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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

generating integer random numbers in c#

I want to generate 4 digit random number and put it in the text box as serial number of something . how do I do that and yet to some extent be sure I wouldn't get duplicated numbers?

If you want to be sure that you don't get duplicates use a Guid:
Guid guid = Guid.NewGuid(); // however a Guid is not a 4 digit integer
If you want a random number use Random:
Random rnd = new Random();
int randomInt = rnd.Next(1000, 10000)
But note that you should not create the random instance in a loop because it is seeded with the current time. Otherwise you get repeating values. So either pas the random instance as parameter to this method, use a field/property or use the same instance in a loop which was created outside.
The easiest approach to get unique random numbers is to create new numbers if one already exists.

List<int> used = new List<int>();
Random random = new Random();
foreach(thing you want to do)
{
int current = random.Next(1000, 9999);
while(used.Contains(current))
current = random.Next(1000, 9999);
//do something
used.Add(current);
}
or some similar variation on this to meet your needs

Best way to do it is with the Random class:
Random random = new Random();
int result = random.Next(1000, 10000);
You could also use the RNGCryptoServiceProvider class, which gives you more secure random numbers:
RNGCryptoServiceProvider csp = new RNGCryptoServiceProvider();
byte[] bytes = new byte[4];
csp.GetBytes(bytes);
int yourNum = bytes.Select(x => x % 10).Aggregate(0, (a, b) => (a * 10) + b);

The Random class is used to create random numbers. (Pseudo-random that is of course.)
Example:
Random rnd = new Random();
int num = rnd.Next(1000, 10000); //creates number from 1000 to 9999
You can use HashSet for store generated number and check duplicate

The easiest way to do this is generate a list of all possible choices in order and shuffle it. Then after shuffling the list just go through the list "in order" and the sequence will be random and non repeating.
List<int> numbers = new List<int>(Enumerable<int>.Range(0,10000);
HelperFunctions.Shuffle(numbers); //There are plenty of examples of how to shuffle the list on this site.
foreach(var number in numbers)
{
Console.WriteLine(number.ToString("D4"); //Displays random numbers from 0000 to 9999 but never repeats
}

Like this:
var random = new Random();
var numbers = new List<int>(0);
for (var i = 0; i < 4; i++)
{
var number = random.Next(1000, 10000);
if (numbers.Contains(number))
{
i--;
continue;
}
numbers.Add(number);
}
tbx1.Text = numbers[0].ToString();
tbx2.Text = numbers[1].ToString();
tbx3.Text = numbers[2].ToString();
tbx4.Text = numbers[3].ToString();

This will give you a 4 digit random number.
Random random= new Random();
int RnNum = random.Next(1000,9999);
If you want to be sure that it doesn't get duplicated, then store the generated random values in Session & compare if its already generated. Of course there are other ways, if you don't want to do it the session way, write to a text file or something

You need to keep a record yourself of the numbers that have been generated in Dictionary or Hashset. Like
HashSet<int> generatedValues = new HashSet<int>();
Random rnd = new Random();
int randomInt = rnd.Next(1000,9999);
while(generatedValues.Contains(randomInt))
{
randomInt = rnd.Next(1000,9999);
}
generatedValues.Add(randomint);

Randomly pick 4 elements from an array using c#

I am trying to use random function in C# to randomly pick four from an array addetailsID which has over six elements.
I am placing these randomly picked into an another array strAdDetailsID:
string[] strAdDetailsID = new string[4];
for (int i = 0; i < 4; i++)
{
Random random = new Random();
int index = random.Next(0, addetailsID.Length);
value = addetailsID[index].ToString();
strAdDetailsID[i] = value;
}
Sometimes, I get two of the same values from the six elements. How can I get all four unique values to be picked?

You might be better off shuffling the array, and then choosing the first 4 elements.

You could do it with LINQ with this method.
List<string> list = new List<string>() { "There", "Are", "Many", "Elements", "To", "Arrays" };
foreach (var item in list.OrderBy(f => Guid.NewGuid()).Distinct().Take(4))
{
Console.WriteLine(item);
}

You have an issue with your placement of Random random ... but I believe you're attacking this the wrong way.
This could be solved by randomly ordering the source and taking the first 4 items.
var result = addetails.OrderBy(x => Guid.NewGuid()).Take(4).ToArray();
Assuming the contents of addetails are unique (like you imply), you will always get 4 unique values here. Using random correctly, it's still possible to get repeats (because it's random).

You need to generate 4 unique indices first, then pull the random values from the source array:
string[] addetailsID = new string[20]; // this is the array I want to index into
// generate the 4 unique indices into addetailsID
Random random = new Random();
List<int> indices = new List<int>();
while (indices.Count < 4)
{
int index = random.Next(0, addetailsID.Length);
if (indices.Count == 0 || !indices.Contains(index))
{
indices.Add(index);
}
}
// now get the 4 random values of interest
string[] strAdDetailsID = new string[4];
for (int i = 0; i < indices.Count; i++)
{
int randomIndex = indices[i];
strAdDetailsID[i] = addetailsID[randomIndex];
}

Put them in a list and remove the selected element from the list once it has been chosen.

