Is my lottery program wrong ? or am I so unlucky? - c#

I made a lottery program : http://yadi.sk/d/bBKefn9g4OC7s
Here is the whole source code : http://yadi.sk/d/rnQMfrry4O8cu
Random rnd = new Random();
int[] loto;
loto = new int[7];
for (int f = 1; f <= 6; f++) {
loto[f] = rnd.Next(1, 50); // Generating random number between 1-49
for (int h = 1; h < f; h++) {
if (loto[f] == loto[h]) { // Check with other numbers for the case of duplicate
loto[f] = rnd.Next(1, 50); // If there is a duplicate create that number again
}
}
}
This section I'm generating random 6 different numbers between 1-49
Also I'm wondering in this example, are nested loops increase the spontaneity ?
I'm getting 3-4 max, this program wrong or am I so unlucky ?
( note that : that's my first program )
For all guys trying to help me : I'm really beginner on programming(c# yesterday | c++ 3 weeks i guess), and if you guys clarify what you mean in codes it'll be great.
And please not give me extreme hard coding examples( I don't wanna quit c# )


Your method looks unsafe, as get value from Random again in the inner loop does not guarantee that it will return unduplicated value. For low value as 1-49, you can use simple random-picking algorithm like this
var numbers = new List<int>();
for (int i = 1; i <= 49; i++) {
numbers.Add(i);
}
Random r = new Random();
var loto = new int[6];
for (int f = 0; f < 6; f++) {
int idx = r.Next(0, numbers.Count);
loto[f] = numbers[idx];
numbers.RemoveAt(idx);
}
Note that this is far from optimal solution in terms of performance, but if you will run it only once in a few seconds or more so it should be fine.


I think it's correct except for the for loop declaration: remember that arrays in C# are zero-based. Thus the loop should look like this:
for (int f = 0; f < 7; f++)
or even better:
for (int f = 0; f < loto.Length; f++)
Update: I cannot comment the other answers (too less reputation), thus I have to post it here:
#Dan: only one loop is not correct as it is not allowed to have the same number twice in Loto. In his inner loop, 1342 checks if the created random number already exists, so it is not correct to leave it out.
#James: As 1342 just started programming, it is not necessary to use a static field in my opinion. I guess that he or she has his whole code in the Main method so there is no benefit using a static variable.


There are a few issues here - you've got one too many loops for a start, and no comments.
See this (over-commented) example below:
// This is static so we don't recreate it every time.
private static Random _rnd;
static void Main(string[] args)
{
_rnd = new Random();
// You can declare and initialise a variable in one statement.
// In this case you want the array size to be 6, not 7!
Int32[] lotoNumbers = new Int32[6];
// Generate 10 sets of 6 random numbers
for (int i = 0; i < 10; i++)
{
// Use a meaningful name for your iteration variable
// In this case I used 'idx' as in 'index'
// Arrays in c# are 0-based, so your lotto array has
// 6 elements - [0] to [5]
for (Int32 idx = 0; idx < 6; idx++)
{
// Keep looping until we have a unique number
int proposedNumber;
do
{
proposedNumber = _rnd.Next(1, 50);
} while (lotoNumbers.Contains(proposedNumber));
// Assign the unique proposed number to your array
lotoNumbers[idx] = proposedNumber;
}
}
}
You should end up with a 6 element long array with 6 random numbers between 1 and 50 in it.
Hope that helps!
Edit:
It's also well worth taking note of James' answer - if you're doing the above in a loop, you'll get the same values every time from Random, due to how the seed is used. Using a static version of Random will give much better results.


You don't want to keep re-creating a new instance of Random each time, that's the likely cause of why you keep getting similar values each time. The better approach is to create a static instance of Random and use that across your entire app - this should give you more realistic results e.g.
using System.Collections.Generic;
using System.Linq;
...
static readonly Random rand = new Random();
...
List<int> lottoNumbers = new List<int>(6);
int drawnNumber = -1;
for (int i = 0; i < lottoNumbers.Count; i++) {
do
{
drawnNumber = rand.Next(1, 50); // generate random number
}
while (lottoNumbers.Contains(drawnNumber)) // keep generating random numbers until we get one which hasn't already been drawn
lottoNumbers[i] = drawnNumber; // set the lotto number
}
// print results
foreach (var n in lottoNumbers)
Console.WriteLine(n);


