https://dotnetfiddle.net/NXYkpU

You can use Linq to create sequence [1..59] and order it by random to shuffle it.
Random rand = new Random();
var winners = Enumerable.Range(1, 59)
.OrderBy(x => rand.Next())
.Take(6)
.OrderBy(x => x)
.ToList();
Console.WriteLine(String.Join(" ", winners));

int temp;
int[] lotto = new int[6];
Random rand = new Random();
int i = 0;
while(i < 6)
{
temp = rand.Next(1, 59);
//check if lotto contains just generated number, if so skip that number
bool alreadyExist = false;
foreach (int item in lotto)
{
if (item == temp)
{
alreadyExist = true;
break;
}
}
if (alreadyExist)
continue;
lotto[i] = temp;
i++;
}
Console.Write($"The lotterry winning numbers are: ");
// Sort array in ascending order.
Array.Sort(lotto);
for (int j = 0; j < 6; j++)
{
Console.Write(lotto[j] + " ");
}
Console.ReadKey();

I would probably do it Dmitri's way because it is quick and obvious and performance isn't that important with an array this size.
But just for fun, this is slightly more efficient.
IEnumerable<int> GetNumbers(int min, int max, int count)
{
var random = new Random();
var size = max - min + 1;
var numbers = Enumerable.Range(min, size).ToArray();
while (count > 0)
{
size--;
var index = random.Next(0, size);
yield return numbers[index];
numbers[index] = numbers[size];
count--;
}
}
This solution creates an array containing all possible values and selects them randomly. Each time a selection is made, the array is "shrunk" by moving the last element to replace the element that was chosen, preventing duplicates.
To use:
var numbers = GetNumbers(1, 59, 6).ToList();
foreach (var number in numbers.OrderBy(x => x))
{
Console.WriteLine(number);
}

How to get number of digits divisible by 5

I want to find the total number of digits divisible by 5 between 1 - 100, in C# windows form, how to proceed from here?
int sum;
private void button2_Click(object sender, EventArgs e)
{
int[] intarray = new int[100];
for (int i = 0; i < 99; i++)
{
intarray[i] = i + 1;
}
foreach (int a in intarray)
{
if (a / 5 == 0)
{
}
}
}

Note than a / 5 == 0 is wrong. For example 10 is divisible by 2, the result is 10/5 = 2, not equal to 0.
if (a % 5 == 0)
{
//then a is divisible by 5. print or store it
}
The modulus operator, also known as Remainder, returns the remainder of the integer division.
Therefore, the full answer:
int nInRange = 0;
foreach (int a in intarray)
if (a % 5 == 0)
nInRange++;

Maybe this is what you want.
public static IEnumerable<int> GetIntsDivisible(int start, int finish, int divisor)
{
for (var i = start; i <= finish; i++)
if (i % divisor == 0)
yield return i;
}
public static void Main()
{
Console.WriteLine(string.Join(", ", GetIntsDivisible(1, 100, 5)));
}
or if you don't want to yield
public static List<int> GetIntsDivisible(int start, int finish, int divisor)
{
var result = new List<int>();
for (var i = start; i <= finish; i++)
if (i % divisor == 0)
result.Add(i);
return result;
}
Output
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100
Full demo here

Initialize a variable as 0
for e.g. Count = 0;
then add statement Count++ in your if block as follows:
if (a % 5 == 0)
{
Count++;
}

Since question is a bit vague in exact requirement, I will write the basic logic that can find the numbers exactly divisible by other number.
There is something called modulus (%) operator. It gives you the remainder of division.e.g. 11%5 will be 1, 13%5 will be 3, whereas 15%5 will be 0
so logic goes like,
for(int i=0;i<=100;i++)
{
if((i%5)==0)
{
\\this is ur number
}
}

Every one has address your question but no one talk about this weird attempt.
I really think you should take 5 minute and read your code because you are just running everywhere.
You should take the pen and paper before going head first into coding.
Here is a simple reading of your code so you understand what you were doing.
Line 1:
int[] intarray = new int[100];
So you start win an array, I guess it's for the result, right ? You will not be storing the number from 1 to 100 for no reason?
Line 2:
for (int i = 0; i < 99; i++)
Now we count from 0 to 98, I though it was form 1 to 100 .. Yes 98 as you are using < instead of <=
Line 3:
intarray[i] = i + 1;
Why ? 3rd line and you are already lost! You are filling the array with number you just iterate. It's like filling a bottle with water, then use it to fill an other bottle because you needed water.
If you iterate from 1 to 100 you could have check if it was divisible.
for (int i = 1; i <= 100; i++)
Line 4:
foreach (int a in intarray)
Again ? We are back counting from 1 to 100..
Line 5:
if (a / 5 == 0)
If this is suppose to tell you if it divisible thats wrong. The correct math operator is the Modulo. The division symbole won't give you the result you expect.
{1,2,3,4} will give you True. Anything else will be false.