The following algorithm works pretty well, and doesn't require extra storage or pre-shuffling. It does change the order of the source array, so if that's not feasible, then the pre-shuffling approach is best.
In pseudo-code:
result = []
For i = 0 to numItemsRequired:
randomIndex = random number between i and source.length - 1
result.add(source[randomIndex])
swap(source[randomIndex], source[i])
In C#:
string[] strAdDetailsID = new string[4];
Random rand = new Random();
for (int i = 0; i < 4; i++)
{
int randIndex = rand.Next(i, adDetailsID.Length);
strAddDetails[i] = adDetailsID[randIndex];
string temp = adDetailsID[randIndex];
adDetailsID[randIndex] = adDetailsID[i];
adDetails[i] = temp;
}

Use a list instead and drop every item that is already used:
List<string> addetailsID = new List<string>{"1","2","3","4","5","6"};
string[] strAdDetailsID = new string[4];
Random random = new Random();
for (int i = 0; i < 4; i++)
{
int index = random.Next(0, addetailsID.Count);
string value = addetailsID[index].ToString();
addetailsID.RemoveAt(index);
strAdDetailsID[i] = value;
}
strAdDetailsID.Dump();

Well, you should select random items one by one, not considering the already-picked items when selecting the next one. This must be faster than shuffling.
If the source list is small, you can just make a copy and remove the chosen items from it. Otherwise, you go like this:
(let n be the number of items in your array)
let S be the set of chosen indices; S = empty at the beginning
choose a random index i from 0 to n-1-size(S)
for each index from S that is smaller than i (starting from the smallest index!), add one to i
now i is a chosen index, add it to S.
return to step 2 until your set contains 4 elements.

generating string from exited characters

I have a character array as shown below :
char[] pwdCharArray = "abcdefghijklmnopqrstuvwxyzABCDEFG" +
"HIJKLMNOPQRSTUVWXYZ0123456789`~!##$%^&*()-_=+[]{}\\|;:'\",<" +
".>/?".ToCharArray();
and from this char array, i want to generate a string of minimum length 7 and all the characters in the string should be from the above char array.
How to do this operation?

What's the maximum length? (You may want to parameterize the methods below to specify a minimum and maximum length.) Here's a simple way of doing it for exactly 7 characters:
Here's the C# version:
public string GeneratePassword(Random rng)
{
char[] chars = new char[7];
for (int i = 0; i < chars.Length; i++)
{
chars[i] = pwdCharArray[rng.Next(pwdCharArray.Length)];
}
return new string(chars);
}
Note that the Random instance should be passed in to avoid the common problem of creating many instances of Random. I have an article describing this problem and ways around it. In essence, you should use one instance of Random per thread - don't create a new instance every time you want to use one, and don't reuse the same instance across multiple threads.
In fact, for a genuine password which is guarding sensitive information, you probably shouldn't be using Random at all, but rather something like RNGCryptoServiceProvider (or less directly, the results of RandomNumberGenerator.Create()). This can be somewhat harder to use, but will give you much more secure random numbers.
In Java it would be pretty similar, but then I'd use SecureRandom (which is fortunately rather easier to use than its .NET counterpart). In this case, you can create a new instance each time:
public String generatePassword() {
char[] chars = new char[7];
SecureRandom rng = new SecureRandom();
for (int i = 0; i < chars.length; i++) {
chars[i] = pwdCharArray[nextInt(pwdCharArray.length)];
}
return new String(chars);
}

Generate 7 random numbers between 0 and yourArray.length -1.
Pick the corresponding char in the array and put it in your final String.
Here is the code with a StringBuilder:
StringBuilder sb = new StringBuilder();
Random random = new Randdom();
for(int i=0; i<7; i++) {
sb.append(pwdCharArray[random.nextInt(0, pwdCharArray.length -1)]);
}
return sb.toString();

(C# example)
Random rand = new Random();
char[] arr = new char[rand.Next(7,15)];
for(int i = 0 ; i < arr.Length ; i++) {
arr[i] = pwdCharArray[rand.Next(pwdCharArray.Length)];
}
string pwd = new string(arr);

Here is a solution in Java, with variable length of the word, and test of working:
import java.util.*;
public class RandomWord {
// Valid characters
private static String VALID_CHARACTERS = "abcdefghijklmnopqrstuvwxyzABCDEFG" +
"HIJKLMNOPQRSTUVWXYZ0123456789`~!##$%^&*()-_=+[]{}\\|;:'\",<.>/?";
// Minimal length thw word can have
private static int MIN_LENGTH = 7;
// Maximal length will be MIN_LENGTH + THRESHOLD - 1
private static int THRESHOLD = 10;
// Generate a random number generator.
// The time in millis is used as seed prevents the numbers to be always the same in same order
Random randomGenerator = new Random(System.currentTimeMillis());
public String generateWord () {
// Actual length of the word (from MIN_LENGTH to MIN_LENGTH + THRESHOLD - 1)
int length = MIN_LENGTH + randomGenerator.nextInt(THRESHOLD);
// Loop for every character
StringBuilder word = new StringBuilder();
for (int i = 0; i < length; i++) {
// Appends one more random char
word.append(VALID_CHARACTERS.charAt(randomGenerator.nextInt(VALID_CHARACTERS.length())));
}
// Returns the random word
return word.toString();
}
// Test the class
public static void main (String[] args) {
// Instantiates and tests the class
RandomWord randomWord = new RandomWord();
for (int i = 0; i < 30; i++) {
String word = randomWord.generateWord();
System.out.println(word + " (" + word.length() + ")");
}
}
}