For easily testing it, I have left the console logs and static void main for you.
You do not need two iterations for this. Also - arrays are 0 based, so either f has to be equal to 0, or less than 7. I went with equal 0 below.
I have created a recursive method which creates a new value and checks if the array contains the value. If it does not contain it, it adds it. But if it does contain it, the method calls itself to find a new value. It will continue to do this until a new value is found.
Recursive methods are methods which call themselves. Don't try and fill an array with an index bigger than 50 with this, as you will get an endless loop.
private static readonly Random Rnd = new Random();
static void Main(string[] args)
{
var loto = new int[7];
for (int f = 0; f <= 6; f++)
{
var randomValue = GetRandomNumberNotInArr(loto);
Console.WriteLine(randomValue);
loto[f] = randomValue;
}
Console.Read();
}
/// <summary>
/// Finds a new random value to insert into arr. If arr already contains this another
///random value will be found.
/// </summary>
/// <param name="arr">arr with already found values</param>
/// <returns></returns>
private static int GetRandomNumberNotInArr(int[] arr)
{
var next = Rnd.Next(1, 50);
return !arr.Contains(next) ? next : GetRandomNumberNotInArr(arr);
}


I can see that you are trying to simulate drawing 6 lottery numbers between 1 and 50.
Your code has some logic errors, but rather than fixing it I'm going to suggest doing it a different way.
There are several ways to approach this; a common one is this:
Create an empty collection of numbers.
while there aren't enough numbers in the collection
let randomNumber = new random number in the appropriate range
if (randomNumber isn't already in the collection)
add randomNumber to the collection
But there's another approach which scales nicely, so I'll demonstrate this one (someone else will probably already have written about the other approach):
Add to a collection all the numbers you want to choose from
Randomly rearrange (shuffle) the numbers in the collection
Draw the required number of items from the collection
This is pretty much what happens in a real-life lottery.
To shuffle a collection we can use the Fisher-Yates Shuffle. Here's an implementation:
/// <summary>Used to shuffle collections.</summary>
public class Shuffler
{
/// <summary>Shuffles the specified array.</summary>
/// <typeparam name="T">The type of the array elements.</typeparam>
/// <param name="array">The array to shuffle.</param>
public void Shuffle<T>(IList<T> array)
{
for (int n = array.Count; n > 1;)
{
int k = _rng.Next(n);
--n;
T temp = array[n];
array[n] = array[k];
array[k] = temp;
}
}
private readonly Random _rng = new Random();
}
Here's a full compilable example. I've avoided using Linq in this example because I don't want to confuse you!
using System;
using System.Collections.Generic;
namespace Demo
{
public static class Program
{
private static void Main()
{
int[] lotoDraw = createDraw();
Shuffler shuffler = new Shuffler();
shuffler.Shuffle(lotoDraw); // Now they are randomly ordered.
// We want 6 numbers, so we just draw the first 6:
int[] loto = draw(lotoDraw, 6);
// Print them out;
foreach (int ball in loto)
Console.WriteLine(ball);
}
private static int[] draw(int[] bag, int n) // Draws the first n items
{ // from the bag
int[] result = new int[n];
for (int i = 0; i < n; ++i)
result[i] = bag[i];
return result;
}
private static int[] createDraw() // Creates a collection of numbers
{ // from 1..50 to draw from.
int[] result = new int[50];
for (int i = 0; i < 50; ++i)
result[i] = i + 1;
return result;
}
}
public class Shuffler
{
public void Shuffle<T>(IList<T> list)
{
for (int n = list.Count; n > 1; )
{
int k = _rng.Next(n);
--n;
T temp = list[n];
list[n] = list[k];
list[k] = temp;
}
}
private readonly Random _rng = new Random();
}
}

Related

lottery vector with for loop without duplicates [duplicate]

I've searched for a while and been struggling to find this, I'm trying to generate several random, unique numbers is C#. I'm using System.Random, and I'm using a DateTime.Now.Ticks seed:
public Random a = new Random(DateTime.Now.Ticks.GetHashCode());
private void NewNumber()
{
MyNumber = a.Next(0, 10);
}
I'm calling NewNumber() regularly, but the problem is I often get repeated numbers. Some people suggested because I was declaring the random every time I did it, it would not produce a random number, so I put the declaration outside my function. Any suggestions or better ways than using System.Random ? Thank you

