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.
Related
I can't fill it with numbers to 0 - 15 then shuffle the array, so that's not the solution
I used this code in C but now in c# it doesn't work, for some reason this code let some numbers pass the do while.
Random r = new Random();
bool unique;
int rand_num;
for (int i = 0; i < 4; i++)
{
for (int j = 0; j < 4; j++)
{
do
{
unique = true;
rand_num = r.Next(16);
for (int k = 0; k < 4; k++)
{
for (int l = 0; l < 4; l++)
{
if (numbers[k, j] == rand_num)
{
unique = false;
}
}
}
} while (!unique);
numbers[i, j] = rand_num;
}
}
}
If the list of possible numbers is small, as in this case, just create the full list and randomise it first, then take the items in the order they appear. In your case, you can put the randomised numbers into a queue, then dequeue as required.
var r = new Random();
var numberQueue = new Queue<int>(Enumerable.Range(0, 16).OrderBy(n => r.NextDouble()));
var numbers = new int[4, 4];
for (var i = 0; i <= numbers.GetUpperBound(0); i++)
{
for (var j = 0; j <= numbers.GetUpperBound(1); j++)
{
numbers[i, j] = numberQueue.Dequeue();
}
}
I suggest you to use the Fisher-Yates algorithm to generate your non-repeatable sequence of random numbers.
It would be very straight-forward to implement a code to fill in a 2d array with those numbers, then.
List<int> seq = Enumerable.Range(0,16).ToList();
int[,] numbers = new int[4,4];
Random r = new();
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 4; j++) {
int n = r.Next(0, seq.Count);
numbers[i,j] = seq[n];
seq.RemoveAt(n);
}
}
The approach you have taken may end up in continuous looping and take lot of time to complete.
Also checking for value in 2D using nested for loop is not efficient.
You can use HashSet to keep track of unique value. Searching in HashSet is fast.
following is the code approach I suggest.
var hashSet = new HashSet<int>();
var r = new Random();
var arr = new int[4, 4];
for(var i = 0;i<4;i++)
{
for(var j = 0;j<4;j++)
{
// Generate random value between 0 and 16.
var v = r.Next(0, 16);
// Check if the hashSet has the newly generated random value.
while(hashSet.Contains(v))
{
// generate new random value if the hashSet has the earlier generated value.
v = r.Next(0, 16);
}
//Add value to the hashSet.
hashSet.Add(v);
// add value to the 2D array.
arr[i, j] = v;
}
}
I hope this will help solving your issue.
The problem with your current approach is that as you get closer to the end of the array, you have to work harder and harder to get the next random value.
Imagine you roll a die, and each time you want to get a unique value. The first time you roll, any result will be unique. The next time, you have a 1/6 chance of getting a number that has already been obtained. And then a 2/6 chance, etc. and in the end most of your rolls will be non-unique.
In your example, you have 16 places that you want to fill with numbers 0 to 15. This is not a case of randomly generating numbers, but randomly placing them. How do we do this with a deck of cards? We shufle them!
My proposal is that you fill the array with unique sequential values and then shuffle them:
Random random = new Random();
int dim1 = array.GetLength(0);
int dim2 = array.GetLength(1);
int length = dim1 * dim2;
for (int i = 0; i < length; ++i)
{
int x = i / dim1;
int y = i % dim1;
array[x, y] = i; // set the initial values for each cell
}
// shuffle the values randomly
for (int i = 0; i < length; ++i)
{
int x1 = i / dim1;
int y1 = i % dim1;
int randPos = random.Next(i, length);
int x2 = randPos / dim1;
int y2 = randPos % dim1;
int tmp = array[x1, y1];
array[x1, y1] = array[x2, y2];
array[x2, y2] = tmp;
}
The shuffle in this code is based on the shuffle found here
int[,] numbers = new int[4, 4];
Random r = new Random();
bool unique;
int rand_num;
List<int> listRandom = new List<int> { };
for ( int i = 0; i < 4; i++ )
{
for ( int j = 0; j < 4; j++ )
{
do
{
unique = false;
if (!listRandom.Contains( rand_num = r.Next( 0, 16 )))
{
listRandom.Add( rand_num );
numbers[i, j] = rand_num;
unique = true;
}
} while ( !unique );
}
}
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:)
I am creating a lotto number generator, with the powerball number at index zero in my array. however I am supposed to print all items in the array with the powerball nuber printing last. If it didn't have to print last I could easily do this with a foreach loop, but I'm stuck on how to do it otherwise.
anyone with any thoughts or ideas.
{
int[] Lotto = new int[6];
int check = 0;
//powerball
Random rand = new Random();
check = rand.Next(5, 64);
Lotto[0] = check;
for (int i = 1; i < Lotto.Length;)
{
check = rand.Next(1, 64);
Lotto[i] = check;
i++;
}
return Lotto;
}
As I commented, to print the first index last, something like…
for (int i = 1; i < Lotto.Length; i++) {
Console.WriteLine(Lotto[i]);
}
Console.WriteLine("PowerBall: " + Lotto[0]);
Or … If order does not matter, then simply loop through the array from the bottom up. Something like…
for (int i = Lotto.Length - 1; i >= 0; i--) {
Console.WriteLine(Lotto[i]);
}
You can use if into for:
{
int[] Lotto = new int[6];
int check = 0;
//powerball
Random rand = new Random();
check = rand.Next(5, 64);
Lotto[0] = check;
for (int i = 1; i < Lotto.Length; i++)
{
check = rand.Next(1, 64);
Lotto[i] = check;
if(Lotto[0] >= 5 ){
Console.WriteLine("The item is "+ Lotto);
}else if(Lotto[0] >= 50 && Lotto[0] <=64){
Console.WriteLine("The item is "+ Lotto);
}
}
//return Lotto;
}
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();
}
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;
}