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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);
}

Find the first uneven number in random array in c#

i must create array and fill it with random numbers , between 1-100. From there, i must find the 1st uneven number and print it.
Also have to print 0 if no uneven numbers are in the array.
Heres what i did:
int[] tab = new int[10];
int[] uneven = new int[tab.Length];
int i;
for (i = 0; i < tab.Length; i++)
tab[i] = new Random().Next(100) + 1;
do
{
uneven[i] = tab[i];
} while (tab[i] % 2 == 1);
Console.WriteLine(uneven[0]);
So my reasoning is that i add uneven numbers in uneven[i] as long as tab[i] is uneven,then print the first element of the array.
However, i get "out of bonds exception".
Thank you in advance for any help.

Your for loop set i to 10 which is outside the bounds of the array. You need to re-set it to 0 before the do loop. Also, you need to increment i.
i = 0;
do
{
uneven[i] = tab[i];
i++;
} while (tab[i] % 2 != 0);

By the time your do loop starts your "i" variable is stuck on 10. Arrays start at 0 so it only goes up to 9 which is why you're seeing the out of bounds exception. Here's a small example of what you're trying to achieve:
int[] tab = new int[10];
var rnd = new Random();
// Create 10 random numbers
for (int i = 0; i < tab.Length; i++)
{
tab[i] = rnd.Next(100) + 1;
}
// Find the first uneven number
bool found = false;
for (int i = 0; i < tab.Length; i++)
{
if (tab[i] % 2 != 0)
{
Console.WriteLine(tab[i]);
found = true;
break;
}
}
// Didn't generate an uneven number?
if (!found)
{
Console.WriteLine("Nothing found");
}

This creates an array[] with random numbers assigned to each element.
The second for loop checks if the number is even/odd then breaks the loop if it is odd.
static void Main(string[] args)
{
int[] numList = new int[100];
var rand = new Random();
Console.WriteLine(rand.Next(101));
for (int i = 0; i <= 99; i++)
{
numList[i] = rand.Next(101);
Console.WriteLine($"Element; {i}: {numList[i]}");
}
for (int i = 0; i <= 99; i++)
{
int num = numList[i];
if (num / 2 != 0)
{
Console.WriteLine($"The first uneven number is: {num} in element: {i}");
break;
}
if(i == numList.Count())
{
Console.WriteLine("All numbers are even");
break;
}
}
Console.ReadLine();
}

How can I fill array with unique random numbers with for-loop&if-statement?

I'm trying to fill one dimensional array with random BUT unique numbers (No single number should be same). As I guess I have a logical error in second for loop, but can't get it right.
P.S I'm not looking for a more "complex" solution - all I know at is this time is while,for,if.
P.P.S I know that it's a really beginner's problem and feel sorry for this kind of question.
int[] x = new int[10];
for (int i = 0; i < x.Length; i++)
{
x[i] = r.Next(9);
for (int j = 0; j <i; j++)
{
if (x[i] == x[j]) break;
}
}
for (int i = 0; i < x.Length; i++)
{
Console.WriteLine(x[i);
}

Here is a solution with your code.
int[] x = new int[10];
for (int i = 0; i < x.Length;)
{
bool stop = false;
x[i] = r.Next(9);
for (int j = 0; j <i; j++)
{
if (x[i] == x[j]) {
stop = true;
break;
}
}
if (!stop)
i++;
}
for (int i = 0; i < x.Length; i++)
{
Console.WriteLine(x[i]);
}