I'm calling NewNumber() regularly, but the problem is I often get
repeated numbers.
Random.Next doesn't guarantee the number to be unique. Also your range is from 0 to 10 and chances are you will get duplicate values. May be you can setup a list of int and insert random numbers in the list after checking if it doesn't contain the duplicate. Something like:
public Random a = new Random(); // replace from new Random(DateTime.Now.Ticks.GetHashCode());
// Since similar code is done in default constructor internally
public List<int> randomList = new List<int>();
int MyNumber = 0;
private void NewNumber()
{
MyNumber = a.Next(0, 10);
if (!randomList.Contains(MyNumber))
randomList.Add(MyNumber);
}

You might try shuffling an array of possible ints if your range is only 0 through 9. This adds the benefit of avoiding any conflicts in the number generation.
var nums = Enumerable.Range(0, 10).ToArray();
var rnd = new Random();
// Shuffle the array
for (int i = 0;i < nums.Length;++i)
{
int randomIndex = rnd.Next(nums.Length);
int temp = nums[randomIndex];
nums[randomIndex] = nums[i];
nums[i] = temp;
}
// Now your array is randomized and you can simply print them in order
for (int i = 0;i < nums.Length;++i)
Console.WriteLine(nums[i]);

NOTE, I dont recommend this :).
Here's a "oneliner" as well:
var result = Enumerable.Range(0,9).OrderBy(g => Guid.NewGuid()).ToArray();

I'm posting a correct implementation of a shuffle algorithm, since the other one posted here doesn't produce a uniform shuffle.
As the other answer states, for small numbers of values to be randomized, you can simply fill an array with those values, shuffle the array, and then use however many of the values that you want.
The following is an implementation of the Fisher-Yates Shuffle (aka the Knuth Shuffle). (Read the "implementation errors" section of that link (search for "always selecting j from the entire range of valid array indices on every iteration") to see some discussion about what is wrong with the other implementation posted here.)
using System;
using System.Collections.Generic;
namespace ConsoleApplication2
{
static class Program
{
static void Main(string[] args)
{
Shuffler shuffler = new Shuffler();
List<int> list = new List<int>{ 1, 2, 3, 4, 5, 6, 7, 8, 9 };
shuffler.Shuffle(list);
foreach (int value in list)
{
Console.WriteLine(value);
}
}
}
/// <summary>Used to shuffle collections.</summary>
public class Shuffler
{
public Shuffler()
{
_rng = new Random();
}
/// <summary>Shuffles the specified array.</summary>
/// <typeparam name="T">The type of the array elements.</typeparam>
/// <param name="array">The array to shuffle.</param>
public void Shuffle<T>(IList<T> array)
{
for (int n = array.Count; n > 1; )
{
int k = _rng.Next(n);
--n;
T temp = array[n];
array[n] = array[k];
array[k] = temp;
}
}
private System.Random _rng;
}
}

This is a unity only answer:
Check this ready-to-use method: Give in a range & count of number you want to get.
public static int[] getUniqueRandomArray(int min, int max, int count) {
int[] result = new int[count];
List<int> numbersInOrder = new List<int>();
for (var x = min; x < max; x++) {
numbersInOrder.Add(x);
}
for (var x = 0; x < count; x++) {
var randomIndex = UnityEngine.Random.Range(0, numbersInOrder.Count);
result[x] = numbersInOrder[randomIndex];
numbersInOrder.RemoveAt(randomIndex);
}
return result;
}

Same as #Habib's answer, but as a function:
List<int> randomList = new List<int>();
int UniqueRandomInt(int min, int max)
{
var rand = new Random();
int myNumber;
do
{
myNumber = rand.Next(min, max);
} while (randomList.Contains(myNumber));
return myNumber;
}
If randomList is a class property, UniqueRandomInt will return unique integers in the context of the same instance of that class. If you want it to be unique globally, you will need to make randomList static.