int count = 0;
for (int i = 1; i <= 100; i++)
{
if ((i % 5) == 0)
{
count++;
textBox1.Text = count.ToString();
}
}
Alternatively,
int num = (100 + 5) / 5 - (1 + 5 - 1) / 5;
//Show result here

Knapsack C# implementation task

I'm trying to write knapsack c# algorithm with given conditions but there's always two problems i encountering. I'm getting "Index was outside the bounds of the array" error or my result is just 0.
I found couple code examples of Knapsack implementation and just can't figure out what I'm doing wrong.
Code examples:
https://www.programmingalgorithms.com/algorithm/knapsack-problem
http://www.csharpstar.com/csharp-knapsack-problem/
My code:
static int Knapsack(int n, int w, int[] s, int[] v)
{
int[,] G = new int[n+1,w+1];
for (int k = 0; k <= n; k++)
{
for (int r = 0; r < w; r++)
{
if (r == 0 || k == 0)
G[k, r] = 0;
else if (s[k] <= r)
G[k, r] = Math.Max(G[k- 1, r], v[k] + G[k - 1, r - s[k]]);
else
G[k, r] = G[k - 1, r];
}
}
return G[n, w];
}
static void Main(string[] args)
{
int[] s = { 60, 100, 120};
int[] v = { 10, 20, 30 };
int w = 50;
int n = s.Length;
Console.WriteLine(Knapsack(n, w, s, v));
}
In this case my result is 0.

The issue with your code is that s is the weights and v is the values and your weights of 60, 100, and 120 will obvious not fit into a capacity of 50, that's why you get a result of 0. The example you pull those values from set's the 60, 100, and 120 as the values and 10, 20, and 30 as the weights which is why it gets a result of 220.
I think this works better if you create a class to handle the related weight and value for the items.
public class Item
{
public int Weight { get; set; }
public int Value { get; set; }
}
Then the method only needs an array of items and the desired capacity. Also using meaningful names can make understanding what's going on easier than a bunch of single letter names.
public static int KnapSack(Item[] items, int capacity)
{
int[,] matrix = new int[items.Length + 1, capacity + 1];
for (int itemIndex = 0; itemIndex <= items.Length; itemIndex++)
{
// This adjusts the itemIndex to be 1 based instead of 0 based
// and in this case 0 is the initial state before an item is
// considered for the knapsack.
var currentItem = itemIndex == 0 ? null : items[itemIndex - 1];
for (int currentCapacity = 0; currentCapacity <= capacity; currentCapacity++)
{
// Set the first row and column of the matrix to all zeros
// This is the state before any items are added and when the
// potential capacity is zero the value would also be zero.
if (currentItem == null || currentCapacity == 0)
{
matrix[itemIndex, currentCapacity] = 0;
}
// If the current items weight is less than the current capacity
// then we should see if adding this item to the knapsack
// results in a greater value than what was determined for
// the previous item at this potential capacity.
else if (currentItem.Weight <= currentCapacity)
{
matrix[itemIndex, currentCapacity] = Math.Max(
currentItem.Value
+ matrix[itemIndex - 1, currentCapacity - currentItem.Weight],
matrix[itemIndex - 1, currentCapacity]);
}
// current item will not fit so just set the value to the
// what it was after handling the previous item.
else
{
matrix[itemIndex, currentCapacity] =
matrix[itemIndex - 1, currentCapacity];
}
}
}
// The solution should be the value determined after considering all
// items at all the intermediate potential capacities.
return matrix[items.Length, capacity];
}
Then running this code
var items = new[]
{
new Item {Value = 60, Weight = 10},
new Item {Value = 100, Weight = 20},
new Item {Value = 120, Weight = 30},
};
Console.WriteLine(KnapSack(items, 50));
results in 220.
Here's a solution that uses recursion.
public static int KnapSackRecursive(Item[] items, int capacity)
{
// If there are no items or capacity is 0 then return 0
if (items.Length == 0 || capacity == 0) return 0;
// If there is one item and it fits then return it's value
// otherwise return 0
if (items.Length == 1)
{
return items[0].Weight < capacity ? items[0].Value : 0;
}
// keep track of the best value seen.
int best = 0;
for (int i = 0; i < items.Length; i++)
{
// This is an array of the other items.
var otherItems = items.Take(i).Concat(items.Skip(i + 1)).ToArray();
// Calculate the best value without using the current item.
int without = KnapSackRecursive(otherItems, capacity);
int with = 0;
// If the current item fits then calculate the best value for
// a capacity less it's weight and with it removed from contention
// and add the current items value to that.
if (items[i].Weight <= capacity)
{
with = KnapSackRecursive(otherItems, capacity - items[i].Weight)
+ items[i].Value;
}
// The current best is the max of the with or without.
int currentBest = Math.Max(without, with);
// determine if the current best is the overall best.
if (currentBest > best)
best = currentBest;
}
return best;
}

Generating random numbers without repeating.C# [duplicate]

This question already has answers here:
Random number generator with no duplicates
(12 answers)
Closed 2 years ago.
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

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);
}

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

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();
}

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

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;
}

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)

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--;
}

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;
}

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

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**);

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());
}
}
}