A simple trace of the posted code reveals some of the issues. To be specific, on the line…
if (x[i] == x[j]) break;
if the random number is “already” in the array, then simply breaking out of the j loop is going to SKIP the current i value into the x array. This means that whenever a duplicate is found, x[i] is going to be 0 (zero) the default value, then skipped.
The outer i loop is obviously looping through the x int array, this is pretty clear and looks ok. However, the second inner loop can’t really be a for loop… and here’s why… basically you need to find a random int, then loop through the existing ints to see if it already exists. Given this, in theory you could grab the same random number “many” times over before getting a unique one. Therefore, in this scenario… you really have NO idea how many times you will loop around before you find this unique number.
With that said, it may help to “break” your problem down. I am guessing a “method” that returns a “unique” int compared to the existing ints in the x array, may come in handy. Create an endless while loop, inside this loop, we would grab a random number, then loop through the “existing” ints. If the random number is not a duplicate, then we can simply return this value. This is all this method does and it may look something like below.
private static int GetNextInt(Random r, int[] x, int numberOfRandsFound) {
int currentRand;
bool itemAlreadyExist = false;
while (true) {
currentRand = r.Next(RandomNumberSize);
itemAlreadyExist = false;
for (int i = 0; i < numberOfRandsFound; i++) {
if (x[i] == currentRand) {
itemAlreadyExist = true;
break;
}
}
if (!itemAlreadyExist) {
return currentRand;
}
}
}
NOTE: Here would be a good time to describe a possible endless loop in this code…
Currently, the random numbers and the size of the array are the same, however, if the array size is “larger” than the random number spread, then the code above will NEVER exit. Example, if the current x array is set to size 11 and the random numbers is left at 10, then you will never be able to set the x[10] item since ALL possible random numbers are already used. I hope that makes sense.
Once we have the method above… the rest should be fairly straight forward.
static int DataSize;
static int RandomNumberSize;
static void Main(string[] args) {
Random random = new Random();
DataSize = 10;
RandomNumberSize = 10;
int numberOfRandsFound = 0;
int[] ArrayOfInts = new int[DataSize];
int currentRand;
for (int i = 0; i < ArrayOfInts.Length; i++) {
currentRand = GetNextInt(random, ArrayOfInts, numberOfRandsFound);
ArrayOfInts[i] = currentRand;
numberOfRandsFound++;
}
for (int i = 0; i < ArrayOfInts.Length; i++) {
Console.WriteLine(ArrayOfInts[i]);
}
Console.ReadKey();
}
Lastly as other have mentioned, this is much easier with a List<int>…
static int DataSize;
static int RandomNumberSize;
static void Main(string[] args) {
Random random = new Random();
DataSize = 10;
RandomNumberSize = 10;
List<int> listOfInts = new List<int>();
bool stillWorking = true;
int currentRand;
while (stillWorking) {
currentRand = random.Next(RandomNumberSize);
if (!listOfInts.Contains(currentRand)) {
listOfInts.Add(currentRand);
if (listOfInts.Count == DataSize)
stillWorking = false;
}
}
for (int i = 0; i < listOfInts.Count; i++) {
Console.WriteLine(i + " - " + listOfInts[i]);
}
Console.ReadKey();
}
Hope this helps ;-)

The typical solution is to generate the entire potential set in sequence (in this case an array with values from 0 to 9). Then shuffle the sequence.
private static Random rng = new Random();
public static void Shuffle(int[] items)
{
int n = list.Length;
while (n > 1) {
n--;
int k = rng.Next(n + 1);
int temp = items[k];
items[k] = items[n];
items[n] = temp;
}
}
static void Main(string[] args)
{
int[] x = new int[10];
for(int i = 0; i<x.Length; i++)
{
x[i] = i;
}
Shuffle(x);
for(int i = 0; i < x.Length; i++)
{
Console.WritLine(x[i]);
}
}
//alternate version of Main()
static void Main(string[] args)
{
var x = Enumerable.Range(0,10).ToArray();
Shuffle(x);
Console.WriteLine(String.Join("\n", x));
}

You can simply do this:
private void AddUniqueNumber()
{
Random r = new Random();
List<int> uniqueList = new List<int>();
int num = 0, count = 10;
for (int i = 0; i < count; i++)
{
num = r.Next(count);
if (!uniqueList.Contains(num))
uniqueList.Add(num);
}
}
Or:
int[] x = new int[10];
Random r1 = new Random();
int num = 0;
for (int i = 0; i < x.Length; i++)
{
num = r1.Next(10);
x[num] = num;
}

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());
}
}
}

generating random number between 1, 16

Hi everyone I have two dimensional array 4*4. I have to fill it with numbers from 1 to 16 inclusive and randomly. Each number should be used only once and all the numbers between 1-16 should be used. I have written the following code, but for some reason that I don't know some items in the array are not filled !!
can anyone tell me what I'm doing wrong
public void generateRandState()
{
bool flag = false;
Random rnd = new Random();
int temp ;
for (int i = 0; i < 4; i++)
{
for (int j = 0; j < 4; j++)
{
temp = rnd.Next(1, 17);
Console.WriteLine(temp);
if (taken[temp] == false)
{
state[i, j] = temp;
taken[temp] = true;
}
else // for repeated rands
{
temp = rnd.Next(1, 17);
Console.WriteLine(temp);
while (!flag)
{
if (taken[temp] == false)
{
flag = true;
break;
}
else
temp = rnd.Next(1, 17);
Console.WriteLine(temp);
}
if (taken[temp] == false)
{
state[i, j] = temp;
taken[temp] = true;
}
}
}
}
}

Each number should be used only once and all the numbers between 1-16 should be used.
This is called a random shuffle. You can use Fisher–Yates shuffle to build it.
This post provides the code for it.
static Random _random = new Random();
public static void Shuffle<T>(T[] array) {
var random = _random;
for (int i = array.Length; i > 1; i--) {
int j = random.Next(i); // 0 <= j <= i-1
T tmp = array[j];
array[j] = array[i - 1];
array[i - 1] = tmp;
}
}

Replace while (!flag) with while (true). The break will already get you out of the while statement. The flag variable is completely unnecessary.
Your algorithm is pretty inefficient. As dasblinkenlight points out, there's better ways to accomplish what you are doing.
You should also know that taken would need to be cleared if you ever call generateRandState more than once.