Depending on what you are really after you can do something like this:
using System;
using System.Collections.Generic;
using System.Linq;
namespace SO14473321
{
class Program
{
static void Main()
{
UniqueRandom u = new UniqueRandom(Enumerable.Range(1,10));
for (int i = 0; i < 10; i++)
{
Console.Write("{0} ",u.Next());
}
}
}
class UniqueRandom
{
private readonly List<int> _currentList;
private readonly Random _random = new Random();
public UniqueRandom(IEnumerable<int> seed)
{
_currentList = new List<int>(seed);
}
public int Next()
{
if (_currentList.Count == 0)
{
throw new ApplicationException("No more numbers");
}
int i = _random.Next(_currentList.Count);
int result = _currentList[i];
_currentList.RemoveAt(i);
return result;
}
}
}

And here my version of finding N random unique numbers using HashSet.
Looks pretty simple, since HashSet can contain only different items.
It's interesting - would it be faster then using List or Shuffler?
using System;
using System.Collections.Generic;
namespace ConsoleApplication1
{
class RnDHash
{
static void Main()
{
HashSet<int> rndIndexes = new HashSet<int>();
Random rng = new Random();
int maxNumber;
Console.Write("Please input Max number: ");
maxNumber = int.Parse(Console.ReadLine());
int iter = 0;
while (rndIndexes.Count != maxNumber)
{
int index = rng.Next(maxNumber);
rndIndexes.Add(index);
iter++;
}
Console.WriteLine("Random numbers were found in {0} iterations: ", iter);
foreach (int num in rndIndexes)
{
Console.WriteLine(num);
}
Console.ReadKey();
}
}
}

I noted that the accepted answer keeps adding int to the list and keeps checking them with if (!randomList.Contains(MyNumber)) and I think this doesn't scale well, especially if you keep asking for new numbers.
I would do the opposite.
Generate the list at startup, linearly
Get a random index from the list
Remove the found int from the list
This would require a slightly bit more time at startup, but will scale much much better.
public class RandomIntGenerator
{
public Random a = new Random();
private List<int> _validNumbers;
private RandomIntGenerator(int desiredAmount, int start = 0)
{
_validNumbers = new List<int>();
for (int i = 0; i < desiredAmount; i++)
_validNumbers.Add(i + start);
}
private int GetRandomInt()
{
if (_validNumbers.Count == 0)
{
//you could throw an exception here
return -1;
}
else
{
var nextIndex = a.Next(0, _validNumbers.Count - 1);
var number = _validNumbers[nextIndex];
_validNumbers.RemoveAt(nextIndex);
return number;
}
}
}

It's may be a little bit late, but here is more suitable code, for example when you need to use loops:
List<int> genered = new List<int>();
Random rnd = new Random();
for(int x = 0; x < files.Length; x++)
{
int value = rnd.Next(0, files.Length - 1);
while (genered.Contains(value))
{
value = rnd.Next(0, files.Length - 1);
}
genered.Add(value);
returnFiles[x] = files[value];
}

with Functional way*
static Func<int> GetNextUniqueIntegerFunc(int min, int max)
{
var list = new List<int>();
var random = new Random();
int getNextValue()
{
while (true)
{
var random_number = random.Next(min, max);
if (!list.Contains(random_number))
{
list.Add(random_number);
return random_number;
}
}
}
return getNextValue;
}

unique random number from 0 to 9
int sum = 0;
int[] hue = new int[10];
for (int i = 0; i < 10; i++)
{
int m;
do
{
m = rand.Next(0, 10);
} while (hue.Contains(m) && sum != 45);
if (!hue.Contains(m))
{
hue[i] = m;
sum = sum + m;
}
}

You could also use a dataTable storing each random value, then simply perform the random method while != values in the dataColumn

randomNumber function return unqiue integer value between 0 to 100000
bool check[] = new bool[100001];
Random r = new Random();
public int randomNumber() {
int num = r.Next(0,100000);
while(check[num] == true) {
num = r.Next(0,100000);
}
check[num] = true;
return num;
}

hi here i posted one video ,and it explains how to generate unique random number
public List<int> random_generator(){
Random random = new Random();
List<int> random_container = new List<int>;
do{
int random_number = random.next(10);
if(!random_container.contains(random_number){
random_container.add(random_number)
}
}
while(random_container.count!=10);
return random_container;
}
here ,,, in random container you will get non repeated 10 numbers starts from 0 to 9(10 numbers) as random.. thank you........

You can use basic Random Functions of C#
Random ran = new Random();
int randomno = ran.Next(0,100);
you can now use the value in the randomno in anything you want but keep in mind that this will generate a random number between 0 and 100 Only and you can extend that to any figure.

Try this:
private void NewNumber()
{
Random a = new Random(Guid.newGuid().GetHashCode());
MyNumber = a.Next(0, 10);
}
Some Explnations:
Guid : base on here : Represents a globally unique identifier (GUID)
Guid.newGuid() produces a unique identifier like "936DA01F-9ABD-4d9d-80C7-02AF85C822A8"
and it will be unique in all over the universe base on here
Hash code here produce a unique integer from our unique identifier
so Guid.newGuid().GetHashCode() gives us a unique number and the random class will produce real random numbers throw this
Sample:
https://rextester.com/ODOXS63244
generated ten random numbers with this approach with result of:
-1541116401
7
-1936409663
3
-804754459
8
1403945863
3
1287118327
1
2112146189
1
1461188435
9
-752742620
4
-175247185
4
1666734552
7
we got two 1s next to each other, but the hash codes do not same.

Initialization of gene in binary array for timetable scheduling with GA in C#

I am currently working on a project for driver scheduling and it's at the initial stage.
I have decided to use GA to generate optimized schedule for a driver, and as most of GA project do, the population shall be represented in binary.
e.g. if a driver is assigned two hours of task and his working duration is 9 hours, the possible population for that specific day looks like 110000000, 011000000, 001100000 and so on.
As GA's initialization, I would like to produce possible gene like 000110000 dynamically with two parameters (working duration of driver, and duty duration).
I managed to get fully random binary code in boolean list (see below) but this is not what I want to represent as the initialization.
This is the partial code which generates random binary string (technically bunch of boolean values) in a list.
private Random Rnd = new Random();
//initial data
private List<bool[]> CreateInitialData()
{
//generate 4 random genes (might be more)
return Enumerable.Range(0, 1).Select(_ =>
{
var array = new bool[GeneLength];
for(int i = 0; i < GeneLength; i++)
{
array[i] = Rnd.Next(0, 2) == 1;
}
return array;
}).ToList();
}
How could I implement the initialization function to produce the binary code which meets the requirement (driver's working hours, estimated duty duration)?
If there is a better way to represent it other than boolean list, please do suggest as well.

Based on 1 hour tasks I came up with this:
private static void Main(string[] args)
{
var genes = GetGenes(9, 2);
}
private static List<bool[]> GetGenes(int workinghours, int estimateddutyduration)
{
// get the base representation
var hours = GetHours(workinghours, estimateddutyduration);
var list = new List<bool[]>();
for (int i = 0; i < (workinghours-estimateddutyduration)+1; i++)
{
// add
list.Add(hours);
// switch
hours = SwitchArray(hours);
}
return list;
}
private static bool[] SwitchArray(bool[] array)
{
// copy the array to a list
var temp = array.ToList();
// insert the last element at the front
temp.Insert(0, temp.Last());
// remove the last
temp.RemoveAt(temp.Count-1);
// return as array
return temp.ToArray();
}
private static bool[] GetHours(int totalhours, int taskduration)
{
// initialise the list
var hours = new List<bool>(totalhours);
// fill the list for the number of working hours
for (int i = 0; i < totalhours; i++)
{
hours.Add(false);
}
// iterate for the task duration and set the hours as working
for (int i = 0; i < taskduration; i++)
{
hours[i] = true;
}
// return as array
return hours.ToArray();
}
for 9, 2 returns
110000000
011000000
001100000
000110000
000011000
000000110
000000011
for 9, 9 returns
111111111
For 9, 4 returns
111100000
011110000
001111000
000111100
000011110
000001111
This code is quite verbose and I have no doubt it can be optimized. But it's the idea I want to convey more than anything.
EDIT: If you want to display the results on the console
private static void ShowGenes(List<bool[]> genes)
{
foreach (var gene in genes)
{
foreach (var bit in gene)
{
Console.Write(bit ? "1" : "0");
}
Console.Write("\n");
}
}

c# unique random number

i'm trying to get X number of random number (where X is a variable) between 0 and 100 and add them to a row in a DataGridView. I'm using the code below, but the problem is i need it to be impossible to have the same number twice. Is there a way make sure i get unique random numbers?
int X = 20;
Random random = new Random();
int randomrow = random.Next(0, 100);
for (int i = 0; i < X; i++)
{
int randomNumber = random.Next(0, 100);
data.Rows[randomrow][3] = randomNumber;
}
Thanks for the help!

Split your problem in two:
Create a list of X random, unique numbers. There are numerous ways to do that:
a. Create a list of all numbers 0-100 and then shuffle them. OR
b. Keep a hash set of all the numbers you already created (in addition to the list) and only add a new one if it has not been added before.
Afterwards, loop through the list and the data rows simultaneously and insert the values into the rows.

Here's the simple way to do it without creating a shuffle method for the List (though there are examples of that). Basically create a list of all possible integers, populate the list, then sort by new GUIDs.
int X = 20;
var RowNumbers = new List<int>();
for (int i = 0; i < X; i++)
RowNumbers.Add(i);
foreach (int i in RowNumbers.OrderBy(f => Guid.NewGuid()))
{
data.Rows[i][3] = i;
}

You would need to compare the numbers you have already used to the next one you get from random.Next. If you have already used it, pick another. I also like Heinzi's answer.

Here is algorithm without shuffling and previous result using:
var max = 100; // max available value
var count = 20; // number of random numbers
var random = new Random();
var rest = (double)max;
for (int i = 0; i < max; i++, rest--)
{
if (count / rest > random.NextDouble())
{
Console.WriteLine(i); // here is your random value
count--;
if (count == 0)
{
break;
}
}
}

How to generate unique random integers that do not duplicate

I have created a short program that creates 3 random integers between 1-9 and stores them in an array, however, I would not like any of them to repeat, that is, I would like each to be unique. Is there an easier way to generate 3 unique integers other than having to iterate through the array and comparing each integer to each other? That just seems so tedious if I were to increase my array to beyond 3 integers.
This is my code to generate 3 random numbers. I saw other code in Java, but I thought maybe C# has a easier and more efficient way to do it.
var number = new Numbers[3];
Random r = new Random();
for ( int i = 0; i < number.Length; i++)
{
number[i] = new Numbers(r.Next(1,9));
}
Console.WriteLine("The Three Random Numbers Are:");
foreach(Numbers num in number)
{
Console.WriteLine("{0}", num.Number);
}

I would do something like this:
var range = Enumerable.Range(1, 8);
var rnd = new Random();
var listInts = range.OrderBy(i => rnd.Next()).Take(3).ToList();

You could make an array or a list of the numbers that might be generated, e.g. 0, 1, 2, 3. Then you generate a number from 0 to this list's length, e.g. 2 and pick list[2] so for the next time you only have 0, 1, 3 in your list.
It takes longer to generate it, especially for long lists but it doesn't repeat numbers.

using System;
using System.Collections.Generic;
public class Test
{
static Random random = new Random();
public static List<int> GenerateRandom(int count)
{
// generate count random values.
HashSet<int> candidates = new HashSet<int>();
// top will overflow to Int32.MinValue at the end of the loop
for (Int32 top = Int32.MaxValue - count + 1; top > 0; top++)
{
// May strike a duplicate.
if (!candidates.Add(random.Next(top))) {
candidates.Add(top);
}
}
// load them in to a list.
List<int> result = new List<int>();
result.AddRange(candidates);
// shuffle the results:
int i = result.Count;
while (i > 1)
{
i--;
int k = random.Next(i + 1);
int value = result[k];
result[k] = result[i];
result[i] = value;
}
return result;
}
public static void Main()
{
List<int> vals = GenerateRandom(10);
Console.WriteLine("Result: " + vals.Count);
vals.ForEach(Console.WriteLine);
}
}
Grate explanation and answers from here
Source http://ideone.com/Zjpzdh

C# - Writing numbers 1 through 10 in random order [duplicate]

This question already has answers here:
Is using Random and OrderBy a good shuffle algorithm? [closed]
(13 answers)
Closed 9 years ago.
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.

/// <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);
}

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

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;
}
}
}

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());
}

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